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math/0006165 | We have to represent an arbitrary vector MATH as MATH for some MATH. It is enough to consider a dense set of vectors MATH (since MATH is always closed). Let MATH and MATH. Clearly, MATH implies MATH in MATH, hence MATH in MATH, therefore MATH in MATH. |
math/0006165 | Let MATH, MATH; we have to prove that MATH weakly. Introduce MATH where MATH is chosen such that for all MATH large enough, MATH in terms of NAME transform it means that MATH such MATH exists due to the asymptotic relation MATH for large MATH (recall REF). The positive definiteness means that MATH for all MATH and then (extending the convolution operator by continuity) for all MATH. Denote by MATH the MATH-neighborhood of MATH, then MATH, and MATH. We have to prove that MATH for every linear functional MATH on MATH. We may restrict ourselves to a dense subset of functionals MATH (since MATH). In particular, we may consider only functionals MATH defined by MATH . Taking into account that MATH we see that the following would be enough: MATH for every MATH. Recalling that MATH and MATH we have MATH . |
math/0006165 | The construction is straightforward, just MATH equidistant intervals of equal length: MATH . We have to find MATH and MATH such that MATH . Still, the construction is straightforward: MATH the proof of REF is more complicated. NAME transform MATH will be used, MATH. Recall that MATH for all MATH, and MATH for large MATH. An elementary calculation gives MATH . Note that MATH . We have MATH . For MATH we note that MATH therefore MATH . For MATH we note that MATH and change the scale, introducing MATH: MATH where MATH. On each period, MATH, we substitute MATH by its maximal value: MATH . Hence MATH. So, REF is verified. Further, MATH for any MATH. It is easy to see that MATH which implies MATH . On the other hand, MATH it tends to MATH uniformly in MATH, when MATH. Choose MATH and MATH such that MATH then MATH which implies REF. |
math/0006165 | Suppose that MATH. REF gives elementary sets MATH such that MATH, and MATH consists of MATH intervals, and the relation MATH is violated. Taking into account that MATH we get MATH by REF . Also, MATH and MATH by REF . So, MATH satisfy REF but violate REF . By REF the product systems are nonisomorphic. |
math/0006165 | Similarly to the proof of REF we have for any MATH . This time, however, the following compactness property holds: MATH which ensures MATH. In order to prove the compactness property we estimate integrals similarly to the proof of REF. We have MATH . For MATH we note that MATH . For MATH, introducing MATH and MATH, we have MATH . |
math/0006165 | Assume the contrary, then there exist MATH such that MATH and MATH whenever MATH. NAME use MATH for MATH only. Introduce a Gaussian type space MATH such that MATH; we have MATH where MATH is the MATH-field generated by the Gaussian subspace MATH. Consider measures MATH; as was explained in REF, MATH is a probability measure on the MATH-field MATH such that MATH . The relation MATH implies MATH . In other words, MATH-fields MATH and MATH are independent with respect to the measure MATH on MATH, and MATH is just the restriction of MATH to MATH. Moreover, MATH-fields MATH are MATH-independent whenever MATH. Every elementary set MATH determines its sub-MATH-field-MATH; as explained in REF, MATH whenever MATH. Note that MATH are MATH-independent whenever MATH. Consider elementary sets MATH and vectors MATH . REF shows that MATH and MATH in MATH. Though, MATH is not a subspace but a quotient space MATH of MATH; anyway, we may choose elements of MATH, denoted again by MATH, such that MATH . The natural map MATH allows us to treat MATH as elements of MATH. Now they are random variables; MATH and MATH in probability. On the other hand, for every MATH, the two random variables MATH are independent (since MATH and MATH are MATH-independent). It follows that MATH is independent of itself, that is, MATH is constant MATH-almost sure. Consider a Gaussian measure MATH. Though, MATH need not be constant MATH-almost sure; however, MATH must be constant on a set of positive probability with respect to MATH. On the other hand, the distribution of MATH with respect to MATH is normal (Gaussian); it cannot have an atom unless it is degenerate, which means that MATH must vanish. However, it does not vanish, which is evident when using NAME transform. A contradiction. |
math/0006168 | First suppose that MATH satisfies REF. Using MATH, we find that MATH . REF follows since MATH. The converse is proved similarly, since, for all MATH, one can always find MATH such that MATH at MATH. |
math/0006168 | It remains to verify the moment map REF , which in this case is MATH . From the definition of MATH we obtain, MATH . These two expressions differ by MATH which vanishes by MATH-invariance of the inner product. |
math/0006168 | Using REF we find that MATH for all MATH. This shows MATH. |
math/0006168 | By REF , MATH takes values in the image of the action map MATH. Equivalently, MATH is tangent to the orbits for the conjugation action. |
math/0006168 | The map MATH is a homomorphism with respect to both the exterior product and the NAME bracket. Moreover, the operator MATH is a generator of the NAME bracket on MATH. Hence, it is sufficient to prove the property MATH on the elements of degree REF. For all MATH, MATH. To compute MATH we consider MATH where MATH denotes an interior product and MATH a NAME derivation. Since MATH is MATH-invariant, MATH, whence MATH, which shows that MATH, for all MATH. |
math/0006168 | Since the NAME is a derivation of the NAME bracket, for each MATH, MATH. Moreover, MATH. The element MATH defines a cycle in NAME algebra homology. Hence, we obtain MATH and MATH. Choosing MATH with MATH a positive MATH-invariant function on MATH, one obtains the new NAME, MATH. The modular vector field also changes, MATH . We conclude that the class of MATH in the quasi-Poisson cohomology is independent of the choice of MATH. |
math/0006168 | Let MATH be the bi-invariant volume form on MATH defined by the basis MATH of the NAME algebra MATH. Then, MATH, where MATH. Applying the NAME differential yields MATH since MATH and MATH commute, and since MATH is both left-and right-invariant and closed. This implies MATH. |
math/0006168 | The trivector field for the MATH-action is the sum of the trivector fields MATH and MATH for the two MATH-factors. By assumption, MATH is MATH-invariant and MATH. Therefore MATH, and MATH . By REF , this implies that MATH where MATH is the image of MATH under the map extending the infinitesimal diagonal action of MATH on MATH. Now suppose that the action is Hamiltonian. For any maps MATH and MATH from MATH to MATH, MATH . This relation together with the moment map property, MATH for MATH, implies, MATH . By equivariance of MATH and therefore, MATH . We obtain MATH which is the moment map condition for the diagonal action. |
math/0006168 | Since the identity map is a moment map for MATH, REF shows that MATH is a moment map for MATH. Hence it is a quasi-Poisson map by REF . |
math/0006168 | Since MATH is MATH-invariant, MATH. To compute MATH we observe that MATH, since MATH is the vector field generating the left action on MATH, and MATH is equivariant with respect to this action. Thus MATH . Finally, since MATH, we obtain MATH which cancels the term MATH. |
math/0006168 | REF follows by equivariance of the moment map, MATH . To prove REF, write MATH as the composition of two maps, MATH where MATH, and MATH is the action map MATH of the second MATH-factor. The tangent map to MATH is defined by MATH . In particular, MATH and MATH, so that MATH . We now use the fact that MATH and MATH and the relation MATH. We find that MATH . On the other hand, MATH is the sum of MATH, of MATH, and of cross-terms which can be found by pairing this bivector with MATH and using the moment map condition. The result is MATH . We next apply MATH to this result. The push-forward of MATH and of MATH are obtained as in the proof of REF , and we obtain MATH . REF is now obtained by subtracting the expression for MATH from that for MATH. |
math/0006168 | Suppose that the action is locally free at the point MATH. To prove that MATH has maximal rank at MATH, we need to show that the equation MATH does not admit a non-trivial solution MATH. Let MATH be any solution of this equation. By equivariance of the moment map, MATH for all MATH. Hence, MATH, and the moment map REF shows that MATH . Since the action is locally free at MATH, this implies MATH. Because MATH is of maximal rank on MATH, MATH is a smooth submanifold of MATH, and MATH is a smooth suborbifold of MATH. To show that MATH is a NAME suborbifold, we have to show that, for all invariant MATH, the Hamiltonian vector field MATH is tangent to MATH. In fact, MATH is tangent to all level sets of MATH, because MATH for all functions MATH. Here we have used the moment map REF and MATH. |
math/0006168 | Let MATH . We want to compute MATH . The first term vanishes since MATH is a NAME structure. For any function MATH, MATH. Using the moment map REF of MATH and the invariance property MATH of MATH, we obtain MATH . To calculate MATH we use the relation MATH . Taking symmetries into account, we obtain MATH . Together with REF, and using the classical dynamical NAME REF , this yields MATH . To prove that MATH is a moment map, we use REF . For all functions MATH, MATH . The moment map REF of MATH follows from this relation together with the moment map map REF of MATH. |
math/0006168 | At MATH, the fiber MATH is the space of covectors MATH with MATH. Let MATH. By the moment map REF , it follows that, MATH for all MATH. |
math/0006168 | Replacing MATH by MATH, we may assume that MATH. It is clear that the moment map condition for MATH is equivalent to that for MATH. We have to show that MATH if and only if MATH . Let MATH denote the extension of MATH to a MATH-invariant bivector field on MATH. Also, let MATH be the MATH-invariant extension of MATH. By definition, MATH, so what we need to show is that MATH . By a calculation similar to that of the term MATH in the proof of REF , MATH . To compute the term MATH, first observe that MATH by MATH-invariance of MATH. Using REF, we obtain MATH . Therefore, MATH . Together with REF , this proves REF. |
