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math/0010245 | We have the NAME representation CITE MATH where MATH is a closed contour containing MATH in its interior. We write MATH . The entries of MATH have NAME series absolutely uniformly convergent in MATH. Hence it follows easily that MATH and MATH, as sums and products of the entries of MATH, have NAME series that are absolutely uniformly convergent in MATH. Furthermore, there are constants MATH such that MATH . Thus by the uniform NAME MATH-theorem CITE we have that MATH has a NAME series that is absolutely uniformly convergent in MATH. This implies that MATH is uniformly bounded in MATH, by MATH, say. Hence MATH where the boundedness of the integral in REF follows from the continuity of MATH. The proof is complete. |
math/0010245 | We have for any MATH that MATH . Here we have used that MATH, and hence MATH, commutes with all relevant time-frequency shifts. Thus the frame operator corresponding to MATH is given by MATH. By the spectral mapping theorem we have that MATH and this gives the first part of the result. Next we compute the tight frame generating window MATH canonically associated to MATH according to MATH . Here we have used that MATH so that MATH and MATH so that MATH. |
math/0010245 | It is sufficient to consider the case MATH. We begin with the observation that for MATH satisfying MATH where MATH and MATH there holds MATH . REF is equivalent to the operator version of the NAME inequality CITE. For the reader's convenience we include a short proof. REF follows from MATH since MATH is a non-negative operator. As a consequence we have now immediately that MATH which is the first inequality in REF. From REF we have MATH and this is the second inequality in REF. |
math/0010246 | The NAME scheme MATH of points in the projective plane exists as a special case of NAME 's construction in CITE, with a universal family MATH having the analogous universal property. We identify MATH as usual with an open subset of MATH, the complement of the projective line MATH ``at infinity." The projection of MATH onto MATH is a closed subset of MATH. Its complement MATH is clearly the largest subset such that the restriction MATH of MATH to MATH is contained in MATH. The required universal property of MATH and MATH now follows immediately from that of MATH and MATH. |
math/0010246 | Let MATH be the universal family over MATH. The fiber MATH of MATH over a point MATH is the disjoint union of closed subschemes MATH and MATH in MATH of lengths MATH and MATH, respectively, with MATH and MATH. Hence over MATH, MATH is the disjoint union of flat families MATH, MATH of degrees MATH and MATH. By the universal property, we get induced morphisms MATH, MATH and MATH. The equations MATH, MATH imply that MATH factors through a morphism MATH of schemes over MATH. Conversely, on MATH, the pullbacks of the universal families from MATH and MATH are disjoint and their union is a flat family of degree MATH. By the universal property there is an induced morphism MATH, which factors through a morphism MATH of schemes over MATH. By construction, the universal families on MATH and MATH pull back to themselves via MATH and MATH, respectively. This implies that MATH is a morphism of schemes over MATH and MATH is a morphism of schemes over MATH. Since they are also morphisms of schemes over MATH, we have MATH and MATH. Hence MATH and MATH induce mutually inverse isomorphisms MATH. |
math/0010246 | Let MATH be the preimage in MATH of the open set MATH of points MATH where the MATH are all distinct. It follows from REF that MATH restricts to an isomorphism MATH, so MATH is irreducible. We are to show that MATH is dense in MATH. Let MATH be a closed point of MATH, which we want to show belongs to the closure MATH of MATH. If MATH with the MATH not all equal, then by REF there is a neighborhood of MATH in MATH isomorphic to an open set in MATH for some MATH. The result then follows by induction, since we may assume MATH and MATH irreducible. If all the MATH are equal, then MATH is the unique point of MATH lying over MATH. Since MATH is finite, MATH is closed. But MATH is dense in MATH, so MATH. Therefore MATH contains a point lying over MATH, which must be MATH. |
math/0010246 | It follows from the cell decomposition of CITE that the closed locus MATH in MATH consisting of points MATH with MATH supported on the MATH-axis MATH in MATH is the union of locally closed affine cells of dimension MATH. The subset MATH is equal to MATH and MATH is finite. |
math/0010246 | By symmetry among the points MATH of MATH we see that MATH, where MATH is the locus in MATH defined by the equations MATH. It follows from REF that MATH is isomorphic to an open set in MATH, where MATH is the closed subset MATH in MATH. As a reduced subscheme of MATH, the latter is isomorphic to MATH, where MATH is the zero fiber in MATH, the factor MATH accounting for the choice of MATH. By REF, MATH is irreducible of dimension MATH, so MATH is irreducible of dimension MATH. Since MATH and MATH is finite, the result follows by descending induction on MATH, starting with MATH. |
math/0010246 | By definition the blowup of MATH at MATH is MATH, where MATH is the NAME algebra. The ring MATH is a homogeneous subring of MATH in an obvious way, and since MATH generates MATH as a MATH-module, we have MATH, that is, MATH for some homogeneous ideal MATH. In geometric terms, using REF and the fact that MATH, this says that MATH is a closed subscheme of the scheme-theoretic fiber product MATH. Since MATH is reduced, it follows that MATH is a closed subscheme of MATH. By REF , MATH is irreducible, and since both MATH and MATH have dimension MATH, it follows that MATH. |
math/0010246 | The possible ideals MATH for the given MATH are the length MATH ideals in the local ring MATH, or equivalently, the one-dimensional subspaces of its socle. The fiber of MATH is therefore the projective space MATH, and we have MATH. First consider the maximum possible dimension of any fiber of MATH. Since both MATH and MATH are projective over MATH, the morphism MATH is projective and its fiber dimension is upper semicontinuous. Since every point of MATH has a monomial ideal MATH in the closure of its MATH-orbit, and since MATH is finite over MATH, every point of MATH must have a pair MATH in the closure of its orbit. The fiber dimension is therefore maximized at some such point. The socle of MATH has dimension equal to the number of corners of the diagram of MATH. If this number is MATH, we clearly have MATH. This implies that for every NAME local ring MATH generated over MATH by two elements, the socle dimension MATH and the length MATH of MATH satisfy MATH. Returning to the original problem, MATH is an NAME local ring of length MATH generated by two elements, with socle dimension MATH, so REF follows. |
math/0010246 | REF gives us isomorphisms on the preimage of MATH between MATH and MATH, and between MATH and MATH. We can identify MATH with the closed subset of MATH consisting of points where MATH are the same in both factors, and MATH contains MATH. On the preimage of MATH, under the isomorphisms above, this corresponds to the closed subset of MATH where MATH, MATH and MATH are the same in both factors and MATH contains MATH. The latter can be identified with MATH. |
math/0010246 | We have the corresponding result for MATH in REF . We can assume by induction that the result for the nested scheme holds for smaller values of MATH for the base case note that MATH. Locally on a neighborhood of any point where MATH are not all equal, the result then follows from REF . The locus where all the MATH are equal is isomorphic to MATH where MATH is the zero fiber in the nested NAME scheme, the factor MATH accounting for the choice of the common point MATH on the MATH-axis MATH. By REF REF we have MATH. |
math/0010246 | Let MATH. Let MATH be the generic locus, that is, the open set of ideals MATH for which MATH is a set of MATH distinct points. The NAME morphism MATH induces an isomorphism of MATH onto its image in MATH, and MATH is the locus where some two of the MATH have the same MATH-coordinate, and another two have the same MATH-coordinate. This locus has codimension MATH. The complement of MATH is the closed subset MATH in REF , which has one irreducible component of dimension MATH. An open set in this component consists of those MATH for which MATH has one point MATH of multiplicity MATH and the rest are distinct. This open set is not contained in MATH, so MATH has codimension at least MATH. This takes care of MATH. If MATH is curvilinear, then the socle of MATH has length MATH for all MATH. Hence the morphism MATH in REF restricts to a bijection on the curvilinear locus. If the complement of MATH in MATH had a codimension MATH component, it would therefore have to be contained in the complement of the curvilinear locus, by the result for MATH. By REF , MATH has fibers of dimension MATH only over MATH for MATH, and it follows from REF that the union of these fibers has codimension at least MATH. For MATH this exceeds MATH. The fibers of dimension MATH over MATH occur only over non-curvilinear points. However, the codimension MATH component of MATH contains all MATH such that MATH has one point of multiplicity MATH and the rest distinct, and such a MATH can be curvilinear. Hence the non-curvilinear locus in MATH has codimension at least MATH and its preimage in MATH has codimension at least MATH. |
math/0010246 | The MATH-form MATH is MATH-invariant and thus defines a MATH-form on the smooth locus in MATH and therefore a rational MATH-form on MATH. If MATH is a point of MATH, then MATH is generated as an ideal in MATH by two polynomials, MATH and MATH where the complex numbers MATH are regular functions of MATH. This is so because the tautological sheaf MATH is free with basis MATH on MATH, and the sections MATH and MATH must be unique linear combinations of the basis sections with regular coefficients. Conversely, for any choice of the parameters MATH, MATH, the polynomials in REF generate an ideal MATH. Hence MATH is an affine MATH-cell with coordinates MATH. On the open set where MATH consists of points MATH with distinct MATH coordinates, the polynomial in REF is MATH. This implies that MATH coincides as a rational function (and as a global regular function) with the MATH-th elementary symmetric function MATH. Likewise, MATH is given as a rational function of MATH and MATH by the coefficient of MATH in the unique interpolating polynomial MATH of degree MATH satisfying MATH for MATH. The equations MATH can be expressed as a matrix identity MATH, where MATH is the NAME matrix in the MATH variables. Modulo terms involving the MATH, this yields the identity of rational MATH-forms on MATH, MATH where MATH is the NAME determinant. For the elementary symmetric functions MATH we have the well-known identity MATH . Together these show that MATH is a nowhere vanishing regular section of MATH on MATH. By symmetry, the same holds on MATH. This shows that we have MATH on MATH and hence everywhere, by REF . |
