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math/9912021
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This follows from the expression of the NAME group action on elements in MATH given by: MATH. If this expression is applied to MATH it gives MATH. On the level of root characters this becomes, by exponentiation of the previous identity, MATH . Now recall that MATH is just MATH evaluated at MATH. Also MATH will be MATH evaluated at MATH. When we evaluate MATH on MATH in order to compute the corresponding j-th sign, we obtain MATH. Therefore the sign of MATH on the MATH is given by the product MATH. Finally we use the fact that the set of all scalars MATH determines MATH. Thus MATH determines the element MATH giving rise to the equation MATH.
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math/9912021
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REF follows from REF but with the observation that in, this case, MATH may fail to be in MATH even if MATH. This happens exactly when MATH and MATH with MATH odd (either MATH or MATH). Under these circumstances MATH (rather than one as required in the definition of MATH). We can fix this problem by factoring MATH as a product MATH where MATH. REF follows from REF and the fact that each MATH is a connected component of MATH in REF . REF follows easily from MATH. Also by the fact that the sign of a root character is constant along a connected component and that we have MATH and MATH is in the new component MATH. The two desired signs have thus been computed in the two connected components and they agree.
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math/9912021
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This follows from REF and the correspondence MATH giving a bijection between MATH and MATH.
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math/9912021
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The two sets MATH and MATH are clearly in bijective correspondence and an element MATH corresponds to the coset MATH. By REF , Notation REF , the actions on these two sets agree. They are both given by REF .
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math/9912021
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We first point out that the exponential map in the NAME groups MATH, MATH gives a diffeomorphism between MATH and the corresponding connected NAME group. Thus the NAME groups MATH, MATH are isomorphic. This takes care of REF on the level of the connected component of the identity. Also the image under MATH of the set MATH gives rise to a basis of the NAME subalgebra of MATH. Thus exponentiating we obtain REF . Recall that MATH and MATH if and only if MATH for all MATH. We have thus MATH for MATH and MATH. The elements MATH with MATH are in the center of MATH. This proves REF . For REF , we proceed by noting that by definition of MATH it must be generated by the NAME group MATH with NAME algebra MATH and the elements MATH with MATH (playing the role that the MATH play in the definition of MATH). This is the same as MATH.
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math/9912021
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Recall that MATH is a NAME group, REF and it thus has defining relations MATH and MATH where MATH is MATH depending on the number of lines joining MATH and MATH in the NAME diagram. The case MATH occurring when MATH and MATH are not connected in the NAME diagram. The only relevant cases then are when MATH, MATH are both colored and connected in the NAME diagram. The only relevant vertices in the NAME diagram are those connected with these two and which are uncolored. We are thus reduced to very few non-trivial possibilities: MATH, MATH, MATH, MATH, MATH; smaller rank cases being very easy cases. We verify only one of these cases , the others being almost identical with no additional difficulties. Consider the case of MATH where MATH, MATH are the two simple roots which are not "endpoints"in the NAME diagram and all the others are uncolored. Then MATH but MATH. For book-keeping purposes it is convenient to temporarily represent red as MATH and blue as MATH and simply follow what happens to these two roots and an orientation MATH. Hence all the information can be encoded in a triple MATH representing the colors of these two roots and the orientation MATH. Now we apply MATH to MATH: MATH .
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math/9912021
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The groups in REF are isomorphic by REF . We can write MATH where MATH is defined by MATH if and only if MATH for all MATH. Now the group MATH is the center of MATH. Intersecting with MATH we obtain the center of MATH. To find this intersection with MATH, we must find all MATH such that MATH is real for all MATH (see the proof of REF ). We find that MATH and that MATH is an integral linear combination of the MATH. As in the proof of REF the elements MATH which are in the center of MATH are those for which MATH. The center of MATH is then the group generated by MATH with MATH and MATH. Since MATH, we have that MATH divided by MATH is isomorphic to MATH. The groups MATH, MATH and those involved in REF or REF all differ by a subgroup of the MATH which is annihilated by MATH. From here and REF , the isomorphism betwen REF follows.
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math/9912021
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We set MATH for simplicity. By REF , we have that each MATH can be written as a union over MATH of sets of the form MATH and thus by REF of MATH, we conclude the statement.
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math/9912021
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We start with the last statement, that MATH is a bijection; we have that MATH is injective because the scalars MATH determine all the root characters MATH, MATH for MATH and these scalars determine MATH in the adjoint group REF . From REF it follows that we can regard MATH as MATH (see REF ). We prove surjectivity by proving first the statement concerning the image MATH . When all the sets MATH are considered then all of MATH will be seen to be in the image of MATH. First consider MATH with MATH and MATH. We now apply MATH. Since MATH we obtain, by exponentiating, MATH. The set MATH becomes the image of the map: MATH given by first defining a map that sends MATH with MATH for MATH. This map is modified so that whenever MATH then the MATH-th coordinate is replaced with MATH. We denote this modified map by MATH. The domain and the image of MATH therefore consists of the set MATH. Together all these MATH give rise to one single map MATH which is surjective.
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math/9912021
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This follows from REF and the fact that MATH for MATH.
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math/9912021
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This follows from REF but is better understood in REF . We omit details.
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math/9912021
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The manifold MATH is the finite union of the chambers as in REF . Since the MATH action on MATH is by continuous transformations, it then suffices to observe that the antidominant chamber is compact. The antidominant chamber MATH of REF can be seen to be compact by describing explicitly its image under MATH. This image is a "box"inside MATH, as can be seen in REF namely the set MATH. The space MATH is now the finite union of the MATH translates of this compact set. That the boundary operators of MATH agree with the boundary operators of MATH will follow from the fact that in the MATH coordinates the MATH is a "box"which is itself part of the boundary of a bigger "box" REF . We start with the set MATH and note that its boundary is combinatorially described by REF or REF . Note that MATH (with MATH) represents the open box. The faces are parametrized by coloring each of the MATH vertices MATH or MATH which then represent opposite faces in the boundary. The signs are just chosen so that MATH for MATH. This description may be best understood by working out REF . All the cells thus appear by taking the faces of a box MATH and then faces of faces etc. By the same process of coloring uncolored vertices MATH or MATH which give rise, each time, to a pair of opposite faces in a box. In each REF or REF correctly describe the process of taking the boundary of a box. Note that it is enough to study what happens when MATH and then consider the MATH translates. We now use REF to conclude that MATH computes integral homology. However each MATH-module appearing in MATH in a fixed degree, can easily be seen to be identical with the corresponding term in MATH. By REF and the agreement of the boundary operators, we obtain that MATH computes integral homology. The non-orientability follows if we use the second cell decomposition described in REF. The unique top cell then has a non-zero boundary (except in the case of MATH).
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math/9912021
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By REF , MATH is regular. Then as in REF, MATH must be conjugate, under an element in MATH, to an element MATH in MATH. Using REF MATH is connected. Since MATH is also a regular element it follows that MATH is a NAME subalgebra and MATH is a NAME subgroup of MATH. Using conjugation by an element in MATH we may conjugate this NAME subgroup if necesary to MATH (REF or REF). We can thus assume that MATH. We obtain that MATH is MATH conjugate to MATH (see Notation REF).
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math/9912021
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The map MATH is the restriction to the NAME group MATH of the diffeomorphism of complex analytic manifolds MATH. We obtain that MATH must be an injective map. We show surjectivity. If MATH then by surjectivity of MATH there is MATH such that MATH and MATH. Thus MATH with MATH and MATH. Therefore MATH. Since MATH we obtain that MATH. But our assumption is that MATH. Hence we have obtained that MATH. By the MATH of MATH we obtain that MATH. Therefore MATH and thus MATH. This proves MATH is a bijection. By REF , MATH is a submanifold of real dimension MATH of MATH. The diffeomorphism MATH restricts to the smooth non-singular map MATH. Since we have shown that MATH is a bijection, then it is a diffeomorphism.
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math/9912021
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First we point out that we already have a smooth manifold MATH and the assumption MATH will simply ensure that we can define the map to the flag manifold. We first show the continuity of MATH: We use the local coordinates, MATH for MATH given in REF . In these local coordinates, we assume that for each MATH, MATH (which equals MATH or zero) converges to a scalar MATH. We note that if MATH with each MATH a non-negative integer, then MATH . We thus let MATH . Let MATH . Then note that MATH if and only if MATH for some MATH. Thus the only MATH which are non-zero correspond to roots in MATH. By the assumption made, we have MATH which can be written in terms of the coordinate functions MATH as MATH . This then converges (by continuity of MATH) to MATH which only involves roots in MATH, and can be written as MATH. Since any MATH is completely determined by the coordinates MATH and some MATH uniquely corresponds to the coordinates MATH, then we can conclude that MATH converges to MATH whenever the argument MATH approaches to MATH in MATH. This proves the continuity of the map MATH. Since MATH is compact REF and MATH contains the orbit MATH , then MATH. From the construction of the map MATH it is easy to see that its image is contained in MATH and thus MATH. The smoothness of the map and that it gives a diffeomorphism follows from the fact that MATH constitutes a coordinate system in the flag manifold.
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math/9912021
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We use REF . The manifold MATH has a MATH action and the only fixed points are the MATH with MATH. These get mapped to the fixed points of the MATH action in MATH, the cosets MATH, MATH. Thus MATH. This also follows directly if if we consider REF with MATH and we just let each MATH go to zero and obtain MATH.
