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math/0005100
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One has MATH . It is easy to see that MATH and one verifies directly that MATH.
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math/0005100
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Put MATH. Then one has MATH .
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math/0005100
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Since MATH it follows from REF that the number of summands of MATH is equal to the rank of MATH. We want to show that the summands of MATH are a strongly exceptional collection. To do this we have to compute the MATH between the summands of MATH. We first compute the MATH's using REF . The result is as follows. MATH . For the square marked MATH we have (using REF ) MATH . It now follows immediately from REF that MATH is zero between the summands of MATH. For the MATH's we find: MATH with the MATH entry given by MATH . So it follows in particular that MATH is defined by a strongly exceptional collection. We are now done by NAME REF.
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math/0005100
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MATH. We have already pointed out that MATH implies MATH CITE, and MATH implies MATH is obvious. MATH. Assume that MATH is finitely generated. If MATH for a finite dimensional hereditary MATH-algebra MATH, then MATH has a tilting object. If MATH where MATH is a sheaf of hereditary orders over MATH, it follows from REF that MATH has a tilting object. Hence we are done using REF .
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math/0005100
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MATH . When MATH has a tilting object MATH, it follows from CITE that MATH is derived equivalent to the finite dimensional algebra End-MATH. MATH . Since the hereditary category MATH is derived equivalent to a finite dimensional algebra MATH, it follows that MATH must have finite global dimension. Hence MATH, and consequently MATH, has a NAME functor CITE. Then it follows that MATH has almost split sequences CITE. Since MATH is finitely generated, it follows that MATH is finitely generated because this property is an invariant of derived equivalence. MATH . Since MATH has almost split sequences and no nonzero projectives or injectives, it follows that MATH has a NAME functor CITE. Since MATH has some object of infinite length, it follows from REF that MATH is of the form MATH where MATH is a sheaf of hereditary MATH-orders. MATH . This follows from REF . MATH . That MATH for a weighted projective line MATH has a tilting object follows from CITE, and MATH follows from CITE.
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math/0005100
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That REF are equivalent follows from CITE. To prove the other equivalences we first review some generalities. First of all if MATH is a MATH-module, then by CITE the NAME dimension of MATH as MATH-module is equal to the NAME dimension of MATH as MATH-module. Furthermore we claim that MATH if and only if MATH for all MATH such that MATH (or equivalently, if and only if MATH for all MATH such that MATH). To see this we may assume that MATH is finitely generated. Then it follows from the theory of associated primes that MATH has a finite filtration (as MATH-module) with subquotients of the form MATH with MATH. The claim is now an immediate verification. The subcategory MATH of MATH is a localizing subcategory. Since MATH is a noetherian ring which is finitely generated as a module over its center, MATH is closed under injective envelopes CITE. Denoting by MATH the associated quotient functor we have that MATH preserves injective objects and injective envelopes by CITE. So if MATH is a minimal injective resolution in MATH, then MATH is a minimal injective resolution in MATH. A similar reasoning shows that for each prime ideal MATH in MATH, we have that MATH is a minimal injective resolution in MATH. CASE: Assume that REF holds, and consider for MATH in MATH a minimal injective resolution MATH in MATH, and the induced minimal injective resolution MATH for a prime ideal MATH in MATH. If MATH and MATH for MATH, we get MATH for MATH and MATH, and hence MATH is in MATH for MATH, so that MATH. Then we have an injective resolution MATH in MATH, which shows MATH. Since MATH is essentially surjective, it follows that MATH. If MATH and MATH for MATH, we get that MATH is in MATH for MATH, so that MATH is an injective resolution of MATH in MATH. Hence we have MATH. CASE: Assume now that MATH is hereditary, and let MATH be a minimal injective resolution of some MATH in MATH. Since then MATH is a minimal injective resolution in MATH, it follows that MATH for MATH. Then MATH is in MATH for MATH, so that we have MATH when MATH. Thus we we have the exact sequence MATH in this case, and hence MATH. Since for a noetherian ring MATH which is finitely generated as a module over its center we have MATH CITE, and furthermore trivially MATH, it follows that MATH, so that MATH. Hence when MATH, we must have MATH or MATH. In the first case we have MATH for MATH, and in the second case MATH for MATH.
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math/0005102
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Since MATH is a closed subgroup of MATH and the open set MATH is MATH invariant, it suffices to show that MATH acts freely on MATH. To do this we must show that the map MATH given on MATH points by MATH is a closed embedding. First we show that the image MATH of MATH is closed in MATH. Let MATH be matrix coordinates on MATH. Expanding the inverse out in terms of the adjoint we see that the image is contained in the subvariety defined by the matrix equations MATH . Suppose that a REFd-tuple of matrices MATH satisfies the matrix equations above. At least one of the MATH and one of the MATH is invertible because we are in MATH. Let MATH. Substituting into our equations we see that MATH for all MATH. Moreover MATH is invertible since MATH is invertible and MATH. Hence every point satisfying the matrix equations is in the image MATH of MATH, so MATH is closed. The variety MATH is covered by open sets of the form MATH . These open sets are isomorphic to MATH, where MATH is the MATH-fold cartesian product of MATH. Hence the image is smooth, in particular, normal. The action of MATH on MATH is set-theoretically free so MATH is a birational bijection. By NAME 's main theorem (compare CITE) a birational bijection of a normal varieties is an isomorphism, so MATH is an isomorphism. Therefore, MATH is a closed embedding.
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math/0005102
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CASE: Because the action of MATH on MATH is completely reducible, we can decompose MATH as MATH-module. As MATH, we have MATH. Hence MATH is a MATH-submodule and we get the desired direct sum decomposition of MATH. CASE: Let MATH be a basis for MATH. The map MATH defines an isomorphism of the span of MATH (denoted MATH) with MATH, and the map MATH, taking the MATH-th component to MATH, is a MATH-module isomorphism. CASE: Decompose the MATH-weight space of MATH into the MATH and MATH eigenspaces of MATH; these are isomorphic to sums of copies of MATH and MATH, respectively.
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math/0005102
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If a group MATH has good representations, then so does MATH, so we may assume that MATH is algebraically closed. Suppose that MATH is a representation of MATH, and let MATH be the complement of a finite set of invariant linear subspaces MATH. We will show that MATH does not act properly on MATH. The strategy of the proof is as follows. Consider the action map MATH. We will find a closed subvariety MATH of MATH whose closed points are of the form MATH whose image is not closed in MATH. Hence MATH is not proper, so the representation is not good. We now carry out the proof. Decompose MATH, where MATH is the MATH-weight space of MATH for MATH. Pick MATH, and write MATH where MATH. Some of the MATH may be MATH; let MATH be the dimension of the space spanned by the nonzero MATH. CASE: If MATH for all MATH with MATH, then MATH. Indeed, suppose not; then MATH for some MATH. For almost all choices MATH of MATH elements of MATH, the vectors MATH are linearly independent. (Here MATH are the indices MATH with MATH.) This follows because the MATH matrix MATH with entries MATH is nonsingular for almost all MATH. (This is because MATH is a sum of monomials, where each monomial is a product of one term from each row and each column. Each monomial has different multi-degree, so MATH is not the zero polynomial.) Therefore, the vectors MATH span the same space as the MATH, so MATH, contradicting our assumption that MATH. We conclude that MATH, as claimed. A similar argument shows that MATH. CASE: There exists an element MATH with MATH for all MATH. To see this, suppose MATH for some MATH. Let MATH be the subspace of vectors of the form MATH. Note that MATH generates MATH as MATH-module. Consider the affine linear subspace MATH . We claim that MATH is nonempty. If it is empty, then because MATH is affine linear and is contained in a finite union of the subspaces in MATH, we see that MATH for some MATH. But then the span of MATH is contained in MATH, so MATH and MATH. Because MATH is MATH-stable, MATH as well; so MATH contradicting MATH. We conclude that MATH is nonempty. Replacing MATH by an element of MATH, which we again call MATH, we do not change MATH for MATH, but we obtain MATH. Iterating this process, we obtain MATH of the desired form. Replacing MATH by MATH, we will assume that MATH for all MATH. From the MATH-module isomorphism of MATH with MATH, we see that for MATH, MATH . Note also that MATH . CASE: Define MATH . For all MATH, both MATH and MATH are in MATH (by REF ). Define MATH to be the closed subvariety of MATH whose points are the pairs MATH . Then MATH . Consider the point MATH . Reasoning as in REF shows that MATH is in the MATH-module generated by MATH or MATH, so if either MATH or MATH were in MATH then MATH would be, but this is impossible as MATH. Hence MATH and MATH are in MATH, so MATH. Also, MATH is not in MATH, but is in the closure of MATH in MATH. We conclude that MATH is not proper, so the representation is not good.
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math/0005102
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Suppose MATH where MATH is reductive and MATH unipotent. If MATH has a good representation then so does MATH. As proved above, this implies that MATH is a torus, so MATH is solvable.
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math/0005102
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For this proof only, we will use ``good" to mean ``set-theoretically freely good". Let MATH be a good representation of MATH, with MATH a finite set of proper invariant subspaces containing the vectors with nontrivial stabilizers. Because MATH is affine CITE, the vector bundle MATH is generated by a finite dimensional space of global sections MATH. We will view sections of the vector bundle as regular functions MATH satisfying MATH, where on the right side we are using the action of MATH on MATH. The action of MATH on the space of sections of the vector bundle corresponds to the left action of MATH on regular functions: MATH. Because the action of MATH on regular functions is locally finite, by enlarging the space MATH, we may assume MATH is stable under the MATH-action. Define MATH to be the subspace of MATH consisting of those elements of MATH which are sections of MATH. Each MATH is a MATH-stable subspace of MATH. Let MATH denote the complement of the MATH in MATH. Let MATH be a good representation of MATH, viewed as a representation of MATH via the map MATH. We claim that MATH is a set-theoretically good representation of MATH. Indeed, let MATH be a finite set of invariant subspaces of MATH containing the vectors with nontrivial stabilizer. It suffices to show that the vectors with nontrivial stabilizer in MATH are contained in the union of the subspaces MATH and MATH. To see this, let MATH be in the complement of these subspaces. so MATH and MATH for any MATH. We must show that MATH is trivial. First, MATH. Let MATH. As above, we will view MATH as a function MATH. Because MATH is not in any MATH, we have MATH is not a section of MATH for any MATH. In other words, the open subsets MATH of MATH are nonempty. Choose MATH in the intersection of these sets, so MATH for any MATH. Our hypothesis implies that MATH. By definition, we have MATH . But MATH, so we conclude MATH as desired.
