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math/9907152 | Consider the case of MATH; the argument for MATH is similar and we will omit it. Suppose that MATH and MATH . Recall the tree structure on the set MATH of MATH-pairs. If neither MATH or MATH is the parent of the other, then we have MATH and MATH, and there is a MATH with MATH and MATH. On the other hand, if MATH is the parent of MATH, then MATH and MATH are in MATH, and we get MATH for which MATH and MATH. For the last statement, just check that in both cases above we have found the only possible MATH for which MATH. |
math/9907152 | First, note that as a consequence of REF , we only need to show that monomials in the MATH's can be expressed in terms of monomials without valleys. Consider the monomial MATH with path MATH, and let MATH be a valley. If MATH, then we can apply REF to obtain another expression for MATH as a monomial of the same length without a valley in the MATH-th place. If MATH, then we have MATH. Now apply REF to express MATH as a product of terms MATH for MATH. Since all these terms are unipotent, we can use the substitution MATH to get an expression for MATH as a linear combination of monomials with length MATH and which only visit nodes MATH. In both cases, the only new valleys created are above MATH. So we can apply this process repeatedly, first getting rid of all valleys for which MATH, then for MATH, and so on. This process will terminate, since any monomials whose paths go outside MATH (respectively, MATH) are zero. |
math/9907152 | Suppose MATH is an irreducible representation. Choose a nonzero vector MATH. Let MATH be the longest path for which the corresponding monomial MATH acts nontrivially on MATH; one exists because of REF . Then for any MATH, we have MATH. If MATH is the irreducible representation for which MATH is one dimensional and all other MATH vanish, then there is a map MATH given by sending a generator of MATH to MATH. This contradicts the irreducibility of MATH unless the path was trivial and MATH. |
math/9907152 | It is enough to show that MATH. Clearly MATH is the image of the map MATH. Diagonal elements of MATH are in the kernel, and the elementary matrix MATH is in MATH unless MATH, MATH, in which case it maps to the elementary matrix with coordinate MATH in MATH. The result now follows from the following lemma. |
math/9907152 | Suppose MATH has MATH for some MATH. If MATH, then MATH, MATH are the standard basis elements corresponding to the column and row of the square MATH. By REF , we have MATH. Then acting on MATH by the matrix MATH kills the entry at MATH. Repeating this argument shows that MATH and MATH always lie in the same NAME cell. |
math/9907152 | The dual cone to MATH is a MATH-invariant irreducible variety, and hence the closure of an orbit MATH. |
math/9907152 | In CITE, Boe and Fu show that the MATH are cut out by conditions on the ranks of submatrices containing MATH and touching, but not crossing, the boundary of MATH. Since the orbits are determined by the ranks of all rectangular submatrices containing MATH, the orbit decomposition is finer. |
math/9907152 | It is not hard to see that MATH can be defined by using only conditions on the ranks of submatrices MATH where MATH is a rectangle with corners at MATH and MATH and where MATH satisfy MATH and MATH. The assumption of this lemma assures that such rectangles touch the boundary of the NAME diagram MATH, so the result follows from the characterization of normal slices to NAME cells in CITE. |
math/9907152 | MATH is spanned by the vectors MATH . A basis for MATH can be obtained by replacing MATH by MATH. Since MATH REF , there is a MATH for which MATH is spanned by the same basis as MATH with MATH replaced by MATH. |
math/9907152 | The first statement follows since MATH and MATH, whereas MATH is the number of pairs MATH with MATH and MATH. For the second statement, we use the following description of the NAME order on MATH (see CITE, page REF). Given MATH and MATH, MATH . Then MATH if and only if MATH for all MATH. If MATH is associated to the matrix MATH, then MATH is just the rank of the submatrix MATH with corners at MATH and MATH, so the ``if" part of the lemma is clear. For the other direction, note that for points MATH we can recursively find MATH: set MATH for all MATH. Then MATH . Thus knowing MATH holds for MATH will imply the same inequality for all MATH, giving the ``only if" part. |
math/9907152 | The first statement follows from REF . To see the second statement, assume first that MATH. Then MATH can be obtained from MATH by interchanging rows MATH and MATH and columns MATH and MATH, and then removing the point MATH. Thus MATH, as required. If MATH, we must have either MATH or MATH and a similar argument holds. For the last part, take MATH, MATH so that MATH is a codimension one subvariety of MATH, and let MATH be the corresponding permutations. By REF we have MATH and MATH. A basic result on reflection groups gives that MATH, where MATH is a reflection. In other words, MATH is obtained from MATH by interchanging two rows and restricting back to MATH. Say the the MATH-th and MATH-th rows are interchanged, and that MATH, MATH. Then there is no point MATH with MATH and MATH, MATH, since if there were, we would have MATH. This gives the required minimality of MATH. |
math/9907152 | Say MATH, and take a maximal element MATH. Suppose MATH, and define a matrix by MATH, MATH, MATH for all other MATH. Then MATH is the required orbit. |
math/9907152 | By REF , we need to show that MATH. Take a point MATH. Let MATH. By REF , we have MATH. If MATH is a MATH-pair, we are done. Otherwise, there are MATH-pairs MATH and MATH. By REF , one of these pairs nests inside the other one. Suppose that MATH. Then MATH satisfies MATH, and MATH, since MATH. Similarly, the other case gives MATH with MATH. |
math/9907152 | The first statement follows from REF . For the second, use REF . |
math/9907152 | If MATH, then MATH gives a normal slice to MATH at MATH. It is easy to construct a two-dimensional torus MATH which fixes MATH, preserves MATH, and induces the normal crossings stratification on MATH. Since there must be at least one codimension one orbit MATH with MATH by REF , the stratification on MATH induced from the orbit stratification of MATH must be either the normal crossings stratification or the stratification by a complete flag. |
math/9907152 | Take a codimension two orbit MATH. Denote the points in MATH by MATH, MATH. There are two cases: CASE: MATH and MATH. Since MATH must be obtained from MATH by two applications of REF , we have MATH for MATH; neither MATH or MATH can be the parent of the other. The dual orbit MATH contains the MATH matrix with support MATH. We can use REF to get MATH so that MATH - if the points in MATH are comparable in the partial order on MATH, this is immediate; otherwise it follows from REF . An easy argument along the lines of REF shows that MATH and MATH. CASE: The points of MATH are in the same row or column. The argument is the same in both cases, so assume Without loss of generality that MATH with MATH. By REF we must have MATH for MATH. Since MATH, we must have MATH, where MATH is the parent of MATH (note that MATH may not be in MATH). The dual orbit MATH contains MATH, so REF gives a partition MATH as before. REF now shows that MATH is the element MATH in the proof of REF . Finally, the analysis in the proof of REF shows that all diamonds arise either by REF or REF . |
math/9907152 | Recalling the definition of MATH, we see that points in MATH are of two types: first, all points MATH with MATH, and second, points MATH where MATH, MATH run over all points in MATH which lie on the diagonal, taking these points in pairs (the first and second, then the third and fourth, and so on). It is clear from this description that there is a MATH matrix MATH with the required support, and that MATH. |
math/9907152 | This follows directly from REF . Also see REF . |