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math/9912021
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This is just REF and the definition of MATH. The two conditions in REF are satisfied by REF (Subsection REF) and by REF in p. REF as noted above in REF .
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math/9912042
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Thanks to CITE MATH is a projective MATH-module of rank MATH. By CITE the NAME group of projective modules over MATH is trivial, in other words MATH . In particular, if MATH is a projective MATH-module whose rank is greater than the NAME dimension of MATH, then MATH is necessarily free, CITE. Since MATH we have MATH, so the proposition follows.
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math/9912042
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Let MATH. In these circumstances CITE states that the NAME locus MATH of MATH coincides with the smooth locus of MATH if MATH is NAME in codimension one, that is the set of points of MATH which are not annihilators of simple MATH-modules of maximal dimension has codimension at least two, see CITE. Since MATH is the centre of a maximal order it is integrally closed, CITE. In particular MATH is smooth in codimension one. This means that the converse of CITE is also true: if MATH coincides with the smooth locus of MATH then MATH is necessarily NAME in codimension one. The map induced by inclusion MATH is surjective with finite fibres. The simple MATH-modules lying over MATH all have dimension MATH and the maximal dimension of a simple MATH-module is MATH. By CITE the variety MATH has dimension MATH, so MATH is stratified by pieces MATH of dimension MATH over which the representation theory is constant. Hence MATH is NAME in codimension one if and only if MATH for all MATH such that MATH. In other words we need only check that MATH for all MATH. By CITE, however, this is equivalent to the condition MATH for all MATH, in other words the involution MATH has no fixed points. This happens if and only if MATH is a NAME matrix of type MATH.
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math/9912042
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CASE: We overline to indicate images modulo MATH in a MATH-module. Now MATH and MATH, with MATH, a unit in MATH when MATH. Thus MATH and MATH, as claimed. CASE: That MATH acts transitively on the simple MATH-modules follows from CITE. Let MATH, and write MATH for MATH. We write MATH for a MATH-module MATH. Since MATH, there is a perfect pairing MATH . Suppose first that MATH. Since MATH via the perfect pairing, MATH. Let MATH and note that MATH is torsion-free since MATH implies that MATH. Hence MATH for some MATH. Thus MATH, where MATH and MATH. Let MATH be such that MATH. Since the integers MATH are prime to MATH (thanks to our continuing hypothesis on MATH given in Paragraph REF) this is equivalent to MATH. Write MATH with MATH. Then for MATH, so that MATH. Since the pairing is perfect, this forces MATH. Hence MATH operates simply transitively on the irreducible MATH-modules by REF . Now suppose that the order of MATH is prime to MATH. Let MATH. Working over MATH, we can argue as above to find a decomposition MATH with MATH-invariant. We claim that MATH. This follows from the observation that MATH, the first equality by MATH-invariance of the elements of MATH, the second by orthogonality and the MATH-invariance of MATH. Thus we have a factorisation MATH . Since the right hand side and the left hand side both have MATH elements this completes the proof of REF . REF is proved entirely similarly.
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math/9912042
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It is clear that MATH factorises as MATH where MATH. It is therefore enough to prove this for the case MATH. By REF we can consider MATH as an affine variety embedded in MATH (with co-ordinate functions MATH). Under this identification MATH takes MATH to MATH. Note that MATH. Indeed the equations MATH show that MATH as required. Define MATH as follows: MATH . Recall that for MATH we define MATH . Then, by definition, the tangent space of MATH at MATH is MATH . Since MATH is homogeneous of degree two it follows that for all MATH which implies that MATH. Now MATH is finite over MATH, so that MATH has dimension REF. Therefore MATH is a singular point.
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math/9912042
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Using REF it is enough to show that any maximal ideal of MATH lying over MATH is singular. Denote MATH by MATH, the fibre product MATH. Claim. Let MATH be the projection map. If MATH is singular then any point of MATH is singular in MATH. Proof of claim. For ease of notation let MATH, MATH and MATH. Thus we are considering MATH where: REF as a MATH-module MATH is finitely generated and free; REF the algebra MATH is smooth; REF all algebras are affine domains. Let MATH be the maximal ideal corresponding to MATH, and suppose MATH lies over MATH. Define MATH and MATH, maximal ideals of MATH and MATH respectively. We first show that MATH . Note that MATH. This means in particular that if MATH and MATH then MATH. So we have an embedding MATH . Let MATH. If MATH were local then MATH would be an invertible element, since MATH is maximal. Therefore the map REF would be an isomorphism, proving the claim. Thus it is enough to show that MATH is a local ring. Observe that by REF the algebra MATH is finite dimensional. Therefore localising this at MATH yields another finite dimensional algebra MATH. NAME 's lemma implies that MATH is finite over MATH. Let MATH. Then MATH is finite dimensional, so NAME 's lemma also implies that MATH is finite over MATH. Therefore MATH is contained in the NAME radical of MATH. However, since MATH is local it follows that MATH is local, as required. So we have proved REF . To complete the claim we must show that MATH has infinite global dimension. By REF has. We have a change of rings spectral sequence MATH for any MATH-module MATH. By NAME reciprocity we have MATH . As MATH is smooth there exists a natural number MATH such that MATH for all MATH and all MATH-modules MATH. If MATH were smooth there would be a natural number MATH such that MATH for all MATH and all MATH-modules MATH. Then by REF we have MATH for all MATH and MATH-modules MATH, contradicting the singularity of MATH. Thus MATH must be singular, proving the claim. The proposition now follows from REF combined with the claim.
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math/9912042
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Let's write MATH for MATH. The equivalence in the first sentence is a consequence of REF . By REF there is an algebra isomorphism MATH . We have isomorphisms MATH . Here MATH is, in the notation of REF , specified by MATH and MATH. Recalling our decomposition in REF , MATH we see that MATH is the tensor product of rings MATH, for MATH, where MATH . Let's describe the possible structure of the rings MATH. Since MATH is NAME it follows from REF that we never have MATH. There are only two cases to consider. CASE: MATH: in this case we have an algebra isomorphism MATH . Indeed sending MATH produces an isomorphism MATH . Since this is a semisimple algebra of dimension MATH, the isomorphisms in REF are clear. CASE: MATH (the case MATH is the same by symmetry): in this case we have an algebra isomorphism MATH . Again sending MATH yields the required isomorphism. To complete the theorem, recall that MATH has exactly MATH simple modules. But MATH has exactly MATH simple modules in REF and a unique simple module in REF . Therefore MATH .
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math/9912042
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For MATH this follows from REF . For MATH with MATH the only point in CITE where the bound MATH was required was to deduce that the algebra MATH is NAME, that is its projective indecomposable modules are uniserial. If we knew this to be so without the bound then the lemma would follow. Let MATH and MATH be unipotent and let MATH. By REF there is an isomorphism MATH . As MATH the algebra MATH is, by REF , a truncated polynomial ring in one variable. In particular it is a NAME algebra. By CITE a skew group extension over MATH of a NAME algebra is again NAME. Therefore MATH is a NAME algebra. By REF so by the first sentence of this proof MATH is a semisimple algebra, implying that the tensor product MATH is a direct product of matrix algebras with coefficients in MATH. Hence MATH is NAME equivalent to a NAME algebra and so must be a NAME algebra itself. This proves the lemma.
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math/9912042
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Suppose we have an inclusion of algebras MATH. Suppose further that MATH has a MATH-bimodule decomposition MATH. Then, by CITE, the representation type of MATH is a lower bound for the representation type of MATH (where finite is less than tame is less than wild). It's clear that MATH is a bimodule direct summand of MATH. The character group of MATH, say MATH, acts naturally on MATH by MATH and by CITE the algebras MATH and MATH are NAME equivalent (this uses the fact that MATH is invertible in MATH). Combining this with the previous paragraph yields the lemma.
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math/9912042
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The proof is based on the fact that the truncated polynomial algebra MATH has wild representation type if MATH. This is a consequence of CITE. In order that MATH we must have one of the following for MATH and MATH: CASE: MATH where MATH (and the symmetric case obtained by exchanging the roles of MATH and MATH); CASE: MATH where MATH; CASE: MATH. It follows from REF that REF are NAME so, by REF , MATH . Therefore MATH, and each of its blocks, has wild representation type. Suppose we are in REF . By REF we can assume without loss of generality that there is an isomorphism MATH where MATH for MATH and MATH unipotent. By REF , it is enough to show that MATH is wild. By REF the algebras MATH and MATH are isomorphic to direct sums of either MATH or MATH . Since MATH it therefore suffices to show that the following algebras are wild: MATH; CASE: MATH. It's clear, however, that the algebra in REF (respectively in REF ) is a skew group ring with coefficient ring MATH and group MATH (respectively MATH). Applying REF again and the comments in the first paragraph of this proof shows that these are indeed wild.
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math/9912042
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By REF MATH is a free MATH-module of rank MATH. Let MATH be a basis for this module and define MATH for MATH by the following equations, MATH . Then for any MATH the structure constants of MATH with respect to the basis MATH are given by MATH. As a result the map MATH defined by MATH, is a morphism of varieties. Let MATH and MATH. By REF MATH is a dense open set of MATH over which all algebras in the family MATH are isomorphic to MATH. It follows from CITE that MATH is a degeneration of MATH.
