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math/9912153
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Recall from CITE that the suspension theorem in morphic cohomology implies that morphic cohomology can be represented by morphisms into spaces of zero cycles in projective spaces. Since zero cycles are given by points in symmetric products this can be interpreted in the following way. Let MATH be the infinite symmetric product of the infinite projective space. Given a projective variety MATH, let MATH denote the colimit of the the algebraic morphism spaces MATH. Let MATH be a colimit of projective varieties. Then MATH . Similarly, MATH represents holomorphic MATH - theory in the sense that MATH . Thus to prove REF we will describe a relationship between the representing spaces MATH and MATH. Using the identification in REF , let MATH correspond to the class in MATH represented by the identity map MATH. There exists a unique class MATH with NAME character MATH. By REF, we know that for every MATH and MATH, MATH is a homotopy equivalence. It follows that by taking limits we have that MATH is a homotopy equivalence. But we also know from CITE that the natural map MATH is an isomorphism. This is true because for the following reasons. CASE: Since products MATH have ``algebraic cell decompositions" in the sense of CITE, its morphic cohomology and singular cohomology coincide, MATH . CASE: Since both morphic cohomology and singular cohomology admit transfer maps CITE there is a natural identification of MATH and MATH with the MATH invariants in MATH and MATH respectively. Since the natural transformation MATH is equivariant, then we get an induced isomorphism on the invariants, MATH . CASE: By taking limits over MATH and MATH we conclude that MATH is an isomorphism. Using this isomorphism and the compatibility of the NAME character maps in holomorphic and topological MATH - theories, to prove this theorem it is sufficient to prove that there exists a unique class MATH with (topological ) NAME character MATH where MATH is the class represented by the identity map. But this follows because the NAME character in topological MATH - theory, MATH is an isomorphism. We now show how the element MATH defined in the above lemma will yield an inverse to the NAME character transformation. The element MATH defines natural transformations MATH such that the composition MATH is equal to the identity. The set of path components of the NAME group completion of a topological monoid is the NAME group completion of the discrete monoid of path components. If we use the notation MATH to mean the NAME group of a discrete monoid MATH, this says that MATH and hence MATH where the subscript MATH denotes the holomorphic mapping space localized at the rationals. This means that MATH can be represented as a difference of classes, MATH where MATH. Now consider the composition pairing MATH which localizes to a pairing MATH . Using this pairing, MATH and MATH each define transformations MATH . Using the fact that MATH is an infinite loop space, then the subtraction map is well defined up to homotopy, MATH . We need the following intermediate result about this construction. For any projective variety (or colimit of varieties) MATH, the map MATH is a map of MATH - spaces. Since the construction of these maps was done at the representing space level, it is sufficient to verify the claim in the case when MATH is a point. That is, we need to verify that the compositions MATH and MATH represent the same elements of MATH, where MATH and MATH are the monoid multiplications in MATH and MATH respectively. But by REF of the last section, this is the same as MATH. Now in the topological category, we know that the class MATH is the inverse to the NAME character and hence induces a rational equivalence of MATH - spaces MATH . This implies that the compositions REF repesent the same elements of MATH, and hence the same elements in MATH. Thus the map MATH is a MATH - map from a MATH operad spaces (as described in REF), to an infinite loop space. But any such rational MATH - map extends in a unique manner up to homotopy, to a map of MATH - spaces of their group completions MATH . This map is natural in the category of colimits of projective varieties MATH. Since any MATH - map between rational infinite loop spaces is homotopic to an infinite loop map, this then defines a natural transformation of rational infinite loop spaces, MATH . So when we apply homotopy groups MATH defines natural transformations MATH . Now notice that if we let MATH in REF , and MATH be the class represented by the identity map, then by definition, one has that MATH represents the class MATH described in REF . Moreover this lemma tells us that MATH. Now as in REF, we view the NAME character as represented by an element MATH which is a map of rational infinite loop spaces, then this lemma tells us that the elements MATH and MATH are both maps of rational infinite loop spaces and lie in the same path component of MATH. But this implies the MATH and MATH define the homotopic natural transformations of rational infinite loop spaces, MATH . When we apply homotopy groups this means that MATH which was the claim in the statement of REF . We now can complete the proof of REF . That is we need to prove that MATH is an isomorphism. By REF we know that MATH is surjective. In order to show that it is injective, we prove the following: The composition of natural transformations MATH is the identity. These transformations are induced on the representing level by maps of rational infinite loop spaces, MATH and MATH . The composition MATH represents an element of rational holomorphic MATH - theory, MATH . Now the fact that MATH is an inverse of the NAME character in topological MATH - theory tells us that MATH where MATH is the class represented by the identity map. But according to the results in REF, we know MATH . So by the compatibility of the NAME characters in holomorphic and topological MATH - theories, we conclude that MATH . This implies that MATH and the identity map MATH induce the same natural transformations MATH. Applying homotopy groups implies that MATH is the identity as claimed. This lemma implies that MATH is injective. As remarked above this was the last remaining fact to be verified in the proof of REF .
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math/9912153
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We first prove that the total NAME class MATH is injective. So suppose that for some MATH, we have that MATH . So each NAME class MATH for MATH. Now recall from REF that in the algebra of operations between MATH and MATH, that the that the NAME classes and NAME character are related by a formula of the form MATH where MATH is some polynomial in the first MATH . NAME classes. So since each MATH then an inductive argument using REF implies that each MATH. Thus the total NAME character MATH. But since the total NAME character is an isomorphism REF , this implies that MATH. This proves that the total NAME class operation is injective. We now prove that MATH is surjective. To do this we will prove that for every MATH and element MATH there is a class MATH with MATH and MATH for MATH. We prove this by induction on MATH. So assume this statement is true for MATH, and we now prove it for MATH. Let MATH. Since the total NAME character is an isomorphism, there is an element MATH with MATH, and MATH for MATH. But REF implies that MATH for MATH, and MATH. Thus the total NAME class has value MATH. This proves that the total NAME class is surjective, and therefore that it is an isomorphism.
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math/9912153
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The space of rational maps in the morphism space MATH stabilizes if and only if the group completion of its space of rational maps is the two fold loop space, MATH . But by definition, the left hand side is equal to MATH . But by Rowland's theorem CITE or by the more general ``projective bundle theorem" proved in CITE we know that the NAME map MATH is a homotopy equivalence. Combining this with REF , we have that the space of rational maps in the morphism space MATH stabilizes if and only if the following composition is a homotopy equivalence MATH .
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math/9912153
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Consider the NAME character defined on the MATH . Now the morphic cohomology of a point is equal to the usual cohomology of a point, MATH, so this group is non zero if and only if MATH. So the NAME character gives an isomorphism MATH . Let MATH be the NAME character of the NAME class, MATH. Since the NAME character is an isomorphism, MATH is a generator. We use this notation for the following reason. Recall the operation in morphic cohomology MATH defined in CITE. Using the fact that MATH is a module over MATH (using the ``join" multiplication in morphic cohomology), then this operation is given by multiplication by a generator of MATH. Therefore up to a rational multiple, this operation on rational morphic cohomology, MATH, is given by multiplication by the element MATH. In CITE it was shown that the natural map from morphic cohomology to singular cohomology MATH makes the following diagram commute: MATH . It also follows from the ``NAME duality theorem" proved in CITE that if MATH is a MATH - dimensional smooth variety, then MATH for MATH. Furthermore for MATH factors as the composition MATH . Let MATH denote the localization of MATH obtained by inverting the transformation MATH. Specifically MATH . Then REF imply we have an isomorphism with singular cohomology, MATH . Again, since rationally the MATH operation is, up to multiplication by a nonzero rational number, given by multiplication by MATH, we can all conclude that when rational morphic cohomology is localized by inverting MATH, we have an isomorphism with singular rational cohomology, MATH . Now since the NAME character isomorphism MATH is an isomorphism of rings, then the following diagram commutes: MATH where the top horizontal map is multiplication by the NAME class MATH, and the bottom horizontal map is multiplication by MATH. Moreover since the NAME character in holomorphic MATH - theory and and that for topological MATH - theory are compatible, this means we get a commutative diagram: MATH . By REF we know that the bottom horizontal map is an isomorphism. Moreover by REF the left hand vertical map is an isomorphism. Of course the right hand vertical map is also a rational isomorphsim. Hence the top horizontal map is a rational isomorphism, MATH .
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math/9912153
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Consider the commutative diagram involving the total NAME character MATH . By REF , if MATH is NAME periodic, then the top horizontal map MATH is an isomorphism. But by REF we know that the two vertical maps in this diagram are isomorphisms. Thus if MATH is NAME periodic, then the bottom horizontal map in this diagram is an isomorphism. That is, MATH is an isomorphism, for every MATH. But as is shown in CITE, MATH, where MATH is the space of algebraic MATH - cycles in MATH up to algebraic (or homological) equivalence. Moreover the image of MATH is the image of the natural map induced by including algebraic cycles in all cycles, MATH, which lies in NAME filtration MATH. Thus MATH is an isomorphism implies that the composition MATH is an isomorphism. In particular this means that MATH.
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math/9912157
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We begin with REF : The map MATH has the RLP with respect to the map MATH if and only if each map MATH factors through MATH, that is, if and only if each MATH-element of MATH is in the image of MATH. Now REF : The map MATH has the RLP with respect to the map MATH if and only if for each MATH-element MATH of MATH whose boundary is the image of a MATH-cycle MATH of MATH, there is a MATH-element MATH of MATH which hits MATH under MATH and MATH under the differential. (In other words, if and only if the natural map MATH induces a surjection of MATH-elements.) So suppose that MATH has the RLP with respect to each map MATH. As a preliminary result, we prove that MATH induces a surjection of MATH-cycles. Suppose we are given a MATH-cycle MATH of MATH. Its boundary is zero and is thus in the image of the MATH-cycle MATH of MATH. Therefore MATH is the image of a MATH-cycle MATH. It follows immediately that MATH induces a surjection in MATH-homology. Now we prove that MATH induces a surjection of MATH-elements. Suppose we are given a MATH-element MATH of MATH. By the above argument, its boundary, which is a MATH-cycle, is the image of a MATH-cycle MATH of MATH. Thus, by the characterization of maps MATH having the RLP, we see that there is a MATH-element MATH which hits MATH. Finally, we prove that MATH induces an injection in MATH-homology. Suppose MATH is a MATH-cycle of MATH whose image under MATH is the boundary of a MATH-element MATH. Then, using the characterization of the RLP, we see that there is a MATH-element MATH whose boundary is MATH. This shows that MATH induces an injection in MATH-homology. We have proved that if MATH has the RLP with respect to the maps MATH, then MATH induces an isomorphism in MATH-homology and a surjection of MATH-elements. The proof of the converse is similar.
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math/9912157
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In this proof we use the terms weak equivalence, (acyclic) fibration, (acyclic) cofibration, and cofibrant as defined in REF . We check the hypotheses of the Recognition Lemma from the previous section. Since MATH is complete and cocomplete, so is MATH; limits and colimits are taken degreewise. The class of weak equivalences is easily seen to be closed under retracts and to satisfy the two-out-of-three condition. The zero chain complex is certainly small relative to MATH. We assumed that each MATH in MATH is small relative to MATH, and it follows that each MATH is small relative to MATH. In more detail: A map is in MATH if and only if it is a degreewise split monomorphism whose cokernel is a transfinite colimit of degreewise split monomorphisms whose cokernels are complexes of MATH-projectives with zero differential. In particular, every map in MATH is a degreewise split monomorphism whose cokernel is a complex of MATH-projectives. Let MATH be an object of MATH. For large enough regular cardinals MATH, MATH is small with respect to MATH-sequences of maps in MATH. Suppose that for such a MATH we have a MATH-sequence MATH of maps in MATH with colimit MATH. A chain map from MATH to MATH is just a map MATH such that MATH. Given such a MATH, it factors through some MATH, since MATH is small relative to MATH, and the sequence MATH is in MATH. Moreover, since MATH, it follows that MATH, at least after passing to MATH for some MATH, again using smallness. Thus the map MATH is surjective. An easier argument shows that it is monic, and therefore MATH is small relative to MATH. Since the projectives in MATH may be used to test whether a map is a fibration or weak equivalence, REF tells us that MATH is the collection of acyclic fibrations and that MATH is the collection of fibrations. Thus we have an equality MATH, giving us two of the inclusions required by the Recognition Lemma. We now prove that MATH. Since MATH, it is clear that MATH, so in particular MATH. We must prove that MATH, that is, that each map which is a transfinite composite of pushouts of coproducts of maps in MATH is a weak equivalence. A map in MATH is of the form MATH for some MATH. Thus a coproduct of maps in MATH is of the form MATH for some contractible complex MATH. A pushout of such a coproduct is of the form MATH, again with MATH a contractible complex. And a transfinite composite of such maps is of the same form as well. In particular, a transfinite composite of pushouts of coproducts of maps from MATH is a weak equivalence. We can now apply the Recognition Lemma and conclude that MATH is a model category with weak equivalences as described in REF , cofibrations given by the maps in MATH and fibrations given by the maps in MATH. We saw above that the maps in MATH are precisely those we called fibrations in REF , and it is clear from this that every object is fibrant. It remains to check that the maps in MATH are as described in REF . We continue use the words ``cofibration" and ``cofibrant" as defined in REF . Suppose that MATH is in MATH. By REF, MATH is a retract of a map MATH which is a transfinite composite of pushouts of coproducts of maps in MATH. As above, one can show that MATH is degreewise split-monic and that the complex MATH of cokernels is cofibrant. It follows that MATH is degreewise split-monic and that the complex of cokernels is a retract of MATH and is therefore cofibrant. That is, MATH is a cofibration. And now it is clear that if the map MATH is in MATH, then MATH is cofibrant. Using that MATH is closed under transfinite compositions and retracts, one can show that every cofibration is in MATH and in particular that if MATH is cofibrant then the map MATH is in MATH.