math/9907152 | This is easy to see when MATH is contractible, since the bundles MATH and MATH can be trivialized. For the general case, cover MATH by contractible open sets containing MATH. Since extensions from MATH to MATH are unique and canonical if they exist, by REF , the extensions over these open sets glue together. |
math/9907152 | Fix orientations of MATH and MATH, and give the dual bundles MATH and MATH the compatible orientations, as in the last section. Using the description of MATH from the last section twice, we see that an object MATH is described by local systems MATH on MATH for MATH, together with maps (making identifications between these spaces as in REF ) MATH; these maps must satisfy relations as in REF . In particular, the monodromies of MATH are completely determined by what these maps do on stalks of the MATH. Applying the NAME transform and using REF , we see that objects in MATH are given by local systems and maps MATH, with relations as before. We can glue these two descriptions using the fact that MATH is a stack to get a description of the category MATH; the key fact is that there are natural identifications MATH, for MATH and MATH. This follows easily from existence of natural isomorphisms REF MATH, where MATH denotes the NAME transform functors MATH, MATH or MATH, and similarly MATH is the NAME transform in the second coordinate. Now fix a basepoint MATH; let MATH be the corresponding points in the other smooth components. The vector space MATH in the quiver REF is the stalk of MATH at MATH, and the map MATH is induced from the map MATH. The maps on stalks determine the monodromies of all four local systems, and the quiver relations arise because the MATH are local systems, and the maps are maps of local systems. The description of the category MATH is standard; see CITE, CITE. |
math/9907152 | The restriction functor MATH is a full embedding of categories, by REF . Thus it is enough to show that every isomorphism class of objects of MATH is in the image of this functor, or in other words that NAME 's conjecture holds for our varieties. We first show that every simple object is in the image. By REF an object MATH in MATH or MATH is simple if and only if MATH. Thus there is one simple object MATH (respectively MATH) for each MATH (respectively, MATH). We show by induction on MATH that MATH. If MATH, then MATH is a point and the result is clear. Now, assuming it holds for all MATH with MATH, we need to show that MATH is irreducible; this is enough, since MATH is one-dimensional. Suppose MATH were not irreducible. Then MATH has a composition series consisting of one copy of MATH and only using MATH for MATH, and in particular there is a nonzero map between MATH and some MATH with MATH. Using the induction hypothesis, REF , and REF , we get a nontrivial morphism between MATH and MATH, a contradiction. Finally, to complete the proof, use REF and induction to show that any quiver object is in the image of MATH. |
math/9907152 | Let MATH be the triple corresponding to the simple perverse sheaf MATH. The NAME local system MATH is trivial, so we need to show that MATH is also trivial. There is an action of the fundamental group MATH on MATH. This action must be trivial, since MATH is simple, and the local system MATH is trivial. The torus MATH acts on MATH and MATH in a manner compatible with the action on the base MATH. The action of MATH on the stalks of MATH must therefore be trivial. Since MATH generates MATH, it is enough to show that MATH has trivial monodromy around a loop in a fiber of the projection MATH. But by REF, this monodromy is MATH, where MATH. The proposition thus follows from the following result, which is a direct consequence of the definition of the relation MATH from REF. |
math/9907152 | Essentially this is just gluing together the presentations of the categories MATH given by REF along the NAME local systems of REF . Given MATH, The vector space MATH in the associated quiver object will be the stalk of the NAME local system MATH at a point MATH. The map MATH is the action of the monodromy around the loop MATH. Since these loops generate MATH, these maps completely describe MATH. Relations REF (respectively, REFs) in the quiver descriptions of MATH and MATH give the relations among these loops, by REF . The categories MATH have quiver descriptions using REF . We choose the orientation of MATH to be MATH times the standard (holomorphic) one, where MATH. This choice will make the description of the codimension two relations simpler - see REF gives maps between the stalks of the local systems MATH and MATH at points MATH, MATH in MATH and MATH. We can identify these stalks with MATH and MATH, by fixing paths from MATH to MATH in MATH and from MATH to MATH in MATH, and using REF . This defines the maps MATH and MATH in the quiver. The fact that the maps in REF are maps of local systems shows that the maps MATH commute with the maps MATH. The quiver relation REF (respectively, REFs) comes from the relations in REF , remembering the calculation of a loop around a point of MATH in REF. This defines the required functor MATH, and in fact shows that it factors through an equivalence of categories MATH, where MATH is the cover of MATH consisting of MATH and the sets MATH defined in REF. The proposition now follows since MATH is a stack. |
math/9907152 | By REF , an object MATH extends to an object in MATH if and only if each MATH extends to an object in MATH; now use REF . |
math/9907152 | Let MATH, so MATH. Consider the diagram MATH . Here the leftmost horizontal maps are the indicated specialization functors, MATH and MATH for MATH, MATH, and all the other functors are the appropriate NAME transforms. There are natural isomorphisms making the left square and the triangle commute by REF . The right quadrilateral commutes by the functoriality of the NAME transform REF . The sheaf MATH is given by taking MATH and mapping down and across the bottom to MATH. Following the upper path, we get a natural isomorphism MATH where MATH is the composition of the natural projection and identification maps. The result follows now from REF . |
math/9907152 | Consider MATH as a vector bundle over the base MATH, so that MATH is a sub-vector bundle. Then the vanishing cycles functor MATH is naturally equivalent to MATH, where MATH is the specialization functor, MATH is the NAME transform MATH, and MATH is given by MATH. Now we do a diagram chase similar to the one in REF Consider the following diagram of functors: MATH where MATH is the subbundle of MATH which annihilates MATH, the map MATH is given by MATH (putting MATH), and all the unmarked arrows are the appropriate NAME transform functors. Arguing as in REF , we see that that there is a natural isomorphism MATH . Now, using the assumption that the vector bundles MATH and MATH are one-dimensional, a simple argument shows that there is a natural isomorphism MATH . But then for any MATH, MATH is naturally isomorphic to MATH, which gives the result. |
math/9907153 | We prove the easier direction first. If MATH is a NAME MATH-space then the orbits are closed by CITE and thus MATH is CCR CITE. Because the action is free the map MATH induces a homeomorphism of MATH onto MATH CITE. Fix MATH and let MATH be an open wandering neighbourhood of MATH in MATH. The action of MATH on MATH is proper by CITE. Because the action is free the algebra MATH has continuous trace CITE, and since MATH is a MATH-invariant ideal of MATH this crossed product embeds naturally as an ideal MATH in MATH CITE. Viewing MATH as a representation of MATH shows that it is a Fell point of the open subset MATH of MATH. Hence MATH is a Fell point of MATH, and MATH is a Fell algebra. Conversely, we show that if MATH is not a NAME MATH-space then MATH is not a Fell algebra. Some of the ideas come from the proof of CITE. Suppose there is a point MATH in MATH with no wandering neighbourhood. To see that MATH is not a Fell algebra it is enough to produce an ideal MATH of the form MATH which is not a Fell algebra. We may assume that MATH has bounded trace, because otherwise it is certainly not a Fell algebra. To prove MATH is not a Fell algebra, we construct a net of irreducible representations MATH