paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9905061 | It is enough to decode the approximate formulas corresponding to MATH . Since MATH, then MATH and MATH. It follows that MATH. Note that for every integer MATH, the constant function MATH belongs to MATH. Hence it follows from the Uniformity Theorem REF that there exists an integer MATH such that: MATH which decoded say... |
math/9905063 | This statement is certainly known. We reproduce a proof for the convenience of the reader. As a first step, the assumption implies that MATH is an open subgroup in MATH. Admit this for the moment. The MATH-rational maximal tori in MATH are given by maximal commutative semisimple subalgebras MATH in MATH which are stabl... |
math/9905063 | Suppose that MATH is a closed point such that the NAME torus MATH is an elliptic maximal torus in MATH over MATH. We know that each elliptic maximal torus MATH in MATH is given by a subfield MATH in MATH of degree MATH stable under MATH, hence MATH acts irreducibly on MATH. So the action of any open subgroup of MATH on... |
math/9905063 | Since the hypothesis does not change if one extends the base field MATH to a bigger number field MATH, it suffices to prove that the abelian variety MATH is simple over MATH. Recall that whether the abelian variety MATH is simple or not is a question on the existence of an endomorphism MATH of MATH such that MATH for s... |
math/9905064 | REF is a result of REF follows from REF. |
math/9905064 | Using REF for MATH and for any MATH proves MATH . REF with MATH shows MATH . Finally by REF , we have MATH . |
math/9905064 | By REF the composition of the linear map MATH and the canonical map MATH which sends MATH to MATH is a surjective homomorphism. Then REF shows that MATH is generated by MATH and MATH for MATH. |
math/9905064 | Note that MATH is spanned by MATH where MATH and MATH are integers. We can assume that MATH is a monomial MATH in MATH . For such MATH we define the length MATH to be MATH . As usual we define MATH. We prove by induction on the length of monomial MATH that MATH . If MATH, then MATH and it is clear. Suppose the lemma is... |
math/9905064 | It is enough to prove MATH. Recall the vector MATH from the proof of REF . We also use induction on the length of the monomial MATH. If MATH, it is clear. Let MATH be a positive integer and suppose that the claim is true for all monomials MATH with lengths strictly less than MATH. Now, consider MATH for MATH with MATH.... |
math/9905064 | Recall the definition of the circle operation MATH . Also recall that MATH and note that if MATH then either MATH or MATH in the sum. This immediately gives MATH for MATH. Thus we have MATH . It is easy to see that MATH . So MATH . Similarly, when considering MATH and MATH respectively, we obtain MATH and MATH . Add RE... |
math/9905064 | We prove the lemma by induction on MATH. When MATH, it is nothing but REF . Let us suppose MATH and decompose MATH as MATH where MATH and MATH with MATH and MATH for MATH . Note that MATH . By the induction hypothesis, MATH . So we have MATH . The proof is complete by the fact that MATH which follows from either REF or... |
math/9905064 | The proof is similar to that of REF . For any MATH and positive integers MATH set MATH . Then MATH is spanned by all possible MATH . We again use induction on MATH for a monomial MATH to show that MATH lies in MATH and MATH which are the left and right MATH-modules generated by MATH respectively. If MATH then by REF , ... |
math/9905064 | REF shows MATH . |
math/9905064 | Set MATH. Since MATH we have MATH . Note that MATH, that is, MATH . Combining REF together proves MATH . Finally, using REF we get the desired result. |
math/9905064 | We prove the lemma by induction on MATH. When MATH, it is clear. Let MATH and MATH with MATH . Then the relation MATH gives MATH which implies MATH . Thus MATH . This shows that MATH . The induction hypothesis then yields that MATH is spanned by MATH . |
math/9905064 | We first prove MATH. It is enough to show that MATH for MATH is expressed as a linear combination of MATH. Note from REF that MATH for any positive integer MATH where MATH . Since MATH is a linear combination of MATH by REF , the element MATH is also a linear combination of MATH for MATH . We also see from REF that the... |
math/9905064 | One could prove the lemma using the fact that MATH is a linear combination of the elements MATH's with a lengthy computation. Here we give a short proof by evaluation method discussed in the proof of REF . Consider MATH or MATH or MATH. Clearly MATH is a linear combination of MATH for MATH modulo MATH and acts triviall... |
