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math/0012191 | Because of the relation REF the second case follows from the first one. Let us restrict to instances REF above. In each of them we can assume that MATH is even using REF. Fix an operator MATH determined by a choice of the functions MATH that is, a choice of admissible values of the complex parameters MATH (see REF). It has the eigenfunction MATH defined in REF MATH compare REF. We need to show that there exists a differential operator MATH having MATH as an eigenfunction, that is MATH for some function MATH . Define the functions MATH . Let us put MATH and consider the operator MATH . It is a regular difference operator with kernel given by MATH . Hence it is related to the operator MATH (recall REF ) by MATH where MATH is the leading coefficient of MATH . REF implies that MATH and MATH are rational functions in MATH . This implies that for MATH and for all MATH are also rational functions in MATH and thus MATH has rational coefficients. In addition REF gives MATH and MATH . Taking into account that MATH we obtain MATH where the factor MATH comes from exchanging the pairs of rows MATH . Set MATH and consider the operator MATH . Because of REF , MATH does not vanish for MATH . Taking into account REF one sees that MATH is related to MATH by MATH . Since MATH is a regular difference operator and MATH does not vanish for MATH (recall REF), there exists a difference operator with rational coefficients MATH such that MATH . From REF it follows that MATH is MATH-invariant. Finally combining this with the MATH-invariance of MATH (see REF) implies the MATH-invariance of the operator MATH . REF now gives MATH . Applying REF we obtain that the function MATH is an eigenfunction of a differential operator MATH for some polynomial MATH . Because of REF our original function MATH is related to MATH by MATH . REF implies that MATH is an eigenfunction of the same operator MATH with eigenvalue MATH . |
math/0012197 | If MATH is not a vertex of MATH, then it is in the convex hull of vertices MATH of MATH. But then MATH would be in the convex hull of MATH. This contradiction proves the first statement, and hence implies that the union of all MATH, MATH forms an order ideal of MATH. This is equivalent to the second statement. |
math/0012197 | Since for any two lattice points MATH we have MATH, a monomial is a standard monomial of MATH if and only if its exponent vector minimizes the linear functional MATH in MATH CITE. Hence the monomial MATH is a standard monomial of MATH if and only if MATH is the minimizer of MATH for MATH for some generic weight vector. But these are precisely the vertices of the fibers of MATH. |
math/0012197 | Suppose MATH is not in MATH. This means that MATH is a vertex of MATH, so there is some MATH such that MATH for all lattice points MATH. But now MATH because MATH means that MATH, and thus MATH is a lattice point in MATH. This implies MATH . This contradiction shows that MATH is in MATH. |
math/0012197 | Let MATH be a minimal generator of MATH. Hence MATH is not a vertex of its fiber, and therefore it is a convex combination MATH of some vertices MATH of MATH, where MATH and MATH. Since MATH is in MATH, we have MATH where MATH are Graver basis elements with MATH and MATH, and MATH. Now clearly MATH, and thus MATH. By REF , MATH is in MATH. But MATH divides MATH, and MATH is a minimal generator, so MATH. |
math/0012197 | If MATH is the support of a positive circuit of MATH, REF implies that MATH is in MATH. And REF says that every minimal generator of MATH is of this form. |
math/0012197 | MATH where MATH is the NAME ideal of MATH. We have the first equality on the second line because taking the radical commutes with intersections, while the second equality follows from REF . The third line is a standard result on NAME ideals, and the last line follows because the complement of the indices of the generators of any full dimensional simplicial cone MATH is involved in some MATH. |
math/0012197 | The NAME ideal of MATH is MATH . The above intersection can be taken over all MATH where MATH is a maximal face. Then since MATH if and only if MATH forms a basis of MATH where MATH is matrix whose columns are a basis for MATH , REF implies that MATH. |
math/0012197 | The NAME series MATH can be written in the form MATH for some polynomial MATH. This means that the NAME function of MATH at MATH, which counts the number of vertices of MATH when MATH, eventually agrees with a quasi-polynomial evaluated at MATH. As there is an upper bound, given by the number of initial ideals of MATH, on the number of vertices of any MATH, this polynomial part of the quasi-polynomial must be a constant. As the period of the quasi-polynomial divides MATH, the result follows. |
math/0012197 | Using REF , the first statement follows from the fact that if two ideals MATH and MATH have minimal primary decompositions MATH and MATH, then MATH is a (not necessarily minimal) primary decomposition of MATH. Minimal primes of a intersection of monomials ideals are always contained in the union of the minomial primes of the ideals. The fact that this containment is an equality in this case follows from the fact, used in REF , that all minimal primes of all initial ideals have the same dimension. |
math/0012197 | The first statement follows from REF and the fact that the origin is a vertex of MATH if and only if MATH is a vertex of MATH. For the second claim we use REF . The statement that MATH is a vertex of MATH, and MATH is a vertex of MATH for all MATH with MATH is equivalent to the statement that the origin is a vertex of MATH and it remains a vertex of MATH for all such MATH. Since MATH, this is the same thing as the origin being the vertex of MATH. Similarly, if for all MATH there exists a MATH with MATH such that the origin fails to be a vertex of MATH, then the origin is also not a vertex of MATH, and hence not a vertex of MATH, and vice versa. |
math/0012197 | The proof of REF implies that MATH where the intersection is taken over all standard pairs MATH such that MATH for all MATH. By REF these standard monomials are in bijection with critical MATH. |
math/0012197 | Suppose not. Clearly we can assume that MATH is two-dimensional and that MATH is the origin. Let MATH and MATH be the two vertices of MATH which are the neighboring vertices of the origin, in the clockwise and counterclockwise directions respectively. We define the pointed cone MATH generated by MATH and MATH, and MATH, the opposite cone generated by MATH and MATH. These constructions are illustrated in REF . We first claim that each edge of MATH has to intersect MATH. Suppose there is an edge MATH, lying on the hyperplane MATH, which does not intersect MATH. Then the convex region MATH does not intersect MATH as well. This is true because if MATH there is a point MATH on the line segment joining MATH to the origin lying on MATH, and MATH would then be in MATH. Since the origin is not a vertex of MATH, either MATH is in the interior of an edge of MATH or it is in the interior of MATH. In the first case there exists two vertices MATH and MATH of MATH such that MATH and MATH with MATH for some MATH. But then MATH, contrary to our assumption. If MATH is an interior point of MATH, then there exist three vertices MATH and MATH of MATH such that MATH with MATH and MATH. Now, either exactly one or exactly two of these vertices are in MATH. In the first case, say MATH and MATH, we have MATH and hence MATH. In the second case, say MATH and MATH, we have MATH, and hence MATH. In both cases we get a contradiction to our assumption that edge MATH does not intersect MATH. This shows that all edges of MATH intersect MATH. Because MATH contains MATH but not MATH, and MATH but not MATH, some edge of MATH must intersect the line segment MATH, and another one the line segment MATH. If we assume that the facets of MATH are labeled going clockwise and the edge MATH is the first edge intersecting the facet of MATH defined by MATH, then edge REF must be the edge intersecting MATH. And if edge MATH is the last edge intersecting the facet of MATH defined by MATH, then edge MATH must be the edge intersecting MATH. NAME REF and edge MATH are the only edges of MATH not lying entirely in MATH, so they need to meet in a common vertex of MATH. But their endpoints outside MATH are on opposite sides of the parallel line segments MATH and MATH, which makes this impossible. |