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math/9912157
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We construct a factorization MATH of the diagonal map MATH in the following way. Let MATH be the chain complex which has MATH in degree MATH. We describe the differential by saying that it sends a generalized element MATH in MATH to MATH. Let MATH be the map which sends MATH to MATH and let MATH be the map which sends MATH to MATH. One can check easily that MATH is a chain complex and that MATH and MATH are chain maps whose composite is MATH. The map MATH is a chain homotopy equivalence with chain homotopy inverse sending MATH to MATH; this implies that it induces a chain homotopy equivalence of generalized elements and is thus a weak equivalence. The map MATH is degreewise split-epi; this implies that it induces a split epimorphism of generalized elements and is thus a fibration. Therefore MATH is a good path object for MATH. It is easy to see that a chain homotopy between two maps MATH is the same as a right homotopy using the path object MATH. By REF , two maps are homotopic if and only if they are right homotopic using MATH. Thus the model category notion of homotopy is the same as the notion of chain homotopy when the source is cofibrant.
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math/9912157
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The group MATH may be calculated by choosing a cofibrant replacement MATH for MATH and computing the homotopy classes of maps from MATH to MATH. (Recall that all objects are fibrant, so there is no need to take a fibrant replacement for MATH.) A MATH-projective MATH-exact resolution MATH of MATH serves as a cofibrant replacement for MATH, and by REF the homotopy relation on MATH is chain homotopy, so it follows that MATH is isomorphic to MATH.
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math/9912157
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That MATH is triangulated follows from NAME 's result, since we have shown that MATH is a self-equivalence. The identification of the triangles uses the definition of fibre sequences from CITE and the construction of the path object from the proof of REF . There is also a dual proof using cofibre sequences and cylinder objects.
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math/9912158
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See CITE and CITE.
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math/9912158
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See CITE .
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math/9912158
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See CITE.
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math/9912158
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See CITE .
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math/9912158
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Suppose MATH, MATH have the same image under REF. We choose representatives MATH, MATH which have closed MATH-orbit. Let us define MATH REF by MATH . Choose complementary subspaces MATH of MATH in MATH. We choose a MATH-parameter subgroup MATH as follows: MATH on MATH and MATH on MATH. Then the limit MATH exists and its restriction to MATH is MATH. Since MATH has a closed orbit, we may assume that the restriction of MATH to MATH is MATH. Note that MATH is a subspace of MATH by the construction. Suppose that there exists MATH such that MATH. We want to construct MATH such that MATH. Since we have MATH, the the restriction of MATH to MATH is invertible. Let MATH be an extension of the restriction MATH to MATH so that MATH is mapped to MATH. Then MATH maps MATH to MATH.
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math/9912158
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See CITE or CITE.
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math/9912158
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See CITE, CITE.
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math/9912158
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See CITE. (REF therein is a misprint of REF .)
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math/9912158
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CASE: See CITE for the first assertion. During the proof of CITE, we have shown the second assertion, using CITE = REF. CASE: Consider the alternating sum of dimensions of the complex MATH. It is equal to the alternating sum of dimensions of cohomology groups. It is nonnegative, if MATH by REF . On the other hand, it is equal to MATH . Thus we have the assertion.
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math/9912158
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Consider MATH satisfying MATH for any MATH, MATH. If we set MATH then MATH is MATH-invariant and contained in MATH by REF. Thus we have MATH by the stability condition.
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math/9912158
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We prove MATH by the induction on MATH. The assertion is trivial when MATH. Take a point in MATH and its representative MATH. As in REF, there exists MATH such that MATH . We decompose MATH into eigenspaces of MATH: MATH . We also decompose MATH into eigenspaces of MATH as MATH. Then REF holds where MATH is replaced by MATH. Choose and fix an eigenvalue MATH of MATH. First suppose MATH for some MATH. Let MATH . Since MATH is not a root of unity, we have MATH for MATH. Hence the above MATH is well-defined. By REF (and MATH from the assumption), we have MATH. By the assumption, we have MATH, and hence MATH again by REF. Then we may assume the restriction of MATH to MATH is MATH as in the proof of REF. Thus the data MATH is defined on the smaller subspace MATH. Thus MATH by the induction hypothesis. If MATH for any MATH, we replace MATH. If we can find a MATH so that MATH for some MATH, we are done. Otherwise, we have MATH for any MATH, MATH, and we have MATH by REF. Then we choose MATH, which may not be an eigenvalue of MATH, so that MATH and repeat the above argument. (This is possible since we may assume MATH.) We have MATH and the data MATH is defined on the smaller subspace MATH as above.
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math/9912158
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The proof is essentially contained in CITE. See also REF for a similar result.
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math/9912158
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We have a surjective homomorphism MATH of codimension MATH over MATH. This gives a morphism from MATH to the fiber product of NAME bundles. By a straightforward modification of the arguments in CITE, one can show that it is an isomorphism. The detail is left to the reader. The assumption MATH is used to distinguish MATH and MATH. Let us prove the remaining REF . First note that MATH . Since we have a MATH-subspace in MATH, we must have MATH . Replacing MATH by MATH, we get REF. Moreover, we have MATH . On the other hand, the dimension REF implies MATH . Since MATH is an open subset of MATH, we get REF.
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math/9912158
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We prove the assertion by induction on MATH, MATH. (The result is trivial when MATH.) We first make a reduction to the case when MATH . Fix a MATH and consider MATH . Since MATH is not a root of unity, we have MATH for MATH. Hence the above MATH is well-defined. Suppose MATH. By REF and the choice of MATH, we have MATH. Let us replace MATH by MATH. Namely we change MATH to MATH and all other data are unchanged. The equation MATH and the stability condition are preserved by the replacement. Thus we have a morphism MATH where MATH is a fixed point subvariety of MATH obtained by the replacement. (This notation will not be used elsewhere. The data MATH is fixed elsewhere.) Conversely, we can put any homomorphism MATH to get a point in MATH starting from a point in MATH. This shows that MATH is the total space of the vector bundle MATH over MATH, where MATH is considered as a trivial bundle. In particular, MATH is (nonempty and) connected if and only if MATH is so. By the induction hypothesis, MATH is connected and we are done. Thus we may assume MATH. Then MATH consists of the last term by the choice of MATH. (Note MATH under the assumption that there is at most one edge joining two vertices of MATH.) Hence we have REF with MATH. Now let us prove the connectedness of MATH under REF. By REF , we have MATH unless MATH for each MATH. Hence it is enough to prove the connectedness of MATH for MATH. Let us consider the map MATH. By REF, MATH becomes smaller for MATH. Hence MATH is connected by the induction hypothesis. Again by REF , MATH is also connected. Since MATH is a fiber product of NAME bundles, MATH is connected.
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math/9912158
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We first show that MATH is surjective. Choose a closed subvariety MATH of MATH so that MATH is a trivial MATH-bundle over MATH. There is a diagram MATH with exact rows by REF. By a diagram chase it suffices to prove the surjectivity for the restrictions of MATH to MATH and to MATH. By repeating the process on MATH, it suffices to prove it for the case MATH is a trivial MATH-equivariant bundle. By NAME isomorphism CITE MATH is an isomorphism if MATH is a MATH-equivariant bundle. Thus we prove the assertion. Let us repeat the same argument for MATH and MATH by replacing REF by REF. By five lemma both MATH are isomorphisms. In particular, we have MATH by assumption. Consider the diagram MATH where the vertical arrows are comparison maps. The left vertical arrow is an isomorphism by assumption. Thus the right vertical arrow is also an isomorphism by the commutativity of the diagram and what we just proved above. REF for MATH can be checked in the same way, and MATH has property MATH. The property MATH can be checked in the same way.
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math/9912158
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By CITE the moment map MATH is a NAME function, and critical manifolds are the fixed point MATH. Let MATH, MATH, be the components of MATH. By CITE, stable and unstable manifolds for the gradient flow of MATH coincide with MATH-attracting sets of NAME decomposition CITE: MATH . These are invariant under the MATH-action since the MATH-action commutes with the MATH-action. Note that results in CITE are stated for compact manifolds, but the argument can be modified to our setting. A difference is that MATH is not MATH unless MATH is compact. On the other hand, MATH is MATH since MATH is proper. As in CITE, we can introduce an ordering on the index set MATH of components of MATH such that MATH is a MATH-partition and MATH is a MATH-partition with respect to the reversed order. By CITE (see also CITE for analytic arguments), the maps MATH are fiber bundles with affine spaces as fibers. Furthermore, MATH (respectively, MATH) is locally isomorphic to a MATH-equivariant vector bundle by the proof. Thus MATH and MATH have properties MATH and MATH by REF. Hence MATH and MATH have properties MATH and MATH by REF . By the argument in CITE, the pairing REF can be identified with a pairing MATH of the form MATH for some pairing MATH such that MATH is the pairing REF for MATH. Since MATH is nondegenerate for all MATH by the assumption, REF is also nondegenerate. The proof of the statement for MATH, MATH is similar. One uses the fact that the intersection pairing MATH is nondegenerate under property MATH.
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math/9912158
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Let MATH denote the projection to the MATH-th factor (MATH). Let MATH be the diagonal embedding MATH. Then we have MATH. Hence MATH . If we substitute REF into the above, we get MATH . In particular, MATH is spanned by MATH's. If MATH for some MATH, then MATH. Hence we have MATH. The above equality REF implies MATH. This means that MATH is torsion-free. Thus we could assume MATH's are linearly independent in REF. Under this assumption, MATH is a basis of MATH, and REF implies that MATH is the dual basis. If we perform the same computation in MATH, we get the same result. In particular, MATH is a basis of MATH. However, MATH, MATH are in MATH, thus we have MATH. We also have MATH is an isomorphism. Thus MATH has property MATH. If MATH has MATH-action and REF holds in the equivariant MATH-group, we do the same calculation in the equivariant MATH-groups. Then the same argument shows that MATH has property MATH. The assertion for NAME groups and homology groups can be proved in the same way.
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math/9912158
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Since MATH is a union of connected components (possibly single component) of the fixed point set of the MATH-action on a nonsingular variety MATH, MATH is nonsingular. Suppose that MATH is a fixed point of the MATH-action. It means that MATH lies in the closed orbit MATH. But MATH converges to MATH as MATH. Hence the closed orbit must be MATH. Since MATH is equivariant, MATH is contained in MATH. In particular, MATH is projective.
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math/9912158
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Let MATH be the structure sheaf of the diagonal considered as a sheaf on MATH. By the above argument, the NAME complex of MATH gives a resolution of MATH: MATH where MATH. Thus we have the following equality in the NAME group MATH . Since MATH is injective and MATH is surjective, we have MATH . Each factor of the right hand side can be written in the form MATH for some MATH. For example, the first factor is equal to MATH where MATH is the weight space of MATH, that is, MATH . The remaining factors have similar description. Thus by REF, MATH has property MATH. Moreover, the above shows that MATH is generated by exterior powers of MATH, MATH and its duals (as a MATH-algebra). Note that these bundles have constant rank on MATH. If MATH have components MATH, MATH, , the structure sheaf of MATH (extended to MATH by setting MATH outside) cannot be represented by MATH, MATH. This contradiction shows that MATH is connected. The assertion for NAME groups can be proved exactly in the same way. By the above argument, the fundamental class MATH is the top NAME class of MATH, which can be represented as MATH for some MATH, MATH.
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math/9912158
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We apply REF. By CITE, the metric on MATH defined in REF is complete. By the construction, it is invariant under MATH, where MATH is the maximal compact subgroup of MATH. (Note that the hyper-Kähler structure is not invariant under the MATH-action, but the metric is invariant.) The moment map for the MATH-actions is given by MATH . This is a proper function on MATH. Thus REF is applicable. Note that we have MATH as in CITE. (Though our MATH-action is different from one in CITE, the same proof works.)
