paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0005025 | Apply the previous lemma and REF to MATH at any MATH, which, by the NAME Fixed Point Theorem, is non-empty since MATH. |
math/0005025 | Let MATH be a rationally smooth MATH-fixed point of MATH. Using the relative order, we may without loss of generality assume that if MATH and MATH, then MATH is nonsingular at MATH. By the relative version of NAME 's Inequality and the fact that the singular locus of MATH has codimension at least two (as MATH is normal), there are at least two good MATH-curves in MATH. Since MATH CITE, the proof is done. |
math/0005027 | REF and the continuity of the concave functions MATH and MATH at MATH imply that MATH for MATH . By the definition of the norm of a NAME space, we have MATH . The NAME space with a given fundamental function is minimal among the symmetric spaces with the same fundamental function [REF, p. REF]; therefore, MATH . Suppose that MATH has not the DSS property. Then there exists a sequence of nonzero disjoint functions MATH such that MATH . By REF , for any MATH there exists a MATH such that MATH for all positive MATH . Choose MATH so that, for MATH where MATH . The subset of finite-valued functions is dense in any NAME space on MATH [REF, p. REF]. Therefore, for each MATH there exists a function MATH for which MATH . Hence MATH and [REF, p. REF] and REF imply that MATH . In addition, by REF , MATH . Thus, MATH . This inequality with a MATH satisfying MATH contradicts REF . |
math/0005027 | The implications MATH and MATH follow from REF and the definition of a DSS operator, respectively. Suppose that REF does not hold, that is, that MATH . Then there exist a sequence MATH and a constant MATH such that we have MATH and MATH for MATH . The functions MATH where MATH are disjoint and MATH satisfy REF . Therefore, REF implies REF ; this completes the proof of REF . |
math/0005027 | Since any NAME space is maximal among all symmetric spaces with the same fundamental function [REF, p. REF], we have MATH . By REF , MATH for MATH therefore, by the definition of a NAME space, MATH . Hence, MATH . Suppose that MATH is not a DSS operator. Then, in particular, there exists a sequence of disjoint functions MATH such that MATH . Choose a MATH for which MATH . Since the functions MATH are disjoint, we can assume that MATH . Therefore, MATH . By REF , MATH as MATH which contradicts REF . This completes the proof of REF . |
math/0005027 | First, MATH and an arbitrary SS MATH is embedded in MATH [REF, p. REF]. If MATH then the function MATH increases, because MATH is concave. Therefore, REF is violated if and only if MATH (that is, if and only if MATH for some MATH and MATH), and MATH . It remains to apply REF . |
math/0005027 | Let the SS MATH be the NAME space MATH with MATH where MATH . It is readily verified that MATH is an increasing concave function on MATH and MATH . Therefore, by [REF, p. REF], MATH . Let us define the space MATH. We put MATH and MATH and define a sequence of numbers MATH by setting MATH and a sequence of functions MATH by setting MATH . Since MATH increases, REF implies that the norms of MATH in MATH satisfy the inequalities: MATH and MATH . Therefore, MATH . Consider MATH and MATH . Let MATH be the set of all functions MATH measurable on MATH and satisfying MATH . Then MATH is an SS on MATH as the intersection of NAME spaces determined by the functions MATH with MATH . In addition, the definition of MATH implies that MATH . Therefore the fundamental functions MATH and MATH of the spaces MATH and MATH satisfy REF . Let us prove that MATH . By REF , it is suffices to show that MATH . Indeed, MATH . Therefore REF and the definition of MATH imply that MATH which proves REF . Next, we put MATH and MATH . The functions MATH are disjoint. Let us show that the norms of MATH and MATH are equivalent on their linear hull. Suppose that MATH . Without loss of generality, we can assume that MATH . Consider MATH . The function MATH monotonically decreases on MATH and MATH . Therefore, by REF , MATH . Since MATH for MATH we obtain MATH . Now, let us estimate MATH from below. By REF , for any MATH we have MATH . Hence REF implies that MATH . Therefore, by REF , MATH . This together with REF means that the norms of the spaces MATH and MATH are equivalent on the linear hull of the set of functions MATH . Hence there exists a MATH such that, for an arbitrary MATH . In other words, the identity inclusion operator MATH has not the DSS property. This completes the proof of REF . |
math/0005027 | Since MATH [REF, p. REF] implies that MATH . Therefore the relation MATH is equivalent to MATH . Since the function MATH is concave, we have MATH hence REF is equivalent to the condition MATH . Put MATH . Then MATH and [REF ] and REF imply that MATH . By the definition of upper dilation index, there exist MATH and MATH such that MATH and MATH for all MATH . Consider the sequence of numbers MATH . Suppose that MATH is a function linear on the intervals MATH for MATH and MATH . Since MATH decreases, MATH also decreases; therefore, the functions MATH and MATH increase on MATH . It follows from REF that, for MATH and MATH . Hence, by REF , MATH . This implies that MATH because MATH . Therefore, according to [REF, p. REF], the function MATH is equivalent to its least concave majorant; we denote this majorant by MATH. The function MATH belongs to MATH and MATH for MATH in addition, since MATH it follows from REF that MATH . This equivalent to MATH the equivalence is proved in the same way as for the function MATH (see REF ). Finally, relation REF gives MATH . This completes the proof of REF . |
math/0005027 | By REF , there exists a function MATH such that MATH . According to REF , the operator MATH has the DSS property; all the more, it has this property when regarded as an operator from MATH into MATH. This proves REF . |
math/0005027 | Let us verify that the assumptions of REF hold, that is, that MATH and that MATH . First, REF implies the existence of MATH and MATH such that MATH whenever MATH and MATH . This and the concavity of MATH give MATH therefore, MATH . Next, since the function MATH decreases, we have MATH . Therefore, to prove REF , it suffices to verify that MATH . By REF , MATH . Hence MATH as MATH and MATH . This proves REF . The above-mentioned extremality of the NAME and NAME spaces in the class of symmetric spaces with the same fundamental function CITE implies that MATH . Therefore, MATH has the DSS property, because it has this property as an operator from MATH into MATH by REF . This completes the proof of REF . |
math/0005033 | The proof is a simple computation which we leave to the interested reader, compare REF. |
math/0005033 | The proof is identical to the proof of REF. |