math/9905064 | Write MATH as a linear combination of the elements MATH and MATH. Then the result is immediate. |
math/9905064 | Since all equalities in the lemma hold on any top levels of the modules MATH and MATH, it suffices to prove by REF that the left hand sides of the equalities belong to MATH. First we consider the case that MATH and MATH are all distinct. Then by REF , we know that MATH is expressed as a linear combination of the elemen... |
math/9905064 | We only consider the case that MATH or MATH here. The case that MATH and MATH are all distinct is treated in Subsection REF; see REF . We first prove that if MATH then MATH is a linear combination of MATH modulo MATH . By REF , it is enough to show that MATH is a linear combination of MATH modulo MATH . Recall that MAT... |
math/9905064 | Let MATH and MATH. REF shows that there exists a scalar MATH such that MATH . If MATH, the left hand side of REF acts trivially on the top level of the module MATH. So the evaluation of REF on this top level shows MATH, which proves MATH. Similarly, MATH. Next, the evaluation of MATH on the top level of the module MATH... |
math/9905064 | In the case MATH, MATH, it is clear. Suppose MATH or MATH, then since MATH by REF , the commutator belongs to MATH. Since MATH acts trivially on the top levels of the known irreducible modules by REF asserts that MATH . The relation MATH is proved by the same argument in the proof of REF . That is, we express MATH as a... |
math/9905064 | We first consider REF . Let MATH and MATH. We can assume MATH without loss of generality. Then by REF we have that MATH with some scalars MATH and MATH. Note that in REF case, element MATH acts trivially on the top levels of the modules MATH, MATH and MATH . This implies that all MATH and MATH are zero. Let us now prov... |
math/9905064 | By REF , MATH . We have already proved in REF that MATH . Using REF again shows that MATH . Similarly, MATH. Since MATH and MATH act trivially on the top levels of MATH and MATH, the corollary follows. |
math/9905064 | Since MATH for all MATH we see that MATH is a polynomial in MATH and MATH (compare REF ). In particular, MATH commutes with MATH. Then by REF , we have MATH . Similarly, MATH. Since MATH commutes with MATH by REF , we also have MATH. Thus the same argument in the previous paragraph shows that MATH. The rest of the equa... |
math/9905064 | Suppose MATH. Then REF or REF shows MATH. If MATH, then MATH. Since MATH lies in the subalgebra generated by MATH, we use REF (also see REF ) to show MATH. Then the evaluation method proves MATH. The rest of the equalities in REF - REF are proved similarly. |
math/9905064 | Since MATH by REF , we have MATH and MATH by REF . |
math/9905064 | Using REF , we see MATH. Similarly we prove MATH by REF . |
math/9905064 | It suffice to prove that MATH and MATH are ideals. But this is clear from REF . |
math/9905064 | First, we see from REF that MATH . Note that MATH and MATH or MATH is nonnegative if MATH. Thus MATH . We now deal with MATH . From REF , MATH . Since the weights of MATH and MATH are less than or equal to REF, we see that MATH with some scalars MATH and polynomials MATH and MATH . Evaluating REF on the top level of th... |
math/9905064 | Recall that MATH . So MATH . Now using MATH (compare REF ), we have MATH and MATH . Then REF shows MATH with some scalars MATH and polynomials MATH and MATH. Now the evaluation of REF on the top level of the module MATH shows MATH . This implies MATH and MATH. Thus we have MATH . Finally the evaluation of REF on the to... |
math/9905070 | Let MATH satisfy REF. Then, MATH . Let MATH and note that MATH . Moreover, note that MATH . Now suppose that MATH for MATH, MATH. If MATH, then MATH. Thus by REF we obtain, MATH by which we obtain the first part of the lemma. However, if MATH then by REF there is a MATH such that MATH . For this case, by REF we obtain ... |
math/9905070 | Except for the conditions given in REF, the claims made in the statement of this lemma are an immediate consequence of REF . To verify the conditions given in REF, we begin by letting MATH where MATH and MATH are defined in REF, and where MATH. Now, let MATH in REF and note that MATH . Moreover, if MATH, then also note... |
math/9905070 | Differentiating REF with respect to MATH and inserting MATH immediately yields REF. |
math/9905070 | With MATH and with MATH defined in REF, MATH satisfies REF. If MATH, then MATH. MATH has rank MATH, hence MATH is nonsingular. As a result, MATH . However, if MATH, then MATH . By REF, MATH; thus, by REF there is a MATH such that MATH. From this, we obtain MATH . Hence for MATH which satisfy REF, MATH . For all MATH an... |
