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math/0006078
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Note that MATH is semisimple by REF . We only need to prove the invertibility of the matrix formed by MATH where MATH are as above, MATH is a basis in the space MATH of characters of MATH (we used above REF for the quantum trace). Observe that the linear map MATH takes any element of the form MATH (that is, MATH), where MATH, into MATH. Indeed, for any such MATH and all MATH we have, using the fact that MATH (this follows from REF ), the properties of MATH and MATH, the relation MATH, and the centrality of MATH : MATH therefore MATH. Since MATH is factorizable, we know from REF that the restriction MATH is a linear isomorphism. Since MATH belongs to the subspace on the left hand side, we have a linear isomorphism between MATH and MATH, hence, there exists an invertible matrix MATH representing the map MATH in the bases of MATH and MATH, that is, such that MATH. Then MATH . Therefore, MATH, where MATH. If MATH is an element from REF then MATH is an invertible central element of MATH and MATH for all MATH. By REF MATH is invertible central, therefore MATH. Hence, MATH for all MATH and MATH is invertible.
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math/0006078
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First let us show that MATH, equipped with a natural involution MATH where MATH, is a MATH-subalgebra of the tensor product MATH-algebra MATH. For this it suffices to show that MATH, that is, MATH, MATH for MATH. For instance, one computes: MATH for all MATH. The right-hand sides of the above equations are equal since MATH . Similarly one gets the other relation. To prove that the comultiplication of MATH is a MATH-homomorphism we compute MATH where we use that MATH for every pair of dual bases. Thus, MATH is a MATH-quantum groupoid and so is MATH (see REF ).
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math/0006078
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Since MATH also implements MATH REF , MATH is central, therefore MATH is also central. Clearly, MATH must commute with MATH. The same Proposition gives MATH, which allows us to compute MATH . REF and the trace property imply that MATH . Since MATH and MATH is central, the above relation means that MATH and, therefore, MATH for any irreducible representation MATH, which shows that that MATH.
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math/0006078
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Relations REF are obvious, REF follows from REF . Let us prove relations REF . On the one hand, for all MATH we have, using the definitions of MATH, REF and the notation MATH for a linear operator MATH and two vectors MATH of a NAME space: MATH . And, on the other hand, using the definition of MATH, we compute : MATH whence the first part of REF follows. To establish the second part, note that for all MATH we have, using the definitions of MATH, REF and the properties of MATH: MATH . On the other hand, using the definition of MATH, we obtain: MATH . The condition MATH for any morphism MATH follows from REF and from the positivity of MATH.
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math/0006078
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The proof follows from REF .
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math/0006078
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CASE: The definition of counit implies : MATH hence MATH. We used here the identity MATH and REF . Similarly, MATH. Using the antipode property, we compute MATH . CASE: Set MATH. Then MATH is central and invertible, MATH, and from REF we conclude that MATH. Next, MATH . Applying the counit to both sides of the last equality, we get MATH, that is, MATH.
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math/0006078
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One verifies that MATH is a homomorphism exactly as in CITE. The properties of MATH and MATH follow directly from definitions. For the counit axiom we have, using the properties of counital maps, REF , and REF : MATH . REF can be verified by a direct computation. Next, we observe that MATH and MATH. The antipode axiom follows from the identity MATH provided by REF : MATH . The anti-multiplicative properties of the antipode follow from the facts that MATH and MATH.
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math/0006081
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Every compact MATH-space MATH has a MATH-invariant point. Indeed, there exists a MATH-map MATH, and since MATH is MATH-invariant, so is MATH. Let MATH be the compact space of all maximal chains of closed subsets of MATH. We saw that MATH is a compact MATH-space. Thus MATH has a MATH-invariant point.
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math/0006081
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NAME 's lemma implies that there exists a minimal element MATH in the set of all closed non-empty subsemigroups of MATH. Fix MATH. We claim that MATH (and hence MATH is a singleton). The set MATH, being a closed subsemigroup of MATH, is equal to MATH. It follows that the closed subsemigroup MATH is non-empty. Hence MATH and MATH for every MATH. In particular, MATH.
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math/0006081
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The composition MATH is a MATH-map of MATH into itself, hence it has the form MATH, where MATH. Since MATH, we have MATH.
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math/0006081
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According to REF , there is MATH such that MATH for all MATH. Since MATH is a compact MATH-space contained in MATH, we have MATH by the minimality of MATH. Thus there exists MATH such that MATH. Let MATH be the MATH-map defined by MATH. Then MATH for every MATH, therefore MATH (the identity map of MATH). We have proved that in the semigroup MATH of all MATH-self-maps of MATH, every element has a right inverse. Hence MATH is a group. (Alternatively, we first deduce from the equality MATH that all elements of MATH are surjective and then, applying this to MATH, we see that MATH is also injective.)
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math/0006082
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There is a bijective correspondence on the level of comples tori, which restricts to abelian varieties. Given a complex torus MATH and a basis in MATH, that is, an isomorphism MATH, from the exact sequence REF a representation MATH is deduced, and this determines a complex torus MATH together with a covering map MATH, unique up to isomorphisms. Then from the exact sequence an isomorphism MATH is also obtained. Conversely, given a complex torus MATH and an isomorphism MATH, by means of the exact sequence an inclusion MATH is deduced, hence a quotient torus MATH, together with an isomorphism MATH. Let MATH be a morphism of complex tori constructed as above. The basis in MATH is symplectic for a unique antisymmetric form MATH of type MATH. This determines a natural MATH-bilinear form MATH on the space MATH such that MATH coincides with MATH on MATH, and one has a polarization if and only if MATH is a positive hermitian form. Similarly the basis in MATH is symplectic for an antisymmetric form MATH of type MATH, which determines a form MATH. The two bases are related through the matrix MATH. The relation REF between the types is the condition so that there is equality MATH of the antisymmetric forms. This implies MATH. As the induced isomorphism MATH is indeed MATH-linear, it follows that MATH is a positive hermitian form if and only if the same holds for MATH.
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math/0006082
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The NAME space MATH may be viewed as a quotient: the space MATH of lattice bases MATH in MATH, subject to the locally closed condition that the alternating form defined by MATH is the imaginary part of some positive hermitian form (a condition that the NAME relations express in terms of the NAME coordinates of MATH), divided by the natural action of MATH. An equivariant isomorphism MATH induces an automorphism of MATH. If the lattice basis MATH defines a torus MATH then the lattice basis MATH determines the complex torus MATH that covers MATH; conversely if MATH defines a torus MATH then MATH defines the quotient torus MATH. This correspondence MATH is an automorphism of the space of all lattice bases. It follows from the preceding proof that the correspondence restricts to an isomorphism MATH.
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math/0006082
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The isogeny is a sum of embeddings if and only if the inclusion MATH is given by a pair of inclusions MATH. If the condition is satisfied, as in the beginning of the section, calling MATH the quotient varieties, the product isogeny MATH factors as MATH . Introducing bases also in the homology of the varieties MATH and MATH, the diagram is obtained. Conversely if a diagram of integer homomorphisms as above exists for MATH then the inclusion of the subgroup into the product variety is given by a pair of inclusions.
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math/0006082
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To an embedding is associated a pair of varieties, as we have seen. Conversely, given two abelian varieties MATH, with symplectic bases for polarizations of types MATH, by means of the isomorphism MATH, from the exact sequence REF an inclusion MATH is deduced. Define the torus MATH and call MATH the quotient isogeny. Then one has an isomorphism MATH, a symplectic basis for an alternating form MATH of type MATH. REF on the types implies that MATH. Using the same argument as in the proof of REF it is then seen that MATH being a polarization implies that MATH is a polarization. Finally because of REF the isogeny MATH is the sum of two embeddings. In particular the restriction MATH is an embedding of the given type.
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math/0006082
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Take MATH, the connected component of REF in the kernel, take MATH, let MATH and MATH be the complementary abelian subvarieties, and let MATH be the restriction of MATH. Uniqueness of the choice is a consequence of uniqueness in the classical reducibility theorem.
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math/0006082
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By hypothesis the matrices of the isogenies satisfy the condition of REF and there are inclusions MATH and MATH as is seen in the proof of the lemma. Using these inclusions it is easily seen that there exists MATH if and only if MATH sends MATH and this happens if and only if MATH sends MATH. If this happens then there is a factorization MATH where MATH. Introducing bases in homology as in the proof of the lemma, the associated matrices satisfy MATH where MATH is the matrix of MATH and where MATH is the matrix of MATH. Conversely if the matrix MATH exists then MATH under MATH.
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math/0006082
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Assume that we are given abelian varieties MATH with symplectic bases for polarizations of types MATH, and an isogeny MATH of type MATH. Because of REF for MATH there are inclusions MATH. Then define the torus MATH. By means of REF an isomorphism MATH is obtained. Because the types MATH are related, this gives on MATH a basis for a polarization of type MATH, as follows from REF . Similarly there are inclusions MATH and defining MATH one has an isomorphism MATH, a basis for a polarization of type MATH. From REF it follows that the morphism MATH sends MATH and induces a morphism MATH. With respect to the bases this morphism is of the given type MATH.
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math/0006083
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Every element in the image of MATH is made by contracting copies of the structure constants of the NAME algebra, which are invariant under MATH.
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math/0006083
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Given a MATH-circus MATH, consider the MATH-circus MATH with an additional double lasso which encircles two knot strands and/or handles of double lassos. The resolution of MATH along the new double lasso is equal to the difference between MATH and a variant MATH in which the two strands have crosses each other. Since MATH has MATH lassos, this difference is zero in the associated graded space MATH so MATH. Likewise framing changes on a strand can be achieved by adding an additional single lasso looping it. These two moves generate homotopy as in the statement.