math/0005033 | The key to the proof rests in the fact that the pair MATH solves the anisotropic averaged NAME equations if and only if MATH is a solution of MATH where MATH . This expression is obtained using REF . Now it is clear that MATH maps MATH vector fields into MATH vector fields since MATH forms a multiplicative algebra, and since the fluctuation tensor MATH at MATH is given by MATH which is MATH. In particular, in the Lagrangian frame, MATH is frozen, so the elliptic operator MATH has MATH coefficients. Thus, the proof of REF gives a unique curve MATH solving REF . That MATH is only MATH in time follows from the fact that the map MATH is only MATH. That MATH is in MATH follows from the regularity of MATH. |
math/0005034 | First observe that the material density of an ideal homogeneous (compressible or incompressible) fluid is constant. Notice also that NAME REF differ only in the potential energy terms. Both these terms are functions of the Jacobian, which is equivariant with respect to the action of volume preserving diffeomorphisms given by REF . Indeed, MATH due to the fact that MATH for a volume preserving diffeomorphism MATH; here MATH can be any function, for example, the stored energy MATH or the constraint MATH. For the same reason, and the fact that REF acts trivially on MATH, the kinetic part of both NAME is also equivariant. |
math/0005036 | The first condition implies that MATH for MATH, and so implies the second condition. The fact that the third condition implies the first is a well-known consequence of the NAME - NAME lemma. So we need only show that the second condition implies the third. The proof is by induction on MATH. First consider the case MATH. We have MATH where we have made the change of variable MATH in the last step. In particular, for MATH on MATH, MATH on MATH for all MATH, so the quantity MATH is independent of MATH. Thus MATH . The hypothesis that this quantity is MATH implies that MATH, that is, that the constant function belongs to MATH. Now we consider the case MATH. We again apply REF, this time for MATH an arbitrary homogeneous polynomial of degree MATH. Then MATH where MATH. Substituting in REF, and invoking the inductive hypothesis that MATH, we get that MATH . Again the last infimum is independent of MATH so we immediately deduce that MATH belongs to MATH. Thus MATH contains all homogeneous polynomials of degree MATH and all polynomials of degree less than MATH (by induction), so it indeed contains all polynomials of degree at most MATH. |
math/0005036 | Again, we need only prove that the second condition implies the third. In analogy to REF, we have MATH where we have made the change of variable MATH in the last step. The proof proceeds by induction on MATH, the case MATH being trivial. For MATH, apply REF with MATH an arbitrary homogeneous polynomial of degree MATH. Substituting REF, and invoking the inductive hypothesis that MATH, we get that MATH . Since we assume that this quantity is MATH, the last infimum must be MATH, so MATH differs from an element MATH by a constant. Thus MATH contains all homogeneous polynomials of degree MATH and all polynomials of degree less than MATH (by induction), so it indeed contains all polynomials of degree at most MATH. |
math/0005036 | The components of MATH are linear functions of MATH and MATH, so if MATH is a polynomial of degree at most MATH, then MATH is of degree at most MATH in MATH and MATH, that is, MATH. Therefore MATH. |
math/0005036 | Assume to the contrary that MATH and MATH. We prove that MATH by induction on MATH. The case MATH being true by assumption, we consider MATH and show that the monomials MATH and MATH belong to MATH for MATH. From the identity MATH we see that for MATH, the monomial MATH is the sum of a polynomial which clearly belongs to MATH (since MATH) and a polynomial in MATH, which belongs to MATH by induction. Thus each of the monomials MATH with MATH belongs to MATH, and, using MATH, we similarly see that all the monomials MATH, MATH belong to MATH. Finally, from REF with MATH, we see that MATH is a linear combination of an element of MATH and monomials MATH with MATH, so it too belongs to MATH. |
math/0005038 | Let MATH be the arc from MATH to MATH along MATH, then back to MATH along MATH. Let MATH be the sum of the winding numbers of MATH around each of the puncture points. If MATH is closer to MATH than MATH along MATH then MATH. If not, then MATH. But MATH for all MATH, so in either case MATH. |
math/0005038 | By definition, MATH is the sign of the volume form MATH at the point MATH in MATH. This is determined by the following three things: CASE: the sign of the intersection of MATH with MATH at MATH, CASE: the sign of the intersection of MATH with MATH at MATH, CASE: which of MATH and MATH is closer to MATH along MATH. These in turn are determined by the following three values respectively: CASE: MATH, CASE: MATH, CASE: MATH. To see this, observe that in general the sign of MATH is determined by whether the two points switch places in the path MATH. This in turn is determined by which of MATH and MATH is closer to MATH along MATH. It follows from the above considerations that MATH is determined by the value of the product MATH evaluated at MATH and MATH. It remains to check that the sign is as claimed. This can be done by calculating MATH in a specific example. |
math/0005038 | Suppose MATH is maximal. Then MATH is maximal among all the integers MATH. By REF it follows that MATH and MATH are maximal among all the integers MATH. Thus MATH. We now show that MATH. Suppose not. Then MATH. Let MATH be an embedded arc from MATH to MATH along MATH. Let MATH be an embedded arc from MATH to MATH along MATH. If MATH does not pass through the point MATH, let MATH and let MATH be the winding number of MATH around MATH. Then MATH. If MATH does pass through MATH, first modify MATH in a small neighborhood of MATH so that MATH lies to its left. Next let MATH and let MATH be the winding number of MATH around MATH. Then MATH. In either case, MATH is greater than zero. Let MATH. Let MATH be the universal cover of MATH. Let MATH be a lift of MATH to MATH. Let MATH be the lift of MATH to MATH which starts at MATH. Let MATH be an embedded arc in MATH from MATH to MATH which intersects MATH and MATH only at its endpoints. Let MATH be the projection of MATH to MATH. Choose MATH so that MATH does not wind around any puncture points. Let MATH be the first point on MATH which intersects MATH (possibly MATH). This lies in the fiber over MATH for some MATH. Let MATH be the initial segment of MATH ending at MATH. Let MATH be the final segment of MATH starting at MATH. Let MATH. Now MATH is a simple closed curve in MATH, so by the NAME curve theorem it must bound a disc MATH. Since MATH passes clockwise around MATH, there is a non-compact region to the right of MATH, so MATH must pass counterclockwise around MATH. Let MATH be the projection of MATH to MATH. Then MATH is equal to the sum of the winding numbers of MATH around each of the points in MATH. This is equal to the number of points in MATH which are lifts of a point on MATH. This must be greater than zero, since otherwise we could isotope MATH so as to have fewer points of intersection with MATH. Thus MATH is greater than zero, contradicting the fact that MATH is maximal among all integers MATH. Therefore our assumption that MATH must have been false, so MATH. The proof that MATH is similar. This completes the proof of the claim. |
