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math/9912101
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By definition of MATH and MATH, MATH, so that, by definition of the torsion: MATH . Taking the scalar product with MATH and using the fact that the MATH are parametrized at unit speed shows that: MATH . Therefore: MATH and we obtain the first equation. Coming back to REF , we see that: MATH and the second equations follows (as well as the derivative of the first).
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math/9912101
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Integrating REF shows that, for MATH: MATH so that: MATH . Let: MATH . For MATH: MATH . Thus there exists MATH such that, if MATH, then: MATH which contradicts the definition of MATH. So MATH, and REF follows. REF is a direct consequence using REF .
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math/9912101
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Let: MATH . REF shows that: MATH while it is easy to check that: MATH because this corresponds to a change in the parametrization of the geodesic starting at MATH in the direction of MATH.
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math/9912101
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Let MATH be the inverse image of MATH by the restriction of MATH to the ball of radius MATH. By the previous corollary and the local convexity of MATH, the restriction of MATH is a diffeomorphism onto its image.
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math/9912101
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REF is a simple consequence of REF , and REF then follows from REF .
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math/9912101
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Let MATH be a small real number; we will see later how small MATH has to be. For MATH, let: MATH . Then define: MATH . For MATH, apply the NAME theorem to an infinitesimal strip bounded by MATH, MATH, MATH and MATH. This shows that: MATH so that: MATH where MATH if MATH is small enough (this last step uses REF applied to the family MATH). Again by REF , it is not hard to check that, again for MATH small enough: MATH . Thus, with REF : MATH . This can be written, for MATH small enough, as: MATH with: MATH . Let: MATH . Then: MATH with: MATH . Thus, by integration: MATH where: MATH with: MATH . The eigenvalues of MATH are the roots of: MATH . If MATH is so small that MATH, those roots can be written as MATH, where: MATH . Therefore, in a well chosen frame, the orbits of MATH are ``spirals" around MATH, with an angular speed which is bounded from below. This already proves, with the upper bound on MATH, that, if MATH is smaller than some MATH, then MATH, so that MATH is not reached and the computations above hold on all of MATH. This proves point REF . Moreover, the trajectories MATH can not remain in a half-plane, so that MATH has to become negative after a time which is bounded in term of MATH (which itself is bounded from below). This leads to point REF.
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math/9912101
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If MATH or MATH is compact, the result is obvious, so we suppose here that neither MATH nor MATH is compact. The proof is by contradiction, so we suppose that MATH. First note that a rather direct smoothing argument shows that, for any MATH, there are smooth curves MATH such that: CASE: MATH bounds a connected closed set MATH which contains MATH; CASE: for each MATH, the curvatures MATH and MATH of MATH at MATH and of MATH at MATH respectively are bounded by: MATH where both curvatures are with respect to the normal oriented towards the interior of MATH; CASE: MATH. For MATH, let: MATH . Thus MATH is not bounded away from MATH near MATH. Choose MATH. There exists MATH with MATH . If MATH is small enough, it is not difficult to show, using REF, that there exists a MATH-geodesic MATH connecting MATH to MATH, of length at most MATH, orthogonal to MATH. For MATH, let MATH be the maximal MATH-geodesic starting at MATH with speed equal to the parallel transport of MATH at MATH along MATH. Let MATH be the distance along MATH between MATH and the first intersection of MATH with MATH, MATH the angle between MATH and MATH, MATH the angle between MATH and MATH. By construction, MATH, while, by REF , MATH is small. Let MATH be again a small real number, for which precisions will come later. Define: MATH . The definition of MATH and the ``almost" convexity of MATH show that MATH, while the same application of the NAME theorem as the one leading to REF shows again that MATH, but with only MATH, while the upper bound is lost because MATH is only ``almost convex" instead of geodesic. Moreover, the same argument as the one leading to REF shows that: MATH again with: MATH . The rest of the proof can now be done just as in the proof of REF , with MATH replaced by MATH, to obtain that there exists MATH (depending on MATH and MATH) such that: CASE: either there exists MATH such that MATH, and this proves the proposition; CASE: or MATH, and in this case the upper bound on the norm of MATH shows that, if MATH has been chosen small enough, then either MATH or MATH. But then, again for MATH small enough, it is not difficult to show that there exists MATH such that there exists MATH such that MATH, so that the proposition holds also in that case.
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math/9912101
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Let MATH and MATH. We can suppose that the integral curve of MATH starting at MATH meets the integral curve of MATH starting from MATH: otherwise, the proposition would fail slightly before the first value of MATH such that the intersection does not exist, because then the length of both MATH and MATH would go to infinity. We call MATH the intersection of the integral curve of MATH starting at MATH with the integral curve of MATH starting from MATH; this intersection has to be unique because MATH is simply connected and MATH and MATH are transverse. Let MATH. Then: MATH . By definition of MATH: MATH so: MATH . Take the scalar product (for MATH) of this equation with MATH to obtain that: MATH which shows, along with REF , that: MATH . The same proof can be used to show also that: MATH . In other terms: MATH . Moreover, MATH and MATH. Integrate REF over MATH to obtain that: MATH . Using REF again leads to: MATH so: MATH . Now integrate this equation to obtain the required upper bound on MATH; the upper bound on MATH is obtained in the same way, exchanging MATH and MATH. Finally, the upper bound on the area comes from the upper bounds on MATH and on MATH which we have found.
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math/9912101
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The intersections between those integral curves remain at bounded distance as long as the lengths of the integral curves of MATH and MATH going from MATH to MATH and to MATH remain below MATH, and the integral curves of MATH and of MATH do not meet MATH because MATH is complete (compare REF).
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math/9912101
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Let MATH be the vector fields and MATH be the functions on MATH such that: MATH . Call MATH and MATH. Then: MATH . But MATH, so that: MATH and therefore: MATH . Now, by definition of MATH: MATH so that: MATH and therefore: MATH . Using this and REF shows that: MATH . Now the definition of a convex map and the bounds on MATH and MATH show that: MATH . This means that, if MATH is large at a point MATH, then it remains large in a neighborhood of MATH in the integral curve of MATH through MATH. REF shows that MATH would then vary a lot on this curve, and this would contradict the definition of a convex map (because MATH).
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math/9912101
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Let MATH be a MATH-quasi-geodesic, with either MATH or MATH and MATH. Note MATH the vector field on MATH defined by: MATH . Let MATH be the angle between MATH and MATH, and MATH the angle between MATH and the parallel transport at MATH of MATH along MATH. By REF : MATH while the definition of a MATH-concave map indicates that: MATH . As a consequence: MATH so that, by definition of a quasi-geodesic, if MATH, then: MATH . We now suppose (without loss of generality) that MATH. Let MATH be the smallest positive number such that: MATH . MATH exists if MATH is small enough (which happens if MATH is small enough). REF indicates that, if MATH at a time MATH, then it remains there until the time MATH where MATH leaves MATH; moreover, MATH reaches MATH before a fixed time MATH (depending on MATH, etc) and, in the interval MATH, it remains above MATH, where MATH is a constant depending also on MATH and MATH. Now it is easy to check that: MATH so that: MATH . As a consequence of the lower bounds on MATH and on MATH: MATH so that MATH must intersect MATH if MATH is small enough.
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math/9912101
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The proof is similar to that of REF ; we call MATH, and we suppose that MATH. Then: MATH . It is the clear that MATH will soon become positive; moreover, if we let MATH and MATH, then: MATH . The second equation indicates that MATH remains positive while MATH, and both equations taken together again show that MATH intersects MATH after time at most MATH for some MATH. The same equations apply for the segment of MATH where MATH; after a bounded time, either MATH will have intersected MATH, or MATH will vanish. The same argument as above then shows that, in both cases, MATH will meet MATH after a time at most MATH for some MATH. This proves the upper bound on the length of MATH. The corresponding lower bound comes from the distance between MATH and the part of MATH that MATH can intersect for MATH. Finally, the case where MATH is parallel to MATH is obtained by applying twice the argument for MATH, which can be used in this case also for MATH because MATH is also directed towards the increasing values of MATH.
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math/9912101
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Note that, if MATH is small enough, then, for any MATH, if MATH is close enough to MATH, then MATH. This comes from REF and from REF . This immediately implies that MATH is a decreasing family of subsets of MATH.
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math/9912101
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Fix MATH. For MATH, let MATH be the maximal integral curve of MATH in MATH containing MATH. We consider two cases. CASE: There exists MATH and a sequence MATH such that, for each MATH, MATH has one end on MATH. If there exists MATH such that MATH, then the proposition is proved. Otherwise, call MATH the connected component of MATH which does not contain MATH in its closure. Let MATH. Since the MATH are integral curves of MATH, they are pairwise disjoint (except when they coincide), so (maybe after taking a subsequence of MATH) MATH is an increasing sequence. Since MATH, it is then not difficult to prove that MATH contains an integral curve of MATH connecting MATH to MATH. CASE: For all MATH, there exists MATH such that, for MATH, MATH remains in MATH and has both ends on MATH. Call MATH a point of MATH where MATH is maximal. Let MATH be the maximal integral curve of MATH (or MATH) with MATH and MATH directed towards the increasing values of MATH. By definition of MATH, MATH is parallel to MATH. Therefore, by REF , the geodesic directed by MATH meets MATH on both sides. With the notations above, this indicates that MATH, so that, by REF , MATH. Thus MATH. The proof then proceeds as in REF above, because the MATH are disjoint and, after taking a sequence MATH, they converge to an integral curve of MATH connecting MATH to MATH.