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math/0006083
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A multiple crossing can be done step by step: MATH . (MATH is the operation of resolving the double lasso.) Modulo Relation REF, we can forget the extra little hooks.
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math/0006083
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MATH .
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math/0006083
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The map from MATH to MATH defined above is clearly surjective. To see that it is injective, we need to check that diagrams differing by cyclic permutations are already equal in MATH. It suffices to check that two NAME diagrams based on an interval differing by moving a single leg from the beginning to the end of the interval are equal modulo the antisymmetry and vertex relations. Since vacuum diagrams play no part in this problem, let us assume that there are none. We can then pick a good routing of the diagram and consider an associated MATH-circus. Now we can expand the first loop on the interval, pass it over the rest of the circus, and shrink it down again; we have moved this leg from the beginning to the end of the interval. See REF . This can also be achieved by a sequence of vertex and antisymmetry relations without referring to topology by replacing the loops in REF by rounded boxes as in REF and sweeping the boxes from one end of the interval to the other, applying vertex and antisymmetry relations along the way.
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math/0006083
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This is a general property of the holonomy of a connection with values in a NAME algebra in which the connection form is primitive. (The usual example is the holonomy of a NAME valued connection, which takes values in the the NAME group or, alternatively, the grouplike elements inside the universal enveloping algebra.)
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math/0006083
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By the definition of a universal finite type invariant MATH, MATH where MATH is the MATH-circus MATH in the complement of MATH. By the Splitting relation, this is MATH.
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math/0006083
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By the above computations, the invariant of the connected cable of a knot MATH is MATH, multiplied by MATH, and closed up with a twist. The conjugating elements MATH and MATH can be swept through the knot and cancel each other. The factor MATH in MATH can be combined with MATH so that we apply MATH to MATH. The twisted closure turns MATH into MATH. The remaining MATH factors of MATH in MATH can be slid around the knot and combined to give MATH . The last factor is absorbed in the ``appropriate choice of framing" in the statement of the proposition.
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math/0006083
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The glued diagrams are the same on the two sides; we either combine the legs of MATH and MATH into one set and then glue with MATH, or we split the legs of MATH into two pieces which are then glued with MATH and MATH. (Note that there are no combinatorial factors to worry about: in both cases, we take the sum over all possibilities.)
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math/0006083
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As before, the diagrams are the same on both sides.
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math/0006083
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We use a sliding argument similar to the one in the proof of REF . Link relations in MATH can be slid over diagrams in MATH, as shown in REF .
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math/0006083
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As advertised, we use the equality of links MATH. Let us see what this equality of links says about the NAME integral of the NAME link. On the MATH side, we see the connected sum of two open NAME links; by REF, the invariant of the connected sum is the connected sum of the invariants. To write this conveniently, let MATH be MATH, with the wire labelled by MATH and the bead labelled by MATH. Then MATH . On the MATH side, we see the disconnected cable of a NAME link. By REF, this becomes the coproduct MATH: MATH . Since the two tangles are isotopic, we have MATH . Now consider the map MATH in other words, in MATH glue the MATH and MATH legs of MATH to MATH and MATH respectively. This descends modulo the two different link relations in MATH by the argument of REF , applied to MATH and MATH separately. We have two different expressions for this map from the two different expressions for MATH. On the MATH side, the gluing does not interact with the connected sum and we have MATH . See REF . For the MATH side, we use REF to see that MATH . See REF . Combining the two, we find MATH .
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math/0006083
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Using REF and noting that gluing with MATH takes the legs of a diagram in MATH and averages over all ways of ordering them, as in the definition of MATH, we see that MATH .
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math/0006083
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MATH . In the second equality, we use REF . This is allowed, since the contraction descends to MATH by the argument of REF . Note that it is crucial that MATH is invariant for this argument.
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math/0006083
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First note that any vacuum diagrams that appear in the MATH's pass through unchanged to the result; let us assume that there are none, so that we can use the vacuum projection MATH without changing the result. By REF , MATH . Let MATH. Each MATH has at least one leg, since if the MATH of MATH eats all the legs of MATH, it also creates a vacuum diagram which is killed by MATH. Then MATH . Let MATH; diagrams in MATH have at least one MATH leg. We see that MATH . Therefore MATH . Each MATH has at least one leg labelled MATH, so the product has at least MATH legs labelled MATH which are the legs in the result.
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math/0006083
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MATH .
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math/0006083
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We need to check that after resolving each vertex like MATH we are left with a collection of MATH double lassos that are trivial in MATH once you forget the knot. The internal vertices can be pulled apart one by one in the specified order: at each stage, at least one side of the encircled lasso ends in a loop with nothing inside.
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math/0006083
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Because MATH is boundary connected, there is a good ordering on the internal vertices. (Order the vertices from the external vertices on the knot inward.) Every good ordering has a comptabile routing.
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math/0006083
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The minimal vertex in the first ordering must have an external vertex as a neighbor. Therefore, there is no obstruction to moving this vertex to the first position in the second ordering by a series of transpositions. We can repeat this for each vertex in turn.
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math/0006083
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We need to check that the two possibilities for the routing at each internal vertex are equivalent. This follows from two applications of the vertex relation: MATH where MATH is younger than MATH and the order of the two vertices in the middle sum is chosen depending on which neighbor of MATH is younger, or is irrelevant if MATH is an external vertex.
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math/0006083
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Pick a good ordering for each routing. We can adjust the routing while keeping the ordering fixed by REF , so we just need to check that we can change the ordering. The two orderings are related by a chain of good orderings related by adjacent transpositions by REF . At each transposition, the two vertices involved must have younger neighbors which are not each other (otherwise we could not exchange them); a routing in which these two neighbors are on distinguished edges is compatible with both orderings.
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math/0006083
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Proceed by induction on MATH. The result is trivial for MATH. MATH .
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math/0006083
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The first relation says that MATH is REF-dimensional. For the second relation, note that both sides, considered as elements of MATH, are multiples of the projection onto the antisymmetric part MATH (which is REF-dimensional). A little computation fixes the constant. (Note that the constant depends on the metric. Here we use MATH, where the trace is taken in the adjoint representation.)
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math/0006083
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Proceed by induction. This is a straightforward computation for MATH. For MATH, compute as follows: MATH .
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math/0006083
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Proceed by induction. The statement is trivial for MATH. For MATH, the two ends of the first strut on the left hand side can either connect to the two ends of a single right hand strut or they can connect to two different struts. These happen in MATH and MATH ways, respectively. (Note that there are MATH ways in all of gluing these two legs.) We therefore have MATH and MATH .
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math/0006083
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By REF , we find MATH . Set MATH. Then by REF , MATH so MATH .
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math/0006085
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Consider the map MATH . It is a smooth fibration with fiber MATH, where MATH. Since the base MATH is contractible, we obtain that the inclusion MATH is a homotopy equivalence. Hence, the integral cohomology ring of MATH coincides with MATH, which we calculate below. REF from CITE describes algebra MATH, where MATH is an arbitrary field. From this description it is clear that the dimension of the cohomology does not depend on field MATH. Therefore, we conclude that the integral cohomology MATH has no torsion; it is a free abelian group of rank one for MATH, where MATH, and it vanishes for all other values of MATH. Let MATH be a point distinct from MATH and MATH. We obtain an inclusion of configuration spaces MATH, where we identify MATH with MATH. The cohomology algebra MATH has generators MATH and the full list of relations was described in REF. From REF we know that the induced map MATH on cohomology with arbitrary field of coefficients MATH is injective. This implies that the induced map on integral cohomology MATH is injective and MATH maps indivisible classes from MATH into indivisible classes in MATH. We claim that for any MATH there exists an indivisible class MATH such that MATH (compare with formulae REF from CITE). Indeed, applying REF from CITE with MATH, we see that the image of the generator of the group MATH under homomorphism MATH equals an integral multiple of the expression in the Right-hand side of REF . Since the classes in Right-hand side of REF are indivisible, and since we know that MATH maps indivisible classes to indivisible classes, we conclude that a there exists generator MATH with the required property. The product formulae REF for classes MATH follow since they hold for the products MATH as can be easily checked using the arguments of the proof of REF from CITE. Now we want to find the action of the reflection MATH on classes MATH. It is clear that MATH, and we need to calculate the sign. Consider the following diagram of natural inclusions MATH (where MATH as above) and the induced diagram of cohomology groups MATH is an isomorphism and MATH is injective. To understand MATH, note that MATH is homotopy equivalent to the cyclic configuration space MATH (compare CITE) and so the cohomology MATH has MATH-dimensional generators MATH which satisfy relations of REF from CITE (we shift indices for convenience). Proof of REF shows that MATH for MATH and MATH. Hence, MATH is an epimorphism with kernel equal the ideal generated by MATH. The reflection MATH acts also on MATH by REF . It is clear that the induced map MATH acts on the generators MATH as follows MATH . Now we may calculate MATH, where MATH. Fix a subsequence MATH (we avoid indices MATH and MATH). Suppose first that MATH is odd. Then MATH contains monomial MATH; therefore MATH contains the same monomial with coefficient REF. Then MATH contains monomial MATH with coefficient MATH. The last monomial appear in MATH with coefficient REF. Since we know that MATH we conclude that MATH. Assume now that MATH is even. Then MATH contains monomial MATH with coefficient MATH . Applying MATH and using REF we see that the monomial MATH appears in MATH with coefficient MATH and in MATH with coefficient MATH . This shows that MATH.