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math/9912158
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In order to relate MATH relative to MATH, MATH, MATH and MATH relative to MATH, MATH, MATH, we replace MATH by MATH factor by factor. CASE: First we want to replace MATH by MATH. We consider the following fiber square: MATH where MATH is the projection. We have MATH where we have used the base change CITE in the second equality and REF in the third equality. If MATH denotes the projection, we have MATH. Hence we get MATH . Similarly, we have MATH where MATH is the projection. Substituting this into REF, we obtain MATH . CASE: Next we replace MATH by MATH. By REF, we have a homomorphism MATH which is nothing but the identity operator. We will consider MATH as an element of MATH or MATH interchangely. We consider the fiber square MATH where MATH is the projection. By base change CITE, we get MATH . By the projection REF , we get MATH . Substituting this into REF, we have MATH where MATH, MATH are the projections, and we have used MATH and MATH. CASE: We finally replace MATH by MATH. By REF, we have a homomorphism MATH which is nothing but the identity operator. We consider the fiber square MATH . By base change CITE, we get MATH . Substituting this into REF, we obtain MATH where we have used REF in the second equality and MATH, MATH in the third equality. Finally, by REF, the homomorphism MATH is nothing but the identity operator MATH. Thus we have the assertion.
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math/9912158
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As in the proof of REF, we replace MATH by MATH factor by factor. CASE: First we replace MATH by MATH. Consider the following fiber square: MATH where MATH is the projection. By base change CITE and REF, we have MATH where MATH and MATH are projections. CASE: Consider the fiber square MATH where MATH is the projection. By base change CITE, we have MATH where we have used REF for MATH. Substituting this into REF, we get MATH where we have used REF in the second equality. Let MATH be the projection. By MATH, we have MATH . We also have MATH where MATH is the projection. Substituting these two equalities into REF, we get MATH . CASE: Consider the fiber square MATH . By base change CITE, we have MATH where we have used REF for MATH in the second equality. Substituting this into REF, we have MATH . By MATH and MATH, we get MATH . This proves our assertion.
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math/9912158
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See CITE.
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math/9912158
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The proof below is modelled on CITE. We give the proof for MATH. Then the same result for MATH follows by MATH, MATH. We consider the complex REF for MATH and MATH: MATH where we put suffixes MATH, MATH to distinguish endomorphisms. We have sections MATH and MATH of MATH and MATH respectively. Identifying these vector bundles and sections with those of pull-backs to MATH, we consider their zero locus MATH and MATH. As in the proof of CITE, we consider the transpose of MATH, MATH via the symplectic form. Their sum gives a vector bundle endomorphism MATH . It is enough to show that the kernel of MATH is zero at MATH. Take representatives MATH of MATH (MATH). Then we have MATH, MATH which satisfy REF for MATH. Suppose that MATH lies in the kernel. Then there exist MATH REF such that MATH . Then we have MATH . Hence we have MATH by the stability condition. Consider REF at the vertex MATH. Since MATH, MATH is an isomorphism. Hence REF implies that MATH is invariant MATH. Since MATH is a codimension MATH subspace, the induced map MATH is a scalar which we denote by MATH. Moreover, there exists a homomorphism MATH such that MATH . For another vertex MATH, MATH is an isomorphism, hence we can define MATH so that the same equation holds also for the vertex MATH. Thus we have MATH . Substituting REF into REF and using the injectivity of MATH, we get MATH . This means that MATH. Substituting REF into REF and noticing MATH is injective, we obtain MATH . Thus MATH is invariant under MATH . Arguing as above, we can find a constant MATH and a homomorphism MATH such that MATH . Substituting this equation into REF, we get MATH . This means that MATH. Hence MATH is injective.
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math/9912158
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Fix a subspace MATH with MATH. Let MATH be the parabolic subgroup of MATH consisting elements which preserve MATH. We also fix a complementary subspace MATH. Thus we have MATH. We will check the assertion for MATH. The assertion for MATH follows if we exchange MATH and MATH. We consider MATH . It is a linear subspace of MATH. Let MATH be the composition of the restriction of the moment map MATH to MATH and the projection MATH. Let MATH denote the set of MATH which is stable. It is preserved under the action of MATH and we have a MATH-equivariant isomorphism MATH . Note that the MATH-part of the moment map MATH vanishes on MATH thanks to the definition of MATH. The assertion follows if we check MATH is surjective at MATH. Here MATH is the natural projection. Thus it is enough to show that MATH is surjective. Suppose that MATH is orthogonal to MATH, namely MATH for any MATH. Hence we have MATH where MATH is the restriction of MATH to MATH. Therefore the image of MATH is invariant under MATH and contained in MATH. By the stability condition, we have MATH. Thus we have proved the assertion.
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math/9912158
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The latter statement follows from the first statement and REF . Thus it is enough to check the first statement. And the first statement follows from the transversality of intersections REF in MATH. Let MATH be the MATH-part of the moment map MATH. It induces a map MATH for MATH. Let us denote it by MATH. Thus we have MATH. Composing MATH with the projection MATH, we have a map MATH for MATH. We denote it also by MATH for brevity. It is enough to show that the restriction of the differential MATH to MATH is surjective on MATH. We consider the homomorphisms MATH, MATH defined in REF where MATH is replaced by MATH (MATH). We denote them by MATH and MATH respectively. Take a point MATH. Then MATH and there exists MATH such that MATH . The tangent space MATH at is isomorphic to the space of MATH such that MATH modulo the image of MATH where MATH and we have used the identification MATH for MATH. Now suppose that MATH is orthogonal to the image of MATH. Putting MATH, we consider MATH as an element of MATH. Then MATH for any MATH. Since the image of REF lies in the kernel of MATH, the above equality holds for any MATH satisfying REF. Taking MATH from MATH, we find MATH . Next taking MATH from the other component (data related to the vertex MATH), we get MATH . Comparing MATH-components, we find MATH . Comparing MATH-components, we have MATH . If we define MATH REF implies that MATH is MATH-invariant and contained in MATH. Thus MATH by the stability condition. In particular, we have MATH. This means that MATH is surjective.
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math/9912158
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The latter statement follows from the former one together with the projection REF . Thus it is enough to prove the former statement. By definition, MATH consists of MATH such that there exists MATH satisfying REF. We fix representatives MATH, MATH. Then the above MATH is uniquely determined. Recall that we have chosen the identification MATH for MATH over MATH. Let us define MATH by MATH . We define a new datum MATH . By definition, we have MATH . Hence MATH is contained in MATH. Moreover, MATH is independent of the choice of the representative MATH. Thus we have defined a map MATH by MATH which is the inverse of the restriction of MATH. In particular, this implies MATH . Since MATH is a principal MATH-bundle, we have MATH . Thus we have proved the first equation. The second equation can be proved in a similar way.
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math/9912158
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As is explained in CITE, the result follows from REF. The factor MATH is introduced to make the differential in the NAME complex equivariant.
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math/9912158
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CASE: Generalizing REF, we have the following formula for MATH: MATH where the summation runs over the set of ordered MATH-tuples MATH such that MATH, MATH for MATH. Choose MATH so that MATH . Consider the following term appeared in the above formula: MATH . By CITE it is equal to MATH where MATH is the NAME polynomial and MATH . Thus we have the assertion. CASE: By REF we have MATH .
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math/9912158
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It is enough to check that MATH, MATH, MATH and the coefficients of MATH are mapped to MATH. For MATH and the coefficients of MATH, the assertion is clear from the definition. For MATH and MATH, we can use a reduction to rank MATH case as in REF. Namely, it is enough to show the assertion when the graph is of type MATH. Now if the graph is of type MATH, REF together with REF shows that MATH is represented by a certain line bundle over MATH extended to MATH by MATH. We leave the proof for MATH as an exercise. The only thing we need is to write down an analogue of REF for MATH. It is straightforward.
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math/9912158
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As in REF, we may assume that the graph is of type MATH. Now REF together with the fact that NAME polynomials form a basis of symmetric polynomials implies the assertion.
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math/9912158
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The following proof is an adaptation of proof of CITE, which was inspired by CITE in turn. We need the following notation: MATH . We prove MATH by induction on the dimension vector MATH. When MATH, the result is trivial since MATH. Consider MATH and suppose that MATH . Take MATH. We want to show MATH. We may assume that the support of MATH is contained in an irreducible component of MATH without loss of generality. In fact, suppose that MATH such that MATH is an irreducible component. Since MATH is a closed subvariety of MATH and since MATH is a closed subvariety of MATH, we have the diagram MATH where the first and the second row are exact by REF. Thus there exists MATH such that MATH. Then MATH, therefore there exists MATH such that MATH. By the induction on the number of irreducible components in the support, we may assume that the support of MATH is contained in an irreducible component, which is denoted by MATH. Let us consider MATH defined in REF. If MATH for all MATH, MATH must be MATH by REF. We have nothing to prove in this case. Thus there exists MATH such that MATH. Set MATH. By the descending induction on MATH, we may assume that MATH . Since MATH is an open subvariety of MATH, we have an exact sequence MATH by REF. Consider MATH. By REF, it is enough to show that MATH . Since MATH, the support of MATH is contained in MATH. We have a map MATH which is the restriction of the map REF. Recall that this map is a NAME bundle (see REF). Let us denote its tautological bundle by MATH. Then MATH can be written as a linear combination of elements of the form MATH where MATH is a tensor product of exterior powers of the tautological bundle MATH and MATH. Since the homomorphism MATH is surjective by REF, there exists MATH such that MATH. Consider MATH. By REF, it is isomorphic to MATH and the map MATH can be identified with the projection to the second factor. Moreover, the tautological bundle MATH is identified with the restriction of the natural vector bundle MATH. Hence we have MATH where MATH is considered as an element of MATH. By REF, MATH can be written as a linear combination of elements MATH where MATH is the projection. By REF, MATH . Hence MATH. Thus we have shown REF.
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math/9912158
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The assertion is proved exactly as CITE. Note that the regularity assumption of MATH is not used here.
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math/9912158
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CASE: If we restrict the complex MATH to MATH, MATH is surjective and MATH is injective by REF. Thus MATH is represented by a genuine MATH-module, and MATH is a polynomial in MATH. The degree is equal to MATH by the definition of MATH. CASE: The first equation is the consequence of MATH, which follows from REF. The remaining equations follows from the definition and REF. CASE: The assertion is proved exactly as in REF. Note that the assumption MATH is used here in order to apply REF.
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math/9912158
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It is enough to show that there exists a simple l-integrable l-highest module with given NAME polynomials MATH. We can construct it as the quotient of the standard module MATH by the unique maximal proper submodule. (The uniqueness can be proved as in the case of NAME modules.) Here the parameter MATH is chosen so that MATH, that is, MATH is a normalization of the characteristic polynomial of MATH.
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math/9912158
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REF follows from REF . We show REF . Note that MATH is mapped to MATH under MATH. And MATH is mapped to the fundamental class MATH under MATH. Combining with the projection REF , we find that the operator MATH is mapped to MATH under the homomorphism REF. Thus as an operator on MATH, it is equal to MATH where MATH is the inclusion. Now, on a connected space MATH, any MATH acts on MATH as a scalar plus nilpotent operator, where the scalar is MATH-part of MATH. In our situation, the MATH-part of REF is given by MATH. (Although we do not prove MATH is connected, the MATH-part is the same on any component.) Furthermore, MATH determines all eigenvalues of the operator MATH acting on MATH. Hence it determines the conjugacy class of the homomorphism MATH. Thus the generalized eigenspace of MATH with the eigenvalue MATH coincides with MATH.
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math/9912158
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By REF, MATH is a geneneralized eigenspace for MATH for a homomorphism MATH. Thus it is enough to study the eigenvalue. We consider MATH, MATH as MATH-modules via MATH as before. Let MATH, MATH be weight space as in REF. By the definition of MATH and MATH, we have MATH . By REF we have MATH where MATH, MATH are defined by REF. Let MATH be the set of eigenvalues of MATH on MATH counted with multiplicities. Then we have MATH . Thus we have MATH . Note that the term for MATH with MATH has the contribution MATH and any other terms are monomials of MATH which are not constant. Moreover, we have MATH by REF. This completes the proof.
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math/9912158
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We have the following equality in MATH: MATH where MATH is (the restriction of) the natural line bundle over MATH. The assertion follows immediately.