math/0010246 | We have tautological sheaves MATH and MATH pulled back from MATH and MATH. The kernel MATH of the canonical surjection MATH is the line bundle with fiber MATH at the point MATH. From REF we have MATH. On the generic locus, the fiber MATH can be identified with the one-dimensional space of functions on MATH that vanish except at MATH. Thus the ratio of two sections of MATH is determined by evaluation at MATH, MATH. Regarding the polynomials in REF as regular functions on MATH, they are the defining equations of the universal family MATH over MATH, as a closed subscheme of the affine scheme MATH. We can use these defining equations to eliminate MATH and MATH, showing that MATH is an affine cell with coordinates MATH. Over the curvilinear locus, the morphism MATH in REF restricts to a bijective morphism of smooth schemes, hence an isomorphism. Under this isomorphism MATH corresponds to the MATH-coordinate MATH of the distinguished point, and modulo MATH we can replace the elementary symmetric functions MATH with MATH, for MATH. As in the proof of REF , we now calculate that a nowhere vanishing regular section of MATH on MATH is given by MATH . By symmetry, MATH is a nowhere vanishing regular section of MATH on MATH. Now, at every point of MATH, the ideal MATH is generated modulo MATH by MATH so this expression represents a nowhere vanishing section MATH of MATH on MATH. Similarly, MATH represents a nowhere vanishing section MATH of MATH on MATH. By the observations in the first paragraph of the proof, the ratio MATH is the rational function MATH on MATH. Since we have nowhere vanishing sections MATH, MATH of MATH on MATH and MATH with MATH it follows that we have MATH on MATH and hence everywhere, by REF . |
math/0010246 | Let MATH be the generic locus, the open set consisting of ideals MATH for which MATH consists of MATH distinct, reduced points. Note that MATH is clearly reduced over MATH, and since MATH is flat over MATH and MATH is dense, MATH is reduced everywhere. (For this argument we do not have to assume that MATH is irreducible, and indeed for MATH it is not: MATH is one of its irreducible components.) Sections of MATH can be identified with regular functions on suitable open subsets of MATH. Since MATH is reduced and irreducible REF , the open set MATH is dense in MATH, and MATH consists of those sections of MATH whose restrictions to MATH define regular functions vanishing on MATH. Let MATH be a section of MATH. For any section MATH of MATH, the section MATH also belongs to MATH. Since it is alternating, MATH vanishes at every point MATH for which two of the MATH coincide. In particular MATH vanishes on MATH and hence on MATH, that is, MATH. This is precisely the condition for MATH to belong to the kernel of MATH. Conversely, if MATH does not vanish on MATH there is a point MATH, with all MATH distinct, where the regular function represented by MATH is non-zero. Multiplying MATH by a suitable MATH, we can arrange that MATH vanishes at every point in the MATH orbit of MATH, except for MATH. Then MATH, so MATH is not in the kernel of MATH. |
math/0010246 | Since MATH and MATH are locally free sheaves, the sheaf homomorphism MATH in REF can be identified with a linear homomorphism of vector bundles over MATH. Let MATH denote the induced map on the fiber at MATH. The rank MATH of the fiber map is a lower semicontinuous function, that is, the set MATH is open for all MATH. If MATH is constant on an open set MATH, then the cokernel of MATH is locally free on MATH, and conversely. When this holds, MATH is also locally free, of rank equal to the constant value of MATH. By REF , we have MATH. The fiber of MATH over a point MATH in the generic locus consists of MATH reduced points, so the generic rank of MATH is MATH. The fiber MATH of the tautological bundle at MATH is MATH. Identifying MATH with MATH, we have a linear map MATH given by composing MATH with the canonical map MATH. It follows from the definition of MATH that MATH if and only if MATH for all MATH, where MATH is the restriction of the canonical map MATH to MATH-alternating elements. In particular, for MATH, we have MATH for all MATH, and MATH spans MATH. Hence, by REF , the kernel of MATH is exactly the ideal MATH in this case. Suppose the MATH conjecture holds for MATH. Then since MATH and MATH have the same image, we have MATH. By lower semicontinuity, since MATH is also the generic rank of MATH, MATH is locally constant and equal to MATH on a neighborhood of MATH. Hence MATH is locally free, MATH is flat of degree MATH, and MATH is locally NAME at MATH. Let MATH be the the maximal ideal of the regular local ring MATH. Since MATH is finite over MATH, the ideal MATH is a parameter ideal. Assuming the MATH conjecture holds for MATH, the NAME ring MATH is NAME if and only if MATH is NAME. We have MATH, so the fiber map MATH factors as MATH and since MATH is locally free, the second homomorphism above is injective. Its image is MATH, so we have MATH, and the latter ring is NAME by REF . Conversely, suppose MATH is locally NAME at MATH. Then MATH is a NAME local ring isomorphic to MATH for some ideal MATH. Since MATH is locally free on a neighborhood of MATH, necessarily of rank MATH, we have MATH. The locally free sheaf MATH is the sheaf of sections of a vector bundle, which is actually a bundle of MATH modules, since MATH acts on MATH as scheme over MATH. The isotypic components of such a bundle are direct summands of it and hence locally free themselves, so the character of MATH on the fibers in constant. In our case, the generic fibers are the coordinate rings of the MATH orbits of points MATH with all MATH distinct. Therefore every fiber affords the regular representation of MATH. The socle of MATH is a one-dimensional MATH-invariant subspace. Since MATH affords the regular representation its only such subspaces are MATH, which consists of the constants, and MATH. The socle must therefore be the latter space (unless MATH, in which case the Proposition is trivial). The factorization of MATH in REF implies that MATH. If MATH then we must have MATH, as the socle is contained in every non-zero ideal. But this would imply MATH and hence MATH, which is absurd. |
math/0010246 | The set MATH of points MATH such that MATH is open and MATH-invariant. From the proof of REF we see that the MATH conjecture implies that MATH contains all the monomial ideals MATH. Since every MATH has a monomial ideal in the closure of its orbit, this implies MATH, so MATH is flat, and MATH is NAME. Let MATH, a locally free sheaf of rank MATH. By REF , the map in REF induces a pairing MATH and MATH factors through the induced homomorphism MATH . Note that the multiplication by a section MATH of MATH in MATH as a sheaf of MATH-modules is given by MATH. By the definition of MATH we have MATH, so MATH is a homomorphism of sheaves of MATH-modules. Since MATH is constant and equal to MATH, which is the rank of both MATH and MATH, MATH is an isomorphism. Now, MATH, so by the duality theorem, MATH is the sheaf of MATH-modules associated to the sheaf of MATH-modules MATH. By REF we have MATH, and we have just shown MATH. Together these imply MATH. |
math/0010246 | Let MATH. By REF , we have MATH, and REF implies that MATH is a free MATH-module. |
math/0010246 | By definition arithmetically normal means that MATH is a normal domain. Since MATH itself is a normal domain, this is equivalent to the ideals MATH being integrally closed ideals for all MATH. The powers of an ideal generated by a regular sequence are integrally closed, as is an intersection of integrally closed ideals, so MATH is integrally closed by REF . |
math/0010246 | The question is local on MATH, so without loss of generality we can assume MATH is affine. Then we are to show that MATH for MATH and that MATH is an isomorphism. REF implies that MATH for MATH. Let MATH and MATH. REF imply that MATH and MATH are both at least MATH. Hence the local cohomology modules MATH and MATH vanish for MATH. By the exact sequence of local cohomology CITE we therefore have MATH and MATH for all MATH. REF yields MATH. Thus we have MATH for all MATH. Since MATH is affine, this shows MATH for all MATH, and MATH. The isomorphism MATH is the canonical homomorphism, since it is determined by its restriction to MATH. |
math/0010246 | We have MATH and MATH. Obviously MATH is a free MATH-module, but there are also the ideals MATH to consider. Necessarily we must have MATH. Then MATH is the union of subspaces in MATH defined by the vanishing of at least MATH of the coordinates MATH. Its ideal MATH is generated by all square-free monomials MATH, where MATH is a subset of MATH of size MATH. The set MATH of all monomials in the MATH coordinates is a free MATH-module basis of MATH, with subsets spanning every ideal generated by monomials in MATH. In particular, each ideal MATH is spanned by a subset of MATH. |
math/0010246 | The points of MATH are exactly the ideals MATH, where MATH is such that MATH. In particular, every such MATH belongs to MATH, for MATH. Since MATH for all such MATH, we have MATH. |
math/0010246 | Let MATH be a point of MATH, where MATH for some MATH. On MATH we have MATH, while on MATH we have MATH. Hence MATH, so MATH. |
math/0010246 | At a point MATH, MATH coincides locally with MATH for a unique MATH. Identifying MATH with MATH, the ideal of the subspace in REF is locally either MATH, if MATH, or MATH, otherwise. It follows that every MATH coincides locally with an ideal in MATH generated by square-free monomials in the variables MATH, and for such an ideal we have MATH. |