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math/0006085
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Let MATH be an orthonormal base. We may assume that MATH. We want to calculate the Hessian of function MATH at a billiard trajectory MATH, where MATH and MATH . Let MATH denote the orthogonal vector to MATH lying in the MATH-plane, that is, MATH . Any tangent vector MATH is determined by numbers MATH, where MATH and MATH, such that the component of MATH in MATH equals MATH . A direct calculation of Hessian MATH of MATH gives the following quadratic form in variables MATH: MATH where in the first sum we understand MATH and in the second sum the symbol MATH denotes the following quadratic form MATH . We see that the Hessian splits as a direct sum of MATH quadratic forms corresponding to different values MATH. The terms involving MATH (the first sum) give a positive definite quadratic form. The rest MATH form are identical and their index and nullity equal the index and nullity of MATH. Hence we conclude that the index and nullity of the Hessian equals MATH times the index and nullity of the form MATH. In order to calculate the index of MATH we observe that the eigenvalues of the following symmetric MATH-matrix MATH are given by MATH and the eigenvector MATH corresponding to MATH is given by MATH . This claim can be checked directly. Therefore the eigenvalues of MATH are MATH and the eigenvectors of MATH are given by REF . Hence the index of MATH equals the number of integers MATH, such that MATH, which is MATH for MATH and MATH if MATH and MATH is odd. Since (according to REF ), MATH, we conclude that index of MATH equals MATH . The special case MATH for MATH odd corresponds to MATH; in this case the index and nullity of MATH equal MATH. From REF we see that the nullity of MATH equals REF for any MATH unless MATH is odd and MATH. The discussion above proves that on any critical submanifold MATH the dimension of the kernel of Hessian of MATH equals the dimension of MATH; hence all submanifolds MATH are nondegenerate in the sense of NAME and their indices are as stated.
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math/0006085
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Let MATH be the equatorial sphere consisting of unit vectors orthogonal to MATH. Any point MATH and an angle REF determine a critical submanifold MATH. Fix an eigenvalue MATH (given by REF ), such that REF is negative. Consider the subbundle MATH of the normal bundle MATH consisting of eigen vectors of the Hessian with eigenvalue MATH. We want to show that MATH is isomorphic to MATH. This would clearly imply the Lemma. Consider a billiard trajectory MATH in the plane of vectors MATH and MATH, where MATH and MATH . Denote by MATH the MATH-dimensional subspace orthogonal to MATH and MATH. We will show that there is an isomorphism between the fiber of MATH over MATH and MATH, which depends continuously on MATH. Let MATH, where MATH, be a sequence of tangent vectors. Using REF we find that a sequence of vectors MATH belongs to the fiber of MATH over the configuration MATH if and only if MATH . We see that the first vector MATH uniquely determines a tangent vector MATH to a configuration MATH in the eigen-deriction MATH. Moreover, MATH can be an arbitrary vector in MATH.
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math/0006085
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Note that the critical value MATH equals MATH . Hence for MATH we have MATH. Choose constants MATH such that MATH . Each MATH is a compact manifold with boundary and we obtain a filtration MATH and the inclusion MATH is a homotopy equivalence (as follows easily from REF). Using REF and the NAME isomorphism we obtain MATH . This holds true also for MATH if we understand MATH. Suppose that MATH is even. Then cohomology group MATH is isomorphic to MATH for MATH and for MATH and vanishes for all other MATH. Comparing with additive structure of MATH given by REF we find MATH which means perfectness of MATH. Suppose now that MATH is odd. Then MATH is MATH for MATH and vanishes for all other values of MATH. If MATH then MATH and thus perfectness REF also holds. Alternatively, for MATH perfectness REF follows without using REF by considering the spectral sequence of filtration MATH and observing that for any of its differential MATH, with MATH, either the source or the target vanish. Therefore, MATH. Moreover, every diagonal MATH of MATH contains at most one nonzero group. If MATH the differential MATH has nonzero source and target, and so the above argument does not work.
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math/0006085
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We will give here a simple proof working for MATH. The case MATH will follow from REF below. Consider filtration MATH as in the proof of REF . Let MATH denote MATH. We obtain a filtration MATH such that the inclusion MATH is a homotopy equivalence and MATH (using the NAME isomorphism). Hence MATH is nonzero (and one-dimensional) only for MATH. The spectral sequence of filtration MATH has MATH for MATH and MATH otherwise. Hence for any differential MATH, where MATH, either the source or the target vanish. Therefore, MATH and our statement follows.
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math/0006085
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As in the proof of REF we obtain that the negative normal bundle MATH splits as a direct sum of MATH vector bundles MATH of rank MATH, one for each negative eigenvalue MATH of the Hessian. Here MATH. Let MATH denote the tangent bundle of MATH. Let MATH be a rank MATH vector bundle over MATH such that its fiber over a line MATH is the orthogonal complement MATH. We claim that MATH . Indeed, this bundle is obtained from the tangent bundle MATH of MATH (compare REF ) by identifying the antipodal points, and under this identification the first vector MATH should be replaced by the last vector MATH (compare REF ). Formulae REF show that MATH and hence the bundle MATH is obtained from MATH by identifying the fibers over points MATH and MATH with a twist MATH. This implies our claim, compare CITE. For given MATH there are equal number of negative eigenvalues MATH of the Hessian on MATH with even and with odd MATH. Therefore the bundle MATH is isomorphic to a direct sum of MATH copies of MATH. The total NAME class of MATH is MATH and the total NAME class of MATH is MATH (compare CITE). Hence the total NAME class of the negative bundle is MATH .
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math/0006085
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By REF the first NAME class of MATH is MATH. This implies our statement.
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math/0006085
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Consider filtration MATH (compare proof of REF ) and the associated spectral sequence MATH . MATH contains a single critical submanifold MATH with index MATH. The normal bundle to MATH is orientable if MATH is even and it is non-orientable if MATH is odd. The NAME isomorphism gives MATH . Here MATH denotes the nontrivial local system of groups MATH over MATH; its monodromy along the generator of MATH is multiplication by MATH. For MATH even we have MATH and MATH . For MATH odd we have MATH . Therefore, in the above spectral sequence holds MATH for MATH. It implies that for MATH, either the source or the target of any differential MATH vanish. Hence for MATH holds MATH and any diagonal MATH contains at most one nonzero group. This proves our statement for MATH. Assume now that MATH and consider the first differential MATH. We have MATH and MATH . We see that MATH vanishes since there are no nonzero homomorphisms MATH for any MATH. The higher differentials MATH vanish by obvious reasons. Hence, the conclusion we made for MATH, holds also for MATH.
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math/0006085
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Consider the universal MATH-bundle MATH and the associated fibration MATH, having MATH as the fiber. The total space MATH is homotopy equivalent to MATH. The NAME spectral sequence of this fibration converges to the cohomology algebra MATH. The initial term is MATH where MATH, the cohomology of the fiber, is understood as a local system over MATH. From REF we know that MATH is either MATH or trivial. There are two types of local systems with fiber MATH over MATH, which we will denote MATH and MATH. Their structure is determined by the monodromy along any noncontractible loop of MATH, which is MATH in the case of MATH and MATH in the case of MATH. Assume first that MATH is odd. From REF we find that MATH where MATH. Hence we find MATH where MATH. As a bigraded algebra, MATH can be identified with the tensor product MATH where MATH and MATH is an exterior algebra with MATH, and MATH. If MATH is the generator, then relation MATH follows from relation MATH (in the notations of REF ). Here we denote MATH the graded subring MATH . The structure of the ring MATH follows from REF . The first nontrivial differential is MATH. Since we know the additive structure of MATH (compare REF ) we find that the differential MATH must be an isomorphism. On the other hand MATH vanishes (since the range is the zero group). It follows that MATH is nonzero iff both MATH and MATH are odd. REF shows the nontrivial differential MATH. The fat circles denote group MATH and the small circles denote MATH. We conclude that the bigraded algebra MATH is isomorphic to the tensor product of algebras MATH where MATH has bidegree MATH and MATH has bidegree MATH. It is clear that all further differentials vanish and hence MATH. Any diagonal MATH contains at most one nonzero group, and hence the algebra MATH coincides with MATH. This proves our statement for MATH odd. Assume now that MATH is even. Recall that we always assume that MATH is even. From REF we find that MATH assuming that MATH. Hence we find MATH . As a bigraded algebra, MATH can be identified with the tensor product MATH where MATH is an exterior algebra with MATH and MATH and MATH is an exterior bigraded algebra with MATH, and MATH. If MATH denotes the generator then MATH follows from relation MATH, compare REF . We denote by MATH the graded subring MATH . Consider now the first nontrivial differential MATH. It is clear that it may be nonzero only for MATH of the form MATH. On the other hand, since we know the additive structure of the limit (compare REF ), we conclude that MATH is onto. Using multiplicative properties of the spectral sequence we find that all the differentials shown on REF are epimorphic. In fact all differentials on REF , except those which start at the MATH axis, are isomorphisms (since they act between isomorphic groups). As before, the fat circles denote MATH and small circles denote MATH. Hence, to the next term MATH, survive classes MATH, and also MATH, MATH and MATH and their products MATH, MATH, and MATH with MATH. It is clear that all further differentials vanish and in each diagonal MATH there is at most one nonzero group. Therefore we conclude that ring MATH is isomorphic to MATH. Its structure coincides with the description given in REF .