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math/0005104
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Our argument closely follows CITE. We can first rule out the case MATH, because for it we only need to show that MATH is not standard. But a standard polyhedron without boundary must have vertices, while MATH. So we proceed assuming that MATH is irreducible. We will now prove that if MATH is not standard then MATH, and that MATH is as in Subsection REF when MATH. To conclude we will later show that if MATH is standard and MATH then MATH and MATH is as prescribed. Suppose then MATH is not standard. First, if MATH is a point then MATH. Suppose now MATH has a REF-dimensional part. So, let MATH be a segment disjoint from REF-dimensional part of MATH. If MATH, looking at the ball MATH, we deduce that there is a properly embedded disc in MATH intersecting MATH in a point of MATH. By irreducibility MATH is then a solid torus, so MATH and MATH is as in REF left. If MATH, looking at the ball MATH again, we see that there is a sphere MATH intersecting MATH in one point of MATH. By irreducibility MATH bounds a ball MATH, and MATH is easily seen to be a spine of MATH. NAME now implies that MATH contains vertices, so MATH is a skeleton of MATH with fewer vertices than MATH. A contradiction. We have shown so far that MATH is quasi-standard unless MATH is MATH or MATH. Since MATH is not standard, either a REF-component MATH is not a disc, or a REF-component is a circle. In the first case, either MATH, or MATH, or MATH contains a simple closed curve MATH which is non-trivial and orientation-preserving in MATH. In the first two cases we have respectively MATH, which is impossible, and MATH, so MATH. The third case is impossible: looking once more at the ball MATH, we deduce that there is a sphere MATH intersecting MATH in MATH, and again MATH. As above, MATH is a spine of MATH. By minimality MATH cannot contain vertices. It follows that MATH is a disc, which contradicts the choice of MATH. Finally, if a REF-component of MATH is a circle but all REF-components are discs, then MATH must be the ``triple hat," a skeleton of MATH. We are left to analyze the case where MATH is standard and MATH, so MATH. Now, if MATH and MATH is a vertex of MATH, then the three faces of MATH incident to MATH are the same as those incident to the other vertex of MATH. Moreover, since MATH, again the same faces are incident to the endpoint of the edge of MATH which starts at MATH. It easily follows that MATH, but MATH by REF , so MATH is either MATH or MATH. It is now a routine matter to check that MATH is respectively MATH or MATH, with MATH as prescribed.
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math/0005104
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There are no closed bricks of complexity zero, since MATH, MATH, MATH, and MATH can be obtained assembling respectively two copies of MATH, two copies of MATH, one copy of MATH and one of MATH, and two copies of MATH. Moreover MATH, MATH, and MATH are not non-trivial assemblings of each other, and the conclusion follows.
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math/0005104
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Let MATH be an edge of MATH, and let MATH be the triple of (possibly not distinct) faces of MATH incident to MATH. The number of MATH's that separate MATH from MATH is even; it follows that MATH is a surface away from MATH. Let MATH be a boundary component of MATH, containing the triod MATH. Since MATH is a disc, which is adjacent either to MATH or to MATH (say MATH), then each MATH-component of MATH incident to MATH (there could be MATH, MATH or REF of them, with multiplicity) has MATH on both sides. So MATH is not adjacent to MATH. Finally, since MATH intersects the link of each vertex either nowhere or in a loop, then MATH is a closed surface. The surface MATH cuts MATH in two components (and is thus orientable, since MATH is) because MATH and MATH lie on opposite sides of MATH.
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math/0005104
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Assume a face MATH of MATH contains no vertices, and let MATH be incident to the triods MATH. Then MATH is a connected component of MATH, but MATH is standard without boundary by REF , so MATH, whence MATH and MATH. A contradiction.
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math/0005104
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Being normal, MATH is determined by an integer attached to each REF-component of MATH. Now we cut MATH open along MATH as explained in CITE: if a REF-component bears an integer MATH we replace the component by MATH parallel ones. We get a polyhedron MATH which contains MATH, such that MATH is the disjoint union of an open ball MATH and an open regular neighbourhood MATH of MATH in MATH. By removing from each torus MATH the open disc MATH we get a polyhedron MATH intersecting MATH in MATH. Now we puncture a MATH-component which separates MATH from MATH and claim that the polyhedron MATH is as desired. Only the inequalities between MATH and MATH are non-obvious. By construction we have MATH. Consider now a vertex MATH of MATH contained in MATH. Of the six germs of REF of MATH at MATH, three are actually the same MATH, so their coefficient in MATH is the same, say MATH. Call MATH, MATH, and MATH the coefficients of the other three germs of REF at MATH. As we cut MATH along MATH we see that MATH disappears if and only if (up to permutation) MATH. If MATH does not disappear then MATH is even. Then we set MATH and note that MATH remains on MATH if and only if MATH. Now let MATH be the other vertex of MATH on MATH. Since the coefficients MATH are the same at MATH, we deduce that either MATH and MATH both disappear, or they both stay on MATH, or they both move to MATH. In the last case, however, one sees that MATH has MATH components parallel to MATH, which is absurd. So both MATH and MATH disappear in MATH (either already in MATH or when we remove MATH). This shows that MATH, so MATH. Suppose now MATH is standard. Then MATH is the union of a quasi-standard polyhedron MATH and some arcs in MATH. REF-components of MATH which separate MATH from MATH are the same as those of MATH, so they give a closed surface MATH by REF . Since no component of MATH is parallel to MATH or to one of the MATH's, REF-component MATH of MATH punctured to get MATH cannot be a closed surface. Now if MATH contains vertices of MATH, we see that MATH, whence the conclusion. Suppose on the contrary that MATH contains a circle MATH with MATH. Note that the process of cutting MATH along MATH allows to define a local injection MATH, and that MATH. Now, if MATH contains some vertex of MATH then this vertex has disappeared in the passage from MATH to MATH, whence the conclusion. If MATH then we consider REF-component MATH of MATH incident to MATH and note that MATH must be a face of MATH without vertices, which is absurd by REF .
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math/0005104
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For MATH it was shown during the proof of REF that MATH is MATH or MATH, so we suppose MATH. By contradiction, assume MATH is not prime and let MATH be a standard minimal skeleton of MATH. Then MATH contains an essential normal sphere MATH. Such a sphere cannot be parallel to the boundary in MATH. Applying REF we get MATH with MATH, MATH, and MATH. Since MATH is an open REF-ball, adding to MATH a generic segment isotopic to MATH we get a skeleton for MATH with as many vertices as MATH. This contradicts minimality of MATH.
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math/0005104
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Let MATH be a minimal skeleton for MATH, which is standard by REF , and let MATH be the minimal skeleton of MATH. Then MATH is a skeleton for MATH with minimal number of vertices, but MATH is not nuclear: there is a face MATH of MATH, glued to the free segment of MATH, which is incident to some vertex of MATH by REF . By collapsing MATH we would get a skeleton with fewer vertices, which is absurd.
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math/0005104
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By REF , MATH is standard. Suppose a face MATH is incident more than once to some MATH. Let MATH be an arc in MATH having endpoints MATH and MATH in two distinct edges of MATH, and let MATH be an essential closed curve in MATH with MATH. Now MATH is cut by MATH into components MATH and MATH. Since MATH is a ball, we can glue to both curves MATH a disc, and the two discs together form a disc MATH with MATH. Since MATH is essential, MATH is a solid torus and MATH.
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math/0005104
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Of course we can assume MATH. Since MATH and the inequality is easy for any non-trivial assembling of MATH, we also assume MATH. Suppose now that a face MATH is incident to two distinct triods MATH. Then there is an arc MATH, properly embedded in MATH, with endpoints MATH, and two essential loops MATH and MATH such that MATH, MATH. Since MATH is a ball, there is an annulus MATH properly embedded in MATH, with MATH and MATH. If some face MATH is incident to the same MATH and to some other MATH, we can construct an annulus MATH in the same way. Moreover MATH with MATH. Irreducibility allows to assume that MATH is just one segment, hence MATH, and then to show that MATH. So MATH, but we are assuming MATH, and the conclusion holds in this case. By REF , MATH has distinct faces MATH incident to MATH. By what already shown we can assume up to permutation that MATH and MATH are not incident to any other triod in MATH. So MATH contains at least MATH distinct faces. By REF we have MATH, therefore MATH.
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math/0005104
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First, MATH is prime and MATH, so MATH by the previous lemma. If MATH were not a brick then it would split as MATH with MATH. In all cases we must have MATH for some MATH, which contradicts the previous lemma.
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math/0005104
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Let MATH be a minimal skeleton of MATH. By REF we have MATH. Now MATH is super-standard, so each edge in MATH determines a different face of MATH. Then MATH, and the conclusion follows.
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math/0005104
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Suppose MATH is an edge of MATH with common endpoints; since MATH is orientable, the regular neighbourhood of MATH in MATH is an annulus, so there is a component MATH with MATH. Then MATH is a length-REF loop; this is impossible by REF , since length-REF loops are never fake.
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math/0005104
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First, MATH is a triod by REF . There are two possibilities for the regular neighbourhood MATH of MATH in MATH, which are shown in REF and lead to a sphere and a torus respectively. In the first case MATH contains three external discs MATH with MATH. By REF all the loops MATH are fake, so MATH for some MATH. In the second case, let MATH be separating, and let MATH and MATH be the manifolds into which MATH separates MATH. Set MATH for MATH. Then MATH is obtained by assembling the manifolds with triods MATH and MATH, where MATH for MATH. Moreover MATH is a skeleton of MATH, which implies that this assembling is sharp unless it is trivial. Since MATH is a brick, the assembling is trivial. Now, MATH and MATH are standard, so MATH and MATH are prime by REF . Therefore, the assembling must be of the first trivial type, namely MATH must be MATH up to permutation. Hence MATH is the unique minimal skeleton of MATH, homeomorphic to MATH. It follows that MATH is parallel to the boundary in MATH.
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math/0005104
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Let MATH be an embedded face with MATH or fewer vertices. A loop in MATH very close to MATH and disjoint from MATH bounds a disc MATH parallel to MATH. Moreover MATH and MATH is not fake since MATH has only one component. Let MATH be a face incident at least twice to an edge MATH of MATH. It follows that there is a length-REF loop MATH intersecting MATH once. NAMEREF loops are never fake, so, by REF , MATH does not bound a disc. Therefore its regular neighbourhood MATH is a NAME strip with one tongue, and MATH is a trace with two vertices of the disconnecting torus in MATH which bounds the regular neighbourhood of MATH in MATH. REF implies that MATH is boundary-parallel, so MATH has no vertices.
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math/0005104
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The proof of REF works away from MATH. We only need to show that MATH is a surface near MATH: let MATH and MATH be the faces other than MATH incident to an edge MATH. Since MATH is orientable, MATH is adjacent to MATH on both sides and MATH is adjacent to MATH on both sides (or the converse). Therefore MATH and MATH are disjoint from MATH, and MATH is a closed surface.