math/0010246 | If MATH then for all MATH we have MATH, just as in the proof of REF . This implies REF . On MATH we have identically MATH, MATH for MATH, so the coordinate ring of MATH is generated by the remaining variables, namely MATH, MATH, MATH, MATH. This implies that the projection on these coordinates is an isomorphism of MATH onto its image, which is clearly MATH. |
math/0010246 | Let MATH be the projection on the coordinates MATH. On MATH we have identically MATH, MATH for MATH. As in the proof of the preceding lemma, this implies that MATH induces an isomorphism of MATH onto its image, which in this case is MATH. Note that MATH is also the image under MATH of the whole of MATH. Let MATH be the ideal on the right-hand side in REF, and let MATH. Then MATH is a closed subscheme of MATH, conceivably non-reduced. Modulo MATH we again have MATH, MATH for MATH, so MATH induces an isomorphism of MATH onto its (scheme-theoretic) image. But we have MATH, and MATH is the image of both MATH and MATH, so it is also the scheme-theoretic image MATH. The isomorphism MATH factors as the isomorphism MATH composed with the closed embedding MATH, so this implies MATH or equivalently MATH. |
math/0010246 | We are to show that every MATH can be uniquely expressed as MATH with MATH and MATH. By hypothesis, this is true for the image of MATH in MATH, with MATH, for all MATH. The local ring MATH is a subring of MATH, and the unique coefficients MATH satisfying REF for any MATH also satisfy REF for MATH. Hence they do not depend on MATH. Since the complement of MATH has codimension MATH, every rational function regular on MATH is regular everywhere. Thus the MATH belong to MATH. Since MATH is torsion-free and REF holds locally on the dense open set MATH, REF holds identically. |
math/0010246 | REF implies that any free MATH-module basis of MATH is also a basis of MATH, so MATH. Similarly MATH. |
math/0010246 | Let MATH be an element of MATH. Since MATH is free with basis MATH we can write MATH with MATH and MATH. Of course this is also the unique expression for MATH in terms of the basis MATH of the MATH-vector space MATH. Hence MATH for MATH, so we have MATH. |
math/0010246 | Let MATH denote the subspace MATH, and let MATH be its ideal and MATH its coordinate ring. By definition, MATH, where MATH is the intersection of the ideals MATH for all MATH. Since the MATH coordinates are independent on each MATH, the coordinate rings MATH are free, and hence torsion-free, MATH-modules. The ring MATH is isomorphic to a subring of the direct sum MATH, so MATH is also a torsion-free MATH-module. Let MATH be the set of pairs MATH with a given MATH, and let MATH be the coordinate ring of the partial union MATH . We have an injective ring homomorphism MATH . By REF , the partial unions MATH have disjoint restrictions to MATH. By REF , this implies that REF localizes to an isomorphism at each point of MATH, and in particular, upon tensoring with MATH. Now the projection of MATH on MATH is an isomorphism, so MATH projects isomorphically on MATH, where MATH is the union of coordinate subspaces MATH in MATH. The coordinate ring of MATH, say MATH, is the face ring of a simplicial complex. It has a homogeneous vector space basis consisting of all monomials MATH such that there is some MATH with MATH for all MATH. The ring MATH in turn is a free MATH module with this same basis. Since MATH, the result follows. |
math/0010246 | The requirement that MATH for all MATH, for a given MATH and some MATH satisfying REF, is equivalent to MATH. |
math/0010246 | We will show that MATH spans MATH as a MATH-vector space. Since the MATH-degree enumerator of MATH is clearly MATH, this implies the result by REF . Let MATH be the subset of MATH containing only the elements defined in REF. By REF we have MATH . It suffices to show that MATH spans MATH. For this it suffices in turn to show that MATH spans MATH and MATH spans MATH. Since MATH in MATH, multiplication by MATH gives a well-defined surjective homomorphism MATH. If MATH spans MATH, then MATH spans MATH, and therefore MATH spans MATH. Hence we need only show that MATH spans MATH. The ideal of MATH contains the ideal MATH . It is a consequence of the argument below that MATH is actually equal to the ideal in REF, but we do not need this result. We only verify that the generators displayed in REF do indeed vanish on MATH. This is clear for the generators in the first term, which is the special case of REF for MATH. The determinants in the second term vanish because on every MATH, either MATH, making the first two rows equal, or MATH, making one of the first two rows equal to the last. From REF it is easy to see that the ideal MATH contains MATH, MATH, MATH, and MATH for all MATH, and MATH for all MATH. With respect to a suitable term ordering, the initial ideal of MATH contains MATH, MATH, MATH, and MATH for all MATH, and MATH for all MATH. Hence MATH is spanned by monomials in MATH not divisible by any of these. In other words, MATH is spanned by monomials MATH where every element of MATH is less than every element of MATH, and MATH if MATH. Let us order the monomials in REF so that those with smaller values of MATH precede those with larger values, and for MATH, those that don't contain MATH as a factor precede those that do. Then it is easy to see that each monomial in REF occurs as the leading term in the reduction of an element of MATH modulo MATH. This implies that MATH spans MATH, as desired. |
math/0010246 | First we verify that the specified elements do belong to MATH. Observe (glancing ahead) that for each MATH, MATH, MATH, the ideal MATH displayed on the right-hand side in REF is generated by polynomials which vanish on MATH, so we have MATH. It is mostly routine now to check case-by-case that the relevant elements MATH belong to MATH. The only tricky case is to show that MATH belongs to MATH for MATH, MATH, and MATH. In this case MATH contains a factor MATH, which is not so obviously in MATH. However, in this case MATH contains MATH, and in MATH we have MATH, so MATH. Using REF , to complete the proof it is enough to show that the specified elements MATH span MATH. Since MATH is a homogeneous basis of MATH, it is equivalent to show that pairs MATH not satisfying REF, counted according to the MATH-degree MATH of MATH, are enumerated by the NAME series of MATH. This is true by REF . |
math/0010246 | In each case the ideal listed on the right-hand side is clearly contained in MATH. Conversely, in the proof of REF we showed that MATH is generated (and even spanned as a MATH-module) by elements belonging to the ideal on the right-hand side. |
math/0010246 | Let MATH be the coordinates in indices MATH, so MATH. Obviously MATH is a free MATH-module with basis MATH the set of all monomials in the variables MATH, and each ideal MATH is spanned by a subset of MATH. By REF , MATH is a free MATH-module with a basis MATH which is a common basis for the ideals MATH. It follows that MATH is a free MATH-module with basis MATH and that every ideal in MATH is spanned by a subset of MATH. This implies that MATH is a distributive lattice, and hence every ideal in MATH is an intersection of ideals of the form MATH, where MATH is a sum of ideals MATH. Now MATH, so we only have to show that MATH is reduced. REF implies that MATH is a free MATH module, and hence a subring of its localization MATH at any point MATH. By REF the latter ring is reduced for MATH (in particular, MATH suffices). |
math/0010246 | We already have the result on MATH by REF . Let MATH be a point of MATH. By REF , MATH coincides locally at MATH with a subspace arrangement isomorphic to MATH, for a set MATH with two elements. Unraveling the definitions of MATH and the local isomorphism, we find that MATH coincides locally, as a reduced closed subscheme of MATH, with a subspace arrangement MATH . Here we set MATH and MATH. The union ranges over MATH and MATH satisfying MATH, where MATH is chosen so that MATH. Note that the set MATH does not depend on the choice of MATH. It follows from REF that the ideals of the subspace arrangements in REF, for MATH, MATH fixed and MATH, MATH, MATH arbitrary, belong to a lattice consisting entirely of radical ideals. |
math/0010246 | A manifestation of translation invariance is that the endomorphism MATH of MATH defined by the substitutions MATH carries the ideals MATH into themselves, as is clear from their definition in REF. We therefore have MATH, where MATH is the defining ideal of MATH. Since MATH, we have an induced ring homomorphism MATH. We see immediately that if MATH is the canonical projection, then MATH is the identity map on MATH. Hence MATH is isomorphic to a subring of MATH. Since MATH is reduced, so is MATH, so we have MATH. |
math/0010246 | By REF , we have MATH, where MATH. By REF , we have MATH where MATH. Replacing MATH with MATH, this reduces the problem to showing that MATH in MATH for MATH. For this MATH we have set-theoretically MATH . By REF , the ideal MATH is equal to its radical, and its zero set is MATH. Applying REF with MATH to the union of subspaces MATH, we see that the ideal MATH is equal to its radical, and hence to MATH. The zero set of this ideal is MATH, which is the first big union in REF. Again by REF , the ideal MATH is equal to its radical, with zero set MATH. Observing that MATH in MATH, it follows from REF that the ideal of MATH, that is, of the second big union in REF, is MATH. Since MATH does not vanish identically on any component of the second big union, the ideal MATH is equal to the ideal of the latter. Hence we have MATH. We now claim that MATH contains MATH, and hence MATH. For the claim, observe that MATH in MATH, which shows that MATH. Observe also that MATH in MATH, which shows that MATH. Finally observe that the determinant MATH from REF reduces modulo MATH to MATH, which shows that MATH for all MATH. Multiplying the identity of ideals MATH by MATH yields MATH, and we showed above that MATH. By the modular law for ideals, since MATH, these imply MATH. |