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math/9912158
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Recall that we have a distinguished vector (we denote it by MATH) in the standard module MATH REF . It has the properties listed in REF. In particular, it is the eigenvector for MATH, and the eigenvalues are given in terms of MATH therein. In the present setting, MATH is equal to MATH. Let MATH . We have MATH. We want to show that any nonzero submodule MATH of MATH is MATH itself. The weight space decomposition (as a MATH-module) REF of MATH induces that of MATH. Since the set of weights of MATH is bounded from MATH with respect to the dominance order, there exists a maximal weight of MATH. Then a vector in the corresponding weight space is killed by all MATH by the maximality. Thus MATH contains a nonzero vector MATH. Hence it is enough to show that MATH since we have already shown that MATH in REF. Let us consider the operator MATH where MATH. If we consider such operators for various MATH, MATH, they form a commuting family. Moreover, MATH is invariant under them since we have the relation MATH where MATH, MATH are tautological bundles over MATH, MATH respectively. (Here MATH.) Thus MATH is a direct sum of generalized eigenspaces for MATH. Let us take a direct summand MATH. By REF, MATH is contained in MATH for some MATH. If we can show MATH, then we get MATH since MATH is an arbitrary direct summand. Since MATH is generic, we have MATH. Hence MATH and MATH is a nonsingular projective variety (having possibly infinitely many components). By the NAME duality, the intersection pairing MATH is nondegenerate. Let MATH denote the transpose of MATH with respect to the pairing MATH, namely MATH . By the definition of the convolution, MATH is equal to MATH where MATH is the map exchanging the first and second factors and MATH is the induced homomorphism on MATH. Let us consider MATH, where MATH is as above and MATH is any other homomorphism. It is just the projection of MATH to the component MATH. We have MATH for some MATH, and MATH, MATH. Those MATH and MATH come from asymmetry in the defintion of MATH, MATH and in the homomorphism REF. We do not give their explicit forms, though it is possible. What we need is that MATH is written by tensor powers of exterior products of MATH for various MATH, MATH. Thus we can write MATH for some MATH by REF. Therefore, for MATH, we have MATH for any MATH, MATH, MATH, MATH. Here we have used MATH, MATH, MATH. Since MATH, we have one of the followings: CASE: MATH for any MATH, CASE: MATH. The first case is excluded by the nondegeneracy of MATH. Thus we have MATH. Let us prove the last assertion. First consider the case MATH for some MATH. If MATH is not a root of unity, MATH is generic for the quiver variety MATH. Hence the above shows that the standard module for MATH is simple, and hence gives an l-fundamental representation. Let us return to the case for general MATH. Let MATH be eigenvalues of MATH counted with multiplicities. By REF, it is enough to show that MATH where MATH is the standard module for MATH with MATH. Since MATH has no odd homology groups REF , we have MATH where MATH denotes the topological NAME number. By a property of the NAME number, we have MATH . If we take a maximal torus MATH of MATH, the fixed point set MATH is isomorphic to MATH . Hence we have MATH . Since we have MATH we get REF.
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math/9912158
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Since MATH is equal to MATH in the NAME group, the assertion follows from the discussion above.
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math/9912158
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We use the transversal slice in REF. The idea to use transversal slices is taken from CITE. Choose and fix a point MATH. Suppose that MATH is contained in a stratum MATH for some MATH. We first show If MATH denote the constant local system on MATH, the corresponding vector space MATH is nonzero. If we restrict MATH to the component MATH, then we have MATH where MATH is a direct summand of MATH. The summation runs over the set of pairs MATH such that MATH is contained in MATH. (In fact, REF was obtained by applying the decomposition theorem to each component MATH and taking direct sum.) If we restrict REF to the open stratum MATH of MATH, the right hand side of REF becomes MATH where the summation runs over the set of isomorphism classes of irreducible local systems MATH on MATH. On the other hand, MATH induces an isomorphism between MATH and MATH by REF. This means that the restriction of the left hand side of REF is the constant local system MATH. Hence we have MATH, and MATH is nonzero. This is the end of the proof of the claim. The claim implies the first assertion of REF . Let us prove the latter assertion of REF . Suppose MATH. Then we have MATH and MATH by REF. By REF, the corresponding l-weight space is nonzero, where the l-weight MATH is given by REF. Furthermore, since MATH can be represented by a genuine MATH-module over a point in MATH by REF, MATH is a polynomial in MATH. Thus MATH is l-dominant. Conversely suppose that we have the l-weight space with the l-weight REF is nonzero. Since MATH is not a root of unity, the MATH-analogue of the NAME matrix MATH is invertible. Hence REF determines MATH. Thus the l-weight space is precisely MATH by REF. In particular, we have MATH, and hence MATH. Furthermore, if we decompose MATH into MATH as in REF, we have MATH since the l-weight REF is l-dominant. By REF , MATH is surjective for any MATH, MATH on a nonempty open subset of MATH. By REF (plus the subsequent remark), MATH is nonempty. This shows the latter half of REF . Let MATH denote the dimension vector corresponding to MATH, that is, MATH. Take a transversal slice to MATH at MATH as in REF. Let MATH be its intersection with MATH. Since the transversal slice in REF can be made MATH-equivariant REF , it is a transversal slice to MATH (at MATH) in MATH. Let MATH. Let MATH, MATH be the inclusions. The stratification MATH induces by restriction a stratification MATH where MATH. Any intersection complex MATH restricts (up to shift) to the intersection complex MATH by transversality. Here MATH is the restriction of MATH to MATH. Taking MATH of REF, we get MATH . Let MATH be the inclusion. It induces two pull-back homomorphisms MATH, MATH, and there is a natural morphism MATH for any MATH. We apply these functors to both hand sides of REF and take cohomology groups. By a property of intersection cohomology sheaves (see CITE), the homomorphism MATH is zero unless MATH (or equivalently MATH), in which case it is a quasi-isomorphism. Thus MATH where the summation runs over isomorphism classes of irreducible local systems on MATH, and MATH is the fiber of the local system MATH at MATH. Moreover, REF is a homomorphism of MATH-modules, and REF is an isomorphism of MATH-modules, where the module structure on the right hand side is given by MATH. On the other hand, we have MATH . As shown in REF, the right hand side is isomorphic to the standard module MATH. Thus the left hand side of REF is a quotient of MATH, and it is indecomposable by REF. Thus the right hand side of REF consists of at most single direct summand. Since we have already shown that MATH in the claim, we get MATH if MATH is a nonconstant irreducible local system. Since MATH was an arbitrary point, we have REF . Let us prove REF . For the proof, we need a further study of REF. By the above discussion, we have MATH . By the base change theorem, we have MATH where MATH is the restriction of MATH to MATH. Further, we have MATH since MATH is a nonsingular submanifold of MATH. Applying the NAME duality, we have MATH . Hence REF becomes MATH where MATH is the dual space of MATH as a complex vector space. Let us introduce a MATH-module on MATH by MATH where MATH denote the dual pairing, MATH is the exchange of two factors of MATH, and MATH is the induced homomorphism on MATH. Then REF is compatible with MATH-module structures (compare CITE). The decomposition REF induces a similar one for MATH: MATH . The homomorphism MATH respects the decomposition, and induces a decomposition on REF. Recall that we have the distinguished vector MATH in MATH. The component MATH of MATH is MATH-dimensional space MATH. (See REF.) By the above discussion, MATH is not annihilated by the above homomorphism MATH. Thus we may consider MATH also as an element of MATH. We want to show that any nonzero MATH-submodule MATH of MATH is MATH itself. Our strategy is the same as the proof of REF. Since we already show that MATH is a quotient of MATH, REF implies MATH. Thus it is enough to show that MATH contains MATH. To show this, consider MATH . By the argument as in the proof of REF, MATH contains a nonzero vector in MATH. Hence it is enough to show that MATH. As in the proof of REF, MATH is a direct sum of generalized eigenspaces for MATH. Let us choose and fix a direct summand MATH contained in MATH. Then MATH satisfies MATH for any MATH, MATH. Since MATH by REF, the above equation implies that MATH. Thus we get MATH as desired. We have shown REF during the above discussion. Let us prove REF . Since MATH is a locally trivial topological fibration on each stratum MATH, MATH and MATH are isomorphic if both MATH and MATH is contained in MATH. Conversely, if MATH and MATH are isomorphic as MATH-modules, the corresponding l-highest weights MATH and MATH are equal. Since MATH determines the homomorphism MATH as in the proof of REF , MATH and MATH are in the same stratum.
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math/9912158
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See CITE and CITE.
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math/9912158
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By the decomposition theorem for a semi-small map CITE, the left hand side of REF decomposes as MATH where MATH is a component of MATH and MATH is the intersection complex associated with an irreducible local system MATH on MATH. Moreover, by CITE, we have MATH where MATH runs over the set of irreducible local systems on the component of MATH containing MATH. But as argued in the proof of REF, the indecomposability of MATH implies that no intersection complex associated with a nontrivial local system appears in the summand. Moreover the left hand side of REF is independent of the choice of the component by REF. Thus we can combine the summation over MATH together as MATH . Our remaining task is to identify the index set of MATH. The fundamental class MATH is nonzero if MATH is nonempty. Thus MATH is nonempty if and only if MATH . By CITE and the construction, MATH is isomorphic to the weight space of weight MATH in MATH. Thus it is nonzero if and only if MATH.
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math/9912167
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Checking that the diagonals are NAME and conelike is complicated but routine; the more significant issue is self-transversality. We describe the minimal strata containing a configuration MATH. Let MATH be the connected components of the inverse images in MATH of points in MATH, not including components which consist of a single point. If MATH lies on some of the diagonals, then this list of subgraphs is non-empty. A subgraph MATH might not be REF-connected. Rather, it is has tree-like structure consisting of maximal REF-connected subgraphs which share cut vertices. We call such a structure a cactus and REF-connected subgraphs lobes. A diagonal MATH is a minimal stratum containing MATH if and only if MATH is a maximal REF-connected subgraph of some MATH. Checking that these strata are mutually transverse is again complicated but routine.
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math/9912167
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Let MATH be the union of all faces of MATH (both finite and infinite) and let MATH be its image in MATH. Consider the cohomology exact sequence of the triple MATH: MATH . On the other hand, MATH since MATH cuts MATH into the pieces of MATH, and on each of these we have a unique top cohomology class, the fundamental class. Thus, MATH is a space of graphs modulo some relations. These graphs are not quite NAME as in the definition of MATH, since the edges are labelled and each edge is oriented. (NAME graph has a global choice of orientations up to sign). But there is a forgetful map MATH from MATH to NAME graphs. To prove the lemma, we only need to check that the relations given by MATH become trivial or the NAME relation under MATH. The space MATH might be rather complicated, but by the same exact sequence it is generated by one cohomology class for each face (of MATH). By cases: CASE: The sum of the six graphs in REF descends to the NAME relation. CASE: We glue together several different graphs which become identical (up to sign) under MATH. Each involution defined in REF negates MATH. For example, if the graph MATH is the single point MATH, then MATH is orientation-reversing, and two edges are reversed. Thus half of the elements of the group generated by these involutions negates MATH, so the total sum vanishes. CASE: Since cohomology is computed relative to these faces, they impose no relation. CASE: Since we reduce the dimension of this face, it imposes no relation.
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math/9912167
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The existence of MATH originates with the geometry of the configuration space MATH. This is a manifold with corners whose interior is MATH, the space of pairs of distinct points in MATH. If MATH, it is clearly homotopy equivalent to MATH. In the general case it has the same homology by a NAME argument. Each of the gluings used to make MATH from MATH is chosen to preserve the second cohomology, although higher cohomology may also appear. The class MATH clearly exists in the absolute cohomology MATH; the question is whether it exists uniquely in cohomology relative to MATH. Observe first that if MATH, then MATH has codimension at least REF. Each allowed graph MATH on MATH with MATH edges (of which there are finitely many) has at most MATH vertices. Thus there are at most MATH degrees of freedom in embedding MATH in MATH. In addition, REF of these degrees of freedom are absorbed by invariance under the homothety group MATH, so MATH has dimension at most MATH. Choose a point MATH . For each MATH, choose a cocycle MATH that represents the class MATH and that is localized at MATH (or for concreteness, a small simplex containing MATH) in the standard sphere MATH. Recall that the space of relative cocycles MATH is a subspace of the space of absolute cocycles MATH. The cocycle MATH exists as a relative cocycle because it avoids MATH. It represents a non-trivial cohomology class because relativization can only diminish the space of boundaries. Thus MATH exists in relative cohomology. To show uniqueness, suppose that MATH is an arc connecting MATH with some point MATH and which is disjoint from MATH: MATH . If MATH represents MATH and is localized at MATH, then there is a REF-cochain MATH localized along MATH which is a homology between MATH and MATH: MATH . In this case MATH is a homology between MATH and MATH, where MATH . Since MATH has codimension REF in MATH, any two points in its complement can be connected by a sequence of moves of this type. Hence MATH is unique.
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math/9912167
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We can measure MATH by pairing it with REF-cycles in MATH. There are several kinds of these, but the only kind that can have non-zero pairing is represented by a torus MATH, where MATH and MATH are two disjoint knots in MATH. In this case MATH measures their linking number in MATH: MATH . Likewise MATH measures their linking number in MATH. The difference is the product of linking numbers with MATH: MATH . This is easy to see when MATH is an unknot, since surgery on MATH has the effect of twisting MATH and MATH about each other without changing MATH, as in REF . Since MATH pairs with homology classes in the same way as MATH, the two cocycles are homologous.
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math/9912167
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Say that a vertex of MATH provides a dollar to the component MATH if it lies in the knot MATH in MATH, and that it provides REF cents if it lies in the NAME surface MATH. By REF , each component MATH, of which there are MATH, needs a dollar in order for MATH to depend on MATH at the configuration MATH. Each vertex, of which there are MATH, can provide at most , and only by lying at the intersection of three NAME surfaces. The components need MATH dollars, which is the most that the vertices can provide. Therefore the vertices lie on the NAME surfaces.