math/0006168 | We show that MATH is integrable near any given point MATH. Let MATH, and MATH be a MATH-invariant slice through MATH, and MATH the corresponding cross-section. Since MATH, it suffices to show that the distribution MATH induced by the quasi-Poisson structure MATH is integrable. However, as explained in the previous section, MATH, where MATH is a MATH-invariant NAME structure in the usual sense, and MATH is a bivector field taking v alues in the second exterior power of the MATH-orbit directions. Moreover, MATH admits a moment map MATH, which implies that the image of MATH contains the MATH-orbit directions. Hence MATH is equal to the distribution defined by the NAME MATH, and therefore integrable by REF and NAME. |
math/0006168 | Given MATH, let MATH be a slice through MATH, and let MATH be the corresponding cross-section. We first evaluate both sides of REF on elements of MATH, and then on orbit directions. REF-form MATH vanishes on MATH, and the bivector MATH vanishes on MATH. Hence MATH, which agrees with the right-hand side of REF since MATH also vanishes on MATH. To complete the proof we evaluate both sides of REF on orbit directions. The moment map properties of MATH and MATH yield MATH and the same result is obtained by applying the right-hand side of REF to MATH. |
math/0006168 | Given MATH, let MATH be a cross-section at MATH, as in the proof of REF , and let MATH. Thus MATH is a non-degenerate Hamiltonian quasi-Poisson MATH-manifold that corresponds to a quasi-Hamiltonian MATH-space, MATH. Let MATH be the unique REF-form on MATH such that MATH is a quasi-Hamiltonian MATH-space, and MATH is the pull-back of MATH. We have to show that MATH satisfies REF . For orbit directions, this follows from the moment map conditions (see the proof of REF ), while, for directions tangent to MATH, it follows by applying the Lemma to MATH. Uniqueness is clear since the equation, MATH the transpose of REF, defines MATH on the image of MATH, while the moment map condition determines MATH on orbit directions. The converse is proved similarly, using the cross-section theorem for Hamiltonian quasi-Poisson manifolds. |
math/0006168 | Choose MATH and a slice MATH containing MATH and let MATH be the cross-section. We observe that in both the quasi-Poisson and the quasi-Hamiltonian settings, the reduction of MATH at MATH is canonically isomorphic to the reduction of MATH at the group unit of MATH. The cross-section MATH carries the NAME bivector MATH and the symplectic form MATH. According to REF , MATH. Hence, the same relation holds for the reduced space. |
math/0006168 | We have to compute MATH . We compute the four terms in the expansion of the right-hand side. By REF , the first term is MATH . Next, using MATH we find that MATH . From MATH we obtain MATH . Finally, MATH . Putting everything together, we get MATH as required. |
math/0006171 | This is equivalent to MATH which is obvious. |
math/0006171 | The sufficiency of the conditions in the lemma is clear. To prove necessity, define the map MATH by MATH where MATH is arbitrary. This map MATH is well defined because MATH for any closed path MATH. The required properties of MATH follow easily from the definitions. |
math/0006171 | The NAME imaginary transformation (see for example, REF) yields MATH . We have MATH . It is obvious that this series, together with all derivatives, is dominated in the MATH limit by only two terms, namely the terms with MATH and MATH which combine into a multiple of MATH. All other terms differ by a factor of at least MATH. |
math/0006171 | Observe that all MATH-functions appear in MATH in the following combinations: either they appear in pairs of the form MATH where MATH is a subset of MATH, or else they appear as the nullwerts MATH . It is clear from REF that the asymptotics in the either case is a polynomial in MATH of degree MATH with coefficients involving MATH and MATH. It remains to observe that in all monomials which appear in the expansion of the determinant in REF the orders of the derivatives sum up to MATH. |
math/0006171 | Follows, as explained in REF , from the corresponding facts for MATH, see REF or REF. |
math/0006171 | Follows from the difference equation satisfied by MATH, see REF or REF. |
math/0006171 | Observe that the binomial theorem and the definition of the function MATH imply that MATH where the sum is over all subsets MATH of MATH and hats mean that the corresponding terms should be omitted. Interchanging the order of summation in the partition MATH and in the subset MATH one obtains MATH where MATH is a partition of the set with MATH elements which is obtained from the partition MATH by mapping MATH to a point. Note that MATH, which according to REF means that MATH and MATH belong to the same block of MATH, implies MATH. Now the obvious identity MATH completes the proof. |
math/0006171 | By induction on MATH. The case MATH is clear, see REF . Suppose MATH. Denote by MATH the right-hand side of REF. We know that MATH satisfies the same difference equation as MATH does. We claim that it also has the same singularities as MATH does. Indeed, MATH is regular at the divisors MATH, MATH, because MATH is regular at MATH provided MATH. It is also clear that on MATH we have MATH and so, by induction hypothesis, MATH and MATH have identical singularities at all divisors MATH. Since they also satisfy the same difference equation, all of their singularities are identical. It follows that the function MATH is regular everywhere. By the difference equation, the induction hypothesis, and REF this function is also periodic in all MATH's with period MATH. From REF we conclude that REF grows at most polynomially as MATH and, therefore, it is a constant. Since both MATH and MATH are regular at MATH, the function REF vanishes there. It follows that it is identically zero. This completes the proof. |
math/0006171 | The NAME coefficients of MATH are certain contour integrals and hence by REF they represent the asymptotics of the corresponding coefficients of MATH. |
math/0006171 | Recall that, by definition, MATH . We have MATH where, we recall REF , MATH means that MATH is a subset of a block of MATH. The first factor in REF is a common factor for all MATH. The NAME series expansion of the second factor in REF can be interpreted as summation over certain graphs MATH with multiple edges MATH where MATH means that MATH for any egde MATH of MATH, no edges from a vertex to itself are allowed, and MATH is a nonnegative integer, called multiplicity, which is assigned to any edge MATH. The NAME inversion in the partially ordered set MATH, see REF , implies that MATH where MATH-connected means that MATH becomes connected after collapsing all blocks of MATH to points, that is, after passing to the quotient MATH . It is now clear that the minimal possible exponent of MATH is MATH and it is achieved by those graphs MATH which have no multiple edges and project onto spanning trees of MATH. That is, MATH where MATH. It is known, see REF, that this sum over spanning trees equals MATH which concludes the proof. |
math/0006171 | By definition, we have MATH . Substituting REF in this sum yields MATH . Interchanging the order of summation we obtain MATH . Using REF we conclude that MATH where dots stand for lower order terms. We know from REF that the exponent MATH takes its minimal value MATH precisely when MATH. This concludes the proof. |
math/0006171 | We can assume that the number of variables MATH is finite and equal to MATH and switch to the variables MATH. It is known, see REF , that MATH where MATH. Expand all fractions in geometric series assuming that MATH . We have MATH . Now let MATH is a partition of weight MATH and let us compute the coefficient of MATH in the above expression. Observe that only the first summand produces positive powers of MATH and, moreover, the monomials of maximal weight come from the expansion of MATH . Clearly, the coefficient of MATH in the expansion of REF equals MATH. By REF this concludes the proof. |
math/0006171 | By definition of the cumulants and of the elementary cumulants, we have MATH and therefore MATH . By REF , for any MATH such that MATH, we have MATH with the equality if and only if MATH. |
math/0006171 | Follows by extracting terms of degree MATH in MATH from the formula MATH . |
math/0006171 | REF specializes to MATH where the summation is over MATH satisfying the conditions described above. In particular, MATH. Recall that the MATH term in REF is to be understood as MATH. This is in agreement with the coefficient of MATH in REF. Therefore, in what follows we can assume that MATH. We know from REF that all MATH appearing in REF satisfy MATH . Let MATH be one such partition and pick the coefficient of MATH in REF. This means that of MATH blocks of MATH we have to chose MATH blocks for which we take MATH so that to produce the factor of MATH. After that, we select MATH blocks of MATH for which we take MATH, and so on. In the remaining MATH parts of MATH we take MATH which results in the factor MATH. This can be imagined as painting the parts of MATH into different colors which we call ``REF", ``REF", ``REF" etc. Observe that the summands in REF depend not on the actual partition MATH but rather on the sizes of blocks of a given color. For any color MATH we can use REF with MATH and this yields the following formula MATH where MATH is the only one of the MATH's that depends on MATH. We have MATH and, therefore, for MATH we obtain MATH . For MATH, introduce the following notation MATH . We compute MATH . Putting it all together using the equalities MATH we obtain MATH . It remains to show that MATH . Recall that the residue of a differential form is independent on the choice of coordinates. Using the change of variables MATH, which implies that MATH, we compute MATH . This concludes proof. |