math/9905070 | With MATH and MATH, we see that MATH by REF. With MATH defined in REF, then by REF , we can define MATH for MATH. As defined, MATH satisfies REF. Then for MATH, REF implies that MATH . This is equivalent to MATH . Given the invertibility of MATH shown in REF , we infer that MATH satisfies REF. The proof of this theorem... |
math/9905070 | Let MATH be the greatest value such that REF holds for MATH. We show that REF must hold for some MATH with MATH, thus proving MATH. The solution of REF, MATH, presumed to be defined on MATH, can be continued onto some MATH with MATH; MATH then satisfies MATH for MATH and for some MATH. For brevity, let MATH denote REF ... |
math/9905070 | Utilizing the standard connection between the explicit exponential solutions of the second-order NAME REF and the NAME REF (in the special case MATH, compare REF ), performing a conformal map of the type REF, and the variable transformations REF then yields the following solution for REF, MATH associated with the gener... |
math/9905070 | Though stated for elements of the NAME disk MATH, the proof of REF shows that the asymptotic expansion given in REF holds, uniformly with respect to MATH for MATH in MATH, for all elements of the NAME disk MATH as noted in REF. Note that the system REF is independent of the reference point MATH. Recall that MATH, defin... |
math/9905070 | In the following let MATH, MATH, and MATH. The existence of an expansion of the type REF is shown as follows. First one considers a matrix NAME integral equation of the type (compare CITE, CITE, CITE, CITE, CITE) MATH and observes that the solution MATH of REF satisfies MATH in accordance with REF. Next, introducing MA... |
math/9905070 | Clearly REF satisfies REF since MATH . Conversely, let MATH be a solution of REF and MATH a fundamental matrix of solutions of REF. Define MATH . Then MATH implies MATH . Thus, there exists a constant MATH matrix MATH (possibly singular), such that MATH with MATH a fundamental matrix of solutions of REF. Hence, MATH. |
math/9905070 | This is obvious from REF. |
math/9905070 | Define MATH by MATH then MATH due to the uniform nature of the asymptotic expansion REF for MATH varying in compact intervals. Next, introduce MATH then MATH . Using MATH a standard NAME iteration argument in REF then yields MATH and hence REF. |
math/9905070 | Define for MATH, MATH . By REF , MATH and hence by REF , MATH where MATH and MATH are fundamental matrices of MATH respectively, with MATH . By REF , MATH as MATH, MATH. Thus, as MATH, MATH, MATH for some constant MATH by REF, and REF. |
math/9905070 | Define MATH and apply REF with MATH, MATH. Then (in obvious notation) MATH as MATH, MATH, and hence the asymptotic expansion REF for MATH in REF coincides with that of MATH (this also applies to the case MATH). |
math/9905075 | Since MATH, MATH, MATH, MATH, we have MATH where MATH . Since the summation MATH is the same as MATH, we have MATH with MATH. Replacing MATH with MATH, the summation turns out be MATH and we have MATH where MATH . Note that from REF , we have MATH . As easily seen from REF , MATH and MATH vanishes if MATH and if MATH r... |
math/9905075 | Since MATH, MATH, MATH and MATH, we see that MATH vanishes except for the following four cases, which have already appeared in REF : CASE: MATH, REF MATH, REF MATH and REF MATH. We will only prove the first case because the other cases are similar. Noting that MATH we have MATH since MATH, MATH, MATH and MATH. Now sinc... |
math/9905075 | From REF , we only have to check that MATH. We have MATH but this coincides with MATH as shown below. We have MATH . On the other hand from CITE, we have MATH . Divided by MATH and taking the limit MATH, we have MATH . Therefore we have MATH completing the proof. |
math/9905075 | It is sufficient to show that MATH. Since MATH, MATH and MATH . But these two coincide since MATH vanishes unless MATH (the charge conservation law), completing the proof. |
math/9905075 | From REF , we have MATH . From the NAME - NAME equation for MATH these two coincide, completing the proof. |
math/9905075 | Since MATH, MATH, MATH, MATH and MATH, we have MATH completing the proof. |
math/9905075 | Noting that MATH and MATH commutes since they are diagonal, the first equality follows immediately from that in REF . The second equality follows from MATH completing the proof. Note that the lemma can also be proved by using CITE. |
