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math/0103087 | We first show that for a general choice of MATH, the ideal MATH is prime and perfect. Similar to what was done in CITE, we consider a new polynomial ring MATH (in the case where MATH), where MATH, and let MATH. Then we can view MATH as the quotient ideal of MATH in the ring MATH, where MATH is the ideal in MATH defined by the MATH minors of MATH, the MATH minors of MATH and the entries of MATH, and MATH . Since, MATH and MATH are both bi-homogeneous, they are in particular also homogenous, so they define subvarieties of certain projective spaces. Let MATH be the subvariety of MATH defined by MATH, and let MATH be the subvariety of MATH defined by MATH. Then MATH is obtained from MATH by cutting MATH with MATH hyperplanes MATH. In other words, MATH is the intersection between MATH and a MATH-codimensional linear subspace of MATH. By NAME 's theorem CITE, we know that MATH is an integral NAME variety. Let MATH be the grassmannian which parameterizes the linear subspaces of codimension MATH of MATH. It follows from NAME 's theorem (compare CITE, CITE) that the subset MATH, such that for any MATH, MATH is again an integral NAME variety, is non-empty and open. We also let MATH be the grassmannian which parameterizes the linear subspaces of codimension MATH of MATH that lie inside the variety defined by the equations MATH. In CITE, the authors actually showed that for a general choice of MATH, the MATH-codimensional linear subspace of MATH given by the hyperplanes MATH is general in MATH. Moreover, consider the element MATH given by MATH linear equations MATH, then MATH is the subvariety of MATH given by the MATH minors of MATH, so it is an integral NAME variety. In other words, MATH. Since MATH is a closed subset of MATH, we deduce that MATH is a non-empty open subset of MATH. All these facts, put together, imply that for a general choice of MATH, the MATH-codimensional linear subspace of MATH given by the hyperplanes MATH is in MATH. In other words, for a general choice of MATH, MATH is an integral NAME subvariety of MATH, which, in turn, implies that MATH is a perfect prime ideal. For MATH, this and REF clearly imply that MATH is the defining ideal of MATH. For MATH, let MATH be the least integer such that MATH. Let MATH be the grassmannian which parameterizes the linear subspaces of codimension MATH of MATH, and let MATH be the grassmannian which parameterizes the linear subspaces of codimension MATH of MATH that lie inside the variety defined by the equations MATH. Again, it follows from NAME 's theorem that the subset MATH, such that for any MATH, MATH is an integral NAME variety, is non-empty and open (for a general choice of MATH, MATH is an integral NAME variety). One can follow a similar argument as above, and consider the element MATH in MATH given by MATH linear equations MATH, to show that MATH is a non-empty open subset of MATH. Moreover, from CITE, it is easy to see that for a general choice of MATH, the MATH-codimensional linear subspace of MATH given by the equations in REF and MATH other general hyperplanes is general in MATH. Thus, for a general choice of MATH, the MATH-codimensional subspace of MATH given by the equations in REF and MATH other general hyperplanes is in MATH, that is, this MATH-codimensional subspace of MATH is general enough to intersect MATH at an integral NAME variety. In particular, the hyperplanes given by the equations in REF are general enough for a general choice of the points in MATH. This and the fact in the previous paragraph show that for a general choice of MATH, the hyperplanes MATH and the hyperplanes defined by the equations in REF are general enough to intersect MATH at an integral NAME variety. Hence, for a general choice of MATH, MATH together with the equations in MATH form a perfect prime ideal, that is, MATH and the equations in MATH form the defining ideal of MATH. The NAME of MATH follows from the perfection of its defining ideal. The theorem is proved. |
math/0103087 | Similar criteria for varieties with negative MATH-invariants were given in CITE. We adopt his argument with a slight modification to prove our result. Let MATH be the maximal homogeneous ideal of MATH. Let MATH and MATH be the ideals in MATH generated by the two sets of variables with respect to the standard bi-gradation of MATH. Then MATH and MATH. Let MATH and MATH be the defining ideal and the coordinate ring of MATH, respectively. Then, MATH. It is also not hard to see that the NAME of MATH is equivalent to the condition that MATH for all MATH, which is the same as the condition that MATH for all MATH. CASE: Suppose that MATH is a. CM. Equivalently, MATH for all MATH. Consider the following NAME sequence of local cohomology: MATH . The condition MATH for all MATH, implies that the homomorphism MATH is an isomorphism for MATH and injective for MATH. Localizing MATH at the maximal ideals of MATH and MATH, respectively, and making use of CITE, we have: MATH . This implies MATH for all MATH. Together with REF , it then follows that MATH for all MATH and MATH. CASE: Suppose now that MATH for all MATH and MATH, and MATH for all MATH. We observe the following. For MATH, this is to say that the homomorphism MATH is surjective. In other words, the homomorphism MATH is an isomorphism. Furthermore, the vanishing of MATH for all MATH, and all MATH, is the same as having MATH for all MATH. This is equivalent to having MATH for MATH. Since MATH, MATH is clearly also REF. Lastly, the vanishing of MATH for all MATH implies that MATH for all MATH, that is, MATH for all MATH. Suppose, in addition, for every MATH, MATH and MATH. We need to show that MATH is a. CM. Let MATH and MATH, then MATH is the canonical projection. For all MATH and MATH, the NAME spectral sequence MATH degenerates. Thus, the edge homomorphisms MATH are just isomorphisms for all MATH and MATH. Let MATH REF be the MATH-module given by MATH, and let MATH be the sheaf associated to MATH on MATH. It is easy to see that for MATH, MATH . Also, MATH. Thus, the canonical homomorphism MATH is an isomorphism for all MATH and MATH. It now follows that the homomorphisms MATH similar to what was mentioned in CITE, are isomorphisms for all MATH, MATH and MATH. Applying the five lemma on the diagram of CITE, it then implies that the homomorphism MATH is an isomorphism for all MATH, MATH and MATH. By symmetry, the homomorphism MATH is also an isomorphism for all MATH, MATH and MATH. Thus, for all MATH, MATH and MATH . Moreover, it is also easy to see that, for all MATH, MATH . Therefore, in the following NAME sequence of local cohomology MATH the homomorphisms MATH are isomorphisms for MATH, and injective for MATH. We get MATH for all MATH. This is equivalent to MATH being NAME, that is, MATH being a. CM. The theorem is proved. |
math/0103087 | Let MATH be the defining ideal of MATH. Then MATH is a bihomogeneous ideal in MATH, so in particular, MATH is a homogeneous ideal in MATH. Let MATH then be the subscheme of MATH defined by MATH. We observe that our discussion only involves the perfection of the defining ideal of MATH (and MATH) which is the same as the defining ideal of MATH (and MATH). Thus, the proposition would remain the same if instead of MATH (and MATH) we look at MATH (and MATH). The proposition now follows from CITE. |
math/0103087 | Let MATH, the least integer at which the difference function of the NAME function of MATH equals REF, and take MATH (note that MATH, so we in fact only need to take MATH). We shall prove that this value of MATH satisfies the requirements of the theorem. Suppose MATH is an arbitrary integer which is bigger than or equal to MATH. We add MATH general smooth points to MATH to obtain a set of points MATH with the generic NAME function up to degree MATH, that is, MATH . We start by showing that the NAME algebra MATH is NAME using induction on the number MATH of points we add into MATH to get MATH. If MATH, then MATH is a set of MATH points in MATH with the generic NAME function. It follows from CITE that the presentation matrix of MATH has linear entries, MATH is generated in degree MATH, and MATH. It now follows from REF that the NAME algebra MATH is NAME. The assertion that the NAME algebra MATH is NAME is true for the base case. Suppose now that our assertion is true for a set of points MATH, and we need to prove the assertion for the set of points MATH. Let MATH and MATH be the defining ideals of MATH and MATH respectively. Since a general point MATH imposes one independent condition at degree MATH, a system of generators for MATH and for MATH may be given by MATH and MATH, respectively. Consider the following rational maps MATH given by MATH and MATH. Let MATH