math/0012197 | Let MATH be a standard pair of MATH. Suppose that MATH is a face of MATH. This means that the origin in MATH is in the convex hull of MATH, where MATH is the MATH-th row of the MATH defined after REF . This follows because positive covectors of (the oriented matroid of) MATH correspond to positive vectors of (the oriented matroid of) MATH (see CITE). So MATH is a polygon. REF now implies that the origin in MATH is a vertex of MATH, but not a vertex of any MATH for MATH. But this is a contradiction to REF . |
math/0012197 | Let MATH be as in the above remark. If we choose MATH as the cost vector we get the following reduced NAME basis: MATH . The corresponding NAME cone is given by MATH which are all facet defining. |
math/0012197 | The initial ideals MATH are all square-free REF and hence MATH is radical. Therefore, by REF , MATH. But the minimal generators of MATH are of the form MATH for some MATH such that MATH is a circuit of MATH. Now REF implies that the Graver basis of MATH is MATH is the support of a circuit MATH. Therefore MATH is the support of a circuit MATH, and hence MATH. |
math/0012197 | Let MATH and suppose MATH, so MATH is not a vertex of its fiber MATH, but MATH. If MATH where MATH, then MATH because MATH and hence MATH. So no such point in MATH exists. Now there must be a vertex MATH of this fiber with MATH, because otherwise MATH would be a vertex. Let MATH be the vertex with MATH such that MATH is the maximum with this property. Let MATH be the line through MATH and MATH, let MATH be the halfspace containing the origin, and let MATH be the other halfspace. If MATH, since MATH is not a vertex of MATH, the line MATH must contain MATH such that MATH and MATH. But then MATH and MATH, which implies MATH. This implies MATH since MATH. Hence we are reduced to the case that MATH is not contained in MATH and MATH so no such MATH. This means that there exists a vertex MATH. Now if MATH, by the construction of MATH we must have MATH. If in addition MATH, the existence of a vector of the form MATH for some MATH means that MATH for MATH which contradicts MATH being a vertex. On the other hand, if MATH, MATH would not be a vertex of MATH. So we conclude that MATH. But now we must have MATH, where the second inequality follows from the assumption that MATH and MATH. Since MATH, it follows that MATH, a contradiction, so MATH. |
math/0012197 | If MATH, there is a unique MATH such that MATH. If MATH, MATH, where MATH for MATH, and MATH, then MATH and MATH, with MATH for MATH. As this contradicts MATH, we conclude that MATH, so MATH. |
math/0012197 | From REF we know that MATH. Let MATH be a minimal generator of MATH. By REF we know that MATH where MATH for MATH and MATH. Now MATH. Writing MATH where MATH and MATH, MATH for all MATH, we see that for MATH. Since MATH divides MATH, it follows that MATH. For the other inclusion, let MATH be a minimal generator of MATH, so MATH for MATH, where MATH and MATH. Let MATH be the preimage of MATH under MATH. We still have MATH, so for MATH, MATH, and thus MATH. The second statement of the proposition follows from the definition of the product ideal, and the observation that if MATH for MATH, we can write MATH as the sum of MATH so that MATH. |
math/0012197 | REF shows that MATH is an equidimensional ideal. Now an associated prime MATH of MATH is a minimal prime if and only if MATH is a full dimensional lattice in MATH. But this is true if and only if there exist MATH such that MATH for all MATH. This happens if and only if MATH for all MATH, which happens exactly whenever MATH is a zero-dimensional ideal, and hence MATH is a minimal prime of MATH. This shows that MATH. The second statement follows from REF and the discussion before it, and the fact that MATH. |
math/0012197 | REF says that MATH when MATH is a minimal prime of MATH (and of MATH). Now REF and the discussion before them imply the result. |
math/0012198 | Let MATH be a subset of MATH. Let MATH be the image of the obvious inclusion of MATH. Let MATH. Note that, as MATH varies over all subsets of MATH, the subschemes MATH stratify MATH. For any subscheme MATH, let MATH respectively, MATH. Thus MATH is the intersection of MATH with the hyperplanes defined by the equations MATH for MATH. Note that MATH, and, as MATH varies over the subsets of MATH, the subschemes MATH stratify MATH. We therefore have, MATH and, by the NAME Principle, MATH . By inspecting the MATH, it is easy to see that MATH and MATH. Dually, if MATH is a forest, MATH (respectively, MATH) where MATH is the graph obtained by contracting each component of MATH to a point. On the other hand, if MATH is not a forest, it is easy to see that MATH is empty. Now, as NAME notes, MATH. Thus MATH through the map MATH. Putting our equations together we obtain the following: MATH . Together, these two equations, the first of which appears (in a different notation) as REF, prove the proposition. |
math/0012198 | Write MATH with MATH and MATH. Since MATH is monic, we can write MATH with MATH and MATH. But this implies that MATH for all MATH which implies that MATH. Thus MATH. |
math/0012198 | Let MATH be the ring generated by the variables MATH for MATH. Let MATH be the ideal generated by the variables MATH for all pairs MATH with MATH. Then MATH with MATH. On the other hand, let MATH is the ring generated by all MATH for MATH, and let MATH be the ideal generated by all expressions of the form MATH for MATH and MATH for MATH and MATH. Then, letting MATH be the determinant of the matrix of MATH's, MATH with MATH. Let MATH be the map MATH . Let MATH be the map MATH . It is easy to verify that MATH, that MATH, and that MATH and MATH give inverse isomorphisms between the rings MATH and MATH. It then follows from the NAME theorem that MATH. Thus MATH and MATH give inverse isomorphisms between the rings MATH and MATH. |
math/0012198 | We first remark that REF follows directly from REF and the fact that MATH. To prove REF we let MATH with MATH. Then MATH. The map MATH then identifies MATH with MATH. (Here MATH denotes the transpose of MATH). |
math/0012198 | REF follows directly from the recursion REF . The necessity of the first two inequalities of REF are obvious for dimension reasons (the rank of a bilinear form can not be greater than the dimension of the ambient space). Necessity of the third inequality follows from REF by induction. We prove the sufficiency of the the inequalities in REF by induction on MATH using REF . We do not actually need this for the rest of the paper so the reader may safely skip the proof. For MATH sufficiency is obvious. For MATH the sufficiency results from the fact that MATH iff MATH when MATH and iff MATH when MATH. Now suppose sufficiency is known for MATH and assume that MATH satisfies the conditions in REF with MATH and MATH. By REF , MATH if there is a MATH such that both CASE: MATH and CASE: MATH. One computes that REF is satisfied whenever MATH. Using the the induction hypothesis, we see that REF is satisfied for MATH . So we need only show that MATH. That MATH only says that MATH which we are of course assuming. And MATH iff MATH which then follows from the fact that MATH. |
math/0012198 | In this proof and the next one we will pick a base field MATH at the beginning and then, for any scheme MATH we encounter, write MATH instead of MATH. Write MATH. For every map MATH let MATH denote the span of the MATH. The map MATH fibers the set MATH over MATH. The fiber over a subspace MATH is then the set MATH of MATH such that MATH. The transitivity of the MATH action on MATH shows that the fibers all have the same number of points. Thus for any given MATH . Now let MATH be the set of MATH such that MATH has rank MATH. This decomposes MATH into disjoint subsets. Consider the map MATH . The fiber of MATH above a given MATH is MATH. Thus MATH . Summing over all the MATH in REF and substituting the result into REF we obtain the desired result. |
math/0012198 | Write MATH and let MATH be the map associating to every MATH the kernel MATH of MATH. The fiber of MATH over a subspace MATH is the set MATH consisting of all MATH with MATH. The transitivity of the action of MATH on MATH show then that MATH . Let MATH and let MATH be the quotient map. MATH reduces in an obvious way to a form MATH on MATH. In fact MATH is a one-one correspondence between bilinear forms on MATH with kernel MATH and non-degenerate bilinear forms on MATH. Now stratify MATH by the dimension of MATH. This stratum corresponding to span MATH maps to MATH by sending MATH to MATH. The fiber above a pair MATH is identified with set of MATH such that MATH and MATH is of dimension MATH. It is elementary linear algebra to verify that this is given by MATH . Hence putting these fibrations together we get the desired result. |