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math/9912168
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Suppose MATH is a section of MATH with MATH. Such a section is the same as a MATH-equivariant function MATH on the universal cover MATH of MATH. Let MATH. Then MATH drops down to a closed real-valued one-form on MATH and MATH where MATH is the lift of MATH to MATH, a path from MATH to MATH. On the other hand, suppose MATH for some closed one-form MATH. Then the pullback of MATH to MATH is the differential of some function MATH and MATH is a MATH-equivariant function on MATH, which gives a non-vanishing section of MATH.
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math/9912168
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Suppose MATH is a section of MATH and an eigenspinor of MATH with eigenvalue MATH. Then MATH so MATH is an eigenspinor of MATH, also with eigenvalue MATH. If MATH is an eigenspinor of MATH then MATH is an eigenspinor of MATH with the same eigenvalue.
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math/9912168
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As before let MATH be a normalized section of MATH and let MATH. Let MATH where MATH and denote by MATH the vectorfield MATH on MATH. The spin structure MATH on MATH induces a spin structure on MATH, the associated spinor bundle is equal to MATH. To shorten the notation we will write this as MATH. Let MATH be a smooth function such that MATH and MATH. Define a connection MATH on sections of the trivial complex line bundle MATH over MATH by MATH where MATH. This connection is metric and the induced connection MATH on MATH interpolates between MATH close to MATH and MATH close to MATH. Let MATH be the NAME operator on MATH constructed using the connection MATH. Close to the boundary component MATH of MATH we have MATH and close to MATH we have MATH. The NAME index theorem CITE tells us that MATH where the index on the left hand side is the NAME index of MATH acting on sections of MATH satisfying the NAME boundary condition. Since MATH acting on sections of MATH is isospectral to MATH acting on sections of MATH we have MATH and MATH . Next we compute MATH and for this we need the curvature MATH of MATH. For MATH we write MATH where as in the proof of REF MATH is a closed integer-valued one-form. For MATH we have MATH so MATH since MATH is closed. Next we have MATH and we conclude that MATH . Thus we find that the NAME character of MATH is MATH and we see that the integral in REF is MATH . When we integrate over the MATH factor only the term with MATH will contribute and we have left MATH . The same calculation can now be used again to see that this integral is the index of a twisted NAME operator. Let MATH be the circle of length MATH. Then MATH . Since MATH there is a complex line bundle MATH with MATH, curvature REF-form MATH, and MATH. This means that REF is equal to MATH . From REF we now get MATH which is a half integer. We have proved REF. To prove REF note that the last term, MATH in REF does not depend on the spin structure MATH. When we take a second difference, MATH these terms will cancel, and REF follows.
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math/9912168
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Since MATH we have from REF MATH . The integrand MATH only contains terms of degrees MATH, MATH so the integral over MATH which has dimension MATH vanishes, and the corollary follows.
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math/9912168
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From REF we have MATH . REF tells us that MATH and using REF we see that MATH . Since MATH has dimension MATH and MATH only contains terms of degrees MATH, MATH, the integral vanishes and we are left with MATH which proves the theorem.
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math/9912171
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Denote an outermost sub-disk of some MATH by MATH and suppose it is cut off by an arc MATH on MATH. By the "Facts" proved in REF pp REF, any such outermost arc MATH must have both end points on a single disk MATH which belongs to some n-float of genus MATH. Furthermore MATH must separate the boundary components of MATH. Assume further that MATH where MATH is an arc on the n-float meeting MATH in exactly two points MATH. On MATH there is a small arc MATH so that MATH is a simple closed curve on the n-float bounding a disk MATH there, since the n-float has no genus (see REF below). Furthermore MATH is a simple closed curve on MATH which together with a boundary component of MATH bounds a sub-annulus of MATH. Hence MATH bounds a disk MATH on the decomposing MATH-sphere of MATH intersecting MATH in a single point. Thus we obtain a MATH-sphere MATH which intersects the knot MATH in a single point. This is a contradiction which finishes REF . For REF , assume that the outermost disk MATH is contained in MATH, say, and that genus MATH is one. As before we have MATH where MATH is an arc on MATH and a small arc MATH so that MATH is a simple closed curve on MATH. If MATH bounds a disk in MATH we have the same proof as in REF . If MATH does not bound a disk on MATH we consider small sub-arcs MATH and MATH of MATH which are respective closed neighborhoods of MATH. These arcs together with a small arc MATH on MATH and MATH bound a small band MATH on MATH. Notice that MATH is an annulus MATH. The annulus MATH together with the sub-annulus MATH of MATH cut off by MATH defines an annulus MATH which determines an isotopy of a meridian curve in MATH to a simple closed curve MATH on MATH. Note that MATH is a solid torus and MATH which is generated by a meridian MATH of MATH. Hence MATH (see REF ). Now we can consider the annulus MATH. If it is non-boundary parallel then since both knots MATH are prime it must be a decomposing annulus which has at least one less disk component intersection than MATH in contradiction to the choice of MATH. If it is boundary parallel, then as above, we have MATH as a decomposing annulus with a smaller number of disks. Again in contradiction to the choice of MATH. So genus MATH cannot be one and this finishes REF .
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math/9912171
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Since the MATH part of an outer-most disk must be contained in a float of genus greater than one we must have a MATH on the float to create the genus. The MATH-handle is the tunnel disjoint from the decomposing annulus MATH.
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math/9912171
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No one of the two knots has a NAME splitting where the meridian is a primitive element since MATH. A primitive meridian would mean that the NAME splittings of the knots will induce a NAME splitting of the connected sum which would make the tunnel number additive or less. Now drill a tunnel in MATH with end points on opposite sides of the meridian curve on MATH for one of the knots MATH and add it as a MATH-tunnel to MATH thus making the meridian primitive at the expense of increasing the genus by MATH. The two NAME splittings will now induce a NAME splitting on the connected sum which is of genus MATH. It is minimal since MATH and weakly reducible by REF .
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math/9912171
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We can assume that the decomposing annulus MATH intersects the NAME splitting MATH as follows: It intersects MATH in two vertical annuli and MATH in one meridional annulus. (This is a consequence of the fact that MATH is induced by the respective NAME splittings). Choose two essential disks MATH and MATH for MATH on both sides of MATH, for example cocore disks for tunnels. Note that MATH and MATH. The handlebody MATH is obtained from MATH and MATH by gluing them along the meridional annulus MATH. Since MATH turns out to be a handlebody MATH must be a primitive annulus in at least one of MATH or MATH, say MATH. So there is at least one essential disk MATH in MATH which is disjoint from MATH and hence is also an essential disk in MATH. But MATH is disjoint from MATH as MATH is also disjoint from MATH and is on the opposite side. Hence the NAME splitting MATH is weakly reducible.
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math/9912171
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Assume in contradiction that MATH is a strongly irreducible NAME splitting for MATH. Let MATH be the NAME surface and let MATH be the decomposing annulus for the connected sum minimizing the intersection with MATH. We can assume (see REF ) that after an isotopy of the annulus MATH is a collection of essential curves on both MATH and MATH. Hence, as we assumed that MATH is the compression body containing MATH then MATH is composed of two vertical annuli MATH and a minimal collection of essential annuli MATH and MATH is composed of a minimal collection of essential annuli MATH. By REF we can find essential disks MATH in MATH respectively which are disjoint from MATH and MATH. Since MATH share a boundary component with MATH and MATH we can conclude that the disks MATH are disjoint from MATH. The annulus MATH splits each of MATH and MATH into two unions of handlebodies MATH and MATH respectively where i = REF depending if the component is in MATH or MATH. If the disks MATH are contained in MATH and MATH respectively, MATH, for different values of MATH and MATH then MATH as both of MATH and MATH are disjoint from MATH. Hence the NAME splitting MATH is weakly reducible in contradiction. So we can assume that both of MATH and MATH are contained in MATH and MATH for the same MATH, say MATH that is, on the same side of MATH. Consider now the components of MATH contained in MATH and MATH. An innermost disk argument shows that each of these components must be incompressible in MATH and MATH as otherwise we obtain a compressing disk MATH disjoint from MATH which is disjoint from both MATH and MATH and hence the NAME splitting MATH is weakly reducible in contradiction. The boundary curves of any component of MATH contained in MATH and MATH are essential curves on the meridional decomposing annulus MATH and hence are isotopic to meridian curves in MATH. Therefore they are isotopic to meridian curves in MATH. Thus these components of MATH are horizontal surfaces. Since we assumed that such surfaces do not exist in MATH we obtain a contradiction to our assumption that MATH is a strongly irreducible NAME splitting of MATH.
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math/9912171
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Assume first that the annuli MATH are simultaneously primitive in MATH. The proof will be by induction on MATH. For MATH we can glue MATH to MATH along MATH and MATH to obtain a manifold MATH. Since the annuli MATH and MATH are incompressible we have that MATH. The generator of the MATH is a primitive element in the free group MATH so MATH is a free group. It now follows from the Loop Theorem that MATH is a handlebody. Assume by induction that MATH is a handlebody. The annulus MATH is disjoint from the annuli MATH and MATH and is still primitive in MATH as the annuli MATH are simultaneously primitive and non-parallel and hence there is an essential disk MATH in MATH which is disjoint from MATH and MATH and which intersects MATH in a single arc. Now MATH is obtained from MATH by gluing the primitive annulus MATH to the annulus MATH. Hence MATH is an HNN extension of the free group MATH where two MATH-subgroups are identified and the generator of one of them is a primitive element. It follows that MATH is a free group and again by the Loop Theorem MATH is a handlebody. For the proof in the other direction: Assume that MATH is a handlebody MATH and let MATH and MATH be as in the theorem. Let MATH be the result of cutting MATH along a maximal set of compression disks of MATH. Note that gluing MATH to MATH along MATH and MATH yields a handlebody. As it is obtained from the handlebody MATH by cutting it along disks which are disjoint from both of MATH and MATH. Up to relabeling we may assume that MATH is the set of annuli in MATH which are a longitudinal annulus of some solid torus component MATH, MATH of MATH containing no other MATH. Denote by MATH, and let MATH. There is no compressing disk in MATH intersecting MATH in a single essential arc. As any such disk would define another torus components MATH containing some annulus MATH, MATH and no other annulus. Let MATH and MATH be the corresponding subsets of MATH. Then MATH can be obtained by gluing MATH to MATH along MATH and MATH to obtain a manifold MATH, and then gluing MATH to MATH along MATH and MATH. The manifold MATH is a handlebody and is homeomorphic to MATH by the definition of MATH and the first part of the theorem. Hence the annuli MATH are still non-primitive annuli on MATH. If MATH is not simultaneously primitive then MATH is non-empty, hence after gluing the remaining components of MATH to MATH, the surface MATH is an essential surface in the handlebody MATH because there is no compressing or boundary compressing disk for this surface, which contradicts the fact that there are no essential non-disk surfaces in a handlebody.