math/0010246 | Note that by REF , MATH is reduced and equal to the union over all MATH of MATH. Thus MATH belongs to the lattice of ideals in REF , as does MATH. Therefore MATH is reduced. For MATH, MATH coincides locally with the arrangement MATH from REF , and we may as well replace MATH with MATH. First suppose MATH. Fixing some MATH such that MATH as in REF , either MATH is locally empty, if MATH, or else MATH coincides locally with MATH. If MATH then MATH for MATH. Otherwise MATH, where MATH, which is reduced by REF . In every case, the scheme-theoretic intersection of MATH and MATH either is empty or coincides locally with a product of reduced schemes, so MATH is locally reduced when MATH. Suppose instead that MATH. Then MATH coincides locally with MATH, where MATH. If MATH we have MATH and the result is immediate, as before. Otherwise it reduces to REF . |
math/0010246 | Define an equivalence relation MATH on MATH by the rule MATH if and only if MATH. It induces an equivalence relation MATH on functions MATH by the rule MATH if and only if MATH for all MATH (this differs from the equivalence relation in REF in that it depends on both the MATH and the MATH coordinates). It is easy to see that the functions MATH such that MATH lies on MATH form an equivalence class, and that the pointwise maximum of two equivalent functions is equivalent to each of them. This shows that MATH lies on MATH. The function MATH takes at most one value in each MATH-equivalence class, namely, the maximum element in that class. Furthermore, that value is only in MATH if the whole equivalence class is contained in MATH. Hence, among all functions in its MATH-equivalence class, MATH maximizes the size of the intersection of MATH with every MATH-equivalence class. The set MATH is a union of MATH-equivalence classes, so it follows that MATH maximizes MATH. But MATH belongs to MATH if and only if the maximum of this quantity over all MATH such that MATH is at least MATH. |
math/0010246 | It is clear that MATH contains the union in REF. Given a point MATH of MATH, let MATH. By the definitions of MATH and MATH, we have MATH for some MATH such that MATH, and MATH for some MATH such that MATH. A priori, MATH and MATH might be different, but REF allows us to take them both to be the maximum MATH for which MATH contains MATH. Hence MATH is contained in the union in REF. |
math/0010246 | Note that MATH is the union of the subspaces MATH, over all MATH such that MATH. Let MATH be the closed subset MATH . Thus MATH is the union of the components of MATH that are not components of MATH. For any subset MATH, it follows that we have MATH . For MATH we show that MATH. Fix a point MATH and let MATH be the maximum MATH such that MATH contains MATH. Note that a component MATH contains MATH if and only if MATH for some MATH containing MATH. Hence MATH is the maximum MATH such that MATH contains MATH. Let MATH. By REF , we have MATH if and only if MATH. Since MATH, we have MATH, and since MATH is maximum we also have MATH. Hence we have MATH and MATH. Note that MATH and that MATH, since MATH. By REF again, we have MATH if and only if MATH. This establishes REF. For MATH, a subtle point arises, because it is not true in this case that MATH. We can show, however, that MATH and MATH have the same intersection with MATH, which suffices to establish REF. Therefore let MATH be a point of MATH, assume MATH and define MATH, MATH and MATH as above. We claim that MATH. Since MATH we certainly have MATH but we could also have MATH for some MATH, in which case MATH would be the largest such MATH. If this occurs, however, then we have MATH, in the notation used in the proof of REF . This implies MATH for all MATH and hence MATH, contrary to assumption. Having established that MATH, we have MATH and MATH. Now we conclude as before that MATH belongs to MATH if and only if MATH belongs to MATH. |
math/0010246 | Let MATH be a point of MATH. The NAME MATH-algebra MATH is a quotient of the polynomial ring MATH, and therefore has a MATH-basis MATH, where MATH. Equivalently, MATH minimally generates MATH as a MATH-module, so with MATH as in the remarks following REF , we have MATH, MATH. This implies MATH, MATH, that is, MATH, MATH. By REF it follows that MATH if and only if MATH. |
math/0010246 | Let MATH, MATH, and MATH, where MATH. Then MATH, so the defining ideal of MATH is MATH. Since MATH is flat, we have MATH. From the definition it then follows that MATH, which is the algebraic equivalent of REF. |
math/0010246 | Let MATH be the leading coefficient of an element MATH with leading form MATH, as in REF. Similarly let MATH be the leading coefficient of MATH with leading form MATH. Then we have MATH, and the leading form of MATH is MATH where MATH. |
math/0010246 | Consider the filtration MATH of MATH, where MATH. Without loss of generality we can assume that MATH, since the hypotheses and conclusion of the Lemma are unaltered if we replace MATH by a larger integer MATH and define the extra subsets MATH for MATH to be empty. We have MATH and therefore, by REF , MATH for all MATH. It follows that MATH is a free MATH-module with basis MATH, and since MATH is the union of these, MATH is a free MATH-module with basis MATH. By REF this implies that MATH is a free MATH-module with basis MATH. By REF , we have MATH for MATH. By the definition of MATH, multiplication by MATH is an isomorphism MATH. The MATH-module MATH is generated by MATH and is therefore equal to MATH, or to MATH, by the modular law. Multiplication by MATH is thus an isomorphism MATH. Hence for MATH we have MATH and therefore MATH. It follows that the map MATH given by multiplication by MATH is surjective. It is injective by the definition of MATH, so it is an isomorphism. Setting MATH, this implies that MATH is a basis of MATH. Since MATH is the union of these, MATH is a basis of MATH. In particular this holds for MATH, which by REF suffices to show that MATH spans MATH. |
math/0010246 | Let MATH be a point of MATH, and let MATH be a point of the fiber. Note that this implies MATH. By REF there is a unique MATH containing MATH and a unique MATH containing MATH. Clearly we must have MATH. By the definition of MATH, we have MATH, and MATH vanishes at MATH and hence at MATH. Note that MATH determines MATH, since MATH projects isomorphically on MATH, and this projection must map MATH to the image of MATH. Conversely, for every MATH such that MATH vanishes at MATH, there is a point MATH lying on MATH, where MATH and MATH. This shows that the number of points in the fiber MATH is equal to the size of the set MATH, where MATH. By REF , MATH consists of those points MATH for which MATH. Since MATH is dense in MATH, the result follows. |
math/0010246 | Let MATH be the projection. If MATH is a point of MATH such that the fiber MATH has at least MATH points then in the notation of REF we have MATH. By REF it follows that MATH. Since MATH is reduced, this implies MATH. |
math/0010246 | By REF we have MATH in MATH. This containment implies equality trivially if MATH or MATH. For MATH we have MATH. For MATH, MATH is empty, so MATH for MATH. Hence we may assume MATH. For MATH, the definition yields that MATH is the kernel of the ring homomorphism MATH corresponding to the projection MATH, that is, the ideal of the image of MATH in MATH. The latter coincides with MATH by construction. Since MATH vanishes on MATH, we have MATH for MATH. If MATH, then MATH vanishes on MATH, and if MATH, then MATH vanishes on MATH. Hence we have MATH for MATH or MATH, and MATH. This leaves only the case MATH, MATH, in which we are to show MATH . We have already seen REF that MATH. Thus we have only to show that MATH, MATH, and the MATH for MATH belong to MATH. For this we need only observe that MATH, MATH, and MATH are the MATH coefficients of the following polynomials, which vanish on MATH: MATH . For the vanishing of the last of these, note that it is congruent modulo MATH to the determinant MATH from REF, which vanishes on MATH. |
math/0010246 | First consider a point MATH. By REF , MATH lies on a unique component MATH, for some MATH, and MATH coincides locally with MATH. Since MATH maps MATH into MATH, every point of MATH lying over MATH belongs to a component MATH with MATH and hence MATH. We can replace MATH with the union MATH of only these components, without changing MATH and therefore without changing the local ideals MATH. Identifying MATH with MATH, MATH is the union of subspaces MATH over all MATH. Thus the defining ideal MATH of MATH contains MATH . From this we see that MATH contains all square-free monomials of degree MATH in the variables MATH. These monomials generate the ideal of the union of coordinate subspaces MATH which coincides locally with MATH at MATH. Hence we have MATH, and we already have the reverse containment by REF . For MATH, the local picture of MATH is given by REF as MATH for some MATH, and MATH. Let MATH be the equivalence class of functions MATH defined in REF , containing all MATH such that MATH. We put MATH and MATH. As before, MATH coincides locally over MATH with the union of its components MATH, where MATH for some MATH. We divide these into two classes. The first consists of components with MATH. These have a point lying over MATH only if MATH belongs to MATH . Note that MATH is well-defined, since MATH depends only on MATH. The union of these components is MATH, where MATH is the union of the subspaces MATH over MATH. Repeating our analysis of the case MATH, we see that for each MATH, MATH contains the ideal MATH generated by all square-free monomials of degree MATH in the variables MATH. Here MATH is the coordinate ring of MATH. The second class consists of components MATH of MATH with MATH and MATH. The union of these components is MATH, where MATH is the special arrangement MATH over MATH, with extra coordinate MATH, and MATH, MATH. By REF we have MATH for all MATH. Note that this is still correct when MATH and MATH is empty. Using REF , we see that for MATH we have MATH. By REF , the ideal MATH is equal to its radical. Here we used the identity MATH, which holds because MATH and MATH are flat over the field MATH. We have MATH . The second equality here follows because the closed subsets MATH and MATH decrease as MATH and MATH increase. Unraveling the local isomorphism of MATH with MATH we see that the union in REF coincides locally at MATH with MATH. This shows that MATH, and we already have the reverse containment by REF , as before. |