of MATH parametrised by a directed family of neighbourhoods MATH of MATH, which converges to its unique cluster point MATH, and a positive element MATH of the NAME ideal of MATH such that MATH does not converge to MATH. The minimality of the NAME ideal among dense ideals of MATH then implies that the ideal MATH described in Equation REF cannot be dense in MATH. It follows from CITE that MATH is not a Fell algebra. The construction of the element MATH and the net MATH have to be carefully intertwined to ensure that we can compare the traces by finding appropriate eigenfunctions. We begin by building some functions in MATH which will later be used to define MATH. Let MATH be a compact neighbourhood of MATH in MATH and MATH an open symmetric neighbourhood of MATH in MATH with compact closure MATH. Choose MATH which is identically one on MATH. For MATH we define MATH. Let MATH be a symmetric compact subset of MATH such that MATH. Choose a non-negative function MATH in MATH such that MATH whenever MATH and MATH. We can round off MATH so that MATH whenever MATH, where MATH is a compact set containing the support of MATH. We would like to use the function MATH to define the element MATH, but this function will in general not be compactly supported; thus we will later multiply it by the function MATH. Let MATH. By CITE there exists an open neighbourhood MATH of MATH, such that for every MATH, the support of MATH is contained in MATH. To see this, let MATH be the map MATH. Observe that MATH is compact and is contained in the open subset MATH of MATH. For each MATH there exist open neighbourhoods MATH of MATH and MATH of MATH such that MATH. Choose a finite subcover MATH and set MATH. Then MATH . Thus for every MATH, the function MATH is well-defined and continuous on MATH. This function will be an eigenvector for the element MATH to be constructed. Without loss of generality we may suppose MATH. Since no neighbourhood of MATH is wandering, for every neighbourhood MATH of MATH there are MATH and MATH such that MATH. Thus we get a net MATH directed by the neighbourhoods of MATH which converges to MATH, and by passing to a subnet if necessary, we may assume that this net is universal CITE and that MATH for all MATH. Because MATH has bounded trace and MATH is a convergent net, the limit set of MATH is discrete in the relative topology CITE. The map MATH induces a homeomorphism of MATH onto MATH, so there exists a neighbourhood MATH of MATH in MATH such that MATH is the unique limit point of the net MATH in MATH. We write MATH for the quotient map and let MATH. Since MATH is open we may assume that MATH. Observe that MATH is the continuous image of a universal net, and hence is universal; thus MATH is properly convergent. We will show that the ideal MATH of MATH is not a Fell algebra. Recall that our strategy to do this is to construct an element MATH in the NAME ideal of MATH such that MATH does not converge to MATH. Let MATH we will obtain the element MATH by pushing enough of MATH into the NAME ideal of MATH. The right NAME measure MATH on MATH given by MATH satisfies the hypothesis of CITE, so MATH is unitarily equivalent to the representation MATH on MATH given by MATH for MATH and MATH. Note that MATH so that MATH is a NAME operator with kernel MATH in MATH. By composing translation on the right by MATH with MATH we see that MATH is a well-defined continuous function on MATH whenever MATH. Note that MATH and MATH have disjoint supports because MATH. The MATH's MATH are linearly independent eigenvectors for MATH; as in the proof of CITE one shows that MATH for all MATH. The eigenvalue corresponding to MATH is MATH . If MATH then MATH because MATH implies MATH and MATH was chosen to be identically one on MATH. It follows that if MATH then MATH is a positive compact operator with eigenvalue MATH greater or equal to MATH of multiplicity at least MATH. On the other hand, MATH is the rank-one operator MATH with eigenvalue MATH. Thus MATH is a rank-one operator with eigenvalue MATH. Finally, we push enough of the operator MATH into the NAME ideal of MATH using the description in CITE. Let MATH be any non-negative function in MATH such that MATH . Now MATH is a positive element of the NAME ideal of MATH. If MATH then MATH and MATH. The existence of such an element implies that MATH and hence MATH are not Fell algebras. |
math/9907153 | If the action is free and MATH is Type I then every closed ideal is of the form MATH for some open MATH-invariant subset of MATH by CITE. Now apply REF to MATH. |
math/9907157 | Let MATH and MATH be two distinct fixed points of MATH. Then MATH. And conversely. |
math/9907157 | Let MATH be the single point that is the image of MATH under the constant map MATH. Then MATH is a fixed point of MATH. Furthermore, if MATH and MATH are fixed points of MATH, then MATH. |
math/9907157 | By induction on MATH. Let MATH be a product of MATH factors from the family MATH which is non-zero (the result being trivial if MATH). Let MATH be the kernel of MATH. MATH is non-zero, of dimension less than MATH, and invariant under MATH, since MATH for any MATH in MATH. By induction, there is a non-zero MATH annihilated by MATH. The family MATH acts on MATH. Lift a strictly upper triangularizing basis of MATH to MATH and let MATH. Let MATH and MATH for MATH. Then MATH for MATH. Thus MATH is a basis of MATH in which all members of MATH are strictly upper triangular. Since the product of any MATH strictly upper triangular matrices is zero, it follows that MATH. |
math/9907157 | CASE: Suppose MATH, with MATH triangular and with MATH for an invertible matrix MATH. Then MATH and the matrices MATH are unit upper triangular. Thus the family MATH, for MATH, is similar to a family of strictly upper triangular matrices, and hence strongly nilpotent. CASE: Let MATH be a composition product of maps MATH. By the chain rule, MATH, where MATH. By induction, MATH is a product of MATH matrices that are Jacobian matrices of the factors MATH evaluated at various points. Factors that are translation-dilations yield scalar factors (multiplication by MATH) in the matrix product, and can be moved to the front. If there are MATH or more factors MATH equal to MATH, the matrix product is zero by strong nilpotence. But MATH implies that MATH is constant. CASE: Let MATH be MATH points in MATH. Let MATH for non-zero MATH. Consider the composition product MATH. By assumption it is constant. Its Jacobian matrix at a point MATH is the product of MATH matrices, MATH of which are scalar multiples of the identity REF and MATH of which have the form MATH for some point MATH. Since MATH and the product of factors corresponding to translation-dilations is a non-zero multiple of the identity matrix, the product of the remaining factors is zero. Now vary the MATH and take a limit in which all MATH tend to zero to obtain MATH - that is, strong nilpotence. |
math/9907157 | Suppose that MATH and MATH are (globally) conjugate. Because the fixed points of MATH are nilpotent (that is, MATH is nilpotent at each fixed point), the fixed point set of MATH consists of isolated points (see remark in REF), hence is countable. If MATH has MATH as an eigenvalue, then its fixed point set is uncountable (as it contains all multiples of an appropriate eigenvector). Thus MATH does not have MATH as an eigenvalue, hence its fixed point set consists of MATH alone, and thus MATH has a unique fixed point. |