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math/9912101
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This is again a consequence of REF , along with REF , which bounds the rate of variation of the direction of MATH along MATH.
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math/9912101
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Since the integral of MATH on MATH is finite, REF shows that: MATH . One can therefore define a parallel vector field on MATH as: MATH and, by REF : MATH . The same works for MATH because MATH; set: MATH and then: MATH . Let MATH; we suppose (without loss of generality) that MATH. The proof will proceed differently according to whether MATH or MATH. If MATH, remark that, since MATH, an elementary argument (as for example, in the proof of REF ) shows that: MATH . Thus, for any fixed MATH and MATH, there exist MATH such that: CASE: MATH; CASE: MATH and MATH; CASE: MATH; CASE: MATH at MATH. Then: MATH . According to REF , MATH is a quasi-geodesic; REF then indicates that, if MATH and MATH are small enough, MATH intersects MATH. Consider now the case where MATH. By REF , there exists MATH so that, for MATH large enough: MATH . But, again: MATH and the same argument as in the case MATH leads to the conclusion, but with REF replaced by REF to shows that MATH, which is a quasi-geodesic, actually intersects MATH if MATH is small enough.
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math/9912101
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By linearity, it is enough to prove this proposition when MATH is positive. Let MATH be a one parameter family of curves such that MATH and that: MATH . To simplify somewhat the notations, we call MATH the speed of MATH, that is: MATH . We also call MATH the unit vector along MATH, and MATH. Therefore: MATH for some function MATH such that MATH. Moreover, MATH. By definition of the torsion MATH of MATH, MATH while, since MATH define a coordinate system on a domain of MATH: MATH . Set MATH and MATH, the previous equation becomes: MATH . Let MATH and MATH. Then: MATH and, since MATH by REF : MATH so that: MATH which is the formula we need.
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math/9912101
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For any MATH, we have: MATH . Moreover, MATH and MATH, so that: MATH and the corollary follows.
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math/9912101
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Let MATH. The relation REF becomes: MATH . Let: MATH . The relation now is: MATH so, for MATH: MATH with: MATH . Now the eigenvalues of MATH are the roots of: MATH . From the NAME theorem, MATH, so MATH, and the eigenvalues are MATH, where: MATH . The associated eigenvectors are MATH. Suppose now that MATH. Then MATH, and MATH remains above MATH until after time MATH. But MATH, so it is clear that, after a time MATH, MATH will become negative, and this provides an upper bound for MATH and for MATH: MATH . The decomposition of MATH on the basis MATH is: MATH and this gives an upper bound for MATH: MATH . Moreover, MATH also has an upper bound because MATH, and this gives a lower bound MATH for MATH. Using the upper bound on MATH REF and on MATH (MATH), we see that, for any MATH: MATH . In addition, MATH, so that MATH for MATH, and it follows that MATH. The function MATH therefore verifies the conclusions of REF .
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math/9912101
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Let MATH, so that MATH by REF . By REF , if MATH is small enough, there exists a function MATH such that, on MATH: MATH so that: MATH . For MATH small enough and MATH, set: MATH where MATH is the unit normal to MATH at MATH towards the convex side of the complement. This defines, for MATH small enough, a smooth curve MATH. Let MATH. REF shows that, for MATH: MATH . Then, by compactness, for MATH small enough, MATH, and the corollary follows.
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math/9912101
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Apply REF recursively to obtain a MATH-concave map MATH such that MATH is a segment of MATH.
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math/9912101
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Let MATH. First apply REF after translating the origin to MATH in MATH, so as to obtain MATH and a solution MATH of REF with: MATH . Apply REF again, now after a translation of the origin to MATH; this provides us with MATH and with a solution MATH of REF with: MATH . Repeat this procedure to find a sequence MATH with MATH and MATH, and functions MATH which are solutions of REF with: MATH . Apply REF once more to find MATH and a solution MATH of REF with: MATH . Do the same to find MATH and a solution MATH of REF with: MATH . Now define MATH as the function whose restriction to MATH is MATH for MATH, and which is zero outside MATH. Note that, if MATH, then MATH vanishes outside MATH. Moreover, it is clear that MATH is a (weak) solution of REF .
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math/9912101
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Let MATH be a sequence of smooth curves, MATH, such that: CASE: MATH; CASE: for MATH, MATH lies entirely on the concave side of MATH; CASE: the curvature of MATH is at least MATH, where MATH. The existence of such an approximating sequence is not too difficult to prove. The MATH are not parametrized at speed one. We suppose (without loss of generality) that, for MATH and MATH, MATH is towards MATH. For MATH and MATH, let: MATH . Apply REF to obtain a sequence of piecewise smooth, MATH-Lipschitz functions MATH with: MATH with MATH when MATH and MATH when MATH. Since the MATH are NAME, we can (by taking a subsequence) suppose that they are MATH-converging to a NAME function MATH. Let MATH be the largest MATH such that, for each MATH and each MATH, MATH is defined. Then MATH by compactness. For MATH and MATH, let: MATH . For MATH and MATH, MATH is a curve which might not be embedded, but which, for MATH small enough, is immersed. It differs from MATH only in MATH. Moreover, REF and a simple compactness argument show that there exist MATH, MATH and MATH such that the curvatures MATH of the curves MATH satisfy: MATH where the left-hand side is a measure and MATH. CASE: Since the MATH are curves and differ from MATH only in a compact set, they separate MATH into several connected components, two of which are non compact; we call MATH the non-compact connected component of MATH whose intersection with the concave side of MATH is (empty or) compact. REF shows that the boundary of MATH in MATH is locally convex, with curvature at least MATH. Finally, for MATH, set: MATH . It is not difficult to check that MATH, with an adequate parametrization, satisfies the conclusion of REF .
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math/9912101
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The underlying idea is to apply REF recursively, to obtain the existence of such a deformation for MATH for some MATH. The formal proof, however, has to be done in a slightly different way. Suppose that such a deformation can not exist for all MATH. Let MATH be the set of couples MATH, where MATH and MATH satisfies the conditions demanded, but only until time MATH. There is a natural order on MATH, with: MATH if MATH and MATH for MATH. MATH has a maximal element, say MATH. For MATH, let: MATH . Because of the convexity of the MATH and because MATH has no concave point, MATH is a convex curve. Thus one can apply REF to MATH, and this contradicts the maximality of MATH.
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math/9912101
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It is similar to the proof of REF . First choose a sequence of smooth curves MATH converging to MATH, such that MATH is in the interior of MATH for MATH and that the curvature MATH of MATH is at least MATH, with MATH. For MATH and MATH, let: MATH where we suppose again that MATH is towards the non-compact side of MATH. Let MATH be the lift of MATH to a function on MATH. Apply REF to MATH, to obtain a piecewise smooth, NAME function MATH which vanishes outside MATH and which is at least MATH on MATH. Let MATH be the function defined by: MATH where only finitely many terms are non-zero by definition of MATH. After multiplying it by a constant, MATH is a NAME, piecewise smooth function from MATH to MATH (for some MATH which depends on MATH and on MATH) which is a solution of REF . The rest of the proof can be done quite like in the proof of REF , so we leave the details to the reader.
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math/9912101
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Like the proof of REF from REF .
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math/9912101
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First note that, by a direct approximation argument, it is enough to prove the result when MATH is smooth, so we suppose that is the case. Let MATH. By REF , there exists MATH (depending only on MATH and MATH), such that MATH is a diffeomorphism from the subset of the ball of radius MATH where it is defined onto its image. Therefore, for all MATH, there exists a unique MATH-geodesic MATH of minimal length between MATH et MATH. For MATH, let MATH be the angle between MATH and MATH, MATH the angle between MATH and MATH, MATH the domain in MATH bounded by MATH and by MATH, and MATH its area. From the NAME theorem, for each MATH: MATH . But it is easy to check, using REF , that, if MATH remains small enough (so that MATH remains smaller than a constant depending only on MATH and on MATH), then MATH is bounded by: MATH . As a consequence of those two equations, if MATH is large enough (larger than a constant depending only on MATH and on MATH), there exists MATH such that: MATH . Then there exists MATH verifying one of the following properties: CASE: either MATH; CASE: or MATH and MATH. CASE: Recall that MATH is injective; therefore, using REF and the upper bound REF on MATH, one sees that, if MATH is small enough (smaller than a constant which this time depends on MATH and on MATH), then: CASE: in the first case, that MATH remains in the domain of MATH bounded by MATH and by MATH; CASE: in the second case, that MATH remains in the domain of MATH bounded by MATH. In both cases, ``half" of MATH remain in a compact domain of MATH, and therefore MATH can not separate MATH in two parts.
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math/9912102
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The natural homomorphism of algebraic groups MATH induces a morphism at the level of NAME (and Spin)-bundles. By REF semistability is preserved under this operation. Thus, taking into account that the spinor norm is preserved under the above group homomorphism and that MATH is a coarse moduli space, we get the above claimed morphisms. The expressions of the associated (half-)spinor bundles are easily deduced from the definitions of the (half-)spinor representations of the Spin groups.
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math/9912102
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First let us prove REF. By REF (MATH), the multiplication map MATH is surjective. Consider the character space decomposition REF of the two spaces. Since the above tensor product is compatible with the linear action of the group MATH, we get a surjective map between the two character spaces of REF corresponding to the zero character,that is, MATH . Considering MATH-eigenspaces proves REF . By REF (MATH), the multiplication map MATH is surjective. Considering the zero character space and MATH-eigenspaces, we will prove REF.