math/0005038 | There exists a disc MATH such that MATH contains MATH, MATH contains MATH for all MATH with MATH, and MATH. Let MATH be the set of unordered pairs of distinct points in MATH. Let MATH be the pre-image of MATH in MATH. Then MATH is a free MATH-module with basis consisting of all MATH with MATH and MATH. But MATH can be considered as an element of MATH, so must be a linear combination of these basis vectors. |
math/0005043 | It is easy to make MATH transversal to MATH. Then MATH is a union of circles. It remains to homotope MATH so that we get only one component. We do this using a standard technique in MATH - manifold topology related to NAME `binding ties' CITE, as in CITE. Suppose MATH has more than one component. Let MATH be an arc joining two components, such that its two end-points have the same image under MATH and so that the images of neighbourhoods of the two endpoints coincide. We shall modify MATH so that MATH represents the trivial element in the fundamental group of MATH. To do this, note that as MATH is a homotopy equivalence, it has degree one, and hence so does its restriction to MATH. Thus the restriction induces a surjection on the fundamental group. In particular, there is a closed loop MATH in MATH whose image MATH in MATH is the inverse of MATH (pushed off along the common image of neighbourhoods of the two endpoints). We replace MATH by its concatenation with MATH and push this off MATH to get the required curve. Now, we first homotope the map on a neighbourhood of the arc, which we identify with MATH so that the image of any point is equal to that of its projection onto the arc, that is, MATH. To do this, first let MATH if MATH and equal to MATH otherwise. Clearly MATH is homotopic to MATH, so we may replace MATH by MATH. Next, by the choice of MATH, we have a homotopy MATH (fixing endpoints) of MATH to a point with MATH disjoint from MATH. Use this to define the map on MATH by MATH and symmetrically on MATH. We identify MATH with MATH. Note that MATH contains the MATH segments MATH, MATH and MATH. Now extend MATH so that there are no further points in the inverse image of MATH. We have reduced the number of components of MATH. By repeating this process we are left with only one component. |
math/0005043 | The image MATH of the NAME polynomial of MATH in MATH is the so called NAME - NAME invariant, which by results of CITE and CITE is known to depend only on the homology class of MATH. Further, results of CITE show that this depends only on the NAME torsion (given an identification of homology groups). But MATH is a simple-homotopy equivalence, and MATH in homology. It follows that MATH, where MATH is the image of the NAME polynomial of MATH in MATH, and the group rings are identified using MATH. As MATH is a core, MATH. |
math/0005043 | Let MATH denote the NAME polynomial of MATH. If the NAME module is cyclic, MATH. Since MATH, there is a NAME polynomial MATH so that MATH. Thus, MATH generates MATH. In the general case, the NAME polynomial is the determinant of the presentation matrix (with respect to some system of generators), say MATH, for the NAME module (which we call MATH). As this is congruent to MATH modulo MATH, the matrix MATH is invertible on reducing to the module MATH. The claim follows immediately, for, if MATH is a word in the generators of the NAME module MATH, then we have: MATH for some word MATH in the generators, or MATH, MATH. But this means that MATH in the module MATH. |
math/0005043 | It follows readily from the previous lemma that MATH can be expressed as MATH, for a lift of some curve MATH in MATH. Now take the band connected sum of MATH with itself (pushed off using, for instance, the MATH - framing) along a curve that goes once around MATH (see REF - here the two dotted arcs are parallel, but may represent any homotopy class and knot type). This represents MATH in MATH. |
math/0005043 | We use a standard construction CITE to construct a knot whose NAME polynomial is given by surgery on an unknot. Namely, we drag a piece of a curve MATH around a knot MATH and then across itself (see REF ). If MATH was a sphere, or a homology sphere, this leads to a change in the homology class of a lift of MATH in the NAME module by MATH. By surgery on an unknot MATH constructed using such transformations, we obtain any NAME polynomial MATH satisfying MATH and MATH. In our situation, this leads to a change in the coboundary by MATH rather than MATH. By first winding around by an element representing MATH, or more generally an arc representing the appropriate element in the NAME module before making the crossing, one can also change the linking by MATH, we can change by an multiple of MATH in the module. Finally, by changing the framing along which the curve is pushed off itself in REF , we can ensure that the total coefficients are the same. |
math/0005043 | Note that one can readily change a crossing between MATH and MATH by dragging an arc of MATH along a closed curve representing MATH and then crossing MATH. This changes the intersection number with MATH by the coefficient MATH of MATH in the boundary MATH of the NAME surface. This also changes the linking of MATH at MATH but as the negative of the previous change, so the intersection number here changes by the negative of the coefficient of MATH in the boundary of the NAME surface. Thus, the linking number with a given NAME surface has been changed by MATH, where MATH is the coefficient of the NAME surface. We shall see that these moves suffice to get the right linking modulo MATH. For, the linking numbers are determined by the homology classes up to a simultaneous change by the coefficient MATH of each NAME surface. But now, as the presentation matrix is invertible modulo MATH, such a change in MATH may be achieved by changing MATH modulo MATH. Thus, we can ensure that MATH has the right linking. |