and MATH be the closure of the graphs of these maps in MATH and MATH, respectively. Clearly, the NAME algebras MATH and MATH are the bi-graded coordinate rings of MATH and MATH, respectively. By induction hypothesis, we know that MATH is a. CM. We need to show that MATH is also a. CM. Let MATH represent the homogeneous coordinates of MATH. Let MATH and MATH be the bi-graded coordinate rings of MATH and MATH, respectively. Let MATH. By REF , we only need to show that MATH, considered as a subscheme of MATH, is a. CM. Clearly, MATH. Let MATH and MATH. By the construction of MATH, we know that MATH and MATH. Therefore, the linear systems MATH and MATH are very ample CITE. Thus, MATH is an isomorphism from MATH onto MATH and from MATH onto MATH. It is easy to see that since the coordinate MATH of MATH is chosen generally, MATH meets each exceptional curve of MATH no more than once. Thus, MATH, restricted to MATH, is at most one point, for every MATH. Since MATH is an isomorphism on MATH, it is also an isomorphism on MATH, whence MATH, restricted to MATH, is also at most one point, for every MATH. Therefore, by CITE, for every MATH, one gets MATH . By REF , to show that MATH is a. CM, it is now enough to show MATH for all MATH, and MATH for all MATH, where MATH is the ideal sheaf of MATH in MATH. It is easy to see that since MATH is the projection of MATH on MATH centered at the point MATH, MATH. Consider the following exact sequence MATH where MATH is the ideal sheaf of MATH considered as a subscheme of MATH. Since MATH is a. CM, taking the cohomology groups, we get MATH . Thus, it suffices to show that MATH . Let MATH be the pull back to MATH of the class of a general line in MATH, and let MATH be the classes of the exceptional divisors corresponding to the blowup at the points MATH, respectively. Let MATH be the images of MATH through MATH, respectively, then they generate the NAME group of MATH. MATH is a hyperplane section of MATH and since the coordinates in MATH are chosen generally, we may assume that MATH belongs to the divisor class MATH. Thus, MATH. We have MATH . Let MATH on MATH. We first prove MATH for all MATH. We shall use double induction on MATH and MATH. For MATH and MATH, we first have MATH. The canonical divisor on MATH is MATH. Let MATH. From CITE, MATH is very ample on MATH. We also have, MATH. Thus, MATH (compare CITE). Suppose now that MATH is true for MATH, we shall show that MATH. Indeed, consider the exact sequence MATH . Since MATH, and since MATH is a rational curve, we have MATH . Thus, MATH. For MATH, we first have MATH. We also have MATH. Thus, MATH (compare CITE). Suppose now that MATH is true for MATH, we shall show that MATH. Indeed, as before, consider the exact sequence MATH . Since MATH, and since MATH is a rational curve, we have MATH . Thus, MATH. Now, suppose that MATH is true for any MATH and some MATH, we shall show that MATH. Indeed, consider the exact sequence MATH . Since MATH, we have MATH. Thus, MATH . Hence, MATH for all MATH. It remains now to prove that MATH for all MATH. Similar to what we did above, we can rewrite MATH where MATH is the ideal sheaf of MATH considered as a subscheme of MATH. It was proved in CITE that MATH for all MATH. That is, MATH for all MATH, where MATH is the ideal sheaf of MATH considered as a subscheme of MATH. Therefore, MATH, considered as a subscheme of MATH, is projective CM (see CITE for definition of projective CM). By considering the exact sequence MATH where MATH and MATH are the ideal sheaves of MATH and MATH in MATH, and from the fact that MATH is projective CM CITE, we deduce that in MATH, the homogeneous coordinate ring of MATH is CM. It is also clear that MATH. Thus, the coordinate ring of MATH in MATH is CM, that is, MATH is a. CM in MATH. Hence, by REF , MATH for all MATH, where MATH is the ideal sheaf of MATH in MATH. By considering the exact sequence MATH and from the hypothesis that MATH is a. CM, we conclude that MATH for all MATH. Hence, MATH . The assertion is proved, that is, MATH is NAME for all MATH. We now proceed to prove that the defining ideal of MATH is generated by quadratics. Again, we use induction on MATH, the number of general smooth points being added to MATH to get MATH. If MATH, then it follows from REF that the defining ideal of MATH is generated by quadratics. Thus, the base case of the assertion that the defining ideal of the NAME algebra MATH is generated by quadratics is proved. Suppose now that this assertion is true for a set of points MATH, and we need to show it for the set of points MATH. Let MATH and MATH be defined as before. By induction hypothesis, MATH is generated by quadratics. We need to show that MATH is also generated by quadratics. Since MATH is a. CM, this amounts to showing that MATH is generated by quadratics. Moreover, since MATH and MATH is generated by quadratics, we only need to show that MATH (the ideal of MATH in the coordinate ring of MATH) is generated in its quadratic degree. It suffices to show that the following multiplication maps are onto: MATH where MATH is the ideal sheaf of MATH in MATH. To prove REF , for each integer MATH, as it was done above, we rewrite MATH . Again, the canonical divisor on MATH is MATH. Let MATH . As before, from CITE, MATH is every ample on MATH. For MATH, we also have MATH . Therefore, it follows from CITE that, for MATH, MATH . Hence, REF follows vacuously. To prove REF , for each integer MATH, as before, we rewrite MATH . Thus, REF is equivalent to MATH . This is indeed true, since MATH is a line in MATH, and hence, is obviously generated in degree REF. Similarly, it can be shown that REF are equivalent to MATH and MATH respectively. Those are again true since MATH is a line on MATH, and so, generated in degree REF. The theorem is proved. |
math/0103091 | The lattice of idempotents has order MATH, with MATH complementary pairs. Consider a sublattice of order four: MATH and any other complementary pair MATH. It must be shown that MATH mod MATH. Now idempotents MATH are complementary, so MATH mod MATH, implying : MATH (mod MATH), thus MATH is idempotent. And MATH mod MATH, so MATH, and similarly MATH. Hence MATH mod MATH, because by MATH the only idempotent covering complementary MATH and MATH is REF. Clearly, for MATH holds MATH so MATH, with carry =REF (base MATH). |
math/0103091 | Notice that MATH, so for each even or odd pair MATH in MATH holds MATH. Hence : MATH , and only if MATH this yields MATH as additive automorphism of MATH. Furthermore, MATH is the lowest odd idempotent, namely the image under MATH of the lowest even idempotent REF in MATH (for squarefree MATH no divisors of REF exist). It is readily verified that this morphism is REF onto, mapping MATH and MATH into each other. |
math/0103091 | Due to the subgroup ordering, a subset of baseprimes disjoint from (complementary to) those in MATH defines a subgroup ordered above or equal to MATH. |
math/0103091 | MATH : Given MATH, notice that MATH, so MATH+REF is identity for MATH, hence MATH for MATH idempotent MATH. Now MATH, and in general expanding MATH, with MATH for all MATH and factoring out MATH, yields: MATH . We need to show MATH for every even idempotent MATH, where MATH is the period of MATH, with corresponding odd idempotent MATH, which equals REF iff MATH. In fact it would suffice if MATH is in a group complementary to MATH in the lattice of MATH. The baseprimes in MATH, which are all necessarily odd, would then complement those in even idempotent MATH. This can be seen as follows: MATH implies MATH, hence MATH, so: MATH. Apparently, the odd baseprimes in MATH complement at least those in MATH because their union is complete (product REF). This implies MATH, independent of the extra factor MATH. So : CASE: MATH, where MATH is the period of MATH+REF in MATH. REF is dual to REF , proven similarly by using MATH REF |
math/0103091 | In short write MATH for MATH. Let MATH be any even idempotent, then multiply MATH REF on both sides by MATH. On the lefthand side this yields MATH which is the max-subgroup on MATH, and on the righthand side MATH, sothat MATH. Using REF yields: MATH for all even MATH, so MATH covers all even residues. |
math/0103091 | The mappings MATH form a morphism because MATH mod MATH, where MATH. |
math/0103091 | By complete inspection. Increments REF in MATH cause successive MATH with carry increment MATH=REF to MATH for MATH. Exclude non-prime REF by including primes REF. Then MATH is covered by pair sums of primes MATH in MATH, extended with primes REF in MATH. |
math/0103091 | Extension sets MATH are disjoint for different carries MATH, and MATH in distinct extension sum sets remain so under any shift MATH. For distinct carrysums MATH with MATH. Their union covers MATH if pair sums MATH cover MATH. Because some MATH missing from MATH implies its translations MATH are also missing from all MATH with MATH. |