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math/9912102
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We will show that the duality between the above vector spaces is given by a reduced NAME divisor, whose set-theoretical support equals MATH . Indeed, as shown in REF of REF, the family MATH of rank MATH vector bundles MATH parametrized by MATH is equipped with a MATH-valued quadratic form. Hence, by REF, we can consider its associated Pfaffian divisor MATH. By REF its support equals the set in REF and an easy computation shows that MATH. Hence we obtain a pairing MATH . First we will show that MATH is an isomorphism. We verify that both sides of REF are MATH-modules REF of level MATH and that MATH is MATH-equivariant. Hence, since MATH is nonzero, it is an isomorphism. To finish the proof we have to analyze how MATH acts on the MATH-eigenspaces of the linear involutions MATH (respectively, MATH). Let us restrict attention to the duality on the component MATH. Consider the rational map, induced by the divisor MATH . We observe that, MATH, the divisor MATH is invariant under the (projective) involution MATH. Consider MATH such that MATH,that is, MATH. Then it follows from the definition of MATH that MATH. By REF and with the notation as above, we have MATH . So we get MATH. Hence MATH. On the other hand, we see that the divisor MATH for MATH is symmetric and, take for example, MATH, MATH. Hence the pairing REF splits as follows MATH . By the same reasoning as above, one gets the other three isomorphisms as stated in the proposition. We leave the details to the reader.
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math/9912102
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Since MATH, we have an isomorphism MATH, which gives rise to a homomorphism MATH. Since both MATH and MATH have degree MATH and since MATH is poly-stable, it follows that MATH is a subbundle of MATH. It remains to determine its supplement bundle. There is a natural homomorphism of MATH into MATH. Using the isomorphism MATH, we get a homomorphism of MATH into MATH. A pointwise check shows that this is a surjective homomorphism and that it is supplementary to MATH.
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math/9912102
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We know that the orthogonal bundle MATH is the tensor product MATH. Let us do the computations for MATH. MATH where we used the fact that MATH is self-dual and the projection formula for the map MATH. Now by example REF, MATH is a semistable orthogonal bundle over MATH, hence this bundle splits MATH. The computations for MATH are similar.
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math/9912102
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It is enough to prove linear independence, which is done by induction on MATH. For the first cases we refer to REF (MATH), REF (MATH), REF (MATH). The statement for MATH will follow by pulling back MATH under the map MATH. We observe that MATH. Since the MATH form a basis, we can conclude. Let us assume that the MATH's are independent. Then the equality MATH implies that the MATH's are also independent.
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math/9912103
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We need to count the number of integer points MATH inside the region MATH which consists of the points in the cube MATH which lie between the hyper-planes MATH . Note that the region MATH is convex and contained in a ball around the origin of radius MATH. By the NAME principle (see CITE) we know that MATH . The distance between the above hyper-planes is MATH thus MATH is contained in a cylinder of height MATH whose base is a MATH dimensional ball of radius MATH. Therefore MATH which together with REF gives the lemma.
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math/9912103
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We first remark that since MATH and the MATH are distinct, the hyper-planes REF with MATH replaced replaced by MATH are not parallel to the hyper-plane given by REF . Moreover, the fact that our sequence is lacunary insures that the angle between these hyper-planes is not small. Thus when we solve for MATH in REF and input the result in REF we get an inequality in MATH variables: MATH in which the Right-hand side is bounded by the largest of the coefficients which appear in the LHS: MATH . Then REF applies to REF, with MATH for MATH and MATH, and we find that the number of vectors MATH having the required properties is MATH as stated.
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math/9912103
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Our proof is by induction on MATH. The case MATH is clear, the number of solutions in this case being zero. Let us assume that the statement holds true for MATH and prove it for MATH. Let MATH be a solution to the system REF. If there exists MATH such that MATH then MATH will be a solution for the same system with MATH replaced by MATH. By the induction assumption the number of solutions of this system is MATH. For each such solution, MATH is free to take values MATH. Therefore the number of solutions to the system REF for which at least one of MATH vanishes is MATH. We now count the solutions to REF with MATH for all MATH. There are MATH possible orders for the MATH. Let us count the solutions for which MATH. Given such a solution MATH we consider the partition of the set MATH as a disjoint union of sets MATH defined as follows. MATH consists of those MATH for which MATH. If MATH is the smallest index not contained in MATH then we put in MATH all those MATH for which MATH, and so on . In the end, if MATH are the smallest indices contained in MATH respectively, then we have: MATH . The number of partitions as above is bounded in terms of MATH. Let us count the number of solutions MATH which correspond to a given partition MATH. We distinguish two cases: MATH and MATH. Let us first treat the case MATH. If we fix MATH then from REF it follows that each of the remaining MATH can take at most MATH values. Hence the number of vectors MATH satisfying REF is MATH. Thus we are done with the case MATH if we show that for any vector MATH as above the number of solutions MATH is MATH. Fix some such MATH and note that by REF one has: MATH . Let us take a solution MATH and look at its first MATH components. These are nonzero integer numbers in the interval MATH satisfying the inequality: MATH . Here we may apply REF with MATH, MATH and MATH replaced by MATH to conclude that the vector MATH can only take MATH values. Let us fix MATH and count the number of solutions MATH whose first MATH components are MATH. We are now interested in those components MATH of MATH for which MATH. Write MATH and use REF to deduce that for any solution MATH, its components MATH with MATH satisfy the inequality: MATH . By REF we know that as MATH varies, the vector formed with the components MATH of MATH for MATH can only take MATH values. We now repeat the above reasoning until we get to the last set of components of MATH, namely the MATH with MATH. The components MATH with MATH being fixed, write MATH and then apply REF (here one uses the assumption that MATH). It follows that the vector formed with the components MATH of MATH can take MATH values only. The number of solutions MATH for a fixed MATH as above is then MATH, which completes the proof in case MATH. Assume now that MATH. Then MATH. In this case we fix MATH only . This can be done in MATH ways. For MATH fixed we apply REF repeatedly to conclude that as the vector MATH varies in the set of solutions, the vector MATH can take MATH values only. Now for MATH fixed, MATH and MATH are uniquely determined from the last two relations in REF (here one uses the fact that MATH). Thus the number of solutions MATH is MATH in case MATH as well, and the lemma is proved.
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math/9912103
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Each solution MATH to REF produces a partition of the set MATH as a disjoint union of subsets MATH, where MATH are the above sets MATH without repetitions. Let us count the number of non-degenerate solutions to REF which correspond to a given partition MATH of the set MATH. For MATH denote MATH, MATH for MATH, then write MATH. If MATH is a non-degenerate solution to REF then not all the numbers MATH vanish. One sees that for any such MATH the pair MATH is a solution of the system: MATH in integers MATH, MATH, MATH distinct, MATH . By REF we know that the number of solutions of the system REF is MATH. Now fix a solution MATH and count the number of non-degenerate solutions MATH to REF which correspond to the above partition MATH and which produce the vector MATH. Clearly MATH is uniquely determined since MATH for any MATH and any MATH. Moreover, for any MATH the number of solutions MATH of the equation MATH is MATH. Hence the number of solutions MATH which correspond to a given pair MATH is MATH and so the total number of non-degenerate solutions to REF is MATH, which completes the proof of REF .
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math/9912103
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We write MATH where MATH we will show that MATH and thus prove our lemma. Fix MATH, and let MATH, MATH. Since MATH, MATH for large MATH. Now divide the range of summation in REF into MATH and MATH: MATH . The second sum is bounded by MATH by our choice of MATH and MATH. As for the first sum, it is bounded by the number of MATH with distinct MATH, and MATH with MATH such that MATH. By REF , this number is MATH. Thus we find that MATH as required.
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math/9912103
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By REF we have MATH . Moreover, MATH . Now summing over all MATH we get MATH . Fix MATH, and choose MATH and MATH sufficiently large in terms of MATH and MATH, say MATH. Also set MATH. We have MATH for large MATH. In REF we break up the sum over MATH into ranges MATH and MATH, and likewise for the sum over MATH. In the range MATH we use the bound MATH, and in the range MATH we use MATH. This gives MATH . The third term in REF is bounded by square of the number of MATH times the square of the sum MATH, giving a total of at most MATH . The second term in REF is bounded by MATH . The first term of REF is bounded by the number of solutions of the equation MATH in variables MATH, MATH distinct, MATH distinct. By REF , this number is at most MATH. Thus we find that MATH and inserting into REF we get MATH .
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math/9912103
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Let MATH. For MATH we write MATH in the form MATH with MATH and MATH. From MATH, with MATH and MATH it follows that MATH and since MATH we get MATH for MATH . For any MATH let MATH . Then for any MATH . In particular one has: MATH . It remains to bound MATH. In order to do this we produce for any MATH an upper bound for MATH in terms of MATH. Let MATH. There is MATH such that MATH. Write: MATH . Then one has : MATH . For a fixed value of MATH the integer MATH may vary in the above interval of length MATH, so it takes at most MATH values. Hence: MATH . Clearly MATH. By multiplying these inequalities we obtain: MATH and therefore MATH . Here we use the assumption that MATH to conclude that MATH which completes the proof of the lemma.