math/0006172 | The nest algebra MATH and the MATH- algebra it generates have a MATH block structure induced by the MATH atoms in the nest for MATH. In this block structure, the diagonal blocks are square matrices of varying size and the off-diagonal blocks are rectangular. Elements of MATH and MATH will be written as MATH matrices, MATH, MATH, where each MATH is a matrix of appropriate size. We may assume that MATH. Consider the star algebra extension of MATH. This is an algebra injection MATH such that the image of each standard matrix unit of MATH is a regular partial isometry. We wish to show that MATH is regular with respect to the MATH by MATH block structure. By the remarks above this is equivalent to showing the following with MATH. Let MATH-with MATH be a matrix unit system for a subalgebra of MATH which is isomorphic to MATH. Suppose moreover that each MATH is regular with respect to the MATH by MATH block structure, so that each block matrix entry MATH of each MATH is itself a partial isometry. We show that there is a matrix unit system for MATH, which consists of (rank one) regular partial isometries, such that each MATH is a sum of some of these matrix units. To see this consider first a product MATH. The MATH block entry is given by the sum MATH . Since MATH is regular, the partial isometries MATH have orthogonal range projections and so the operators of the sum have orthogonal range projections. For similar reasons the domain projections are pairwise orthogonal. (We are not assuming here that these products are partial isometries.) Since, by hypothesis, the product MATH is a regular partial isometry, it follows that the sum above is a partial isometry, and therefore, by the orthogonality of domain and range projections, each of the individual products MATH is a partial isometry. Now, since, for example, MATH is a partial isometry it follows that the range projection of MATH commutes with the domain projection of MATH. Abusing notation somewhat, and regarding the entry operators MATH as identified with operators in MATH, it follows, by considering other block entries, that for all MATH the range projection of MATH commutes with the domain projection of MATH. Note also that the domain projections and the range projections commute amongst themselves. Furthermore it is clear that these projections commute with the projections in the centre of the block diagonal subalgebra of MATH. Choose now a maximal family of rank one projections MATH which commute with all these projections and are dominated by MATH. Then for each such projection the set of operators MATH satisfy the relations of a matrix unit system. The projections MATH are pairwise orthogonal and decompose MATH as a direct sum of multiplicity one embeddings, as desired. |
math/0006172 | We first prove the theorem for the special case in which every atom of MATH has rank one (that is, MATH is some MATH). Select a matrix unit system for MATH compatible with MATH; for each pair of (rank one) atoms MATH and MATH, let MATH denote the matrix unit with initial projection MATH and final projection MATH. Note that MATH is a matrix whose columns are indexed by the atoms of MATH and whose rows are indexed by the atoms of MATH. The entry in the column indexed by the atom MATH of MATH and the row indexed by the atom MATH of MATH is the non-negative integer MATH. We assume the order of the rows and columns of this matrix reflect the usual MATH ordering of the atoms of MATH and MATH. Let MATH be an atom of MATH with the property that the first non-zero entry in the column of MATH indexed by MATH is less than or equal to the first non-zero entry in any other column. Let MATH be the atom which indexes the row containing the first non zero entry in column MATH. Now assume that MATH is some other atom of MATH. First, assume MATH. Let MATH be the atom of MATH which gives the first non-zero entry in column MATH. Let MATH and MATH. The fact that MATH is locally order conserving ensures that MATH. To see this note that for an appropriate choice of matrix units in MATH, MATH induces an order conserving bijection from the rank one matrix unit subprojections of MATH to the rank one matrix unit subprojections of MATH. Since MATH and MATH correspond to the first block occurrences of these matrix unit projections and since, by the choice of MATH, MATH has at least as many of them as MATH does, it follows that MATH conjugates MATH to a subprojection of MATH. Let MATH. If MATH, we argue in a similar fashion to obtain a projection MATH which is a subprojection of the index atom for the row containing the first non-zero entry in column MATH and which satisfies MATH. The fact that MATH is an embedding implies that, for any two atoms MATH, MATH. Note that if MATH and MATH are distinct atoms of MATH, then the projections MATH and MATH are orthogonal. This is true even if MATH and MATH are subprojections of the same atom of MATH, since MATH, MATH and MATH and MATH are orthogonal projections. Let MATH, where the sum runs over all atoms of MATH. Let MATH. It follows from the way that MATH is defined that MATH and MATH are locally order conserving embeddings of MATH into MATH and that MATH. The reason that MATH and MATH are locally order conserving is that any subpartial isometry of an order conserving partial isometry which is obtained by left and right multiplication by a block diagonal projection is again order conserving. Note that MATH is obtained from MATH by subtracting from the first non-zero entry in each column of MATH the integer in column MATH, row MATH. Since MATH, we may construct a projection MATH which has the same properties with respect to MATH that MATH has with respect to MATH. MATH is a sum of projections, one for each atom of MATH; the projection in the sum associated with an atom MATH has the same rank as MATH and is a subprojection of the same atom of MATH as MATH. Consequently, there is a unitary element of the diagonal MATH such that if MATH, then MATH and MATH is an embedding. We may now repeat this procedure with the two embeddings, MATH and MATH. When we do so, we obtain a projection MATH which is orthogonal to MATH and an embedding MATH which is inner conjugate to MATH such that MATH. Furthermore, because we are really working in essence on MATH, we can select the unitary which implements the conjugacy so that it is the identity on MATH. With this done, we also have MATH and hence MATH. It is now clear that if we continue in this fashion, after finitely many steps we obtain an inner conjugacy between MATH and MATH. Slight modifications of the argument given above yield the proposition when MATH is a general finite dimensional nest algebra. Alternatively, we can deduce the general version of the proposition from the special case above by the following `general principles' technique. If MATH is a general nest algebra, select a subalgebra MATH of MATH which is isomorphic to some MATH such that each block of MATH corresponding to an atom of MATH contains exactly one rank one diagonal projection from MATH. Note that MATH is the algebra generated by MATH and MATH. Let MATH and MATH be two locally order conserving embeddings such that MATH. Since MATH and MATH are star extendible, they are determined (up to inner equivalence) by their restrictions to MATH. But MATH. By the argument above, MATH and MATH are inner equivalent; it follows immediately that MATH and MATH are inner equivalent. |
math/0006172 | Assume that MATH is not MATH-degenerate. Let MATH have atomic interval projections MATH and let MATH have atomic interval projections MATH. The hypothesis for MATH implies that there exists a matrix unit system MATH, MATH, MATH, for a copy of MATH in MATH such that MATH is supported in MATH the MATH block subspace of MATH. In particular note that for MATH there are rank two order conserving partial isometries in MATH of the form MATH. Similarly, there are rank two order conserving partial isometries supported in a single block column. Let MATH be the multiplicity one decomposition of MATH, with MATH . Note first that since MATH is order conserving, any subsum, such as MATH, is also order conserving. Secondly, observe that for each MATH, the partial isometry MATH is supported in a single off-diagonal block. For if not, then MATH for some MATH and MATH, and so the range of MATH lies in MATH. On the other hand MATH is order conserving and it follows readily that MATH must be an order summand of MATH, and hence equal to MATH by order irreducibility, contrary to the assumption MATH. Consider an index MATH for which MATH is supported in the MATH block subspace and is such that MATH for all MATH satisfying MATH and MATH. We complete the proof of the lemma by showing that for all MATH the partial isometry MATH is supported in the same block subspace as MATH. In particular all the summands MATH are inner conjugate and MATH is a refinement embedding. Note first that MATH is supported in the MATH block subspace and that MATH is supported in the MATH block subspace. Given MATH, there are various a priori possibilities for the support projections MATH of MATH. The first possibility, MATH, is suggested by the following diagram: MATH . In this case we can deduce from the order conserving nature of MATH that MATH is supported in a block MATH with MATH. Indeed, if such a MATH is greater than MATH, then it follows that MATH and MATH have support in the MATH block, MATH and MATH have support in the MATH block and so the support of MATH is not of staircase type. Since MATH is order conserving, this is a contradiction. We have shown that if MATH then MATH is an ordered sum. Now suppose that MATH. Since MATH is the support projection for MATH, it follows immediately (without the need for the order conservation of MATH) that MATH is an ordered sum. Now consider the case MATH. Then MATH has support in the MATH block and the MATH block and so is not of staircase form. Since MATH is order conserving, we have a contradiction once again. Thus the only possible value for MATH is MATH. In summary, we have shown that if MATH, then MATH is supported in the same block MATH as MATH. By the original assumption on MATH, we must have that MATH is supported in the MATH block, since the only alternative is that it is supported in a MATH with MATH. But this would imply that MATH is not order conserving. For similar reasons it follows that for MATH, the projections MATH and MATH are equivalent. For suppose that these projections are supported in the blocks for MATH and MATH respectively, with MATH. Then MATH has support in the MATH block while MATH has support in the MATH block and hence MATH is not order conserving. Similarly, MATH is not possible. We have shown that for all MATH and for all MATH and MATH the projections MATH and MATH lie in the same block subspace and so are inner equivalent. It follows that MATH is a refinement type embedding. |