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math/9912171
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Let MATH be the NAME splitting of MATH determined by the minimal tunnel system which intersects the decomposing annulus MATH in a single point. We can therefore assume that MATH. The once punctured annulus MATH has two boundary components coming from the vertical annuli MATH and denoted by MATH respectively and one boundary component MATH coming from the tunnel. As MATH intersects MATH minimally MATH is an incompressible planar surface in a handlebody and hence is boundary compressible. A boundary compression cannot be on an arc connecting MATH to MATH as then we could use the compressing disk to isotope the tunnel off MATH. Such an arc will be called of Type I. Furthermore a boundary compression cannot be on an arc connecting MATH to MATH as then MATH will be boundary parallel in contradiction. Hence the boundary compressing arc will connect MATH to itself and since it is non-trivial it must separate MATH and MATH. Such an arc will be called of Type II. Choose a meridional system of disks MATH for MATH. Each disk in MATH must intersect MATH as otherwise the NAME splitting will be weakly reducible and we are done. An outermost arc of intersection MATH on some MATH separates a boundary compressing sub-disk MATH and from the previous paragraph MATH is an arc of type II on MATH . We can boundary compress MATH along MATH or alternatively isotope MATH along MATH. Doing the second operation does not change MATH or the isotopy class of the NAME splitting MATH, but does change the intersection of the "new" NAME surface, also denoted by MATH, with MATH. The result is that now MATH, where MATH is an essential sub-annulus of MATH which contains the disk MATH. The intersection MATH, where MATH are also essential sub-annuli of MATH (as in REF ). CASE: Let MATH denote the components of MATH and MATH denote the components of MATH. Assume that the disk MATH is contained in MATH or MATH. Note that isotoping the NAME surface MATH along MATH changes the induced NAME splitting only on the knot complement containing MATH, (that is, on MATH only!). On the induced NAME splitting of MATH this isotopy is equivalent to cutting MATH along MATH to obtain MATH and adding the MATH-handle MATH to MATH to obtain MATH. It is possible that in this case MATH might not be a handlebody. It is also possible that MATH is a separating disk in MATH and in this case, MATH might have two components MATH and MATH. The annuli MATH are essential annuli contained in MATH which together separate MATH. Hence, when we cut MATH along them we obtain a handlebody MATH, and if neither of MATH or MATH is separating a handlebody MATH. If one of MATH or MATH is separating then MATH splits into two handlebodies MATH and MATH . This is the situation corresponding to the disk MATH being a separating disk in MATH. Denote the "traces" of MATH and MATH on MATH by MATH. Since MATH and only one tunnel gets split into two arcs by cutting along MATH it follows that after cutting MATH along MATH there are two possibilities: The induced NAME splitting MATH of MATH is of minimal genus and MATH of MATH is of minimal genus plus one or vice versa Up to relabeling the knots we assume that MATH is of minimal genus. CASE: If one of MATH or MATH is primitive in MATH then MATH has a weakly reducible NAME splitting of minimal genus. Proof REF : If the disk MATH is contained in MATH then MATH is a NAME splitting of minimal genus for MATH. So, if either MATH or MATH is a primitive annulus on MATH (which ,in this case, is equal to MATH less a collar ) we will treat an isotopic image of the primitive annulus MATH or MATH respectively on MATH as a decomposing annulus. Now glue MATH to MATH along this annulus to obtain a NAME splitting of MATH which is of minimal genus (as that of MATH) and is weakly reducible by REF . If, on the other hand, the disk MATH is contained in MATH then recall that we obtain MATH from MATH by identifying together the two "traces" (copies) of the disk MATH on MATH, that is, adding a MATH-handle to these traces. These traces intersect both of MATH and MATH in a single arc each. Hence if one of MATH or MATH is primitive in MATH it would also be primitive in MATH, regardless of whether MATH and MATH are separating or not. We now use the same argument as above to obtain a weakly reducible NAME splitting of the same genus as that of MATH of MATH. Thus we can assume that both of the annuli MATH and MATH are not primitive in MATH. Since MATH is a handlebody it follows that MATH and MATH must be primitive in MATH: By setting MATH, MATH and MATH, MATH we satisfy the conditions of REF and can conclude that MATH and MATH are simultaneously primitive in MATH. If it happens that MATH has more than one component we certainly have disjoint annuli intersecting disjoint disks in a single arc. We will refer to this situation as the annuli being extended simultaneously primitive. CASE: If MATH are simultaneously primitive or extended simultaneously primitive on MATH the complement with the non-minimal genus NAME splitting, then MATH is a weakly reducible NAME splitting of minimal genus. Proof REF : The induced NAME splitting of MATH is of genus at least three since it is induced by a tunnel system containing at least two tunnels, that is, one interior tunnel (by REF ) and the "half" tunnel coming from the split tunnel crossing MATH. Assume that the disk MATH is contained in MATH so after cutting MATH along MATH we obtain either a handlebody of genus at least two with two simultaneously primitive annuli on it or a disjoint union of two handebodies one of which has at least genus two with two extended simultaneously primitive annuli on them. Thus in both cases there is at least one essential disk MATH in MATH, (a separating disk in the first case), which is disjoint from MATH and MATH and hence from MATH. Since MATH, as before, the disk MATH is an essential disk in MATH which is disjoint from MATH and hence from the essential disk MATH which is the image of the disk MATH pushed slightly into MATH. Thus the NAME splitting MATH of MATH is weakly reducible and we are done (see REF ). CASE: Assume therefore that the disk MATH is contained in MATH. If MATH then MATH is a genus two handlebody and after cutting MATH along MATH we obtain either one or two solid tori (depending if MATH is separating or not) embedded in MATH with non-primitive annuli on their boundary. Extend these annuli all the way to MATH. When attaching disks to these meridional annuli one obtains a Lens space contained in MATH, which is a contradiction. If MATH then MATH is a genus three handlebody and after cutting MATH along MATH we obtain either one solid torus component with one or two non-primitive annuli on its boundary (if MATH is separating) or a genus two handlebody with two non-primitive annuli on its boundary (if MATH is non-separating). In both cases the components are embedded in MATH. The first case is dealt with as in the previous paragraph. In the second case, then after adding two meridional disks along these annuli we obtain a MATH-sphere MATH which intersects MATH in four points. In particular MATH bounds a MATH-ball on both sides. If we change the order of cutting along MATH and adding disks by first adding the two meridional disks to the meridional annuli on MATH we obtain a solid torus MATH with MATH as its unique meridional disk. Since adding MATH to the complement of MATH gives us a MATH-ball then MATH is also a solid torus MATH. The solid torus MATH can also be obtained from the genus three compression body MATH as follows: Fill MATH with MATH to get a pair MATH. Now cut the pair MATH along meridional disk corresponding to the meridional annul.i on MATH. These annuli are not parallel on MATH so we get a solid torus MATH whose unique meridian disk MATH is a cocore disk of one of the MATH-handles of MATH. Now since we obtained MATH from MATH and MATH the disks MATH and MATH are a canceling pair. But this implies that the minimal genus NAME splitting MATH is reducible in contradiction. Hence this case cannot happen and the proof is complete.
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math/9912171
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Since MATH has a genus two NAME splitting (as MATH is a tunnel number one knot) and is irreducible, the NAME splitting is strongly irreducible. Otherwise we could compress the NAME surface to both sides and obtain an essential MATH-sphere in contradiction. Similarly any NAME splitting of minimal genus three of a hyperbolic knot is strongly irreducible: As the knot complement is irreducible we can compress at most twice (once to each side). But then, by compressing the NAME surface we obtain an incompressible non-boundary parallel torus in contradiction to the fact that the knot is hyperbolic (see REF ). The knots MATH are alternating knots and not torus knots so by REF of CITE they do no contain incompressible non-boundary parallel tori and hence are hyperbolic. Note that MATH induces minimal genus NAME splittings, of genus two and three respectively, on both of MATH and MATH. By slightly abusing notation we will denote the components of MATH by MATH and MATH. As in REF let MATH denote the cocore disk of the tunnel MATH which intersects the decomposing annulus MATH and let MATH denote the cocore disk of MATH the tunnel interior to MATH. We can choose the disks MATH as a meridional system of disks for the compression body MATH. Note also that MATH minimizes the intersection with MATH as if MATH the NAME genus of MATH would be additive and equal to three. Let MATH be the NAME splitting surface MATH, and let MATH, and MATH. For each essential disk MATH, MATH in MATH and MATH respectively, we choose a representative in their isotopy class so that MATH is minimal; in particular, each component of MATH, MATH, is an essential circle or essential arc on MATH, and each component of MATH is an arc. Claim : Let MATH be an essential disk in MATH then: CASE: If MATH then the outermost sub-disk MATH of MATH is an essential disk in the components MATH or MATH of MATH, depending on which side of MATH contains MATH. If it is in MATH then MATH where MATH is an inessential arc on one of the vertical annuli MATH and MATH is an arc on MATH as indicated in REF . CASE: If MATH and MATH is contained in the MATH component then MATH is parallel to MATH. CASE: Note that MATH is the union of two arcs MATH and MATH. If MATH is inessential we could isotope MATH off MATH. This is a contradiction to the choice of MATH. Assume now that MATH is contained in MATH. If MATH then MATH is isotopic to a curve which represents a power of the meridian in MATH which is a contradiction as the meridian has infinite order in MATH. So MATH. Consider now the disk MATH which is the intersection of MATH with the component of MATH contained in MATH. If MATH then since MATH the disk MATH is an inessential disk in this component of MATH which is a solid torus. If MATH then since this solid torus is irreducible we can reduce the intersection by isotoping MATH off the neighborhood of the half tunnel until MATH is isotopic to MATH. CASE: If MATH is contained in the component MATH then, as above, since it is in the component of MATH which is a solid torus and cannot intersect MATH it is isotopic to MATH which is parallel to MATH. Assume in contradiction that the NAME splitting-MATH is weakly reducible and let MATH be a pair of essential disks in MATH and MATH respectively, so that MATH. As MATH contains no interior tunnel it follows from REF that all outermost disks of MATH are in MATH. If the disk MATH then it is either contained in MATH or parallel to the disk MATH: As if it is not in MATH it must be a non-essential disk in the solid torus MATH and these are parallel to MATH. In the first case it is essential in the strongly irreducible induced NAME splitting on MATH and so must intersect the outermost sub-disks of any essential disk MATH: Note that all outermost sub-disks of MATH which are contained in MATH are essential disks in the strongly irreducible NAME splitting induced on MATH. In the second, case as all outermost sub-disks of MATH intersect the parallel copy of MATH it follows that the correspomding disks of MATH must run through the annulus MATH and intersect MATH. If the disk MATH then assume first, that the outermost sub-disk MATH is in the MATH component of MATH. By the above claim MATH is an essential disk there. Since the induced NAME splitting MATH is strongly irreducible any two outermost sub-disks of MATH and MATH in MATH must intersect. If the outermost sub-disks of MATH are in MATH then by the claim above if we cut this component of MATH along MATH we obtain two components one of which is a solid torus and the other is a MATH-ball MATH (see REF ). CASE: Consider now an essential disk MATH in MATH. If MATH then MATH is an essential disk in MATH or MATH, the two components of MATH, depending on which side of MATH the disk MATH is. Hence MATH is an essential disk in the handlebody part of the induced NAME splitting on either MATH or MATH. However these NAME splittings are strongly irreducible so MATH must intersect MATH the cocore disk of MATH as it is an essential disk in the corresponding MATH or MATH. This implies that MATH must intersect the decomposing annulus which is a contradiction. Hence MATH is non-empty. Let MATH be a sub-disk, which is outermost among all sub-disks of MATH which are contained in the MATH component of MATH. Let MATH be the components of MATH, then for all but one, say MATH, the arcs MATH are ourtermost arcs of MATH and hence are of type II (as in the proof of REF ). Hence MATH have both end points on MATH, the cocore disk of the tunnel MATH. The arc MATH may be of type II or type I in which case it has one end point on one of MATH or MATH, and one on MATH. CASE: Since we are assuming that MATH, in both cases MATH is a set of arcs contained in the annular sub-surface of MATH depicted in REF with all but at most one endpoint on MATH. Since by assumption all these arcs must be essential in MATH, it follows that MATH and MATH is of type I. But this contradicts the fact that an outermost arc of intersection cannot be of type I as then we can reduce the intersection of MATH with MATH in contradiction to the choice of MATH. Thus we have showed that any two essential disks in MATH and MATH must intersect and hence the NAME splitting MATH is strongly irreducible.
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math/9912173
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The independence of the definition with respect to the ordering of the classical crossings and the invariance of MATH under NAME moves of type II and III follows exactly as in CITE, the only difference being an exchange of the variable MATH for MATH. The behaviour under NAME moves of type I is depicted in REF . Since the changes of sign corresponding to NAME moves of type I are compensated by the factor MATH in MATH and since the definition of MATH and MATH does not depend on the virtual crossings of the diagram, the statement on MATH follows immediately. MATH .
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math/9912173
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REF follows as in CITE. REF is an immediate consequence of the definition of the matrix MATH. MATH .
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math/9912173
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The formula can easily be checked by verifying the corresponding relation for every state MATH in the state sum model used in CITE to define the "partition function" MATH. MATH .
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math/9912173
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REF follow from the fact that summing up the columns (rows) of the determinant belonging to MATH REF gives the trivial column (row) vector. REF is an immediate consequence of setting MATH in REF (or in the definition). MATH .
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math/9912173
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Assume the above skein relation has a non-trivial solution. Consider the skein triples MATH where MATH is a classical knot diagram with one crossing and MATH where MATH is a standard diagram, with arbitrary orientations of the components, of the NAME link (the latter triple corresponding to an arbitrary of the diagram's two crossings). Then the related NAME polynomials have values, up to normalization, as follows. MATH . Inserting these values, each multiplied by a factor MATH, into the skein relation immediately shows that, up to normalization, MATH and MATH. Thus the skein relation is equivalent to the classical NAME skein relation: MATH . Finally, let MATH be the virtual link diagram arising from MATH by changing an arbitrary classical crossing to a virtual crossing. Then MATH and MATH. Inserting these values, each multiplied by a factor MATH, into the NAME skein relation yields a contradiction. MATH .
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math/9912174
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To set up notation, let MATH intersect MATH in points MATH, MATH, MATH, and MATH, where MATH is joined to MATH and MATH is joined to MATH via oriented arcs MATH and MATH in MATH. Similarly, MATH is joined to MATH and MATH is joined to MATH-via arcs MATH and MATH in MATH. Fix a pair of arcs on MATH, MATH and MATH, joining MATH to MATH and MATH to MATH on MATH such that MATH. Then the pair of oriented circles MATH and MATH is the oriented boundary of a properly embedded oriented connected surface MATH in MATH. The proof is a standard argument in relative obstruction theory, considering maps of the complement of the arcs (MATH and MATH) to MATH and letting MATH be the preimage of a regular value. Similarly, the union of arcs MATH forms a circle, and this circle is the oriented boundary of a properly embedded oriented connected surface MATH in MATH. The union MATH is an oriented NAME surface for MATH. Since MATH is orientation preserving and MATH, MATH and MATH can similarly be joined to form an oriented NAME surface MATH for MATH. Pick oriented bases for MATH and MATH. A basis for MATH can be formed as the union of these, along with one more element that can be represented by a curve on MATH that intersects MATH and MATH each exactly once. It is now clear that there will be a corresponding basis for MATH. Furthermore, working directly from the definition of the NAME matrix, one sees that the corresponding NAME matrices are identical as follows. The pairings between elements of the basis represented by curves that miss MATH are clearly unchanged. Let MATH be a curve representing the extra element of MATH that intersects MATH and MATH each exactly once. It is clear that the linking number of MATH with the positive push-off of any curve missing MATH is unchanged (here the fact that the mutation is positive is essential). The self-linking of MATH with its positive push-off is unchanged (regardless of the sign of the mutation).