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math/0005104
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By contradiction let MATH with MATH non-separating, and put MATH. The co-disconnecting surface MATH is by REF a closed orientable surface, which is non-empty since MATH disconnects MATH, whereas MATH does not. We assume that MATH is minimal among all mimimal skeleta of MATH for which there exists a non-separating torus whose trace is a triod. We focus now on a component MATH of MATH. Choosing a transverse orientation for MATH as in REF , we can trace on MATH two trivalent graphs MATH and MATH which intersect transversely. These graphs represent the way the rest of MATH glues to MATH, and the sign MATH or MATH depends on whether MATH locally lies on the positive or on the negative side of MATH. We show now several properties of the triple MATH which do not require the bound REF on complexity. Only later we will use this bound. CASE: MATH consists of planar surfaces. Given a point MATH of MATH there are two points MATH of MATH closest to MATH, with MATH on the positive side of MATH and MATH on the negative side. It is not hard to show that the map MATH extends to a homeomorphism of MATH onto an open subset of MATH, and similarly for MATH. CASE: The components of MATH bound discs in MATH. This follows from the same argument just explained. CASE: MATH consists of discs. This is because MATH, MATH, and MATH is standard. CASE: If a component of MATH is not a disc then its boundary loops are essential in MATH. We refer to MATH. If one of them is not, it is very easy to see that there is a disc MATH in MATH such that MATH but MATH, so in particular MATH contains vertices of MATH. The move suggested in REF then contradicts minimality of MATH. CASE: Not all the components of MATH are planar. Again we refer to MATH. By contradiction, from points REF and the irreducibility of MATH, we would readily get that MATH bounds a handlebody, but MATH is non-separating. CASE: Every component of MATH intersects MATH, and conversely. Otherwise, since MATH is connected, there would exist a component of MATH with disconnected boundary, contradicting point REF. CASE: MATH contains at least two points. Assume there is only one point MATH (a crossing between MATH and MATH). If a face MATH of MATH is incident to MATH, then it must be multiply incident, because faces contain an even number of quadrivalent vertices (with multiplicity). If two instances of MATH are adjacent to each other at MATH, we find in the closure of MATH a length-REF loop bounding an external disc, which contradicts minimality. If two instances of MATH are opposite at MATH, then for the same reason there is another face MATH doubly incident to MATH, and MATH. Now in the closure of MATH we can easily find a length-REF loop bounding an external disc which meets edges opposite at MATH. By minimality the loop must be fake, so these edges must actually be the same. NAME of MATH then implies that MATH: a contradiction. CASE: If a component of MATH is a circle then it intersects MATH in at least REF points, and conversely. This readily follows from REF and minimality, because this circle is precisely the boundary of a face of MATH. CASE: No squares as in REF left occur in MATH. If one such square exists, we can correspondingly apply to MATH one move as in REF right. The result is a new minimal skeleton MATH on which MATH still has trace MATH, but MATH. A contradiction. We show now how to conclude, using the fact that MATH. It follows from point REF that both MATH and MATH have vertices. Being trivalent, they have an even number of them, and the total is at most MATH by point REF. So up to permutation we can assume that MATH has REF vertices. In particular MATH has only one non-planar component, which is homeomorphic to a punctured torus (with a component of MATH sitting as a triod in this torus). From point REF we deduce that MATH can have at most one circular component, and it is now easy to deduce from point REF that MATH indeed is a torus. Point REF then implies that MATH consists of the triod only. In the rest of our proof we will always depict MATH cut open along MATH. So MATH appears as a hexagon, and we denote by MATH its interior. To conclude the proof we will first show that MATH also has REF vertices, and then that it appears in one of the two shapes shown in REF . This indeed yields a contradiction to the fact that MATH is a brick, since MATH consists of two points, so cutting MATH along MATH we see that MATH can be obtained via a sharp-assembling. So, let MATH have REF vertices. We claim that all the components of MATH are trees. If one of them is not then there is a face of MATH inside MATH and bounded by MATH. Then either this face has MATH vertices, which contradicts REF , or it is a square of the first forbidden type. Our claim is proved. Now note that if MATH has MATH components then it has MATH free endpoints, which give MATH vertices in MATH. Since MATH has REF vertices and MATH has REF, we deduce that MATH and that MATH has no vertices. Moreover MATH is connected and standard, and MATH. It is not hard to show that with these constraints the only possibility for MATH is as shown in REF , so MATH has two components. In addition, also MATH consists of discs (as MATH), and we get a contradiction because MATH should then be a sphere with some holes. We can now assume that MATH has two vertices, and show that it appears as in REF . Knowing already that MATH is a disc, it is enough to show that MATH is connected. Suppose by contradiction that MATH is disconnected. Then there exists an arc MATH properly embedded in MATH which separates two components of MATH. Let us consider the endpoints of MATH. By minimality of MATH, they cannot belong to the same edge of MATH, nor to two adjacent ones, otherwise we could make MATH slide on MATH and reduce the number of vertices, as in REF . The ends of MATH also cannot belong to two edges adjacent to one and the same edge, as in REF left. To see this, consider how many vertices of MATH can lie in MATH. If there are no vertices at all, then either a face of MATH contained in MATH has less than REF vertices or there is a square of the second forbidden type. If MATH has both vertices in MATH, then again MATH contains either a small face or a forbidden square. These cases are excluded, so there is one vertex of MATH in MATH, and the only possible case is shown in REF center. Now we let MATH slide over MATH as shown in REF right. The result is a new minimal skeleton MATH on which MATH still has trace MATH, but MATH now contains one of the forbidden squares of REF , which contradicts minimality of MATH. We are left to show that the endpoints of MATH also cannot belong to opposite edges of MATH (REF left). Denote by MATH and MATH the number of ends of MATH on MATH and on MATH respectively. If MATH then MATH can be isotoped so to give rise to a length-REF loop in MATH bounding an external disc: a contradiction. If MATH or MATH then we can replace MATH by a curve disjoint from MATH and having ends on edges of MATH which are not opposite, so we get back to a case already ruled out. So up to permutation we can assume that MATH. Now the face of MATH containing the portion of arc MATH shown in REF right must meet another edge of MATH, otherwise it is either small or forbidden (recall that MATH has REF vertices only). So MATH extends to a properly embedded arc disjoint from MATH. Either MATH belongs to a case already ruled out, or the corresponding MATH is smaller, and a contradiction is reached anyway. This eventually shows that MATH is connected, and the proof is complete.
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math/0005104
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The trace MATH is obtained from MATH as shown in REF ; it follows from the figure that if MATH then MATH and if MATH then MATH or MATH. By REF the edges of MATH have distinct ends. Using this fact one easily sees that MATH if MATH and MATH if MATH or MATH, and the conclusion follows.
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math/0005104
|
The condition that MATH and MATH lie on the same of side of MATH means that MATH, during its transformation into MATH, is pushed towards MATH, and the conclusion is obvious.
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math/0005104
|
It is enough to show that one of the following must hold: CASE: MATH is a non-separating torus, and MATH has a triod as a trace on some MATH; CASE: MATH bounds one of the polyhedra of type REF. So we assume REF does not hold and show REF . Our argument is long and organized in many steps. We first describe the overall scheme stating without proof REF assertions. Later we will provide full proofs. Let MATH be a component having lowest MATH. CASE: If MATH then MATH bounds a polyhedron of type REF, or REF Suppose then that MATH. Since MATH is trivalent it has REF vertices, so MATH, where MATH is the number of components of MATH. Each component is incident to at least REF vertices, so MATH, whence MATH. It easily follows that MATH is a torus and MATH. Then MATH consists of two discs MATH and MATH, both good by REF . Recalling from REF that all edges of MATH have distinct endpoints one easily sees that only the types MATH and MATH for MATH are possible. The restriction that MATH then implies that up to homeomorphism there is only one possible configuration MATH and only one MATH, as shown in REF . If MATH is of type A we have MATH, otherwise we have MATH, and the two discs of MATH are completely symmetric. REF also contains notation used throughout the proof (note that MATH are the edges in MATH both in case MATH and in case MATH). Let MATH be the face of MATH incident to MATH. Moreover, let MATH be the face of MATH incident to MATH. Since MATH is good, we have MATH. Finally, let MATH be the edge of MATH which contains MATH. CASE: Either the faces MATH are all distinct or MATH bounds a polyhedron of type REF or REF. Assuming that MATH does not bound a polyhedron of type REF or REF, it follows that the segments MATH for MATH are distinct. Then let MATH be the endpoint of MATH not lying on MATH. CASE: Up to symmetry we have MATH in case MATH and either MATH or MATH in case MATH. Let us now set MATH in case MATH, and either MATH or MATH in case MATH, depending on whether MATH or MATH, so there are two edges of MATH which start at the endpoints of MATH and both end at MATH. These edges are MATH and MATH, with MATH depending on the case. Recall now that if two edges end at the same vertex then one face incident to the first edge is also incident to the second one. Since we are assuming that the MATH's and MATH's are distinct, we deduce that MATH bounds a disc of MATH, which is a triangle, that is, MATH. Following REF we can then perform a MATH-move to which REF apply. Denoting by MATH the disc corresponding to MATH after the move, we have MATH, and equality can hold only if MATH is of type MATH. CASE: If MATH then MATH bounds a polyhedron of type REF or REF . If MATH then MATH bounds a polyhedron of type REF. This establishes the theorem. We now prove our assertions. Proof of REF . By REF the loop MATH is fake, and we can perform a move MATH as explained in Subsection REF. The result is a trace MATH with REF vertices of a surface MATH isotopic to MATH. By REF either MATH is a non-separating torus, or MATH is boundary-parallel, or we have MATH for some MATH. In the first case, up to isotoping MATH back to MATH, getting an isotopic copy MATH of MATH, we get a contradiction to our initial assumption. In the other cases we have to see which polyhedra can result from an inverse MATH move applied to MATH or to MATH. It is now rather easy to examine all possibilities and check the assertion. Proof of REF . Of course no MATH can be equal to a MATH, because MATH and MATH. Let us first show that if two MATH's coincide then MATH bounds a polyhedron of type REF or REF. We refer to REF for the notation. Two adjacent MATH's cannot coincide because of REF . Up to symmetry, the only cases we are left to deal with are MATH, MATH, and MATH. In all cases we will show that MATH bounds a polyhedron of type REF or REF The key point will be to exhibit two loops that must be fake because of REF . Case MATH is examined in REF left: since MATH and MATH are fake, one sees quite easily that MATH, where MATH is a NAME strip with two tongues (REF right). Case MATH is similar (REF left); we have MATH, where MATH is an annulus with two tongues on opposite sides (REF right). In case MATH we consider the loops of REF left. Since MATH and MATH are fake we deduce that all the edges MATH end at the same vertex MATH, such that MATH. We can then apply a move MATH whose effect on MATH is shown in REF right. The result is a trace MATH which falls into case MATH. So MATH with MATH a NAME strip with two tongues. Recalling that the inverse of a MATH-move is again a MATH-move, we only need to consider which such moves can be applied to MATH. The move is determined by the edge of MATH which disappears during the move: of REF edges of MATH, REF lead to a situation in which MATH, so we exclude them. The other REF edges are actually symmetric, and the result is of type REF To conclude the proof of REF we must show that