math/0006172 | Fix matrix units for MATH and MATH so that MATH maps matrix units of MATH to sums of matrix units in MATH, and then choose matrix units for MATH such that MATH maps matrix units in MATH to sums of matrix units. Let MATH be the bimodule over MATH generated by MATH. Since MATH is, in particular, a bimodule over the diagonal matrices, MATH is the linear span of the matrix units which it contains. Let MATH be a matrix unit in MATH. Then there are atoms MATH and MATH for MATH such that MATH. If MATH and MATH are matrix units satisfying MATH and MATH, then MATH. It follows that MATH. But any matrix unit MATH for which MATH can be written in this form. Thus, if MATH then MATH. This enables us to identify the bimodule over MATH generated by MATH: it consists of all elements of MATH which are supported in the collection of matrix blocks which contain non-zero entries for elements of MATH. Now suppose that MATH and MATH are two matrix units in the same block in MATH and suppose, further, that MATH is order conserving. By the first assumption, there are matrix units MATH and MATH in diagonal blocks of MATH such that MATH. Since MATH is order conserving, its support has `staircase' form in MATH; since MATH and MATH lie in MATH, MATH also has `staircase' form; that is, MATH is also order conserving. This argument works equally well if MATH is a rank one partial isometry in the same block as MATH rather than a matrix unit. (The only modification is that MATH and MATH are merely rank one partial isometries in the appropriate diagonal blocks.) As a consequence of these observations, we see that we can prove the lemma by proving it in the special case in which MATH is a subordinate of a partial isometry in the range, MATH, of MATH, If we make this additional hypothesis, there is a rank one regular partial isometry MATH in MATH such that MATH. Because of our matrix unit choice for MATH, it follows that the matrix support of MATH contains the matrix support of MATH. However, the block matrix support of MATH is of staircase type, since MATH is locally order conserving. It follows that MATH is an order conserving partial isometry. |
math/0006172 | The proof that REF implies REF is immediate. For the converse, let MATH be a star extendible isomorphism. Since each algebra in the systems MATH and MATH is finitely generated, there is a commuting diagram isomorphism of the systems with crossover maps MATH, MATH: MATH . Since MATH, the range of MATH is contained in MATH. By hypothesis, MATH is locally order conserving and so locally regular and hence regular. If MATH is a rank one regular partial isometry, then MATH is a regular, order conserving, partial isometry in MATH. Consequently, MATH is a sum, MATH, of regular rank one partial isometries in MATH. Since MATH is locally order conserving, each partial isometry MATH is order conserving, by REF , and hence regular in MATH. Therefore MATH is regular and so MATH is locally regular. By REF , MATH is a regular embedding. The same argument shows that MATH is regular; continuing in this fashion, we may replace the initial commuting diagram isomorphism with one whose crossover maps are all regular. This shows that REF implies REF . |
math/0006172 | The proof of this theorem follows easily from REF below. If MATH and MATH are star extendibly isomorphic, then REF gives a regular system isomorphism in which any adjacent pair of crossover maps have a composition which is order preserving or order conserving, as appropriate. From the lemma, any consecutive triple of crossover maps has composition which is order preserving or order conserving; a sequence of such triple compositions yields a system isomorphism with order preserving or order conserving crossover maps. |
math/0006172 | The proof of the Lemma in the order preserving context differs considerably from the proof in the order conserving context, so we present the two arguments separately. We start with the the order preserving context, where the argument is simpler. First, suppose that MATH is a rank one partial isometry in MATH such that MATH is not order preserving. Then there are multiplicty one summands MATH and MATH of MATH, MATH and MATH of MATH, and MATH and MATH of MATH such that MATH is not order preserving. Observe that MATH is a partial isometry which is not order preserving. It is a partial isometry since MATH is an embedding; if it were order preserving, then its image under MATH would be order preserving. But MATH is a subordinate of this image and is not order preserving. Since MATH (a sum of multiplicity one summands of MATH) is order preserving, it follows that MATH is order preserving. Consequently, MATH and MATH are in the same block. This, in turn, implies that MATH and MATH are in the same block. This means that MATH and MATH have the same block structure - either both are order preserving or both are not order preserving. But the second one is not order preserving by assumption while the first one is the image of the rank one partial isometry MATH under the order preserving embedding MATH, a contradiction. The other possibility which we need to consider is that there are rank one partial isometries MATH and MATH in MATH such MATH is an order preserving partial isometry but MATH is not order preserving. Again, there are multiplicty one summands MATH and MATH of MATH, MATH and MATH of MATH, and MATH and MATH of MATH such that MATH is not order preserving. Observe that MATH is a partial isometry which is not order preserving. It is a partial isometry since it is a subordinate of MATH; if it were order preserving, then its image under MATH would be order preserving. But MATH is a subordinate of this image and is not order preserving. Since MATH is order preserving, it follows that MATH is order preserving. Consequently, MATH and MATH are in the same block, whence MATH and MATH are in the same block. This means that MATH and MATH have the same block structure But MATH is not order preserving by assumption while MATH is a subordinate of the image of the order preserving partial isometry MATH under the order preserving embedding MATH, a contradiction. (MATH is order preserving since any multiplicity one embedding is order preserving and MATH is an order preserving partial isometry.) Having completed the order preserving context, we move on to the order conserving context. Let MATH be an order conserving partial isometry such that the block support of MATH has staircase form. Let MATH be a rank two subordinate of MATH which is supported in two of these block subspaces, denoted MATH and MATH. The orientation of these blocks in the block decomposition of MATH can be indicated diagramatically as one of six types, namely the triple, MATH together with the corresponding triple with the letters reversed. Suppose the MATH is not order conserving. Then there exist two blocks of MATH, MATH and MATH say, which have nonstaircase orientation MATH and are such that at least one of four possibilities occurs: CASE: MATH has support meeting MATH and REF MATH has support meeting MATH and REF MATH and MATH have support meeting MATH and MATH respectively REF MATH and MATH have support meeting MATH and MATH respectively. Consider the companion blocks MATH, MATH for MATH, MATH indicated by the diagram MATH . Since the composition MATH is order conserving, if MATH is a multiplicity one summand of MATH, the composition MATH is order conserving. In particular such a map MATH must `correct' the MATH, MATH support of MATH (indicated in one of the four possibilities REF to REF ) by a `fusion' of blocks, as indicated in each of the three diagrams MATH . This means that, in the first case for example, the image under MATH of an element in MATH with support in the blocks MATH, MATH or in the blocks MATH, MATH has support in a single block of MATH. (The appearance of MATH and MATH, for example, in the same block in the diagram indicates that MATH.) These types of fusion may vary among the various summands MATH of MATH. The comments preceding the statement of the lemma preclude a correction of the MATH, MATH support of MATH of the form MATH or the similar correction with MATH and MATH interchanged. Let MATH be rank two partial isometries with support in the blocks MATH, MATH and MATH, MATH respectively. We now show that MATH is order conserving. The idea for this argument is that the fusion of blocks `binds' the supports of MATH and MATH and the latter is order conserving by the hypotheses, since MATH is order conserving. More precisely let MATH be the rank one decomposition and suppose that MATH is not order conserving. Then there exist two blocks MATH in MATH with nonstaircase form MATH and there exist multiplicity one summands MATH, MATH of MATH and MATH, MATH of MATH such that at least one of the following possibilities occurs. CASE: MATH and MATH meet MATH and MATH respectively. CASE: MATH and MATH meet MATH and MATH respectively. CASE: MATH and MATH meet MATH and MATH respectively. CASE: MATH and MATH meet MATH and MATH respectively. Let MATH be the rank one decomposition of MATH. If the first possibility REF occurs, then noting that MATH fuses MATH and MATH, or MATH and MATH, and that MATH also fuses MATH and MATH, or MATH and MATH, it follows that MATH (or MATH) meets MATH and similarly that MATH (or MATH) meets MATH. For each of these four alternatives MATH is not order conserving, and this contradicts the fact that MATH is order conserving. The other three possibilities REF also lead to contradictions in the same manner. It has been shown then that MATH is order conserving or, more intuitively, that the map MATH `corrects' the MATH, MATH block structure of MATH and MATH. We now wish to deduce that since every such block pair is corrected by MATH then in fact MATH is order conserving. This will complete the proof. To see this, suppose that MATH is not order conserving and that it has support in two blocks MATH, MATH of MATH which are not in staircase form. Then there are multiplicity one summands MATH and MATH of MATH (where MATH, MATH are multiplicity one summands of MATH; MATH, MATH are multiplicity one summands of MATH; and MATH, MATH are multiplicity one summands of MATH) such that MATH meets MATH and MATH meets MATH. Thus there are rank one subordinates MATH of MATH such that MATH meets MATH and MATH meets MATH. It follows that the partial isometry MATH is not order conserving (since the map MATH is order conserving). However the argument above shows that MATH is order conserving, which is the desired contradiction. |
math/0006172 | It will be enough to show that the assertion of the lemma follows if it is assumed that the lemma is true in the case of matrices of sizes MATH, MATH and MATH, where MATH, MATH, MATH, MATH, and at least one of these inequalities is strict. We may assume that all the entries MATH, MATH and MATH, MATH are non-zero. The reason for this is that if a staircase form matrix has its first entry equal to zero, then either the first row or the first column of the matrix is zero; the assertion then follows from the induction hypothesis. Let MATH. Let MATH be a subprojection of MATH which has rank MATH; let MATH; and let MATH be a regular partial isometry of rank MATH with initial projection MATH and with MATH rank distribution matrix of the form: MATH . Set MATH (MATH), so that MATH has a similar rank distribution matrix, of size MATH. Consider now the matrices MATH, MATH, and MATH, which are obtained from MATH, MATH, and MATH by subtracting MATH form the first entry, together with the partial isometries MATH and notice that these satisfy the hypotheses of the lemma. By the induction hypothesis, we may assume that there is a lifting MATH for MATH with MATH, for all MATH, such that MATH has rank distribution MATH. Now MATH is a partial isometry with the properties needed to prove the lemma. |