math/0012198 | To get a non-zero contribution corresponding to MATH in the previous theorem need CASE: MATH. CASE: MATH. CASE: MATH. CASE: MATH. In the case of the corollary s=k, so we get MATH and MATH. Hence MATH, and the formula reduces to exactly the above. |
math/0012198 | Let MATH. Let MATH with MATH a MATH-dimensional subspace. The span of MATH is either a MATH or a MATH dimensional subspace. If MATH, counting the possibilities for MATH proves the lemma. |
math/0012198 | REF . Apply REF to obtain an expression for MATH as a MATH-linear combination of terms of the form MATH with MATH. Then apply REF to obtain an expression for each MATH as a MATH-linear combination of terms of the form MATH. CASE: To see that the MATH span, note that REF implies that MATH modulo MATH. To see that the the MATH span, use the fact that MATH for MATH, a consequence of REF . |
math/0012198 | By REF , MATH . Now applying REF to MATH and expanding out MATH in terms of MATH gives the result. The polynomials MATH and MATH in the theorem are clearly in MATH as long as they are nonzero. Inspection shows that this is the case under the assumption that MATH. |
math/0012198 | The obvious map MATH restricting MATH from MATH to MATH has fiber MATH. |
math/0012198 | For the proof, let MATH be the MATH-module spanned by the functions MATH for MATH. Since MATH, MATH. Thus it follows from REF that MATH. To prove that MATH we use REF and an inductive argument. By REF , it will be enough to show that MATH for all MATH. Since MATH this is obvious for MATH. Now by REF MATH with MATH. (The first term on the right hand side of REF vanishes because MATH is empty). We know that MATH and MATH are in MATH. Thus MATH. We then assume inductively that MATH for all MATH and for all graphs MATH. Another application of REF shows us that MATH . By induction, the left-hand side and the two first terms on the right hand side are in MATH. Thus, as MATH, MATH as well. |
math/0012198 | The proof is by induction on the cardinality of MATH. If MATH is empty, MATH. Thus the result follows from REF . Now assume the result holds for all graphs MATH and all MATH such that MATH. Let MATH be a set of subsets with MATH elements, let MATH and let MATH. Let MATH be a partially defined rank function, and let MATH be the extension of MATH to MATH such that MATH. Clearly, any partially defined rank function with domain MATH is of the form MATH for some MATH and some MATH. Now for each MATH we define a graph MATH as follows: MATH is the graph obtained from MATH by adjoining MATH disjoint vertices MATH and connecting each of the MATH by edges only to the vertices in MATH. Thus MATH where MATH, and MATH . Since MATH, MATH. Thus we can consider MATH as a partially defined rank function for MATH. The result will follow from the following equation: MATH . To see that the equation holds, note that we can stratify the MATH points of MATH according to the dimension of the span of MATH. Let MATH be the stratum where this dimension is MATH. This stratum maps to MATH by restricting MATH from MATH to MATH. The fiber of map above any point MATH is an affine space MATH. This is because the only condition on the MATH is that they be orthogonal to the span of MATH. Thus, as the bilinear form MATH is always nondegenerate, they must lie in a linear subspace of dimension MATH. To complete the proof, note that by varying the MATH from MATH to MATH we obtain a system of equations for the MATH in terms of the MATH. Solving this system for the MATH using NAME 's rule, we have to invert a NAME determinant which lies in MATH. Thus, as we assumed by induction that MATH lies in MATH, it follows that each MATH lies in MATH as well. |
math/0012198 | The proof follows essentially from the following observations CASE: MATH elements in MATH such that any MATH are linearly independent can by a unique automorphism of MATH in PGL REF , be assumed to be MATH and MATH. CASE: if given two points MATH and MATH on the MATH, then by drawing lines alone through REF points above and these two points, we can locate MATH,MATH,MATH. Finding intersection of lines can be translated as a vector which lies on both lines, and hence as a condition on linear dependence. These constructions can be found for example in the proof of REF . CASE: Iterating these constructions, given MATH we can determine the points MATH, where f is a polynomial with integer coefficients by just drawing lines starting from the configuration of the four given points and the points MATH. Setting MATH either equal to zero or not equal to zero is just another spanning condition: A condition on whether MATH is linearly independent or not. CASE: The cone over any quasi-projective scheme/MATH can be written as a set of equalities and a set of nonequalities in a finite set of variables MATH. Note that we can also have conditions of the form MATH in the list. |
math/0012198 | For REF let MATH be the vertex in MATH. Then the map MATH given by MATH is an isomorphism. For REF assume first that MATH. Then tracing through the definitions one sees that MATH. The rest of REF follows by induction from REF . For REF we work over MATH and consider the map MATH given by MATH. The fiber of MATH above a point MATH depends on whether MATH is MATH or not. Let MATH (respectively, MATH) be the set where MATH (respectively, MATH). Above a point MATH the fiber of MATH will have MATH points. Above a point MATH the fiber will have MATH points since MATH is non-degenerate. Thus MATH . The result now follows from the observation that MATH. |
math/0012198 | An easy induction using REF shows that MATH for any MATH. Thus MATH. It follows from REF that MATH. But this implies that MATH by REF . |
math/0012201 | The implication MATH follows from the straightforward formula MATH for all MATH. For MATH , assume that MATH. It suffices to show that MATH . Indeed, MATH, since MATH holds for all MATH and MATH. To simplify notation, put MATH and let MATH denote a NAME MATH-subgroup of MATH, where MATH is the characteristic of the commutative domain MATH. (Here MATH if MATH.) Then our desired conclusion, MATH, is equivalent with MATH . Furthermore, our assumption MATH entails that MATH, because MATH. Thus, leaving MATH for MATH, we may assume that MATH is a MATH-subgroup of MATH. Let MATH denote the decomposition group of MATH; so MATH. We claim that MATH . To see this, choose MATH so that MATH for all MATH but MATH. Then MATH also belongs to MATH but not to MATH and, in addition, MATH. Now assume that, contrary to our claim, there exists an element MATH so that MATH. Then MATH, and hence MATH, a contradiction. By the claim, we may replace MATH by MATH, thereby reducing to the case where MATH is MATH-stable. (Note that MATH is unaffected by this replacement.) So MATH acts on MATH with kernel MATH, MATH is a MATH-subgroup of MATH, and MATH. Thus, MATH holds for all MATH. Our desired conclusion, MATH, will follow if we can show that MATH holds for some MATH. But MATH induces a nonzero endomorphism on MATH, by linear independence of automorphisms of MATH; so MATH holds for some MATH. Putting MATH, we have MATH and MATH. Since MATH is MATH or a power of MATH, we obtain MATH, as required. |
math/0012201 | One has MATH where MATH runs over the prime ideals of MATH containing MATH and MATH runs over the primes of MATH containing MATH. Here, the first equality is just the definition of height, while the second equality is a consequence of the standard relations between the primes of MATH and MATH; see, for example, CITE. By REF , MATH . Since MATH belongs to MATH for MATH, the latter condition just says that the NAME MATH-subgroups of MATH do not belong to MATH or, equivalently, some MATH-subgroup MATH does not belong to MATH. Therefore, MATH which implies the asserted height formula. |
math/0012201 | The action of MATH on MATH can also be interpreted as coming from the cup product MATH where the map denoted by MATH comes from the MATH-equivariant map MATH, MATH; see, for example, CITE. Furthermore, the relative trace map MATH is identical with the corestriction map MATH; compare CITE. Thus, the transfer formula for cup products CITE gives, for MATH and MATH, MATH . Therefore, if MATH then MATH. |
math/0012201 | Let MATH denote the splitting data of MATH, that is, MATH. By REF , MATH, and by REF , MATH. The proposition follows. |