math/9907157 | Since MATH is bounded, so is its closure. Let MATH be a disc (closed ball) large enough so that the closure of MATH is contained in its interior. Let MATH be a concentric larger disk. Let MATH be the closed annulus between the MATH spheres MATH and MATH. Then MATH is homeomorphic to MATH. The restriction of MATH to MATH maps the MATH sphere MATH into the contractible space MATH (because MATH contains the entire image of MATH). So it is homotopic to a constant map. Interpret the homotopy as a map of MATH into MATH which coincides with MATH on MATH and is constant on MATH. Let MATH be the constant terminal value of the homotopy on MATH. Now define a map MATH on MATH as follows: it coincides with MATH on MATH, it is the selected homotopy on MATH, and it is the constant MATH on MATH. By construction, MATH is continuous (and it could be made MATH as well), and MATH. Since MATH, the only fixed points of MATH are in MATH, hence they are those of MATH. The homology groups of MATH with integral coefficients are MATH and MATH, with all others MATH. The induced map MATH on MATH is the identity. On MATH it is multiplication by the degree of MATH. Since MATH factors through a map to MATH, which is contractible, MATH is MATH. Therefore the NAME number of MATH is MATH. So MATH has exactly one fixed point, and thus so does MATH. |
math/9907157 | Consider the map MATH. Then MATH has no zero eigenvalues anywhere, and so the Jacobian determinant of MATH vanishes nowhere. The leading forms of the components of MATH and of MATH are negatives of each other. By NAME 's Theorem, MATH is a diffeomorphism, so the equation MATH has a unique solution, which is also the unique solution to MATH. |
math/9907157 | If the algebraic degree, MATH, of MATH is zero, then MATH is constant, and hence it has a unique fixed point. If MATH, then the Jacobian matrix of MATH is constant, with no eigenvalue of MATH. Thus MATH is an affine map with a nonsingular linear part, so it is bijective, and hence MATH has a unique fixed point. If MATH, the fact that MATH has no zeros at infinity implies that the same is true for MATH. Though MATH may have some components MATH of degree MATH, NAME 's condition must be satisfied, since the leading forms of the components of degree MATH have no common non-trivial zero. So MATH is a diffeomorphism, and MATH has a unique fixed point. |
math/9907157 | Obvious induction. The last claim uses MATH. |
math/9907157 | The matrix in question consists of the first MATH rows and columns of MATH. Temporarily fix the values of the parameters. With MATH fixed, apply the chain rule to see that the matrix is MATH, where the leading principal submatrix of rank MATH of MATH is nilpotent of index MATH (by induction) and the last row of MATH is MATH. It follows that MATH and hence the same is true for MATH. Since that is true for every fixed value of MATH, the desired result holds. |
math/9907157 | MATH, where MATH and MATH. By induction, and the fact that all the parameters are also parameters for MATH, it follows MATH and hence that MATH has the desired form. |
math/9907157 | Let MATH. MATH, where MATH and MATH. MATH with MATH and MATH. By the previous lemma, and the fact that parameters for MATH are parameters for MATH, it follows that MATH. By induction, the result of raising this transformation to the composition power MATH depends only on the parameters for MATH - that is, on MATH and MATH. Denote the power by MATH. Then MATH. But MATH since all components of MATH are zero from the MATH-th on. So MATH depends only on MATH. |
math/9907157 | Let MATH. By REF - REF , the map MATH is in MATH. By REF , its MATH-th composition power is constant. Since this is true for every MATH, it follows from REF that MATH (composition power). |
math/9907166 | We calculate by using REF that MATH . |
math/9907166 | First assume that MATH is the character of a representation MATH of MATH. We calculate the character value of MATH at MATH first. Take MATH where MATH and MATH is a MATH-cycle, say MATH. Denote by MATH a basis of MATH, and we write MATH, MATH . It follows that MATH in which the coefficient of MATH is MATH . Thus we obtain MATH since MATH is in the conjugacy class MATH which means MATH lies in MATH. Since the sign character of MATH takes value MATH at MATH, we have MATH . Given MATH, where MATH and MATH, we clearly have MATH . Thus it follows that if MATH is of type MATH, then MATH where MATH. Putting REF into a generating function, we have MATH . In a similar manner we can prove REF by using REF . Note that REF can be obtained from REF by substituting MATH with MATH and MATH with MATH. Let MATH be the characters of two representations of MATH. It follows from REF that MATH . Therefore the proposition holds for MATH, and so for any element MATH. |
math/9907166 | By REF , the character value of MATH at an element MATH of type MATH is MATH . Thus it follows by definition of the weighted bilinear form MATH . Here MATH denotes the conjugacy class in MATH of type MATH. By REF , we see that MATH . Since MATH consist a MATH-basis of MATH, we have established that MATH is an isometry. |
math/9907166 | The first and second identities were essentially established in REF together with REF . The only minor difference is that the components appearing in these two identities are regarded as operators acting on MATH and MATH. Note that by definition the adjoints of MATH and MATH with respect to the bilinear form MATH are MATH and MATH, respectively. The third and fourth identities are obtained by applying the adjoint functor to the first two identities. |
math/9907166 | We first compute MATH . The second identity above uses the formula MATH since MATH here commutes with MATH and MATH. On the other hand, we have MATH . Combining with REF we compute MATH . |
math/9907166 | The identity for the product of two vertex operators in REF implies the commutation relations between the components of vertex operators either by formal calculus or the NAME residue formula (see CITE). Taking into account REF one deduces that the correspondence defined in REF indeed realizes a representation of the toroidal algebra MATH, (compare CITE). |
math/9907166 | The NAME algebra structure REF implies that the nonzero elements MATH of distinct indices generate a spanning set in the space MATH. To see that they correspond to NAME functions we compute in a way similar to REF that MATH . The method in CITE implies that this is exactly the generating function of NAME functions under the isomorphism MATH. Since the vertex operators corresponding to different MATH commute up to a sign it follows that the isometry MATH maps the basis vector MATH to MATH . The orthonormality follows readily from the NAME algebra commutation relations in REF . |
math/9907167 | Let MATH. Write MATH, MATH, where MATH. By REF we get MATH . The proof is finished. |
math/9907167 | Iterating REF we get MATH . So, MATH . Fix now MATH and take a point MATH where the function MATH takes on its maximum. In view of REF , for every MATH we have MATH . Hence, iterating REF as before, MATH . So, MATH. The proof is finished. |
math/9907167 | Since by REF MATH and consequently MATH for all MATH, we have MATH for all MATH and all continuous functions MATH. Since this equality extends to all bounded measurable functions MATH, we get MATH for all MATH, all MATH, and all NAME sets MATH. Now, for each MATH set MATH. Then MATH for all MATH. Thus apllying REF to the function MATH and later to the function MATH, we obtain MATH . Hence MATH. The proof is complete. |
math/9907167 | Given MATH by REF there exits MATH such that MATH. It then follows from REF that for every MATH, MATH. Similarly by REF there exists MATH such that MATH. It then follows from REF that for every MATH, MATH. The proof is finished. |
math/9907167 | In view of REF MATH for all MATH and therefore one can define a NAME probability measure MATH on MATH, the algebra generated by the cylinder sets of the form MATH, MATH, putting MATH. Hence, applying REF again we get for all MATH. MATH and therefore in view of NAME 's extension theorem there exists a unique probability measure MATH on MATH such that MATH for all MATH. The proof is complete. |
math/9907167 | Let MATH be an arbitrary closed subset of MATH and for every MATH let MATH. In view of REF applied to the characteristic function MATH we have for all MATH . Since the family of sets MATH is descending and MATH we therefore get MATH. Since the limit set MATH is a metric space , using the NAME classification of NAME sets we easily see that this inequality extends to the family of all NAME subsets of MATH. Since both measures MATH and MATH are probabilistic we get MATH. The proof is finished. |