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math/0006085
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Let MATH denote the space of orbits MATH. Function MATH determines a continuous function MATH. We want to show that any orbit MATH, representing points MATH with MATH, is not a critical point of MATH in the sense of REF . This would imply that the number of critical orbits of MATH is at least the number of critical points of MATH; the later can be estimated from below by MATH by REF . We will assume that MATH is supplied with a MATH-invariant Riemannian metric. Let MATH be a point with MATH. We want to construct a smooth vector field MATH in a neighborhood of the orbit of MATH having the following properties: CASE: MATH; CASE: the norm of vector MATH equals REF; CASE: MATH is MATH-invariant; CASE: in case MATH belongs the boundary MATH, the vector MATH points inside MATH. To construct such vector field MATH one first finds a vector MATH for each point MATH of the orbit so that REF are satisfied. It is then possible to extend the vectors MATH to form a smooth vector field MATH in a neighborhood of the orbit of MATH with REF . Then REF can be achieved by averaging. The flow determined by the vector field MATH, gives a deformation of a ball around the point MATH, which represents the orbit of MATH, showing that the slope MATH is positive.
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math/0006085
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Consider fibration MATH where the image of a cyclic configuration MATH under projection MATH is given by MATH. The fiber of MATH is the configuration space MATH. Consider the NAME 's spectral sequence of this fibration. The cohomology of the fiber MATH is described by REF ; it has generators MATH, which multiply according to REF . This spectral sequence may have only one nonzero differential MATH. We will show that this differential vanishes MATH. This would clearly imply our statement. Since we may write MATH, it is enough to show that MATH. Vanishing MATH follows from the fact that fibration REF admits a continuous section MATH and thus the transgression is trivial. To construct MATH, fix a nowhere zero tangent vector field MATH on the sphere MATH (recall that MATH is odd). For MATH, the tangent vector MATH determines a half circle starting at MATH, tangent to MATH and ending at the antipodal point MATH. Then the section MATH can be defined by MATH where MATH and the points MATH are situated on the half circle making equal angles as shown on the following picture. Analytically we may write MATH .
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math/0006085
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First we will assume that MATH; the case MATH will be treated separately later. We will describe the additive structure of MATH, using approach of the NAME theory. Let MATH be the unit sphere. Consider the total length function MATH where for MATH we have MATH . The critical points of MATH are MATH-periodic billiard trajectories in the unit sphere; hence the critical configurations are regular MATH-gons lying in two-dimensional central sections of the sphere. A regular MATH-gon is determined by two first vectors MATH, which must make an angle of the form MATH . Recall, that we assume that MATH is odd. Fixing MATH, we obtain a variety of critical configurations, which we will denote by MATH. Each MATH has dimension MATH and is diffeomorphic to the NAME manifold of pairs of mutually orthogonal vectors in MATH. Since we assume that MATH is even and the characteristic of MATH is MATH, we have MATH. Note also that MATH is simply connected (since MATH). NAME has shown (compare REF ) that function MATH is nondegenerate in the sense of NAME and the index of each critical submanifold MATH equals MATH. Moreover, it is clear that MATH for MATH. Fix MATH small enough and consider the submanifold MATH, where MATH . If MATH is sufficiently small then (according to REF) MATH is a compact manifold with boundary containing all the critical points of MATH and such that the inclusion MATH is a MATH-equivariant homotopy equivalence. Moreover, at every point of the boundary MATH the gradient of MATH has the outward direction. Choose constants MATH such that MATH for MATH and MATH. Let MATH . We obtain a filtration MATH . Since the inclusion MATH is a homotopy equivalence, we may use the spectral sequence of this filtration in order to calculate the cohomology of MATH. We claim that this filtration is perfect, that is, the NAME polynomial of the cyclic configuration space MATH equals the sum of the NAME polynomials of the pairs MATH. The initial term of the spectral sequence is MATH . NAME the NAME isomorphism (recall that MATH is simply connected), we find that MATH is isomorphic to MATH; hence MATH is one-dimensional for MATH and for MATH and vanishes for all other values of MATH. This follows since in MATH there is a single non-degenerate critical submanifold MATH, which has index MATH and MATH is diffeomorphic to the NAME manifold MATH. The gradient of MATH at points of the boundary MATH has the outward direction and hence the points of the boundary do not contribute to the usual statements of the NAME critical point theory. For a given MATH there are precisely two values of MATH such that MATH is nonzero (MATH and MATH). From the geometry of the differentials we see that all the differentials MATH, MATH, must vanish if MATH. This proves that the cohomology MATH is one dimensional for MATH and MATH, where MATH and MATH vanishes for other values of MATH. Having recovered the additive structure of MATH, we may use REF to find its multiplicative structure. The mapping MATH is a NAME fibration MATH; its fiber is MATH. The NAME spectral sequence has only two nonzero columns and MATH is the only differential which could be nonzero. In the MATH-th columns we have classes MATH in dimensions MATH, and in the MATH-th column we have classes MATH having dimension MATH, compare REF . Since we already know the additive structure of MATH, we conclude that the differential MATH is an isomorphism for MATH odd and vanishes for MATH even. Hence the classes MATH and MATH survive. Now, we set MATH, and the conclude that MATH has the multiplicative structure as stated in REF . For MATH the above argument, based on the spectral sequence of the filtration MATH is not sufficient, since, in principle, this spectral sequence could have a nonzero differential, shown on the picture. Also, for MATH the critical submanifolds MATH are not simply connected and so the NAME isomorphisms for the negative normal bundles of the Hessian may require additional twists by flat line bundles (depending on the orientability of the negative normal bundles of the critical submanifolds). However, in the case MATH a different argument can be applied. Consider the action of MATH on MATH arising from the standard action of MATH on MATH. Fix a point MATH and consider MATH as being canonically embedded in MATH. We obtain the map MATH given by applying an orthogonal matrix MATH to a configuration of points on the sphere MATH. It is easy to see that REF is a fibration with fiber MATH. If MATH is a configuration of points on MATH such that MATH, MATH for MATH and MATH, then the fiber of fibration REF over MATH consists of the space of all pairs MATH, where MATH denotes the rotation by angle MATH about MATH. The cohomology algebra of the total space of this fibration MATH is given by REF . It has a generator MATH, with MATH (coming from a generator of MATH) and also classes MATH, where MATH, with MATH, which are pullbacks of the generators of MATH, compare REF . We have the relation MATH and each product MATH equals a multiple of MATH, the coefficient indicated in REF . Let us show that the restriction map from the total space to the fiber MATH is onto. Since MATH, our statement is equivalent to the following. Let MATH be a fixed configuration. We obtain an embedding MATH given by MATH, where MATH. We claim that the induced map MATH is onto. In other words, we want to show that the cohomology class MATH is nonzero. We may assume that the antipode MATH of MATH does not appear in the configuration MATH. Identify MATH with MATH using the stereographic projection with MATH as a center; this leads to the following commutative diagram MATH where MATH is given by rotations of a fixed configuration MATH of points on the plane, MATH, around the origin MATH. Clearly, the space MATH is homotopy equivalent to MATH and thus the cohomology algebra MATH, as given by REF, has REF-dimensional generators MATH, which satisfy the relations MATH for MATH and also a relation of degree MATH, compare REF. From REF we obtain that MATH . Let MATH denote the generator corresponding to the usual anti-clockwise orientation of the circle. Then MATH . Indeed, MATH, where MATH is the degree of the following map MATH and hence it is clear that MATH. Here MATH denotes the plane rotation by angle MATH. Comparing REF we obtain MATH where we have used the assumption that MATH is odd. (Note that for MATH even the above arguments give MATH.) Let us examine the NAME spectral sequence of fibration REF . It has two rows and may have one nontrivial differential. Since we know that the fundamental class of the fiber MATH survives, that is, application of the differential to it gives zero, it follows that all the differentials in the NAME spectral sequence vanish. We conclude that the cohomology algebra of the base MATH is the factor of MATH with respect to the ideal generated by class MATH. Since MATH, we obtain that MATH has generators MATH with MATH and MATH, with MATH, where MATH, which satisfy relations REF .
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math/0006086
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Suppose that MATH and that MATH vanishes on all the elements in the claim. Since MATH vanishes on MATH for MATH, it can be written as a linear combination of the MATH. Since the subdiagram of the NAME diagram for MATH spanned by the vertices of MATH is the NAME diagram of a semisimple group, it follows that MATH is zero. This shows that the given elements span MATH. Since the number of them is equal to the dimension of MATH, it follows that they are a basis.
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math/0006086
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Let MATH. Let MATH be the saturation of MATH in MATH, and define MATH similarly. Since MATH is a direct summand of MATH, there is an induced injection MATH. Suppose that MATH. Then MATH vanishes on MATH. Of course, it takes rational values on MATH and integral values on MATH. Thus, MATH determines a homomorphism MATH to MATH, and thus by restriction a homomorphism MATH. In fact, an easy argument shows that the sequence MATH is exact, where the map on the right is the one induced by MATH. The subspace MATH is a complementary subspace in MATH to MATH. Thus projection from MATH to MATH defines a homomorphism from MATH to MATH, whose kernel is MATH. Summarizing, we have a commutative diagram with exact columns and exact first row: MATH . The point MATH is the image of MATH under the homomorphism MATH. It follows that MATH determines MATH up to an element of MATH. Thus, two distinct points MATH with the same image in MATH have distinct images in MATH. This shows that given MATH, there is at most one lift of MATH to a point MATH whose image in MATH is MATH, and hence that MATH and MATH (or equivalently the topological type of MATH) determine the topological type of MATH as a MATH-bundle. Of course, the topological type of MATH as a MATH-bundle determines the topological type of MATH as a MATH-bundle. Next let us show that the NAME point MATH satisfies the three conditions stated in the lemma. By construction it lies in the NAME algebra of the center of MATH. Clearly, since MATH maps to MATH in MATH, the congruence given in the second item holds. Lastly, since MATH for every character MATH of MATH, and since under the inclusion of MATH the element MATH maps to MATH, it follows that MATH for every character of MATH. Conversely, fix a point MATH satisfying the three conditions above. We shall show that MATH is the NAME point of a reduction to MATH of the MATH bundle MATH. We claim that there is a MATH-bundle MATH such that MATH projects to MATH and such that MATH. It suffices to show that there is an element MATH which projects to MATH and whose image in MATH is MATH, for then there is a MATH-bundle MATH with MATH, and this bundle has the required properties. Choose MATH lifting MATH. For all MATH, MATH. Thus there exists a MATH such that MATH for all MATH. Let MATH be the image of MATH in MATH. Then MATH maps to MATH, and the image MATH of MATH satisfies: MATH for all MATH. Clearly, for every character MATH of MATH, MATH. By REF , MATH, and so MATH is the required element. This completes the proof of REF .