math/0006172 | The usual scheme of proof that REF. implies REF. can be completed as follows. With the aid of REF , obtain a lifting of a suitable restriction MATH to a locally order conserving star extendible homomorphism MATH. In the same way, the restriction of MATH to MATH may be lifted to a locally order conserving embedding MATH. Since MATH, where MATH is a composition of the given embeddings for MATH, it folows that MATH is actually locally order conserving. From REF we can replace MATH by an inner unitary conjugate to obtain MATH. Continuing in this way, we obtain a commuting diagram for the desired isomorphism. |
math/0006172 | In view of REF , it will be sufficient to show that if MATH are star extendibly isomorphic then not only is the induced isomorphism of MATH regular, but the commuting diagram isomorphism may be implemented by order conserving embeddings. However this follows immediately from REF . |
math/0006174 | This is just the statement that MATH. |
math/0006174 | With MATH defined as above, MATH is in the kernel of all simple roots MATH distinct from MATH, and thus, since MATH is primitive, MATH is an embedding of MATH into the center of MATH. Also, since MATH, if MATH, then MATH if and only if MATH, if and only if MATH. Thus MATH is the cyclic subgroup of order MATH in MATH, and so MATH where the image of MATH in the first factor lies in the center of MATH, and corresponds to the element MATH. To describe this central element, let MATH be a root of MATH. Then MATH if MATH and MATH. Thus MATH is the central element of MATH given by MATH. |
math/0006174 | We have MATH . Then MATH is the subgroup of MATH which is in the kernel of the character MATH. The restriction of MATH to MATH is non-trivial, and hence surjective, and for an element of MATH, written as MATH, we clearly have MATH. Thus, for each MATH there is an element MATH of the form MATH. This element MATH is in the center of MATH, since it is in the kernel of MATH for all MATH, as well as in MATH. It follows that an arbitrary MATH is of the form MATH, where MATH, as claimed. To see the second statement, note that MATH. Since MATH is in the kernel of all of the remaining roots, MATH lies in the center of MATH if and only if MATH. |
math/0006174 | Let MATH be the fundamental weight corresponding to the simple root MATH. The unique primitive dominant character of MATH is MATH, viewed as a character on MATH, and MATH. |
math/0006174 | Since MATH is primitive MATH for some integer MATH. Since the character MATH of MATH is given by MATH, the character MATH of MATH is given by MATH. The action of MATH is diagonal with respect to the decomposition of MATH as a sum of root spaces, and the character on the one-dimensional subspace spanned by a root MATH is simply the restriction of MATH to MATH. The space MATH is the subspace of MATH spanned by the set of all roots MATH with the property that MATH. Recall that MATH is in the kernel of all the simple roots except MATH. Thus, to compute the character of MATH given by the product of MATH over all MATH such that the coefficient of MATH in MATH is positive, we may as well take the product over all of the positive roots MATH. In other words, the character of MATH which gives the degree of the top exterior power is simply the character MATH. We can rewrite this expression as MATH. Thus, the character that we are computing is MATH. Recalling that the embedding of MATH is given by MATH, we see that the character MATH is given by raising to the power MATH. Thus MATH. |
math/0006174 | By definition, MATH as additive characters on MATH. By the previous lemma, MATH. On the other hand, MATH, where MATH lies in the MATH-span of the simple coroots MATH, MATH, and hence in the NAME algebra of the derived group MATH. Thus MATH and so MATH. |
math/0006174 | As before, MATH for some integer MATH. To compute MATH, note that, if MATH, then MATH acts on MATH via raising to the power MATH. Since MATH, we must have MATH . Hence, MATH. |
math/0006174 | Let MATH be the subset of MATH consisting of roots MATH such that MATH divides MATH. Clearly MATH is again a root system. Let MATH be the real span of MATH and let MATH be the subspace of MATH spanned by MATH. Then clearly MATH is the set of all roots which are linear combinations of elements of MATH. Thus MATH is a root system with simple roots MATH. By CITE, since MATH, there exists a set of simple roots for MATH containing MATH. In fact, the proof of this proposition shows that there are at least two different sets of simple roots, each of the form MATH. Then MATH, and since MATH, in fact MATH. Suppose for example that MATH. If MATH, then MATH and MATH. Thus MATH and MATH for all MATH, so that MATH is a lowest root for MATH. Suppose now that MATH is also a set of simple roots for MATH, where MATH. It follows that MATH. In this case, it is easy to check that MATH is a highest root for MATH. |
math/0006174 | Since MATH, MATH. Suppose inductively we have shown that MATH for all MATH. If MATH, then MATH is not a root. If MATH is not a root, then MATH is the highest root and MATH. Otherwise, MATH. |
math/0006174 | Clearly MATH has the property that MATH is a set of simple roots for MATH. By the proof of REF , MATH, proving REF . To see REF , note that MATH, and thus MATH. To see REF , write MATH, where MATH. Then MATH . On the other hand, MATH, and plugging this back in gives the first part of REF . The second part is proved in a very similar way. |
math/0006174 | The first equality follows since, by REF , MATH . The second follows since MATH. The third is an easy consequence of the fact that MATH, using MATH. |
math/0006174 | By REF , it suffices to prove that MATH. By REF , since MATH, MATH. Since MATH exchanges positive and negative roots in MATH, it follows that MATH is a permutation of MATH. Clearly, given MATH, MATH . Thus, since MATH permutes MATH, we have MATH . Next we claim: MATH . Suppose that MATH, and consider the MATH-string defined by MATH, say MATH. Since MATH, it is easy to see that MATH. If MATH is a negative root, then every root in the MATH-string is negative and thus none of them appears in the right hand side of the above equality. After reindexing, we can assume that MATH is the origin of the MATH-string. If MATH, then MATH for all MATH, and so the total contribution to the right hand side from the sum over the MATH-string is zero. If MATH, then the contribution to the sum on the right hand side is MATH. The remaining possibility for the right hand side is MATH, and in this case MATH. Thus we see that the right hand side in REF is equal to the left hand side. MATH . First note that MATH . We consider as before the MATH-strings defined by a root MATH which lie in MATH. If the origin of such a string lies in MATH, then so does every MATH lying in the string, and the sum over all such MATH of MATH is zero. Next we claim: If a MATH-string meets MATH but is not contained in MATH, then either: CASE: The origin of the string lies in MATH and all other elements of the string lie in MATH. CASE: The extremity of the string lies in MATH, and all other elements lie in MATH. Moreover, there is a length-preserving bijection between strings of types REF above. First note that, if MATH, then MATH cannot be a root. For then MATH would also be a root, necessarily negative, and then MATH would be an element of MATH higher than MATH. Thus, every MATH-string meeting MATH but not contained in it must either begin or terminate in MATH. Clearly, the only possibilities are REF above, and the bijection is given by sending the origin of a string of type REF to its negative, which is the extremity of a string of type REF . Returning to the proof of REF , the only nonzero contributions to the sum MATH come from CASE: MATH-strings whose origin is MATH, and these contribute MATH; CASE: MATH-strings whose extremity is MATH, and these contribute MATH; CASE: the root MATH and this contributes MATH. Summing these up, we see that MATH . Dividing by MATH gives the final formula of REF . To complete the proof of REF , we have MATH as claimed. |
math/0006174 | In the notation of REF , set MATH and let MATH, with simple roots MATH. Although MATH need not be irreducible, we can still define the integer MATH with respect to the root system MATH as in REF . By REF , which holds even if MATH is reducible, MATH, where MATH is the sum of the fundamental weights of the root system MATH. Applying REF , and using the notation introduced in its statement, we have MATH where MATH is the coefficient of MATH in MATH. Next we compute MATH. Write MATH, where MATH. Denote the fundamental weights of MATH by MATH, MATH, and MATH. Then it is easy to check that MATH . Thus MATH . Since MATH and MATH for all MATH, we see that MATH . Now an argument similar to the calculation of MATH above shows that MATH, the fundamental coweight for MATH dual to MATH, is given by MATH. Also, MATH for all MATH. Thus MATH for all positive integers MATH. It follows that MATH. Putting this together with the above gives MATH, which is the statement of the proposition. |
math/0006174 | For MATH, we have MATH and MATH, and the corollary is clear. |
math/0006174 | If MATH, it is easy to check that the above conditions follow from REF and the fact that MATH. The remaining cases are: MATH with MATH the root such that MATH, MATH with MATH the root such that MATH, or MATH with MATH a root such that MATH. These cases may be checked by hand. |