math/9907167 | First notice that, using REF , for each MATH and each MATH we have MATH and MATH . Let now MATH be a NAME limit defined on the NAME space of all bounded sequences of real numbers. We define MATH. Hence MATH and therefore it is not difficult to check that the formula MATH defines a finite non-zero finitely additive measure on NAME sets of MATH satisfying MATH. Using now a theorem of Calderon (REF of CITE) and its proof one constructs a NAME probability (MATH-additive) measure MATH on MATH satisfying the formula MATH for every NAME set MATH with, perhaps, a larger constant MATH. Thus, to complete the proof of our theorem we only need to show total ergodicity of MATH or equivalently of MATH. Toward this end take a NAME set MATH with MATH. Since the nested family of sets MATH generates the NAME MATH-algebra on MATH, for every MATH and every MATH we can find a subfamily MATH of MATH consisting of mutually incomparable words and such that MATH and MATH, where MATH. Then MATH . Therefore MATH . Hence for every NAME set MATH with MATH, for every MATH, and for every MATH we get MATH . In order to conclude the proof of total ergodicity of MATH suppose that MATH for some integer MATH and some NAME set MATH with MATH. Put MATH. Note that MATH. In view of REF , for every MATH we get MATH. Take now MATH so small that MATH and choose a subfamily MATH of MATH consisting of mutually incomparable words and such that MATH and MATH. Then MATH. This contradiction finishes the proof. |
math/9907167 | Suppose that MATH . It means that MATH and consequently MATH . Thus MATH . Now suppose that MATH. We shall show that MATH. So, MATH . But MATH, so it suffices to show that MATH . But MATH . But MATH, so it suffices to show that MATH. And indeed, using REF we get MATH . Since MATH, it is enough to show that MATH . And indeed, MATH . But in view of REF MATH are negative everywhere for all MATH large enough, say MATH. Then using REF again we get MATH which is finite due to our assumption. Hence, MATH. Finally suppose that MATH. We need to show that MATH . We have MATH . Hence MATH and therefore MATH . The proof is complete. |
math/9907167 | It follows from REF that MATH. To show that MATH is an equilibrium state of the potential MATH consider MATH, the partition of MATH into initial cylinders of length one. By REF , MATH. Applying the NAME theorem and the NAME ergodic theorem we therefore get for MATH-a.e. MATH . Hence MATH, which in view of the variational principle (see REF in CITE), implies that MATH is an equilibrium state for the potential MATH. The proof is finished. |
math/9907168 | Consider the pull-back diagram MATH . The middle column splits, by the assumptions on MATH and MATH, and hence the middle row yields the required sequence. |
math/9907168 | CASE: From REF we obtain exact sequences MATH and MATH; so MATH. Similarly, MATH, and hence MATH, by transitivity of MATH. CASE: Say MATH for permutation lattices MATH and MATH. Then we have exact sequences MATH and MATH, proving MATH. |
math/9907168 | First, MATH implies that MATH and MATH are stably isomorphic as MATH-fields over MATH. Indeed, given the exact sequences REF, the usual ``NAME descent" argument (for example, CITE) shows that MATH is generated over MATH by algebraically independent MATH-invariant elements, and similarly for MATH. Conversely, if MATH and MATH are stably isomorphic as MATH-fields over MATH then, for any MATH-field MATH containing MATH, MATH and MATH are stably isomorphic as MATH-fields over MATH. Indeed, for suitable MATH and MATH, the fields MATH and MATH are isomorphic as MATH-fields over MATH, and hence MATH and MATH are isomorphic as MATH-fields over MATH. As noted in REF, this implies that the invariant fields MATH and MATH are stably isomorphic over MATH. In particular, choosing MATH so that MATH acts faithfully on MATH, we deduce from CITE (or CITE) that MATH. Since the statement about fixed fields has already been explained in REF, the proof is complete. |
math/9907168 | CASE: The assertion about MATH is clear. If MATH is a fixed ordering of MATH then MATH has MATH-basis MATH as MATH-sets. Thus, MATH . Also, MATH has MATH-basis MATH and MATH permutes this basis up to MATH. The description of MATH follows. CASE: The MATH-map MATH, MATH, gives rise to a map of MATH-lattices MATH which passes down to a map of MATH-lattices MATH . One easily checks that MATH. Since MATH is spanned by the elements of the form MATH for MATH, and MATH we have MATH. Now, MATH is a pure sublattice of MATH and MATH . This implies that, in fact, MATH. Thus, we obtain an exact sequence of MATH-lattices MATH . Using the above decomposition MATH and noting that MATH, we deduce the following exact sequence MATH . The sum MATH is direct, as follows by counting ranks. Furthermore, using the notation MATH for reduction mod MATH, we have MATH. Therefore, we obtain an exact sequence of MATH-modules MATH proving REF . CASE: Consider the map (a piece of the NAME complex) MATH . This map is MATH-equivariant and surjective, and one checks that MATH is contained in the kernel. Counting ranks, one sees that, in fact, MATH, which completes the proof. |
math/9907168 | Continuing with the notation used in the proof of REF , we first show that MATH is coflasque. Inasmuch as the permutation MATH-lattice MATH is certainly coflasque, the sequence REF reduces the assertion to the following statement about fixed points: MATH . Now MATH, and MATH has MATH-basis the orbit sums MATH, where MATH is a MATH-orbit in MATH. Therefore, MATH is spanned by the orbit sums MATH, where MATH has even length, together with the elements MATH with MATH and MATH both of odd length. Both types can be written as MATH, where MATH is a MATH-invariant subset of even size. For any MATH-invariant MATH, say MATH, put MATH, a MATH-invariant element of MATH (note that MATH). The image MATH is MATH . Thus, if MATH is even then MATH, and so REF is proved. If there is an odd length MATH-orbit MATH then MATH. Thus, we can extend the map MATH in REF to a MATH-equivariant surjection MATH by sending MATH to MATH. This leads to an exact sequence of MATH-modules MATH . Since MATH and MATH are permutation lattices, REF (with MATH) yields an exact sequence MATH . Finally, since MATH is coflasque, this sequence splits which proves that MATH is stably permutation. In view of the sequence REF, the assertion about MATH now follows from REF . |
math/9907168 | CASE: First, MATH, because MATH is additive on direct sums and MATH is detected by restrictions to cyclic subgroups. The isomorphism MATH now follows from the commutative exact diagram below, obtained from the cohomology sequences that are associated with the given exact sequence: CASE: Let MATH be a flasque resolution of MATH; so MATH. By periodicity of cohomology for cyclic groups, we have MATH for all MATH, because MATH is flasque. Thus, MATH and the isomorphism MATH follows from REF . CASE: Since the property of being equivalent to a direct summand of a quasi-permutation lattice is inherited by restrictions to subgroups, it suffices to show that MATH or, equivalently, MATH holds whenever MATH is quasi-permutation. But MATH is easily seen to be additive, as is MATH. So MATH, because MATH. This entails that MATH, as desired. |
math/9907168 | From the cohomology sequence that is associated with the augmentation sequence MATH, one obtains the exact sequence MATH which implies the asserted description of MATH. For surjectivity of restriction maps, consider the commutative diagram of cohomology sequences MATH . |
math/9907168 | Since monomial lattices are quasi-permutation, we know by REF that MATH; so we may concentrate on MATH and MATH. As both functors are additive on direct sums, we may assume that MATH with MATH nontrivial (otherwise, MATH is a permutation module and the assertions are clear). Then MATH where the first isomorphism comes from NAME 's Lemma and the second is given by inflation; see CITE. This proves the asserted isomorphism for MATH. The NAME isomorphism MATH is equal to the composite MATH where MATH is the projection onto the direct summand MATH of MATH. Fixing MATH with MATH, the restriction map MATH is an isomorphism, and hence so is the map MATH. This proves that MATH is injective, whence MATH. |