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math/0006086
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Since every MATH bundle has a semistable holomorphic structure, this is clear from the previous lemma and REF .
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math/0006086
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REF shows that the parabolic subgroup and the topological type of the MATH-bundle MATH are determined by MATH and the topological type of MATH. Given two connections MATH, MATH on MATH which define semistable holomorphic structures, an open dense set of the line in MATH joining them will also define semistable holomorphic structures. We can use this line to join the corresponding MATH-connections MATH and MATH on MATH. The space of holomorphic MATH-bundles with a given MATH-reduction is an affine space, by CITE. After choosing a MATH trivialization of these spaces, we can find holomorphic, connected families of MATH-connections joining MATH to MATH and MATH to MATH.
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math/0006086
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Since the NAME group acts trivially on MATH, it suffices to divide out by MATH. We may thus assume that MATH is semisimple. By CITE, for MATH, MATH if and only if MATH for all MATH. But the simplicial cone MATH is spanned over MATH by the elements MATH, MATH, and MATH for some positive constant MATH. Thus MATH if and only if MATH for every MATH.
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math/0006086
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Suppose that MATH is a point of NAME type with MATH. Let MATH be a MATH-connection in MATH. Let MATH be the holomorphic MATH-bundle determined by MATH. According to REF there is an arbitrarily small deformation of MATH to a holomorphic MATH-bundle MATH contained in MATH. We can extend MATH to a MATH-connection on this deformation. A MATH trivialization of the deformation allows us to view all the MATH-connections in the family as connections on MATH. The resulting MATH-connection determining MATH is then arbitrarily close to MATH in the space of MATH-connections. Thus, every neighborhood of MATH in MATH contains a MATH-connection in MATH. Hence, MATH is contained in the closure of MATH. This proves that the subspace MATH, where MATH ranges over points in MATH of NAME type for MATH, is contained in the closure of MATH. The NAME result is that the closure of MATH is contained in this union. Hence, the closure of MATH is equal to this union.
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math/0006086
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As we have seen, MATH is superharmonic at MATH if and only if MATH. Consequently, MATH is superharmonic if and only if MATH.
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math/0006086
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Since the simplicial cone MATH is spanned by MATH, the first statement is clear. Since MATH for all MATH, by for example, CITE, it follows that, if MATH is superharmonic, then MATH. Finally, if MATH is harmonic, then MATH and MATH are both superharmonic, so that MATH.
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math/0006086
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Since the negative of a harmonic function is harmonic and the negative of a superharmonic function is subharmonic, the result for a subharmonic function MATH with MATH follows from that for a superharmonic function MATH with MATH. After replacing MATH by MATH, we may assume that MATH is superharmonic except at MATH and that MATH, and wish to show that MATH. Suppose that MATH for MATH with MATH, where the MATH are given nonnegative real numbers for MATH. The function MATH is superharmonic except at MATH if and only if MATH for all MATH if and only MATH say, where the MATH are given nonnegative real numbers since MATH. It suffices to show that MATH for all MATH. This follows since the inverse of the NAME matrix for the subdiagram corresponding to MATH has nonnegative entries.
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math/0006086
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By REF appplied to the dual root system, the set MATH is a basis for MATH. Thus if MATH for MATH, then MATH is uniquely determined by the conditions MATH for MATH and MATH for MATH. The positivity statement is the special case of the previous lemma, applied to the inequality MATH on MATH and viewing MATH as superharmonic except at MATH and MATH as harmonic.
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math/0006086
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The first statement is immediate from REF . Suppose MATH. Then MATH in the NAME ordering if and only if MATH, if and only if MATH. Since MATH, it follows from the first statement that this holds if and only if MATH which is clearly equivalent to the statement MATH.
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math/0006086
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The condition MATH is equivalent to the condition that for each character MATH of MATH we have MATH. The condition that MATH is harmonic at MATH means that MATH. The condition that MATH is equivalent to MATH. Thus, by REF , MATH is of NAME type for MATH. The final statement follows from REF .
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math/0006086
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There are only finitely many points MATH of NAME type for MATH with MATH. Thus, it suffices to show that if MATH, and if there is no point MATH of NAME type for MATH with MATH, then a finite sequence of the moves listed can be applied to MATH, each one not increasing MATH, to produce MATH. First suppose that MATH. Then choose MATH and define a function MATH which is harmonic on MATH with MATH for all MATH and MATH. By REF there is a unique such function. Clearly, by REF , since MATH and MATH are of NAME type for MATH, the resulting function MATH is determined by a point MATH in MATH such that MATH of NAME type for MATH. Since MATH, it also follows from REF that MATH. Since MATH is harmonic at MATH, it follows easily from the definitions, REF , and the fact that MATH that MATH is subharmonic at MATH. Suppose that MATH. Then there exists a MATH such that MATH. Since MATH and MATH are both harmonic except at MATH, by REF there exists a MATH with MATH. Since MATH, it follows that MATH. Now perform the same construction with MATH replacing MATH, producing a function MATH which is harmonic except at MATH and subharmonic at MATH. Let MATH be the point of NAME type for MATH for which MATH. Since MATH and since MATH is subharmonic except at MATH and harmonic except at MATH, it follows from REF that MATH. Continuing in this way, we can eventually choose MATH and a corresponding point MATH with MATH of NAME type for MATH such that the corresponding function MATH has the property that MATH. Clearly with this choice, MATH is obtained from MATH by a move of Type REF and MATH. It follows from our hypothesis that either MATH or MATH. Thus, either this move replaces MATH by MATH or it leaves MATH unchanged and decreases the cardinality of MATH. We can continue applying moves of Type REF in this manner until MATH. Now suppose that MATH. If for some MATH we have MATH, define the function MATH by setting MATH for all MATH, MATH, and requiring that MATH be harmonic elsewhere. Again it is clear by REF that MATH for a point MATH of NAME type for MATH. Since MATH and MATH are harmonic except at MATH, by REF MATH. Since MATH, both MATH and MATH are harmonic except at MATH. Since MATH, it follows that MATH. Applying REF once again, we see that MATH. We can then obtain MATH from MATH by by a move of Type REF . Since MATH, it follows from our hypothesis that MATH. Thus, MATH is obtained from MATH by a single move of Type REF . Thus, this allows us to arrange that MATH and that MATH. Then, we can obtain MATH from MATH by a single move of Type REF .