math/0012201 | Put MATH. Then MATH for MATH, and so the MATH-sequence in REF implies that MATH for MATH. Therefore, the MATH-sequence satisfies MATH . Furthermore, MATH; so MATH . Finally, MATH . To prove REF , assume that MATH. Then MATH, by REF, and MATH for MATH, MATH, MATH. Recall that the differential MATH of MATH has bidegree MATH. Thus, MATH has no nontrivial boundaries and consists entirely of cycles. This shows that MATH, and hence MATH. Thus, REF is proved. For REF , we check that MATH. Our hypotheses imply that, at position MATH, all incoming differentials MATH are MATH as well as all outgoing MATH. Therefore, MATH and MATH. The former implies that MATH, by hypothesis in MATH, and the latter shows that MATH is injective, because MATH by REF. Thus, MATH embeds MATH into MATH, forcing the latter to be nonzero. Hence, MATH is nonzero as well, as desired. |
math/0012201 | By hypothesis, MATH for some MATH. Let MATH denote the localization of MATH at the multiplicative subset MATH. Then the MATH-action on MATH extends to MATH and MATH; see CITE. Similarly, MATH; so MATH is NAME. By choice of MATH the relative trace map MATH is onto. Fix an element MATH so that MATH and define MATH by MATH. This map is a ``NAME operator", that is, MATH is MATH-linear and restricts to the identity on MATH. Since MATH is integral over MATH, a result of NAME and NAME (CITE or CITE) implies that MATH is NAME, which proves the proposition. |
math/0012201 | By CITE, the trace map MATH is surjective for NAME actions and MATH is finitely generated projective as MATH-module. Thus, MATH is faithfully flat as MATH-module. Moreover, for any prime MATH of MATH and MATH, the fibre MATH has dimension MATH. Therefore, by CITE, MATH is NAME if and only if MATH is. |
math/0012201 | Note that MATH entails that MATH. Further, MATH, because MATH is NAME. The lemma follows. |
math/0012201 | Let MATH be an ideal of MATH with MATH. Our hypothesis on MATH entails that the value of MATH in REF satisfies MATH. Also, MATH, by REF . Thus, REF gives MATH, as required. |
math/0012201 | Our hypothesis MATH for some MATH is equivalent with MATH; so MATH. The asserted equality is trivial for MATH, since MATH holds in this case. Thus we assume that MATH. Then, in the notation of REF , we have MATH, and REF gives the inequality MATH. To prove the reverse inequality, note that REF gives MATH. Therefore, it suffices to show that MATH if MATH. For this, we quote REF with MATH and MATH (so MATH). |
math/0012201 | Let MATH be given and put MATH. Then, by REF , MATH . Since MATH is NAME, MATH. Finally, REF with MATH gives MATH. Thus, there exists a MATH-subgroup MATH of MATH with MATH and MATH. Note that both the condition MATH and the value of MATH are preserved upon replacing MATH by a conjugate MATH with MATH. Therefore, we may assume that MATH, which proves the proposition. |
math/0012201 | The implication MATH follows from REF with MATH. For the converse, let MATH be NAME and assume, without loss, that MATH. Then REF implies that there is a subgroup MATH with MATH. On the other hand, by hypothesis on the MATH-action, MATH; so MATH. |
math/0012201 | First note that MATH is a normal subgroup of MATH and MATH is a MATH-group. Thus, if MATH or, equivalently, MATH then also MATH; see CITE. In view of the exact sequence MATH (see CITE) we further obtain MATH. Thus, MATH holds in REF and every MATH consists of bireflections. Therefore, MATH and REF implies that MATH is injective, contradicting the above exact sequence. Therefore, we must have MATH. |
math/0012201 | Put MATH, MATH, and MATH. In order to prove that MATH, we use the fact that MATH holds for MATH; see CITE. If MATH then MATH and so MATH. Moreover, MATH equals MATH in all degrees. This proves the assertion for MATH; so we assume MATH odd from now on. In this case, MATH with MATH and MATH; see CITE. Moreover, identifying MATH with MATH, the action of MATH on MATH becomes scalar multiplication, MATH, MATH, where MATH. Taking MATH to be a generator for the subgroup of MATH corresponding to MATH, we see that MATH see CITE. The smallest positive degree where MATH does not vanish is therefore indeed MATH. Now assume that MATH is a MATH-module direct summand of MATH and MATH and MATH are both NAME. The former hypothesis implies that MATH and hence MATH. Moreover, our hypothesis on MATH implies that MATH holds in REF , because otherwise MATH would consist of the identity subgroup alone. Therefore, MATH, as desired. |
math/0012201 | By definition, the ideal MATH of MATH is generated by the elements MATH for MATH, MATH. Thus, MATH, where we have put MATH. Consequently, MATH. Finally, since the group algebra MATH is semisimple, MATH; so MATH. |
math/0012201 | Our hypothesis MATH is equivalent with MATH; so MATH holds as well, by REF. Assuming, MATH to be NAME, REF and the subsequent remark imply that MATH. Since all MATH-elements of MATH belong to MATH, it follows that MATH is generated by the images of the bireflections in MATH. Since MATH is cyclic, it follows that MATH is generated by a bireflection. Now, MATH acts faithfully on the lattice MATH of rank at most MATH. Thus, MATH is isomorphic to a cyclic MATH-group of MATH, and these are easily seen to have orders MATH, MATH, or MATH. The converse follows from the more general Lemma below which does not depend on cyclicity of MATH or nontriviality of MATH. |
math/0012201 | By REF , it suffices to show that MATH is NAME; so we may assume that MATH is a MATH-group. Note that MATH acts faithfully on MATH. If MATH acts as a reflection group on MATH then it does so on MATH as well, and hence the invariants MATH will be NAME; see REF. Thus we may assume that MATH has rank MATH and MATH acts on MATH as a non-reflection MATH-group. By the well-known classification of finite subgroups of MATH (for example, CITE), this leaves the cases MATH with MATH or MATH to consider. The cases MATH or MATH can be dealt with along similar lines. Indeed, for both values of MATH, the only indecomposable MATH-lattices, up to isomorphism, are MATH, MATH, and MATH, where MATH; see CITE. Thus, MATH, and MATH, where we have put MATH. Since MATH is NAME if and only if MATH is, we may assume that MATH. Now, MATH; so MATH. When MATH, this leads to either MATH, MATH or MATH, MATH. In the former case, MATH and so MATH is surely NAME, being a normal domain of dimension MATH. If MATH, MATH then MATH is a MATH-permutation lattice of rank MATH. Hence, MATH is a localization of the symmetric algebra MATH, and likewise for the subalgebras of invariants. Since linear invariants of dimension MATH are known to be NAME (for example, CITE), MATH is NAME in this case as well. For MATH, there are three cases to consider, one of which (MATH, MATH) leads to an invariant algebra of dimension MATH which is clearly NAME. Thus, we are left with the possibilities MATH, MATH and MATH, MATH. Explicitly, after an obvious choice of basis, MATH acts as one of the following groups on MATH: CASE: MATH; CASE: MATH. For MATH, MATH is a permutation lattice. Hence, as above, it suffices to check that the linear invariant algebra MATH for MATH is NAME which is indeed the case, by CITE, since MATH. For MATH, one can proceed as follows: Embed MATH into MATH and denote the corresponding basis of MATH by MATH; so MATH, MATH, and MATH. One easily checks that MATH, where MATH, MATH, and MATH. Furthermore, MATH. With this, the invariant subalgebra MATH is easily determined; the result (for MATH) is MATH which is indeed NAME. This completes the proof for MATH or MATH. We now sketch the remaining case, MATH. The action on MATH can then be described by MATH; so MATH. With this, one calculates MATH. Thus, there is exactly one (up to isomorphism) non-split extension of MATH-modules MATH. A suitable module MATH is MATH. Furthermore, one calculates MATH. Consequently, either MATH or MATH, and hence either MATH which is NAME because MATH has dimension REF, or MATH which is NAME precisely if MATH is. This reduces the problem to the case where MATH which can be handled by direct calculation, taking advantage of the fact that a conjugate of group MATH is contained in MATH. We leave the details to the reader. |
math/0012209 | See CITE. |
math/0012209 | The result follows immediately from CITE and REF above. |
math/0012209 | Initially choose MATH for MATH defined in REF , so that MATH. Hence, we have MATH at all iterations of Procedure IDREF. We now estimate MATH using REF . We have directly from the constraints REF that MATH . For the vector MATH, we have for MATH that MATH and MATH, and so MATH . Meanwhile for MATH, we have MATH and MATH, and so MATH . By substituting these estimates into REF , and using the equivalence of MATH and the Euclidean norm and the result of REF , we have that there is a constant MATH such that MATH . Using REF again, we have MATH giving the result. |