math/0103091 | By inspection REF MATH holds for MATH (MATH=REF). Now assume MATH to fail for an even primepair sum MATH in MATH, to guarentee prime summands (by REF ). Then REF : MATH mod MATH implies MATH to be missing as primepair sum from MATH, making it incomplete by REF . This in turn, by MATH mod MATH, reduces to incomplete MATH, etcetera, down to incomplete MATH, contradicting REF . Notice that MATH contains only primepair sums, by REF . Combining this with the overlap (by NAME 's Postulate) of prime summand intervals for successive MATH, hence also of pairsum intervals MATH for MATH, it follows that NAME 's Conjecture holds. |
math/0103092 | We will prove the first identity. Other identities are obtained similarly. If suffices to check for a monomial MATH. But in this case one has MATH which is the first identity. |
math/0103092 | As MATH preserves the principal gradation, we may assume MATH for some MATH. Since MATH, we have MATH by using REF . It follows by comparing REF with REF that MATH . The solutions of REF are MATH or MATH. The first two solutions give rise to what we have listed in REF. The last two solutions are easily ruled out since an involution can not send MATH to MATH or MATH. |
math/0103092 | We may assume that MATH . Since MATH, we have MATH . On the other hand we have MATH . If follows by comparing REF that MATH. Similar calculation by using MATH shows that MATH. If follows from the following identity MATH that MATH. Since both MATH and MATH are polynomials, MATH is some nonzero constant MATH while MATH is MATH. |
math/0103092 | Assume that MATH . Then we have MATH . As we may rewrite MATH, it follows from REF that MATH . This implies that MATH or MATH. In the first two cases we get MATH. The last two solutions for MATH give rise to MATH. We first claim that these last two cases cannot happen. For example let us take MATH (the case MATH is similar). On one hand by REF we have MATH . So MATH. On the other hand, MATH since MATH. This is a contradiction. We claim that MATH cannot occur either. Indeed if MATH then by REF MATH which is a contradiction. |
math/0103092 | Assume that MATH for some MATH. Then we have MATH . Thus MATH. This implies MATH for some MATH or MATH. We have MATH as MATH. On the other hand, we can compute directly that MATH . It follows that MATH. In the case that MATH, MATH. Thus MATH. We claim that MATH cannot occur. Indeed if MATH, then MATH and MATH. Together with REF and the fact that MATH this implies that MATH . This contradicts the fact that MATH. |
math/0103092 | Assume that MATH . Since MATH, we have MATH . On the other hand, we also have MATH . If follows by comparing REF that MATH. Similar calculation by using MATH shows that MATH. If follows from MATH that MATH. Since both MATH and MATH are polynomials, MATH is some nonzero constant MATH while MATH is MATH. Noting that MATH, we have MATH. |
math/0103092 | Assume that MATH . We calculate MATH . As we may rewrite MATH, it follows that MATH . This implies that MATH and MATH are constants. Let us put MATH then MATH and so MATH. It follows from MATH that MATH. On the other hand we have by REF MATH . This implies MATH that is, MATH. Thus MATH . |
math/0103092 | Assume that MATH for some MATH. We have MATH . It follows that MATH . This implies that MATH and thus MATH for some MATH or MATH. From MATH we conclude that MATH. We calculate MATH . Thus MATH. We claim that MATH is impossible. Indeed if MATH, then MATH. So by REF (to be definite we choose MATH, the case MATH is the same) it follows that MATH which is a contradiction. Thus MATH and so MATH. |
math/0103092 | Assume that MATH and MATH belong to MATH. MATH . It follows by comparing with MATH that MATH . Define MATH and MATH. We see that MATH. On the other hand, we calculate that MATH . By comparing with MATH we have MATH . We observe that these two equations are equivalent to each other. |
math/0103092 | Assume that MATH and MATH belong to MATH. MATH . We obtain by comparing with MATH that MATH . The two equations are clearly equivalent. On the other hand, we have MATH . We obtain by comparing with MATH that MATH . If we let MATH and MATH, we see that MATH. |
math/0103092 | Assume that MATH and MATH belong to MATH. We have MATH . We obtain by comparing with MATH that MATH . These two equations are equivalent. On the other hand, we have MATH . We obtain by comparing with MATH that MATH . If we let MATH and MATH, we see that MATH. |
math/0103092 | Of course if MATH is quasifinite, then MATH is necessarily finite-dimensional for all MATH. Conversely suppose that MATH is finite-dimensional for all MATH. We need to show that for any positive integers MATH the space MATH is finite-dimensional. We will do this by induction on MATH, with MATH being the hypothesis of the proposition. Now MATH is finite-dimensional, hence there exist finitely many elements MATH such that MATH is spanned by MATH . This implies that MATH lies in the span of MATH, for MATH. Thus it is enough to prove that, for each MATH, each of these MATH is finite-dimensional. Now we have MATH . By induction MATH is a finite-dimensional vector space and hence MATH is finite-dimensional. Thus it is enough to prove that MATH is finite-dimensional as well. But MATH . However each of these spaces on the right-hand side lies in some finite-dimensional vector space by induction hypothesis. Thus MATH is also finite-dimensional. |
math/0103092 | Consider MATH so that MATH. Take any MATH, for some MATH. We compute their bracket MATH . Thus MATH, for any MATH, that is, MATH is invariant under the multiplication of any odd polynomial. But then it follows that MATH is invariant under the multiplication of any polynomial, and hence is an ideal of MATH. Again take any MATH so that MATH, and any element MATH for some MATH and MATH. We compute MATH where MATH. As MATH ranges over all odd polynomials, MATH ranges over all even polynomials. From this it follows that MATH is a module over MATH. Replacing MATH with MATH in the above calculation one shows similarly that MATH is a module over MATH. |
math/0103092 | We will first prove REF . Suppose that MATH. Let MATH so that MATH. We compute MATH where MATH. Since MATH and MATH can be taken to be an arbitrary non-zero element of MATH, it follows that MATH. The claim that MATH implies that MATH is proved similarly and thus omitted. Conversely suppose that MATH. Let MATH and MATH and compute MATH where MATH and MATH. Now MATH if and only if MATH. Now since MATH is an ideal of MATH, we may replace MATH above by MATH. This implies then that MATH as well. But this is impossible unless MATH, which is a contradiction. Hence MATH. Similarly MATH. This proves REF . Next we prove REF . For this suppose MATH and MATH. Then MATH and hence MATH. Similarly MATH implies MATH. Finally suppose that MATH and MATH and consider MATH where MATH and MATH. Now the same argument proving MATH implying MATH can be used to derive that MATH, for MATH. |
math/0103092 | The identity follows by applying successively the differential operators of the form MATH, MATH,MATH, MATH to the identity in REF. |
math/0103092 | From the definition of the determinant it is easy to see that the simple root vectors in REF kill each MATH. Hence they kill MATH. Also it is clear that MATH is annihilated by REF , since such an expression acts on MATH by replacing the row MATH by MATH, where MATH. As the latter row already appears in MATH, the resulting determinant is zero. Hence it remains to show that REF also annihilates MATH. But this now can be easily seen using REF . |
math/0103092 | The proof is similar to that of REF. Again it is easy to show that the expression is annihilated by operators of the form REF . The fact that it is annihilated by operators of the form REF is again a consequence of REF . |
math/0103092 | The second part is a variation in our infinite-dimensional setting of REF. Since MATH is a rational representation of MATH, it follows from the second part that all irreducible representations of MATH appears in the decomposition of MATH. The multiplicity-freeness follows from REF. |
math/0103092 | The generating functions above can be directly calculated by using the embedding MATH given in REF. The irreducibility follows from REF. |
math/0103092 | We will prove the last identity only and the proof of the others is similar. Using the embedding MATH we calculate MATH . |