math/0005043 | Pick a NAME surface for MATH. We shall perform MATH surgeries along curves disjoint from the NAME surface, which must thus be homologically trivial. Namely, it is well known that there are such curves that form a dual basis to a basis of curves on the NAME surface with respect to the linking pairing. Hence, one can find curves with any desired combination of linking numbers. Surgery along a curve changes the linking number between a pair of other curves by an amount determined by the linking with the surgery locus. As the entries of the NAME matrix are linking numbers between curves of the NAME surface pushed off in two directions, it is easy to see that surgeries on such curves can be used to transform the NAME matrix, and hence the NAME polynomial, to that of an unknot. For, surgery on a curve linked once with each of a pair of basis curves on the surface (and unlinked from others) changes their linking, while surgery on a curve linked with just one basis curve changes framing. |
math/0005044 | Because MATH is self-dual and contains MATH, it is the whole MATH. One has to show (REF , page REF) that the restriction of MATH to MATH is split (the uniqueness is obvious). Since MATH is reduced, it is enough to define the splitting at the level of MATH-points. Let MATH. Since MATH and MATH is divisible, there exists a unique MATH over MATH satisfying MATH. Since MATH again, the restriction of MATH to MATH is trivial and therefore MATH is isotropic for MATH. This implies that any two elements MATH commute and the map MATH is a morphism of group schemes. |
math/0005044 | Because MATH is the pull-back of MATH by MATH, the restriction of MATH to both MATH splits (see REF , page REF again). Observe however that the splitting is determined not only by MATH but by MATH. For the splitting over MATH, proceed as in REF . |
math/0005044 | Let us construct geometrically the basis MATH. We define MATH by MATH . By construction, one has MATH . Because MATH comes from MATH, it is MATH invariant. The relations REF follow. Because MATH is irreducible, MATH is a basis. If we have another such basis MATH, the endomorphism given by MATH is MATH-equivariant and therefore a scalar by NAME 's lemma, which proves REF . The sections MATH live in MATH by REF follows. |
math/0005044 | REF follows immediately from REF is a special case of NAME 's singularity theorem (see for example, CITE). The differential at the origin of the separable isogeny MATH is an isomorphism MATH, which identifies with the NAME map MATH. Given MATH, it will be enough to compute the tangent space to the divisor MATH at the origin. Let MATH be the canonical Theta basis of MATH (Apply REF to MATH with MATH). Then, by the isogeny REF , we have (up to a scalar) MATH. Let MATH be a tangent vector to MATH at the origin. Then we compute, using MATH, MATH . Form this we deduce that MATH is singular at MATH if and only if MATH and, assuming MATH smooth, the equation of MATH. The second description of MATH is a consequence of CITE. |
math/0005044 | In order to show commutativity of REF it suffices to show that the image of the injection MATH is contained in the image of the multiplication map MATH . Let MATH and MATH be the canonical Theta bases of MATH and MATH and consider the pull-back of equality REF, written for MATH, by the morphism MATH, MATH, MATH . If MATH, we get MATH with MATH, and define MATH. If MATH, we see that both members of REF are zero. In order to get a meaningful statement we restrict to a first infinitesimal neighbourhood of MATH. The notation is as in REF. We pull-back the morphisms MATH and MATH to MATH replacing MATH by MATH and keeping the same notation for the object over MATH and its pull-back to MATH. Pulling back equality REF by MATH, for MATH, we get the following two elements in MATH where MATH is a scalar which vanishes if and only if MATH is singular at the point MATH REF , and MATH where MATH runs over a set of representatives of MATH. Since these elements are equal MATH and there exists MATH such that MATH, we obtain (up to a multiplicative non-zero scalar) MATH . In order to complete the proof it suffices to show that for a general curve MATH the MATH's are non-zero, for all MATH. Assume that the contrary holds. Then the equality in MATH (which we leave as an exercise) MATH where MATH is any index MATH subgroup of MATH not containing MATH and the MATH are the quadrics REF, shows that either MATH or MATH is zero, since the Right-hand side is mapped to zero in MATH. But this is impossible, since MATH is a basis. Hence for any curve MATH there exists MATH such that MATH and we conclude by a monodromy argument found in CITE. |
math/0005044 | Let MATH be the inclusion in the first factor MATH. A standard computation modelled on REF , which involves REF, gives the equality in MATH . We observe that the composite MATH is the diagonal map, hence the LHS can be rewritten as MATH. Applying the isogeny REF , we get MATH . Comparing with REF we can conclude. Finally we observe that REF is well-defined since MATH such that MATH we have MATH (MATH is symmetric). |
math/0005044 | First we observe that the image MATH is invariant under the Theta group MATH. Hence the equation of MATH has to be MATH-invariant. Since MATH and MATH (MATH is separable), we have MATH. The unique MATH-invariant quadric is MATH. Since MATH is non-degenerate, we see that it is not contained in a quadric. A straightforward computation shows that a basis of MATH-invariant quartics is given by the MATH polynomials MATH . Let us denote by MATH the equation of MATH, where MATH need to be determined. First MATH, where MATH is the origin, implies that MATH. Since the image is reduced, MATH. Next, the equation of MATH is given in a Theta basis by MATH . Since MATH is not contained in a quadric, the rational map MATH REF is uniquely defined. Since MATH, there exists a MATH-invariant quartic MATH, such that (up to a non-zero scalar) MATH . We shall use the Theta coordinates MATH on both spaces. In order to determine the equation of MATH, we restrict equality REF to the hyperplane MATH and write the expressions as degree MATH polynomials in MATH. A straightforward computation, which we omit, leads to MATH . Suppose that MATH, that is, MATH. Then at least one of the factors MATH has to divide MATH. Again elementary computation shows that this can only happen when MATH, which is impossible. Hence MATH is a multiple of MATH and, since MATH is also MATH-invariant, we get MATH. Moreover MATH implies that there exists a MATH such that MATH for all MATH. It remains to determine the constant MATH. We compute the degree MATH polynomial MATH (again we omit details) MATH . This expression being equal to MATH, we get the equation mentioned in the proposition. |