math/0012209 | Since we know the procedure terminates finitely, we need show only that MATH at all iterations of the procedure. Initially set MATH, so that MATH and the result of REF holds. Suppose for contradiction there is an index MATH such that MATH either is not included in the initial index set MATH or else is deleted from MATH at some iteration of Procedure IDREF. Suppose first that MATH is not included in MATH. Then we must have MATH, which by REF implies that MATH . However, by decreasing MATH and using MATH, we can ensure that REF does not hold whenever MATH. Hence, MATH is included in MATH. Suppose now that MATH is deleted from MATH at some subsequent iteration. For this to happen, the subproblem REF must have a solution MATH with MATH for some MATH. Hence from REF , we have that MATH . By combining the result of REF with REF , we have that MATH . However, this inequality cannot hold when MATH is smaller than MATH. Therefore, by decreasing MATH if necessary, we have a contradiction in this case also. |
math/0012209 | Given any MATH, we have for sufficiently small choice of MATH that MATH. We prove the result by showing that Procedure IDREF cannot terminate with MATH. We initially set MATH, where MATH is the constant from REF . (We reduce it as necessary, but maintain MATH, in the course of the proof.) For contradiction, assume that there is MATH such that MATH at all iterations of Procedure IDREF, including the iteration on which the procedure terminates and sets MATH. Recalling REF of MATH, we use compactness of MATH to choose MATH such that MATH. In particular, we have MATH for our chosen index MATH. We claim that, by reducing MATH if necessary, we can ensure that MATH is feasible for REF whenever MATH. Obviously, since MATH by REF , MATH is feasible with respect to REF . Since MATH and MATH we have MATH for some constant MATH that depends on the norms of MATH and MATH, MATH in the neighborhood of MATH and on a bound on the set MATH (which is bounded, because of MFCQ). Since MATH and since MATH, we can reduce MATH if necessary to ensure that MATH whenever MATH, thereby ensuring that the constraints REF are satisfied by MATH. Since MATH is feasible for REF , a lower bound on the optimal objective is MATH . However, since Procedure IDREF terminates with MATH, we must have that MATH for the solution MATH of REF with this particular choice of MATH. But we can have MATH only if MATH for all MATH, which means that the optimal objective is no greater than MATH. But since MATH, we can reduce MATH if necessary to ensure that MATH whenever MATH. This gives a contradiction, so that MATH (which is set by Procedure IDREF to the final MATH) can contain no indices MATH. Since MATH whenever MATH, we must therefore have MATH, as claimed. |
math/0012209 | Let MATH be chosen so that MATH. We show first that MATH is feasible for REF , thereby proving that this linear program is feasible and that the optimum objective value is at least MATH. Initially we set MATH. By REF , the constraint REF is satisfied by MATH. Since MATH, we have from REF that MATH, so that REF also holds. Satisfaction of REF follows from REF , by choice of MATH. Moreover, it is clear from MATH that the optimal MATH will satisfy MATH. We now show that REF is bounded for MATH sufficiently small. Let MATH be the vector in REF , and decrease MATH if necessary so that we can choose a number MATH such that MATH . From the constraints REF and the triangle inequality, we have that MATH . However, from REF and MATH, MATH, we have that MATH . By combining these bounds, we obtain that MATH whenever MATH, so that the feasible region for REF is bounded, as claimed. To prove our final claim that MATH for some MATH, we use REF . We have from REF and the cited theorem that MATH . For MATH, we have from MATH and MATH that MATH . For MATH, we have MATH and MATH, and so MATH . By substituting the last three bounds into REF and applying REF , we obtain the result. |
math/0012209 | We prove the result by showing that MATH defined by REF satisfies REF for some choice of MATH. For contradiction, suppose that no such choice of MATH is possible, so that for each MATH, there is a starting point MATH with MATH such that the sequence MATH generated from this starting point in the manner prescribed by REF with MATH eventually comes across an index MATH such that this choice of MATH violates REF , that is, one of the following two conditions holds: MATH . Assume that MATH is the first such index for which the violation REF occurs. By REF , we have that MATH . Therefore by REF , we have for MATH sufficiently large that MATH . Hence, taking limits as MATH, we have that MATH . Dividing both sides of REF by MATH, we conclude from finiteness of MATH that REF is impossible. By using REF again together with REF , we obtain MATH and therefore MATH as MATH. Hence, REF cannot occur either, and the proof is complete. |
math/0012209 | For each MATH, we use REF to obtain positive constants MATH (sufficiently large), MATH, MATH, and MATH, using the argument MATH for each constant to emphasize the dependence on the choice of multiplier MATH. In the same vein, let MATH be the constant from REF . Now choose MATH for each MATH in such a way that MATH and consider the following open cover of MATH: MATH . By compactness of MATH, we can find a finite subcover defined by points MATH as follows: MATH . MATH is an open neighborhood of MATH. Now define MATH . Also, choose a quantity MATH with the following properties: MATH . Now consider MATH with MATH . We have MATH, and so MATH. It follows that for some MATH, we have MATH . Moreover, since MATH, we have from REF that MATH where the final inequality follows from REF . Application of REF now ensures that the stabilized SQP sequence starting at MATH with MATH chosen according to REF yields a sequence MATH satisfying MATH where we used REF to obtain the final inequality. To prove REF , we have from REF , the bound REF , and the stabilizing parameter choice REF that MATH where in the last line we use MATH. Therefore, the result REF follows by setting MATH. Finally, we have from REF (with MATH) and REF that MATH . Therefore, we have MATH verifying REF and completing the proof. |
math/0012209 | Our result follows from REF . Choose MATH in REF , and let MATH, MATH, and MATH be as defined there. Using also MATH and MATH from REF and MATH defined in REF , we choose MATH as follows: MATH . Now let MATH satisfy MATH, and let MATH be calculated from REF . From REF , we have that MATH and MATH . Since MATH is closed, there is a vector MATH such that MATH . From REF , we have that MATH so that MATH for MATH. We therefore have from REF , and REF that MATH . From here on, we set MATH, as in Algorithm sSQPa. Because of the last bound, we can apply REF to MATH. We use this result to prove the following claims. First, MATH . Second, MATH . We prove both claims by induction. For MATH in REF , we have from REF and MATH in REF that MATH. Assume that the first MATH inequalities in REF have been verified. From REF , we have that MATH so that the next inequality in the chain is also satisfied. For REF , we have from REF , and REF that MATH where the last bound follows from REF . Hence, REF is verified, so that the condition in the ``if" statement of Algorithm sSQPa is satisfied for all MATH. NAME convergence with NAME MATH follows from REF . |
math/0012210 | When MATH, every MATH-spin structure of type MATH naturally gives a MATH-spin structure of type MATH simply by MATH . |
math/0012210 | Flatness follows from the valuative criterion of flatness CITE, which states that it is enough to check flatness of MATH over each MATH-valued point MATH, where MATH is a discrete valuation ring. Since the completion MATH of MATH is faithfully flat over MATH, it suffices to check this for each complete discrete valuation ring. But in this case, the results of CITE show that the universal deformation (relative to the universal stable map MATH) of a spin structure over the central fiber of MATH corresponds to the ring homomorphism MATH, where MATH is a uniformizing parameter for MATH. In particular, MATH is a free MATH-module, and thus is flat over MATH. Since the universal deformation is faithfully flat (étale) over MATH, this shows that MATH is also flat. Properness also follows by the valuative criterion in exactly the same manner as was proved in CITE for spin structures on stable curves. Nothing in that proof required the underlying curves to be stable - only prestable. |