math/0010246 | By REF , the coordinate ring MATH of MATH is a torsion-free MATH-module. By REF we have MATH for all MATH and all MATH. Thus the hypotheses of REF are satisfied and we have REF immediately. For REF also to yield us the last conclusion we need to have MATH for MATH and MATH, where MATH is the scheme-theoretic preimage MATH in MATH, that is, MATH where MATH. As in the proof of REF , if MATH is a point of MATH belonging to the unique component MATH, then MATH contains MATH distinct points lying over MATH, where MATH. If MATH belongs to MATH, then (since MATH is unique) we have MATH. If MATH also belongs to MATH then the points lying over MATH belong to MATH, which shows that MATH as sets. By REF , the last intersection is scheme-theoretically reduced, so we have MATH for all MATH. Let MATH be the defining ideal of MATH, so MATH. We defined MATH as an ideal in MATH, but we can also regard it as an ideal in MATH which contains MATH. Obviously MATH is contained in the defining ideal MATH of MATH. This implies MATH and hence MATH. For the reverse containment, we clearly have MATH for MATH, and we have MATH because MATH and MATH contains MATH. |
math/0010246 | REF implies that MATH is a free MATH-module and therefore a subring of MATH. Let MATH be the ideal of MATH as a subscheme of MATH, so MATH. Both MATH and MATH belong to the lattice in REF . Moreover MATH for all MATH, as follows easily from REF . Hence REF implies that MATH is equal to its radical for all MATH, and by REF this implies that MATH is a reduced ring for all MATH. In particular, taking MATH, MATH is a reduced ring. |
math/0010246 | For each basis element MATH, let MATH be the unique non-negative integer MATH such that MATH belongs to MATH but not to MATH. Similarly, let MATH be the unique MATH such that MATH belongs to MATH but not to MATH. Thus by REF , the set of elements MATH is a free MATH-module basis of MATH and the set MATH is a free MATH-module basis of MATH. From the definition we have MATH, hence MATH. Therefore we have MATH for all MATH. For each MATH such that MATH, the element MATH belongs to MATH but not to MATH. There is, however, a unique MATH-linear combination MATH of the elements in MATH such that MATH. Set MATH, so we have MATH, and MATH is congruent to MATH modulo MATH. Let MATH be the set of all such elements MATH. The union MATH is again a basis of MATH, while MATH is a basis of MATH and MATH is contained in MATH. This implies that MATH spans MATH, that is, MATH is a free MATH-module with basis MATH. By REF , MATH is spanned by the subset MATH. Consider an element MATH with MATH, and the corresponding element MATH constructed as above. By REF applied to MATH, the set MATH spans MATH. Hence there is a MATH-linear combination of the elements of MATH congruent to MATH modulo MATH. By uniqueness this linear combination must be MATH, so MATH belongs to MATH. Let MATH be the set of these elements MATH. As before, MATH spans MATH, while MATH is a basis of MATH and MATH is contained in MATH. This implies that MATH spans MATH. |
math/0010246 | We can assume we are given a common ideal basis of MATH, represented by a set of homogeneous polynomials MATH not involving the variables MATH, MATH. We will show that MATH is the required basis of MATH. Let MATH. From the definition it is immediate that MATH, where MATH. This implies that MATH is a free MATH-module with basis MATH. It follows from REF that for MATH, the image of MATH under the isomorphism is MATH, so MATH. This shows that MATH is a common ideal basis as far as the ideals MATH with MATH are concerned, and those with MATH are trivial. |
math/0010246 | Choosing an arbitrary representative MATH of each MATH-coset belonging to MATH, we can assume that MATH is given as a subset of MATH. Then MATH is a free MATH-module with basis MATH. This basis need not be a common basis of the ideals MATH; to construct MATH we have to modify it. Given MATH, let MATH be the set of indices MATH for which MATH contains MATH, and let MATH. Then MATH contains MATH. For MATH we have MATH by REF , hence MATH. Thus there exists an element MATH satisfying MATH . These conditions determine MATH and MATH uniquely modulo MATH. Now, it follows from the hypotheses of the Lemma that if MATH is the subset consisting of elements not belonging to MATH, then MATH is a free MATH-module with basis MATH. Expanding the unique element MATH satisfying REF in terms of the basis MATH, we have MATH with MATH and MATH. The coefficients MATH are uniquely determined by REF, and the solution for any MATH is also the solution for MATH. As in the proof of REF , it follows that as elements of MATH, the coefficients MATH do not depend on MATH and hence belong to MATH. In particular, the polynomial MATH defined in REF is an element of MATH satisfying REF for all MATH. We now define MATH to be the union of MATH and the set of elements MATH constructed as above, for all MATH. By REF we have MATH. This implies that the elements MATH form a basis of MATH, so MATH is a basis of MATH. The hypothesis that MATH is a torsion-free MATH-module means that the canonical map MATH is injective. Since MATH for MATH, and we have MATH, this implies MATH for all MATH. For each MATH, the elements MATH span the MATH-module MATH by hypothesis, and therefore span the vector space MATH. Since the corresponding elements MATH belong to MATH and satisfy MATH, it follows from the canonical isomorphism MATH that they span MATH. By hypothesis the elements of MATH span MATH, so the elements of MATH all together span MATH. By REF , this implies they span MATH as a MATH-module. |
math/0010246 | Let MATH. By definition, MATH, so MATH. By REF , MATH. Since MATH does not vanish identically on any MATH, it is a non-zero-divisor in MATH. Hence multiplication by MATH is an isomorphism of MATH onto MATH. We may assume we are given a homogeneous common ideal basis MATH of MATH. Suppose that in a given MATH-degree MATH, we can find a free MATH-module basis MATH of MATH such that every MATH is spanned by a subset of MATH. We claim that MATH is then a basis of MATH with subsets spanning each MATH. Granting this claim for the moment, we can apply REF with MATH and MATH the degree MATH part of the given basis MATH to obtain a free MATH-module basis MATH of MATH which is a common basis for every MATH. In degree zero, we can take MATH, since MATH. We can then construct MATH by induction for all degrees MATH, obtaining a common ideal basis MATH of MATH. It remains to prove the claim. If MATH is a free module basis of MATH then so is MATH, and therefore MATH is a basis of MATH, since multiplication by MATH is an isomorphism of MATH onto MATH. To complete the proof, observe that for any ideal MATH we have MATH. If MATH is radical, then so is MATH, and MATH is the union of those components of MATH on which MATH does not vanish identically. Applying this to MATH for MATH, we get MATH, and hence MATH. For MATH, this shows that if MATH has a subset spanning MATH, then MATH has a subset spanning MATH. This suffices, since MATH is trivially equal to MATH or MATH for MATH. |
math/0010246 | We caution the reader immediately that the intersections in REF are not scheme-theoretically reduced in general, so it is not enough to check the result set-theoretically. Instead, we must use our knowledge of the local picture on MATH to write down equations. If MATH is a point of the intersection on either side of REF, then we have MATH for some MATH, MATH with MATH, MATH, so MATH is not in MATH. Thus we need only consider points MATH. The schemes in question are subschemes of MATH. Fix a point MATH, and let MATH, MATH, MATH, MATH, and MATH be as in REF . We may replace MATH with MATH without changing any of the local ideals at MATH. First consider the case MATH. If MATH we must have MATH. But if MATH or MATH, we must have MATH. Hence both intersections in REF are locally empty at MATH. For the case MATH we may assume MATH, and hence MATH, as otherwise MATH and the result is trivial. Under the local isomorphism of MATH with MATH, MATH coincides locally with MATH, where MATH is the subspace arrangement MATH over MATH such that MATH, MATH for MATH. Similarly, MATH coincides locally with MATH, where MATH, and MATH coincides with MATH, where MATH over MATH such that MATH, MATH for MATH. We are to show that MATH as subschemes of MATH. Renaming the indices so that MATH becomes MATH, MATH becomes MATH, and (hence) MATH becomes MATH, our subschemes become subschemes of MATH: MATH . Let MATH be given by MATH . By REF , MATH and MATH, where MATH. By REF or REF , we have MATH in MATH, which shows that MATH is a radical ideal. Since we clearly have MATH set-theoretically, it follows that MATH . By symmetry, we have MATH and by REF we have MATH . Now we see immediately that both MATH and MATH contain MATH and hence they both reduce to MATH . |
math/0010246 | To simplify notation let MATH and let MATH be the ideal MATH in MATH. We first prove that MATH is a free MATH-module. The ideal MATH is isomorphic to MATH. Since MATH in MATH, we have MATH, and the latter is a free MATH-module by hypothesis. The locus MATH is isomorphic to the preimage MATH in the special arrangement MATH over MATH, where MATH. The isomorphism is given by transposing the indices MATH and MATH in MATH. REF then implies that MATH is a free MATH-module. Let MATH be a free MATH-module basis of MATH and let MATH be a basis of MATH. By REF we have MATH for MATH, so MATH is a basis of MATH. Then MATH is a free MATH-module by REF . Now take MATH , MATH, and MATH as in the preamble to REF , so MATH and MATH are special arrangements over MATH, where MATH. Set MATH, MATH, and let MATH be the ideal of MATH as a closed subscheme of MATH. By REF we have MATH in MATH so the canonical map MATH is an isomorphism. Similarly, applying MATH to REF, we have MATH in MATH, hence MATH is an isomorphism as well. Since MATH and MATH are both free MATH-modules, finitely generated and MATH-graded in each MATH-degree, it follows that MATH is a free MATH-module. Hence MATH is a free MATH-module. By REF , MATH is a free MATH-module with a common basis MATH for the ideals MATH, where MATH. In particular, this implies that MATH and MATH are free MATH-modules. By REF , MATH is a torsion-free MATH-module. By REF , the free submodules MATH and MATH of MATH coincide when localized at MATH, so by REF they are equal. Let MATH. With MATH as in the previous paragraph, MATH is a free MATH-module basis of the ideal MATH in MATH, with subsets spanning the ideals MATH. We have canonical isomorphisms MATH. Let MATH be the image of MATH under the composite isomorphism MATH. The canonical isomorphism MATH is given by restriction to MATH of functions on MATH vanishing on MATH. REF implies that each ideal MATH is equal to its radical, namely the ideal MATH of the reduced preimage of MATH in MATH. Therefore, since every function MATH vanishes on MATH, MATH belongs to MATH if and only if its restriction to MATH vanishes on MATH. Similarly, writing MATH for the coordinate projection MATH, a function MATH belongs to the ideal MATH in MATH if and only if the same criterion holds (note that MATH and MATH have the same restriction to MATH). This shows that the isomorphism MATH carries MATH onto MATH. Hence the latter ideals are spanned by subsets of MATH. Given MATH, MATH and MATH, set MATH, MATH if MATH, or else MATH, MATH if MATH. By REF , we have MATH. The ideal in MATH of MATH is MATH, and we have shown above that it is spanned by a subset of MATH. But this ideal is equal to the ideal of MATH, namely, MATH. This shows that all the ideals MATH are spanned by subsets of MATH, the cases with MATH being trivial. The conclusion now follows from REF . |