if the MATH's are distinct then MATH. If MATH is of type MATH then MATH has a certain component of MATH on both sides, and MATH has the other one, so MATH. Assume in case MATH that MATH. Referring to REF let MATH be the midpoint of MATH, and join MATH to MATH by an arc MATH in MATH. There are REF distinct arcs MATH having endpoints MATH and MATH and intersecting MATH twice. For two of them the polyhedron MATH is an annulus with MATH tongues on the same side. Then some MATH is fake, which is in contrast with the fact that the MATH's are distinct. Proof of REF . We start with case MATH. Assume that MATH, and put MATH. If MATH is incident to MATH different vertices of MATH then MATH. Since MATH is minimal we have MATH. On the other hand MATH is incident to MATH and MATH, so MATH. Now REF shows a triod MATH in MATH, trace of a torus parallel to MATH. By REF , either MATH is non-separating or MATH is boundary-parallel. In the first case we get a contradiction to the initial assumption. In the second case we deduce that MATH is incident to MATH and MATH, so MATH. So either MATH, or MATH, or MATH. In all cases but the last one the conclusion is the desired one up to symmetry. Concentrating on the last case, we note that MATH is a surface near MATH and MATH, and that the MATH's and MATH's are all distinct. From these facts it is not hard to deduce that MATH and MATH appear as in REF . The figure readily implies that MATH: a contradiction. The proof in case MATH is similar, except that MATH cannot be used directly: a perturbed version MATH as in REF left must be employed. We are again supposing here that MATH, so MATH is incident to MATH vertices of MATH, but now MATH. Since MATH is minimal we have MATH, so MATH. We first claim that we can suppose MATH up to symmetry. By contradiction, assume that both MATH and MATH are incident to exactly REF vertices. We deduce that the situation is as in REF right, where we also show a face MATH incident twice to an edge, which is absurd by REF . Our claim that MATH up to symmetry is proved, so MATH and MATH is minimal too. A figure very similar to REF shows that a triod must exist in MATH, and allows to conclude as above that either MATH is separating or MATH is incident to MATH and MATH. So either MATH, which gives the desired conclusion up to symmetry, or MATH (recall that MATH is incident to exactly REF vertices). If MATH or MATH we get the desired conclusion. Otherwise we can assume up to symmetry that MATH. So MATH and MATH have a common vertex in MATH, which implies that there is a face incident to both. But MATH is adjacent to MATH and MATH is adjacent to MATH, and the MATH's and MATH's are distinct, so we get a contradiction. Proof of REF . If MATH is of type MATH, then MATH, so by REF (and its proof) MATH bounds a polyhedron MATH of type REF or of type REF, but the latter is impossible because MATH is the trace of a torus. We only need to consider which MATH-moves can be applied to a MATH of type REF. By REF the move actually takes place towards the exterior of MATH (that is, its result contains REF vertices of MATH). The move is determined by the edge of MATH which disappears during the move: of REF edges in MATH, REF lead to a situation in which MATH, so we exclude them. The other REF edges are actually symmetric, and the result is one of the polyhedra of type REF. If MATH is of type MATH, then MATH must be an edge in MATH (otherwise MATH), so MATH is of type MATH. Moreover MATH is the trace of a torus. Combining REF and the part of REF already established we see that MATH with MATH either of type REF or a NAME strip with two tongues (type REF). However, if we denote by MATH the faces of MATH incident to MATH, by REF we have MATH up to permutation, so the MATH's are distinct. This shows that type REF is impossible, and again we are left to analyze what can we get from a MATH of type REF by a move MATH which takes place towards the exterior. Of REF edges of MATH, REF lead to a situation in which MATH, so we exclude them. The other REF edges are actually symmetric, and the result is type REF. Proof of REF . The first step of our proof is the extension of the move MATH to a flow MATH of MATH-moves. As mentioned in the proof of REF we must have MATH in this case, so we assume up to symmetry that MATH, and we note that REF apply. The situation is described in REF . One easily sees that the faces of MATH incident to MATH are MATH and two new ones (one of which is contained in MATH), which we denote by MATH. If MATH are not distinct, the flow is reduced to MATH, and we move to the next step. Otherwise let MATH be the ends of MATH (see REF left). If MATH then again the flow is reduced to MATH. Assume on the contrary that MATH, and consider REF right. Then either MATH or MATH is contained in MATH, but certainly MATH is not, for otherwise MATH would contain an embedded face with two vertices, which is absurd by REF . Setting MATH, we are now in a position to apply a move MATH along the triangle determined by MATH and MATH, getting from MATH to MATH. We proceed in a similar way and note that the process must come to an end because MATH contains one vertex less than MATH by REF . Our second step is to understand the final stage MATH of our flow. By construction either MATH are not distinct or MATH. In the first case, since at each step only REF face not contained in the previous one is inserted (and REF is deleted), precisely REF of MATH are distinct. We know by REF (and its proof) that MATH (which is of type MATH) bounds a polyhedron MATH which is either an annulus with REF tongues on opposite sides, or of type REF The first case is excluded by what just said about the MATH's. By REF , MATH bounds MATH. Since at each step of the construction of our flow the choice of move MATH was forced, the polyhedron MATH is defined unambiguously (it depends on MATH only). We only need to explain which edge of MATH determines the MATH-move which glues MATH to MATH. Of REF edges, REF lead to a trace of type MATH, REF give rise to an embedded face with REF edges (excluded by REF ) and the other REF are symmetric, so MATH also depends on MATH only. It is now a routine matter to check that indeed MATH is the polyhedron of type REF with MATH vertices. Having understood the case where MATH are not distinct, we assume that they are. The rest of the proof is devoted to showing that it is actually impossible that MATH. Let us first assume that MATH. By REF we then have MATH up to symmetry, and we can apply a move MATH which reduces MATH. REF shows that MATH bounds a polyhedron MATH of type REF or REF, but MATH is of type MATH, so it must be of type REF. Once again we must analyze the possible results of a move MATH, towards the exterior of a MATH of type REF. Of REF edges of MATH, REF lead to a trace of type MATH, and therefore are excluded. REF other edges come in REF symmetric pairs. For one type, the result of the move MATH contains an embedded face with REF vertices, which is absurd by REF . For the other type, the result contains an embedded face with REF vertices. We can then apply a disc-replacement move as in REF , getting a new minimal skeleton MATH of MATH. The evolution of the singular set is shown in REF , where the two white dots lie on some MATH, the black dots are vertices, and the gray dots lie on MATH. Since the edges leaving MATH end at the same vertex, a MATH-move transforms MATH into a triod which is not boundary-parallel. This contradicts REF . We are left to deal with the case where MATH are distinct, MATH, and MATH. In this case we can perform a MATH-move along either MATH or MATH, and we can proceed just as above, constructing a flow MATH. During this process the faces MATH, and the vertices MATH remain unaffected, while MATH get transformed into MATH. As above, we have at the end of the sequence either that MATH are not distinct or that MATH. In the first case, REF implies that MATH bounds a polyhedron of type REF or REF Such a polyhedron has at most REF vertex, but MATH contains at least MATH, and we get a contradiction. In the second case we are precisely in the situation MATH previously considered, and again we get a contradiction.
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math/0005104
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Take points MATH; we have MATH. Let MATH be a face of MATH incident to some MATH. The gluing path of MATH to MATH can be split into arcs MATH, meeting at points MATH, where MATH and MATH for all MATH, and each MATH is glued to one MATH. The map MATH is not necessarily injective, since MATH can be multiply incident to an edge MATH. We can give the points MATH alternating (red and black) colors. Since MATH is super-standard, MATH can intersect at most one loop MATH among those in MATH. Now take MATH pairwise disjoint segments MATH, properly embedded in MATH, such that MATH. We can ask the MATH's to be disjoint from MATH, since the points on MATH are separated into two even subsets by MATH. It is easy to see that the two endpoints of each MATH automatically have distinct colours. If we do this for each face MATH incident to some MATH, the union of all the chosen segments is a trace MATH disjoint from MATH and hence contained in MATH. We claim that MATH has a product regular neighbourhood in MATH: take for MATH a vector MATH at MATH, tangent to MATH and directed towards MATH. Each segment of MATH is a MATH, properly embedded in a face MATH such that MATH consists of points with distinct colors. It follows that the vectors at the ends of MATH extend along MATH to a non-vanishing field tangent to MATH. The existence of such a field on MATH easily implies that MATH is orientable and that MATH cuts MATH into two components.
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math/0005104
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The first assertion is easy and taken for granted. By construction MATH sits in MATH and it is simple, so we only need to show that MATH is an open MATH-ball. To this end we note that MATH is a ball MATH. Moreover MATH consists of two discs MATH and MATH, and MATH consists of two balls MATH and MATH, with MATH and MATH. So MATH is a ball.
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math/0005106
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CASE: The left module operation on MATH is determined by MATH and the right module operation by MATH. This proves uniqueness. To prove existence, we must show that the right module operation is well-defined REF and satisfies the right module REF , furthermore that MATH is equivariant REF . NAME and axioms of the left module operation, the bimodule axiom and the NAME rule clearly hold true. CASE: If MATH, that is, MATH, then MATH since MATH is a right ideal of MATH. Hence, the specified right module operation is well-defined. CASE: We calculate that MATH and, since MATH, MATH. CASE: The map MATH is given by MATH. We still have to show the assertions about MATH. Since MATH and MATH, the maps MATH and MATH are well-defined, and MATH with MATH is a coalgebra. If MATH, then MATH, thus MATH. Because MATH is a right ideal of MATH, MATH for all MATH, so MATH is a right ideal of MATH. From MATH follows MATH, thus MATH. Let MATH with MATH, that is, MATH. Then MATH . This implies that MATH, therefore MATH with MATH. Hence, MATH, while MATH follows from MATH. CASE: If MATH, then MATH and MATH . Therefore, MATH is a right ideal of MATH. If MATH is equivariant and MATH, then MATH for any MATH-linear functional MATH on MATH. This implies that MATH. From this and MATH we conclude that MATH. CASE: Directly from the definitions, we have *[] REF MATH for right ideals MATH, MATH of MATH, and *[] REF MATH for derivations MATH, MATH of MATH. We show that, in addition, *[] REF MATH, if MATH is a right ideal of MATH, and *[] REF MATH, if MATH is an equivariant derivation of MATH. CASE: Let MATH with MATH, that is, MATH with the canonical projection MATH. Application of MATH leads to MATH, therefore MATH. CASE: If MATH, then MATH for any MATH-linear functional MATH on MATH, since MATH is equivariant. Thus MATH and MATH, where MATH is the canonical projection. This implies that MATH. If MATH is an equivariant derivation of MATH, then MATH by REF . If MATH is a right ideal of MATH, then MATH by REF . This proves REF .
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math/0005106
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If MATH is an equivariant derivation of MATH, then MATH according to REF , so it remains to show that *[] REF MATH, if MATH is an equivariant derivation of MATH, and *[] REF MATH, if MATH is a right ideal of MATH with MATH. CASE: Let MATH be an equivariant derivation of MATH. Then MATH is an object of MATH. According to Takeuchi's Theorem, the map MATH is bijective. The kernel of MATH is MATH: If MATH, then MATH, and if MATH, then MATH exist such that MATH, therefore MATH. The MATH-linear map MATH is injective and surjective, thus MATH. CASE: Let MATH be a right ideal of MATH with MATH. Then MATH is an object of MATH. Furthermore MATH, which follows from MATH, where MATH is the canonical projection, and MATH. According to Takeuchi's Theorem, the map MATH is injective. The kernel of MATH is MATH, see the proof of REF , thus MATH. Since MATH, we obtain MATH.