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math/9912042
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The first part follows directly from REF . Indeed, if MATH then the algebra MATH is semisimple whilst if MATH then MATH is NAME equivalent to a direct sum of truncated polynomial algebras in one variable, hence NAME. Let MATH denote the NAME order on MATH. For the second part note that if MATH then there exists MATH such that MATH, MATH and MATH. Arguing exactly as in CITE it follows that MATH so MATH. Let MATH. Then by REF MATH is wild and by REF MATH is a degeneration of MATH. Therefore, by REF , MATH must be wild.
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math/9912042
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The first part follows from REF and the observation that MATH has finite representation type by REF . For REF , arguing as in the proof of REF we see that it is sufficient to show that MATH is wild in the case that MATH, that is MATH for MATH. Let MATH be such that MATH where MATH and MATH are unipotent. By REF we have an algebra isomorphism MATH . Since MATH we have, by REF , an isomorphism MATH . Therefore MATH . So MATH has the same representation type as MATH and hence, by REF , as MATH. Now apply REF .
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math/9912042
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The second claim is obvious. For the first we can assume that MATH is concentrated in the MATH position. Then MATH where the right hand side is concentrated in the MATH position. On the other hand, since MATH is non-zero only in the MATH position, we have MATH as required.
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math/9912042
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Let MATH send MATH to MATH. We must check this is well-defined. First note that, for MATH. Thus we have, for MATH and MATH, MATH . Moreover MATH is a MATH-module isomorphism, since, for MATH, MATH . Since MATH is natural in MATH it follows that MATH is naturally isomorphic to the identity functor. One shows similarly that MATH is also naturally isomorphic to the identity functor.
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math/9912042
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The discussion above shows that MATH is a skew group ring MATH, and that we may reduce to the case where MATH is a basic algebra. By CITE MATH. By Lying Over for MATH, CITE, we deduce that MATH is a basic algebra. Thus MATH is a finite direct sum of copies of MATH. Commutativity of MATH forces the action of MATH on MATH to be trivial. Since there are exactly MATH simple MATH-modules it follows that MATH is (scalar) local.
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math/9912042
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CASE: Since MATH by CITE, MATH, so that MATH is basic and has quiver with vertices labelled by MATH. By definition (see for example . CITE), to find the arrows of the quiver of MATH we may assume without loss that MATH. The orthogonality relations for MATH show that the primitive idempotents of MATH (and hence of MATH) are MATH. Taking a basis for MATH consisting of eigenvectors MATH with respect to its structure REF as MATH-module, and noting that MATH is non-zero if and only if MATH, one finds MATH is non-zero if and only if MATH, proving REF . CASE: It's clear from REF that two vertices MATH and MATH are in the same connected component of the quiver if and only if MATH and MATH are in the same coset of MATH in MATH. The final equality is obvious, since MATH. CASE: Immediate from REF . CASE: By REF . CASE: By definition of MATH. CASE: It's easy to show that any choice of lifts to MATH of a basis of MATH generate MATH as a MATH-module - see for example CITE. Thus, if REF holds then MATH operates unipotently on MATH and hence on MATH. The assumption on the characteristic of MATH ensures that the action of MATH on MATH is completely reducible. So REF follows. CASE: Trivial.
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math/9912042
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CASE: By REF the winding automorphisms afforded by MATH permute the irreducible MATH-modules transitively. Thanks to the definition of the coproduct and of the winding automorphisms (see REF ), the latter act trivially on MATH. Hence, if a simple MATH-module does not feature in a component of any one simple MATH-module, then that simple MATH-module doesn't feature in any of the simple MATH-modules. But this would imply that MATH, contradicting REF . So each simple MATH -module is faithful for MATH. CASE: Suppose MATH were a non-zero proper MATH-invariant ideal of MATH. Then, thanks to REF , MATH would be a proper (two-sided) ideal of MATH , and MATH would then annihilate a simple MATH-module MATH. Thus MATH contradicting REF .
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math/9912042
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See CITE.
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math/9912042
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It is clear that MATH since by construction MATH is a combination of the simple roots MATH for MATH. We prove the opposite inclusion by induction on MATH, the case MATH being trivial. Let MATH so that MATH, and let MATH be the set of positive roots corresponding to MATH. We have MATH and MATH for MATH and MATH. Therefore MATH and so MATH. By induction MATH so we have MATH as required.
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math/9912042
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Only REF and the final sentence of REF are not immediate from REF . By REF , MATH where the sum is taken over the set MATH of all MATH such that MATH does not appear in a reduced expression of MATH. Thus each such MATH is fixed by MATH and we have MATH. The same argument shows that MATH. That MATH is clear from REF . Finally, note that for MATH, the corresponding unit MATH of MATH is just a scalar, since MATH commutes with MATH. Since these scalars have trivial action on MATH, the final sentence of REF follows.
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math/9912042
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Recall that MATH is the minimal length of MATH when written as a product of arbitrary reflections. Suppose that MATH. Then MATH so MATH. Conversely suppose that MATH for all MATH. We prove that MATH by induction on MATH, the case of MATH being clear. Let MATH so that MATH. Suppose that MATH. Since a reduced expression of MATH does not contain the simple reflection MATH we deduce that if MATH then MATH. As MATH we have MATH . As a result MATH, implying that MATH. Thus MATH and so we are in the case where MATH, (see CITE), so that MATH. This proves the lemma.
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math/9912042
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By REF , under the hypothesis of the corollary MATH has cardinality MATH. By REF , MATH, proving the corollary, in the light of REF .
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math/9912042
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CASE: By CITE the set MATH is closed in MATH. Thus, by REF the algebra MATH has no more blocks than MATH. CASE: For MATH in the non-empty open set MATH of MATH, MATH is semisimple with MATH simple modules, by REF . The first part of the claim now follows from REF . CASE: MATH . Since MATH is integrally closed, MATH if and only if these algebras have distinct quotient fields, and this happens if and only if, for a generic maximal ideal MATH of MATH, MATH. Since the generic maximal ideal of MATH is contained in a maximal ideal of MATH unramified over MATH, we can conclude that MATH if and only if there is a maximal ideal of MATH contained in at least two maximal ideals of MATH. So the equivalence now follows from REF . MATH: By REF . MATH: This can be read off from CITE.
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math/9912042
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We freely use the notation of the proof of REF . By REF it is sufficient to show that MATH has exactly MATH maximal ideals. By the proof of REF MATH where MATH is as in REF . Let MATH and MATH. We have already seen in REF that if MATH then MATH is isomorphic to MATH, whilst if MATH and MATH then MATH is isomorphic to MATH by REF . It remains to consider the case MATH. In this case inclusion provides an isomorphism MATH where MATH is the ideal generated by MATH for all MATH and MATH. Since this is manifestly a local ring, the result follows from these calculations and REF .
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math/9912042
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Let's write MATH for the regular function MATH, and similarly MATH for MATH. Let MATH be an arbitrary element of MATH. We have MATH for MATH since MATH is a highest weight vector. Hence MATH is not identically zero if and only if MATH is a lowest weight vector, and in this case it is non-zero for all MATH. Note that this happens if and only if MATH. A similar analysis applied to MATH shows MATH for some MATH if and only if MATH for all MATH if and only if MATH is a highest weight vector. That is, if and only if MATH. We conclude from the previous paragraph that if MATH then MATH and MATH if and only if MATH. The lemma now follows from REF .
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math/9912042
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By REF different blocks arise from the different maximal ideals of MATH lying over MATH, so we need to see how these are permuted by the right winding automorphisms. By REF such maximal ideals are determined by the central elements MATH for MATH. Hence we need only study the action of the winding automorphisms on MATH and MATH. By REF where both MATH and MATH are highest weight vectors. Therefore MATH where MATH and MATH are dual bases of MATH and MATH respectively. Let MATH. We have MATH . Therefore, letting MATH denote the right winding automorphism of MATH defined analogously to those for MATH in REF, MATH. Similarly, we find that MATH. We have shown that MATH . As in the proof of REF the maximal ideals of MATH lying over MATH are obtained by piecing together the maximal ideals of the algebras MATH for MATH (notation of the proof of REF ). These maximal ideals in turn depend on the vanishing behaviour of MATH and MATH. Specifically if MATH or MATH then MATH has a unique maximal ideal whilst if MATH then MATH is semisimple with exactly MATH maximal ideals. In the second case it follows from the previous paragraph that the winding automorphisms permute the primitive idempotents of MATH non-trivially unless MATH. Hence the same is true of the maximal ideals. We deduce that the normaliser in MATH of a block of MATH is simply the subgroup MATH . By the proof of REF this equals the subgroup MATH . The theorem follows from REF .
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math/9912043
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Using REF , the set of isomorphism classes of holomorphic principal MATH-bundles is equal to the set of isomorphism classes of the category MATH. Two isomorphic extensions as in REF give isomorphic pairs MATH and MATH, because REF implies that the image of MATH in MATH under the map induced by MATH is MATH. Conversely, two isomorphic pairs give two extensions (unique up to noncanonical isomorphism) that are isomorphic in the category MATH.