math/0010246 | We prove the theorem by induction on MATH and MATH. The base case MATH is given by REF . The base case for MATH is MATH. Note that for MATH we have MATH, for all MATH, and that the only non-trivial MATH is MATH, which already appears in MATH. Thus the case MATH is included in the case MATH. For MATH and MATH we can assume by induction that MATH has a homogeneous common ideal basis. Then by REF , so does MATH. We can also assume by induction that MATH has a homogeneous common ideal basis. Applying REF repeatedly, with MATH descending from MATH to MATH, we conclude that MATH has a homogeneous common ideal basis. Then by REF , so does MATH. |
math/0010246 | Set MATH, and let MATH be the polygraph over MATH, a subspace arrangement in MATH. Let MATH be the Cartesian product of MATH copies of the symmetric group MATH, acting on MATH by permuting the factors in MATH in MATH consecutive blocks of length MATH. In other words each MATH fixes the coordinates MATH on MATH, and for each MATH permutes the coordinate pairs MATH through MATH among themselves. Let MATH be the coordinate ring of MATH. By REF , MATH is a free MATH-module. By the symmetry of its definition, MATH is a MATH-invariant ideal, so MATH acts on MATH. We claim that MATH is isomorphic as a MATH-module to the space MATH of MATH-alternating elements of MATH. Each MATH-degree homogeneous component of MATH is a finitely generated MATH-graded free MATH-module. Since MATH is a graded direct summand of MATH, it is a free MATH-module, so the claim proves the Lemma. Let MATH be defined by MATH for all MATH, MATH. Restriction of regular functions from MATH to its component subspace MATH is given by the MATH-algebra homomorphism MATH mapping MATH to MATH. Observe that MATH maps MATH surjectively onto MATH. Let MATH be an arbitrary element of MATH. Since MATH is MATH-alternating, MATH vanishes on MATH if MATH for some MATH and some MATH. Thus the regular function defined by MATH on MATH is determined by its restriction to those components MATH such that for each MATH, the sequence MATH is a permutation of MATH. Moreover, for every such MATH there is an element MATH carrying MATH onto MATH. Hence MATH is determined by its restriction to MATH. This shows that MATH vanishes on MATH if MATH, that is, the kernel of the map MATH is zero. |
math/0010246 | By the preceding remarks, the Theorem holds for MATH. NAME the exact sequence MATH over MATH with MATH, we get an exact sequence MATH with MATH a free MATH-module. Since MATH for all MATH, and MATH by REF, we have MATH. Let MATH. Note that MATH is the affine open set MATH, so inverting MATH is the same thing as localizing to MATH. By REF , therefore, we have MATH, and hence MATH. By REF is the kernel of the canonical homomorphism MATH . Hence inverting MATH in REF gives MATH. Now MATH is a non-zero-divisor in MATH, so it is also a non-zero-divisor on the free MATH-module MATH and its submodule MATH. But since MATH, every element of MATH is annihilated by a power of MATH. Therefore we have MATH and MATH. |
math/0010247 | If MATH, this was already pointed out by CITE. Now suppose MATH. Let MATH be a MATH-tree of degree MATH. If we choose any univalent vertex to be the root of MATH, then we can associate to this rooted MATH-tree an element in MATH. For example, if MATH is the MATH-graph in REF , with the counter-clockwise orientation at each vertex and root at MATH, then the associated element would be MATH, where MATH is a lift of MATH into MATH. Now we associate to MATH the sum of the associated elements of MATH as the roots range over all the univalent vertices of MATH. It is not hard to see that this element actually lies in MATH and that, in fact, MATH is generated by the elements associated to all possible MATH-trees of degree MATH. |
math/0010247 | We construct a manifold MATH from MATH by adjoining handles to MATH along the framed link MATH and by removing tubular neighborhoods of properly imbedded disjoint disks MATH in MATH obtained by pushing MATH into the interior of MATH. Then MATH, so it only remains to observe that the pair MATH is acyclic. Let MATH. It is easy to see that the only non-zero homology group of MATH is MATH, which is freely generated by the meridians of MATH. By considering the triple MATH we find there is an exact sequence: MATH . Now MATH is the only non-zero homology group of MATH and it is freely generated by the disks adjoined along MATH. Since the homomorphism MATH is represented by the matrix of intersection numbers of the components of MATH with those of MATH, it follows that MATH is acyclic if and only if REF is satisfied. |
math/0010247 | Recall that for each edge of MATH there are two components of MATH, one at each end of the edge (see REF ). We will call them companion components. We will construct MATH and MATH by assigning, for each edge of MATH, one of the associated companions to MATH and the other to MATH. This choice can be represented by an orientation of the edge pointing from the end associated to the companion in MATH toward the end associated to MATH. Our aim will be to make these choices satisfy: CASE: An edge with a univalent vertex ( a leaf of MATH) is oriented toward the leaf, that is, outward (thus the leaves of MATH will all belong to MATH), CASE: No trivalent vertex is a source, that is, not all the incident edges are oriented away from the vertex. We will see that these conditions can be satisfied if and only if MATH is not a tree. But for now note that if these conditions are satisfied then the decomposition of MATH will satisfy REF . Since the three components of MATH associated to any trivalent vertex are a Borromean rings, any two of the components bound disjoint disks. Thus we can choose disjoint disks MATH bounded by each component of MATH, the disks from components associated to different vertices will be disjoint. The only problem would be if these components were associated to the same edge, but this is ruled out. Now each disk from a component of MATH will intersect the companion component of MATH once - the only other intersections will be with one of the components of MATH which is associated to the same trivalent vertex, but the intersection number will be MATH. Thus the intersection matrix of the components of MATH with those of MATH will be the identity matrix. See REF for an example. Suppose that MATH is a unitrivalent tree, with an orientation prescribed for each leaf edge, not all outward. Then we can extend this to an orientation of all the edges of MATH which satisfies REF . A similar fact is proved in CITE. Choose one of the leaves MATH of MATH which is oriented inward. Now orient every edge of MATH which is not a leaf so that it points away from MATH, that is, if we travel along any non-singular edge path which begins at MATH and ends at a non-leaf, then the orientations of all the edges in the path point in the direction of travel. Then it is clear that any trivalent vertex will have at least one of its incident edges oriented toward that vertex. Thus REF will be satisfied whatever the orientations of the other leaves. See REF for an example. We can now complete the proof of REF . Since MATH has a cycle we can make one or more cuts in edges of MATH to create a tree. Each edge of MATH which is cut will create two new leaves in MATH. We now choose arbitrary orientations of each cut edge of MATH, which will induce orientations of the new leaf edges of MATH. Note that one of each pair of new leaves will be oriented inwards. Thus the outward orientations of the leaves of MATH together with these orientations of the new leaves of MATH provides orientations of all the leaves of MATH which satisfies the hypothesis of REF . Applying this lemma gives an orientation of MATH satisfying REF . But now we can glue the cut edges back together and we get an orientation of MATH satisfying REF , thus proving the Theorem. See REF for an example. |
math/0010247 | First note that it follows from REF that the composition MATH factors through MATH. This yields the commutative REF , assuming for the moment that MATH. To see that MATH for MATH, and, at the same time, identify the composition of the maps in the bottom line of the diagram, we first consider a MATH-clasper MATH in MATH associated to a connected MATH-tree MATH. If we choose a root of MATH and the element of MATH associated to this rooted MATH-tree is MATH, then, by a sequence of NAME moves, we can convert MATH into a MATH-component link MATH, where MATH is the leaf of MATH corresponding to the root, and so represents MATH, and MATH represents MATH . For example, REF explains this for a rooted MATH-tree of degree MATH. Now suppose MATH and MATH. To compute MATH we push a curve MATH representing MATH down MATH and observe how it changes every time we cross MATH. When it crosses the leaf MATH, the effect may be computed by using the NAME link MATH instead of MATH. But then it is not hard to see that the effect is to add a copy of MATH to MATH. Thus the contribution of all the crossings of MATH with the leaf MATH is MATH, where MATH denotes the homological intersection number in MATH. Under the canonical isomorphism MATH this corresponds to MATH. The total change in MATH is then the sum of these contributions over all the univalent vertices of MATH, which is exactly the element of MATH corresponding to MATH. This shows that MATH, at least for the elements represented by trees in MATH, and that the composition MATH associates to any MATH-tree the corresponding element of MATH under the isomorphism of REF . The proof of the commutativity of REF is completed by noting that if MATH is any MATH-graph with a cycle (and an ordering of its univalent vertices), then we can cut open some edges to create a tree MATH, where the new vertices are labeled by MATH. Now a clasper representing MATH is also a clasper representing MATH and the above argument applies. Since some of the labels are MATH, the image in MATH is MATH. To prove that MATH is an isomorphism first note that the map MATH is onto, since MATH is onto. So if MATH it must also be one-one, since the composition to MATH is an isomorphism. Thus all the maps in the bottom row must be isomorphisms. |