math/0006174 | By CITE, with MATH and MATH as above, every pair MATH such that MATH is conjugate to such a pair with MATH and MATH. This proves that the map is surjective. Clearly, it is finite-to-one. If MATH is in the interior of MATH, then it is regular, and the only further possible conjugation is via an element MATH, which acts on MATH via MATH. Thus, a fundamental domain for the map MATH is given by MATH, where MATH is a fundamental domain for the quotient map MATH. It follows that the degree of the map MATH is the product of the degree of the map from MATH to MATH with the ratio MATH, where volume is computed with respect to any NAME invariant metric. We consider these two integers separately. Let MATH. Then MATH . Moreover, the set MATH is an integral basis for MATH. Finally, for each orbit MATH, choose MATH and let MATH be the image of MATH in MATH. Then MATH . Proof. There is an exact sequence MATH where MATH. The homomorphisms in this sequence are equivariant with respect to the action of MATH. Moreover, MATH acts on MATH by a permutation of the basis. The proof of the lemma follows easily by considering the associated long exact sequence MATH . The torsion subgroup MATH, and MATH . The order of MATH is MATH. The natural map MATH is finite and surjective of degree MATH. Beginning with the short exact sequence of MATH-modules MATH we get a long exact sequence MATH . The quotient MATH. Since MATH is a finite group and MATH is torsion free, the induced map MATH is an isomorphism from MATH to MATH, and hence MATH. Moreover, it is clear that the degree of the map from MATH to MATH is the index of the image of MATH in MATH. By REF , it suffices to compute the order of the quotient of MATH by the relations MATH and MATH. Since MATH, it is clear that the quotient has order MATH. Now we compute MATH using the volume determined by the inner product MATH. The alcove MATH has vertices equal to MATH and MATH, MATH. By REF , the vertices are MATH and MATH, where the MATH are the dual basis with respect to MATH to the basis MATH of MATH. Now we have the following elementary lemma, whose proof is left to the reader: Let MATH be a simplex in MATH with vertices MATH. Let MATH be an affine linear transformation of MATH which acts via a permutation of the vertices of MATH. Suppose that the orbits of MATH on the vertices are MATH, with MATH. If MATH is the order of the orbit MATH, set MATH . Then the fixed set of MATH for the action of MATH on MATH is a simplex with vertices MATH . Applying the lemma to MATH, we see that MATH is a translate of the simplex in MATH spanned by MATH and MATH, where MATH and MATH is any representative for MATH. It follows by REF that MATH is an integral basis for MATH, where MATH. Since MATH we see that MATH and MATH are dual bases for the restriction of MATH to MATH. Now MATH where MATH is the parallelepiped spanned by the basis MATH and as usual MATH is any representative for MATH. On the other hand, MATH, where MATH is the parallelepiped spanned by the dual basis MATH. Thus MATH . To complete the proof of REF , the degree in question is the product MATH as claimed. |
math/0006174 | The dominant character MATH lifts to the character MATH on MATH. By CITE, MATH is the unique point MATH in the center of MATH such that MATH. Thus MATH, showing REF . The congruence condition MATH follows from CITE. REF follows from CITE. The remaining statements are clear. |
math/0006174 | On the level of NAME algebras, there is a direct sum decomposition MATH, where MATH, and the proof follows. |
math/0006174 | The first two statements are immediate from NAME on MATH. The final one follows from REF . |
math/0006174 | We compute the degree of the line bundle MATH. Let MATH be the dominant character for MATH. The line bundle MATH is associated to MATH by the character MATH. By REF , MATH lifts to the character MATH of MATH. Since MATH lifts to MATH on MATH and since the line bundle associated to MATH by the character MATH has degree MATH, the degree of MATH is MATH. A similar argument (or duality) handles the case of MATH. |
math/0006174 | By REF applied to MATH, the character of MATH defined by the determinant on MATH lifts to the character MATH on MATH. The degree of MATH is thus MATH. It follows that the slope of MATH is MATH. |
math/0006174 | It follows from the definition of a special root that MATH. By REF , there is an isomorphism MATH where the image of MATH is mapped to MATH and to MATH. From this, we must have MATH. The map from MATH to MATH which is the natural inclusion MATH and which maps MATH to MATH then factors to give an induced homomorphism MATH. It is clear from the construction that this induced homomorphism is injective and that its image is the subgroup of matrices of equal determinant. Let MATH denote the value of any of these determinants under the inverse isomorphism. For MATH, we see that MATH, and hence MATH. |
math/0006174 | If the center of MATH is trivial, then MATH is a primitive element of MATH, and hence MATH. This handles the cases MATH. Next suppose that MATH is simply laced and not of type MATH, so that the NAME diagram of MATH is a MATH diagram, with MATH or MATH with MATH. Let MATH, so that MATH, MATH in the respective cases above. In particular MATH. There exists a labeling of the roots as MATH, where MATH are ends of the diagram, MATH, MATH and similarly for the MATH and MATH, and MATH meet MATH, such that MATH . It follows that MATH is integral, and hence MATH, unless MATH and MATH is odd, in which case MATH. A similar argument handles the case of MATH. In case MATH, if we number the roots as in CITE beginning at the long end of the NAME diagram, then MATH and MATH . Thus MATH if MATH is odd and MATH if MATH is even. Finally, for the case of MATH, again numbering the roots in order as in CITE beginning at a short root, so that MATH, we have MATH . Thus MATH. |
math/0006174 | Recall that, for every MATH, there is a unique stable vector bundle MATH of rank MATH over MATH such that MATH. Given the structure of MATH as in REF , it is clear that there is a unique principal MATH-bundle, up to isomorphism, satisfying REF above, with MATH for every MATH. Since the vector bundles MATH in REF are simple, the automorphism group of each of these is isomorphic to the center of MATH acting by multiplication. It then follows that the MATH-automorphisms of MATH are given by the action of the center of MATH acting by multiplication. |
math/0006174 | The cases MATH and MATH follow easily from the explicit descriptions of the maximal parabolic subgroups and are left to the reader. In case MATH, the corresponding maximal parabolic of MATH is the set of MATH preserving an isotropic subspace of MATH of dimension MATH. The corresponding NAME factor MATH is the subgroup of matrices in MATH with equal determinant. If MATH is the representation of MATH induced by the standard representation of MATH on MATH and MATH are the two representations of MATH induced by the standard representations of the second and third factors of MATH on MATH, then it is easy to check that the representation of MATH on MATH which is the restriction of the standard representation of MATH is just MATH . The vector bundle associated to MATH is thus MATH . Moreover this is an orthogonal direct sum with respect to the induced form and MATH and MATH are isotropic subspaces. Furthermore, by a result of CITE, MATH, and since the each line bundle summand of MATH is not isomorphic to the dual of any other summand, the direct sum decomposition of MATH must be orthogonal with respect to the quadratic form and thus as described above. The case of MATH is similar. |
math/0006174 | First, by REF , MATH since MATH. Next we show that MATH has the following minimality property: If MATH is special, then MATH. If MATH is not special, then MATH. To see that MATH, it suffices by REF to show that MATH. First, by CITE, MATH is always a coroot. Clearly, MATH for all MATH, MATH if MATH is an end of the NAME diagram and MATH can only be a coroot if MATH is an end of the NAME diagram. These properties say that, if MATH, then MATH is not a coroot. Thus MATH. Now suppose that MATH is not of MATH-type, and hence that MATH is not of MATH-type. Then MATH is not the highest coroot of MATH. Thus there exists a simple root MATH such that MATH is again a coroot. By what we have just seen, we must have MATH. It follows that, for MATH, MATH is equal to MATH plus a sum of simple coroots. Hence MATH. Thus, we have proved the first statement in the theorem. To see the second, first assume that the NAME parabolic subgroup for MATH is maximal. In this case, we can assume that the NAME parabolic for MATH is MATH for some MATH. Thus MATH is isomorphic to MATH, where MATH is a semistable bundle of negative degree MATH on MATH. Now MATH with equality holding if and only if MATH, MATH, and MATH is special. In this last case, it follows from REF that MATH is a translate of MATH. Now suppose that the NAME parabolic for MATH is not maximal. There exists a maximal parabolic subgroup MATH such that MATH has a reduction to a MATH-bundle MATH, where MATH has degree MATH. By REF , MATH is unstable, for otherwise the NAME parabolic for MATH would be MATH, which is maximal. Hence, the associated MATH-bundle MATH is also unstable. Thus the vector bundle MATH is unstable of degree zero, and hence contains a semistable summand of negative degree. It follows that MATH. Applying REF , we see that MATH . This completes the proof in case the NAME parabolic for MATH is not maximal. |
math/0006174 | The group MATH acts on MATH with weight MATH. By REF , MATH . Thus, it suffices to show that MATH. If we define MATH then it clearly suffices to show that, for all MATH, MATH, in the notation of REF . By REF , the integers MATH have the circular symmetry property with respect to MATH and MATH, since MATH is a long root, and by REF , MATH. By the proof of REF, the integers MATH have the circular symmetry property with respect to MATH and MATH, where MATH. By REF, MATH. (We will give another proof of this fact in REF.) Clearly MATH. By REF , MATH for all MATH. |