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math/0005110
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Let MATH be a maximal family of orthogonal projections in MATH which reduces MATH . Then MATH, where MATH, and this is an indecomposable decomposition. Suppose that MATH is another indecomposable decomposition, and let MATH . Then the projections MATH belong to MATH and reduce MATH . It follows that the projections in the finite dimensional C*-algebra MATH belong to MATH and reduce MATH. Thus, from the maximality of the family MATH it follows that for each pair MATH the positive operator MATH is a scalar multiple of MATH, since otherwise the generated C*-algebra contains a nonzero projection strictly less than MATH. If MATH is also a maximal family then for each pair MATH we have MATH for some MATH (depending on the pair MATH). It follows there is a permutation MATH such that MATH and MATH are unitarily equivalent in MATH and the theorem follows.
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math/0005110
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It will be enough to establish the proposition when MATH and MATH. Let MATH be the identity embedding. Then MATH and so MATH which is to say that MATH and hence that the induced maps MATH are unitarily equivalent. From this it follows that MATH are unitarily equivalent.
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math/0005110
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Let MATH be the given presentations. Consider the MATH-module homomorphism MATH which is the composition MATH where MATH is the natural map. Suppose first that MATH and let MATH be the map MATH where MATH is a rank one projection. Then MATH, the class of MATH in MATH, has image MATH in MATH which in turn coincides with the image of a class MATH from MATH, for some MATH, under the natural map MATH. The representative MATH of MATH is a map MATH, for some MATH, which in turn gives an induced map MATH from MATH to MATH. Suppose first that MATH and MATH are stable algebras, so that MATH and MATH. In particular this means that for any positive integers MATH and MATH one can find MATH so that if MATH is the given embedding then the induced map MATH is inner equivalent to a map MATH. Increasing MATH if necessary it follows that we can replace MATH by an inner equivalent map MATH such that the associated class MATH in MATH has image MATH in MATH. It now follows that we have the factorisation MATH . Indeed, since MATH and MATH are MATH-module maps we have, for MATH in MATH, MATH whilst the image in MATH of MATH is the image of MATH and by construction, MATH has image MATH. In other words, the map MATH is a lifting for MATH. The case when MATH has more than one summand now follows by combining the liftings of partial embeddings. Repeat the argument above for the MATH-module homomorphism MATH which is the composition MATH where MATH is the natural map, to obtain a map MATH which is a lifting of MATH. Since MATH is equal to the given map from MATH it follows from REF that we can replace MATH by a unitarily equivalent map to obtain a commuting triangle. Since the process can be repeated we obtain an infinite commuting diagram of maps between the two given direct systems from which it follows that MATH and MATH are star-extendibly isomorphic. Suppose now that MATH and MATH are not necessarily stable and that MATH preserves the scales. Once again consider first the single summand case MATH. If MATH is as above note that MATH lies in MATH and so we may choose MATH large enough so that MATH is in MATH . It follows that there is an extension MATH and the proof may be completed as before.
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math/0005110
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Let MATH be the given presentations. Consider the MATH-module homomorphism MATH which is the composition MATH where MATH is the natural map. Suppose first that MATH and let MATH, be the map MATH where MATH is a rank one projection. Then MATH, the class of MATH in MATH, has image MATH in MATH. Let MATH . Choose MATH in MATH, for large enough MATH, so that MATH . View MATH as a partially defined map from MATH to MATH, defined on MATH. Consider first the case of stable algebras. Then we may increase MATH if necessary to obtain a unique extension MATH of MATH, with MATH. In particular we have MATH as classes in MATH and MATH. By MATH-action preservation, for MATH in MATH, MATH which is MATH-close to MATH where we identify MATH with their image classes in MATH. This is true for all MATH in MATH and so MATH is an approximate lifting of MATH in the sense that the maps MATH and MATH are MATH-close as metric space maps from MATH to MATH. The case when MATH has more than one summand now follows by combining the liftings of partial embeddings. Repeat the argument above for MATH and the map MATH from MATH to MATH to obtain a MATH-approximate lifting MATH from MATH to MATH. Now MATH and so, increasing MATH we may obtain MATH and so we may choose a representative MATH in MATH so that MATH. Continue in this way, for a suitabe sequence MATH, to obtain an approximately commuting diagram and the desired isomorphism. Suppose now that MATH are not necessarily stable. Once again consider first the single summand case MATH and MATH . Noting that MATH lies in MATH choose MATH large enough so that MATH and MATH are in MATH and MATH . Since MATH lies in the scale there is an extension MATH and the proof may be completed as before.
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math/0005110
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It is clear that MATH and MATH are star-extendibly isomorphic then there is an induced isomorphism MATH of the invariants. The sufficiency direction will follow from REF once we show that MATH is naturally isomorphic to MATH with corresponding identification of scales. Let MATH be the morphism for which MATH, where MATH and MATH denotes the image in MATH of the class MATH of MATH in MATH, and where MATH denotes the class of MATH where MATH is the natural injection. If MATH, then there is a sequence MATH of unitaries in MATH for which MATH as MATH, which is to say that the morphisms MATH from MATH to MATH are approximately unitarily equivalent and so MATH. For the same reasons we see that the metric MATH on MATH coincides under MATH with the metric on MATH . Extending MATH by continuity we obtain a continuous isometric injection MATH . In view of the fact that MATH is stable this map is also surjective and thus bijective. It also follows from the stability of MATH that the scale of MATH corresponds to MATH.
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math/0005110
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Let MATH be presentations with partial embeddings in MATH. For each MATH choose MATH large enough so that MATH . By uniform stability we assume that MATH is chosen so that we may obtain in MATH a star-extendible embedding MATH with MATH. The map MATH induces a map MATH, MATH . Furthermore, this map is graded by multiplicity in the sense that if MATH is the multiplicity of MATH then for each MATH . The maps MATH are equicontinuous metric space maps. For suppose that MATH . Then MATH . Consider the equicontinuous family of maps between the compact metric spaces MATH and MATH given by the family of restrictions MATH . By the NAME theorem there is a uniformly convergent subsequence, MATH say. Consider next the restrictions MATH and similarly obtain a uniformly convergent subsequence MATH. Continue in this way and select a diagonal subsequence MATH which converges uniformly on MATH for all MATH. The limit map, MATH say, inherits the properties of the maps MATH in being a semiring homomorphism which respects the right action of MATH. Furthermore MATH is contractive and graded and determines a scale respecting commuting diagram MATH where the horizontal maps are the given embeddings. It follows that MATH determines a contractive homomorphism of invariants MATH . One could appeal to the lifting arguments of REF at this point to construct the desired injection but there is a direct shortcut which makes use of the commuting diagrams above. By uniqueness it follows that for any lifting MATH of the restriction MATH there is a lifting MATH of MATH which is an extension of MATH. Thus we may obtain a sequence MATH in this manner which determines the desired star extendible isomorphism .
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math/0005110
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We begin the proof by putting the map MATH into a standard form. First replace MATH by a unitary conjugate MATH, where the unitary MATH has the form MATH to arrange that MATH and MATH have standard orthogonal final projections of the form MATH where MATH are the matrix units of MATH and MATH is the multiplicity of the embedding. Next replace MATH (the new MATH) by a further unitary conjugate, where the unitary has the form MATH to arrange also that the initial projections of MATH and MATH are also standard orthogonal projections. In this way arrange the resulting standardisation of MATH to have partitioned matrix MATH where MATH are sums of matrix units, MATH and MATH. Since MATH has initial projection orthogonal to that of MATH and since MATH and MATH have the same final projection it follows that MATH and MATH have the induced partitioned matrices MATH . Note that the projections MATH and MATH are block diagonal, supported in the MATH block only, with partitioned matrices MATH when MATH is a MATH matrix and MATH is a MATH matrix. To simplify notation, assume without loss of generality, that MATH, so that the last row and last column of the matrices above are not present. Assuming that MATH satisfies the rank equality conditions in the statement of the proposition we may also assume that MATH is standardised so that MATH. We claim that MATH and MATH are inner conjugate if and only if the pair of projections MATH in MATH is unitarily equivalent in MATH to the pair MATH, when MATH. In fact the necessity of this condition is elementary so assume that these pairs are unitarily equivalent, by a unitary MATH in MATH. Since MATH it follows that MATH is block diagonal with respect to MATH, so that MATH . Thus, conjugating MATH by the unitary MATH we obtain a unitarily equivalent embedding, also in standard form, with quadruple MATH satisfying MATH (as well as MATH). Of necessity MATH, since, by star extendibility MATH . Finally, observe that we may now conjugate MATH by a unitary of the form MATH to obtain a map MATH with quadruple MATH. Thus, with the choice MATH and the proof is complete.
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math/0005110
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Let MATH have quadruple MATH as before, let MATH have quadruple MATH and consider the composition MATH. The images of MATH and MATH give rise to the quadruple MATH for this composition. Note that MATH where MATH with a similar formula for MATH. Thus we may compute MATH . By orthogonality MATH . Thus MATH is equal to MATH . Note that each of the bracketed terms vanishes and so we obtain MATH as the sum of the three positive operators MATH . The first operator has nonzero spectral distribution MATH with appropriate multiplicity whilst the sum of the second two operators which is orthogonal to the first has nonzero spectral distribution MATH which is the same distribution. Appealing now to REF the proof is complete.
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math/0005110
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In view of REF if the sequences are asymptotically equivalent then one can construct an asymptotically commuting diagram of unital star-extendible embeddings. From this it follows that MATH and MATH are star extendibly isomorphic. On the other hand by the stability of MATH-algebras if MATH and MATH are star extendibly isomorphic then there is an approximately commuting diagram that implements this isomorphism. From this it follows that the sequences are asymptotically equivalent.