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math/9912103
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Given MATH we choose a positive integer MATH, depending on MATH only, whose precise value will be given later. Let MATH and MATH. There exists MATH such that the set MATH has more than MATH elements. Arrange the elements of MATH in increasing order: MATH and pick from this set the first element MATH, then ignore the next MATH elements, pick the next one, ignore again MATH elements, and so on. We get a set of ``well spaced" integers MATH with MATH, such that MATH and (since MATH): MATH . Now look at the sequence of fractional parts MATH. They all fall in an interval of length MATH centered in MATH. We cut this interval in MATH intervals MATH having the same length: MATH. By the box principle, one of these intervals, MATH say, will contain at least MATH elements of MATH, that is, MATH will contain at least MATH elements of MATH. So let MATH be MATH elements of MATH for which the fractional parts MATH belong to MATH. Then clearly one has: MATH for MATH sufficiently large in terms of MATH and MATH. Note also that since the MATH are still well-spaced, by REF one has: MATH . Let MATH be given by: MATH . By REF we see that for MATH one has: MATH while REF says that MATH . From REF we see that one may apply REF to the vector MATH, with MATH replaced by MATH. In the terminology of that Lemma, MATH belongs to MATH. Since for each MATH there is such a vector MATH it follows that MATH . By REF we derive: MATH . Now each vector MATH as above is uniquely determined by a MATH-tuple MATH of positive integers MATH. The number of such MATH-tuples is MATH. It follows that MATH . We now let MATH and the lemma is proved.
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math/9912103
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We use the representation of MATH as in REF: MATH where MATH. Note that MATH so we may assume MATH. Now fix MATH, and set MATH; then for MATH to lie in the support of MATH, we need MATH. By REF , for almost all MATH there are at most MATH integers MATH satisfying this. Similarly, we need MATH for all MATH which forces the number of possible MATH contributing to the sum to be at most MATH. Now summing over the MATH possible MATH's gives MATH.
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math/9912105
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Induction in MATH. For MATH the identity MATH is true. Now let MATH. Using the commutation of MATH with each of MATH we rewrite the left hand side of REF as: MATH . Applying the to above expression the basic relation REF written in the form MATH, we see that the left hand side of REF equals: MATH . Finally, applying the inductive REF with MATH, we obtain MATH . Lemma is proved.
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math/9912105
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Denote MATH. Note that MATH. Then the identity REF can be rewritten as MATH . This identity is true because the morphism MATH is right MATH-equivariant. Lemma is proved.
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math/9912105
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It suffices to take MATH. In this REF implies that for MATH we have: MATH where MATH. Lemma is proved.
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math/9912105
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Clear.
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math/9912105
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It suffices to prove the proposition in the assumption that MATH, that is, when MATH is semisimple of types MATH and MATH respectively. REF follows. Let us prove REF . It suffices to analyze the case when MATH. Due to REF it suffices to prove the statement only for MATH, MATH. In this case MATH, and we set MATH in the standard way. It is easy to see that MATH for MATH, where MATH . Furthermore, is easy to see that the actions MATH are given by REF : MATH for MATH. Define the functions MATH on MATH by: MATH where MATH . It is easy to see that MATH . This identity implies REF . Furthermore, it is easy to see that MATH for MATH and MATH for MATH. This implies REF . REF is proved. REF easily follows from REF .
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math/9912105
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It is easy to see that MATH satisfies REF . Since the centralizer in MATH of any element MATH is trivial we see that MATH does not depend on a choice of REF . Lemma is proved.
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math/9912105
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It suffices to show that MATH. By definition, we have: MATH for MATH. Lemma is is proved.
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math/9912105
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Note that MATH for all MATH. So we prove the lemma in the assumption that MATH. Let MATH. Express MATH as MATH, where MATH, MATH. Then MATH because MATH. Furthermore, for each MATH we have: MATH, where MATH. Finally, MATH . Lemma is proved.
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math/9912105
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Throughout the proof we denote for shortness MATH. In this case we have MATH, MATH (where MATH) which implies that MATH. The following result is obvious. Let MATH. then for any MATH and MATH we have: MATH . Furthermore, let MATH, MATH, MATH. Then MATH . Applying REF twice, we obtain MATH . By by REF , we have MATH, and MATH. Taking into account that MATH we obtain MATH . REF is proved.
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math/9912105
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Taking into the account that MATH, we rewrite REF: MATH . REF is proved.
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math/9912105
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It is easy to see that for any MATH and MATH we have MATH. It is also clear that for any MATH, MATH we have: MATH where MATH as in REF. Furthermore, by definition of MATH, one has MATH . Composing MATH with MATH, we obtain for MATH, MATH: MATH . Lemma is proved.
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math/9912105
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We proceed by the induction in MATH. If MATH, that is, MATH for some MATH then MATH . Let us now assume that MATH. Using REF and the fact that MATH is a group homomorphism, we obtain for any MATH: MATH REF is proved.
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math/9912105
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It follows from REF that for any MATH satisfying MATH we have: MATH . This proves REF . Let us prove REF now. Denote MATH. We will proceed by induction in MATH. If MATH, the statement is obvious. Let now MATH. Let us express MATH, where MATH, MATH and MATH. Thus, the inductive hypothesis implies that MATH is dense in MATH, and MATH is dense in MATH. Then REF implies that MATH . But MATH is a dense subset of MATH. Finally, REF implies that the latter set contains a dense subset of the variety MATH. This proves REF . Proposition is proved.
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math/9912105
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Prove REF . The multiplication in MATH induces the open inclusions MATH, MATH. Furthermore, MATH . Therefore, it follows from REF that MATH . REF is proved. Let us prove REF now. Recall that MATH is MATH-equivariant, and MATH. Note that MATH intersects MATH non-trivially. Thus MATH . Recall that the action MATH is defined in the proof of REF . Clearly, both MATH and the reduced NAME cell MATH are invariant under this action, and the morphism MATH commutes with this action. Thus applying REF to the above identity, we obtain MATH . REF is proved. REF follows. REF is proved.
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math/9912105
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Prove REF . Indeed, MATH. Thus MATH is a section of MATH because MATH is MATH-equivariant. Prove REF . It is easy to see that MATH. Thus, for MATH and any MATH, MATH we have MATH. On the other hand, MATH. This implies that MATH. Prove REF . We proceed by the contradiction. Assume that MATH is not injective. That is, we assume that there are elements MATH such that MATH and MATH for some MATH and MATH. In other words, MATH . Denote MATH. Clearly, MATH because MATH. Furthermore, let us factorize MATH, where MATH, and MATH. Thus, the above identity can be rewritten as MATH . This implies that MATH, MATH, MATH, and MATH. In particular, MATH. In its turn, this implies that MATH which contradicts to the original assumption.
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math/9912105
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Follows from REF .
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math/9912105
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Let us describe the sub-torus MATH of MATH. We have MATH . That is, MATH can be described in MATH by the equations CASE : MATH . On the other hand, for any MATH one has MATH . REF is proved.
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math/9912105
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For any MATH, MATH let MATH be the representative of MATH defined by MATH for any MATH such that MATH. We need the following obvious refinement of REF . MATH. For any MATH choose an element MATH such that MATH. By REF , we have: MATH where MATH. Clearly, MATH, and MATH. It is also clear that MATH commutes with each MATH whenever MATH, where MATH. Denote MATH for MATH. The only element of MATH is equal to MATH, and MATH . Let MATH. Clearly, MATH. Next, let us rewrite the equation from REF : MATH or, equivalently, MATH, where MATH. Clearly, MATH. Thus, MATH. REF is proved. Since MATH is MATH-equivariant it suffices to prove REF in the case when MATH. Using REF we obtain: MATH . Let us compute MATH using the fact that MATH for MATH: MATH . The last equality holds because because MATH for MATH, and MATH if MATH. Proposition is proved.
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math/9912106
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It suffices to prove that the following diagram commutes. MATH . Recall that MATH, where MATH and MATH. Recall further that the minimality condition on MATH implies that the linear part of its differential vanishes. The linear part of MATH is the linear map MATH defined by the condition MATH. Recall that MATH and MATH REF . The model MATH extends to a morphism of constructible acyclic closures REF MATH by REF. Since MATH is MATH-minimal, MATH. By REF, MATH is equivalent to MATH in MATH. Apply MATH to MATH to get a MATH-morphism MATH. Let MATH and MATH be the projections. The maps MATH and MATH fit into the diagram MATH . For MATH, REF states that MATH has total wordlength at least two. It follows that MATH has MATH-wordlength at least two, so MATH, so REF commutes. Dualize and pass to homology to get REF .
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math/9912106
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It suffices to prove the dual statement, namely that MATH factors over the surjection MATH to induce a differential in MATH. But MATH consists of products along with elements of the form MATH for MATH, MATH. Since MATH is a MATH-derivation, MATH is a product. It follows that MATH, completing the proof.