math/0006174 | The minimally unstable points MATH are listed in REF. From this list, it is easy to check that MATH and MATH. To prove the remaining statements, we make a case-by-case analysis. MATH: We choose an identification of MATH with MATH in such a way that MATH with MATH with MATH being the standard unit vector in the MATH-coordinate direction. We write MATH as the image of an element of the form MATH for some MATH. We factor MATH where MATH and MATH. Then MATH is an element of order MATH. By CITE, the MATH-special roots are those MATH for which MATH. There are exactly MATH such roots. Suppose MATH. Then MATH. In particular, MATH. Since MATH is a special root for MATH, MATH, and hence MATH. On the other hand, since MATH acts freely on the NAME diagram for MATH we see that MATH. Let us consider the group MATH. By REF , MATH is isomorphic to the subgroup of MATH matrices of equal determinant. Hence MATH. The map MATH induces an identification MATH and the projection MATH sends MATH to the element MATH where MATH, respectively, MATH generates MATH, respectively, MATH. Direct computation shows that MATH and that the natural map MATH is multiplication by MATH. Thus, we have an identification MATH. Under this identification the element MATH is MATH. MATH and projects to MATH. Since MATH, the projection of this element to either factor generates that factor. MATH, MATH. The MATH-special root is the unique short root MATH in the NAME diagram. Direct inspection shows that MATH, and that MATH is a generator of the fundamental group. Thus, MATH and MATH generates MATH. Furthermore, MATH so that MATH. Since MATH acts on the extended NAME diagram for MATH with one free orbit and MATH fixed points, we see that MATH. MATH, MATH: Suppose first that MATH is odd. Then there is a unique MATH-special root, the unique long root MATH. In this case, MATH and MATH generates the fundamental group of MATH. Hence, MATH and MATH is the square of a generator for this group. Since MATH is odd, MATH is a generator of MATH. Since MATH is special for the simply connected form of the group, MATH. In this case the element MATH acts freely on the nodes of the extended NAME diagram so that MATH. Now suppose that MATH is even. Then there is a unique MATH-special root, the unique short simple root MATH which is not orthogonal to the unique long simple root. Direct computation shows that MATH is isomorphic to MATH and that MATH is the diagonal element MATH. Thus, MATH and MATH is a generator of this group. Furthermore, MATH and the map MATH is onto. Thus, the image MATH of MATH generates MATH, and hence its projection to each factor generates the fundamental group of that factor. Lastly, direct computation shows that MATH. Since MATH acts on the extended NAME diagram for MATH with one fixed point and MATH free orbits, we see that MATH. MATH, MATH: For MATH we identify MATH with the even integral lattice inside MATH. Let MATH be the standard unit vector in the MATH-coordinate direction. Then MATH where MATH for MATH and MATH. There are two MATH-special roots MATH and MATH. (Of course, these elements are interchanged by an outer automorphism of MATH.) Let MATH be one of the MATH-special roots. Then MATH and MATH is a generator of MATH. Thus, MATH and MATH is a generator of this group. Direct computation shows that MATH. Since MATH acts on the NAME diagram for MATH with two free orbits and MATH fixed points, we see that MATH. MATH, MATH and MATH is an element of order MATH: There is one MATH-special root. It is the simple root MATH corresponding to the ``ear" of the NAME diagram (that is, either MATH or MATH) with the property that MATH. In this case MATH. Hence MATH and MATH is a generator. Under the projection to MATH the image MATH of MATH is the square of the usual generator. This is clearly still a generator. Lastly, as above MATH whereas MATH. Thus, MATH. MATH, MATH and MATH is an element of order two not contained in MATH: There is one MATH-special root. It is the simple root MATH corresponding to the node of the ``long" arm of the NAME diagram next to the trivalent node. Thus, MATH is isomorphic to MATH where the cyclic group is embedded in the standard way in MATH and the usual generator maps to the standard generator of MATH and to the element of order MATH in MATH. Thus, we can identify MATH with MATH. The element MATH is the image of the element MATH where MATH and MATH are central elements of order MATH in MATH and MATH under the inclusion of MATH. Thus, MATH is isomorphic MATH. Hence MATH is a generator of MATH, the image MATH of MATH is a generator for MATH, and hence the projection of MATH into either factor generates the fundamental group of that factor. Direct computation shows that MATH. Since the action of MATH on the extended NAME diagram for MATH has one fixed point and MATH free orbits, we see that MATH. MATH: There is one MATH-special root. It is a simple root MATH corresponding to the node next to the trivalent node on one of the arms of length MATH with the property MATH. In this case MATH is isomorphic to MATH where the element in MATH that maps to MATH maps to the generator in MATH and to the square of the usual generator in MATH. Hence, MATH is isomorphic to MATH where the element in MATH that maps to MATH maps to the generator in MATH and to the usual generator in MATH. Thus, MATH, MATH is cyclic and MATH is a generator of this group. It follows that MATH generates and hence the image of MATH under projection to either factor is a generator of the fundamental group of that factor. Direct computation shows that MATH. Since the action of MATH on the extended NAME diagram of MATH has two free orbits and one fixed point, we see that MATH. MATH: There is one MATH-special root. It corresponds to the node of the NAME diagram next to the trivalent node on the arm of length MATH. In this case MATH is isomorphic to MATH, where the element in MATH that maps to MATH maps to the usual generator of MATH and the square of the usual generator of MATH. Thus, MATH where the element in MATH that maps to MATH maps to the inverse of the usual generator of MATH and the inverse of the usual generator of MATH. Thus, MATH, MATH is isomorphic to MATH and MATH is a generator. Consequently, MATH is a generator of MATH. Direct computation shows that MATH. The action of MATH on the extended NAME diagram of MATH has two fixed points and MATH free orbits so that MATH. |
math/0006174 | If MATH is cyclic and MATH is a generator, then MATH. The remaining cases are MATH and MATH, and these can be checked directly. |
math/0006174 | The weight for the action of MATH on MATH is MATH. By REF , the dimension of MATH is MATH. By REF , the integers MATH have the circular symmetry property with respect to MATH and MATH. By REF , MATH. If we define MATH then the integers MATH satisfy: MATH, and the MATH have the circular symmetry property with respect to MATH and MATH, where MATH is the maximum value of the MATH. It follows by inspection or from CITE that MATH, that is, MATH. By inspection, MATH. Thus MATH, and the proof follows. |
math/0006174 | Let MATH be the closed subgroup of MATH whose NAME algebra is MATH. Then the filtration MATH is a decreasing filtration of MATH by normal, MATH-invariant subgroups such that MATH is in the center of MATH for every MATH, and MATH. By REF of the appendix, it suffices to check that MATH. The second statement is clear since MATH, and the first follows from REF , which implies that MATH for every MATH. |
math/0006174 | Let MATH be represented by the MATH-cocycle MATH, where MATH is an open cover of MATH and MATH is a morphism. Then MATH acts on the cocycle MATH. Define morphisms MATH as follows: MATH . There is a MATH-equivariant morphism from the unipotent subgroup MATH to the affine space MATH (see for example CITE). Using this MATH-equivariant isomorphism, and the fact that all of the MATH-weights on MATH are positive, it is easy to check that the MATH are morphisms and so define a MATH-cocycle for the sheaf MATH over MATH. Thus, they define a bundle MATH over MATH, reducing to MATH mod MATH and such that MATH. By the functorial property of MATH, there is a morphism from MATH to MATH corresponding to MATH. Clearly, the image of MATH is the origin of MATH, and the image of MATH is exactly the MATH-orbit of the cocycle MATH. This proves REF . |
math/0006174 | Let MATH denote the dual numbers. The space MATH is the set of maps from MATH to MATH such that the closed point is mapped to the origin. By the functorial interpretation of MATH, such a morphism corresponds to a MATH-bundle MATH over MATH, which is the pullback of MATH, together with an isomorphism from MATH to MATH, and such that MATH restricts to MATH over the closed point. The second condition says that MATH is a first order deformation of the MATH-bundle MATH. Such deformations are classified by MATH . The first condition says that the corresponding first order deformation of the MATH-bundle MATH is trivial, or equivalently that the projection of the NAME class of MATH to MATH is zero. Thus we have defined a canonical map from MATH to MATH. Conversely, by reversing this construction, every element of MATH defines a first order deformation of MATH which reduces to MATH mod MATH, so that in fact the map from MATH to MATH is an isomorphism. Since this isomorphism is canonical, it is easily seen to be MATH-equivariant. The last statement is clear by construction. |
math/0006174 | Let MATH be the affine coordinate ring of MATH, where MATH is defined by MATH, and let MATH be a basis for the linear functions on MATH. The finite-dimensional subspace of MATH spanned by the MATH is contained in a finite-dimensional MATH-invariant subspace MATH of MATH, by the NAME lemma CITE (or by using the grading on MATH induced by the MATH-action). The map MATH is a MATH-equivariant map from MATH to MATH, and hence restricts to a MATH-equivariant map from MATH to MATH. Choosing a MATH-equivariant splitting of the map MATH defines a MATH-equivariant map MATH and thus a MATH-equivariant homomorphism MATH. Let MATH be the corresponding morphism. By construction, MATH has an invertible differential at the origin and is MATH-equivariant. Thus, MATH is injective in a neighborhood MATH of the origin, and the image of MATH contains an open set MATH about the origin. Since the weights on MATH are positive, every point of MATH lies in the MATH-orbit of some point of MATH. Thus MATH is surjective. Likewise, MATH is injective: if MATH, then since the closures of the MATH-orbits of MATH and MATH contain the origin, and the weights of the action on the tangent space at MATH are all positive, it follows that there is a MATH such that MATH and MATH both lie in MATH. By REF. But since MATH is injective on MATH, MATH, and hence MATH. It follows that MATH is a MATH-equivariant bijection from MATH to MATH and thus it is an isomorphism. |
math/0006174 | The MATH-weights are of the form MATH for MATH, and so the lemma is clear. |
math/0006174 | Let MATH be the set of points of MATH where MATH acts freely and effectively. We have seen that there is a lifted action of MATH on MATH, which in fact is free. The action of the isotropy group MATH of a point in the base on the fiber is via multiplication by elements of the center of MATH, and thus there is an induced MATH-bundle on the MATH-quotient of MATH with the desired properties. |