math/0005044 | We proceed in two steps. First we consider a flat family of curves MATH, where MATH is a discrete valuation ring such that its residue field at the closed point MATH is MATH, its field of fraction MATH is of characteristic zero, and MATH. We consider the moduli scheme MATH of semi-stable rank MATH vector bundles of trivial determinant over the family MATH. Let MATH be the fibre of MATH over MATH. Then, by GIT over arbitrary base CITE, we have a canonical bijective morphism MATH. Moreover on the open set of stable points MATH is an isomorphism since the action of the projective group on the Quot scheme is free. We conclude that MATH is an isomorphism by NAME 's Main Theorem. Secondly, we extend the morphism MATH CITE to the family MATH,that is, we construct a morphism over the base MATH such that MATH. In order to show the proposition, it will be enough (again by NAME 's Main Theorem) to show that MATH is birational and bijective. We consider the fibre MATH over the generic point MATH. Working over an algebraic closure of MATH (of characteristic MATH), we see CITE that MATH is an isomorphism. Hence MATH is birational. It remains to show that MATH is bijective. NAME is obvious. Let MATH be the hyperplane MATH of divisors passing through MATH. The inverse image of MATH is MATH showing MATH. It follows that MATH is finite because MATH is ample. Let MATH be the relatively ample line bundle over MATH. Consider the canonical inclusion of MATH in MATH with cokernel MATH. MATH . We twist by MATH. If MATH is large enough, we have MATH and MATH, MATH. Hence, since MATH is flat, we see that MATH is constant. If MATH, this number is zero CITE. Hence MATH and MATH. So MATH and MATH is injective. |
math/0005044 | CASE: This follows immediately from CITE and REF . The base locus of MATH is given by the intersection MATH, which turns out (after some elementary computations) to be a unique point with projective coordinates MATH. In terms of vector bundles this point, denoted MATH, corresponds to the direct image MATH. Indeed, MATH is stable: a nonzero map MATH is equivalent, by adjunction, to a non-zero map MATH, hence MATH. Moreover, since we have a canonical nonzero map MATH, this bundle is destabilized. Uniqueness can also be proved without the use of the equations: consider any stable bundle MATH, which is destabilized by MATH. By REF there exists a theta-characteristic MATH which appears as a quotient MATH. By adjunction this quotient map induces a map MATH. Since the two bundles are stable with the same slope, we deduce that they have to be isomorphic. The determinant of MATH being trivial, we get MATH and we are done. CASE: First we observe that the hyperplane MATH is mapped into the hyperplane MATH. After fixing an isomorphism MATH, we straightforwardly compute that a point in the image MATH satisfies the equations MATH which is precisely (after squaring) REF 's quartic surface REF restricted to the hyperplane MATH. Moreover MATH and by adjunction MATH which proves the first assertion. CASE: Given a point MATH with projective coordinates MATH, we have to solve the system of quadratic equations MATH where the quadrics MATH's are defined in REF. We write MATH. Adding any two MATH's with MATH, we find MATH and similarly MATH and MATH with MATH. We let MATH. Then REF imply that MATH (respectively, MATH) are the roots of the polynomial MATH NAME MATH with MATH and MATH respectively in the equation MATH, we find MATH . Assuming MATH,that is, MATH, we see that the fiber MATH consists of the point MATH with projective coordinates MATH given by MATH plus the other MATH points obtained by switching MATH and MATH as well as MATH and MATH. Since MATH, one easily sees that these MATH points are distinct. Assume MATH, that is, MATH. We get MATH. Hence MATH. Assume that MATH,that is, MATH. Because of REF the system REF does not have a solution. If MATH, that is, MATH, it is easy to check that MATH is the MATH, intersection of the two hyperplanes MATH . |
math/0005044 | It will be enough to show that any point MATH is in the image, since, twisting by a degree zero line bundle, we can always assume the determinant to be trivial. For any non-zero MATH, we choose a MATH such that MATH and we denote by MATH the moduli of rank MATH vector bundles with fixed determinant equal to MATH. Then we have a commutative diagram MATH . Now since the vertical maps are isomorphisms, the image of the first horizontal map contains the complement to the hyperplane MATH. Since the hyperplanes MATH have no common point in MATH we get surjectivity. |
math/0005044 | We write MATH as an extension of line bundles MATH for some line bundle MATH with MATH and MATH. The extension class MATH of the exact sequence, gotten by pull-back under MATH, that is, MATH is obtained from MATH via the linear map MATH . The last map coincides with the induced map on cohomology of the canonical exact sequence MATH . In order to prove that MATH, we tensorize REF with MATH. If MATH, we are done. Therefore we assume MATH. It follows from REF that MATH . On the other hand, we see that MATH if and only if the symmetric coboundary map MATH is degenerate. We write MATH. It is well-known that the linear map MATH is the dual of the multiplication map of global sections MATH. Let us denote by MATH the space of divided powers of MATH,that is, the subspace of tensors invariant under the involution MATH. We have MATH. We denote by MATH the subspace of MATH generated by the squares MATH with MATH. Dually, the kernel of the surjection MATH coincides with the space of alternating maps MATH. The main point, which will be used later, is that, since, by NAME, MATH is odd, any map in MATH is degenerate. By NAME duality we have MATH. Put MATH. Then we have a MATH-linear isomorphism MATH defined as follows: we have MATH and we put MATH. We also note that MATH. Again by NAME duality, we observe that the linear maps REF coincide, that is, we have a commutative diagram MATH where the vertical maps are MATH-linear isomorphisms. Now we can conclude as follows: by commutativity, any extension class MATH, with MATH, is mapped by MATH into MATH. Hence the corresponding coboundary map is degenerate. |
math/0005044 | Since the proof is in the same spirit as the proof of REF , we just give a sketch. Let MATH be the vector field on MATH associated to MATH. We also denote by MATH the endomorphism of MATH obtained via the canonical MATH-integrable connection MATH REF . We observe that we have a commutative diagram MATH . The first horizontal map is the canonical projection and MATH is the NAME map associated to MATH REF. We consider the map MATH defined in REF and compose with MATH. For MATH general, the composite is non-degenerate and the induced (injective) map on global sections coincides with MATH CITE. Using REF we can now deduce the equations of MATH. The last assertion can be proved as in REF . |
math/0005044 | Since MATH, the action of MATH on MATH factorizes as MATH . Since MATH, we see that in terms of the canonical Theta coordinates MATH . Let MATH be the subfield of MATH generated by the constants MATH and MATH, for MATH, be the finite field extension of MATH generated by the coordinates MATH such that MATH with MATH. We obviously have a tower of extensions MATH and, by the computations carried out in the proof of REF , we see that MATH is a power of MATH. Hence, by induction, any element MATH has coodinates MATH, which lie in an extension of MATH of degree MATH for some MATH. But elements of odd degree over MATH are evidently dense in MATH. |