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math/9912106
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We proceed by induction. For MATH, let MATH. Since MATH, MATH, establishing the first statement. For the second statement we may take MATH to be the identity map. Let MATH be the homotopy NAME algebra of MATH. Apply REF to the minimal algebra MATH to get a graded NAME algebra isomorphism MATH. Since MATH in MATH, MATH (reduced modulo MATH). Thus because MATH is a MATH-derivation, so is MATH. Since MATH is the MATH-algebra dual to MATH it follows by REF that MATH and so MATH. Now suppose the three statements are established for MATH. Let MATH be a constructible acyclic closure for MATH. By REF, there is a chain isomorphism MATH. Fix a well-ordered basis MATH of MATH; this determines a dual basis MATH of MATH. The isomorphism MATH identifies the NAME - NAME - NAME basis element MATH of MATH as a dual basis to the basis element MATH of MATH. It follows that MATH factors as the composition of dgc isomorphisms MATH. Since MATH and MATH are finite type, we dualize to obtain the dga isomorphism MATH. Let MATH be a minimal model. Let MATH be a constructible acyclic closure of MATH REF. Since MATH is MATH-minimal, MATH, so by REF, the differential in MATH is zero. By REF , MATH induces a MATH-morphism MATH . Since MATH is a field, by REF, we may identify MATH with MATH, where MATH is the differential torsion functor CITE. Therefore since MATH is a quasi-isomorphism, MATH is an isomorphism. Let MATH be the composition of algebra isomorphisms MATH to establish the first statement. Note that MATH will be the isomorphism dual to MATH in the next stage of the induction. Setting MATH, we get the commutative diagram MATH and it follows from the definitions that MATH is the identity on MATH. By the inductive hypothesis, there exists a MATH-morphism MATH such that MATH whenever MATH, MATH satisfy MATH. We now show that MATH satisfies REF . For MATH choose MATH so that MATH. Then MATH, hence MATH and MATH is defined. Since MATH induces the identity in homology, MATH for some MATH. Thus MATH, so MATH is a boundary in MATH, whence MATH. This establishes the second statement. The model MATH determines an isomorphism MATH, where MATH is the homotopy NAME algebra for the model MATH. As a graded vector space, MATH. Let MATH, and suppose for some MATH that MATH. Then MATH, so MATH for some MATH. Thus MATH. Since MATH and MATH are MATH-morphisms, MATH so MATH. Furthermore, MATH so MATH . By REF , this establishes the third statement, completing the inductive step and the proof.
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math/9912106
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Let MATH be a minimal model. Recall that the underlying algebra of MATH is MATH, where MATH. Let MATH and MATH be constructible acyclic closures for MATH and MATH, respectively. The model MATH determines a MATH-morphism MATH where MATH is an isomorphism. The composition MATH induces an isomorphism of NAME spectral sequences, establishing the first statement. The reduced minimal model MATH has homotopy NAME algebra MATH, so by REF , MATH. Suppose that MATH. Let MATH be the inclusion. Then MATH. The homotopy NAME algebra of the minimal model MATH is MATH, so REF states that MATH, completing the induction and the proof.
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math/9912106
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NAME in CITE proves that there is a dgl MATH and a dgh quasi-isomorphism MATH. Thus as NAME algebras, for MATH, MATH and MATH. The result follows by applying REF to the dgl MATH.
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math/9912106
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Observe that, for MATH, the sequence of functors MATH identifies MATH as the natural dual of the free MATH-algebra MATH, with MATH naturally dual to MATH, and the inclusion MATH naturally dual to the projection MATH. Therefore if MATH, then MATH is a MATH-morphism. Conversely, if MATH is a MATH-morphism, then MATH, so MATH.
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math/9912106
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Straightforward.
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math/9912109
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The first part contains standard facts in the theory of toric varieties. The only point to be made is that, since the vectors MATH belong to all maximal simplices in the triangulation MATH no monomial in MATH corresponding to MATH will contain any of the elements MATH . The second part is a direct consequence of the way MATH was obtained from MATH . It implies that the following relation holds in MATH . REF implies then the desired statement. REF . is a simple consequence of the fact that MATH is a complete intersection. For REF ., consider a smooth refinement fan MATH of the simplicial fan MATH . Since MATH does not intersect the singular locus of MATH the proper transform MATH of MATH under the induced map MATH does not intersect the exceptional locus in MATH . This means that for any toric divisor MATH in MATH corresponding to a new ray that is in MATH but not in MATH we have that MATH . Moreover, since we only deal with smooth objects in MATH we have that MATH . But the restriction MATH is an isomorphism of smooth varieties, so MATH and the statement follows.
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math/9912109
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Indeed, for any MATH we have that MATH for some holomorphic function MATH (we have used the classical identity MATH). REF is then an immediate consequence of the relations REF in the definition of the NAME - Reisner ring MATH . To check REF ., assume that for some MATH we have that MATH is a simplex in MATH . Then there exists a simplex MATH such that MATH . It follows that MATH for all MATH so REF follows directly from MATH .
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math/9912109
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The proof is by induction. For MATH the operator MATH is just the identity operator. REF follow directly from REF . To prove the induction step, we consider two adjacent cones MATH and MATH and let MATH and MATH be the corresponding regular triangulations. Assume that the analytic continuation operator MATH corresponding to the cone MATH exists and has the expected properties. REF shows that there exists a circuit MATH in MATH such that the two triangulations are supported on MATH and MATH . According to REF . the hyperplane in MATH separating the cones MATH and MATH is defined by an element MATH with MATH where MATH are the elements of the NAME transform of MATH for MATH . We choose the element MATH to be minimal in the lattice MATH that is so that no rational non - zero multiple of MATH with smaller length is in MATH . According to REF there exists a class MATH such that MATH where MATH are the toric divisors on the ambient toric variety MATH defined by the simplicial fan MATH . As usual, define the subsets MATH and MATH so that the triangulation MATH contains all the simplices of type Conv-MATH and the triangulation MATH contains all the simplices of type Conv-MATH . As mentioned in the remarks after REF , any saturated MATH can be included as an element of an integral basis for the lattice MATH . The induced decomposition MATH implies that any MATH can be written as MATH . Since the parameters MATH are coordinates on MATH the decomposition REF leads to some choice of coordinates on the linear subspace MATH such that any MATH can be written as MATH . Each MATH is naturally identified with an element in MATH and as a consequence of REF the parameter MATH represents the class in MATH that is dual to MATH . Since the big cones MATH and MATH have a common face, their dual cones MATH and MATH in MATH project according to the decomposition REF onto the same cone MATH in MATH . The existence of such a cone is a consequence of the important result of NAME, NAME and NAME which states that the secondary fan is the normal fan of some convex polytope called the secondary polytope. The hyperplane that contains the common face of the cones MATH and MATH corresponds to an edge of the secondary polytope joining the vertices corresponding to the regular triangulations MATH and MATH . The cone MATH can be taken to be the cone obtained by intersecting the secondary polytope with a section normal to the edge. Let MATH the set of elements in MATH that are generators of the cone MATH which is the common face of the of the cones MATH and MATH. For any element MATH we have that MATH and the following facts hold MATH . Hence there exist two integers MATH and MATH such that for any MATH, decomposed as in REF with fixed MATH . We emphasize here that, in general, the integers MATH and MATH depend (piecewise) linearly on the value MATH . We have seen in the previous section that the series MATH (see REF ) is convergent in the neighborhood of the point MATH for any complex parameters MATH such that MATH . The induction hypothesis implies that for any evaluation MATH the function MATH gives the solutions to the GKZ system in a neighborhood of the point MATH . We have that MATH where MATH . The coordinate MATH is well defined in the complex plane considered with a branch cut and can be viewed as the coordinate on the rational curve joining the points MATH and MATH in the toric variety MATH determined by the secondary fan (see also REF). According to our discussion in the previous section, the series MATH is absolutely convergent in a neighborhood of the point MATH and there we can operate without restraint any change in the order of the summation. For notational reasons, in REF we applied the change MATH which amounts to having MATH vary in a a cone denoted by MATH obtained by a translation by a constant integer vector of the cone MATH . In general, the series MATH is convergent in an annular region with a branch cut. As mentioned before, the branch cut can be replaced by a restriction on the real part of the MATH - parameters. To pursue the analytic continuation we use REF - NAME integral representations. The following property holds (see CITE, page REF, and REF). Consider the integral MATH with MATH all real. The path of integration is parallel to the imaginary axis for MATH - large, but it can be curved elsewhere so that it avoids the poles of the integrand. Introduce the following notations MATH and MATH . Then the absolute value of the integrand has the asymptotic form MATH when MATH is large (this is a consequence of NAME 's formula). Therefore, if MATH the integral is absolutely convergent (and defines an analytic function of MATH) in any domain contained in MATH . Moreover, if MATH the integral is equal to the sum of the residues on the right of the contour for MATH and to the sum of the residues on the left of the contour for MATH (these facts are obtained by closing the contour to the right, respectively to the left, with a semicircle of radius MATH). To make use of the lemma, we write MATH . In order to analytically continue the function MATH from a neighborhood of MATH to a neighborhood of MATH we consider the following NAME - NAME integral (compare to the last line in the previous formula) MATH with MATH . The path of integration avoids all the poles of the integrand, is parallel to the imaginary axis at large MATH and indented such that the poles of the function MATH lie on the left of the path. Such a choice of the contour with a ``finite" indentation is possible, given REF that define MATH . Although the integer MATH depends (piecewise) linearly on MATH we can choose indentations that are ``similar" for all MATH . With such a choice, the change of variable MATH allows us to rewrite the previous integral as MATH where the contour of integration is now independent of MATH and coincides with the imaginary axis for large MATH . The relation MATH combined with REF , give that MATH . This shows that, in the notation of REF , MATH and MATH. So the integral REF is absolutely convergent in any domain included in MATH and this defines the branch cut in the MATH - plane. We also conclude that the path of integration can can be closed both to the right for MATH small, and to the left for MATH big. An important remark to make is that the