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math/0006086
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We shall consider each type of move separately. We begin with two general lemmas. Let MATH and MATH be connected linear algebraic groups over MATH. Suppose that MATH is a surjective homomorphism. Let MATH be a holomorphic MATH-bundle over a curve MATH of arbitrary genus and let MATH be the holomorphic MATH bundle MATH. Then any small deformation of MATH can be covered by a small deformation of MATH. The tangent space to the deformations of MATH is given by MATH. The tangent space to the space of deformations of MATH is given by MATH. The natural map of vector bundles MATH is surjective, and hence the map MATH is surjective. From this the result follows. The following lemma will be used for holomorphic bundles, but of course a similar result holds in the MATH category. Suppose that MATH are complex NAME groups and that MATH are homomorphisms with MATH surjective. Let MATH be a holomorphic principal MATH-bundle over a complex manifold MATH, and suppose that MATH is isomorphic to a bundle of the form MATH, where MATH is a holomorphic MATH-bundle. Then there exists a holomorphic MATH-bundle MATH, such that MATH is isomorphic to MATH. Choose an open cover MATH of MATH and trivializations of MATH and MATH over each MATH. Suppose that MATH are the transition functions for MATH and MATH are those for MATH. There exists a MATH-cochain MATH with values in MATH such that MATH. After shrinking the open cover, we can assume since MATH is surjective that the functions MATH lift to functions MATH with MATH. Using the MATH-cochain MATH to modify MATH, we may then assume that the MATH satisfy: MATH. Thus, MATH. Clearly, MATH is a MATH-cocycle, and the bundle MATH which it defines satisfies: MATH. Returning to the proof of REF , we begin with moves of Type REF , the simplest to describe. This case relies on the following elementary lemma which is implicit in CITE and CITE: Let MATH be a reductive group. Every holomorphic MATH-bundle over MATH has an arbitrarily small deformation to a semistable MATH-bundle. Fix a MATH-bundle MATH over MATH. The MATH-connections on MATH which determine a semistable holomorphic MATH bundle structure are a non-empty open subset of the space of all MATH-connections on MATH and hence form a dense open subset. For a move of Type REF MATH and MATH. Thus, MATH and MATH. By REF the MATH-bundle MATH has an arbitrarily small deformation to a semistable MATH-bundle MATH. The NAME point of MATH is MATH. Now apply REF to the surjection MATH and the bundle MATH to produce an arbitrarily small deformation of the MATH-bundle MATH to a MATH-bundle MATH with MATH isomorphic to MATH. Viewing this deformation of MATH-bundles as giving a deformation of MATH-bundles exhibits the required deformation for a move of Type REF . Now we turn to a move of Type REF . This time MATH, so that MATH and MATH. Let MATH be the maximal parabolic MATH. Its NAME factor is MATH. Denote by MATH its unipotent radical. First let us consider the case where MATH is strictly subharmonic at MATH. This means that if MATH is a semistable MATH-bundle with NAME point MATH, then the NAME parabolic for the MATH-bundle MATH is MATH, the opposite parabolic to MATH. The relevant lemma for this case is the following. Suppose that MATH. Let MATH be a reductive group and let MATH be a semistable MATH-bundle over MATH. Let MATH be a parabolic subgroup with NAME factor MATH and unipotent radical MATH. Let MATH be the opposite parabolic in MATH, and let MATH be its unipotent radical. Suppose that MATH is a semistable MATH-bundle over MATH such that the NAME parabolic of MATH is MATH, and such that MATH and MATH are MATH isomorphic. Then there is an arbitrarily small deformation of MATH to a bundle of the form MATH where MATH is a holomorphic MATH-bundle such that MATH is semistable and is MATH-isomorphic to MATH. Let MATH, MATH, MATH, MATH be the NAME algebras of MATH, MATH, MATH, and MATH respectively. The direct sum decomposition MATH is preserved by the action of MATH. Thus MATH . Since MATH is the NAME parabolic for MATH, the bundle MATH is a direct sum of semistable bundles of positive degrees. Since MATH, it follows from stability and NAME duality that MATH. Thus the natural map MATH is surjective. Since MATH is a curve, all deformations are unobstructed, and the map from the deformation space of the MATH-bundle MATH to that of the MATH-bundle MATH is a submersion. Thus every arbitrarily small deformation of the MATH-bundle MATH arises from an arbitrarily small deformation of the MATH-bundle MATH. In particular, there is an arbitrarily small deformation of the MATH-bundle MATH whose associated MATH-bundle is semistable. The set of all semistable MATH-bundles MATH isomorphic to MATH may be parametrized by an irreducible scheme, in the sense that there exists an irreducible scheme MATH and a MATH-bundle over MATH whose restriction to every slice MATH is semistable, and moreover such that every semistable MATH-bundle MATH isomorphic to MATH arises in this way. By the results of CITE, since by NAME MATH is independent of MATH, there is an irreducible scheme MATH, fibered in affine spaces over MATH, which parametrizes all MATH-bundle deformations of MATH. By the above, there is a nonempty open subset MATH of MATH and a dominant morphism from MATH to the moduli space of all semistable MATH-bundles of the same topological type as MATH, and this moduli space is irreducible. From this, the result follows. Applying this lemma with MATH, MATH, MATH, we see that there is an arbitrarily small deformation of the MATH-bundle MATH to a bundle of the form MATH where MATH is a MATH-bundle and MATH is a semistable MATH-bundle with NAME point MATH. Using REF for the surjection MATH, we produce an arbitrarily small deformation of MATH to a bundle MATH whose reduction modulo MATH is isomorphic to MATH. Since MATH is the preimage of MATH under the natural projection MATH, and hence MATH, it follows from REF that MATH can be written as MATH for some MATH-bundle MATH whose reduction modulo MATH is MATH. This completes the discussion of the move of Type REF in the case when MATH is strictly subharmonic at MATH. Now assume that MATH is harmonic at MATH. This means that MATH. Suppose that MATH. Let MATH be a reductive group and let MATH be a semistable MATH-bundle over MATH. Let MATH be a parabolic subgroup with NAME factor MATH and unipotent radical MATH. Then there is an arbitrarily small deformation of MATH to a bundle of the form MATH where MATH is a semistable MATH-bundle whose NAME point is equal to that of MATH under the inclusion of the center of MATH into the center of MATH. Let MATH be the NAME algebra of MATH and MATH the NAME algebra of the opposite unipotent radical. Let MATH denote the identity component of the center of MATH. The MATH-module MATH decomposes as a direct sum MATH where MATH are characters vanishing on the identity component of the center of MATH. No MATH is trivial, since MATH is a direct sum of root spaces MATH corresponding to roots which are not trivial on MATH. Since MATH is the center of MATH, there is an action of the group of holomorphic MATH-bundles on the space of holomorphic MATH-bundles. We denote this action by MATH. If MATH and if MATH is semistable, then MATH is a semistable holomorphic MATH-bundle which is MATH-isomorphic to MATH. Fix MATH a semistable MATH-bundle whose NAME point is that of MATH under the inclusion MATH. For any MATH trivial on MATH, consider the vector bundle MATH. This is a semistable vector bundle of degree zero over the genus one curve MATH. Of course, MATH. For generic choices of MATH with MATH, the cohomology group MATH vanishes for all characters MATH, and hence for such generic MATH we have MATH. By duality, MATH for generic such MATH. For such MATH, MATH . Consider the set of all pairs MATH for which MATH. As in the proof of the previous lemma, there is an irreducible scheme MATH parametrizing (possibly many-to-one) the isomorphism classes of such pairs and a dominant map from MATH to the moduli space of semistable MATH-bundles MATH-isomorphic to MATH. Hence, the generic such MATH-bundle can be written as MATH with MATH a semistable MATH-bundle with the same NAME point as MATH. Applying this lemma with MATH, MATH, MATH, we see that there is an arbitrarily small deformation of the MATH-bundle MATH to a bundle of the form MATH where MATH is a semistable MATH-bundle with NAME point MATH. Applying REF as before completes the discussion in this case. Notice that we did not use the hypothesis that there were no NAME points of type MATH strictly between MATH and MATH. We include this hypothesis in order to handle the case of genus greater than one below. Now we consider a move of Type REF . The relevant result is the following. Let MATH be a reductive group and let MATH be a simple root of MATH. Let MATH be the corresponding maximal parabolic, MATH its unipotent radical and MATH its NAME factor. Suppose that MATH is a semistable MATH-bundle with NAME point MATH with MATH. Then there is an arbitrarily small MATH-deformation of MATH to a MATH-bundle of the form MATH where MATH is a MATH-bundle with MATH a semistable MATH-bundle whose NAME point MATH satisfies MATH. The proof of this theorem involves some new ideas and we postpone it to the next section. We show how this result implies REF for moves of Type REF . Set MATH and MATH and let MATH be the unipotent radical of MATH. We let MATH and MATH. Since MATH is a positive integer, applying REF repeatedly produces an arbitrarily small deformation of the MATH-bundle MATH to a bundle of the form MATH where MATH is a semistable MATH-bundle such that the NAME point of MATH is MATH. Applying REF produces an arbitrarily small deformation of the MATH-bundle MATH to a MATH-bundle of the form MATH where MATH is isomorphic to MATH. It follows as before from REF that MATH reduces to a MATH-bundle MATH with MATH isomorphic to MATH. This concludes the proof of REF .
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math/0006086
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The proof given above for moves of Types REF applies equally well for curves of higher genus. For the case of moves of Type REF we further divided into the case when MATH was strictly subharmonic at MATH and the case when MATH was harmonic at MATH. The proof in this case when MATH is harmonic reduces to the following elementary fact. Given two reductive groups MATH and a semistable MATH-bundle MATH whose NAME point lies in the center of MATH, the MATH-bundle MATH is semistable. The remaining case is where MATH is strictly subharmonic at MATH. Let MATH. Let MATH be a MATH-bundle with NAME point MATH and such that MATH. The set MATH is a set of simple roots for MATH and determines a fundamental NAME chamber MATH in MATH for the NAME group of MATH. This chamber contains MATH but will be strictly larger than it if MATH. Since MATH and MATH are subharmonic at MATH and harmonic on MATH, it follows that MATH lie in MATH. Our hypothesis is that there is no point MATH of NAME type for MATH (and the group MATH) with MATH. Hence there is no point MATH of NAME type for MATH (and the group MATH) with MATH. Of course, under the NAME ordering for MATH we have MATH since MATH indexes the stratum of semistable MATH-bundles. According to REF , applied to the group MATH, this implies that there is no point of NAME type for MATH with MATH in the NAME ordering. Let MATH be the maximal parabolic subgroup whose NAME factor is MATH, and let MATH be its unipotent radical and MATH its NAME algebra. We denote by MATH the opposite parabolic, by MATH its unipotent radical and by MATH the NAME algebra of this opposite unipotent radical. Begin with a semistable MATH-bundle MATH with NAME point MATH and such that MATH is MATH isomorphic to MATH. The NAME parabolic for MATH is MATH. Thus the tangent space to the stratum containing MATH is MATH and the normal space is MATH. The bundle MATH is a direct sum of semistable vector bundles of negative degrees, and so MATH. Thus, by NAME MATH. This means that there is an arbitrarily small MATH-deformation MATH of MATH such that MATH is not contained in the stratum containing MATH. Let MATH. According to REF, this means that the NAME point MATH in the NAME ordering. Of course, since MATH is the NAME point of semistable MATH-bundles we have MATH. It now follows from the discussion in the previous paragraph that MATH, that is, that MATH is a semistable MATH-bundle of the topological type of MATH. Thus, we have produced a MATH-bundle MATH which is semistable and whose NAME point is MATH which reduces to a MATH-bundle MATH whose associated MATH-bundle MATH is isomorphic to MATH and hence is semistable with NAME point MATH. This completes the proof of the theorem in this last case.
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math/0006086
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It suffices to show that, for all characters MATH of MATH, we have MATH, where MATH is the integer such that MATH. But clearly MATH, and we are reduced to the case of a line bundle, that is, MATH and MATH. In this case, it is easy to see that, if MATH is the line bundle over MATH corresponding to MATH, then the line bundle corresponding to MATH is MATH, and then the statement about degrees is clear.