math/9905075 | In fact we can show that MATH is of the form MATH by using REF repeatedly to `push' MATH and MATH from left to right (See the proof of REF ). Details are omitted. |
math/9905075 | This lemma comes from REF and the following three facts CITE (see also CITE). CASE: the NAME polynomial of a non-trivial torus knot is not trivial. CASE: the NAME polynomial is multiplicative under the connect sum. Therefore if MATH and MATH are non-trivial, then MATH is also non-trivial. CASE: if MATH is a knot obtain... |
math/9905075 | First note that every coefficient of both colored NAME polynomial and NAME polynomial as a power series in MATH is a NAME invariant. So a knot MATH with every NAME invariant trivial has the trivial colored NAME polynomial for any color and the trivial NAME polynomial. In particular MATH for any MATH. Therefore assuming... |
math/9905075 | We only prove the equality for MATH since the other case is similar. We use the following quantized NAME relation. MATH . Then since MATH we have the following recursive formula. MATH . Now the required formula follows since MATH for any integer MATH. |
math/9905076 | Suppose that MATH with MATH, or REF. We begin with MATH. Suppose that MATH splits off REF times from MATH. Then the residual system MATH. MATH is disjoint from MATH only if MATH, and so MATH implies that MATH. Thus we have the class MATH with virtual dimension MATH, but dimension MATH. Suppose that MATH splits off REF ... |
math/9905076 | If either MATH or MATH is empty then MATH is empty as well since the kernel systems are empty. If MATH and MATH are not empty, then MATH . The first equality follows from REF . The second is true because the systems are non-special and not empty. The third equality is REF a. The final inequality holds by REF. Therefore... |
math/9905076 | The proof relies on the identities from REF . We claim that with the given hypotheses, MATH. There are three possibilities. If both MATH and MATH are empty, then MATH since MATH. If both systems are non-empty and non special, then MATH and MATH. Then using the three identities we get MATH . The inequality holds since M... |
math/9905076 | We may assume that MATH and that MATH. The proof is by induction on MATH. Assume the theorem is true for smaller values of MATH. Then assume that MATH is not (-REF) special, and prove that it is non-special. Begin with the case MATH. We perform a MATH degeneration with MATH. This gives the following relevant systems. M... |
math/9905079 | Write each of the terms in REF as a multiple of MATH to get the equation MATH where MATH . It suffices to show that MATH. But this follows from the standard NAME identities MATH and MATH. |
math/9905079 | First we show that MATH is an integer. We use the well known fact that if MATH is even and MATH is odd, then MATH is even. If MATH is even, then obviously MATH is an integer, so assume that MATH is odd. Now if MATH is also odd, then MATH is even, so we may assume that MATH is even. Now one of MATH and MATH is even. REF... |
math/9905079 | Let MATH so that MATH is the MATH-entry of MATH. Then MATH satisfies the recurrence MATH . The preceeding recurrence was found by NAME. The theorem will follow if we can establish the correct values of MATH, MATH, and MATH. CASE: Maple computes MATH, and it computes MATH . Now MATH, and with MATH as input, the function... |
math/9905079 | First we show that each summand of the sum which defines each entry is an integer. It is well known that if MATH, MATH, MATH, then MATH, MATH, and MATH are all divisible by REF. Using this fact, we find that one of the terms MATH, MATH, or MATH is divisible by REF unless MATH and MATH. But now MATH, MATH, and MATH. Thu... |
math/9905080 | The proof consists to establish the differentiability of the following MATH-linear map MATH uniquely defined by: MATH . The map MATH is a formal sum of linear maps MATH with MATH being the identity map on MATH. We will show that the MATH's are differential operators. By REF for MATH and MATH. It is easy to see from REF... |
math/9905080 | Any star-product MATH on MATH is determined by the quantities MATH, MATH. Suppose that MATH is a MATH-covariant NAME star-product, then the star-exponential of MATH defined by: MATH coincides with the usual exponential MATH. The covariance property of MATH allows to use the NAME formula: MATH where MATH is the usual NA... |
math/9905080 | We just need to see what kind of graphs contribute to MATH, MATH. The graphs for MATH must be such that the vertices MATH and MATH receive only one edge, respectively. For MATH, we simply have the NAME bracket MATH. If MATH, we need to draw MATH edges in such a way that each vertex MATH, MATH receives at most one edge ... |