math/0012210 | Given a MATH-valued point MATH for a representable MATH, we must show that the stack MATH of coherent nets of MATH-th roots of MATH on the associated family MATH of prestable curves is a NAME stack, finite over MATH. In particular, we need to construct a smooth cover of MATH and show that the diagonal MATH is representable, unramified, and proper. These facts are all straightforward generalizations of their counterparts over the stack MATH of stable curves as described in CITE. The only real difference is that we are now working with a specific family of prestable curves over MATH, as opposed to working with the universal family of stable curves (over MATH), but that changes nothing of substance in the proof. The proof of properness is also an easy generalization of the case of stable MATH-spin curves, and the morphism is obviously quasi-finite, hence finite. |
math/0012210 | As mentioned above, it is straightforward to check that the maps MATH and the sheaves MATH are MATH-flat and produce a MATH-spin structure of type MATH on each fiber of MATH (this will also follow from the computations below). Thus we only need to verify that MATH (which implies that it commutes with base change), and that the maps and sheaves meet the local conditions outlined in Subsection REF for being a coherent net on the family of curves MATH, provided the original sheaves MATH and maps MATH form a coherent net on the family MATH. Let us fix a point MATH of a geometric fiber MATH of MATH. There are three cases to consider. First is the case when the point MATH is not the image of a contracted component (that is, MATH is a single point). Second is the case when MATH is a smooth point of the fiber MATH, but MATH is the image of a whole irreducible component of the fiber MATH of MATH; that is, MATH contracts a MATH-curve to the point MATH. Third is the case that MATH is a node of the fiber MATH containing it, and it is the image of a contracted component of MATH; that is, MATH is the image of a MATH-curve MATH. CASE: The first case is easy, since when MATH is a single point, then MATH is an isomorphism in a neighborhood of MATH (or of MATH). In particular, MATH is an isomorphism, MATH, and MATH is a MATH-th power map near MATH. The second and third cases are more involved. Before we attack them, we note that the conditions we must verify are local (and analytic) on the base MATH, so it suffices to check the result when MATH is affine and is the spectrum of a complete local ring MATH. Moreover, the conditions are analytic on MATH; that is, the conditions are all determined by restricting to the completion of the local ring of MATH near the point MATH. To simplify, we will make the calculations in the case of MATH, but all other values of MATH (dividing MATH) are similar. CASE: In the second case (MATH contracts a MATH-curve of MATH to the point MATH) we will show that the induced sheaves MATH are locally free at MATH, and the maps MATH are all isomorphisms; thus the local coordinate and power map conditions are automatically fulfilled. The fiber MATH over MATH has one irreducible component MATH lying over MATH, and MATH contains at most one marked point MATH, labeled with an integer MATH, where MATH. This is indicated in REF . On MATH, the sheaf MATH is isomorphic to MATH, where MATH is the point of MATH which maps to the node MATH attaching MATH to the rest of MATH. Moreover, MATH must divide MATH, so either MATH, which implies that MATH is locally free REF near MATH, or MATH, which implies that MATH is not locally free (it is NAME) at MATH. In either case, MATH has degree MATH and thus has no global sections. This also gives MATH, since this is true on each fiber. Now, in the NAME case, the sheaf MATH is simply the sheaf MATH restricted (modulo torsion) to the rest of the prestable fiber MATH. But MATH on MATH is a MATH-th root of MATH, where MATH is the other side of the node defined by MATH. The actual value of MATH is determined by the relation MATH, which implies that MATH. In the NAME case, the vanishing of the global sections of MATH implies that MATH is MATH, so it is a MATH-th root of MATH. In both the NAME and NAME cases, the new marked point MATH of MATH is labeled with MATH, just as the old marked point MATH was marked with MATH on MATH. If no point was marked on MATH, then the point MATH remains unmarked (and MATH). Finally, MATH is MATH-flat and MATH vanishes, so we have that MATH commutes with base change, and the calculations above on the fibers all hold globally on the family MATH. Thus MATH is invertible near MATH, and MATH is an isomorphism near MATH. In particular, MATH is a MATH-power map. A similar argument holds for each MATH and each MATH near MATH. CASE: The third case is that of a point MATH which is the image of a MATH-curve MATH of MATH. It is easy to see that, just as in REF , on the unstable (contracted) MATH-curve, the degree of the bundle is MATH. Also, we have MATH; the sheaf MATH is MATH-flat and commutes with base change; and on the fibers, the induced collection of sheaves and bundles forms a MATH-spin structure of type MATH. We have still to check that the induced sheaves have the necessary family structure for spin curves (existence of a local coordinate of suitable type, with respect to which the sheaves have the standard presentation - see REF ), and that the induced maps are power maps, as described in REF . For simplicity we will assume that the orders MATH, MATH, MATH, and MATH of the MATH-spin map MATH along the two nodes MATH and MATH where the MATH-curve intersects the rest of the fiber have the property that MATH. The case with common divisors larger than MATH is similar. It is shown in CITE that MATH is locally isomorphic to MATH where MATH, and MATH and MATH are elements of the maximal ideal MATH of MATH with MATH. This shows the existence of the special local coordinate. We next show that MATH has a presentation of the form MATH . If we let MATH and MATH, then near the exceptional MATH-curve MATH the curve MATH is covered by two open sets, MATH and MATH . Since MATH is a MATH-spin structure, we can describe MATH on MATH by MATH, and on MATH by MATH, where MATH. On the exceptional curve MATH the sheaf MATH is isomorphic to MATH, and degree considerations show that MATH, so MATH and MATH. Moreover, in a neighborhood of MATH, if MATH is the image of the MATH-th section MATH, the invertible sheaf MATH is trivial and is generated by the element MATH. The MATH-th power map MATH is an isomorphism away from the nodes of MATH, and since it is a power map (changing the isomorphisms MATH and MATH, if necessary), it maps the generators MATH and MATH as follows: MATH and MATH . Since MATH is an isomorphism away from the nodes, we have MATH, or MATH, for some MATH-th root of unity MATH. Changing the isomorphism MATH by MATH, we may assume MATH . On MATH we also have MATH . So global sections of MATH are of the form MATH . We claim that the MATH-module MATH is isomorphic to MATH via MATH . The map is clearly a MATH-module homomorphism. Moreover, for any section MATH we may assume that MATH and MATH. Likewise, we may assume that MATH and MATH . Consequently, we have MATH or MATH . Thus MATH and MATH are completely determined by MATH . We may, therefore, map MATH to MATH via MATH and it is easy to check that this homomorphism is the inverse of the first. An identical argument shows that MATH is isomorphic to MATH, where MATH and MATH. This shows the existence of the desired presentation for MATH. It remains to show that the maps MATH are power maps REF . Again, since the arguments are essentially identical for each pair MATH and MATH, it suffices to prove this in the case of MATH for some MATH dividing MATH. As above, we have MATH. Let MATH be a divisor of MATH, and MATH. Let MATH be the smallest non-negative integer congruent to MATH modulo MATH and MATH be the smallest non-negative integer congruent to MATH modulo MATH. Define integers MATH and MATH as MATH . The module MATH is generated by MATH and MATH with MATH and MATH. Further, MATH may be defined on MATH by MATH, MATH and on MATH by MATH, so we may describe MATH as above: the module MATH is isomorphic to MATH and is generated by MATH and MATH. We must show that MATH maps, via MATH, to MATH for MATH and to MATH when MATH. We will do the first case - the second case is similar. The element MATH is of the form MATH so on MATH, this element MATH maps as MATH . On MATH, the element MATH maps as MATH . It is straightforward to check that these are the same on MATH. But this is exactly the canonical MATH-th power map REF for MATH, as desired. |