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math/9912043
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MATH . This follows from holonomy and REF . MATH . Take and object in MATH, that is, an extension MATH with a flat connection MATH preserving MATH. Take a MATH splitting MATH. This is given by a smooth morphism MATH (that is, a smooth section of MATH) with MATH. The fact that MATH preserves MATH and that it induces the trivial connection MATH on MATH means that, using this splitting, MATH can be written as MATH where MATH is the connection induced on MATH and MATH is a smooth REF-form with values in MATH. Note that MATH depends on the splitting (that is, the smooth section MATH chosen) but MATH doesn't. Flatness of MATH translates into MATH . Since by REF is semisimple, as we have already explained it has a harmonic metric and then there is a polystable NAME bundle MATH (this is independent of the chosen section MATH, since MATH was). If we take a different splitting, that is, change the section MATH to a new smooth section MATH of MATH (again with MATH), the diffeomorphism between the old and the new splitting is given by a matrix of the form MATH . Note that the image of MATH is in MATH (because MATH), and then this matrix makes sense. In the new splitting the connection MATH has the form MATH . Then REF-form associated to the new splitting is MATH. In other words, we obtain an element of MATH (independent of the splitting). To show that this construction gives a bijection, we now construct an inverse. Take a triple MATH where MATH is a polystable NAME bundle and MATH. Since the NAME bundle is polystable, there is a harmonic metric that gives a flat connection MATH. Let MATH be a representative of MATH and define on MATH a connection MATH . MATH . Let MATH be an object of MATH. Let MATH be the element corresponding to MATH under the natural isomorphism. Let MATH be a MATH-closed smooth REF-form representing MATH. Then MATH . Consider the vector bundle MATH. Using this splitting, let MATH . REF imply that MATH, and then this defines a structure of holomorphic vector bundle MATH. Let MATH . Then MATH and MATH again by REF , and then MATH is a NAME bundle. It is easy to check that it is semistable (but not polystable if MATH). Conversely, given an object of MATH, let MATH be the underlying smooth vector bundle of MATH, and let MATH be the corresponding MATH-operator. It can be written as MATH with MATH a smooth MATH -form. The fact that MATH is MATH-invariant and that it induces the zero NAME bundle on MATH imply that MATH can be written as MATH with MATH a NAME field on MATH, and MATH a smooth MATH -form. Let MATH. Since MATH is a NAME bundle, we have MATH, MATH and MATH, and this implies that MATH, where MATH. Then MATH defines a class MATH, and under the natural isomorphism we obtain an element MATH.
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math/9912043
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Since MATH, MATH and then MATH .
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math/9912043
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MATH . This is given by holonomy. MATH . Follows from CITE (see the remarks after NAME 's proof). MATH . Note that since the flat filtration induces semisimple flat connections on the quotients MATH, we get polystable NAME bundles MATH, and as we have already discussed, the flat connection MATH gives a well defined element of MATH. It is easy to check that a weakly isomorphic flat filtration gives isomorphic NAME bundles and element MATH. Conversely, given NAME bundles MATH and MATH, define the filtration MATH . Take a representative MATH of MATH, define the connection MATH and this defines a flat filtration.
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math/9912048
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Let MATH be a stable set of MATH such that MATH, and MATH. If MATH, then MATH is adjacent to some MATH, otherwise MATH is a stable set larger than MATH, which contradicts the maximality of MATH. Hence, MATH, and MATH. Therefore, using this exchange procedure, after a finite number of steps, we have to obtain a maximum stable set including MATH.
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math/9912048
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Suppose, on the contrary, that MATH. Hence, any MATH has MATH. Since MATH is a tree and MATH is a stable set of size MATH, it yields the following contradiction: MATH . Consequently, we infer that MATH is not empty. We can assert now that there exists some MATH, such that MATH. CASE: There exist two vertices from MATH at distance two apart. CASE: Any two vertices of MATH are at distance at least three. Let us denote MATH. According to the hypothesis of the case, all the vertices MATH are different. Assume, on the contrary, that for any MATH and any MATH, the distance between them is greater than two. Hence, any MATH has MATH. Let MATH, be the connected components of the subgraph MATH, of orders MATH, respectively. Since, by our assumption, no MATH is connected to any MATH, we infer that for every MATH there exists an edge joining this component to a vertex from MATH. Hence, it yields the following contradiction: MATH . Consequently, there must exist some MATH and MATH such that the distance between them equals two.
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math/9912048
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Let MATH be a bipartition of MATH. Both MATH and MATH are maximum stable sets, because MATH are stable and MATH. Hence, REF implies that both MATH-and-MATH, that supports the conclusion.
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math/9912048
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CASE: If MATH, then REF implies MATH. The equivalence REF is clear. CASE: Without loss of generality, we may suppose that MATH. Since MATH is also a stable set, REF ensures that MATH, and, consequently, MATH. Let MATH and assume that MATH. Then MATH and, hence, MATH. Since no vertex in MATH is adjacent to any vertex in MATH, it follows that at least one tree, say MATH, of the forest MATH, has MATH. Consequently, by REF , it proves that MATH has at least one pendant vertex, say MATH, in MATH. Since MATH, we infer that some pendant vertex of MATH must be in MATH, in contradiction with MATH. Therefore, every maximum stable set of MATH is a subset of MATH. Since MATH is stable, it follows that, in fact, MATH, that is, MATH is a strong unique independent tree.
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math/9912048
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We prove that if a stable set MATH of MATH satisfies MATH, then MATH. If MATH and MATH, then MATH is the bipartition of the forest MATH, while MATH is the bipartition of the forest MATH. Since MATH and MATH, it follows that MATH. Consequently, by REF , every connected component of MATH or MATH, which is different from an isolated vertex, is a strong unique independent tree. Moreover, every isolated vertex of MATH belongs to MATH, and every isolated vertex of MATH belongs to MATH. Therefore, both MATH and MATH. Hence, we get that MATH which completes the proof of the first assertion. Now, let MATH be a stable set such that MATH and let MATH, MATH. Any MATH has MATH and MATH. Since MATH we see that MATH. Therefore, it follows that some tree MATH of the forest MATH must have MATH, where MATH. According to REF , it follows that MATH, which implies that MATH, because MATH. In addition, if MATH, then there is MATH and the distance between MATH equals MATH, while if MATH, then, according to REF , there exist MATH and MATH, such that the distance between them is MATH. In both cases, we may conclude that there are MATH and MATH such that the distance between them equals two.
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math/9912048
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CASE: Let MATH. Then MATH is stable in MATH, and, therefore, it follows that MATH, for each MATH. Hence, we get that MATH . CASE: There are MATH, such that MATH. Then MATH is stable in MATH and MATH and this implies that MATH. CASE: There are MATH, such that MATH, that is, MATH, (or MATH). Hence, MATH is stable in MATH, and MATH and this leads to the following contradiction with the hypothesis MATH: MATH and MATH. Thus, we may conclude that MATH. CASE: If MATH, then REF explicitly means that MATH. Conversely, let MATH, and assume that there is MATH, such that MATH. Then, the set MATH is stable in MATH and MATH. Clearly, MATH is stable in MATH, and because MATH, we have that MATH. Hence, MATH and this contradicts the choice MATH. CASE: According to REF , we have MATH, and REF ensures that MATH. Let MATH and MATH. Then MATH, and, therefore, MATH. Since MATH is an arbitrary set from MATH, we get that MATH . Similarly, one can show that if MATH, then MATH. Therefore, we may conclude that MATH . Conversely, let MATH, and suppose there is MATH, such that MATH. Let us denote MATH, for MATH. Since MATH, it follows that MATH. Hence, we get a contradiction: MATH . Consequently, MATH is also valid, and this completes the proof.
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math/9912048
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According to REF , we infer that MATH. Since MATH is a tree, it follows that MATH. To prove the theorem we use induction on MATH. The result is clearly true for MATH. Let MATH be a tree with MATH, and suppose that the assertion is valid for any tree with fewer vertices. If MATH, the result is clear. If MATH, let MATH and MATH be two trees such that MATH. A partition of MATH in two non-empty sets gives rise to a corresponding division of MATH into MATH and MATH. According to REF , MATH. Hence, REF implies that MATH. By the induction hypothesis, each MATH has at least two pendant vertices belonging to MATH. REF ensures that MATH, and, therefore, MATH itself has at least two pendant vertices in MATH.
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math/9912048
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Let us partition MATH into MATH subsets MATH, each one having at least two vertices. Then we can write MATH where MATH is the subtree of MATH containing MATH as the neighborhood of MATH. Since MATH is pendant in no MATH, we get MATH. By REF , it follows that MATH. According to REF , the vertex MATH, and, consequently, REF implies that: MATH and this completes the proof.