math/9907168 | We first deal with MATH. By REF , it suffices to show that MATH holds for some subgroup MATH. To this end, we use the exact sequence of REF which now takes the following form (compare REF ) MATH where MATH denotes reduction mod MATH. By REF , this sequence implies MATH, and hence the issue is to show that MATH holds for some subgroup MATH. Now, suppose we can find a subgroup MATH satisfying CASE: MATH is a MATH-group; CASE: MATH has no fixed point in MATH, but every MATH does have a fixed point. Then the cohomology sequence that is associated with the exact sequence MATH yields the following commutative diagram MATH . Here, the first MATH results from the fact that MATH is spanned by the orbit sums of the MATH-orbits in MATH, all of which have positive even length, while the second MATH is a consequence of the existence of MATH-fixed points, which implies surjectivity of MATH for all MATH. Thus, we conclude that the (nonzero) image of MATH in MATH does indeed belong to MATH, forcing this group to be non-trivial as required. The actual construction of a suitable subgroup MATH is a simple matter. For example, we can take the group MATH, with MATH, that was used in the proof of REF for MATH; see REF . This proves the assertion about MATH. In order to show that MATH, it suffices to exhibit a subgroup MATH such that MATH . In fact, the above MATH will do. For, by REF (see also REF ), we know that MATH. On the other hand, MATH. Thus, it suffices to show that MATH holds for every MATH. Consider the exact sequence MATH from REF and the resulting map MATH. Here, the last equality holds since MATH is monomial; see REF . Thus, in order to show that MATH, it suffices to show that MATH is mono or, equivalently, that MATH . But, as MATH-lattices (and hence as MATH-lattices), MATH contains MATH as a direct summand, and MATH contains MATH as a sublattice, with MATH . Moreover, MATH under MATH. Therefore, restricting MATH to the subgroup MATH, we obtain MATH. This completes the proof. |
math/9907171 | It is a straightforward computation; see CITE for details. The associativity of the MATH - product follows from the associativity of the operator product. It can be also verified directly from the integral REF using the NAME theorem; all issues regarding convergence, etc., are standard. The same applies to the cyclic trace property. Finally, the property of the unit follows from the definition of the NAME kernel. |
math/9907171 | It is another straightforward computation, based on the following expression of MATH-product of covariant symbols through corresponding contravariant ones, MATH . |
math/9907171 | It follows from the definition of the NAME function that MATH . Therefore, both sides of the associativity equation are equal to the same formal integral MATH provided that one has the formal analog of the NAME theorem: ``double formal integral equals to the repeated formal integral". The latter statement can be proved as follows. First, it is clearly valid for convergent integrals, provided that all sufficient conditions for the uniform convergence are satisfied. These conditions are in the form of inequalities for the coefficients MATH and for the NAME coefficients of MATH. Since the associativity equation can be written as a system of polynomial equations for these coefficients, these equations hold for arbitrary values of the coefficients involved - ``the nonessential property of algebraic identities" principle. |
math/9907171 | It is an easy computation, based on REF . |
math/9907171 | As it follows from the previous lemma, MATH so that the correspondence principle is valid. The symmetry with respect to the complex conjugation follows from the definition of the MATH - product. |
math/9907171 | It follows from the definitions that MATH is given by the NAME expansion around MATH of the following formal double integral: MATH which is symmetric with respect to MATH and MATH. |
math/9907171 | The integral REF is invariant with respect to a local change of coordinates MATH since MATH is a well-defined function on MATH. The same applies to the power series expansion of REF, which is ``the only thing that actually exists". |
math/9907171 | It is an immediate consequence of REF . Consider an open covering of MATH, MATH with the property that for every MATH the forms MATH admit a representation REF- REF with some NAME potential MATH. On every MATH define the MATH - product MATH of MATH as before. Though NAME potentials MATH on MATH do not necessarily agree on the intersections MATH, the corresponding NAME functions MATH on MATH, which exist for real-analytic MATH, combine to a well-defined function in some neighborhood of the diagonal in MATH. Applying the previous theorem in the real-analytic case, we conclude that MATH. Since the coefficients of the MATH - product are polynomials in elements of MATH and MATH and their partial derivatives, the same conclusion holds for the smooth case. |
math/9907171 | The property MATH, for all MATH, follows from the definition. The strong form REF of the correspondence principle follows from the computation of MATH, given in REF . The proof of the trace property is the same as in REF . |
math/9907171 | It is clear that in the real-analytic category the integral REF , when it converges, is a holomorphic function. For the asymptotics of the integral, this is an algebraic property of coefficients. Therefore, the same is true for the formal power series with the coefficients in infinite jets in the MATH-category. |
math/9907171 | This follows from the identity between two types of formal integrals MATH where MATH, proved in Appendix. Namely, for MATH and MATH set MATH. Using the property of the unit and that MATH is holomorphic in MATH, we get MATH . |
math/9907171 | This is a formal analog of REF . Using NAME theorem for formal integrals and the equality of two types of formal integrals proved in the Appendix, we have MATH where MATH. The same arguments also show that MATH and the statement follows. |
math/9907171 | We will start with case MATH; the precise definition of the formal integral MATH will become clear in a due course. First, assume that MATH, that the NAME function has a single critical point at MATH, and, assuming that the integral MATH exists for MATH, consider its the asymptotic behavior as MATH. Replacement of the domain of integration by a neighborhood MATH of MATH (with certain MATH depending on MATH and MATH in the support of MATH) will result in an exponentially small error term as MATH. (It would be also sufficient to consider terms that decay faster than any power MATH with MATH). Therefore, MATH where the symbol MATH stands for equality modulo exponentially small terms as MATH. Next, for fixed MATH, consider the following representation (compare CITE) MATH where MATH and satisfies the equation MATH so that MATH . Applying NAME 's formula, we get MATH where MATH, and we have used that the integral over the circle of radius MATH around MATH is exponentially small as MATH. Repeating this procedure, we obtain MATH where MATH. Finally, expanding MATH into power series in MATH with coefficients MATH that have singularities MATH as MATH, and using the generalized NAME formula MATH which is valid for any MATH, we get an asymptotic expansion for MATH in terms of a power series in MATH. Clearly, the coefficients of this expansion are given by the infinite sums, convergent when the function MATH and NAME potential MATH are real-analytic. This expansion, extended to the formal category (with respect to MATH) with coefficients given by the global sections of the bundle of infinite jets over MATH, is the definition of a formal integral MATH. Now proposition follows by the same arguments as in the proof of REF . The proof for the case MATH is similar and is left to the reader. |