math/0012210 | The degree of the sheaf MATH is an integer and is given by MATH hence when MATH we have MATH . The dimension MATH of MATH is MATH . If MATH we have MATH, which implies MATH, and we immediately have MATH. If MATH, then since MATH, we have MATH; and hence MATH is the only solution to the congruence MATH. Consequently, MATH . If any of the MATH are equal to MATH, then the argument in the proof of REF shows that MATH must be zero. |
math/0012210 | For MATH, this is true by definition. In the case that MATH, since MATH, the MATH-spin virtual class MATH is the top NAME class MATH of the first cohomology of the MATH-th root MATH on the universal curve MATH, by the convexity axiom of REF. We have the following commutative diagram. MATH . Here MATH is the natural map induced by MATH and stabilization of MATH. If MATH is the MATH-th root on MATH, then by REF and the universality of the sheaves involved, MATH is isomorphic to the pullback MATH of the MATH-th root MATH from MATH, and MATH. By the NAME spectral sequence we have MATH . But MATH is flat (it is the composition of flat morphisms - see the commutative REF ), so that MATH . |
math/0012210 | We will give the proof on the level of (operational) NAME groups MATH with notation as in CITE. From CITE it will follow then that such results also hold for MATH. In particular, if we denote the identity maps on MATH, MATH, MATH, MATH, and MATH by MATH, MATH, MATH, MATH, and MATH, respectively, then we have MATH, and MATH. We take MATH in MATH, so that MATH is in MATH. Also, we have MATH. Finally, by MATH we mean MATH . As in CITE, for any morphism MATH, we define MATH to be MATH where MATH and MATH is an arbitrary morphism, and MATH. We also define, for any proper, flat morphism MATH of NAME stacks MATH and MATH, the proper flat pushforward MATH to be MATH where MATH is an arbitrary morphism, MATH is an element of MATH, and MATH is an element of MATH. Note that REF of NAME 's definition in CITE of the operational NAME ring MATH for the identity morphism MATH states that elements of MATH only need to commute with pullback along representable, flat morphisms of NAME, despite the fact that standard definitions of general operational NAME rings require that these elements commute with pullback along all flat morphisms of NAME (see CITE and CITE). In what we do below, we will need the definition of MATH that requires commutativity with all flat pullbacks; that is, we require the following. Let MATH be a flat morphism of NAME stacks, which is not necessarily representable, and let MATH be an arbitrary morphism of NAME stacks. For any MATH and MATH, we have MATH . This seemingly minor difference in the definition of MATH allows us to prove a projection formula for non-representable morphisms. CASE: Let MATH be a proper, flat morphism of NAME stacks (which is not necessarily representable), and let MATH be an arbitrary morphism of NAME stacks. We have MATH . CASE: (Projection formula for MATH) Let MATH be a proper, flat morphism of NAME stacks (which is not necessarily representable). For any MATH and MATH we have MATH . For REF, the same proof as given by NAME for this equation CITE works exactly for our case, too; nowhere is the representability of MATH used in NAME 's proof. For REF , again NAME 's proof of the projection formula CITE works for non-representable morphisms, the only change needed is that CITE (commutativity with flat, representable pullbacks) must be replaced by our REF for non-representable, flat pullbacks. One more fact we will need in the proof of REF is the commutativity with proper pushforwards required by the definition of MATH (compare CITE); namely, if MATH is proper, and MATH is an arbitrary morphism, then by definition of MATH, for any MATH and for any MATH we have MATH . Now we may proceed with the proof of REF . We will refer throughout the proof to the notation of the commutative REF . Since MATH is a birational map, it is a splitting morphism (that is, MATH, as a map on MATH). Moreover, the morphism MATH is flat CITE and proper, MATH is flat CITE and proper CITE, and MATH is flat and proper by REF . We have the following relations: MATH . This completes the proof of REF . |
math/0012210 | This is an immediate consequence of REF . |
math/0012210 | All axioms follow immediately from REF . |
math/0012210 | REF shows that the intersection numbers MATH are completely determined by the classes MATH and MATH if MATH is stable. We must still address the unstable cases - when MATH. But by REF , these are always of dimension zero. Let MATH where MATH are the connected components of MATH, and let MATH be the morphisms forgetting the MATH-spin structure. Furthermore, let MATH be MATH restricted to MATH and let us assume that MATH is zero dimensional. For all MATH in MATH we have MATH where MATH denotes the (orbifold) degree of MATH. This completes the proof. |
math/0012210 | The proof is identical to the case of MATH in CITE. It follows from the long exact sequence associated to the short exact sequence MATH and the fact that MATH for all MATH, which follows from an immediate generalization of REF from CITE. |
math/0012210 | This was proved in the case where MATH is a point in CITE. The same proof goes through here using the definition of MATH (which is now defined in the unstable range) and the fact that MATH. |
math/0012210 | This follows from the Mapping to a Point property. |
math/0012216 | As was mentioned above, the former part of the lemma is a direct consequence of REF. For the latter part, let MATH denote the specialization of MATH obtained by putting MATH. Note that MATH is defined over MATH so that its representation space is MATH. Let MATH and MATH denote the matrix form of the action of MATH and MATH, respectively, on MATH with respect to the basis MATH above. Then the assertion follows from the existence of MATH such that MATH and MATH. A direct calculation shows that the following matrix gives a solution: MATH . |
math/0012216 | A direct computation implies the following equality in MATH: MATH . This implies immediately that MATH. Furthermore, in view of the fact that MATH, the degree MATH part of MATH, is trivial on MATH REF , we obtain the required equality by taking REF modulo MATH. |
math/0012216 | It suffices to show that MATH for MATH, MATH. By definition, we have MATH. Thus taking the mod MATH classes, we have MATH as elements of MATH. In view of REF, this is nothing but MATH. |
math/0012217 | This is a trivial check. |
math/0012217 | We have to show that the following square commutes: MATH . To do so we consider this square as the left-hand face of the cubical diagram MATH . The top and bottom squares commute by REF . The front and back squares are the same and commute by assumption. The right-handsquare commutes as well because MATH is a homomorphism. This forces the left-hand square to commute and we are done. |
math/0012217 | If MATH is a localization, all three conditions have to be satisfied. Indeed given an automorphism MATH, REF tells us that there exists a unique group homomorphism MATH such that MATH. Since MATH is finite and simple, MATH is an automorphism. We set then MATH and REF follows. Given a subgroup MATH and an isomorphism MATH, a similar argument with REF ensures the existence of an automorphism MATH such that MATH. Thus REF holds. Finally REF is also valid since the unique extension of MATH to MATH is the identity. Assume now that all three conditions hold. For any given homomorphism MATH, we need a unique homomorphism MATH such that MATH. If MATH is trivial, we choose of course the trivial homomorphism MATH. It is unique since MATH is in the kernel of MATH, which must be equal to MATH by simplicity. Hence, we can suppose that MATH is not trivial. Since MATH is simple we have that MATH and MATH. By REF there is an automorphism MATH such that MATH, or equivalently by REF , MATH. Therefore the composite map MATH is some automorphism MATH of MATH. By REF this automorphism of MATH extends to an automorphism MATH. That is, the following square commutes: MATH . The homomorphism MATH extends MATH as desired. We prove now it is unique. Suppose that MATH is a homomorphism such that MATH. Then, since MATH is simple, MATH. The composite MATH is an element in the centralizer MATH, which is trivial by REF . This finishes the proof of the theorem. |
math/0012217 | Since MATH is a maximal subgroup of MATH, MATH. The corollary is now a direct consequence of REF taking into account REF about the number of conjugacy classes of subgroups of MATH isomorphic to MATH. |