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math/9912048
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Let MATH be the bipartition of MATH. Notice that the distance between two vertices is even if and only if both of them belong to one set of the bipartition. To prove the theorem we use induction on MATH. If MATH, then MATH and the assertion is true. Let now MATH be a tree with MATH vertices. According to REF , MATH yields MATH. CASE: MATH. Then we get MATH . Hence, at least two vertices of MATH belong to one set of the bipartition, that is, the distance between them is even. CASE: MATH. REF shows that such trees exist. If MATH, then the distance between them is two, which is both even and different from four. Suppose now that MATH, and let MATH. Since MATH and MATH belong to all maximum stable sets of MATH, we conclude that neither MATH nor MATH are contained in any maximum stable set of MATH. Hence MATH. Consequently, MATH and MATH. Suppose that MATH consists of MATH trees MATH. Since MATH, at least one tree, say MATH, has MATH. By the induction hypothesis, there exist two distinct vertices MATH such that the distance between them in MATH is even. We claim that MATH. Otherwise, if, for instance, MATH, then MATH and this contradicts the fact that MATH. Further, if MATH or MATH, then MATH is not a tree, since MATH or MATH, respectively, builds a new path connecting MATH and MATH in addition to the unique path between MATH and MATH in MATH (together the two paths create a cycle, which is forbidden in trees). Suppose that MATH and MATH. Then MATH, because, otherwise, MATH can not be a tree. No edge from the set MATH exists in MATH, since the vertices MATH and MATH are pendant in MATH. One can find an example of such a situation in REF . The vertices MATH and MATH are not adjacent in MATH, because they are pendant in MATH. Therefore, MATH, and the shortest path between MATH and MATH goes through the vertices MATH, at least. Thus, the distance between MATH in MATH is greater than the distance between MATH and MATH in MATH by four, and consequently, it is even and, moreover, different from four, because the vertices MATH and MATH are distinct.
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math/9912051
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REF is CITE. MATH follows from the fact that a very ample divisor plus a divisor generated by its sections is a very ample divisor. Given any divisor MATH and a very ample divisor MATH, choose MATH such that MATH is generated by global sections for MATH. Then MATH is very ample for MATH. REF is trivial. CASE: For any ample divisor MATH and any MATH, one can choose MATH so that for MATH, we have MATH is an ample divisor. Then REF follows immediately from REF .
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math/9912051
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Let MATH be right MATH-ample. Then for any coherent sheaf MATH, there exists a MATH such that MATH for MATH and MATH. Since cohomology is preserved under automorphisms, pulling back by MATH, we have MATH for MATH and MATH. So MATH is left MATH-ample.
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math/9912051
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The first claim is CITE. For the second claim, the property of MATH not being quasi-unipotent is reduced to saying MATH has an eigenvalue of absolute value not MATH. Since MATH has determinant MATH, MATH has an eigenvalue of absolute value greater than MATH.
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math/9912051
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Let MATH be the cone generated by numerically effective divisors in MATH. In the terminology of CITE, MATH is a solid cone since it has a non-empty interior CITE. Since MATH maps MATH to MATH, the spectral radius MATH is an eigenvalue of MATH and MATH has an eigenvector MATH CITE. Since MATH, there exists a curve MATH with MATH. Given an ample divisor MATH, there is a positive MATH so that MATH is in the ample cone CITE. Thus MATH . Taking MATH, we have the lemma.
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math/9912051
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Suppose that MATH is a right MATH-ample divisor. Let MATH. By REF , we may replace MATH with MATH and MATH with MATH and assume that MATH is ample. Let MATH be the action of MATH on MATH. Suppose MATH is non-quasi-unipotent with spectral radius MATH and choose an integral curve MATH as in REF . Let MATH be the ideal sheaf defining MATH in MATH. Since MATH is right MATH-ample, the higher cohomologies of MATH and MATH vanish for MATH. So one has an exact sequence MATH . For MATH, the NAME formula for curves CITE gives MATH . Thus using the exact sequence and the previous lemma, there exists MATH so that MATH for MATH. Thus the associated twisted homogenous coordinate ring has exponential growth and hence is not (right or left) noetherian CITE. So MATH cannot be right MATH-ample, by CITE.
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math/9912051
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Again choose a curve MATH as in REF with ideal sheaf MATH. By REF , there exists MATH such that for all MATH and MATH, MATH for any ample divisor MATH. In particular, the above cohomologies vanishes for MATH where MATH. Then repeating the last paragraph of the proof of REF shows that MATH has exponential growth.
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math/9912051
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CITE constructs a family of MATH surfaces whose general member MATH has MATH . (That is, MATH is the free product of two cyclic groups of order REF.) The ample generators MATH and MATH of MATH have intersection numbers MATH . MATH has two generators MATH whose actions on MATH can be represented as two quasi-unipotent matrices MATH . However, the action of MATH has eigenvalues MATH. So MATH has no MATH-ample divisor. Note that by REF below, any ample divisor is MATH-ample and MATH-ample.
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math/9912051
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Since MATH, there exists a divisor MATH and curve MATH such that MATH. Choose MATH so that MATH is ample. By REF , the intersection numbers MATH are given by a polynomial in MATH with leading coefficient MATH. Since this polynomial must have positive values for all MATH, we must have MATH.
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math/9912051
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Choose MATH such that MATH is ample. Let MATH . Then MATH is ample and thus MATH is ample.
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math/9912051
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Let MATH be the span of MATH. MATH is a MATH-dimensional vector space by REF . By REF , it contains the real cone MATH generated by MATH. Using REF, any element of MATH can be written as a linear combination of MATH elements of MATH with non-negative real coefficients. Thus for all MATH, MATH where MATH and MATH. Expanding the MATH above and comparing the coefficient of MATH with REF , one finds that MATH . Since MATH, for each MATH, there must be some MATH such that MATH. Now choose MATH such that MATH is ample and MATH such that, MATH. Then MATH is in the ample cone for the given MATH. Set MATH. The other MATH are non-negative. Then MATH is in the ample cone as it is a sum of elements in the ample cone. Since it is a divisor, it is ample CITE. We now prove the lemma by induction on MATH, the smallest positive integer such that MATH. Since MATH is nilpotent, there is such a MATH for any MATH. The case MATH is handled by the previous lemma. Now as MATH, we know MATH is killed by MATH. So there is a MATH so that MATH is ample. Then as MATH fixes the ample cone MATH is ample.
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math/9912051
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Let MATH be right MATH-ample. By REF , MATH is quasi-unipotent and MATH is ample for some MATH. Then MATH is quasi-unipotent and MATH is ample. Applying the theorem again, we have that MATH is right MATH-ample. Thus MATH is left MATH-ample by REF . The same lemma gives the second statement of the corollary.
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math/9912051
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In the first case, for MATH to be ample or minus-ample, it must be a NAME divisor. Thus the intersection numbers MATH are defined, where MATH is a curve. Since MATH must be fixed by MATH, some power of MATH must be numerically equivalent to the identity by REF . In the second case, the action of MATH itself must be numerically equivalent to the identity.
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math/9912051
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Take MATH such that MATH is ample and MATH is respectively ample, generated by global sections, or numerically effective. Then MATH is ample and we again apply the main theorem.
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math/9912051
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As MATH and ampleness depends only on the numerical equivalence class of a divisor, the corollary follows from our main theorem.
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math/9912051
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Since MATH is unipotent and intersection numbers only depend on numerical equivalence classes, we may replace MATH by the divisor on the right hand side of REF . As noted below that equation, it is not a problem to treat the MATH as divisors. Since the intersection form is multilinear and integer valued on divisors, MATH must be a numerical polynomial. By the NAME criterion for ampleness CITE, the function is positive for all positive MATH (since MATH is ample) and hence has a positive leading coefficient. Thus REF is proven. Now for some fixed MATH, we know that MATH is ample. Hence MATH for all MATH. Thus MATH and by symmetry the two degrees are equal. We can continue this argument, replacing each MATH with MATH, so MATH. By also noting that MATH is ample, one can similarly replace each MATH with MATH. Thus the second claim is proven. Now let MATH be a closed subvariety with MATH. One has MATH by CITE. We claim that for some fixed MATH, the intersection number of MATH with any collection of MATH ample divisors is positive. This is well-known if MATH is normal so MATH is a NAME divisor, so for some MATH, the NAME divisor MATH is effective CITE. The general case can be seen by pulling back to the normalization of MATH. Since normalization is a finite, birational morphism, ampleness CITE and intersection numbers CITE are both preserved under pull-back. Thus the claim is proven. An argument similar to the proof of REF proves the third claim of the lemma. For REF shows that the leading coefficient of MATH is a sum of terms MATH where MATH is an integer. So any leading coefficient times MATH is a positive integer. Thus given any set of ample divisors MATH, there is a MATH in that set such that MATH has the smallest leading coefficient. Now let MATH be a natural number such that MATH has the smallest leading coefficient of all MATH. Then for any MATH, MATH is a rational function with limit, as MATH, greater than or equal to MATH. So given any natural number MATH, MATH . Since this is true for any MATH, the limit must be MATH. So MATH . Examining the proof of REF , we see the right hand side equals MATH, proving REF . Finally, for REF , find a chain of subvarieties MATH. Then REF combined with REF proves the claim for each part of the chain.
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math/9912051
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If the left hand side is larger than the right hand side, then by the discussion before the lemma, there is a subvariety MATH with MATH where MATH is an irreducible component properly containing MATH. This cannot happen by REF . On the other hand, if the right hand side is larger, then there exists a subvariety MATH with MATH in the notation of REF and MATH . The earlier discussion shows that MATH for each MATH. Hence MATH is properly contained in some irreducible component. But again this cannot happen by REF .
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math/9912051
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By REF we may pick an irreducible component MATH of MATH with MATH . Then MATH has an irreducible component MATH with MATH. Combining REF , the claim is proven.
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math/9912051
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The first claim is obvious since the intersection numbers in REF are numerical polynomials, as noted in REF . The independence of the degree comes from REF . If MATH is numerically equivalent to the identity, then MATH. So MATH has degree MATH.