math/0010247 | REF are proved above. To prove REF first consider MATH. Suppose that MATH and MATH. Then MATH and so MATH. For MATH we proceed by induction, using the following commutative diagram. MATH . The columns are exact and, by induction, the middle row is exact. It then follows that the top row is exact. |
math/0010247 | By REF MATH and so MATH. |
math/0010247 | Let MATH denote the subset (it is not a subgroup) of MATH consisting of all MATH such that MATH is an isomorphism, where MATH is the inclusion and MATH is the projection with MATH. Note that MATH. We will define a ``retraction" MATH, which is not a homomorphism, but will satisfy MATH and so prove the theorem. We identify MATH with the complement of the trivial framed string link. Now consider an imbedding of MATH into MATH defined by removing a thin collar of the entire boundary of MATH. Thus we obtain an imbedding of MATH into the interior of MATH. Now let MATH denote the solid handlebody of genus MATH whose boundary is MATH. We will make this identification so that the MATH represent a basis for MATH, which we identify with MATH, and the MATH correspond to the meridians of the handles of MATH. We can imbed MATH into MATH as a thickening of the imbedded MATH and so that MATH is the imbedding which we used to define MATH. Now suppose MATH represents an element of MATH. Then we can cut open MATH along the imbedded MATH and insert a copy of MATH so that MATH. If we identify MATH with the complement of the trivial framed string link then our newly constructed manifold is identified with the complement of some string link which lies in a homology MATH-ball precisely when MATH. We take this to represent MATH. It is not hard to see that MATH. |
math/0010247 | CASE: Suppose MATH. Then MATH and, therefore, by NAME 's lemma, if MATH is the meridian disk of MATH corresponding to MATH, MATH bounds a disk MATH . By standard cut and paste techniques we can assume that the MATH are disjoint. We can now extend MATH over each MATH by mapping it onto MATH. Since the complement of MATH is a MATH-ball, as is the complement of MATH, we can extend over MATH. CASE: Suppose MATH. Let MATH be a homology bordism from MATH to itself, with MATH, where MATH are two copies of MATH. Let MATH, where MATH is attached to MATH along MATH. Then MATH is a homology bordism between MATH, the complement of the trivial string link (triv in REF ), and MATH, where MATH is the string link constructed as the representative of MATH. Thus we can fill in MATH with product strings to yield a concordance from the trivial string link to MATH. See REF Conversely suppose MATH and MATH. Let MATH be the complement of a concordance between the trivial string link and MATH, the constructed representative of MATH. If MATH is the complement of MATH, then we can decompose MATH. Thus MATH . In this way we can see that MATH is a homology bordism from MATH to itself which exhibits MATH as an element of MATH. |
math/0010247 | MATH is defined by MATH (see REF ). To see that MATH is one-one we only need note that MATH, which follows from REF and that MATH. To prove that MATH is onto we need: If MATH, then MATH. The idea is given by the schematic pictures below. The first picture shows the complement of MATH. MATH is the complement of the string link MATH and the shaded region is a product MATH. Note that MATH can be constructed by stacking MATH on MATH along MATH. The second picture gives an alternative view of the first picture which can then be recognized as the complement of MATH. Filling in the strings completes the proof. MATH . Now suppose MATH. Then MATH, by REF , and MATH. |
math/0010247 | If MATH, then we can lift MATH to an endomorphism MATH of MATH and we can realize MATH by an imbedding MATH. If MATH denotes the imbedded copy of MATH, then define MATH, attaching MATH to MATH. MATH is a homology bordism between MATH and MATH and clearly maps to MATH under MATH. |
math/0010247 | CASE: Write MATH, where MATH represents the MATH disjoint handles. Then the imbedding MATH which defines MATH can be regarded as the restriction of an identification of MATH with the upper hemisphere MATH of MATH - the holes in MATH which are removed to obtain MATH are the disks along which one end of each handle of MATH is attached to MATH. Now consider MATH and a framed string link MATH. We create a manifold MATH by attaching MATH to MATH by attaching one end of each handle to MATH and the other to the trivial framed string link in MATH. It is not difficult to see that this is a homology bordism from MATH to itself which extends MATH. CASE: This follows directly from REF , since MATH. |
math/0010247 | CASE: Since MATH for all MATH, this follows immediately from the fact that MATH is onto CITE. CASE: Note that MATH iff. MATH for all MATH, where MATH, the normal closure of MATH. To see this first note that MATH iff. MATH for every MATH. Secondly note that MATH iff. MATH for all MATH - but this is equivalent to MATH for all MATH, since, if a symplectic matrix has the form MATH, it follows that MATH iff. MATH. Now note that, as a consequence of CITE, MATH iff. MATH for all MATH. Thus, since MATH and MATH, REF will follow from MATH. By NAME 's theorem the quotient MATH. But MATH, which is a free group and so MATH. CASE: If MATH, then MATH, for all MATH. We can thus apply the argument in the proof of REF . |
math/0010247 | CASE: If MATH then MATH. We will abuse notation and allow ourselves to denote the induced bases of MATH and MATH by MATH and MATH, respectively. If MATH, then MATH in MATH and so MATH in MATH , since MATH. CASE: Let MATH be a lift of MATH. Then we can write MATH for some MATH and MATH, where the latter inclusion is some splitting of MATH. Since MATH we can choose MATH. If MATH is the homology class of MATH in MATH, then MATH and so MATH represents MATH. If MATH represents MATH, then MATH. Thus MATH represents MATH. From these observations we conclude that MATH, which is represented by MATH is given by REF This is immediate. |
math/0010247 | Recall the definition of MATH from REF . It is well-known that MATH, the MATH-th term of the lower central series, if and only if every MATH - see for example, CITE. We therefore have MATH and the induced MATH is injective. That MATH follows from REF . It also follows directly from the definitions that the composite MATH is as claimed. |
math/0010247 | The first assertion is clear from the definitions. The rest of the theorem follows from the observation that the following diagram is commutative MATH and the fact that MATH is onto. |
math/0010247 | We need to show that MATH. First note that MATH - this follows from REF . For the ontoness we can apply the ontoness argument of REF with the extra fact that MATH. This follows from the observation that the following diagram is commutative MATH where MATH is defined by MATH. |
math/0010249 | From the isomorphism given by REF and recalling that the transform MATH preserves the MATH groups of WIT sheaves, it is enough to show that the same claim holds for the morphism REF . By the very definition of the NAME map, the tangent map to MATH may be identified with the map MATH obtained in the following way. Let MATH be a sequence representing an element of MATH. If we apply the functor MATH to the above sequence we obtain the exact sequence MATH . One checks immediately that the map MATH is well defined. If MATH then MATH is represented by the extension MATH . Now applying the functor MATH to the above sequence and noting that MATH for every vector bundle MATH on MATH we obtain the split exact sequence MATH . Therefore MATH implies MATH and MATH is injective. |
math/0010249 | Since MATH is smooth, and MATH is injective and is an immersion, it is also an embedding. Now, let MATH. It is enough to show that the NAME product MATH vanishes when applied to pairs MATH of elements in MATH where MATH and MATH are represented, respectively, by the sequences MATH with MATH. It is enough to remark that the product of the classes of the sequences of sheaves on MATH is zero for dimensional reasons, and apply the functor MATH. In the case MATH the moduli space is smooth by the results in CITE; moreover, MATH . |
math/0010249 | Since MATH is prime, every sheaf in MATH is properly stable. Let MATH and assume that the subsheaf MATH destabilizes MATH. Let MATH. Standard computations show that if MATH is not MATH-stable then MATH . Setting MATH we have MATH, with MATH and MATH. This is impossible whenever MATH is prime. The statement about local freeness follows from the NAME inequality. |
math/0010251 | Let MATH be MATH-stable and assume that MATH is a proper sub MATH-module of dimension vector MATH. Restricting MATH to a representation of MATH we see that MATH and is a subrepresentation of MATH. Because MATH we have MATH, which is impossible as MATH is MATH-stable. Let MATH be a simple MATH representation and assume that MATH is a proper subrepresentation of dimension vector MATH with MATH. If MATH then MATH whence MATH has a kernel but this contradicts the fact that MATH is invertible. Hence, MATH but then MATH is a proper subrepresentation of MATH contradicting simplicity. |