math/0005046 | REF is proved by CITE for simply laced groups, and by CITE in the general case. REF follows from REF since MATH for MATH, MATH. It remains to prove REF . Since the conjugation action of MATH on MATH is trivial, its action on any connected component of MATH is scalar multiplication on the fibers by some character for MATH. To compute the weight for this action at MATH, let MATH be the left-invariant connection REF-form defined by the splitting MATH. The weight MATH for the MATH-action on the fiber over MATH is given by MATH where MATH is any lift of MATH. In left trivialization of the cotangent bundle of MATH, MATH is the constant map from MATH to MATH. This shows that its contraction with the left-invariant vector field generated by MATH is zero. To compute its contractions with the right-invariant vector field generated by MATH, we note that under the right action of MATH, MATH . Since MATH is the difference between left and right invariant vector fields, we find that MATH, proving MATH. |
math/0005046 | We begin by showing that MATH is symplectic. Since MATH is a regular element, MATH. Let MATH, MATH, and consider the splitting of the tangent space MATH where MATH and the second summand is embedded by the generating vector fields. By CITE, the splitting is MATH-orthogonal and the restriction of MATH to MATH is symplectic. Since the action of MATH on MATH preserves REF-form, the subspace MATH is symplectic as well. This shows that MATH is non-degenerate. It is closed since MATH. The moment map condition for MATH follows from that for MATH. This proves REF . Clearly MATH is MATH-invariant, and its image under the moment map is contained in MATH. Also MATH. To prove the reverse inclusion, suppose MATH. Its pre-image under the map MATH meets MATH. We need to show that any pre-image MATH is fixed under MATH. By definition of the action on MATH, any pre-image satisfies MATH for some MATH. By equivariance and since MATH acts trivially on MATH, this means MATH. Since MATH acts freely, we conclude MATH and therefore MATH. Viewing MATH as a principal MATH-bundle, MATH is the pull-back bundle under the map MATH. From the constructions, it follows that MATH is the Hamiltonian MATH-manifold associated to MATH, proving REF . Now view MATH as a MATH-principal sub-bundle of MATH. We have MATH since every MATH-orbit in MATH passes through a unique point in MATH. Then REF follows since MATH and MATH are pull-back bundles with respect to MATH, and from the fact that the form MATH vanishes if pulled back to MATH. |
math/0005046 | Note first of all that the right hand side of REF is well-defined. Indeed, if MATH is replaced by MATH with MATH, the factor MATH does not change because MATH is fixed under the level MATH action of any element of MATH fixing MATH, and MATH changes by an even number. Given faces MATH of the alcove and any MATH, the symplectic normal bundle of MATH inside MATH is MATH-equivariantly isomorphic to MATH, with MATH acting via the isomorphism MATH induced by MATH. Using the sign convention from REF, the square root of the eigenvalue for the action of MATH on MATH is given by MATH, and similarly for the action on MATH. Therefore, MATH . Since MATH, we have shown that the right hand sides of REF patch together to a well-defined locally constant function on MATH. The action of MATH amounts to replacing MATH by MATH. This does not change MATH, and changes MATH by an even number. The factor MATH changes by MATH. |
math/0005046 | The projection map MATH restricts to an equivariant diffeomorphism MATH. Let MATH be MATH-line bundle corresponding to MATH. Then MATH . The normal bundle of MATH in MATH splits MATH-equivariantly into the normal bundle MATH in MATH and the constant bundle MATH. Using MATH we obtain, MATH . Let MATH be the square root for the action on MATH. We have MATH . |
math/0005046 | By the NAME character formula, MATH . Given MATH let MATH. We claim that the sum over MATH is just MATH . Indeed, by REF and since MATH, we have MATH . But MATH since MATH. Hence MATH proving the claim. |
math/0005046 | For all MATH, with MATH we have MATH. Hence, ``quantization commutes with reduction" REF together with REF shows that MATH . Let MATH be the characteristic function of MATH. Using the NAME formula MATH the alternating sum over MATH equals MATH. |
math/0005046 | The first part follows by observing that the integral can be re-written as an integral over the cut space MATH of MATH, that is over the image of MATH in MATH: MATH . Here MATH is the ``cut" of MATH as explained in REF, and similarly for MATH and MATH. REF is expressing the integral over MATH as a sum of integrals over all pieces MATH in the decomposition. |
math/0005046 | The proof is an extension of the argument given in CITE. We may assume MATH. Observe that all of the characteristic forms in REF admit MATH-equivariant extensions. Hence we can write MATH where MATH . Let us apply the NAME localization formula for orbifolds (compare CITE) to this expression. Let MATH be a fixed point manifold for the MATH-action on MATH. Letting MATH be the map induced by MATH, it follows that MATH is a point. Let MATH be the unique open face containing MATH, and let MATH be the unique connected component with MATH. Recall that by REF the restriction of MATH to MATH is the tangent bundle of MATH plus a trivial bundle. Similarly the restriction of MATH is the tangent bundle to MATH plus a trivial bundle, and the restriction of MATH is the normal bundle MATH to MATH in MATH. On the other hand, the normal bundle of MATH in MATH is the pull-back of the normal bundle MATH of MATH in MATH. We therefore obtain the formula MATH . We obtain similar formulas for all of the cut spaces MATH: MATH where the sum is over all connected components MATH of the MATH-fixed point set of MATH, for all MATH. If MATH is a fixed point component for the MATH-action on MATH, then MATH is contained in the interior of a unique top-dimensional polyhedron MATH. Hence, the fixed point contribution appears exactly once as a MATH-fixed point contribution of the sum MATH. We must show that the remaining fixed point contributions cancel. These other integrals are over connected components MATH of MATH-fixed point sets of MATH, where MATH is defined as in REF. These fixed point components can be organized as follows. Consider the finite subset of points of the form MATH. Given any such point, there exists a unique open face MATH of MATH containing it. If MATH, then MATH is simply one of the MATH-fixed point orbifolds for MATH, and the cancellation of the corresponding fixed point contributions is just the gluing formula CITE. In case MATH is a proper face of MATH, the argument from CITE carries over without essential change, the reason being that the fixed point contributions look exactly like fixed point contributions for MATH, except for the extra factor MATH which appears in all of these integrals. |