residues at poles MATH of the function MATH can be discarded for the purposes of the analytic continuation, since the corresponding terms are canceled out by the operator MATH. Indeed, the poles MATH have residues which give factors of the form MATH . The inequality MATH and REF show that the function MATH can be written as MATH with MATH . We know that the triangulation MATH contains simplices of type MATH so MATH is not a simplex in MATH . Clearly MATH is saturated, so the induction hypothesis implies then that MATH . This shows the residues at integer points MATH do not contribute to the result of the analytic continuation of the solutions to the NAME - NAME equations, and this is true independently of the location of the poles on the left or right side of the path of integration. We now study the evaluation of the integral REF using residues, when we close the contour to the right (for MATH), or to the left (for MATH). Assume first that all the poles MATH of the function MATH with MATH are located on the right of the contour of integration. This means that all the poles MATH of MATH are located on the right of the contour of integration in the integral REF , and the sum of all their residues is MATH which is exactly the last line in REF . The series MATH written as in REF is absolutely convergent in the neighborhood MATH of the point MATH corresponding to MATH - small. NAME 's dominated convergence theorem shows that, for MATH (that is, MATH - small), we can express MATH by the following absolutely convergent integral MATH with MATH . Consider the coordinates MATH in MATH such that the rational curve joining the points MATH and MATH is given by MATH with the point MATH given by MATH (see CITE for a detailed discussion of coordinates). We argue now that the integral in REF is in fact absolutely convergent in a ``tubular" neighborhood of the rational curve given by MATH . Indeed, choose some MATH such that MATH contains a product of MATH complex disks with radii equal to MATH . For an arbitrary coordinate MATH choose a real number MATH such that MATH . We can write MATH where the integers MATH are determined by the integers MATH . The important fact to remember is that the integer MATH depends (piecewise) linearly on the integers MATH that is, on the integers MATH . This means that we can choose the coordinates MATH to be small in absolute value, such that there exist complex numbers MATH with MATH for MATH and MATH . If MATH are the MATH - coordinates of the point MATH it follows that for MATH we have that MATH . Since the function MATH is a convergent series of absolutely integrable functions, NAME 's dominated convergence theorem implies indeed that the integral MATH is absolutely convergent in a tubular neighborhood of the curve joining MATH and MATH in MATH . This means that we can indeed perform the analytic continuation using the above integral, so that we can cross over to the convergence region MATH around the point MATH . There we can close the contour of integration of the integral REF (or REF ) to the left. As shown above, any residues of the function MATH can be discarded, so we only have to account for the poles of the functions MATH . Hence, for MATH fixed and MATH large, the last line in REF becomes MATH . We now analyze the case when the assumption does not hold, so there exist poles of the function MATH on the left of the contour in the integral REF . As shown above, we only have to account for the poles MATH . For fixed MATH only a finite number of them can be on the left of the contour. The function obtained after evaluating the residues at these poles is merely a polynomial in MATH (of degree depending on MATH though), and this means that the corresponding piece of MATH can be manipulated freely in the MATH - direction, that is in a ``tubular" neighborhood of the rational curve joining MATH and MATH in MATH . This shows that the residues at all the poles of the function MATH whether located on the right or on the left of the contour, can be added together to obtain the series MATH . It follows that the analytic continuation of the series MATH is given by MATH . The symbol MATH will be explained in detail below. The characterization REF of the elements MATH leads us to conclude that the analytic continuation of MATH is MATH with MATH defined by MATH where the choice of the generators MATH is determined by the combinatorial structure of the transition from the cone MATH to the cone MATH (these are just different choices of coordinates on the linear subspace MATH). In what follows we show that this is indeed the analytic continuation, which means that MATH gives the solutions of the GKZ system in a neighborhood of the point MATH. The operators MATH will not depend on the various choices of coordinates involved. They describe the analytic continuation procedure for the solutions to the holonomic GKZ system, and the process is unique given the path defined by the sequence of adjacent cones in the secondary fan. Such a path always exists, because it is a well known fact that the singularities of the GKZ system have real codimension MATH. One needs to analytically continue along paths which are consistent with the the branch cuts mentioned above. In terms of the MATH - coordinates, these branch cuts impose conditions on the real part of the parameters MATH . Care is required in defining the symbol Res-MATH used in REF . A direct calculation shows that MATH where MATH is the integer defined by MATH . The poles that we have to consider are the poles of the function given by REF other than MATH, such that we avoid over-counting in REF , that is the complex values MATH with MATH . The fact that, according to REF , MATH is the largest integer such that MATH for some MATH, implies that by taking residues at MATH with MATH and summing over all integers MATH we have taken care of all the required poles of the functions involved in REF without over-counting. In order to deal with the issue of residues, we need the following lemma. For any function MATH we have that MATH for some function MATH . The underlying message of this lemma is that the sum over the residues ``around" the origin of MATH (which is an integration over some appropriate contour, as we will see below) has the feature of ``curing" the poles of the integrand around the origin. Fundamentally, this is a consequence of the classical NAME 's theorem. (of the lemma) The main point of this lemma is to show that the sum of residues MATH defined as above is an analytic function in MATH in a neighborhood of the origin in MATH . In order to see this, assume first that the complex numbers MATH are fixed and close to zero. The fact that MATH implies that the values of MATH where the function MATH has residues (for fixed MATH) are non - zero, so we can choose MATH small in an open neighborhood of the origin such that MATH is analytic in that open set. Assume first that MATH . That means that the sum of residues MATH can be replaced by a contour of integration MATH which leaves outside the value MATH but encloses all the required poles in MATH . If the function MATH has poles given by the affine form MATH for some MATH, the contour will enclose those that are in MATH but will avoid all the others. We can write: MATH where the contour MATH is chosen such that it encloses the contour MATH and the value MATH but no other poles of the functions involved. The fact that some values in the set MATH might coincide among themselves, or with MATH does not change the analyticity of the function defined by the integral over the contour MATH . In other words, the function of MATH defined for fixed MATH by the integral over MATH is analytic also for MATH . Since for small values of MATH the point MATH is not a residue of MATH the last term in REF is simply equal to MATH . Recall that we fixed the complex numbers MATH . However small variations of them around the origin have no effect on the contours of integration involved. This shows indeed that the equality REF holds in a neighborhood of the origin in MATH and MATH defined by MATH is analytic around the origin. The integral over MATH can be analytically continued by an appropriate continuous ``movement" of the contour. However, this procedure will fail for certain values MATH . NAME 's theorem shows that the function of MATH and MATH defined by integration over the contour MATH will have again poles of finite order, and, combinatorially we see that these poles are included in the hyperplanes of the form MATH for non - zero integers MATH . We have constructed the operator MATH which has all the required properties with the exception of the first property in the statement of the theorem. We turn our attention to this question next. Assume that MATH is saturated and does not define a simplex in the triangulation MATH . A possible case is that MATH defines a simplex in the triangulation MATH . We know that the change in simplices between MATH and MATH is given by the circuit MATH and the only way in which the simplex MATH can be in MATH but not in MATH is if it contains a simplex of type MATH for some MATH. But MATH so the simplex MATH contains the simplex MATH. REF of the operator MATH shows that MATH with MATH's written in the appropriate coordinates. But MATH so all the poles in REF are canceled, and the result is zero. Note that in the analytic continuation procedure terms that are canceled immediately in this way show up when we perform a transition from MATH back to MATH . Assume now that MATH does not define a simplex neither in MATH nor in MATH. By using REF we can write that MATH for some function MATH . The second term in REF is canceled by MATH according to the induction hypothesis, since MATH is saturated and it is not a simplex in MATH . We show now that in fact MATH cancels the first term in REF , too. The two corresponding cones MATH and MATH in MATH have a common face which is a cone MATH of dimension MATH and let's denote by MATH the set of vectors of the NAME transform included in this cone and by MATH the hyperplane in MATH containing the cone MATH . Assume first that for any element MATH the corresponding element in the NAME transform MATH is not contained in the hyperplane MATH . This means that MATH, and, since MATH is not a simplex in either one of MATH and MATH the only possibility is that MATH contains MATH. But now we are back to the previous case, and REF proves again the required statement. Assume now that MATH contains an element MATH such that the corresponding vector of the NAME transform has the property MATH . Moreover, assume that MATH is not contained in any cone of type MATH (see REF ), for MATH . Equivalently this means that MATH belongs to any simplex of the triangulation MATH (otherwise, there would be a simplex MATH such that MATH that is MATH which contradicts the assumption). So the element MATH will not be missing from any of the simplices of MATH and because MATH is not a simplex in MATH we obtain again that MATH contains MATH so again we can apply REF to prove the desired statement. REF that was imposed in the beginning of this section shows that the last case we have to consider is the case when MATH contains an element MATH such that MATH is a generator of the cone MATH but there is no other larger cone in the hyperplane MATH that includes the cone MATH as a proper subset. Denote by MATH the set of elements MATH such that MATH for some positive real number MATH . The point here is then that for any simplex MATH (which means MATH) the cone MATH has to contain at least one of the vectors MATH for some MATH (equivalently MATH). Hence MATH is not included in any maximal dimensional simplex of MATH so it cannot be a simplex in MATH . Now we can apply REF . In particular, for any MATH the element MATH is independent of MATH (the integration parameter), so it remains unchanged after integration and we can write that MATH for some function MATH . Again, because neither MATH nor MATH are simplices in MATH and they are saturated, the induction hypothesis implies that MATH cancels both terms in REF .
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math/9912109
|
We make repeated use of the NAME - NAME - NAME formula and of the properties of the NAME map in cohomology. For MATH we have that MATH is given by MATH . The last expression is exactly MATH which proves the lemma.