math/0010251 | Let MATH be a MATH-stable representation of MATH of dimension vector MATH and consider the MATH-dimensional representation MATH . It is clear from the definition that MATH is the local quiver setting corresponding to MATH. As before, there is a semi-invariant MATH such that MATH and MATH is a semi-simple representation of the universal localization MATH. If there are MATH-stable representations of dimension MATH, then there is an open subset of MATH consisting of MATH-stable representations. But we have seen that they become simple representations of MATH. This means that every NAME neighborhood of MATH contains simple MATH-dimensional representations. By the étale local isomorphism there are MATH-dimensional simple representations of the quiver MATH. Conversely, as any NAME neighborhood of the zero representation in MATH contains simple representations, then so does any neighborhood of MATH. We have seen before that MATH consists of MATH-semistable representations and that the MATH-stables correspond to the simple representations of MATH, whence MATH has MATH-stable representations of dimension vector MATH. |
math/0010251 | Let MATH such that MATH and MATH a subrepresentation of dimension vector MATH. Then, MATH for if MATH then MATH and the composed map MATH must have a kernel contradicting the fact that MATH is an isomorphism. If MATH, then MATH and the restriction of MATH to MATH gives an linear isomorphism MATH. Finally, the decomposition MATH determines the action of MATH on a MATH-dimensional space MATH and MATH the action of MATH on MATH, that is, MATH for a representation MATH of MATH. |
math/0010251 | Assume that MATH is the dimension vector of a MATH-stable representation of MATH, then the set of irreducible representations of MATH is an open subset of MATH. Assume that MATH and consider the vertex spaces (the eigenspaces) MATH and MATH as subspaces of the MATH-dimensional representation MATH (via the identification given by the matrix MATH). Then MATH is a subrepresentation of MATH of dimension vector MATH for some MATH. But MATH so MATH cannot be MATH-stable, a contradiction. Conversely, assume the numerical condition is satisfied and consider the direct sum of MATH-stable representations MATH . We note that MATH . Let MATH be the MATH identity matrix and let MATH be the MATH matrix of the form MATH . Then the NAME form of the local quiver MATH is the symmetric matrix MATH . When MATH, we have that MATH is obviously a MATH-stable representation. When MATH, we notice that MATH is the dimension vector of a simple representation if and only if MATH . Now, consider MATH and consider the dimension vector MATH. We have to verify that MATH is the dimension vector of a simple representation of MATH which, by symmetry of MATH, amounts to checking that MATH for all MATH and MATH where MATH are the standard base vectors. Computing the left hand term this is equivalent to MATH or MATH . NAME the values of MATH and MATH in this expression we see that this is equivalent to MATH . Therefore, the condition is satisfied if for all MATH and MATH we have MATH . |
math/0010252 | As in the proof of REF , we divide the diagram MATH into rectangular blocks MATH, MATH , and compute the weight within each block MATH. Clearly choosing a pair MATH in MATH is equivalent to, for each MATH, first choose the MATH left-most (respectively, the rest MATH) columns of length MATH (respectively, MATH) for MATH in MATH, and then choose MATH (respectively, MATH) left-most columns of length MATH (respectively, MATH) for MATH among the MATH columns of MATH, also choose MATH (respectively, the rest MATH) left-most columns of length MATH ( respectively, MATH) for MATH. Thus the corresponding weight is MATH if MATH and, for each MATH, MATH . Multiplying up over all MATH we get the desired formula. |
math/0010252 | Recall that MATH. Suppose there exists a pair MATH in MATH, then there should be an integer MATH such that MATH and MATH for any MATH. In view of the definition of MATH we must have MATH and MATH, for MATH is a horizontal strip. It follows that MATH and MATH. Furthermore, if MATH is also in MATH, we must have MATH. Summarizing, we have the following equivalence: MATH . It is easy to see that the last condition is equivalent to MATH or MATH. In what follows we shall assume that MATH. Thus we can define a unique partition MATH such that MATH. Graphically, the diagram MATH can be obtained by deleting, successively for MATH, the MATH-th and MATH-th columns and shift all the cells on the right of MATH-th column of MATH to left by two units. For MATH, if we apply the same graphical operation to the MATH and MATH, we get a pair MATH. For example, in the previous example, if MATH, then MATH. The corresponding triples MATH with MATH and MATH with MATH are illustrated as follows : Since the weight corresponding to the deleted MATH-th and MATH-th columns of MATH, MATH and MATH is MATH for each MATH, we have MATH . Therefore MATH . The lemma follows then immediately from REF . |
math/0010254 | Distinguish one point MATH. Draw the segments MATH. Some of them may be obstructing, but by splitting these into smaller ones, one gets a planar graph connecting all points in MATH and whose edges are non obstructing. The result then follows from the main theorem in CITE and its reformulation in the appendix of CITE. |
math/0010254 | This is a consequence of one of the NAME relations (see CITE or CITE, théorème A. REF) namely that MATH . The proof of this relation is by induction on MATH. The case MATH is checked by a direct computation, and for other MATH we have MATH where the first equality is by induction and the second by the case MATH. |
math/0010254 | The convex hulls of MATH and MATH cannot intersect, thus the generators corresponding to their edges commute pairwise. |
math/0010254 | We prove REF by induction on the cardinality of MATH. With the notations of REF, let MATH, MATH so that MATH, and let MATH, so that MATH. Then by REF applied to MATH we have MATH. By induction hypothesis, we have MATH. As MATH, and as MATH commutes to MATH (since MATH), we get the result by induction. REF : denote by MATH the only non trivial part of MATH, and by MATH the non trivial parts of MATH. To prove the result, the essential step is to notice that since for MATH we have MATH, each MATH lies inside a single connected component of the complement of MATH in MATH, thus each is included in a part of MATH. Thus REF follows by induction on the number of parts of MATH from REF . The uniqueness comes from the fact that the identity MATH is valid in MATH, and MATH is a group. Finally, REF is easily obtained by applying REF to all the non trivial parts of MATH. |
math/0010254 | REF is easy. REF comes from REF and the following remark: when MATH and MATH are in the same orbit, multiplying by MATH splits this orbit into two orbits, thus increasing by MATH the number of orbits; when MATH and MATH are in different orbits, multiplying by MATH merges their orbits, thus decreasing by MATH the number of orbits. REF is an easy induction from REF and its proof: as the decomposition is reduced, the relation MATH can only be achieved if the successive multiplications by the MATH merge orbits. REF are easy. |
math/0010254 | CASE: Let MATH. As the converse implication is trivial, we only have to check that MATH . Let MATH such that MATH. Consider the image in MATH of this identity: MATH. By REF , we have MATH and MATH, and, using REF , we have MATH and thus MATH. Consequently, when concatenating reduced decompositions for MATH and MATH, one get a reduced decomposition for MATH. By REF , this implies that the orbit decomposition for MATH is finer than the one for MATH, that is, MATH. We conclude by REF . CASE: By REF , the map MATH is a poset morphism. It is injective: from MATH, one recovers MATH by considering the orbit decomposition of MATH. The fact that the inverse map is a poset morphism, that is, MATH, has already been obtained in our proof of REF . |
math/0010254 | By REF can be transformed into MATH by a finite rewriting process MATH in which, at each step, the elementary transformation MATH is CASE: either of the type MATH CASE: or of the type MATH where MATH are such that their product is defined in MATH and equal to MATH. Suppose the length MATH of the rewriting process is minimal. Suppose some of the transformations are of the second type, and chose MATH maximal such that MATH is of the second type MATH . As all further steps are of the first type, we have MATH and the product MATH is defined in MATH (and is equal to MATH). Choosing a bracketing starting by MATH, we see that the product MATH must also be defined in MATH (and equal to MATH). But this yields a rewriting of length MATH and we have a contradiction. Thus there are no transformation of the second type in a minimal rewriting. The result follows. |
math/0010254 | The statement follows CITE, and the proof is exactly the same. |
math/0010254 | We can follow the proof of CITE, replacing REF by REF . |
math/0010254 | We apply REF to the set MATH of MATH such that MATH and MATH. To check the assumptions of that lemma, we need that if MATH and if MATH then MATH exists and MATH (then we have MATH). Since MATH and MATH divide MATH, by cancellability and REF MATH and MATH have a common multiple in MATH, so by REF MATH exists and by REF MATH. |
math/0010254 | The statement is CITE and the proof is the same. |
math/0010254 | By REF the products of both sides with MATH are equal. By REF we will be done if we show that MATH is in MATH. By REF there exists MATH such that MATH and MATH. As MATH and MATH we have MATH for some MATH (by definition of MATH and cancellability in MATH). Hence MATH and MATH. So MATH. The result follows since any divisor of MATH is in MATH by REF. |
math/0010254 | Assume that all elements of the family MATH divide MATH. If we can apply REF to the set MATH of elements of MATH which divide all common multiples of the MATH, it will give the result. Let us check the assumption of REF. This set MATH is finite as it is included in the set of divisors of MATH. The first assumption of REF is clearly satisfied. The second assumption is a consequence of the fact that if MATH are such that MATH and MATH divide some element MATH, then MATH and MATH divide MATH, so MATH exists and divides MATH, whence MATH divides MATH. |
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