math/9905080 | As the vertex MATH can receive at most one edge, we distinguish two cases. CASE: The vertex MATH receives no edge. We will see that the vertex MATH must receive exactly MATH edges. If there are strictly more than MATH edges ending at vertex MATH, then there must be a vertex MATH, MATH, such that the edges MATH and MATH... |
math/9905080 | The form MATH, where MATH, is MATH. This easily follows from a simple recurrence using explicit expressions for the forms MATH. |
math/9905080 | The first statement follows directly from REF , and REF. The bidifferential operator for the graph MATH is MATH clearly it has constant coefficients and using REF for MATH, we see that the previous equation can be written as a trace of adjoint maps. |
math/9905080 | Recall that MATH is defined by MATH, MATH, MATH. It was shown that the MATH's in MATH are differential operators. Here we have only to solve the recurrence relation for MATH appearing in the proof of REF : MATH where MATH. According to REF , there exist differential operators MATH such that MATH. From REF it follows th... |
math/9905081 | Write MATH. Then MATH, so we have surjective maps MATH . The first map is MATH and the composition is MATH, proving the first inclusion. A similar argument works for MATH and MATH. |
math/9905081 | If MATH is a connected subgroup of the upper triangular matrices then MATH with MATH a torus and MATH unipotent CITE. An argument similar to that used in CITE implies that MATH so we assume MATH is a torus. Let MATH and MATH. We will show that the filtrations of MATH given by MATH and MATH give the same topology. CASE:... |
math/9905081 | The NAME group of MATH is MATH over any field, so by CITE MATH. In this case it is well known that MATH and MATH where MATH is the MATH-th elementary symmetric polynomial in the variables MATH. This fact obviously implies REF . REF follows from CITE applied to the maximal compact subgroups MATH and MATH. |
math/9905081 | By induction it suffices to prove the lemma when MATH. Let MATH be the disjoint union of MATH and MATH. Then MATH. The finite surjective map MATH gives a map of localization exact sequences MATH A diagram-chase shows that the map MATH is surjective. |
math/9905081 | Let MATH and MATH be the projections. By the homotopy property of equivariant MATH-theory the smooth pullbacks MATH and MATH are isomorphisms of MATH-modules. Since MATH is a regular embedding and MATH we have, by the compatibility of flat and l.c.i. pullbacks CITE, that MATH. Thus, the pullback MATH is an isomorphism ... |
math/9905081 | Let MATH be a separated algebraic space. Following CITE we define a NAME envelope MATH to be a proper morphism from a quasi-projective scheme MATH, such that for every integral subspace MATH, there is a subvariety MATH of MATH such that MATH maps MATH birationally to MATH. Using NAME 's lemma for algebraic spaces CITE,... |
math/9905081 | Let MATH be the quotient algebraic space. The proof of CITE extends to algebraic spaces and shows that MATH is a MATH-principal bundle. Thus, there is an equivalence between the categories of coherent sheaves on MATH and MATH-equivariant sheaves on MATH; that is, MATH. Under this isomorphism, MATH. The lemma now follow... |
math/9905081 | We prove only REF . The corresponding result about NAME groups has essentially the same proof. To show that the filtrations of MATH by the submodules MATH and by powers of the ideal MATH generate the same topology, there are two steps. CASE: We must show that given any pair MATH, there is an integer MATH such that MATH... |
math/9905081 | It suffices to show that given MATH and MATH with MATH and MATH greater than MATH the construction using MATH and MATH agrees with that using MATH and an open subset MATH of MATH. We can choose the open subset of MATH arbitrarily, provided the codimension is sufficiently large, so we take the open subset MATH. The cons... |
math/9905081 | REF follow from the definition of MATH and the non-equivariant NAME theorem of CITE. REF follows because, as in REF , the diagram MATH commutes. If MATH is the augmentation ideal of MATH, then MATH. Thus, by REF , MATH . Thus, MATH restricts to a map MATH . Taking the limit as MATH gives the desired factorization MATH ... |
math/9905081 | If MATH is a model for calculating MATH, then the model MATH for calculating MATH is a MATH bundle over MATH. Since MATH, the projection formula implies that MATH. Now if MATH is in MATH, then we can write (using the relation MATH) MATH, so MATH which is the desired formula. |