math/9907172 | Let MATH have a NAME presentation of nonnegative deficiency MATH. Then the group MATH has a presentation with MATH generators MATH and MATH relators MATH. Define a REF-dimensional CW complex MATH as follows: REF-skeleton of MATH is a one-point union of MATH circles, each of which represents each generator MATH, and MATH REF-disks are attached to REF-skeleton along the relators MATH. By the NAME theorem, MATH is the fundamental group of MATH. Thus, MATH and rank-MATH. The cellular chain complex of MATH is MATH, where MATH is a zero map, and thus MATH rank-MATH rank-MATH. In conclusion, the group MATH of a NAME diagram MATH (hence any virtual knot group) has a NAME presentation of deficiency REF or REF. By rank arguments similar to the one above, MATH is zero and infinite cyclic if the deficiency is REF, respectively. Continuing attaching higher cells to MATH we get an NAME space MATH whose REF-skeleton is MATH. Thus, the group homology MATH, which is defined as MATH, is cyclic since it is a quotient of MATH. |
math/9907172 | Let MATH be a NAME presentation of deficiency REF or REF, say, MATH, where MATH or MATH. If MATH, by doubling the relator MATH, we may assume MATH. By the arguments prior to this lemma, it suffices to construct a cyclic NAME presentation from MATH. Let MATH be a graph with MATH vertices MATH and MATH edges corresponding to relators in the way that an edge has end vertices MATH and MATH if and only if there is a relator of the form MATH. Such an edge is denoted by MATH. Since all MATH are conjugate, the graph MATH is connected. If two edges MATH and MATH meet at a vertex MATH, then we have two relators MATH and MATH, or MATH and MATH. This implies MATH. We now remove the relator MATH and add MATH to get a new presentation. It is obvious that the new presentation presents the same group. This operation corresponds to an operation on the graph MATH of deleting an edge MATH and adding an edge MATH. Recall that a cycle of a graph is a simply closed path on the graph. Using the above operation on the graph MATH we will construct a cycle from MATH. Since MATH has betti number MATH, it has one and only one cycle MATH. We will use induction on the length MATH of MATH. If MATH, then MATH and thus MATH is a cycle. Suppose MATH. Then there is a vertex MATH which is not on MATH. Since MATH is connected, there is a path from MATH to a vertex of MATH. On this path, there is an edge MATH such that MATH is not in MATH but MATH is in MATH. By the operation described above, we can construct MATH with a cycle MATH containing all vertices of MATH and MATH. Since the length of MATH is greater than the length of MATH, we keep doing this process to get a cycle. The corresponding operations on the NAME presentation of MATH give a cyclic NAME presentation. This completes the proof. |
math/9907172 | By REF , we can assume that a realizable NAME presentation with MATH generators MATH and MATH relators MATH is given. We start with a circle. Choose MATH points on the circle, which divide the circle into MATH arcs labeled MATH, successively counterclockwise. For each MATH, if MATH, attach an oriented chord with sign MATH from a point on the arc labeled MATH to the point dividing MATH and MATH. By definition, the group of this NAME diagram has the given presentation. |
math/9907172 | Iterating all relators MATH, we have MATH, or MATH commutes with the longitude. This then shows that the peripheral subgroup of MATH is an abelian group generated by the meridian and longitude. |
math/9907172 | Suppose that a group MATH has a NAME presentation of deficiency REF. Then adding a redundant relator MATH to the presentation gives a NAME presentation of deficiency REF presenting the same group. We can easily see that the longitude obtained from the new presentation is trivial. Moreover, in this case, if MATH is a commutator element commuting with MATH in MATH, adding MATH to the original presentation, we have longitude MATH for the virtual knot obtained from the presentation. |
math/9907172 | If MATH and MATH are realizable, then a connected sum can be used to show MATH is realizable. To realize MATH, use the NAME diagram used to realize MATH with orientation and all signs of chords reversed. That the set is nonempty follows from REF since a knot group is a virtual knot group. It follows from a method of modifying group representations similar to the ones given in REF as well. |
math/9907172 | It has been proved in REF that MATH. To show the converse, let MATH be an element in MATH. Let MATH be a knot in MATH and let MATH be a representation with MATH as seen in the remark prior to this proof. The knot group MATH has a NAME presentation MATH, where the last relator is redundant, the longitude of MATH is MATH, MATH and MATH. Since MATH is onto and hence the restriction of MATH to MATH is onto MATH, there is an element MATH in MATH with MATH. Let MATH. Then MATH is a quotient group of MATH with an extra relator MATH. Thus MATH has a NAME presentation of deficiency REF. Since MATH, the group MATH can be realized as the group of a virtual knot with longitude MATH. Define MATH by MATH, MATH. Since MATH, MATH is a well-defined surjective homomorphism and the longitude maps to MATH. This implies that MATH is realizable. |
math/9907172 | Let MATH be a representative NAME diagram of a virtual knot. We first compute the second homology of MATH. Consider the cellular chain complex of MATH where MATH, MATH, and MATH are the free abelian groups generated by the rectangles MATH's, the circles MATH's, and the vertex, respectively. The map MATH is a zero map and the map MATH is represented by a matrix MATH where MATH is generated by the element MATH. Let MATH be the quotient map from MATH to MATH. Then MATH is the element MATH in MATH, where MATH denotes the generator of MATH. Observe that a meridian of the NAME diagram MATH is MATH and its longitude can be exactly read off from the top side of the square of MATH. Since the meridian and the longitude of MATH map to MATH and the top side of the square via the quotient map MATH, respectively, MATH in MATH is the NAME product of the meridian and longitude of the diagram MATH and hence it generates MATH and MATH. |
math/9907174 | The semi - invariants are invariants for MATH and conversely, MATH - invariant polynomials that are homogeneous with respect to the MATH - grading are also semi - invariant for MATH. We may therefore use NAME 's description of the homogeneous multilinear invariants for MATH and hence for MATH. Given a homogeneous multilinear MATH - invariant MATH it factors as MATH for a suitable linear map MATH. Since MATH as MATH - representation, we may write MATH . Moreover, MATH as MATH - representation is a tensor product of covariant and contravariant vectors for MATH. Thus we may re - write MATH and MATH for MATH - invariant linear maps: MATH . Roughly speaking CITE showed that there are MATH basic linear semi - invariant functions on tensor products of covariant and contravariant vectors for MATH. Firstly, there is the linear map from MATH to MATH given by MATH; this is just the trace function on MATH. Secondly, there is the linear map from MATH to MATH determined by MATH. The third case is similar to the second; there is a linear map from MATH to MATH again given by the determinant. A spanning set for the linear semi - invariant functions on a general tensor product of covariant and contravariant vectors is constructed from these next. A spanning set for the MATH - invariant linear maps from MATH to MATH is obtained in the following way. We take three disjoint indexing sets MATH: we have surjective functions MATH such that MATH and MATH have one element each for MATH, and MATH and MATH have MATH elements each for MATH. We label this data by MATH. To MATH, we associate a MATH - invariant linear map MATH where MATH. Note that MATH is determined only up to sign since we have not specified an ordering of MATH or MATH. A spanning set for MATH - invariant linear maps from MATH is therefore determined by giving quintuples MATH where MATH and surjective maps MATH where MATH and MATH have one element each, MATH and MATH have MATH elements each for MATH. Then MATH determines data MATH for each MATH and we define MATH . We show that