math/0103092 | From the second part we see that all irreducible modules of MATH that do not factor through MATH appear in the NAME space. The first part follows from this and the fact that MATH and MATH form a dual pair. In order to show that REF is a highest weight vector it is enough to check that it is annihilated by MATH, for all MATH, due to REF, which is an easy computation. The proof of REF is somewhat more tedious, and we will omit the details, as it is also a rather straightforward calculation. We only remark that the crucial identity used in the computation is again REF. Finally, once the highest weight vectors are explicitly given, the computation of weights of these vectors is straightforward. |
math/0103092 | We will only give a proof of the first identity, as proofs for the other two are obtained similarly. MATH . |
math/0103092 | The generating functions for MATH can be computed using REF. Irreducibility is a consequence of REF. |
math/0103092 | We will only give a proof of the second identity, as proofs for the other two are obtained similarly. MATH . |
math/0103092 | We will give a proof of the second identity only. MATH . |
math/0103094 | When MATH, the result is clear, since the NAME characteristic of the smooth quadric MATH in MATH is equal to MATH. So assume MATH and choose an hyperplane MATH in MATH. Denote by MATH the arrangement in MATH obtained by deleting MATH from MATH and by MATH the arrangement in MATH obtained by intersecting MATH with the hyperplanes in MATH. Since MATH and MATH, the result follows by induction on the number of hyperplanes. |
math/0103094 | Since a NAME group acts transitively on the corresponding set of chambers (compare REF ), this follows from REF . |
math/0103094 | Note that MATH, since MATH is homogeneous. Thus we can calculate MATH by applying REF to the resolution MATH defined in REF. Since the NAME characteristic of a complex algebraic variety with a free MATH-action is zero (see, for example, CITE), we have to sum in the formula of REF for MATH only over strata lying inside the strict transform in MATH of the exceptional divisor MATH of the blow up MATH of MATH in MATH. Let MATH be the complement in MATH of the strict transform in MATH of the locus of MATH, and let MATH be a partition of MATH into constructible subsets MATH on which MATH and MATH are constant. Then the above discussion yields MATH where MATH is any point in MATH. A similar formula holds also for MATH, where MATH is defined below. Applying now REF to the resolution MATH defined in REF, and using REF , we obtain the relation MATH where, for any subset MATH of MATH, MATH is the group generated by the reflections with respect to the walls containing MATH. Here by stratum, we mean non dense stratum of the arrangement associated to MATH. Using REF , we obtain MATH where by ``face" we mean any face of codimension MATH of some chamber of the arrangement associated to MATH. The faces which are contained in a stratum MATH are indeed exactly the chambers of the set of real points of MATH. Let MATH be a fixed chamber. For each face MATH, there exists a unique face MATH of MATH such that there exists MATH in MATH with MATH (see REF, Remarque REF). We say that the faces MATH and MATH are related. We have MATH with MATH the number of faces related to MATH. Note that MATH is equal to the total number of chambers divided by the number of chambers in the arrangement associated to MATH, since every chamber has exactly one face related to MATH, and the number of chambers containing the same face MATH is equal to the number of chambers in the arrangement associated to MATH. Hence, since the number of chambers of a NAME arrangement is equal to the order of the group, we obtain MATH . Associating to MATH the subgraph of the NAME diagram MATH of MATH whose vertices correspond to the walls of MATH that contain MATH, we get MATH . Since, by REF, which is given in CITE a direct self contained proof, MATH when MATH is not connected, the result follows. |
math/0103094 | For any element MATH of MATH, the monodromy zeta function MATH of MATH is equal to MATH, where MATH is as in REF (see, for example, CITE, p. REF). Hence, we have to prove that MATH for every MATH in MATH. Since both functions MATH and MATH are homogeneous of degree MATH, the map MATH, MATH induces the monodromy action MATH on the elements of MATH appearing in REF . Hence the traces in REF are equal to the NAME characteristic of the fixed point manifold of MATH restricted to the generic fibre of respectively MATH, MATH and MATH (see, for example, REF). Thus, we may write MATH and MATH . Hence it is enough to prove that MATH . This follows from the fact that the variety MATH is isomorphic to the variety MATH through the morphism MATH, whose inverse is given by the same formula, since the two varieties are étale covers of degree MATH of the two varieties occuring in REF . |
math/0103094 | Here again we apply REF to the resolution MATH, defined in REF, to compute MATH. And similarly, replacing MATH by MATH, one computes MATH. Direct observation shows that both are equal. To prove the equality MATH, one remarks that MATH induces a MATH-equivariant fibration MATH, hence MATH and MATH are already isomorphic as complexes of sheaves with MATH-action. |
math/0103094 | Follows directly from REF . |
math/0103094 | Denote by MATH the part of MATH on which the MATH-action is given by multiplication by the determinant. By a direct adaptation of the proof of REF we get an isomorphism MATH . Thus, it suffices to show that MATH . Now consider the compact set MATH and set MATH. Since MATH is compact and MATH is constant on MATH, the set MATH determines a cycle class MATH in MATH which is non zero since MATH for every MATH. We remark that, for MATH in MATH, we have MATH because MATH is induced by an element of MATH on which MATH acts by multiplication with the determinant. For almost all MATH, the canonical morphism MATH is an isomorphism by REF. Note that MATH has rank one, for almost all MATH, because of REF . Thus, when MATH is general enough, the cycle class MATH is a generator of MATH. Choose a topological generator MATH of MATH. It is enough to prove that MATH . Now remark that the map MATH, with MATH is a realization of the monodromy of MATH which is induced by MATH. Let MATH be the character sending MATH to MATH. Since MATH, we obtain that MATH and the result follows. |
math/0103094 | Since MATH is homogeneous of degree REF and MATH is homogeneous, the map MATH induces the monodromy action on MATH. This proves the existence of MATH and MATH in REF . That MATH follows directly form REF , since the virtual rank of MATH equals MATH. To prove REF we first show that MATH . To show REF , we use the fact that REF, and REF for MATH, remain valid when the constant sheaf is replaced by a local system on MATH with MATH-action (see Remarque REFEF). Moreover, this remains valid in the more general situation of REF . We apply this to the rational map MATH induced by MATH. In this case, using the notation of REF, we have MATH where MATH is the strict transform in MATH of the exceptional divisor MATH of the blow up MATH of MATH in MATH, and MATH is the strict transform in MATH of the locus of MATH in MATH. Since the NAME characteristic of a complex algebraic variety with a free MATH-action is zero (see, for example, CITE), we may sum in REF only over strata lying inside MATH. In fact, because of REF, we may even only sum over strata lying inside MATH, where MATH is the strict transform in MATH of the locus of MATH. Equality REF follows now directly from the above discussion and the fact that MATH and MATH are locally isomorphic, as local systems with MATH-action, on a neighbourhood of MATH in MATH. But MATH, hence MATH by the projection formula, and MATH which shows that REF follows from REF . Since MATH, REF follows from REF . |
math/0103094 | Let us check that REF are verified for almost all MATH. Consider the open immersion MATH. The canonical morphism MATH is an isomorphism for almost all MATH, hence the canonical morphism MATH is an isomorphism for almost all MATH. Since MATH is affine, MATH is zero for MATH, hence, by NAME duality, it follows that REF is verified for almost all MATH. For such a MATH, the rank of MATH is equal MATH times the NAME characteristic of MATH, so REF follows from REF . Now the result follows by putting together REF . |
math/0103094 | We have MATH and MATH . Hence we deduce from REF , that MATH . The result follows now from NAME 's formula MATH and the relation MATH. |
math/0103094 | Indeed, this follows from REF and the remark following it in CITE. |
math/0103094 | We first write MATH considering the morphism MATH sending MATH to MATH with MATH. By the proof of REF and by REF , we may apply REF to MATH, MATH and MATH the morphism induced by MATH. Comparing with REF , one deduces the relation (remind the duplication formula MATH) MATH . The result follows by rewriting everything in terms of virtual representations of MATH, using the fact that, by REF , MATH is the zeta function of the monodromy action around MATH on the cohomology of the NAME fiber of MATH. |