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math/0005110
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In the proof we will indicate the set of elementary compression type maps for MATH and MATH by MATH and MATH respectively. Suppose first that MATH is the elementary compression type homomorphism MATH where MATH where MATH is a compression type embedding determined by an interval projection of MATH of rank MATH and where MATH is a multiplicity one star-extendible injection mapping matrix units to matrix units. We wish to obtain a C*-algebra extension MATH for the algebra homomorphism MATH. Define first the restriciton MATH, where MATH now denotes the summand of MATH corresponding to MATH, to be the map MATH, viewed as an algebra homomorphism from the summand MATH to the largest summand MATH of MATH, where MATH. Since MATH is star-extendible so too is this partial embedding MATH, with extension given by MATH where MATH is the star extension of MATH. We now want to fully define MATH on all the other summands of C* MATH so that MATH is a C*-algebra homomorphism and MATH for MATH in MATH. In view of the definition of MATH, a matrix MATH in MATH is determined by its largest summand, that is, by the MATH matrix summand MATH. Thus the element MATH is a direct sum of various compressions MATH of the MATH matrix MATH. The key point to note is that each such compression of MATH, which is determined by an interval of MATH, can be viewed as the image of a summand MATH of MATH under a star-extendible map. Let us denote this star-extension as MATH ; it is a multiplicity one C*-algebra homomorphism from MATH to MATH. (If MATH, the largest compression, then MATH.) The map MATH is the required extension. We have shown that for any elementary compression type embedding MATH the induced algebra homomorphism MATH is star-extendible. It now follows that REF is implied by REF . Consider now a star-extendible algebra homomorphism MATH. This determines a contractive algebra homomorphism MATH which may be viewed as the composition MATH where MATH is the restriction map for the maximal dimension summand of MATH. Let MATH be a partial isometry. Then it is straightforward to see that MATH is a unimodular sum of matrix units and that MATH where MATH is a unimodular sum of matrix units. Indeed, the maximal matrix summand of MATH is the matrix MATH which is a partial isometry along with each compression summand. From this it follows that MATH is a regular partial isometry in MATH. Since MATH maps partial isometries to partial isometries, being star-extendible, it follows that MATH is a regular partial isometry in MATH. Thus we conclude that MATH is a contractive regular homomorphism from MATH to MATH, and that REF holds.
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math/0005110
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The equivalence of REF is elementary and we have already noted the equivalnece of REF . Plainly REF implies REF . Assume REF . Let MATH have MATH block decomposition MATH-with MATH. By assumption each MATH is a partial isometry. Consider a product MATH. The MATH block entry is given by the sum MATH . Since MATH is regular, the partial isometries MATH have orthogonal range projections and so the operators of the sum have orthogonal range projections. For similar reasons the domain projections are pairwise orthogonal. Since, by hypothesis, the product MATH is a regular partial isometry, it follows that the sum above is a partial isometry, and therefore, by the orthogonality of domain and range projections, each of the individual products MATH is a partial isometry. Since, for example, MATH is a partial isometry it follows that the range projection of MATH commutes with the domain projection of MATH. Regarding the entry operators MATH as identified with operators in MATH, it follows, by considering other block entries, that for all MATH the range projection of MATH commutes with the domain projection of MATH. Note also that the domain projections and the range projections commute amongst themselves. Furthermore it is clear that these projections commute with the projections in the centre of the block diagonal subalgebra of MATH. Let MATH be a rank one projection which is dominated by MATH. By the commutativity there is a maximal family MATH of rank one projections satisfying MATH. The projection MATH commutes with MATH and determines an elementary compression type embedding summand of MATH. Now REF follows from induction.
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math/0005110
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The first part of the theorem follows from the arguments above which show that the indecomposable maps MATH are labelled by the partially defined order preserving functions MATH whose domains are intervals. The second assertion follows from the combinatorial discussion.
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math/0005110
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The sufficiency direction follows from REF . The necessity of the condition, that is, the fact that MATH is an invariant, will follow from REF once we show the stability of MATH. However this follows readily form the perturbational stability of MATH. Let MATH be star extendible and suppose that MATH . Then consider the map MATH given by MATH where MATH is the projection onto the largest summand. Since MATH is an almost inclusion it follows that MATH . By NAME 's theorem, for MATH sufficiently small MATH is close to a star extendible map MATH. Now it follows that the map MATH is close to MATH.
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math/0005110
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We may assume that MATH. First note that if MATH is irregular then for at least one of the marix units MATH of the triple MATH the partial isometry MATH has the block form MATH where the operator MATH is not a partial isometry. To see this we argue by contradiction and assume otherwise. Since MATH and MATH are block diagonal, the operators MATH and MATH are partial isometries. By assumption, MATH is a partial isometry and so it follows that the operator MATH is a partial isometry and also has block diagonal initial and final projections. But this implies that MATH and MATH are partial isometries. We deduce then that each operator MATH, is a regular partial isometry, which is to say that MATH is a locally regular map. By our remarks above this implies that MATH is regular, contrary to hypothesis. Since the entry MATH of MATH is not a partial isometry it follows, by reasoning as in the last paragraph, that MATH and MATH are not partial isometries and in particular are nonzero operators. Without loss of generality assume that matrix units for MATH are chosen so the restriction of the map MATH to the self-adjoint subalgebra has the form MATH where the projections MATH and MATH have the same rank. More precisely, we can remove the rows and columns of MATH corresponding to the projection MATH and obtain the (typical) image MATH in the operator matrix form MATH . We shall now show that the map MATH is locally regular. Since the composition MATH is assumed to be regular the matrix above, for the element MATH, is a regular partial isometry. In particular the MATH block entry and the MATH block entry have orthogonal ranges and so MATH . Thus MATH, and so MATH. If MATH is a rank one projection in MATH then the partial isometry MATH has block diagonal initial and final projections and, after removal of block rows and columns of zeros, has the MATH block matrix form MATH, where MATH for MATH. Since MATH it follows from the block diagonality of the range projections that MATH and MATH are partial isometries. Reasoning as before it follows that MATH is a regular partial isometry. Similarly, since the MATH block and the MATH block of MATH have orthogonal initial projections it follows that MATH . In particular MATH and so it follows, as before, that MATH is a regular partial isometry. If MATH and MATH are regular partial isometries with MATH then it need not be the case that the partial isometry MATH is regular. Thus we need additional argument in order to see that MATH is a regular partial isometry. Returning once more to the regularity of MATH and the orthogonality of the range projections of the MATH and the MATH block entries, we see that MATH . Since MATH, by the regularity of the partial isometry MATH it follows, since MATH, that MATH. Since MATH has block diagonal final projection this implies that MATH and MATH are partial isometries, and so, as before, MATH is a regular partial isometry. Thus, by our earlier remarks, MATH is regular. Suppose now that MATH is an indecomposable summand of MATH which is necessarily of multiplicity one. By the NAME theorem, REF , indecomposable decompositions are unique and from this it follows that since MATH is regular so too is MATH. If MATH is not MATH-degenerate then MATH is not a regular partial isometry, contrary to the regularity of MATH. It follows then that MATH is of MATH-character.
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math/0005110
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We first note the immediate consequence of REF that if MATH and MATH are MATH-algebra systems for which none of the embeddings MATH, MATH and their system compositions are of MATH-character then a commuting diagram isomorphism between MATH and MATH is necessarily regular. In general consider the commuting diagram MATH with MATH star extendible for all MATH. Suppose moreover that infinitely many of the maps MATH are irregular. Replacing the systems by subsystems we may assume that all these maps are irregular. By the lemma all the maps MATH are of MATH-character. Since MATH is regular it must be that the range of the regular map MATH does not meet those off-diagonal blocks which contain rank one matrix units MATH for which MATH is an irregular partial isometry. This implies that MATH is locally regular. Indeed the range of MATH meets more blocks of MATH than does the range of (the regular map) MATH. Since the triple composition is locally regular it is regular and it now follows that there is a commuting subdiagram of regular maps. Thus MATH is functorial and hence an isoclassic family. The argument is the same for MATH.
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math/0005110
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Note that MATH and that MATH . Thus we have the following calculation. MATH where MATH is a unimodular multiple of the rank one partial isometry MATH . We now want to show that the composition MATH is not merely locally regular, which is what the above calculation shows, but that it is regular, that is, REF-decomposable. (A star extendible locally regular map not be regular, as we have seen.) Let MATH in MATH be the basis with MATH so that MATH where MATH indicates the rank one operator for which MATH. Observe that the embedding MATH for which MATH is a regular star extendible embedding unitarily equivalent to a map MATH determined by two cyclic shifts as indicated above. Thus MATH is the composition MATH where MATH is the linear NAME product map given by MATH . Although these unimodular coefficients do not form a cocycle, that is, MATH is not realisable as a diagonal unitary conjugation, the restriction of MATH to the span of the matrix units MATH is a cocycle. This may be checked directly. Alternatively note that since MATH, the map MATH is the orthogonal direct sum of MATH star extendible embeddings MATH. Since MATH is star extendible so too is each map MATH. That MATH is diagonally implementable on the ranges of the multiplicity one maps MATH follows from the fact that an isometric NAME product map on the bipartite graph is diagonally implementable. (In fact this property is shown to hold for any digraph algebra in CITE.) Since there is a diagonal partition of the identity operator which dominates the ranges of the maps MATH it follows that MATH is diagonally conjugate to MATH, as desired.
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math/0005110
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Let MATH, be the maps MATH . Then for all MATH the compositions MATH are regular by the last lemma. Also, since MATH is also a primitive root of unity the lemma shows that the compositions MATH are regular. Thus the maps MATH provide a commuting diagram between two regular systems in MATH consisting of irregular maps. Moreover it is clear that the crossover maps of any subdiagram are necessarily irregular, and so the theorem is established.
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math/0005113
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The proper subgroup MATH maps onto MATH (for instance under MATH), so MATH. Then the proper subgroup MATH of MATH maps onto MATH (under MATH), so MATH; etc.
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math/0005113
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MATH is a subgroup of MATH, which clearly is residually finite: it is approximated by its finite quotients given by the action on MATH, for any MATH.
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math/0005113
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First, we will prove that if MATH is central then MATH must be in MATH. Let MATH where MATH and MATH. If MATH and MATH for some MATH then MATH does not commute with the elements MATH such that MATH. If MATH then choose MATH with MATH and consider MATH . It is clear that MATH since MATH, which, along with the fact that MATH, gives MATH and therefore MATH. Now, we proceed by induction on the length of the elements and we prove the statement for all MATH simultaneously. From the above discussion it is clear that no MATH-generator and no MATH-generator outside of MATH is in the center. Consider a generator MATH and choose an element MATH with MATH. Then MATH. Therefore no element in MATH is in the center and we have completed the basis of the induction. Consider an element MATH of length MATH. If MATH we already know that MATH is not in the center. Let MATH. At least one of the projections, say MATH, is not trivial. Since MATH has strictly shorter length than MATH, we obtain that MATH is not in the center of MATH so that MATH is not in the center of MATH (and therefore not in the center of MATH).
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math/0005113
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Take any MATH, and consider the coordinates of MATH. They all belong to MATH, whence MATH is a finite group of order at most MATH, and MATH is finite, since MATH is of finite index in MATH.
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math/0005113
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The proof is by induction on the length of MATH and it will be done for all MATH simultaneously. The statement is clear for the empty word. Next, no word of length MATH represents the identity in any group MATH. Now assume that the claim is true for all words of length less than MATH, with MATH and let MATH be a word of length MATH representing the identity in MATH. If MATH is not reduced, then we reduce it to a shorter word MATH. Since the reduced word MATH is in MATH if and only if the original word MATH is in MATH the claim follows in this case from our inductive hypothesis. Assume MATH is reduced. Since MATH must be in MATH the decomposition of MATH of depth MATH is well defined. The length of each of the (possibly not reduced) words MATH is at most MATH (since MATH), so that, by the inductive hypothesis, each of the words MATH is in MATH. The set MATH is clearly closed under concatenation, so MATH is in MATH. Assume MATH; then each of the MATH-letters MATH from MATH appears exactly once in some word MATH, and therefore MATH and MATH represent the same element in MATH, namely, the identity. We conclude that MATH is in MATH.