math/0012217 | We show that the conditions of REF are satisfied, starting with REF . Since MATH is maximal, it is self-normalizing and therefore the action of MATH on the cosets of MATH is isomorphic to the conjugation action of MATH on the set of conjugates of MATH. By our second assumption, this set is left invariant under MATH. Thus the action of MATH extends to MATH and this yields the desired extension MATH. To check REF , let MATH be a subgroup of MATH which is isomorphic to MATH and denote by MATH an isomorphism. Let MATH be the stabilizer of a point in MATH under the action of MATH. Since the orbit of this point has cardinality MATH, the index of MATH is at most MATH, hence equal to MATH by our first assumption. Thus MATH acts transitively. So MATH has a second transitive action via MATH and the action of MATH. For this action, the stabilizer of a point is a subgroup of MATH of index MATH, hence conjugate to MATH by assumption. So MATH is also the stabilizer of a point for this second action and this shows that this action of MATH is isomorphic to the permutation action of MATH on the cosets of MATH, that is, to the first action. It follows that the permutation representation MATH is conjugate in MATH to MATH. Finally, since MATH is a transitive subgroup of MATH with maximal stabilizer, the centralizer MATH is trivial by CITE and thus REF is satisfied. |
math/0012217 | In each case, it suffices to check in the ATLAS CITE that the conditions of REF are satisfied. It is however necessary to check the complete list of maximal subgroups in CITE for the NAME group MATH and CITE for the NAME group MATH. |
math/0012217 | We prove both statements at the same time. The group MATH acts on the projective line, whereas MATH acts on the set of isotropic points in the projective plane. In both cases, let MATH be the stabilizer of a point for this action (NAME subgroup). Let us also denote by MATH either MATH or MATH, where MATH is a prime power as specified above, and MATH is MATH, or MATH respectively. Then MATH is a subgroup of MATH of index MATH by CITE and CITE. By CITE, which is REF when MATH is a prime, the group MATH has no non-trivial permutation representation of degree less than MATH if MATH. The same holds for MATH by CITE if MATH. Thus MATH satisfies REF . It remains to show that REF is also satisfied. The subgroup MATH is the normalizer of a NAME MATH-subgroup MATH, and MATH, where MATH is a complement of MATH in MATH. If MATH denotes the normalizer of MATH in MATH, we know that MATH. This is the NAME decomposition (for more details see CITE). We are now ready to prove that any subgroup of MATH of index MATH is conjugate to MATH. Let MATH be such a subgroup. It contains a NAME MATH-subgroup, and we can thus assume it actually contains MATH. Since MATH is generated by MATH and MATH, the subgroup MATH is generated by MATH and MATH. Assume MATH contains an element MATH. The NAME group MATH is cyclic of order two, generated by the class of MATH (the linear group MATH is a NAME group of type MATH and the unitary group MATH is a twisted NAME group of type MATH). Moreover MATH by CITE and both MATH and its conjugate MATH are contained in MATH. This is impossible because MATH, so MATH. It follows that MATH is contained in MATH. But MATH and MATH have the same order and therefore MATH. |
math/0012217 | The order of MATH is MATH, which is larger than MATH. So the main theorem in CITE implies that any maximal sugbroup of MATH containing MATH has to be one of CITE or a parabolic subgroup. Of course MATH has to divide MATH. Therefore MATH can not be parabolic because the order of a parabolic subgroup of MATH is MATH (see CITE). When MATH there are only two maximal subgroups left to deal with. When MATH, we could also have used the ATLAS CITE. In both cases the only maximal subgroup whose order is divisible by MATH is isomorphic to MATH. |
math/0012217 | The strategy is to verify REF , or rather REF , and REF . Let us first review some facts from CITE. There exist two subgroups MATH with MATH a subgroup of index REF in MATH. This is the inclusion we consider here. Moreover MATH (this is REF), and MATH REF . Thus MATH and REF obviously holds. To check REF consider a subgroup MATH with MATH. By the above lemma MATH must be contained in a maximal subgroup isomorphic to MATH. When MATH the group MATH is complete and CITE shows that MATH is contained in some conjugate of MATH. Therefore MATH is conjugate to MATH in MATH. When MATH, the group MATH is not complete. The group MATH is cyclic of order REF as there is a ``graph automorphism" MATH which is not an inner automorphism (we follow the notation from CITE). Then MATH and MATH are not conjugate in MATH, but of course they are in MATH. Again Theorem A in CITE shows that MATH is conjugate either to MATH or to MATH in MATH and REF is also satisfied. Finally REF is valid since MATH is cyclic of order REF. The number of conjugacy classes of subgroups isomorphic to MATH in MATH is REF when MATH and REF when MATH. |
math/0012217 | The proof is similar to that of the preceding proposition. Apply also CITE for the subgroup MATH. |
math/0012217 | We have MATH while for an odd prime MATH, MATH is complete. On the other hand MATH is cyclic of order REF. By CITE there are exactly MATH conjugacy classes of subgroups isomorphic to MATH in MATH, fused by an automorphism if MATH. Applying REF , we see that the inclusion MATH given by CITE is a localization. |
math/0012218 | We will use only the case where the fibres of MATH are contractible in this paper, when one can apply fiber by fiber the homotopy formula for compactly supported cohomology (see for example, CITE). (The general case then follows from a ``NAME complex" argument.) |
math/0012218 | In the circumstances of the NAME transform as we have been discussing, there is a short exact sequence of vector bundles on MATH where MATH denotes the vector bundle of forms of type MATH in the complex structure of the fibres of MATH (so that MATH). The result follows by taking top exterior powers. |
math/0012218 | Note that by the preceding Lemma, MATH and so MATH . For each fibre MATH the cohomology groups MATH are NAME and so we can identify MATH . Now split the pairing integral on MATH into a fibre integral, which is precisely the NAME duality pairing, followed by an integral over the base. |
math/0012218 | Note first that MATH. The proof uses the fact that the filtrations of the spectral sequences are induced by a sub-bundle of the MATH. One can check directly that MATH descends to the corresponding terms of the MATH-level of the spectral sequences. |
math/0012231 | Observe that if MATH is any root lattice and if MATH is the root system of a lattice MATH then MATH, so for MATH we have MATH . Thus MATH. We may assume the MATH's are ordered so that their dimensions are non-decreasing, and so that among those with the same dimension the determinants are non-increasing. With this ordering, MATH whenever MATH, so MATH is upper triangular. Since each diagonal element MATH is positive, we have MATH so MATH. |
math/0012231 | There are REF root systems MATH for which MATH and MATH. Of these, the root system MATH and REF root systems containing MATH correspond to decomposable lattices. The remaining REF root systems correspond to indecomposable lattices. |
math/0012231 | Each such lattice has at least two automorphisms, so the number of such lattices is at least MATH . We will have more to say about lattices without roots is REF. |
math/0012231 | Let MATH be a REF-dimensional unimodular lattice with no roots, and MATH the corresponding REF-dimensional even unimodular lattice. As in CITE, MATH is equal to MATH, where MATH is the sublattice of MATH consisting of vectors of even norm; MATH and MATH are the cosets of MATH in its dual MATH; and MATH. The only ways the vector MATH in MATH can have norm REF is if MATH and MATH with MATH, if MATH and MATH with MATH, or if MATH and MATH with MATH or MATH. In the first two cases MATH would be in MATH which would contradict MATH having no roots. Thus MATH has a root if and only if there is a vector MATH of norm REF/REF in MATH. But in this case MATH is in MATH - the so-called shadow of MATH - and MATH has norm REF. Since the parity vectors of MATH are precisely twice the shadow vectors CITE, MATH has a parity vector of norm REF if and only if MATH has a root. (REF-dimensional unimodular lattices with no roots and no parity vectors of norm REF must have parity vectors of norm REF by CITE.) |
math/0012231 | See CITE, and note that the product of local densities MATH is equal to MATH when MATH is an even unimodular lattice of dimension MATH CITE. |
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