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math/9912174
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The idea of the proof is simple and follows the approach of CITE or CITE. It is actually implicit in CITE. Basically, if MATH denotes a REF - manifold used in the computation of MATH then a REF - manifold that can be used to compute MATH is built from MATH by adding copies of a REF - manifold MATH along the lifts of MATH. The manifold MATH is a REF - manifold with boundary REF - surgery on MATH, MATH, for which the natural map of MATH extends. But it is this manifold MATH that can be used to compute the classical signatures of MATH. Notice that since MATH the sign issue disappears.
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math/9912174
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The idea of the proof is related to that of the previous theorem. In the definition of MATH one considers the infinite cyclic cover of MATH. The infinite cyclic cover of MATH is built from that of MATH by removing MATH copies of MATH and replacing it with MATH - copies of the infinite cyclic cover of MATH. Hence it is expected that MATH factors of the NAME polynomial of MATH should appear. The MATH appear because of the twisting in the homology. That the product appears as it does follows from a NAME type argument. Details are presented in CITE.
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math/9912174
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First note that a subgroup MATH of MATH, a torsion group with nonsingular linking pairing, is a metabolizer if and only if the linking form vanishes on MATH and the order of MATH is the square root of the order of MATH. (From the exact sequence MATH we have that MATH. If the linking form vanishes on MATH, then MATH.) It is clear that the linking form for MATH vanishes on MATH. All that we need to show is that the order of MATH is the square of the order of MATH. To simplify notation we will view MATH and MATH as subgroups of MATH in the natural way. Let MATH denote the projection from MATH to MATH. Finally, define MATH. We have the two exact sequences: MATH . Notice that MATH pairs trivially against both MATH and MATH. Hence it pairs trivially against the sum, MATH. In particular, since the pairing is nonsingular, we have that the orders satisfy MATH. Notice also that the order of the sum of groups is equal to the product of their orders, divided by the order of the intersection. Hence MATH . From the second exact sequence we have MATH . From the first exact sequence it follows that MATH . Using the inequality (along with the fact that MATH) applied to the denominator, we have MATH . Since each metabolizer has order the square root of the order of its ambient group, and since no self - annihilating subgroup can have order greater than the square root of the order of the ambient group, the result follows.
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math/9912174
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If MATH is cg - slice, let MATH be the appropriate metabolizer. Similarly, let MATH be the metabolizer that exists because MATH is cg - slice. Then the previous theorem provides a metabolizer MATH. For any MATH there is a MATH such that MATH. Hence, MATH. Hence, MATH is cg - slice as desired.
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math/9912174
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The quotient of MATH under the MATH - action of MATH is a MATH - fold branched cover of the quotient of MATH under the action of MATH. This quotient is a REF - sphere and the branch set has quotient consisting of REF points. The MATH - fold branched cover of a MATH over REF points is again a REF - sphere, so the quotient of MATH under MATH is a REF - sphere. Since MATH is an involution, MATH splits as a direct sum of the MATH and MATH eigenspaces. A transfer argument (see CITE) shows that the +REF eigenspace is isomorphic to the first homology of the quotient, which is trivial since the quotient is a sphere. The theorem follows.
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math/9912174
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The cover of MATH decomposes as MATH, but note that the cyclic action of the deck transformation restricted to MATH is the inverse of the action that arises when we consider the cover of MATH, since the character MATH has been replaced by MATH. That the MATH action is well defined on the union implies that MATH. Consider the restriction of MATH and MATH to the preimage of a fixed point of MATH, a set with MATH elements. Then MATH acts as a MATH - cycle and MATH commutes with MATH. In the symmetric group on MATH letters the only elements that commute with a MATH - cycle MATH are powers of MATH. But MATH has a fixed point and the only power of MATH with a fixed point is the identity. We now have the relations MATH, and MATH. Since neither MATH or MATH are trivial, the result follows.
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math/9912174
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The action of MATH and MATH generate a dihedral action on the rational vector space MATH. It follows from representation theory that any such rational representation of the dihedral group splits as the direct sum of three types of representations CITE. The first, denoted simply as MATH, is the trivial action on MATH. The second, denoted MATH, is the action on MATH for which MATH acts trivially and MATH acts by multiplication by MATH. The last is denoted MATH, where MATH is a primitive MATH-th root of unity, MATH acts by multiplication by MATH, and MATH acts by conjugation, that is, MATH. NAME (that is, tensoring with MATH over MATH) these representations we see that the only one for which MATH has a MATH eigenspace is for MATH, where that eigenspace is generated by MATH. Now the action of MATH is to conjugate; again, this map sends MATH to MATH. When we complexify the representation, the action of complex conjugation fixes the MATH and maps MATH to MATH. In either case we see the actions are the same. This completes the proof.
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math/9912174
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As expected, the proof begins by decomposing the manifolds used to compute the NAME - NAME invariants. Hence: CASE: Decomposing MATH . We begin by describing how to construct a knot MATH as the union of MATH and a trivial twisted tangle, MATH, (see REF , illustrating a trivial twisted tangle) so that the restricted character MATH on REF - fold branched cover of MATH extends to the branched cover of MATH. There is a representation of MATH to the dihedral group, MATH, so that its restriction to the index REF subgroup gives the character MATH. (This observation concerning dihedral representations is well known; we present the details in the more complicated setting of REF - fold covers in the proof of REF .) The restriction of this representation to MATH extends naturally to the complement of some standard twisted tangle, which we denote MATH. REF : A trivial twisted tangle Notice that the same tangle MATH can be glued to the tangle MATH to yield the knot MATH for which the character also extends in the same way over the cover of MATH. We let MATH, MATH, and MATH be the induced branched covers of MATH, MATH, and MATH. Notice that MATH extends to an involution of MATH, and lifts to give involutions MATH. We can form MATH, the MATH - fold branched cover of MATH. Similarly, form MATH, the MATH - fold branched cover of MATH. Let MATH and MATH be the induced characters. CASE: NAME of the signature We now want to relate the NAME - NAME invariants of both MATH and MATH to those of MATH and MATH. As usual, we let MATH copies of MATH and MATH bound REF - manifolds MATH and MATH over MATH and MATH, respectively. The desired manifolds MATH and MATH needed to compute the NAME - NAME invariants for MATH and MATH are obtained by gluing MATH and MATH along MATH using either the identity map or the extension of MATH to MATH. We then want to apply additivity of the signature, both on the level of the base manifold and for the eigenspace signature on the MATH - fold cyclic covers. However signatures do not add in this setting; the necessary result is NAME 's ``non-additivity" formula for signature CITE, which we now summarize. Suppose that a REF - manifold MATH is formed as the union of MATH and MATH along a REF - dimensional submanifold MATH contained in both boundaries and the boundary of MATH is the surface MATH. Then MATH where MATH is a function defined by NAME and NAME, and MATH are the kernels in MATH of the inclusion of MATH into MATH, MATH, and MATH respectively. CASE: Applying non-additivity From this most steps of the theorem follow readily. Using the notation of REF, we have MATH . Now, by identifying MATH (and MATH) with the boundary of MATH, we see that all these terms are in fact identical, except perhaps the terms MATH and MATH. The first is the kernel of MATH and the second is the kernel of this same inclusion, precomposed with MATH. But since MATH acts by multiplication by MATH, these are the same as well. On the eigenspace level the situation is more difficult. First, one needs to observe that NAME 's theorem applies on the level of eigenspaces. At the conclusion of CITE, NAME notes that the results of his paper hold in the setting of MATH - manifolds and for the MATH - signature. Specifically, MATH where MATH denotes the MATH - signature and MATH is an invariant depending only on the subspaces MATH of MATH. Since the eigenspace signatures and the MATH-signatures are equivalent via a NAME transform by CITE, it follows that (again using the notation of REF, so, for example, MATH means the MATH eigenspace of MATH) we have MATH where MATH is a function depending only on the subspaces MATH of MATH. Alternatively, one can check line-by-line that NAME 's arguments go through in the setting of eigenspaces and that hence these formulas hold. By identifying MATH with the boundary of MATH, we have that all the terms in the sum are identical, except for two pairs that we now deal with. CASE: MATH . These might be different because the second term is obtained after replacing MATH with MATH on MATH. This has the effect of inverting the MATH action on MATH, and thus interchanges the eigenspaces. However, complex conjugation also induces an interchange of eigenspaces, and as we noted in the previous section, this preserves signatures, so the terms are in fact equal. CASE: This is the most difficult and delicate step. We must prove that the action of the lift of MATH, MATH, to the MATH - fold cyclic cover preserves the kernels of the inclusions on the eigenspaces. More precisely, what must be shown is that the kernel of MATH and the kernel of MATH are the same. Here the subscript MATH denotes the MATH eigenspace. As mentioned earlier, there is a natural isomorphism from MATH to MATH induced by complex conjugation and this second eigenspace is isomorphic to MATH since the MATH denotes the groups associated with MATH for which the action of the deck transformation is inverted. Also, notice that complex conjugation preserves the kernel of inclusions, since the complex kernel is simply the real kernel tensored with MATH. In the last section we proved that MATH induces an isomorphism of MATH to MATH that agrees with the one induced by complex conjugation. REF follows.
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math/9912174
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We have already seen that there is a correspondence between characters on MATH and MATH and that, by REF , MATH is invariant under this correspondence. It remains to check that metabolizers are preserved under the correspondence. To see this we need to be look a little more closely at the homology and the linking form. Via a NAME argument, MATH and MATH where MATH is the subgroup generated by elements of the form MATH, and MATH is the subgroup generated by the set of elements of the form MATH. An isomorphism is given by the map MATH. To check that the isomorphism is an isometry of linking forms, we need only to check that the linking number of classes of the form MATH or MATH is preserved. Recall that the linking form MATH is defined geometrically by letting a multiple of MATH bound a chain and computing the intersection number MATH with that chain, and then dividing by the order of the multiple. To check that the linking number of a class of the form MATH and a class of the form MATH is preserved, let MATH bound a chain MATH in MATH. We can assume that MATH is transverse to MATH and intersects MATH in a MATH invariant REF - chain. It follows that MATH can then be cut and reglued to give a chain MATH in MATH. Notice though that since MATH acts by MATH on MATH, the portion of MATH in MATH has its orientation reversed in the construction. Hence, the intersection number of MATH with MATH in MATH is the same as the intersection number of MATH with MATH in MATH. A similar argument applies to the cases of linking numbers of classes of the form MATH and MATH or of classes of the form MATH and MATH. Since the isomorphism of MATH with MATH preserves linking, it takes metabolizers to metabolizers.
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math/9912174
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The NAME form for MATH is given by MATH . A symplectic basis for the form is given by the pair of elements MATH. In this basis the NAME form has a MATH in the upper left-hand entry, that is, the vector MATH is a metabolizing vector for the NAME form. In particular MATH is algebraically slice, and hence lies in the kernel of MATH; its image in MATH lies in the kernel of MATH. By REF MATH is non-trivial in MATH if MATH is not cg-slice. Representing the elements MATH and MATH by curves that meet once on the NAME surface, and taking a neighborhood of the pair, one can describe the given NAME surface as a disk with two bands added, one with core the curve representing MATH and the other with core representing the element MATH. One can easily check that the first band is tied in a MATH - torus knot, which we will denote for now by MATH. Finally note that tying a copy of MATH in the band yields a slice knot, since it now bounds a genus one NAME surface for which one band has REF framing and has a core that is slice. We let MATH denote this slice knot. REF - fold branched cover of MATH (and MATH) has homology MATH. If MATH is prime, there is a unique metabolizer for this form, and hence a unique (up to multiple) character vanishing on that metabolizer. By REF , tying the knot in the band changes the value of MATH by the sequence MATH. In other words, for a character MATH in the metabolizer, MATH for some MATH. The number MATH is determined by the value of MATH on the curve refered to as MATH in the statement of REF . A careful analysis of the branched cover of MATH shows that MATH will be nonzero modulo MATH. Details of such a calculation are presented in CITE or CITE. They are based on the presentation of REF - fold cover of MATH using the methods developed in CITE. This result must be REF in the group of sequence modulo bounded sequences since MATH is slice and since the metabolizer is unique. Therefore the knot MATH will have MATH. A proof of the following lemma is given in the next section. Let MATH denote the MATH - torus knot with MATH. Then the signature function MATH is positive for MATH . Since MATH is non-zero modulo MATH, we may assume MATH so that MATH . Therefore REF implies that MATH is positive and so in particular MATH is not cg-slice. Thus REF implies that MATH is non-trivial in MATH, and hence MATH is a non-trivial knot in the kernel of MATH. To construct the infinite family of linearly independent algebraically slice knots in MATH, let MATH be an increasing sequence of positive integers so that the sequence MATH is a sequence of distinct primes. We will show the sequence MATH is linearly independent. Given a sequence of integers MATH we must show that the knot MATH is not slice. By reindexing, and replacing MATH with MATH if necessary, we may assume that MATH is positive. We will show that for any character MATH on REF - fold branched cover of MATH taking values in MATH the value of MATH is nontrivial for any MATH. Since REF - fold branched cover of MATH is a connected sum, we can write MATH as a sum MATH where MATH is the restriction of MATH to the cover of MATH. But since REF - fold branched cover of MATH has no MATH homology for MATH, we have that MATH is trivial for MATH and therefore MATH for MATH. Hence by the additivity of MATH, MATH . We will show that MATH is nonzero. Using additivity again this is the sum of values of MATH where MATH denotes the restriction of MATH to the MATH-th factor of MATH. Explicitly MATH . Since MATH is nontrivial, at least one of the MATH is nontrivial. For each such MATH, MATH is non-zero modulo MATH as before, and REF implies that the corresponding MATH is positive. For those MATH such that MATH is trivial, MATH is zero. Therefore, MATH is non-zero. Since MATH, MATH is not cg-slice and so represents a non-trivial element in MATH by REF . This concludes the proof of REF .