math/0103095 | For any function MATH, nowhere equal to MATH, define the modified connection, MATH . Using the NAME formula (see CITE) we have MATH and one can easily compute MATH . Then, assuming MATH on MATH, we can choose MATH so that MATH . Inserting REF , and since the complement of MATH in MATH is of zero-measure, we conclude by observing that the left member of REF is nonnegative. |
math/0103095 | Suppose the limiting case holds in REF , then the right hand side has to be constant on MATH, and MATH . Note that equality holds in REF which yields MATH. Hence MATH is an eigenspinor for the operator MATH with eigenvalue MATH. Using REF , we can show that MATH must be constant on MATH (therefore, MATH) and compute MATH . Then, by REF and the fact that MATH, MATH . Since in the equality case, MATH, we can deduce the relation: MATH . With respect to the isomorphism MATH, we can rewrite REF as an intrinsic equation on MATH: MATH . Note that if there exists two smooth real functions MATH and MATH on MATH and a non-zero section MATH of MATH satisfying for all vector field MATH on MATH then, by computing the action of the curvature tensor on MATH, we see that necessarily MATH which implies MATH . Moreover, in the equality case, the fact that MATH is constant implies that MATH is constant. |
math/0103095 | For any real function MATH that never vanishes, consider the modified covariant derivative defined on MATH by MATH . As in the proof of REF , we compute MATH . To finish the proof of REF , if MATH, we take MATH and then observe that by the NAME inequality, we have MATH . |
math/0103095 | For MATH an eigenspinor of MATH with eigenvalue MATH, let MATH. Then REF gives MATH . Recall that MATH and MATH . Now, it is straightforward to get MATH, hence, MATH . REF , which is also true on MATH, applied to MATH yields MATH . Since MATH, and MATH, we have MATH . As in the proof of REF , we finally take MATH and use the NAME REF . |
math/0103096 | Recall that for a hypergeometric function REF , one has MATH with MATH . Let us replace MATH by the degree operator MATH in the coefficients MATH; REF can be therefore rewritten as MATH. From the recursion relation REF , one readily obtains MATH completing the proof. |
math/0103096 | The quantum operator REF is obtained, using REF - REF , by a direct computation. However, REF turns out to be a particular case of REF since the NAME connection is projectively flat. |
math/0103096 | Straightforward computation leads to MATH and REF therefore yields the result. |
math/0103097 | We have MATH . The total residue is zero outside the homogeneous component of degree MATH. This gives the formula of the lemma. |
math/0103097 | If MATH is in MATH, then for any MATH, the element MATH is in MATH, and it is easy to see that the map above is injective. To prove that it is surjective, observe first that for any set of elements MATH: MATH . If MATH is a subset of MATH, we denote by MATH. Then MATH . If MATH is a basis of MATH, then for every MATH, MATH is a basis of MATH. The lemma follows. |
math/0103097 | Let MATH be a subset of MATH spanning a MATH-dimensional vector space. If MATH is contained in the set MATH, the hyperplane spanned by MATH is the hyperplane MATH. If not, then MATH contains a vector MATH, and we conclude by induction. |
math/0103097 | The set MATH is a small chamber since it doesn't meet any hyperplane. But MATH is the simplicial cone generated by the elements MATH so that MATH is also a big chamber. |
math/0103097 | Let MATH and write MATH. Using the dual base, we compute MATH and we find that the linear form MATH is the sum over all elements MATH such that MATH of the linear form MATH. |
math/0103097 | As we just observed, the linear form MATH is the sum over all elements MATH such that MATH of the linear form MATH. So, to prove the lemma, it is enough to prove that if MATH, then MATH. Assume that MATH . We let MATH be the smallest integer such that MATH. In particular if MATH then MATH which forces MATH, and therefore, if MATH, then MATH is not contained in MATH . |
math/0103097 | Due to REF , the linear form MATH is the iterated residue MATH. Thus MATH. By REF , the form MATH is a signed sum of iterated residues MATH over the elements MATH in MATH and we obtain the lemma. |
math/0103097 | It is immediate to check that MATH is a non negative integer. We will give a combinatorial interpretation of this integer in REF. |
math/0103097 | We have MATH . The function MATH is the iterated residue of MATH. If we start by taking the residue in MATH, (use the first expression), we have to develop the term MATH up to order MATH, so we obtain the second property. On the other hand, MATH is considered as generic up to the last step. Consider the function MATH . This function is of the form MATH, and MATH with MATH. Thus, expanding the exponential, we obtain MATH . This establishes the first property . |
math/0103097 | Let MATH . We have MATH . We first take the residue in MATH. We obtain MATH . This shows already that MATH is independent of MATH proceed to take the residue in MATH. There is a double pole in MATH, so that the dependence in MATH is of degree at most MATH. More precisely, MATH as MATH annihilates functions MATH. Residues vanishes on derivatives, so that we obtain MATH . On the other hand, the residue computation of MATH gives MATH as the step MATH as well as the step MATH involves only simple poles, and we obtain the divisibility property announced. |
math/0103097 | If MATH, the remaining roots MATH occurring in the denominator of MATH are contained in the hyperplane MATH. So the total residue of MATH is MATH. The same argument shows that MATH is MATH . We thus may assume that MATH. We first show that the function MATH is proportional to the function MATH modulo MATH. Thus the functions MATH and MATH will be proportional too, modulo the vector space MATH, their total residues will be proportional and we will get the first recursive relations for the constant MATH . In the following, we use MATH instead of MATH, etc. as it is a more familiar notation for computing derivatives. We write MATH so that MATH. We compute MATH . This is equal to MATH . Using MATH, this is also equal to MATH . Assume first that MATH is odd, so that MATH is anti-invariant by the group MATH. Let us antisymmetrize over permutations. We obtain MATH . To compute MATH, we first sum over the transposition MATH. If MATH, we use MATH . If MATH, we use MATH . We obtain that MATH is equal to MATH . Thus finally, we obtain MATH . If MATH is even, we also obtain MATH . Thus we see that MATH modulo derivatives with respect to MATH, MATH,., MATH. In particular the total residue of the function MATH is proportional to the total residue of the function MATH. This proves the first property. We proceed to the proof of the second property. We return to the notation MATH. To avoid confusion in the following argument we will write explicitly the dependence by the parameters of MATH, that is we will write, whenever necessary, MATH . We now compute the total residue of MATH with MATH . We have: MATH with MATH . Consider the subgroup MATH generated by the circular permutation of MATH. Then MATH . Assume MATH odd. Thus MATH as MATH is anti-invariant under MATH. Remark that: MATH so that MATH . It follows that we have MATH . We now use REF to compute the total residue of the last term in the equality. We write the vector space MATH as MATH, and we consider MATH . Using the decomposition MATH we write MATH . Thus MATH belongs to the vector space MATH and we can easily compute the total residue using REF, as well as the obvious calculation for a one dimensional space. Precisely MATH . Only the term MATH gives a non zero residue, so we obtain MATH . Now MATH . So we obtain MATH . We apply induction hypothesis on MATH. We have MATH . By REF , as the total residue commutes with the action of MATH, we obtain: MATH . But MATH . Thus we obtain the second relation. The case MATH even is completely analogous, so the proof of the first and second relation is complete. The symmetry property in MATH is obvious. Let us check MATH . More precisely, we have the following exact formula ( without projection on MATH). MATH . Indeed, by reduction to the same denominator, the right hand side can be written as MATH . From invariance consideration, MATH has to be anti-invariant under MATH , so is divisible by MATH . From degree consideration, we obtain the desired equality, and MATH. |