math/9905081 | Let MATH be a representation of MATH and let MATH be an open set on which MATH acts freely. Then MATH is a MATH-equivariant vector bundle on MATH. The group MATH acts freely on the open set MATH which surjects onto MATH. Identifying MATH with MATH we see that the maps MATH and MATH are the same. Using pairs of the form... |
math/9905081 | We will only prove REF for NAME groups; the arguments for NAME groups are similar. Let MATH be a MATH-space. Define a MATH-action on MATH by: MATH . We also define a MATH-action on MATH by: MATH . The projection MATH is MATH-equivariant, and moreover a MATH-principal bundle (MATH-torsor). The projection MATH is MATH-eq... |
math/9905081 | This follows from the argument of CITE, using the fact that MATH is a tower of projective bundles and the projective bundle theorem of CITE. |
math/9905081 | In general, if MATH is a smooth proper morphism with MATH, then the NAME theorem implies that MATH for all MATH. (Proof: By the projection formula we may assume MATH. Then by NAME MATH.) Applying this to the case where MATH and MATH are the mixed spaces MATH and MATH yields the result. |
math/9905081 | In view of the previous REF follows from REF from REF |
math/9905081 | The rings MATH are local, so MATH. Thus, MATH . Hence MATH. Since MATH, MATH is MATH-adically complete, proving the result. |
math/9905081 | The proof proceeds as in previous proofs, by building up from a torus to MATH to a general group. CASE: MATH is a torus. By NAME 's generic slice theorem CITE, there is a MATH-equivariant open subset MATH such that MATH is equivariantly isomorphic to MATH where MATH for a diagonalizable subgroup MATH, and MATH acts tri... |
math/9905081 | By the generalization of CITE to algebraic spaces there is a finite cover MATH on which MATH acts freely. Since MATH acts freely, MATH is generated by invariant cycles CITE. On the other hand, the proper pushforward MATH is surjective because MATH is finite and surjective. Therefore, MATH is generated by invariant cycl... |
math/9905081 | The kernel is just the kernel of the localization map MATH. |
math/9905081 | By the localization theorem for diagonalizable group schemes MATH CITE. Since MATH acts trivially on MATH we have CITE MATH . Let MATH be the augmentation ideal of MATH. Since MATH, MATH . Thus, by the NAME isomorphism of REF we have MATH . (Here the NAME character MATH makes MATH into a MATH-module.) The first stateme... |
math/9905081 | CASE: To show that the filtrations induced by powers of the ideals MATH and MATH induce the same topology, we must check two things. First, we must show that for any MATH, there exists a MATH such that MATH. This is clear because under the map MATH, the image of MATH is contained in MATH, so we can take MATH. Second, w... |
math/9905081 | By embedding MATH it suffices to prove the result for MATH. The corollary now follows by applying the theorem when MATH. |
math/9905081 | The action of MATH on MATH factors through the map MATH. Since MATH, REF implies that the ideals MATH and the MATH generate the same topology. Thus, MATH proving REF . By the projection formula applied to the commutative triangle MATH. Hence, MATH is continuous with respect to the MATH-adic topology, proving REF . |
math/9905083 | The cases MATH, MATH, and MATH are given in CITE; more precisely, MATH is given there as REF , while MATH and MATH are given in REF . (Note that if MATH, then MATH is a fixed-point-free involution with decreasing subsequences corresponding to the increasing subsequence of MATH.) We also give new, elementary, proofs bel... |
math/9905083 | For MATH, we have MATH . But MATH for MATH odd, so this is MATH as required. The calculations for MATH and MATH are analogous. |
math/9905083 | Using the NAME integration formula for the unitary group, we have MATH . The result follows from the standard theory of NAME determinants, or by the classic formula for the integral of a product of two (generalized) NAME determinants (see, for instance, CITE): MATH for any measure MATH on any set MATH. |
math/9905083 | As observed in REF, integrals over the orthogonal and symplectic groups can be expressed as NAME determinants; thus, for instance, MATH for any polynomials MATH with MATH. In particular, this must be true when MATH (NAME polynomials). In that case, noting that MATH, the MATH coefficient of the determinant is MATH . The... |
math/9905083 | Use MATH which follows from REF below. |
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