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math/0006086
|
We may assume that the cover MATH was chosen so that each MATH, MATH, lifts to MATH. By hypothesis, MATH lifts to MATH. Thus, we have lifted the transition functions MATH of MATH to a collection MATH. By definition, MATH is the coboundary of MATH, viewed as an element of MATH, and similarly for MATH. But clearly, since there are no triple intersections of the MATH involving MATH, we have MATH.
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math/0006086
|
By REF , there is an arbitrarily small deformation of MATH which is semistable. If MATH are the transition functions for MATH as defined above, this means that we can find transition functions MATH, depending on MATH in a small disk about MATH in MATH, such that MATH, MATH, and such that the bundle whose transition functions are MATH is semistable for all MATH. Define MATH for MATH, and MATH. Clearly the functions MATH satisfy the cocycle condition. For MATH, MATH, and MATH. Thus the functions MATH define a deformation MATH of MATH, possibly trivial, and MATH is semistable for all MATH.
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math/0006086
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That we can arrange REF after an arbitrarily small deformation of MATH follows from REF . Since MATH lifts to MATH, REF follows from REF follows from REF .
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math/0006086
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Since there are no nonempty triple intersections involving MATH and MATH, MATH is vacuously a MATH-cocycle. To prove the rest of the result, we shall show: for all MATH, there exists MATH and, for all MATH, there exists MATH such that MATH . For this says that the MATH-cocycles MATH and MATH are cohomologous. Clearly, the cocycle MATH defines a MATH-bundle MATH whose associated MATH-bundle is MATH as required. We can rewrite the above condition as MATH if MATH and MATH, or in other words MATH. This condition is automatically satisfied if MATH for MATH by setting MATH. Thus for example we can take MATH for MATH and set MATH for MATH. For MATH we get the single condition MATH for some MATH. Assuming that we have chosen MATH, we seek a function MATH such that MATH . Clearly, it suffices to solve this equation in the MATH which is a subgroup of the universal cover of MATH and whose NAME algebra is MATH. In this copy of MATH, MATH and we may assume that MATH. Taking MATH, we have the equation MATH as required.
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math/0006086
|
By definition, MATH, and so it suffices to show that MATH. Choose MATH, and let MATH be define by MATH and MATH is harmonic outside of MATH. By REF , MATH and MATH. Since MATH is harmonic except at MATH it follows from REF that MATH is superharmonic. Clearly, by REF MATH for some point MATH of NAME type for MATH and MATH, so that MATH in the NAME ordering. By minimality, MATH. Thus MATH. This proves the first statement in the lemma. To see the second, if MATH, then there is a unique function MATH which is harmonic except at MATH and such that MATH. But then MATH is associated to a point MATH of NAME type for MATH such that MATH, MATH, and MATH. This contradicts the choice of MATH.
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math/0006086
|
Clearly, if MATH, then MATH. To prove the converse, we use our previous results on harmonic and superharmonic functions. The NAME diagram of MATH is a union of MATH connected components MATH. For MATH, let MATH be the set of the vertices of MATH together with MATH, and assume that MATH. For MATH, both MATH and MATH are harmonic on MATH, and MATH. It follows from REF that MATH for all MATH. Now consider the restrictions of MATH and MATH to MATH. The function MATH is harmonic except at MATH, and the function MATH is superharmonic on MATH. Since MATH, REF implies that MATH for all MATH. Hence, for all MATH, MATH. This concludes the proof of REF .
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math/0006086
|
The assumption that MATH is long implies that MATH for every simple coroot MATH which is not orthogonal to MATH. Thus the condition that MATH is harmonic except at MATH and MATH implies that MATH for all MATH with MATH. Hence MATH is constant. If in addition MATH, then MATH and so MATH for all MATH. Thus MATH. Hence, for MATH, the slope of MATH is MATH, where MATH is the cardinality of MATH.
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math/0006086
|
It suffices to prove this for a superharmonic function MATH which is harmonic except at a single coroot MATH. If MATH, then we have seen that MATH is linear and increasing toward MATH on each subset MATH with its natural ordering. Now suppose that MATH. Then in particular MATH does not correspond to a trivalent vertex of the NAME diagram, so that MATH has at most two connected components MATH and MATH. Let MATH be the set of all simple coroots in MATH together with MATH and let MATH be the set of all simple coroots in MATH together with MATH. We suppose that MATH. In particular, the NAME diagram of MATH is a simply laced chain (possibly consisting of a single element). Since MATH is harmonic except at the endpoint MATH, it is linear and increases toward MATH. Thus the maximum value of MATH is MATH. The restriction MATH is superharmonic, and is harmonic except MATH, which corresponds to an end vertex. As before, we write MATH, where each MATH is a chain containing MATH. We may assume that MATH. By REF , MATH is linear, and, for MATH, MATH increases up to MATH. Moreover the slope of MATH is MATH. First suppose that MATH is trivalent and meets MATH, where MATH. The condition that MATH is harmonic at MATH says that MATH. For MATH, MATH is the slope of MATH, and hence is at most MATH. It follows that the slope MATH of MATH is nonnegative. Thus MATH is also increasing toward MATH. Since MATH increases toward MATH, which is the end vertex of MATH, it follows that MATH increases toward MATH. Next suppose that MATH is not simply laced and that MATH is the long simple root which meets a short root. If MATH does not correspond to an end of the NAME diagram, then MATH, with MATH, and MATH. Moreover MATH is linear with slope MATH. If MATH are not orthogonal to MATH, then the harmonic condition at MATH says that MATH where MATH with MATH. Hence MATH and thus MATH . Since MATH, MATH is also increasing toward MATH, and the proof concludes as in the trivalent case. The remaining case is where MATH corresponds to an end vertex of the NAME diagram. In this case, if MATH is the unique simple coroot not orthogonal to MATH, the harmonic condition reads: MATH . Once again, MATH is increasing toward MATH, and the proof concludes as before.
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math/0006086
|
In case MATH is not of type MATH, the first statement follows from REF and the second from REF . In case MATH is of type MATH, the first statement is trivially true since every vertex is special, and the second again follows from REF .
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math/0006086
|
Since MATH is the highest root, MATH for all MATH, and so MATH is superharmonic. The result is then immediate from the previous corollary.
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math/0006086
|
It suffices to show that MATH. By REF applied to the superharmonic function MATH, we see that MATH. Thus by REF , MATH. Since MATH, in fact MATH.
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math/0006086
|
By REF any point MATH which is of NAME type for the trivial bundle and indexes a minimally unstable stratum must be of the form MATH for some MATH. By REF , only MATH for MATH special can index a minimally unstable stratum.
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math/0006089
|
We proceed by induction, the case MATH being true by inspection. For MATH we surely have MATH . Notice that from REF of the MATH, MATH and therefore MATH . Now using the fact (easily proven by your favorite calculus student) that MATH for MATH, we conclude that MATH as desired.
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math/0006089
|
To show that the product REF converges, we must show that the corresponding sum MATH converges. But by REF we certainly have MATH for MATH, and therefore MATH . Similarly, using the fact that MATH for any real numbers MATH, we see that for MATH . REF for MATH follow from those for MATH, as it is easily verified by hand that MATH.
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math/0006089
|
Writing MATH, we have from the binomial theorem MATH . Since all of these binomial coefficients MATH are integers, each term in this last sum is visibly divisible by MATH.
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math/0006089
|
If MATH is any prime power dividing MATH, a MATH-fold application of REF shows us that MATH is divisible by MATH. Then, since MATH we see that MATH is also divisible by MATH. In particular, since MATH was an arbitrary prime power dividing MATH, we see that MATH is divisible by every prime power that divides MATH. This is enough to verify that MATH is divisible by MATH itself.
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math/0006089
|
We proceed by induction on MATH, the cases MATH and MATH being evident by inspection. For the inductive step, suppose (as our induction hypothesis) that MATH is indeed an integer for a given MATH. We may write MATH by REF of MATH and MATH, respectively. The second factor MATH is an integer by the induction hypothesis. On the other hand, we may rewrite MATH . Applying REF with MATH, we see that this expression is divisible by MATH. Therefore the fraction on the right-hand side of REF is in fact an integer, and so MATH is itself an integer, which completes the proof.
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math/0006089
|
We shall prove the contrapositive, that no algebraic number can satisfy the NAME REF for all positive MATH. Suppose that MATH is algebraic. Without loss of generality, we may suppose that MATH by adding an appropriate integer. Let MATH be the minimal polynomial for MATH, where the coefficients MATH are integers. Now suppose that MATH is a rational approximation to MATH, say MATH. Then MATH . Since neither MATH nor MATH exceeds REF in absolute value, we see that MATH where we have defined the constant MATH . On the other hand, MATH is a rational number with denominator at most MATH, and it is nonzero since MATH is irreducible. Therefore MATH . Together, REF imply that MATH which precludes REF from holding when MATH is large enough.
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math/0006089
|
It is immediate that MATH where we have used REF for the second inequality.
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math/0006089
|
We can easily show show that the MATH provide the very close rational approximations needed in REF to make MATH a NAME number. Indeed, MATH can be written as a fraction whose denominator is MATH by REF , while REF tells us that for MATH where the last inequality is by REF . Since MATH tends to infinity with MATH, this shows that MATH is a NAME number (and hence transcendental by REF ).