math/0103097 | The polytope MATH is isomorphic to the polytope MATH. The linear isomorphism described above using the map MATH preserves the volume, as it preserves the corresponding lattices. We thus compute the relative volume of MATH. Consider MATH . It is an element of MATH of homogeneous degree MATH. We choose the big chamber MATH in MATH consisting of elements MATH, with MATH, for MATH. The element MATH is in the closure of this big chamber. NAME formula is: MATH . The function MATH up to renumbering has been introduced in REF. This is the function MATH. Recall the formula for its total residue given in REF . Let MATH be the particular element of MATH obtained by taking the inverse of the product of simple roots. Then MATH . The linear form MATH has been computed in REF. We write MATH . It is the function MATH restricted at elements MATH . Then MATH . As iterated residues form a dual basis to the basis MATH of MATH (with MATH), we obtain MATH. |
math/0103097 | We use the iterated residue formula for MATH and the iterated constant term formula REF for MATH . Indeed MATH . |
math/0103097 | Let MATH. Then MATH is the differential of some MATH-form MATH, with MATH. The vector space MATH with MATH is stable by the action of MATH on differential forms. As MATH commutes with MATH, MATH is the differential of some MATH-form MATH, with MATH. We have MATH . We now analyze the differential form MATH with MATH. The function MATH is the inverse of the product of MATH linear forms MATH. As MATH preserves MATH, we have MATH where MATH is another linear form in the system MATH and MATH is holomorphic at MATH. Thus MATH is again in MATH. Furthermore, we see that MATH is an element of MATH of degree MATH. Thus MATH is an element of MATH of degree MATH. It follows that MATH is the differential of some MATH-form MATH, with MATH. Adding these two informations, we see that for any MATH, MATH is the differential of some MATH-form MATH, with MATH. This implies the formula of the lemma. |
math/0103097 | We realize MATH in MATH as the system MATH. Then MATH . We use REF for changing variables in residues. Let MATH in MATH. Define MATH. This change of variables preserves the vector space MATH . Indeed MATH is divisible by MATH. The differential of MATH at the origin is the identity. We have MATH . We write MATH instead of MATH. Let MATH be integers and let MATH . Then the function MATH is equal to MATH . Thus, we obtain from REF : MATH . To compute the total residue of the last expression, as the denominator MATH is homogeneous of degree MATH, we have to seek for the term in the numerator which is homogeneous of degree MATH, thus we seek the coefficient of each term of the form MATH with MATH. We thus obtain the first part of REF . The second is proved the same way, using MATH and the function MATH . |
math/0103097 | Indeed, on the big chamber MATH, we have seen in REF that that multiindices MATH such that MATH is non zero are such that MATH. We have MATH, thus the corollary follows. |
math/0103097 | The proof is almost identical to the proof of REF . Let MATH . We have: MATH . We write MATH. We first take the residue of MATH in MATH. We obtain MATH . This shows already that MATH is independent of MATH. We proceed now to take the residue in MATH. There is a double pole in MATH, so that the dependence in MATH is of degree at most MATH. More precisely, a simple calculation shows that MATH . As residue vanishes on derivatives, we obtain MATH . On the other hand, the residue computation of MATH gives MATH as the step MATH as well as the step MATH involves only simple poles, and we obtain the divisibility property announced. |
math/0103097 | We prove it by induction on MATH. If MATH, the set MATH is an interval. So it is either equal to MATH or to MATH. Thus a transformation MATH of MATH gives rise to two transformations of MATH namely MATH and MATH. |
math/0103097 | If MATH, we have MATH, as elements MATH form a basis of linear forms on MATH. Let, for MATH, MATH . The coefficient MATH is equal to MATH. We need to prove CASE: If MATH is not in MATH, then MATH CASE: If MATH, then MATH with MATH. Let MATH be in MATH. The set MATH consists on MATH linearly independent elements of MATH. The partial residue MATH is non zero if and only one of the elements MATH in the denominator of MATH is proportional to MATH. Indeed, in computing MATH, we replace MATH by MATH, when MATH, and MATH by MATH. Thus the only factor creating a pole on MATH is MATH. It follows that MATH is zero unless there exists MATH, MATH, such that MATH is proportional to the linear form MATH . If MATH, then MATH and we see that necessarily MATH and MATH. Then MATH since MATH and that the basis MATH, with MATH is dual to the basis of iterated residues on MATH, as seen in REF. On the other hand MATH and thus the case MATH is completed. Let MATH. We proceed by induction on MATH. Let MATH and MATH such that MATH. Then, with MATH, MATH and MATH . The function MATH has only two simple poles in MATH and in MATH thus MATH iff MATH, or MATH . Precisely CASE: if MATH, then MATH CASE: and if MATH, then MATH . In particular we see that if MATH is not an interval, then MATH is zero, so a fortiori MATH is equal to MATH. If MATH is an interval, then we check on the preeceding formula that CASE: if MATH, then MATH . CASE: if MATH, then MATH . We conclude by induction, using REF . |
math/0103097 | We argue on the cardinal on the underlying set MATH to MATH. If the cardinal of MATH is minimum, then MATH is a basis of MATH. If not, there is a linear relation MATH between elements MATH belonging to MATH. Then MATH . We conclude by induction. |
math/0103100 | A derivation MATH is determined by its values on the generators MATH, and if MATH, then MATH . Thus if MATH, then MATH, which implies REF . Conversely, given MATH, one can define a function MATH using REF . REF ensures that this is well-defined. The construction of MATH ensures that it is a derivation. |
math/0103100 | For MATH one applies REF with the set MATH of MATH for which MATH is a MATH-module homomorphism, for the module structures given by MATH and MATH. For MATH one uses the set MATH of MATH for which MATH satisfies REF using the module structures MATH and MATH to turn MATH into a MATH-bimodule. This is clearly a closed subset. Finally, a sum of upper semicontinuous functions is upper semicontinuous, so the assertion for MATH follows from REF . |
math/0103100 | The subsets of MATH on which the dimensions of MATH and MATH take their minimum values are nonempty open subsets of MATH, and so they must meet. Away from this intersection MATH must have non-minimal dimension. |
math/0103100 | We show first that if MATH is a deformation of MATH and for some MATH the lower triangular block of MATH has coefficients in MATH for all MATH, then there is a matrix MATH where MATH, such that the lower triangular block of MATH has coefficients in MATH for all MATH. Write MATH for some maps MATH, and write each MATH in block form MATH . Thus MATH where MATH. By assumption the constant term satisfies MATH, MATH, MATH and MATH, and we are assuming that MATH for MATH. The fact that MATH is an algebra homomorphism implies that MATH for all MATH. Compare for example CITE, p. REF. Now since MATH for all MATH, we have MATH . Identifying MATH with MATH, and considering this as a MATH-bimodule using the actions of MATH on MATH and MATH, it follows that MATH defines a derivation MATH. Now by REF and our assumption that MATH, every derivation is inner. Therefore, there is a matrix MATH with MATH for all MATH. Let MATH be the matrix given by REF . Then MATH and so the lower triangular MATH block of MATH is equal to MATH . This has entries in MATH, so MATH has the required property. We now prove the lemma. Let MATH be a deformation of MATH. For each MATH, the lower triangular block of MATH has entries in MATH, so by the construction above there is a matrix MATH with the property that if MATH is defined by MATH then the lower triangular block of MATH has entries in MATH. Thus there is a MATH such that the lower triangular block of MATH has entries in MATH. Repeating in this way gives deformations MATH of MATH and matrices MATH of the form REF . Define MATH and define MATH by MATH. We have MATH where the notation MATH means we ignore powers MATH and above. Thus MATH . Thus, for all MATH, the lower triangular block of MATH has entries in MATH. Thus it is zero. |