these specific semi - invariants lie in the linear span of the homogeneous components of determinantal semi - invariants. First, we treat the case where MATH is empty. Let MATH. We have two expressions for MATH: MATH . To each arrow MATH, we have a pair MATH associated. To this data, we associate a map in MATH in the following way. We consider a map MATH whose MATH - entry is MATH . Given MATH is a MATH by MATH matrix which we may regard as a partitioned matrix where the rows are indexed by MATH and the columns by MATH, there are MATH rows having index MATH, MATH columns having index MATH and the block having index MATH is MATH . We claim that MATH up to sign where MATH. To prove this it will be convenient to define a new quiver MATH whose vertices are given by MATH and whose edges are the same as those of MATH. The initial and terminal vertex of MATH are given by MATH. On MATH we define data MATH which is defined by the same quintuple MATH as MATH, but with has different decompositions MATH, MATH. In fact MATH . We define a functor MATH by MATH, MATH, MATH where MATH, MATH, MATH. Since MATH and MATH have the same edges, the action of MATH on MATH lifts canonically to an action of MATH on MATH. Put MATH. Using REF we then find MATH and similarly MATH. Finally one also verifies that MATH. Hence to prove that MATH we may replace the triple MATH by MATH. We do this now. In order to prove that MATH, we need only check that the two functions agree on the image of MATH in MATH. Let MATH be the character given by MATH . Then one checks that both MATH and MATH are semi-invariants on MATH with character MATH. Now we claim that on MATH we have MATH, up to sign. To prove this we use the fact that MATH has an open orbit on MATH. For vertices MATH and MATH, we let MATH and MATH be the sets of arrows incident with MATH and MATH respectively. So MATH . We take the point MATH whose MATH - th entry is the MATH - th standard column vector in MATH; that is its MATH-th entry is MATH and whose MATH - th entry is the MATH - th standard row vector in MATH. The MATH-orbit of this point is open in MATH. Hence to show that MATH on MATH, it suffices to do this in the point MATH. Now MATH. Furthermore we have MATH . In particular MATH is a permutation matrix and thus MATH. Hence indeed MATH (on MATH). Note that the non-zero entries of MATH are naturally indexed by MATH. To prove that MATH on MATH it is now sufficient to prove that MATH on MATH. To this end we lift the MATH action on MATH to MATH by defining MATH (this is just some convenient choice). Now we have to show that MATH is itself homogeneous with character MATH when restricted to MATH. Since the action of MATH commutes with the MATH-action it suffices to do this in the point MATH. Now if we put MATH then MATH is obtained from MATH by multiplying by MATH for all MATH the non-zero entry in MATH indexed by MATH. Therefore MATH is multiplied by MATH. This proves what we want. It remains to deal with the case where MATH is non - empty. Roughly speaking one of two things happens here; if we have two distinct arrows MATH and MATH for some MATH then this element of MATH corresponds to replacing MATH and MATH by their composition; if on the other hand MATH then MATH and this element of MATH corresponds to taking the trace of MATH. We associate a quiver MATH with vertex set MATH and arrow set MATH; given MATH, MATH. This quiver has very little to do with the quiver MATH; it is a temporary notational convenience. The connected components of MATH are of three types: either they are oriented cycles, open paths or isolated points. The vertices of components of the first type are arrows of MATH that also form an oriented cycle, those of the second type are arrows that compose to a path in the MATH (which can in fact also be an oriented cycle); the isolated points we shall treat in the same way to the second type. We label the oriented cycles MATH and the open paths MATH. To an oriented cycle MATH in MATH, we associate the invariant MATH where MATH is the path around the oriented cycle in MATH. This is independent of our choice of starting point on the loop. To the open path MATH in MATH, we associate the path MATH, the corresponding path in MATH. We consider the adjusted quiver MATH with vertex set MATH and arrow set MATH where MATH and MATH are defined as usual. Define the functor MATH as follows: on vertices MATH is the identity, and on edges MATH. If MATH, we define MATH . Then MATH are surjective functions giving data MATH on MATH. One checks directly that MATH for MATH and MATH of weight MATH on each arrow for the quivers MATH and MATH respectively, which completes our proof. |
math/9907174 | It is enough to consider semi - invariants homogeneous with respect to the MATH - grading of weight MATH where MATH. We may also assume that no component of the MATH - grade on the semi - invariant is MATH since we may always restrict to a smaller quiver. To MATH, we associate a new quiver MATH with vertex set MATH and arrow set MATH where MATH where MATH. We have functors MATH . Given a semi - invariant MATH for MATH, we define MATH to be the MATH - component of MATH where MATH. So MATH is homogeneous multilinear. Then one checks that MATH. However, by the previous corollary, MATH lies in the linear span of determinantal semi - invariants; therefore MATH and so MATH . Hence, in characteristic REF, semi - invariants homogeneous with respect to the MATH - grading and hence all semi - invariants lie in the linear span of the determinantal semi - invariants as required. |
math/9907174 | Since the determinantal semi - invariant polynomial functions span all semi - invariant polynomial functions, there must exist some MATH with the properties MATH, MATH is not constant and MATH is injective. Then MATH. If MATH then by REF MATH is nowhere vanishing, and hence constant by the NAME. This is a contradiction, whence we may take MATH. |
math/9907175 | We compute directly that MATH . |
math/9907175 | A simple fact of finite group theory says that MATH . Assume that MATH. Observe that MATH and the equality holds if and only if MATH. Let MATH. Note that for any faithful representation MATH of MATH we have that MATH . Then it follows that MATH . According to NAME (see for example, CITE), MATH is positive semi-definite which generalizes NAME 's observation in the case of MATH. This implies that the eigenvalues of MATH are MATH. Thus the matrix MATH is positive-definite when MATH and MATH. |
math/9907175 | Let MATH be the matrix of the bilinear form MATH and let MATH be a MATH-th root of unity. Then MATH and MATH. From this and REF we see that as a matrix over MATH the eigenvalues of MATH are MATH, MATH. The function MATH is non-zero except when MATH and MATH. |
math/9907175 | This is proved by a direct computation using REF . MATH . |
math/9907175 | This is verified by the definition of REF and the character values of MATH defined above. |
math/9907175 | We first let MATH be an element of MATH such that MATH is a cycle of length MATH, say MATH. Let MATH be a basis of MATH, and MATH is afforded by the action: MATH, where MATH. We then have MATH . It follows that MATH . Given MATH where MATH and MATH, by REF we clearly have MATH . This immediately implies the formula. |
math/9907175 | It follows from definition of REF that MATH . Similarly we can prove REF using the following identity MATH . The same argument as in the classical case (compare CITE) by using REF will show that the proposition holds for linear combination of simple characters such as MATH, and thus it is true for any element MATH, where MATH. |
math/9907175 | It follows immediately from the definition of the comultiplication in the both NAME algebras (compare REF ). |
math/9907175 | By REF , the character value of MATH at an element MATH of type MATH is MATH . Thus it follows from definition that MATH . By REF , we see that MATH . Since MATH form a MATH-basis of MATH, we have shown that MATH is an isometry. |
math/9907175 | This is similarly proved as for the classical setting in CITE. |
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