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math/9912109
|
According to the previous lemma, the induced action on MATH of the NAME - NAME functor determined by the complex MATH is identical to the action NAME - NAME functor determined the same complex viewed as a complex of sheaves on MATH . The lemma shows then that, for MATH this action is given by MATH where the sign is determined by the parity of MATH . But the map MATH embeds MATH in MATH as a complete intersection of the divisors MATH so again, by NAME - NAME - NAME, it follows that the action of the complex MATH on MATH can be written as MATH . This formula is implied by REF and by the fact that the complex MATH is equivalent to the sheaf MATH placed at the MATH-th position. The proposition follows as a direct consequence of REF .
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math/9912109
|
According to the last part of REF . we have MATH . We will need the following lemma. For any cohomology class MATH with the function MATH defined by MATH and the contours MATH give the residues at MATH and MATH respectively. (of the lemma) For any series MATH around the origin MATH we denote by MATH the coefficient of the monomial MATH . Using the relations REF in MATH we can write that MATH . We can write that MATH and MATH . The lemma follows after pointing out that the last term in REF is in fact zero, since the factor MATH cancels the pole at MATH of the integrand. We proceed with the proof of the proposition. We first investigate the action of MATH on MATH . The loop MATH is exactly the loop involved in the statement of REF corresponding to the transition MATH . The subtle point here is that the curve MATH is not the rational curve connecting the toric fixed points MATH and MATH in MATH . Indeed as marked in REF , if the coordinates around the point MATH are MATH then the coordinates around the point MATH are MATH . As a consequence of this fact, the analytic continuation of the series MATH from a neighborhood of MATH to a neighborhood of MATH will be a series, which, when viewed in the curve MATH will not converge in the neighborhood of the point MATH . Hence, the loop obtained as in REF will only go around the point MATH and not around the point of intersection of MATH with MATH . Using REF , as well as REF , we can write explicitly the effect of the transformation MATH on the series MATH (given by REF ) MATH where MATH is the NAME delta function. But MATH which means that the action of the transformation MATH on the MATH - valued series MATH is given by MATH . We now turn our attention to the computation of the action on MATH induced by the loop MATH . As in the one - parameter case we have to work with the series MATH with the property compare REF MATH with MATH given by MATH . The results in REF refer to the MATH - series, and not directly to the MATH - series. We point out that this fact had no relevance for the computation of the transformation MATH simply because the variable MATH was not essentially involved in that computation. The expressions for the transformations associated to the paths MATH and MATH are obtained by applying REF to the transitions MATH and MATH respectively, while the action induced by the loop MATH is given by applying REF to the transition MATH . In the following group of formulae, we suppress the variables MATH and MATH . MATH . Since we are interested in studying the composed transformation on the series MATH we use REF and write that the action of the loop MATH on the series MATH is given by MATH . A first remark is that using the delta function is equivalent to ignoring the loop MATH . So the delta function gives a transformation corresponding to the analytic continuation along a contractible loop MATH followed by the loop MATH . The contour MATH is chosen according to REF such that it encloses all the values MATH for MATH but not the value MATH (we dealt with the subtleties associated with this choice in the proof of REF ). We first deform the contour such that it encloses the value MATH and then operate the change of variable MATH . The computation can be finished if we can estimate the sum of the residues of the newly obtained function at MATH . As a first step, it is useful to study the residue MATH by using the MATH - functions instead of MATH - functions, that is by using the formulae REF . The part of the transformation that we are interested in is given by MATH where MATH denotes the action of the contractible loop with the same name. But this action is trivial, and it is given exactly by the the residue at MATH in the sequence of changes of variables MATH (this was the detail that made the proof of REF work). We conclude that the partial transformation obtained from the residue at MATH is MATH with the contour MATH chosen to only enclose the pole MATH . Finally, we can analyze the last piece of the transformation. According to REF , the contour MATH needs to be chosen such that it encloses only the poles MATH for MATH and the value MATH . But our REF guarantees that these points are not poles of the integrand (they are canceled out by the product in the numerator), with the only possible exceptions MATH and MATH . So we can assume that the contour MATH encloses only the residues MATH . Since the contour MATH can be chosen to be a small circle with radius MATH around the origin, we can write MATH . This means that in REF we can change the order of integration. We need to calculate the last part of the transformation, that is MATH . The change of variable MATH gives MATH . Since MATH direct power counting shows that MATH is not a pole. By computing the value of the residue at MATH the value of the integral is obtained to be MATH and, after using REF , the transformation REF can be written as MATH . By adding up the partial transformations REF , and REF , we conclude that the action of the loop MATH on the MATH - valued series MATH is given indeed by (check with REF ) MATH which ends the proof of the proposition.
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math/9912109
|
The NAME divisors (spanned by global sections) MATH come from a nef - partition of the vectors in MATH . As a consequence of the second part of REF , for any divisor MATH with MATH we have that MATH . This shows that any curve that is contracted under the map MATH does not intersect the generic divisors MATH for MATH . REF implies then that MATH for any MATH . Hence, for any MATH we have that MATH . Note also that, for MATH and the result follows.
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math/9912109
|
The fan defining the weighted projective space MATH has the property that the multiplicity index in MATH of any maximal cone generated by the vectors MATH is equal to MATH . Since the vectors MATH do not belong to any maximal cone of the fan MATH we have that MATH . For each MATH the closed toric orbits MATH given by the cone with generators MATH is mapped under the morphism MATH to the closed toric orbit in MATH represented by MATH itself. REF shows that, in MATH we have that MATH . We obtain that, for any MATH . According to REF, any class MATH can be written as a sum of ``monomials" without repetitions of classes corresponding to toric divisors. To prove the result, we only have to consider the case when MATH is such a monomial. Assume that MATH where MATH and the divisors MATH correspond to vectors in MATH other than MATH . For any MATH, there is no danger of confusion if we also denote by MATH the class of the toric divisor in MATH whose pull-back is the class MATH in MATH . Hence MATH . If MATH then MATH and relation REF shows that the only case when the result is non-zero is the case when MATH for some MATH . In that case, we have MATH . Consider now a system of generators of MATH corresponding to a basis of the space of linear relations among the vectors in MATH chosen such that MATH denotes the generator corresponding to the relation REF . We can write that MATH while the expressions for the classes MATH do not involve MATH . If the monomial MATH involves all the classes MATH then the integrand in REF does not have any residue, so the value of the integral is indeed zero. Also, if the subset MATH in REF has strictly less than MATH elements, the sum of the residues at finite poles calculated by the right hand side of REF is again zero, simply because the residue at MATH is zero. The remaining case is again MATH for some MATH . But then MATH which ends the proof of the proposition.
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math/9912109
|
Indeed, we have that MATH so MATH .
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math/9912109
|
According to REF , the automorphism MATH acts on MATH by MATH with the canonical maps (see REF ) MATH . REF shows that MATH where, for MATH we denote by MATH the pull-back hypersurfaces MATH (see REF ). It follows that MATH . According to the construction of REF , the subvarieties MATH and MATH are complete intersections in MATH and MATH respectively. Hence, in MATH we can write MATH . REF gives then that, for a class MATH we have MATH where MATH is the inclusion MATH . Finally, REF applied to the toric fibration MATH shows that MATH can be written in terms of the chosen generators for MATH as MATH where MATH is a contour enclosing all the poles MATH such that MATH for some MATH . The result follows after combining this last formula with REF , since for MATH .
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math/9912109
|
The proof reduces to a calculation very similar to the one performed in the beginning of REF and it is designed to match the formulae of REF . Recall REF that the solutions to the GKZ system are given by the series MATH where the parameter MATH satisfies REF , and MATH is the dual of the cone MATH given by MATH . As explained in REF, the choice of generators for MATH implies that the product MATH is written as MATH for some integers MATH . Consider now the series MATH . In terms of explicit generators, each coefficient MATH is written as MATH . A very similar calculation to REF shows that, possibly after some change of the coordinates MATH to some new coordinates MATH differing only by complex phase changes, the series MATH and MATH are related by MATH . REF implies that in fact, the MATH-series gives all the periods corresponding to the mirror NAME - NAME MATH as the coefficients of the monomials in the series given by the generators of MATH . The monodromy formula for the series MATH is obtained from the monodromy formula for the series MATH given by REF . Up to a conjugation in the fundamental group (or, equivalently a change in the coordinates MATH), the calculation from REF mentioned before shows that the analytic continuation of MATH around the prescribed loop is MATH . A change of variable which replaces any generator of MATH of the form MATH by the generator MATH ends the proof of the theorem.
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math/9912111
|
It is sufficient to show only the implications MATH . Let MATH be a log resolution of MATH, MATH exceptional divisors and MATH, MATH proper transforms of MATH and MATH, respectively. By NAME 's theorem, MATH is a simple normal crossing divisor, so MATH is also a log resolution of MATH. We can choose MATH so that MATH does not contain MATH. Thus we have MATH. This implies the first part of our proposition. In the second part we can use remark in REF.
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math/9912111
|
Let us prove, for example, the first inequality. Write MATH, where each MATH is the proper transform of MATH and MATH is the (reduced) exceptional divisor. Then MATH. From the exact sequence MATH (compare CITE) we have MATH .
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math/9912111
|
We have MATH . From this by NAME Vanishing Theorem CITE, MATH . Applying MATH to an exact sequence MATH we get the surjectivity of the map MATH . Let MATH be a component MATH. Then either MATH is MATH-exceptional or MATH is the proper transform of some MATH whose coefficient MATH. Thus MATH is MATH-exceptional and MATH . Assume that in a neighborhood of some fiber MATH, MATH the set MATH has two connected components MATH and MATH. Then MATH and both terms do not vanish. Hence MATH cannot be a quotient of the cyclic module MATH.