math/0103100 | Recall that the points of a scheme are in REF correspondence with irreducible closed subsets, with the point MATH corresponding to the closure MATH. Since MATH has finite type over MATH, any nonempty constructible subset of MATH must contain a closed point. Since MATH is a constructible subset of MATH, to show it is open it suffices to show that MATH is closed under generization, that is, if MATH, MATH and MATH, then MATH. By the remark above, it is sufficient to prove this in the case when MATH is a closed point. For in general, since MATH, we know that MATH is nonempty, so since it is constructible it contains a closed point MATH. Then MATH, and hence MATH. Thus suppose that MATH, let MATH, let MATH be a closed point in MATH, and assume for a contradiction that MATH. Since MATH does not contain MATH, this set must be contained in a proper closed subset MATH of MATH. Choose a closed point MATH. Now it is known that it is possible to join any two points of an irreducible variety by a curve, see CITE, p. REF. Thus there is a morphism MATH, where MATH is a curve, and with the image containing MATH and MATH. By passing to the normalization if necessary, we may assume that MATH is smooth. Let MATH be a closed point of MATH which is sent to MATH. The complete local ring of MATH at MATH is isomorphic to MATH, and this gives a morphism MATH sending the closed point of MATH to MATH. We complete this to a commutative square as in the statement of the lemma using the morphism MATH corresponding to a closed point MATH with MATH. Thus the hypotheses of the lemma give a morphism MATH. Let MATH be the generic point of MATH. Then MATH is the generic point of the closure of the image of MATH. Thus MATH contains MATH. On the other hand MATH, and MATH, so MATH. Thus MATH, which does not contain MATH. This is a contradiction. |
math/0103102 | We prove the result by showing that MATH and then applying REF . Since MATH, the first part of the vector satisfies the required estimate. For the second part, we have from REF that MATH and hence that MATH . Because of REF , we have that MATH and MATH. It follows immediately that MATH for MATH. Hence, by substitution into REF , we obtain MATH . We conclude that the second part of the vector in REF is of size MATH, so the proof is complete. |
math/0103102 | By boundedness of MATH REF , we have for all MATH sufficiently close to MATH that MATH . Given MATH satisfying REF , let MATH be the projection of MATH onto the set MATH, and let MATH be some strictly complementary optimal multiplier (for which REF is satisfied). From REF we obtain MATH . Using this observation together with smoothness of MATH and MATH, we have for the gradient of MATH that MATH . Since MATH and MATH are both in MATH, we find from REF that the last term vanishes. From REF and MATH, the second-to-last term has size MATH. For the remaining term, we have MATH, and MATH. By assembling all these observations, and using MATH, we obtain MATH . Using again that MATH, we have from REF that MATH . By gathering the estimates REF , and REF , we obtain MATH . By substituting from REF and using REF , we obtain MATH and therefore MATH . By using REF , and REF , we obtain MATH . Since MATH and MATH, all terms MATH and MATH, MATH are nonnegative, so there is a constant MATH such that MATH . For all MATH, we have MATH by our strictly complementary choice of MATH, and so MATH . On the other hand, we have by boundedness of MATH and our REF that MATH for some constant MATH. By combining REF , we have that MATH . For the MATH component, we have that MATH . Hence, by combining with REF , we obtain that MATH . This completes the proof of REF . We omit the proof of REF , which is similar. |
math/0103102 | From REF , we have directly that MATH. We have from REF that MATH so that MATH whenever MATH is sufficiently small. For the remaining components, we have MATH . By substituting REF , and REF into REF , we obtain the result. |
math/0103102 | If we define MATH we have MATH . Using this notation, we can rewrite REF as MATH where we have defined MATH and MATH is defined in REF . From REF , we have MATH . The fact that MATH annihilates MATH is crucial, because it causes the term with MATH to disappear from the last component of the right-hand side of REF , which becomes MATH . From REF , the perturbation bound REF , our assumptions that MATH and MATH, compactness of MATH, and the fact that MATH is NAME continuous, we have that MATH for some MATH. Using these same facts, we have likewise that MATH so by substituting from REF , we have that MATH . Similarly, from the definition of MATH, we have MATH . For the MATH block, we have from REF that MATH where MATH has all its singular values of size MATH, so that MATH . Using these estimates together with REF , we can rewrite REF as MATH where MATH while MATH and MATH . For purposes of specifying the required conditions on MATH in REF , we define MATH to be a constant such that expressions of size MATH and MATH that arise in the perturbation terms in the coefficient matrix in REF can be bounded by MATH and MATH, respectively. For example, we suppose that the perturbations in MATH, MATH, and MATH can bounded as follows: MATH and that MATH . From REF , we have that MATH . It is therefore easy to show that if MATH can be chosen large enough that MATH (while still satisfying REF ), then MATH . Similarly, we can show that MATH . Note, too, that because of NAME continuity of MATH and compactness of MATH, and the bounds REF , the norms of MATH, MATH, MATH, MATH, and MATH are all MATH. Hence the matrix MATH is itself invertible, and we have MATH . Noting that MATH we examine the size of MATH. Note first from REF together with REF , and REF that MATH . By forming the product of REF with REF and using the bounds in REF , we can show that the norm of MATH can be made less than MATH provided that MATH in REF is sufficiently large. The MATH block of MATH, for instance, has the form MATH . Because of REF , its norm can be bounded by a quantity of the form MATH (for some MATH that depends on MATH), which in turn because of REF is bounded by the following quantity: MATH . Provided that MATH is large enough that this and the other blocks of MATH can be bounded appropriately, we have that MATH, and therefore from REF we have MATH . Our conclusion is that for MATH satisfying the conditions outlined in this paragraph, the inverse of the MATH block of the matrix in REF can be bounded in terms of MATH, which because of REF , and REF can in turn be bounded by a finite quantity that depends only on the problem data and not on MATH. Returning to REF , and using REF , we have that MATH . Meanwhile, for the second block row of REF , we obtain MATH . Since from REF , and REF , we have MATH it follows from REF that MATH . By substituting from REF , we obtain MATH and therefore MATH as claimed. The estimate for MATH is obtained by substituting into REF . |
math/0103102 | From REF and the assumed bound REF on the size of MATH, we have that MATH . By eliminating MATH from REF , we obtain the reduced system MATH where from REF , we obtain MATH . These perturbation matrices satisfy the assumptions of REF , which can be applied to give MATH . From the last block row of REF , and using REF , we obtain MATH . |
math/0103102 | Note first that REF holds trivially in this case for MATH sufficiently small, because our assumption of exact computations is equivalent to setting MATH. We prove the result by identifying the system REF with REF and then applying REF . For the right-hand side, we note first that, because of smoothness of MATH, NAME 's theorem, REF of MATH, and REF , MATH . We now identify the right-hand side of REF with REF by setting MATH . The sizes of these vectors can be estimated by using REF , and REF on the size of MATH to obtain MATH . (By choosing MATH and MATH in this way, we ensure that the terms involving MATH in the estimates of the solution components in REF are not grossly larger than the other terms in these expressions.) We complete the identification of REF with REF by setting all the perturbation matrices MATH to zero and by identifying the solution vector components MATH, MATH, and MATH with MATH, MATH, and MATH, respectively. By directly applying REF , substituting the estimates REF , and setting MATH (since we are assuming exact computations), we have that MATH . To show that the remaining solution component MATH of REF is also of size MATH, we write the second block row in REF as MATH from which the desired estimate follows immediately by substituting from REF and MATH. |
math/0103102 | By combining REF with REF , we obtain MATH . From the bounds on the perturbations MATH in REF and the result of REF , we have for the right-hand side of this expression that MATH . Using these estimates, we can simply apply REF to REF (with MATH) to obtain the result. |