math/0005047 | Since MATH is obtained from a direct product of MATH copies of MATH by passing to diagonal actions for some of the MATH-factors, it suffices to prove REF for MATH. By REF, an element MATH is fixed by MATH if and only if MATH . Both MATH and MATH belong to the exponential of the alcove MATH. Since each conjugacy class meet MATH only once, REF holds if and only if MATH. |
math/0005047 | The line bundle MATH is MATH-equivariant at levels MATH. Since MATH carries up to isomorphism a unique line bundle at every level CITE, it follows that MATH is the pre-quantum line bundle for the symplectic structure defined by MATH. Hence MATH is a pre-quantum line bundle for the corresponding symplectic structure on MATH (compare CITE, Subsection REFEF), and MATH is the symplectic volume MATH for REF-form defined using MATH. We claim that the symplectic volume coincides with the Riemannian volume, which will complete the proof since MATH by REF from REF. By our description of MATH as a fusion product, the fixed point manifold MATH is obtained from the fixed point manifold MATH (viewed as a group valued Hamiltonian MATH-space) by fusion: MATH and MATH with MATH factors MATH and MATH factors MATH. REF from REF says that the symplectic volume of group valued Hamiltonian torus spaces does not change under fusion. Hence MATH. Finally, REF for REF-form on MATH shows that MATH coincides with the Riemannian volume of MATH with respect to MATH. |
math/0005047 | The point MATH lies in identity level set of MATH, and its stabilizer in MATH is the image of the diagonal embedding of MATH. REF-form MATH restricts to a symplectic form on the tangent space MATH. By REF, MATH can be computed in terms of the symplectomorphism MATH of MATH defined by MATH: Choose a MATH-invariant compatible complex structure on MATH to view MATH as a unitary transformation, and let MATH be the unique square root having all its eigenvalues in the set MATH. Then MATH. We first apply this recipe for REF-holed sphere MATH, so that MATH. REF shows that MATH is the standard REF-form on MATH, given by the inner product MATH. A compatible complex structure is given by the endomorphism MATH. Thus, as a complex MATH-representation MATH is just the complexification MATH. It follows that the eigenvalues of MATH (other than MATH) come in complex conjugate pairs MATH and the corresponding eigenvalues of MATH are MATH and MATH. Hence MATH . Now consider the case MATH arbitrary. The tangent space is MATH, but because of the fusion terms the symplectic form is not the standard symplectic form defined by the inner product on MATH. However, by REF it is equivariantly and symplectically isotopic to the standard symplectic form. Since the phase factor MATH is a root of unity, it is invariant under equivariant symplectic isotopies, and we conclude as before that MATH. |
math/0005047 | We first show that the map MATH is surjective. Present MATH as a quotient of a MATH-gon MATH, with sides identified according to the word MATH . Then MATH is obtained as a similar quotient of MATH minus a disk in MATH. The sides MATH map to generators of MATH, which we also denote MATH. Given MATH, choose continuous maps MATH such that MATH at the base point and such that the loop MATH represents MATH. Since MATH is abelian, the concatenation of loops MATH is homotopically trivial. Hence the maps MATH extend to a continuous map MATH trivial on MATH. By a MATH-small perturbation, MATH can be changed to a smooth gauge transformation, which still vanishes on MATH. We next show exactness of REF at MATH. Since MATH is simply connected, a necessary and sufficient condition for an element MATH to admit a lift MATH is that the induced maps on fundamental groups be trivial. Thus MATH is in the image of the map MATH if and only if it is in the kernel of the map MATH. Injectivity of the map MATH is obvious. This proves REF and the proof of REF is similar. Finally we prove REF. Let MATH. Since MATH is homotopic to MATH, the restriction of MATH to the boundary defines a contractible loop in MATH. Hence MATH. |
math/0005047 | The image MATH of an element MATH is fixed under MATH if and only if MATH. In this case, MATH fixes the maximal torus MATH pointwise, hence MATH since MATH is a regular element. Therefore MATH and MATH. |
math/0005047 | We describe the inverse map. Let MATH. Given MATH, the element MATH lies in MATH. The equation MATH means, by definition of the action of MATH on MATH, that MATH. |
math/0005047 | We have MATH, and the fixed point set is just a fusion product of the fixed point sets of the factors. Because of REF it is enough to consider the case MATH. Thus MATH is the MATH-valued Hamiltonian MATH-space MATH with MATH acting by conjugation, moment map MATH, and REF-form MATH . Using left-trivialization of the tangent bundle to identify MATH, and using the metric MATH to identify skew-symmetric REF-forms with skew-symmetric matrices, REF-form MATH is given at MATH by the block matrix, MATH where MATH are the following endomorphisms of MATH, MATH . Suppose now that MATH is the fixed point manifold labeled by MATH. REF-form MATH on MATH is given at any point MATH by the endomorphism MATH, with MATH becoming the NAME group action of MATH on MATH. Using that MATH commute one verifies that MATH showing MATH. Since the top degree part of MATH is equal to the standard volume form on MATH, times the Pfaffian of MATH, this shows MATH. |
math/0005047 | It suffices to consider the case of REF-punctured torus MATH. Let MATH be given. By REF from REF, it is possible to choose commuting lifts MATH of MATH. Then MATH and MATH. |
math/0005047 | Recall that MATH carries a unique equivariant line bundle MATH at level MATH CITE. Moreover, this line bundle carries a connection MATH whose curvature is the two-form defined by MATH. The pull-back of MATH to MATH is isomorphic to MATH. The average of MATH over MATH descends to a connection on MATH. For REF , we may assume MATH is a line bundle at level MATH. The quotient MATH is a MATH-equivariant line bundle over MATH. Since MATH, every MATH-line bundle over MATH is isomorphic to a flat line bundle defined by a character of MATH (see for example, CITE). The discussion in Subsection REF (respectively, REF) shows that there exists a level MATH line bundle over MATH if MATH is a multiple of MATH (respectively, MATH) with the property given in REF. The theorem follows by taking tensor products and pull-backs under the map REF. |