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math/9912051
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Given an ample divisor MATH, one has MATH by REF . So there exists a curve MATH such that MATH. Since MATH for all MATH and in particular for MATH, MATH. However, if MATH is odd, REF implies that the leading term of MATH is MATH where MATH. Then MATH for large MATH, which cannot occur. For the lower bound, note MATH. Constructing a chain of subvarieties between MATH and MATH, REF shows that MATH.
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math/9912051
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If MATH the lemma is trivial. So assume that MATH. Let MATH. Expanding MATH gives terms of the form MATH where MATH and MATH. We will show that if MATH then MATH. Order MATH in the following way: MATH if the right-most non-zero entry of MATH is positive. We proceed by descending induction on this ordering. The largest MATH-tuple in this ordering is MATH. Since MATH, MATH exists (taking MATH) so MATH and hence MATH. Now suppose MATH is such that MATH and we have proven our claim for all larger MATH. Since MATH, we have MATH so examine MATH . A typical term in the right-hand side is of the form MATH where MATH. The terms with MATH where MATH are all higher in the ordering than MATH and hence are zero. This only leaves MATH and so MATH.
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math/9912053
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There is a standard bijection between lozenge tilings and families of nonintersecting lattice paths. This bijection is explained in REF (see in particular REF ). Thus, the problem of enumerating lozenge tilings is converted to the problem of counting certain families of nonintersecting lattice paths. By the NAME - NAME - NAME theorem (stated in REF ), the number of such families of paths can be expressed as a determinant (see REF ). Thus, in principle, we would be done once we evaluate this determinant, given in REF. However, REF applies only if the size MATH of the core is even. We show, in REF, that it suffices to address this case, by proving that the number of lozenge tilings that we are interested in is a polynomial in MATH. The evaluation of the determinant REF for even MATH is carried out in REF (see REF ).
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math/9912053
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The first steps are identical with those in the preceding proof: the lozenge tilings are converted into nonintersecting lattice paths, in the way that is described in REF. Therefore, REF yields a determinant for the MATH-enumeration that we are interested in. Unlike in the previous proof, this provides a determinant for our weighted count only if the size MATH of the core is odd (see REF ). Again, the considerations in REF show that this number is a polynomial in MATH, so it suffices to evaluate the determinant REF for odd MATH. This is done in REF (see REF ).
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math/9912053
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Again, we use the strategy from the proof of REF . We convert the lozenge tilings into families of nonintersecting lattice paths as described in REF. The starting and ending points are slightly different from the ones used before. They are given in REF yields a determinant for the number we are interested in for even MATH (see REF ). The considerations of REF still apply, so the number of lozenge tilings is a polynomial in MATH and it suffices to evaluate the determinant REF for even MATH. This is accomplished in REF (see REF ).
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math/9912053
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We proceed analogously to the proof of REF . The lozenge tilings are converted into nonintersecting lattice paths, in the way that is described in REF. Therefore, REF yields a determinant for the MATH-enumeration in the case of odd MATH (see REF ). Again, the considerations in REF show that this number is a polynomial in MATH, so that it suffices to evaluate the determinant REF for odd MATH. This is worked out in REF (see REF ).
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math/9912053
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We follow the arguments of the proof of REF , as given in CITE. Suppose we are given a cyclically symmetric lozenge tiling MATH of our cored hexagon MATH. It is completely determined by its restriction to a fundamental region, the lower-left fundamental region, say. (In the example in REF , the lower-left fundamental region is framed.) Some of the lozenges are cut in two by the borders of the fundamental region. (In REF these are the shaded lozenges.) We draw lattice paths which connect these ``cut" lozenges, by ``following" along the other lozenges, as is indicated in REF by the dashed lines. To be precise, in each lozenge in the interior of the fundamental region, we connect the midpoints of the sides that run up-diagonal, in case the lozenge possesses such sides. Clearly, these paths are nonintersecting, by which we mean that no two paths have a common vertex. Since they determine completely the cyclically symmetric lozenge tiling, we may as well count all these families of nonintersecting lattice paths, with respect to the corresponding weight. In fact, as is easy to see, because of the cyclic symmetry, the statistic MATH is exactly equal to MATH minus the number of paths. If we fix the ``cut" lozenges, say in positions MATH (counted from inside out, beginning with MATH; thus, in REF , the ``cut" lozenges have positions MATH and MATH), then, according to REF , the number of families of nonintersecting lattice paths connecting the fixed ``cut" lozenges is given by the corresponding NAME - NAME - NAME determinant (the left-hand side of REF). This determinant turns out to be the minor of MATH consisting of rows and columns with indices MATH. This number must be multiplied by the common weight MATH of these families of nonintersecting lattice paths. Therefore, in order to obtain the total weighted count that we are interested in, we have to sum all these quantities, that is, take the sum of MATH over all MATH and MATH. Clearly, this sum is exactly equal to MATH, which equals the left-hand side of REF if MATH, the left-hand side of REF if MATH is a primitive third root of unity, and the left-hand side of REF if MATH a primitive sixth root of unity. The respective right-hand sides provide therefore the solution to our enumeration problem.
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math/9912053
|
We adapt the arguments used in the proof of REF . (Clearly, here we want to count the same objects, but with respect to a different weight.) So, again, we draw paths that connect the lozenges which are cut in two by the borders of the fundamental region. This time, we choose the top-right region as the fundamental region. REF shows an example. There, the top-right fundamental region is framed. As in REF , paths are indicated by dashed lines. (In the example in REF there is just one path.) If we slightly distort the underlying lattice, we get orthogonal paths with positive horizontal and negative vertical steps. REF shows the orthogonal path corresponding to the path in REF . The manner in which we have chosen the coordinate system ensures that possible starting points of paths are the points MATH, MATH, and possible ending points are the points MATH, MATH. Now, as before, we fix the positions of the ``cut" lozenges. Then a weighted version of the NAME - NAME - NAME theorem (see CITE or CITE) can be used to express the weighted count of the corresponding families of nonintersecting lattice paths in form of a determinant. In fact, this weighted version just says that REF remains true when the number MATH of paths from MATH to MATH is replaced everywhere by the weighted count MATH of all paths MATH from MATH to MATH, where MATH is some weight function on the edges of the square lattice and the weight MATH of a path is the product of the weights of its steps. Thus, if we repeat the subsequent arguments in the proof of REF , then we obtain the determinant MATH for the weighted count of our families of nonintersecting lattice paths. We now choose the weight function MATH so that the weight of the family of nonintersecting lattice paths corresponding to a tiling MATH is equal to MATH. To do this, it will be convenient to stick on an extra initial horizontal step at the beginning of each path, so that now it starts on the line MATH. Weight the vertical steps on this line by REF, all the remaining vertical steps by REF, and weight horizontal steps at height MATH by MATH. Since the height of a horizontal step is equal to the distance of the corresponding horizontal lozenge to our reference line in the tiling, the weight of a family MATH of nonintersecting lattice paths is equal to MATH, where MATH is the corresponding tiling. On the other hand, it is clearly equal to MATH, where MATH denotes the area between a path MATH and the MATH-axis. To find an expression for the entries of the NAME - NAME - NAME matrix we use the well-known fact (see CITE) that the weighted count MATH, summed over all lattice paths MATH from MATH to MATH, is equal to MATH, where MATH is the standard MATH-binomial coefficient, MATH . Thus, the determinant REF becomes (see also CITE) MATH . From the MATH-binomial theorem (see CITE), MATH it is straightforward to extract that MATH . We have to compute the determinant REF. Let us denote it by MATH. We have to distinguish between four cases, depending on the parities of MATH and MATH. First, let MATH be even. We reorder rows and columns simultaneously, so that the even-numbered rows and columns come before the odd-numbered, respectively. If MATH is even, then we obtain for MATH the block determinant MATH where MATH is the MATH identity matrix and MATH is the MATH matrix MATH. By a few simple manipulations, this determinant can be factored into a product of two determinants, MATH where MATH is a primitive sixth root of unity, each of which can be computed by application of REF . The result is the first expression in REF. On the other hand, if MATH is odd, then analogous arguments yield MATH where MATH is the MATH matrix which arises from MATH by deleting its last column, while MATH is the MATH matrix which arises from MATH by deleting its last row. It is easy to check that MATH where MATH is the MATH-matrix with MATH-entry MATH, MATH. (So the first column of MATH is zero). We expand MATH with respect to the first column and get MATH. Therefore, in the case of even MATH and odd MATH, we have MATH . Both determinants can be evaluated by means of REF . The result is the second expression in REF. Now let MATH be odd. We proceed analogously. If MATH is even, then reordering rows and columns according to the parity of the indices gives MATH . The first determinant is evaluated by means of REF , while the second is evaluated by means of REF . The result is the third expression in REF. Finally, if MATH is odd we get MATH . Again, the first determinant is evaluated by means of REF , while the second is evaluated by means of REF . The result is the fourth expression in REF.