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math/0005113
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First, note that MATH is generated as a subgroup by the set MATH . Indeed, conjugation of any generator in MATH by an element from MATH gives another generator, conjugation by MATH is unimportant since MATH and, for MATH, MATH . The subgroup generated by MATH is thus normal. On the other hand MATH and MATH so that MATH and MATH. Since MATH is generated by MATH together with the elements of the form MATH, MATH, MATH, we note that MATH. Let us prove that MATH. Assume MATH. Since MATH we can consider MATH. Since MATH we have MATH, for all MATH, and MATH. On the other hand, since MATH we have MATH, for all MATH. Therefore MATH for all MATH, so MATH. Consider MATH. Clearly MATH. On the other hand, given MATH we have MATH where MATH with MATH. Therefore MATH and since the index of MATH in MATH is MATH we obtain the result.
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math/0005113
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Define MATH to be the number of MATH-letters from MATH appearing in the words at the level MATH and MATH to be the number of simple reductions performed to get the words MATH on the level MATH from their unreduced versions MATH. A reduced word MATH of length MATH has at most MATH-letters. Every MATH-letter in MATH that is in MATH contributes one MATH-letter and no MATH-letters to the unreduced words MATH. The MATH-letters in MATH that are not in MATH (there are at most MATH such letters) contribute one MATH-letter and one MATH-letter. Finally, the MATH simple reductions reduce the number of letters on level REF by at least MATH. Therefore, MATH . In the same manner, each of the MATH-letters on level MATH that is from MATH contributes at most one MATH-letter to the words on level MATH and the other MATH-letters (at most MATH of them) contribute at most MATH letters, so MATH . Proceeding in the same manner, we obtain the estimate MATH . If MATH, then the claim of the lemma follows. Assume therefore MATH . For MATH, define MATH to be the number of MATH-letters from MATH appearing in the words at the level MATH. Clearly, MATH is the number of MATH-letters in MATH and MATH . Going from the level MATH to the level MATH, each MATH-letter contributes one MATH letter of the same type. Thus, the words MATH from the first level before the reduction takes place have exactly MATH letters that come from MATH. Since we lose at most MATH letters due to the simple reductions, we obtain MATH . Next we go from level MATH to level MATH. There are MATH-letters on level MATH that come from MATH, so there are exactly MATH-letters from MATH in the words MATH and then we lose at most MATH-letters due to the reductions. We get MATH and, by proceeding in a similar manner, MATH . Since MATH is complete, we have MATH and REF give MATH which implies our claim.
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math/0005113
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Let MATH denote the growth of MATH with respect to word length. We will obtain MATH for some positive constants MATH. Then, MATH applications of REF yield, neglecting the (unimportant) constant MATH, MATH, from which the theorem's claim follows. See CITE for a similar proof. We now prove REF . Choose some MATH with MATH, let MATH be a (set) retraction MATH of MATH, and consider the (set) map MATH defined on reduced words over MATH. Note that MATH does not in general induce a group homomorphism, though this is the case for the first NAME group, where it is traditionally called MATH. We may, however, naturally consider MATH. Given any reduced word MATH representing MATH, we obtain an element MATH of MATH, that has the following properties: MATH . In case MATH, we restrict our consideration to words MATH such that at most MATH of their MATH-letters are not in MATH. It then follows in all cases that all coordinates except the MATH-th of MATH are bounded. To prove the only non-trivial case, suppose MATH and MATH, where the MATH are words over MATH. Then MATH where MATH. By REF , this last group is finite, so each MATH is bounded (say of length at most MATH); then MATH is bounded, of length at most MATH. Given words MATH each of length at most MATH, we wish to construct a word MATH with MATH where MATH and MATH belong to a finite set MATH. Let us choose elements MATH, for MATH, such that MATH. We take MATH . In every coordinate MATH, we get MATH (as MATH), and other words, each of which is bounded. We may therefore take for MATH the set of words of length at most MATH. Note now that for any word MATH we have MATH, so MATH. Also, MATH has at most MATH-letters not in MATH, namely MATH. Finally, MATH determines MATH up to the choice of MATH and MATH, so REF holds with MATH and MATH .
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math/0005113
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Let a minimal form of MATH be MATH . Then MATH can be written in the form MATH and rewritten in the form MATH where MATH. Clearly, MATH, which yields MATH . Now, observe that if the MATH-generator MATH is of weight MATH with MATH then MATH has as components one MATH-generator of weight MATH and one MATH-generator (of weight MATH of course) with the rest of the components trivial. Thus, such a MATH (from REF ) contributes at most MATH to the sum MATH. On the other hand, if MATH is a MATH-generator of weight MATH then MATH has as components one MATH-generator of weight at most MATH, and the rest of the components are trivial. Such a MATH contributes at most MATH to the sum MATH. Therefore MATH and the claim of the lemma follows by combining REF .
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math/0005113
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Let MATH. Choose MATH so that MATH is big enough in order that MATH be satisfied for all MATH and all MATH. Such a choice is possible because MATH for MATH and the two expressions are equal for MATH. Define a function MATH on MATH by MATH . We prove, by induction on MATH, that MATH. If the weight of MATH is MATH, the portrait has MATH leaf and MATH. Otherwise, the portrait of MATH is made up of those of MATH. Let the weights of these MATH elements be MATH. By REF we have MATH, so by induction the number of leaves in the portrait of MATH is at most MATH, MATH and the number of leaves in the portrait of MATH is, therefore, at most MATH. Suppose that MATH of the numbers MATH are no greater than MATH and the other MATH are greater than MATH, where MATH. Without loss of generality we may assume MATH. Using NAME 's inequality, REF , the fact that MATH and MATH, and direct calculation, we see that MATH where the last used inequality holds by the choice of MATH. In case none of MATH is greater than MATH, we have MATH which is no greater than MATH, again, by the choice of MATH.
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math/0005113
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Let MATH be a MATH-factorable sequence, factored in complete words of lengths MATH, with all MATH. We can define a modification of the portrait of an element by requiring that whenever we ``blow up" a leaf on the level MATH because its size is too big, we expand it MATH levels down (that is, the original word is expanded MATH levels down, the words on the level MATH are expanded MATH levels, the words on the level MATH are expanded MATH levels, etc.) and obtain MATH new leaves. An analog of REF still holds, that is, if the length of MATH is at most MATH then sum of the lengths of the elements at the newly obtained MATH leaves is at most MATH (indeed, the fact that we multiply by an element from MATH here and there does not increase the MATH-length of the words in question, so that the sum of the lengths at the newly obtained leaves is still at most MATH plus a constant). Proceeding as before completes the proof of the theorem.
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math/0005113
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For MATH, denote by MATH, MATH and MATH the number of MATH-letters, MATH-letters and MATH-letters from MATH, respectively, in the words on the level MATH of the decomposition of depth MATH of MATH. Clearly, MATH. Also, MATH since every MATH-letter from the level MATH contributes at most one MATH-letter to the next level, except for those MATH-letters that are in MATH. This gives MATH and therefore MATH since MATH. It is easy to see that MATH . Indeed, any change (up or down) in the number of letters in MATH going from the level MATH to the level MATH is due to simple reductions involving letters from MATH, but each such simple reduction also changes (always down) the total number of MATH-letters by the same amount. Then, by telescoping, MATH whenever MATH. Combining REF gives MATH which implies our result since MATH and MATH.
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math/0005113
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We will prove that the order of any element MATH in MATH divides some power of MATH. The proof is by induction on the length MATH of MATH and it will be done for all MATH simultaneously. The statement is clear for MATH and MATH. Assume that it is true for all words of length less than MATH, where MATH, and consider an element MATH of length MATH. If MATH is odd the element MATH is conjugate to an element of smaller length and we are done by the inductive hypothesis. Assume then that MATH is even. Clearly, MATH is conjugate to an element that can be represented by a word of the form MATH . In this case MATH divides MATH and if all the words MATH have length shorter than MATH we are done by the inductive hypothesis. Assume that some of the words MATH have length MATH. This is possible only when MATH does not have any MATH-letters from MATH. Also, the words MATH corresponding to the words MATH of length MATH must be reduced, so that the words MATH having length MATH have the same MATH-letters as MATH does. For each of these finitely many words we repeat the discussion above; namely, for each such MATH of length MATH we construct the period decomposition. Either all of the constructed words MATH are strictly shorter than MATH, and we get the result by induction; or some have length MATH, but the MATH-letters appearing in them do not come from MATH. This procedure cannot go on forever since MATH holds for some MATH. Therefore at some stage we get a shortening in all the words and we conclude that the order of MATH is a divisor of some power of MATH.
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math/0005113
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MATH, yielding MATH. Using an argument similar to that in REF and the observations on the structure of the words MATH given above, we conclude that MATH .
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math/0005113
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As usual, we use induction on MATH and we prove the statement simultaneously for all MATH-homogeneous MATH. We will prove that MATH where MATH. The statement is obvious for MATH. Consider an element MATH of length MATH, MATH and let MATH be its period sequence. We know that MATH because the word MATH is MATH-homogeneous. If MATH has length MATH then it has order MATH and MATH. Consider the case when MATH has (even) length greater than MATH and MATH. In that case MATH for some MATH in MATH that has length at most half the length of MATH (the word MATH is one of the leaves of the period decomposition of MATH). In case MATH we have MATH which is no greater than MATH by the induction hypothesis. Let MATH and MATH. Since MATH we know that the length of MATH is at most MATH so that the length of MATH is at most MATH and we have MATH which is no greater than MATH by the induction hypothesis and the fact that MATH holds in this case. In case MATH and MATH, the length of MATH is at most MATH and we have MATH which is no greater than MATH by the induction hypothesis and the fact that MATH holds in this case. Therefore, in each case MATH, which proves our claim.