math/0103102 | From REF , we have that MATH . We deduce that the required REF will hold provided that MATH . Since by definition we have that MATH for some positive constant MATH, we find that a sufficient condition for the required inequality is that MATH which is equivalent to REF for an obvious definition of MATH. Identical logic can be applied to MATH to yield a similar condition on MATH. For REF , we have from REF that MATH . Hence, REF holds provided that MATH . Similar logic can be applied to this inequality to derive a bound of the type REF , after a possible adjustment of MATH. Finally, we obtain REF by substituting MATH into REF and applying REF . (Note that, despite the relaxation of the centrality REF , the result of REF still holds; we simply modify the proof to replace MATH by MATH in REF , and MATH by MATH in REF .) |
math/0103102 | The proof follows from REF when we show that the perturbations to REF from all sources - evaluation of the matrix and right-hand side as well as the factorization/solution procedure - satisfy the bounds required by this earlier result. Because of REF , perturbations arising from the factorization/solution procedure satisfy the bounds REF . REF show that the errors arising from evaluation of MATH, MATH, MATH, and MATH are all of size MATH, and hence they too satisfy the required bounds. Similarly to REF , evaluation of MATH and MATH yields errors of relative size MATH, that is, MATH where MATH and MATH are diagonal matrices. We now obtain all the estimates in REF by a direct application of REF , with the exception of the estimate for MATH. Since the expressions for recovering MATH are identical in procedures condensed and augmented, we can apply REF from REF to deduce that the desired estimate holds for this component as well. |
math/0103106 | We know that a compact NAME surface MATH with boundary admits a PALF. We may also assume that the regular fiber MATH has only one boundary component. The fibration induces an open book decomposition of MATH with connected binding MATH. First we enlarge MATH to MATH by attaching a REF-handle along MATH with REF-framing. Note that MATH is a MATH-bundle over MATH, where MATH denotes the closed surface obtained by capping off the surface MATH by gluing a REF-disk along its boundary. Also MATH admits a positive NAME fibration over MATH with regular fiber MATH. Let Map-MATH denote the mapping class group of the closed surface MATH. The second author learned the proof of the next lemma from NAME. Any element in Map-MATH can be expressed as a product of nonseparating positive NAME twists. Let the curves MATH on a genus MATH surface MATH be drawn as in REF and write MATH and MATH for the positive NAME twists about MATH and MATH, respectively. The following is a standard word in the mapping class group Map-MATH. (compare CITE). MATH . Hence we conclude that MATH is a product of nonseparating positive NAME twists. Therefore any negative nonseparating NAME twist is a product of nonseparating positive NAME twists since any two negative nonseparating NAME twists are equivalent. This finishes the proof of the lemma combined with the fact that Map-MATH is generated by (positive and negative) nonseparating NAME twists. We use REF to extend MATH into a positive NAME fibration MATH over MATH with regular fiber MATH as follows. Let MATH be the global monodromy of the PALF on MATH, where MATH denotes the positive NAME twist along a simple closed curve MATH on MATH. Then this product (after capping off the boundary component) can be viewed as a product in Map-MATH. We clearly have MATH . By REF we can replace every negative twist by a product of positive twists to obtain MATH. MATH admits a symplectic structure with symplectic regular fibers by REF. Consequently, MATH is emdedded naturally into a closed symplectic REF-manifold MATH. Conversely suppose that MATH is a symplectic NAME fibration over MATH with a section. Then if we remove a neighborhood of this section union a regular fiber we get a NAME surface. This is clear from our description of NAME surfaces as PALF's. |
math/0103106 | MATH is minimal since MATH is minimal. MATH is simply-connected since MATH and MATH is simply-connected. MATH admits a symplectic structure by REF. Suppose that MATH is NAME. We can embed MATH into a complex surface MATH of general type by REF and CITE. We will show that we can assume MATH. Suppose that MATH. First we extend MATH into MATH by attaching a REF-handle with framing MATH along a Legendrian knot MATH satisfying MATH and contained in a standard REF-ball MATH in MATH. (Here MATH denotes the NAME invariant). Because of the framing of REF-handle, MATH is also a NAME surface. So we can embed MATH into a complex surface MATH of general type. (compare REF ). Thus MATH, since we created a second homology class with positive self-intersection in MATH. Also note that MATH. This gives a contradiction to the following result of NAME and NAME, hence showing that MATH is not NAME. CITE If MATH is a complex surface of general type, and MATH, then MATH admits no splitting along an embedded copy of MATH of the form MATH with MATH. |
math/0103106 | We use the same trick as in the proof of REF . |
math/0103106 | First note that MATH. Suppose that MATH. Then we can apply REF to conclude that MATH is not NAME since MATH. |
math/0103106 | Let Y denote the NAME homology sphere oriented as the boundary of the MATH plumbing as shown in REF . Note that MATH has a metric with positive scalar curvature. Suppose MATH is a NAME filling of MATH. Embed MATH into a complex surface MATH of general type with MATH as in the proof of REF . Since MATH has non-vanishing NAME invariants and MATH has a metric with positive scalar curvature, we have MATH. (see CITE for a proof). Hence MATH is a closed, smooth and negative definite REF-manifold , which can not exist by REF. |
math/0103106 | We apply REF as many times as needed to obtain MATH at the beginning of the given product. |
math/0103106 | Fix an integer MATH. Define MATH where MATH . Map-MATH is the map in REF . Consider the closed oriented REF-manifold MATH . In other words, MATH has a positive open book decomposition with binding MATH, page MATH and monodromy MATH which is expressed as a product of nonseparating positive NAME twists. By the results in CITE, MATH admits a NAME filling MATH which is a PALF (positive allowable NAME fibration over MATH with bounded fibers) of genus MATH with MATH singular fibers. We can calculate the NAME characteristic MATH as follows: MATH . The following is a standard relation in the mapping class group. (compare CITE). MATH . Using this relation and REF , we can write the map MATH as follows: MATH . This last product gives another NAME filling MATH of MATH as a PALF with MATH less singular fibers. Thus MATH . We can iterate the substitution above MATH-times to get MATH mutually non-homeomorphic NAME fillings MATH of MATH such that MATH . It is easy to see that all the NAME fillings are simply-connected. Next we show that MATH, which implies in particular, that MATH is irreducible. |
math/0103106 | Fix an integer MATH. Let MATH denote a genus MATH surface with one boundary component and let MATH denote the closed surface obtained by capping off the surface MATH by gluing a REF-disk along its boundary. Then there is a natural map MATH . So the relation in REF induces a word in MATH. Consequently this gives a positive NAME fibration over MATH with a REF sphere of square MATH and regular fiber MATH. Now consider a neighborhood MATH of a regular fiber union this section. First we observe that MATH. Moreover MATH is obtained by plumbing a disk bundle over MATH and a disk bundle over MATH. We can blow down the MATH sphere to get a disk bundle over MATH with euler number MATH. We prove our result by reversing the orientations. |
math/0103106 | Let MATH denote a closed, connected and oriented surface of genus MATH. Let MATH denote the MATH-bundle over MATH with monodromy MATH . Map-MATH. We can express MATH as a product MATH of (nonseparating) positive NAME twists by REF . Using this product we can fill in the MATH-bundle over MATH with a positive NAME fibration MATH over MATH with regular fiber MATH. We trivially have MATH . Now replace each negative twist in this word by a product of (nonseparating) positive NAME twists again by REF . Hence we get a positive NAME fibration MATH over MATH which is a union of MATH and MATH. Suppose that MATH for some MATH. (Otherwise the theorem is proved). Let MATH denote the genus MATH . NAME fibration over MATH given by the word MATH . These higher genus NAME fibrations can be considered as the generalization of the elliptic fibration on MATH. We have MATH since MATH has a symplectic structure by REF. Define MATH where MATH denotes the fiber sum along a regular fiber. It is easy to see that MATH for MATH. Hence MATH is a closed symplectic REF-manifold which is a union of MATH and MATH (with MATH), glued along the MATH-bundle MATH. |
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