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math/9912053
|
Let us denote the determinant in REF by MATH. We proceed in several steps. An outline is as follows. The determinant MATH is obviously a polynomial in MATH and MATH. In REF - REF we show that the right-hand side of REF divides MATH as a polynomial in MATH and MATH. In REF we show that the degree of MATH as a polynomial in MATH is at most MATH. Of course, the same is true for the degree in MATH. On the other hand, the degree of the right-hand side of REF as a polynomial in MATH is exactly MATH. It follows that MATH must equal the right-hand side of REF times a quantity which does not depend on MATH. This quantity must be polynomial in MATH. But, in fact, it cannot depend on MATH as well, because, as we just observed, the degree in MATH of the right-hand side of REF is already equal to the maximal degree in MATH of MATH. Thus, this quantity is a constant with respect to MATH and MATH. That this constant is equal to REF is finally shown in REF , by evaluating the determinant MATH for MATH. Before we begin with the detailed description of the individual steps, we should explain the odd looking occurrences of ``MATH mod REF" below (for example, in REF - REF ). Clearly, in the present context this means ``MATH mod REF", as MATH is even by assumption. However, REF - REF will also serve as a model for the proofs of the subsequent REF - REF . Consequently, formulations are chosen so that they remain valid without change at the corresponding places. In particular, in the context of the proofs of REF , the statement ``MATH mod REF" will mean ``MATH mod REF". CASE: MATH divides the determinant. The original determinant is symmetric in MATH and MATH for combinatorial reasons. The factors which were taken out of the determinant in REF are also symmetric in MATH and MATH (this can be seen by reversing all the products involving MATH). Therefore it suffices to check that the linear factors involving MATH divide MATH, that is, that the product MATH divides MATH. We distinguish between four subcases, labeled below as REF , and REF . CASE: MATH divides MATH for MATH, MATH mod REF: This follows from the easily verified fact that MATH is a factor of each entry in the first MATH columns of MATH. CASE: MATH divides MATH for MATH, MATH mod REF: We prove this by finding MATH ``different" linear combinations of the columns of MATH which vanish for MATH. By the term ``different" we mean that these linear combinations are themselves linearly independent. (Equivalently, we find MATH linearly independent vectors in the kernel of the linear operator defined by the matrix underlying MATH.) See REF, and in particular the Lemma in that section, for a formal justification of this procedure. To be precise, we claim that the following equation holds for MATH, MATH . Since the entries of MATH have a split definition (see REF), for the proof of the above equation we have to distinguish between two cases. If we restrict REF to the MATH-th row, MATH, then REF becomes MATH whereas on restriction to the MATH-th row, MATH, REF becomes MATH . First, let MATH. Here and in the following, we make use of the usual hypergeometric notation MATH . In this notation, the sum on the left-hand side of REF reads MATH . Next we use a transformation formula due to CITE (see also CITE), MATH where MATH is a nonnegative integer. This gives MATH . The factor MATH vanishes for MATH and the denominator is never zero, so the sum in REF equals zero, as desired. We proceed similarly in order to prove REF for MATH. The hypergeometric form of the sum in REF is MATH . Using the transformation REF again, we get MATH . This expression is zero, because the factor MATH vanishes for MATH (it is here where we need MATH mod REF, because this guarantees that MATH is an integer). So the sum in REF equals zero, as desired. CASE: MATH divides MATH for MATH, MATH mod REF: Proceeding in the spirit of REF , we prove this by finding MATH linear combinations of the columns of MATH which vanish for MATH. To be precise, we claim that the following equation holds for MATH: MATH . In order to prove this equation, we first restrict it to the MATH-th row, MATH. Then, in hypergeometric notation, the left-hand side reads MATH . We apply the transformation REF and get MATH . This expression is zero because the factor MATH vanishes. If instead we restrict the left-hand side of REF to the MATH-th row, MATH, and convert it into hypergeometric form, then we obtain MATH . We apply again the transformation REF . This gives MATH . This expression is zero because the factor MATH vanishes for MATH. So the sum in REF equals zero, as desired. CASE: MATH divides MATH for MATH, MATH mod REF: Still proceeding in the spirit of REF , this time we find MATH linear combinations of the rows of MATH which vanish for MATH. To be precise, we claim that the following equation holds for MATH: MATH . In the sum, it is only the first MATH rows which are involved, whereas the extra term is a row out of the last MATH rows of the determinant. Therefore, by restriction to the MATH-th column, we see that it is equivalent to MATH . We treat the cases MATH and MATH separately. For MATH the factor MATH, which appears in the sum, is zero for all the summands, as well is the factor MATH, which appears in the extra term in REF. For MATH we convert the sum in REF into hypergeometric form and get MATH . We can evaluate the MATH-series by the NAME - NAME summation formula (see CITE), MATH where MATH is a nonnegative integer. Thus we get MATH . It is easily seen that adding the extra term in REF gives zero. CASE: MATH divides the determinant. We find MATH linear combinations of the rows of MATH which vanish for MATH. To be precise, we claim that the following equation holds for MATH, MATH: MATH . Restricted to the MATH-th column, and converted into hypergeometric notation, the sum in REF reads MATH . Here we use the NAME - NAME summation formula (see CITE) MATH where MATH is a nonnegative integer. Thus we get MATH . It is easily verified that adding the MATH-th coordinate of the extra term in REF gives zero, as desired. For now, we need REF only for even MATH. CASE: MATH divides the determinant. We find MATH linear combinations of the columns of MATH which vanish for MATH. To be precise, we claim that the following equation holds for MATH, MATH mod REF, and MATH: MATH . Restricted to the MATH-th row, MATH, and converted into hypergeometric notation, the left-hand side sum in REF reads MATH . This is summable by the NAME - NAME summation REF . We get MATH . This expression equals zero because the factor MATH vanishes. On the other hand, if MATH, the left-hand side sum in REF, restricted to the MATH-th row and converted into hypergeometric from, reads MATH . The NAME - NAME summation REF turns this expression into MATH . This expression is zero because the factor MATH vanishes for MATH mod MATH. So the sum in REF is zero, as desired. CASE: MATH divides the determinant. We find MATH linear combinations of the columns of MATH which vanish for MATH. To be precise, we claim that the following equation holds for MATH and MATH: MATH . Restricted to the MATH-th row, MATH, and converted into hypergeometric notation, the left-hand side sum in REF reads MATH . The result after application of the NAME - NAME summation REF is MATH . This expression equals zero because the factor MATH vanishes. On the other hand, if MATH, the left-hand side sum in REF, restricted to the MATH-th row and converted into hypergeometric from, reads MATH . NAME summation REF yields MATH . This expression is zero because the factor MATH vanishes. So the sum in REF is zero, as desired. CASE: MATH divides the determinant. We find MATH linear combinations of the rows of MATH which vanish for MATH. To be precise, we claim that the following equation holds for MATH and MATH: MATH . Restricted to the MATH-th row, and converted into hypergeometric notation, the left-hand side sum in REF reads MATH . After applying NAME - NAME summation REF again, we obtain MATH . This expression equals zero because the factor MATH vanishes. So the sum in REF is zero, as desired. CASE: Determination of the degree of MATH as a polynomial in MATH. Obviously the degree of the MATH-entry of MATH as a polynomial in MATH is MATH. Therefore, if we expand the determinant MATH according to its definition as a sum over permutations, each term in this expansion has degree MATH in MATH. Hence, MATH itself has degree at most MATH in MATH. The unique lozenge tiling for MATH REF . Computation of the multiplicative constant. As we observed at the beginning of this proof, REF - REF show that the determinant MATH is equal to the right-hand side of REF up to multiplication by a constant. To determine this constant, it suffices to compute MATH for some particular values of MATH and MATH. We choose MATH. The value of MATH is most easily determined by going back, via REF , to the origin of the determinant MATH, which is enumeration of lozenge tilings. REF shows the typical situation for MATH. As the figure illustrates, there is exactly one lozenge tiling of the region. Hence, by REF , it follows that the determinant REF must be equal to MATH for MATH. If we substitute this into REF, we have evaluated MATH, which is the determinant on the right-hand side of REF, for MATH. It is then a routine task to check that the result agrees exactly with the right-hand side of REF for MATH. This completes the proof of the theorem.
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math/9912053
|
We proceed analogously to the proof of REF . The only difference is the parity of MATH, so we have to read through the proof of REF and find the places where we used the fact that MATH is even. As it turns out, the arguments in REF - REF in the proof of REF can be used here, practically without change, to establish that the right-hand side of REF divides the determinant on the left-hand side of REF as a polynomial in MATH and MATH. Differences arise only in the products corresponding to each subcase (for example, the arguments in REF of the proof of REF prove that MATH divides the determinant MATH if MATH is even, while for odd MATH they prove that MATH divides MATH), and in the fact that in REF we are now interested in the factors corresponding to odd values of MATH, MATH (because here the factors with even MATH are covered by REF ). Also REF , the determination of a degree bound on the determinant, can be used verbatim. A lozenge tiling and the corresponding path family for MATH, MATH . For the determination of the multiplicative constant relating the right-hand and the left-hand side of REF, we have to modify however the arguments in REF of the proof of REF . We determine the constant by computing the determinant for MATH. Again, this value is most conveniently found by going back, via REF , to the combinatorial root of the determinant, which is enumeration of lozenge tilings. We claim that the number of lozenge tilings for MATH, MATH odd and MATH even, equals MATH . This can be read off REF , which shows a typical example of REF The path starting at MATH (see the labeling in REF ; it is derived from the labeling of starting points of paths in REF ) must pass either to the right or to the left of the triangle. Since the hexagon is symmetric, we can count those path families where the path passes to the right, and in the end multiply the resulting number by two. For those path families, the paths starting at points to the right of MATH are fixed. The paths to the left have all exactly one NAME step. Suppose that the NAME step of the path which starts in MATH, MATH, occurs as the MATH-th step. Then we must have MATH . So we just have to count monotonously decreasing sequences of MATH numbers between REF and MATH. The number is exactly the binomial coefficient in REF. It is then a routine task to check that, on substitution in REF, the result agrees exactly with the right-hand side of REF for MATH.