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math/0005113
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Let MATH be the abelianization map. Recall that MATH is the elementary MATH-group of rank MATH. We recast the construction of the period sequence as follows: in the graph below, nodes correspond to images of elements MATH under MATH; arrows indicate taking a projection, MATH or MATH. Double arrows indicate a squaring was applied before taking the projection (because MATH was not yet in MATH). Also, recall that in the squaring case the obtained projections MATH and MATH are conjugate. A condition labeling an edge indicates that such an edge can exist only if the condition is satisfied. One step of the algorithm for calculating the order of an element We proved in the previous section that all double arrows are as described. Let us complete the proof for single arrows. According to REF , the commutator of a NAME group is MATH . Let us take an arbitrary element MATH with MATH and assume that MATH is in the kernel MATH, but it is not in any kernel with smaller index. We follow our squaring procedure MATH times obtaining an element MATH and set MATH. We know that MATH, MATH and we wish to compute MATH for MATH. For this purpose, write MATH for some MATH. Then MATH . Now, note that MATH, MATH, MATH and MATH, MATH, MATH, MATH, MATH, so that all MATH-letters appear in pairs in the expression MATH . Of course, the element represented by MATH is in MATH and can be rewritten in the form MATH, where MATH and MATH. Let MATH and MATH denote the product of the MATH-letters in MATH preceded by an odd and even number, respectively, of MATH's. Those MATH-letters preceded by odd number of MATH's will appear conjugated by MATH when we rewrite MATH in the form MATH and those preceded by even number of MATH's will appear without conjugation. It is not difficult to see that we have either MATH or MATH. Indeed, if the number of MATH's in the expression MATH is odd, that is, MATH is not in MATH then both the number of MATH factors and the number of MATH factors in both MATH and MATH are even so they cancel out and the number of MATH factors in MATH and MATH is equal, so that their product is REF or MATH. Similarly, if the number of MATH's in the expression MATH is even then the number of MATH factors in both MATH and MATH is even so the MATH's cancel out and the number of MATH factors and MATH factors in MATH and MATH is equal and even so their product is REF or MATH. Considering the extra MATH in the expression for MATH we may suppose, up to a permutation of the indices MATH and MATH, that MATH and MATH. Then MATH . The same argument works when we start with MATH and set MATH; we then obtain either MATH whence MATH for MATH, or MATH, where MATH is the only MATH-letter in the kernel MATH, whence MATH . Thus, we enter the graph above at the vertex MATH, loop MATH times at MATH, then in REF move to the vertex MATH and either use two consecutive single arrows to exit the graph and lend into an element of MATH or we proceed to the vertex MATH, follow another double (squaring) arrow to MATH and then either leave the graph through two consecutive single arrows and lend into an element of MATH or we lend in an element in MATH with projection MATH. The order of MATH will depend on the exit point, that is, on the number of squarings performed and the various length reductions that occurred during the trip through the graph. Recall that each time we follow a non-squaring arrow we may claim a length reduction by a factor of MATH and in case we follow MATH consecutive squaring arrows (that is, MATH) we may claim an additional length reduction by a factor of MATH. As usual, we can now use induction on MATH and prove the statement simultaneously for all MATH-homogeneous MATH (just like in the proof of REF ).
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math/0005113
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Assume that MATH lies in MATH and that MATH contains an element MATH satisfying the following conditions: CASE: MATH is of order MATH; CASE: MATH has a representation of length MATH with MATH odd; CASE: this representation is of the form MATH, with all the MATH in MATH except for one, which is in MATH. We shall construct an element MATH of MATH of order at least MATH, having a representation of length MATH satisfying REF (with MATH instead of MATH). We restrict our attention to words of the form MATH, with MATH. A word-set is such a word, but where the MATH's are non-empty subsets of MATH. An instance of a word-set is a word (or group element) obtained by choosing an element in each set. For all MATH, there is a map from word-sets in MATH to word-sets in MATH, defined as follows: MATH . Any instance MATH of MATH satisfies MATH where MATH is an instance of MATH. Let us now consider MATH of order MATH. Set first MATH and note that MATH, where each MATH represents an element in MATH and we are not interested in their actual value. Choose MATH such that MATH, where MATH is the only letter from MATH in MATH and replace MATH by the word-set MATH. Note that MATH so that the choice indicated in the previous sentence can be done. Also note that this transformation does not change the first coordinate of MATH. We claim that MATH has an instance that, as a word, is a square, say of MATH. Then since MATH, we will have constructed a word MATH satisfying the required conditions. Let us compute the lengths. Before the substitution of the element in MATH, MATH is a word-set of length MATH. The substitution of the element in MATH increases the length by MATH. Thus, the length of the word set MATH is MATH. Also, the number of appearances of MATH in MATH is REF. Write MATH in REF lines of length MATH. Each line will be of the form MATH, with the MATH's elements or subsets of MATH. Our goal is to choose an instance of MATH such that the two lines are identical, that is, we want to choose identical elements in each column. Half of the columns will consist of MATH's and the other MATH columns (an odd number of them!) will have one of the following: CASE: MATH and MATH; CASE: MATH and MATH (MATH columns); CASE: MATH, and MATH (two columns); CASE: MATH and MATH (one column). In each case the two elements in the column can be chosen identical. The above construction is a single step in an inductive construction in which, starting with an element MATH where MATH, MATH, MATH repetitions of the step give an element in MATH of order at least MATH and length at most MATH, so that MATH.
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math/0005119
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The proof is standard. Let us consider the following exact sequence of functors from MATH to the category of MATH-linear spaces. MATH . Here MATH, MATH, and MATH, MATH, MATH are natural transformations given as follows MATH where MATH. The two middle terms in the exact sequence REF are exact functors (both with respect to the first and to the second arguments). Thus MATH is right exact, which proves REF. The equality REF follows from evaluation of dimensions in REF.
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math/0005119
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All the statements except the functoriality of MATH are obvious. The functoriality of MATH follows from properties of the NAME characteristic (namely, additivity with respect to algebraic stratifications, and multiplicativity for fiber bundles).
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math/0005119
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Associativity follows from functorial REF of the maps MATH.
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math/0005119
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REF is a simple calculation using the definition of the MATH-product, all the rest is obvious.
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math/0005119
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REF have been proven by Ch. CITE. REF follows from REF follows from REF and from the fact that MATH for any MATH.
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math/0005119
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Given MATH let MATH, where MATH . We fix an object of the form MATH (respectively, MATH, MATH) in the isomorphism class of MATH for each MATH (respectively, MATH for each MATH). Using MATH and MATH we construct a functor MATH. Namely the action of MATH on objects is given by REF, and the action on morphisms is the natural one. It is easy to see that MATH is exact, MATH is isomorphic to MATH for any MATH, and the natural map MATH is a bijection for any MATH, MATH. Now one can deduce the statement of the proposition by induction on length of NAME series of an object of MATH.
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math/0005119
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Follows from definitions. Note that being a NAME functor MATH induces algebraic maps between varieties MATH, MATH, MATH (used in REF) for the quiver MATH and the corresponding varieties for the quiver MATH.
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math/0005119
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The proposition follows from REF and from the fact that MATH is full, faithful, exact, and has épaisse image.
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math/0005119
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An easy calculation using the definition of MATH. See also REF .
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math/0005119
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Follows from the definition.
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math/0005119
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REF follow from the definition of MATH, REF Let MATH be a sink. Then there is the following split exact sequence MATH for any MATH. Here MATH, where MATH . If MATH then MATH. The proof for MATH being a source is analogous. REF follows from REF follows from the Snake Lemma.
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math/0005119
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One has to check that the bracket is skew-symmetric and satisfies the NAME identity. Both statements follow from the definition of MATH.
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math/0005119
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Induction using the fact that for any MATH there exists MATH such that MATH CITE.
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math/0005119
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Note that MATH because MATH is simply laced. Now the proposition follows from REF, and REF. The action of MATH on the generators of MATH is given by MATH where the notation is as in REF.
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math/0005119
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Follows from the fact that MATH and from REF.
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math/0005119
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One has to check that the bracket is skew-symmetric and satisfies the NAME identity. Both statements follow (after lengthy, but straightforward calculations) from the definition of MATH. A more conceptual proof is to consider MATH as embedded into an algebra of vertex operators, associated with the quiver MATH (see CITE).
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math/0005119
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Induction on MATH, where MATH for a root MATH.
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math/0005119
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An element MATH of MATH is, by definition, a MATH matrix such that MATH. It follows that MATH for some set MATH.
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math/0005119
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REF follow from the fact that MATH is projective. REF follow from the fact that MATH is injective. Let us prove REF. We need to find the value of the function MATH on the isomorphism class of the representation MATH. We give below the calculation for MATH. The case of arbitrary MATH is completely analogous, and, moreover, the result does not depend on MATH. The value of MATH on the isomorphism class of MATH is equal, by the definition of the MATH-product, to the NAME characteristic of the variety MATH of all subrepresentations MATH of MATH such that MATH is isomorphic to MATH and MATH is isomorphic to MATH. Since MATH it follows that MATH coincides with the set of all lines MATH such that MATH is indecomposable, where MATH is the projection. The object MATH is indecomposable if and only if MATH. Therefore MATH, and we get REF is dual to REF. Let us prove REF. We need to find the value of the function MATH on the isomorphism class of the representation MATH. This value is equal, by the definition of the MATH-product, to the NAME characteristic of a variety MATH of all subrepresentations MATH of MATH such that MATH is isomorphic to MATH and MATH is isomorphic to MATH for some MATH. Similar to REF one can use the MATH map to prove that the variety MATH is a fibration over MATH with the fiber over MATH equal to the variety MATH of all subrepresentations MATH of MATH such that MATH is isomorphic to MATH and MATH is isomorphic to MATH. Let us consider the case MATH. Since MATH, where MATH and MATH are as in REF, and MATH, it follows that MATH. Therefore MATH is a point. The calculation for arbitrary MATH is completely analogous, and, moreover, the variety MATH does not depend on MATH. Thus MATH, which is REF is dual to REF.
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math/0005119
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We prove the proposition by induction. It follows from the definitions that MATH . Suppose that MATH . Then using REF we have MATH which gives MATH, and MATH which provides the induction step. One can avoid some of these calculations using an argument similar to the proof of REF to get MATH for MATH (note that the indecomposable object with dimension MATH can be obtained by repeated applications of reflection functors to a simple object). However, one would still need explicit calculations with the MATH-product to get the statement of the theorem in the case of an imaginary root.
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math/0005119
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Follows from REF .
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math/0005119
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An exercise in (graded) linear algebra (see, for example, CITE).
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math/0005119
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Follows from the fact that if MATH is a subobject of MATH then MATH is isomorphic to MATH for some MATH and MATH, and MATH is isomorphic to MATH.
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math/0005119
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Let us temporarily denote the map given by the formulas in the formulation of the proposition by MATH. It follows from REF that MATH is a homomorphism of NAME algebras, whose value on generators MATH of MATH coincides with the value of MATH. Thus MATH, which proves the proposition. Another possible proof is an induction on length MATH in MATH similar to the one used in the proof of REF .
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math/0005119
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Follows from REF .
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math/0005119
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The proposition follows from REF , and REF .
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math/0005119
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REF follow from REF follow from REF follows from the definition of root, REF, and REF follows from REF, and REF. Let us prove REF. Let MATH be an extending vertex (that is, MATH). It follows from the following equality MATH that either MATH or there exists MATH such that MATH, which proves REF.
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math/0005119
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The theorem follows from REF.
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math/0005119
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Given MATH the statement is true for one MATH, say MATH, because of REF . Any other MATH can be obtained from MATH by repeated applications of the NAME element (see REF). Then one can use the sequence of reflection functors corresponding to the sequence of reflections in the NAME element, and reason similarly to the proof of REF to get the first equality in the statement of the proposition for any MATH. The second equality in the statement of the proposition follows from the fact that MATH.
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math/0005119
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The proposition follows from REF, and REF.
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math/0005119
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Note that MATH is linearly independent with MATH because MATH whereas MATH. Now the proposition follows from REF.
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math/0005119
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The map MATH is surjective by construction. REF implies that MATH.
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