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math/0006089
|
We have just shown in REF that MATH is irrational. As for proving that MATH is absolutely abnormal, the idea is that for every integer base MATH, the number MATH is just a tiny bit less than the MATH-adic fraction MATH. Since the MATH-ary expansion of MATH terminates in an infinite string of zeros, the slightly smaller number MATH will have a long string of digits equal to MATH before resuming a more random behavior. (This is evident in the decimal expansion REF of MATH, as MATH is a REF-adic fraction as well as a REF-adic fraction.) This happens more than once, as each of MATH, MATH, MATH, and so on is a MATH-adic fraction. Consequently, the MATH-ary expansion of MATH will have increasingly long strings consisting solely of the digit MATH, which will prevent it from being even simply normal to the base MATH. More quantitatively, let MATH and MATH be positive integers. Since MATH is an integer by REF , and since MATH by definition, the MATH-ary expansion of MATH terminates after at most MATH nonzero digits. On the other hand, by REF we know that MATH is less than MATH but by no more than MATH. Therefore, when we subtract this small difference from MATH, the resulting MATH-ary expansion will have occurrences of the digit MATH beginning at the MATH-th digit at the latest, and continuing through at least the MATH-th digit since the difference will start to show only in the MATH-th digit at the soonest. Using the notation defined in REF , this implies that MATH . At this point we can calculate that MATH . Using REF , we see that MATH . In particular, the frequency MATH defined in REF either does not exist or else equals REF, either of which precludes MATH from being simply normal to the base MATH. Since MATH was arbitrary, this shows that MATH is absolutely abnormal.
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math/0006093
|
Let hence REF generate solutions of REF and let for a second TLM system REF the deflection MATH be defined by REF . Then for every sequence of states MATH incident at a cell the total states of the second system are MATH . This substituted for MATH in REF yields MATH . By virtue of the additivity of MATH, MATH follows MATH REF states that MATH above vanishes and inspection of the remaining terms yields the proposition.
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math/0006094
|
Define the vector MATH in the MATH plane as the shift of the first collision point. By a direct computation one finds MATH . Since all the incoming shock of the linearly degenerate family have the same speed MATH, by simple geometrical considerations it follows that the vector MATH is constant during all interactions REF . REF follows easily.
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math/0006094
|
If MATH, MATH are the shift rates before interaction, and MATH, MATH after interaction, then REF follows easily from the conservation relation MATH because by assumptions no waves of other families are generated. By REF the conclusion follows.
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math/0006094
|
We consider only the case of linearly degenerate family MATH, since in the other case the proof is exactly the one given in CITE. Let MATH, MATH, be the position of the shock MATH of the MATH-th family in involution, and let MATH be the value of the NAME coordinate at MATH. For MATH, define MATH as the projection of MATH on the hyperplane MATH, and MATH. We choose the shift rates such that MATH where MATH is a scalar function different from MATH only in MATH, and we recall that MATH. By imposing the value MATH, that is, MATH, MATH, we need to prove that REF can be satisfied at time MATH. We have two cases. CASE: If the jump MATH belongs to the MATH-th family and is inside MATH, then set MATH. CASE: If the jump MATH belong to the MATH-th family with MATH, then by REF and by REF there exists a unique shift MATH and a unique constant MATH such that MATH . Since we assume that the shocks are in involution, setting MATH we have that REF holds at time MATH: in fact the last jump has size MATH. We now show that this property is conserved for all MATH. This follows easily from conservation and REF . The proof is exactly the same as in CITE: we repeat it for completeness. Consider the interaction between two shocks MATH and MATH in the point MATH, see REF . By inductive assumption, we have for the states MATH, MATH and MATH that MATH . Using conservation we have MATH so that for the new middle state MATH we have MATH and using REF we conclude MATH . The same relation proves that they vanish outside MATH: in fact, assume for example that MATH and MATH. Then from REF we get MATH which implies that MATH. This concludes the proof.
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math/0006094
|
The theorem will be proved outside the times of interaction, because the NAME dependence in MATH of the approximate semigroup implies the validity of REF for all MATH. If is sufficient to show that MATH is piecewise constant, with jumps only at the points MATH where MATH has a shock MATH, and the following relation holds: MATH where MATH is the shift rate of MATH, located in MATH. Note that by REF the second equality of REF is trivially satisfied. By linearity in the shift rates MATH, we can consider the case in which a single shock is shifted, let us say MATH at MATH: REF becomes MATH . REF follows from the following considerations: consider a wave front pattern, REF , where for simplicity we assume that MATH. The states MATH, MATH are computed considering the NAME problem generated by adding to the MATH-jump MATH in MATH all the MATH-waves starting from the left of MATH and ending in the right of MATH and all the MATH-waves, with MATH, starting from the right of MATH and ending in the left of MATH. The jump MATH is a single wave of the MATH-th family formed by adding all the MATH-waves between MATH and MATH. Using the definition of MATH given REF, one obtains easily the second case of REF: in fact the shift rates of the shocks in the left of MATH is given by the shift rates of the jumps of the NAME problem MATH ending in the left of MATH, MATH, plus the shift rate of the shock MATH, MATH. Since only the MATH-waves with MATH are present in MATH, then MATH. The other cases can be computed in a similar way: in this case one solves the NAME problem MATH in MATH, and consider the MATH-wave MATH starting in the left of MATH and ending in the right of MATH. From the above considerations it is clear that MATH is piecewise constant, with jumps only when in MATH there is a MATH-shock MATH: in fact otherwise the wave front pattern used to compute MATH remains the same. Let MATH be the set of the starting points of all shocks arriving in MATH, and define MATH . We consider two cases: CASE: the shocks arriving in MATH start on both sides of MATH: MATH. In this case, MATH is the shift rate of the MATH-shock starting in the NAME problem MATH if MATH (MATH if MATH) which collides with a MATH-shock MATH (MATH if MATH): in fact the only difference is that in MATH there is a shock of the MATH-th family starting in MATH, and MATH is genuinely nonlinear. Finally, using MATH and REF , one can change position to the MATH-wave and the remaining MATH-wave MATH, whose strength does not change. If MATH, there are no MATH-shocks starting on the right (left) of MATH and ending on the right (left) of MATH, so that MATH is the shift rate of the MATH-shock of the NAME problem MATH. CASE: the shocks of the MATH-th family arriving in MATH start either in MATH or MATH: assume for definiteness that MATH. In this case the difference MATH is the shift rate of the shock MATH colliding with the shifted shocks of the NAME problem MATH in MATH, crossing the jump MATH, and finally overtaking MATH. In fact one can use REF (and MATH if MATH is genuinely nonlinear) to obtain the wave pattern of REF . The various cases will be proved in the following lemmas. Assume that MATH, that is, REF . If the shock MATH is of the MATH-th family, then its shift MATH is MATH . We follow closely the method of CITE. Assume for definiteness MATH, the other cases being similar. The basic idea is to reduce the computation to the single NAME problem MATH, with eventually a single MATH-wave MATH. Consider REF . By REF , we can simplify the wave configuration considering only the fronts crossing starting in the right of MATH and ending in the left of MATH: in fact we can move the other fronts to MATH without changing the shift rate of MATH. We can now shift the initial position of the waves of the MATH-th family merging in MATH such that their initial position coincide with MATH, without changing the shift rate MATH. This operation can be repeated for all shocks of genuinely nonlinear families. Finally, we can move the shocks of the linearly degenerate families such that they have the same sequence of interaction with the other shocks. This means that, if MATH is the position of the MATH-th shock of the MATH-th linearly degenerate family, the only interactions among shocks occurring in the sector MATH are the one involving one MATH-th wave and one MATH-th wave, with MATH. Using REF , we can at this point substitute them with a single shock, whose strength is the sum of the strengths of the MATH-waves. Finally we move their position at MATH such that it coincides with MATH: we obtain the wave patterns of REF . To conclude, we just need to prove that the NAME problem obtained in this way is exactly MATH and that the remaining MATH-wave is MATH. By the previous argument, the strength of the shock of the MATH-th family MATH, MATH, is given by the MATH-waves starting in the right of MATH and ending in the left of MATH: since they are the only MATH-wave crossing the segment MATH, it follows MATH . The other relations for MATH and MATH follows in the same way. Finally, for MATH the jump is MATH. Note that the wave pattern is the same obtained in REF. We consider only the case MATH, since the other is entirely similar. Assume that MATH. Then the shift MATH of MATH is given by MATH . The hypothesis implies that the MATH-th shocks ending at MATH starts in the right of MATH. With the same simplification considered in REF , we reduce to the NAME problem MATH in MATH, such that the waves of the MATH-th families, MATH, generated at MATH collide with the MATH-wave in MATH (see REF ), after overtaking the MATH-wave MATH. The conclusion follows easily, since the wave pattern is the same considered in REF. This concludes the proof of REF .
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math/0006094
|
For genuinely nonlinear fields, the proof is the same as in CITE. We then restrict the proof to the case of a linearly degenerate fields MATH. Assume that there exists two jumps MATH, MATH of the MATH-th family, with positions MATH, such that MATH . For definiteness, assume MATH, and the following conditions is satisfied: MATH . Let MATH, MATH be the jumps of linearly degenerate family MATH in the strip MATH: if we define MATH it is easy to verify that the shocks MATH, MATH, are in involution. By REF , we can then moving the jumps to the left until either MATH meets the wave fronts MATH, or MATH coincides with another shock of the MATH-th family REF . It is clear that we can repeat the same procedure also in the following cases: CASE: MATH and MATH; CASE: MATH and MATH; CASE: MATH and MATH. It is now easy to prove that the total variation of the jumps of the MATH-th family satisfying REF can at most be MATH. Since MATH, MATH divide the lines MATH and MATH in three regions, the total variation of MATH is bounded by MATH.
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math/0006094
|
The first inequality is an easy consequence of the MATH continuous dependence for front tracking solutions, see CITE. For the second one, note that all the shocks different from MATH have size uniformly bigger than MATH, so that their position is shifted of the order MATH. Thus the second inequality follows by standard ODE perturbation estimates, see CITE.
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