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math/9912053
|
We proceed analogously to the proof of REF . The only difference is the parity of MATH, so we have to check the places in the proof of REF where we used the fact that MATH is even. Again, REF - REF can be reused verbatim, except that the products corresponding to the individual subcases are slightly different, and in REF we are now interested in the factors corresponding to odd values of MATH, MATH (because the factors with even MATH are covered by REF ). The computation of the multiplicative constant relating the right-hand and the left-hand side of REF is done analogously to REF in the proof of REF . That is, we compute the determinant for MATH by going back, via REF , to the lozenge tiling interpretation of the determinant. We already concluded in the proof of REF that for MATH there is just one lozenge tiling (see REF ). By definition, the statistic MATH attains the value MATH on this lozenge tiling, so that its weight is MATH. It is then not difficult to verify that, on substitution of this in REF, the result agrees exactly with the right-hand side of REF for MATH.
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math/9912053
|
Analogously to the previous cases, we can show that the product MATH divides the determinant as a polynomial in MATH and MATH. Although not completely obvious, this is implied by the linear combinations of REF - REF . The degree in MATH of this product is MATH which is larger than the maximal degree MATH of the determinant viewed as a polynomial in MATH. So the determinant must be zero.
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math/9912053
|
Let us denote the determinant in REF by MATH. We will again proceed in the spirit of the proof of REF . That is, we first show, in REF - REF below, that the right-hand side of REF divides MATH as a polynomial in MATH and MATH. Then, in REF , we show that the degree of MATH as a polynomial in MATH is at most MATH, the same being true for the degree in MATH. Analogously to the proof of REF , we conclude that MATH must equal the right-hand side of REF, times a constant with respect to MATH and MATH. That this constant is equal to REF is finally shown in REF , by evaluating the determinant MATH for MATH. In order to prove (in REF - REF ) that the right-hand side of REF divides MATH, for each linear factor of REF we exhibit again sufficiently many linear combinations of columns or rows which vanish. These linear combinations are almost identical (sometimes they are even identical) with the corresponding linear combinations in the proof of REF . Consequently, we will merely state these linear combinations here, but will not bother to supply their verifications, because these parallel the verifications in the proof of REF . MATH divides the determinant. Unlike in the case of the previous determinant MATH (see REF), here it is not possible to infer symmetry of MATH in MATH and MATH directly from the definition. Therefore it will be necessary to prove separately that the factors involving MATH, respectively MATH, divide the determinant. Again, we distinguish between four subcases, labeled below as REF , and REF . CASE: MATH divides MATH for MATH, MATH mod REF: This follows from the easily verified fact that MATH is a factor of each entry in the first MATH columns of MATH, respectively, that MATH is a factor of each entry in the last MATH columns of MATH. CASE: MATH divides MATH for MATH, MATH mod REF: The following equations hold for MATH: MATH and MATH . CASE: MATH divides MATH for MATH, MATH mod REF: The following equations hold for MATH: MATH and MATH . CASE: MATH divides MATH for MATH, MATH mod REF: The following equations hold for MATH: MATH and MATH . CASE: MATH divides the determinant. The following equation holds for MATH, MATH: MATH . Here, we need REF only for even MATH. CASE: MATH divides the determinant. The following equation holds for MATH, MATH mod REF, and MATH: MATH . CASE: MATH divides the determinant. The following equation holds for MATH and MATH: MATH . CASE: MATH divides the determinant. The following equation holds for MATH and MATH: MATH . CASE: Determination of the degree of MATH as a polynomial in MATH. This is clearly the same degree as for MATH, that is, MATH. CASE: Computation of the multiplicative constant. In analogy to the proof of REF , we evaluate the determinant for MATH. Again, we do this by going back, via REF , to the combinatorial origin of the determinant, which is enumeration of lozenge tilings. We can still use REF for our considerations. The number of lozenge tilings is easily seen to be equal to MATH. It is then a routine computation to verify that this does indeed give the multiplicative constant as claimed in REF.
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math/9912053
|
We proceed analogously to the proof of REF . The only difference is the parity of MATH, so we have to check the places in the proof of REF where we used the fact that MATH is even. REF - REF can be reused verbatim, but the corresponding products are slightly different. In REF we are now interested in the factors corresponding to odd values of MATH (MATH), because the factors with even MATH are covered by REF can be reused verbatim. The computation of the multiplicative constant is done analogously to REF in the proof of REF . Again using REF , we see that the number of lozenge tilings, related to our determinant via REF , for MATH equals REF. It is then a routine computation to verify that this gives the multiplicative constant as claimed in REF.
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math/9912053
|
We proceed analogously to the proof of REF . The only difference is the parity of MATH, so we have to check the places in the proof of REF where we used the fact that MATH is even. REF - REF can be reused verbatim, but the corresponding products are slightly different. In REF we are now interested in the factors corresponding to odd values of MATH (MATH), because the factors with even MATH are covered by REF can be reused verbatim. The computation of the multiplicative constant is done analogously to REF in the proof of REF . Using again REF , we see that the MATH-enumeration of lozenge tilings, related to our determinant via REF , for MATH equals MATH. It is then a routine computation to verify that this gives the multiplicative constant as claimed in REF.
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math/9912053
|
We proceed analogously to the proof of REF . The parameters MATH and MATH are odd, so we have to check the places in the proof of REF where we used the fact that MATH or MATH is even. REF - REF can be reused verbatim, but the corresponding products are slightly different. The computation of the multiplicative constant is done analogously to REF in the proof of REF . Again using REF , we see that the MATH-enumeration of lozenge tilings, related to our determinant via REF , for MATH equals MATH. It is then a routine computation to verify that this gives the multiplicative constant as claimed in REF.
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math/9912053
|
As in the proof of REF, define matrices MATH, MATH and MATH and MATH, MATH . Thus, MATH equals MATH. Now, as in CITE, multiply MATH on the left by MATH and on the right by MATH. Subsequently do the manipulations given in CITE (which amount to applying the NAME - NAME summation formula several times). The result is that MATH where MATH where MATH and MATH are restricted to be MATH or MATH, as in CITE. It is straightforward to check that MATH. Hence, each entry of the matrix MATH in an even-numbered row and even-numbered column is REF. This implies that MATH must be REF whenever the size of the matrix, MATH, is odd. In the case that MATH is even it implies the factorization MATH . As is easily verified, this equation is exactly equivalent to REF.
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math/9912053
|
Now choose MATH, MATH, MATH in REF . Then all the sums appearing in REF can be evaluated by means of the NAME - NAME summation REF. The result is MATH . Both determinants on the right-hand side of this identity can be evaluated by means of REF, which reads MATH . This completes the proof of the theorem.
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math/9912059
|
By the following short exact sequence of chain complexes MATH the assumption MATH for all MATH is equivalent to the acyclicity of the chain complex MATH (notice that MATH). Now if Conjecture REF holds, then take an element MATH which is a cycle. Then MATH where MATH and MATH. Then MATH is a cycle in MATH and a linear combination of thin elements. Therefore MATH is a cycle in MATH. By Conjecture REF, MATH where MATH. Therefore MATH. Conversely, suppose that the sub-complex generated by the thin elements is acyclic. Take a cycle MATH of MATH which is a linear combination of thin elements. Then MATH is a cycle of MATH, therefore there exists MATH and MATH such that MATH.
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math/9912059
|
We give only a sketch of proof. By definition, MATH for MATH. Because of the hypothesis on MATH, every element MATH of MATH is determined by the values of the MATH where MATH runs over the set of words on the alphabet MATH. It turns out that there is a bijective correspondence between MATH and the word of length MATH on the alphabet MATH : if MATH is an element of MATH, the associated word of length MATH is the word MATH such that MATH and if MATH, then MATH. It is then straightforward to check that the simplicial structure of MATH is exactly the same as the simplicial structure of MATH in strictly positive dimension .
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math/9912059
|
By REF or by REF , the simplicial nerve of the MATH-category of any composable pasting scheme is contractible.
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math/9912059
|
It is obvious for MATH and for MATH, MATH. But MATH, therefore it suffices to notice that the MATH-simplex is contractible.
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math/9912059
|
It suffices to calculate the simplicial homology of a simplicial set homotopic to a MATH-sphere.
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math/9912059
|
It is clear that MATH. For MATH, MATH by REF . But MATH is the free MATH-category generated by the permutohedron, and with REF , one gets MATH for MATH.
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math/9912059
|
The induction equations define a fillable MATH-shell (see REF ).
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math/9912059
|
Obvious for MATH and MATH. Recall that MATH is defined by MATH . Let us suppose that MATH and let us proceed by induction on MATH. Since MATH by the previous proposition, then the evaluation map MATH from MATH to MATH is surjective. Now let us prove that MATH and MATH and MATH imply MATH. Since MATH and MATH are in MATH, then one sees immediately that the four elements MATH and MATH are in MATH. Since all other MATH and MATH are thin, then MATH and MATH. By induction hypothesis, MATH and MATH. By hypothesis, one can set MATH and MATH for MATH. And one gets MATH. Therefore MATH for all MATH and all MATH. By REF , one gets MATH.
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