math/0006174 | By REF , if MATH is unstable and MATH, then MATH. As we have noted above, the same holds for a semistable MATH-bundle MATH which is not regular. Thus, a MATH-bundle MATH is semistable and regular if and only if MATH. To prove REF , we shall show that, for all MATH, MATH. Let MATH. We have the inclusion of the NAME algebra MATH in MATH. Clearly, viewing MATH as a representation of MATH, the vector bundle MATH is the same as MATH. Moreover MATH. Thus there is an exact sequence of vector bundles MATH . Now replacing MATH by MATH gives the corresponding exact sequence MATH since MATH. Furthermore, by REF in the simply connected case and REF in the non-simply connected case, MATH has dimension MATH. By semicontinuity, there is a neighborhood MATH of the origin in MATH such that, if MATH corresponds to a MATH, then MATH as well. As every point of MATH is MATH-equivalent to such a MATH, we must have MATH for all possible MATH. Next consider the exact sequence of NAME algebras MATH . There is the associated bundle sequence MATH . Since MATH is a direct sum of semistable bundles of negative degrees, MATH. It follows as before from semicontinuity and MATH-equivariance that MATH for all MATH. So MATH. Thus MATH with equality holding if and only if the map MATH is surjective. This can only happen if the natural homomorphism MATH is surjective on the connected component of the identity, which would say that every MATH in MATH lifts to an element of MATH. But then MATH must be a fixed point for the MATH-action, and hence MATH is the origin. Conversely, if MATH, then MATH and MATH, and as we have seen above, this statement implies REF . |
math/0006174 | As in the proof of REF , the tangent space MATH can be identified with bundles MATH over MATH which restrict to MATH over the closed fiber and reduce mod MATH to MATH. If MATH is given by the MATH-cocycle MATH, where the MATH take values in MATH, and MATH by the MATH-cocycle MATH, where the MATH take values in MATH, then it is easy to see that MATH is given by a MATH-cocycle MATH, where the MATH are also MATH-valued. Moreover MATH defines an element of MATH. In this way, we identify MATH with MATH. There is a long exact sequence MATH . The natural map MATH is the NAME map for deformations of the MATH-bundle MATH, and its kernel is the image of the coboundary map MATH. This kernel also contains the image of the differential of the action of MATH on MATH at MATH, which has dimension equal to MATH. Thus it follows by hypothesis that MATH is injective. The Killing form identifies the vector bundle MATH with the dual of MATH. In particular, MATH has a filtration whose successive quotients are stable bundles of positive degrees. Hence MATH and the natural map MATH is surjective. Now consider the commutative diagram MATH where the middle row and column are exact. A diagram chase shows that, if the map MATH is surjective, then so is the map MATH which is the statement of the theorem. The map MATH is given by the composition MATH and the above sequence is NAME dual to the sequence MATH . By the naturality of the connecting homomorphisms associated to the following commutative diagram of short exact sequences MATH the composition MATH is just the homomorphism MATH, which is injective by hypothesis. Hence by duality MATH is surjective, which completes the proof. |
math/0006174 | Since the MATH-action is via strictly positive weights, the differential of the action is injective at every nonzero point MATH, so the hypothesis of the previous theorem is satisfied. This proves the first statement. To see the second statement, the induced map from MATH modulo the tangent space to MATH to MATH is surjective. The dimension of MATH modulo the tangent space to MATH is MATH. Since, for every MATH, MATH, equality must hold, giving a new proof of REF , and the induced map from MATH modulo the tangent space to MATH to MATH is an isomorphism. |
math/0006174 | It is an elementary exercise to check that, if MATH acts linearly and with positive weights on two affine spaces MATH and MATH and if MATH is a MATH-equivariant morphism whose differential at the origin is injective, then there exist coordinates on MATH for which MATH acts linearly and such that MATH is a linear embedding, and the MATH-weights for the image of MATH can be determined from the differential of MATH at the origin. The final statement then follows from the differential computation. The differential of the morphism MATH is given by the inclusion MATH . Hence the image of the differential of MATH at the origin is the span of the eigenvectors of the MATH-action on MATH whose weights are divisible by MATH. Thus, the same ids true for MATH. |
math/0006174 | REF follows easily from the explicit description of the special root. To see REF , it follows from REF that, if MATH is a nonzero element of MATH, then the image of MATH in MATH is also nonzero. In particular, if MATH is the corresponding MATH-bundle, then MATH is semistable. It is easy to check that, in this case, the MATH-bundle corresponding to MATH is again semistable. Thus, MATH is a minimally unstable MATH-bundle, and so MATH is a MATH-special simple root for MATH. |
math/0006174 | We must show that the MATH-linearization on MATH is given by the character which is raising to the power MATH. To compute the MATH-linearization, it suffices to compute the action of MATH on the fiber of MATH over the origin, which is a fixed point for the MATH-action on MATH. The fiber over MATH is canonically MATH . Now, by REF , MATH . Since MATH is contained in the center of MATH, the action of MATH on MATH is trivial. By REF , MATH acts on MATH with weights MATH. Thus the action on MATH is via MATH. A similar argument (or duality) handles the case of the MATH-action on MATH. Putting these together, we get the power MATH. |
math/0006174 | The morphism MATH defined by MATH is MATH-equivariant, where MATH acts with all weights equal to MATH on the domain and with weights MATH on the range. Thus there is an induced cover MATH, and it is easy to check that the degree of this cover is MATH. There is always a natural inclusion MATH, and one checks that this inclusion is an isomorphism if MATH for every MATH. In this case, MATH . This proves the formula of REF . |
math/0006174 | This is immediate from REF , with MATH, MATH, and MATH. |
math/0006174 | There is a universal MATH-bundle MATH over MATH, which in fact arises from a universal MATH-bundle, which we shall also denote by MATH. One can describe MATH as follows. A MATH-bundle over MATH is the same thing as an element of MATH. The inclusion MATH induces an inclusion MATH and we take the image of the element MATH. As vector bundles over MATH, MATH where MATH is the pullback to MATH of the NAME bundle MATH over MATH, via the morphism induced from MATH from MATH to MATH. Thus, by functorial properties of determinant line bundles (compare CITE), it will suffice to show that, over MATH, MATH, where as before MATH. It is clear in any case that the inverse of MATH is represented by an effective divisor supported at MATH, and the only question is the length of MATH. A standard calculation using the NAME theorem shows that this length is one (compare CITE for the case of the trivial line bundle). |
math/0006174 | By REF , it clearly suffices to show that, for every MATH, we have an equality (under the obvious identifications) MATH . As such, this equality is a general fact about lattices MATH: suppose that MATH is a lattice and MATH is a homomorphism. There is an associated morphism MATH which we shall denote by MATH. We can define the divisor MATH for MATH, as well as the cohomology class MATH. To prove REF , it is enough to prove: The class of the divisor MATH is equal to MATH. Proof of REF . First assume that MATH is primitive. Then after a suitable choice of a basis of MATH we can assume that MATH and that MATH is projection onto the last factor. In this case, MATH and MATH, where MATH is a positive generator of MATH. Clearly, equality holds in this case. If MATH is not primitive, we can write MATH, where MATH is a nonnegative integer and MATH is primitive. In this case, MATH factors as the morphism MATH followed by multiplication by MATH on MATH, and so MATH is cohomologous to MATH copies of MATH. Likewise MATH and so the claim follows from the case where MATH is primitive. |
math/0006174 | The proof is similar to that in the simply connected case for REF , and we shall be a little sketchy. Suppose that MATH is a flat MATH-bundle corresponding to the MATH-pair MATH. Let MATH be a fixed, general line bundle of degree zero on MATH. We compute when MATH is in the support of MATH. As we have seen in REF, MATH where the MATH are the orbits for the action of MATH on MATH. Here MATH is a sum of certain torsion line bundles, MATH is a line bundle with holonomy MATH for any fixed choice of MATH, and MATH is the sum of the root spaces MATH, with the action defined by MATH. It follows that MATH is a direct sum of distinct line bundles of degree zero. Given a MATH-orbit MATH, let MATH be a choice of MATH. Next we construct a universal bundle MATH as in the simply connected case, along the lines of CITE. The construction outlined there shows that MATH where MATH is the projection onto the first factor, MATH is a fixed MATH-pair and, as in the simply connected case, MATH is the pullback to MATH of the NAME bundle MATH on MATH via the morphism MATH induced by MATH. The proof of REF then shows that the divisor MATH is reduced. For a general choice of MATH, there exist MATH and MATH depending only on MATH such that MATH is in the support of MATH if and only if there exists a MATH such that MATH . Thus, in cohomology, MATH corresponds to the element MATH . Now every MATH has the same restriction to MATH as MATH. Thus the above sum become MATH where MATH is the order of MATH. On the other hand, we have MATH. Clearly, the restriction of this form to MATH is MATH, and this completes the proof of REF . |
math/0006174 | Let MATH be the MATH-form corresponding to MATH. First suppose that MATH is diagonalizable with respect to some MATH-basis for MATH, corresponding to a given isomorphism MATH. Then MATH is of the form MATH where MATH, say, MATH is projection onto the MATH factor, and MATH is the positive generator for MATH. In this case, we can write MATH, where MATH is the generator on the MATH factor of MATH. Clearly MATH . In general, the statement makes sense for MATH-coefficients. Note that MATH and a basis element MATH is given by choosing the standard positive generator for MATH, together with a MATH-basis for MATH. Changing the MATH-basis for MATH to some new MATH-basis changes the element MATH by MATH, where MATH is the change of basis matrix. In particular, if MATH has determinant MATH, then MATH is unchanged. Now every quadratic form on MATH can be diagonalized via a MATH-basis such that the change of basis matrix relating the new MATH-basis to a MATH-basis has determinant MATH. Thus we may reduce to the case where MATH is diagonalizable, where we have already checked the result. |
math/0006174 | Let MATH be the degree of the covering MATH. We see by REF that it suffices to prove that MATH which we can rewrite as MATH . This is exactly the statement of REF . |
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