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math/9912111
|
Consider a log resolution MATH. We have MATH where MATH is the proper transform MATH on MATH and MATH, MATH are effective exceptional MATH-divisors without common components. Then MATH is a boundary and MATH is dlt. Apply log MMP to MATH over MATH. We get a birational contraction MATH from a normal MATH-factorial variety MATH. Denote by MATH the proper transform of MATH on MATH. Then MATH is dlt and MATH-nef. It is also obvious that MATH. We prove REF . Since the inverse to the birational map MATH does not contract divisors, MATH . On the other hand, by numerical properties of contractions (see for example, CITE) in the last formula all the coefficients of MATH should be nonpositive, that is, all of them are equal to zero. Finally, we consider the case MATH. If MATH, then MATH. From this MATH for some MATH. Then MATH and by REF, MATH and MATH. Hence MATH is a MATH-curve and steps of log MMP over MATH are contractions of such curves. Continuing the process, we obtain a smooth surface MATH. This proves the statement.
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math/9912111
|
Take a sufficiently small MATH and put MATH . Then MATH . Next, run MATH-MMP over MATH. By REF, each extremal ray is negative with respect to the proper transform of MATH, an effective divisor. Such a divisor can be nef only if it is trivial. Hence the process terminates when the proper transform of MATH becomes zero.
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math/9912111
|
As a first approximation to MATH we take a minimal log terminal modification MATH as in REF. Then MATH, where MATH is an integral reduced divisor and MATH is the proper transform of MATH. In particular, MATH is the number of components of MATH. Since MATH has only klt singularities, MATH, where the MATH are components of MATH and MATH for all MATH. Therefore MATH cannot be MATH-nef by numerical properties of contractions CITE. Run MATH-MMP over MATH. At each step, as above, MATH cannot be nef over MATH. Hence at the last step we get a divisorial extremal contraction MATH, negative with respect to MATH and such that MATH. Since at each step MATH is numerically trivial over MATH, the log divisor MATH is lc (see REF ). Obviously, MATH is irreducible and MATH. Then MATH is plt and MATH is lc. By REF , MATH is plt.
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math/9912111
|
The relation in REF follows from the equality MATH . REF can be proved similarly.
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math/9912111
|
Let MATH be a minimal log terminal modification (see REF). Since the fibers of MATH are connected and MATH, it is sufficient to prove the assertion for MATH. Let MATH be the composition map. Set MATH and MATH. Then MATH. Assume that MATH is disconnected. Run MATH-MMP over MATH. If the fiber MATH is reducible, then there is its component MATH meeting MATH. Then MATH is an extremal curve. Let MATH be its contraction. Since MATH and MATH, the dlt property of MATH is preserved (see REF). On the other hand, by Connectedness REF , MATH meets only one connected component of MATH. Hence the number of connected components of MATH remains the same. Continuing the process, we obtain a contraction MATH with irreducible fiber MATH. Since MATH is nef, for a general fiber MATH of MATH we have MATH. By our assumption, the fiber MATH does not contain MATH. Hence MATH has exactly two connected components, which are sections of MATH. It is also clear that MATH is plt. The components of MATH cannot be contractible over MATH. Hence MATH is the identity map. This proves the statement.
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math/9912111
|
As in the proof of REF, MATH a minimal log terminal modification. Again set MATH and MATH. Assume that MATH is disconnected. Run MATH-MMP. All intermediate contractions is MATH-nonpositive. Therefore the log canonical property of MATH is preserved (see REF). Since at each step MATH is klt, MATH is klt outside of MATH and MATH. By Connectedness REF , each contractible curve meets only one connected component of MATH. Therefore the number of connected components of MATH is preserved. At the last step there are two possibilities: CASE: MATH, then irreducible components of MATH are intersect each other and gives only one connected component of MATH; CASE: MATH and there is a nonbirational contraction MATH onto a curve. Here we can apply REF .
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math/9912111
|
Let us proof, for example, REF . Write MATH, where MATH and MATH, MATH, MATH. Then MATH . In both cases MATH.
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math/9912111
|
Let MATH. Then MATH for some MATH. This yields MATH and MATH. Hence MATH.
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math/9912111
|
REF is trivial. As for REF we notice that MATH. Hence by REF , MATH and we may assume that MATH and MATH. Put MATH. It is sufficient to show that there exists MATH such that MATH . This is equivalent to MATH . Taking into account that MATH, we have MATH for some MATH. So there exists MATH such that MATH . This proves the lemma.
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math/9912111
|
Take MATH and apply REF.
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math/9912111
|
Consider the crepant pull back MATH . Write MATH, where MATH is reduced, MATH, MATH have no common components, and MATH. We claim that MATH is a MATH-complement of MATH. The only thing we need to check is that MATH. From REF we have MATH. This gives that MATH (because MATH is MATH-nef; see CITE or CITE). Finally, by Monotonicity REF and because MATH is an integral divisor, we have MATH .
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math/9912111
|
Let MATH be a log resolution. Write MATH, where MATH is the proper transform of MATH on MATH and MATH. By Inversion of Adjunction, MATH is normal and MATH is plt. In particular, MATH is a birational contraction. Therefore we have MATH . Note that MATH, because MATH is smooth. By REF we see that MATH. So we can apply REF to MATH. We get a MATH-complement MATH of MATH. In particular, by REF, there exists MATH such that MATH . By NAME Vanishing, MATH . From the exact sequence MATH we get surjectivity of the restriction map MATH . Therefore there exists a divisor MATH such that MATH. Set MATH . Then MATH and MATH. Note that we cannot apply Inversion of Adjunction on MATH because MATH can have negative coefficients. So we put MATH. Again we have MATH and MATH. We have to show only that MATH is lc. Assume that MATH is not lc. Then MATH is also not lc for some MATH. It is clear that MATH is nef and big over MATH. By Inversion of Adjunction, MATH is plt near MATH. Hence MATH near MATH. On the other hand, by Connectedness Lemma, MATH is connected near MATH. Thus MATH is plt. This contradiction proves the proposition.
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math/9912111
|
Similar to the proof of REF. By REF , the curve MATH is nodal. Further, we can take a log resolution MATH so that MATH.
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math/9912111
|
By REF the number of singular points is MATH. Assume that MATH has exactly six singular points MATH. Then by REF we have MATH. This means that MATH are ordinary double points. In particular, MATH is NAME. Applying NAME 's formula to the minimal resolution MATH of MATH, we obtain MATH. Let MATH be a MATH-curve and MATH its image. Then MATH. Since MATH, we have MATH, so MATH. On the other hand, MATH is NAME, a contradiction.
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math/9912111
|
Replacing MATH with a log terminal modification, we may assume that MATH is dlt. Then MATH. In this situation we can apply REF . The last statement follows by Connectedness Lemma.
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math/9912111
|
If MATH is not klt, the assertion follows by REF . Thus we may assume that MATH is klt (in particular, MATH). Let MATH . As in REF put MATH . Note that MATH (because MATH, see REF). If MATH is lc, then this is a regular complement. Assume that MATH is not lc. Take MATH so that MATH is maximally lc. It is clear that MATH and MATH is nef and big. Now we can apply REF to MATH to get the desired regular complement.
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math/9912111
|
NAME gives that MATH is sufficiently large, where MATH is divisible enough and MATH. Then standard arguments show that MATH is not klt for some MATH and MATH (see for example, CITE or the proof of REF ).
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math/9912111
|
First note that MATH, because MATH is not klt and MATH is connected by REF. REF follows by REF . To prove REF we note that there exists an extremal MATH-negative contraction MATH. By REF , MATH is not birational. Hence MATH is a curve of genus MATH and fibers of MATH are irreducible. If MATH is contained in fibers of MATH, then MATH is nef and big. By REF , MATH is rational in this case. So we assume that there is a component MATH such that MATH. Thus MATH. Let MATH be a general fiber. Then MATH. It follows that MATH. Further, MATH . From this we have that MATH is smooth elliptic curve, MATH, MATH is contained in the smooth part of MATH and does not intersect other components of MATH. In particular, MATH. Since MATH is the section, MATH has no multiple fibers. Therefore MATH is smooth and MATH, where MATH is a rank two vector bundle on MATH. From MATH and MATH we have MATH. As for REF , run MATH-MMP MATH. At the end we obtain MATH with an extremal contraction MATH as in REF . By REF MATH cannot be contracted to a point on MATH. As in REF , by Adjunction we have MATH . This yields MATH and MATH. Hence MATH is a smooth elliptic curve, MATH is smooth along MATH and MATH does not intersect other components of MATH. Finally, REF follows by REF because MATH.
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math/9912111
|
Let MATH be an effective NAME divisor on MATH containing MATH and let MATH. First we take the MATH such that MATH is maximally lc (see REF) and replace MATH with MATH. This gives that MATH. Next we replace MATH with a log terminal modification. So we may assume that MATH is dlt and MATH. Then REF give us that there exists a regular complement MATH of MATH. By construction, MATH. If MATH is not exceptional, then there exists a MATH-complement MATH of MATH and at least two divisors with discrepancy MATH. Then we can replace MATH with MATH. Taking a log terminal blowup, we obtain that MATH is reducible. The rest follows by REF .
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