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math/9912174
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The argument begins with some basic computations that follow readily from the techniques of CITE and CITE. Details are presented in CITE. The knot MATH is built from MATH by removing neighborhoods of curves MATH and MATH linking the bands and replacing them with the complements of MATH and MATH. REF - fold branched cover of MATH, MATH, has first homology MATH, and if we consider a map of MATH to MATH taking value MATH on a lift of MATH, it takes value MATH on the other lift, and takes values MATH and MATH on the lifts of MATH. The homology of REF - fold cover of MATH is MATH, and a careful examination of the linking form reveals that for any metabolizer, one of the characters that vanishes on that metabolizer will take value REF on a generator of the first MATH summand and MATH on the corresponding generator of the other summand. (In brief, a metabolizing element MATH must satisfy MATH mod REF.) Hence, a calculation similar to the one for doubled knots given in the previous section shows that for this character MATH . An explicit calculation for the MATH - torus knot, MATH, gives that MATH and MATH. Hence, if we let MATH denote the connected sum of MATH - torus knots, one computes MATH . Clearly, this last sequence is unbounded unless MATH.
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math/9912174
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First we note that MATH and MATH cobound an annulus MATH embedded in the complement in MATH of the slice disk for MATH just constructed. To see this, perform the band move joining MATH and MATH along the arc MATH. The resulting circle is unknotted and unlinked from the pair of circles formed during the slicing of MATH, so that circle can be capped off with a disk in the complement of the slice disk to yield MATH. We also need to note that MATH is trivial when viewed as an annulus in MATH. That is, the pair MATH is equivalent to the pair MATH, where MATH is the unknot. This is easily seen by ignoring MATH and the slice disk during the construction of MATH. It is clear that if one removes from MATH a neighborhood of MATH and replaces it with MATH then MATH results. The boundary, MATH, is obtained from the original MATH by removing neighborhoods of MATH and MATH and replacing them with the complement of MATH and MATH respectively, as desired. During this construction, the slice disk for MATH becomes the slice disk for MATH.
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math/9912174
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The NAME polynomial of MATH is MATH. This can be quickly seen by noticing that changing one of the two rightmost crossings on MATH in REF changes MATH into the connected sum of two copies of REF - twisted double of the unknot, which has NAME polynomial MATH. The NAME skein relation shows that changing this crossing does not change the NAME polynomial, since the link obtained by smoothing the crossing is a unlink. It follows that the order of MATH equals MATH (see, for example, CITE). The assertion that MATH will follow once we show that MATH. The cube roots of MATH in MATH are MATH and MATH. The deck transformation MATH acting on MATH satisfies MATH. Hence, MATH. If MATH denotes the fixed cohomology, then the composite MATH is just multiplication by MATH, where MATH denotes the transfer and MATH is the branched cover (see for example, CITE); hence MATH. Thus MATH is injective on the finite group MATH and hence an isomorphism. It follows that MATH. Let MATH and MATH . For any MATH, MATH so MATH. If MATH then MATH, so MATH which implies MATH since MATH is relatively prime to MATH. Hence, the cohomology does split as the direct sum of an REF - and a REF - eigenspace, MATH. We will show that these are each REF - dimensional (that is, cyclic of order MATH), by identifying elements in each eigenspace with certain metabelian representations of MATH. Define the metacyclic group MATH . Consider the elements of MATH as homomorphisms MATH. For each MATH, we construct a representation MATH taking value MATH on meridians, as follows. Pick a basepoint for MATH on the boundary of a meridinal disk for MATH and choose the basepoint of MATH to lie above it, and hence on the boundary of a meridinal disk for the branch set in MATH. We can then lift a loop MATH to a loop MATH in MATH by letting MATH be the lift of MATH followed (if necessary) by an arc back to the basepoint in the meridinal disk in MATH. Let MATH be the homomorphism taking the meridian of MATH to MATH. To a class MATH we assign a representation MATH on MATH by the formula MATH . If MATH denotes the meridian of MATH then MATH. The projection induces an inclusion MATH and so we can recover MATH by restricting MATH to MATH; its image lies in the MATH subgroup of MATH generated by MATH. To turn this into a one-to-one correspondence, one needs to identify representations of MATH to MATH that are equivalent on the MATH subgroup. These are precisely the conjugacy classes of representations. REF below describes a representation of MATH to MATH. The edges of the knot diagram (that is, the NAME generators of MATH) are labeled with numbers in MATH, where the label MATH means the corresponding NAME generator is sent to the element MATH in MATH. To give a well defined representation, the NAME relation MATH at a crossing implies that the relation MATH must hold, where MATH denotes the label on the MATH-th NAME generator. This relation is satisfied if and only if the labels satisfy MATH at each crossing. Observe that conjugation by MATH takes MATH to MATH and so in enumerating conjugacy classes of representations (and hence MATH), we can assume that any one label of the diagram is MATH. Or, put otherwise, two labelings determine conjugate representations if all corresponding labels differ by the same amount. It is a simple exercise to check that all labelings are obtained from the labeling of REF by either multiplying each label by MATH or by adding the same number to each. It follows that MATH is isomorphic to MATH, generated by the character MATH determined by the labeling in REF . A similar computation applies for MATH. Hence MATH. To prove REF one needs only compute the value of the representation MATH on the MATH. In REF we have chosen based loops to represent the curves MATH and MATH. The basing determines specific lifts MATH, and MATH of these curves to MATH. For the representation given by the labeling in REF it is easy to compute that MATH, MATH, and MATH. Hence the corresponding character MATH satisfies MATH, MATH, and MATH. A similar argument applies to MATH.
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math/9912174
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The curves MATH each lift to three curves in MATH, and the NAME sequence shows that replacing the six solid tori lying above these curves with knot complements, MATH or MATH, does not affect the cohomology.
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math/9912174
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Each irreducible factor of MATH is of the form MATH for some MATH and MATH, since the irreducibility of a polynomial MATH implies the irreducibility of MATH. Each of these factors, MATH, satisfies MATH. Hence, if MATH for some MATH, then the exponent of each irreducible factor of MATH must be even. It remains to observe that the set of polynomials MATH are distinct. Suppose that MATH. Then a change of variable shows that MATH. Since the MATH are integer polynomials, the only way these can be equal is if MATH. (This is where we use the fact that the MATH are not polynomials in MATH.) But this results in the equation MATH, implying that MATH.
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math/9912174
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Using REF one sees that since the character MATH takes the value MATH on the lift MATH of MATH, it takes the value MATH on the translate MATH and takes the value MATH on MATH. Similarly MATH takes the value MATH on MATH and hence takes the value MATH on MATH and MATH on MATH. The lemma now follows from REF .
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math/9912174
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Given MATH, the set MATH equals MATH if MATH, and equals MATH if MATH. Suppose MATH. Then by REF and the choice of MATH . Where MATH. By REF MATH is a norm. Using the hypotheses on MATH, REF , and the fact that MATH is odd, we see that, modulo norms, MATH equals a product MATH where MATH is the set of those MATH so that the set MATH contains an odd number of elements. Since MATH is odd, MATH is non-empty and so by REF the expression in REF is not a norm. The same argument applies to MATH.
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math/9912174
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Suppose that MATH has dimension MATH. Choose some basis for MATH. Each basis element is written MATH which we consider as the MATH-th row of a MATH matrix MATH. Elementary row operations, and interchanging of columns, which does not affect whether a row is odd (that is, has an odd number of non-zero entries) transforms the matrix MATH to a matrix of the form MATH, where MATH is a MATH identity matrix and MATH is a MATH matrix. If some row of MATH is not odd, then the corresponding row of MATH is odd and we have the desired odd character. Thus assume every row of MATH is odd. Since MATH has more rows than columns, some non-trivial linear combination of the rows of MATH gives the zero vector, say MATH. If an odd number of the MATH are non-zero, then MATH is odd. Finally, if an even number of the MATH are nonzero, choose MATH so that MATH, and pick MATH non-zero and different from MATH. Then MATH is odd since MATH-row-MATH is odd.
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math/9912174
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Since MATH for MATH, it follows that MATH takes MATH eigenvectors to MATH eigenvectors and vice-versa. Hence MATH and so MATH since MATH. Similarly MATH. The off - diagonal entries are equal since the form is symmetric, and MATH must be a unit since the form is non-singular.
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math/9912174
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Consider the columns of MATH as a basis MATH of MATH. Let MATH denote the standard basis of MATH. Let MATH-Hom-MATH. The condition that every linear combination of the rows of MATH is even can be restated by saying that for each MATH, the sets MATH have the same cardinality modulo MATH. Another way to state this is as follows. Let MATH be the function MATH and let MATH be the product MATH. Similarly define MATH by MATH and MATH be the product MATH. Then the hypothesis is equivalent to the statement that MATH. Let MATH be the NAME transform MATH . Define MATH similarly using MATH. We compute MATH as follows. Let MATH denote the dual basis to the MATH. Then MATH . From its definition one sees that MATH, so that MATH . Substituting this into REF and simplifying one gets MATH . Now MATH . Notice that MATH . Thus for each MATH, MATH where MATH is the cardinality of the set MATH. Similarly MATH where MATH is the cardinality of the set MATH. Since MATH, MATH, and since MATH, MATH . This implies that MATH since MATH (recall that MATH is prime). Now MATH and therefore MATH. Hence for each MATH there exists a unique MATH so that MATH. Since MATH is non-singular it follows that MATH is a permutation, and that MATH. In other words, MATH is obtained from a diagonal matrix by permuting the columns.
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math/9912176
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This is an immediate consequence of the last corollary.
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math/9912176
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By induction we easily prove that MATH . NAME the possible orders of elements of MATH are MATH and MATH . In particular the order of the element MATH is MATH. Let MATH be the subgroup generated by MATH . Assume MATH admits an action of genus zero on some NAME surface MATH. Then there exists a short exact sequence MATH as in REF , where MATH for some MATH and MATH for some choice of non-negative integers MATH . These integers must be chosen so that MATH is a well defined epimorphism with torsion free kernel. In particular this means that MATH . The restriction MATH follows from the fact that MATH and therefore there is an epimorphism MATH . The NAME formula becomes MATH . The action of the subgroup MATH on the NAME surface MATH has genus-zero and therefore, according to REF , there are only MATH possibilities for MATH: MATH . We will analyze each case separately and conclude that the group MATH does not admit an action of genus zero. CASE: Assume MATH . Then necessarily MATH . The left hand side will be larger than what we get by putting MATH and using MATH . This leads to the inequality MATH, and so MATH . But then MATH will not be an epimorphism. CASE: Assume MATH . Then the NAME equation becomes MATH which can be rewritten in the form MATH . From this equation it follows that MATH, for otherwise the left hand side would be greater than MATH . Recall that MATH . Thus the equation becomes MATH which is equivalent to MATH . Now it is simple to see that there are no solutions. CASE: The last case to consider is MATH . By arguments similar to the second case there are no solutions of the NAME equation.
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math/9912176
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We need only show that the group MATH presented in REF does not have genus zero. The order MATH of MATH modulo MATH will be MATH for some MATH, MATH . If MATH then MATH commutes with MATH and since MATH it follows that MATH has a subgroup isomorphic to MATH namely the subgroup MATH . But this contradicts REF . Thus suppose MATH . Then consider the subgroup MATH generated by MATH and MATH . This is a ZM group with presentation MATH . But now the order of MATH is MATH, and again we arrive at a contradiction. Therefore a ZM group of odd order does not admit actions of genus zero.
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