math/0005047 | Let MATH be the image of MATH in MATH, and MATH the connected component containing MATH. By REF, the square root MATH coincides with the square root of the eigenvalue of the action of MATH on the symplectic vector space MATH. As in the simply connected case, it is enough to consider the case of REF-punctured torus MATH. We adopt the notation and REF from the proof of REF . Suppose MATH is invariant under both of the commuting operators MATH and MATH. From the formula for MATH it follows that MATH as a MATH-orthogonal direct sum. Take MATH to be the direct sum of joint eigenspaces for MATH with eigenvalues MATH, MATH or MATH. Then MATH . Hence MATH restricts to minus the standard symplectic form on MATH. Therefore the square root for the action on the canonical bundle MATH is MATH. Consider on the other hand the MATH contribution. We claim that the symplectic form MATH on MATH is homotopic to the standard symplectic form MATH, by a homotopy of symplectic forms which are invariant under the action of MATH. This will imply that the square root for the action on MATH is MATH, and finally MATH . For the rest of this paragraph we consider the restriction MATH of MATH to MATH. For MATH let MATH. It suffices to show that MATH for MATH. We calculate MATH . For MATH one has MATH. Hence it suffices to show that all eigenvalues of the symmetric operator MATH are strictly smaller than MATH. But MATH . On the joint eigenspace of MATH with eigenvalue MATH, this is strictly negative. On the orthogonal complement of the eigenspaces for eigenvalue pairs MATH and MATH, the operators MATH are represented by REF rotations by angles MATH not both of which are multiples of MATH. On any such REF-plane, the operator MATH becomes MATH . The claim follows since MATH. |
math/0005047 | Let MATH be the surface obtained from MATH by capping off the boundary component. Let MATH denote the kernel of the restriction map MATH. As explained in CITE, it suffices to show that the cocycle is trivial over the subgroup MATH of MATH. We would like to define a map MATH with coboundary condition MATH . Given MATH, let MATH be its image under the map REF. The element MATH defines a covering MATH of MATH by MATH, which is a possibly disconnected surface with MATH boundary components. Choose a base point of MATH mapping to the base point of MATH, and let MATH be the covering projection. The pull-back MATH admits a unique lift MATH, with MATH at the base point and MATH for all MATH, MATH. Note that MATH is constant along each of the boundary circles of MATH. The covering MATH extends to a covering MATH over the capped-off surface. Extend MATH first to MATH by the constant extension on the capping disks, and then further to a map MATH, in such a way that the extension is trivial on MATH. Define MATH . Since MATH by our assumption on the level, REF is independent of the choice of MATH. It remains to verify the coboundary condition. Let MATH and MATH their images. Define MATH as the fibered product of MATH and MATH. Let MATH denote the projections onto MATH respectively. Let MATH. We have canonical lifts MATH and MATH. Choose extensions MATH and MATH as above. Then both MATH and MATH are equal on MATH. Using REF, MATH . For the last term we compute, using the property of REF-form MATH under group multiplication MATH, MATH (where MATH are projections to the respective MATH-factor), MATH . This shows that the cocycle condition holds and completes the proof. |
math/0005047 | If MATH is contained in the identity component of the stabilizer MATH, then the formula follows from the pre-quantum condition. Our strategy is to reduce to this case, using finite covers. Since MATH fixes MATH, there exists a flat connection MATH mapping to MATH and a gauge transformation MATH restricting to MATH on the boundary such that MATH. The eigenvalue for the action of MATH on MATH is equal to the eigenvalue for the action of MATH on the fiber MATH of the level MATH pre-quantum line bundle over MATH. Let MATH be the image of MATH under the map MATH. As in the proof of REF let MATH the covering defined by MATH, and MATH be the lift of the pull-back MATH. Clearly MATH fixes MATH. The pull-back map MATH lifts to the central extensions if one changes the level. Indeed, the cocycle for the central extension at level MATH pulls back to the cocycle for the extension at level MATH. The map is compatible with the given trivializations over MATH and MATH, hence they define a commutative diagram MATH . It follows that the eigenvalue of MATH on the fiber MATH is equal to that of MATH on the fiber MATH. By REF below, MATH is contained in the identity component of MATH. Hence its action on the line-bundle is determined by the moment map (NAME 's formula), and hence is trivial since MATH pulls back to the zero connection on MATH. |
math/0005047 | It is well-known that evaluation at MATH, MATH restricts to an injective map MATH. The image MATH of this map is the centralizer of the image MATH of the homomorphism MATH defined by parallel transport using MATH. Since MATH and MATH is a torus, MATH. It follows that MATH is contained in the identity component of MATH. |
math/0005047 | Let MATH, and MATH the unique element projecting to MATH. By REF , the weight for the action of MATH on MATH is MATH. On the other hand, by REF , MATH, which completes the proof. |
math/0005047 | Recall MATH. Since MATH is fixed by MATH, the stabilizer of MATH is equal to the stabilizer of MATH, using the embedding MATH induced by MATH. By REF, the fixed point components for the action of MATH on MATH are the sub-manifolds MATH with MATH. Since MATH is diffeomorphic to MATH, we have MATH. The normal bundle MATH is MATH-equivariantly isomorphic to the constant bundle with fiber MATH. Indeed, left translation by any MATH on MATH commutes with the action of MATH, and induces a MATH-equivariant isomorphism of MATH with MATH. Therefore, MATH . Since the line bundle MATH is pre-quantum at level MATH, the integral MATH is equal to the symplectic volume with respect to the inner product MATH. By REF the symplectic volume is equal to the Riemannian volume. Since MATH, this shows MATH . Miraculously, the contribution is independent of MATH. The fixed point formula hence gives the following expression for the MATH-index of a symplectic quotient MATH, for MATH: MATH . By the computation of the phase factor in REF , MATH which completes the proof. |
math/0005047 | This is a special case of a result for non-abelian groups proved in CITE. In the abelian case, the following much simpler argument is available. Notice that MATH with diagonal MATH-action has moment map MATH not only for the fusion form MATH but also for the original symplectic REF-form MATH. Suppose MATH is a (weakly) regular value of MATH, so that MATH is a smooth sub-manifold and MATH is an orbifold. Since the pull-back of REF-form MATH to MATH vanishes, the reduced symplectic forms are the same: MATH. It follows that the two NAME measures MATH and MATH coincide. Since the symplectic volume is the integral of the NAME measure, the proof is complete. |
math/0005047 | We may assume that MATH is the exponential of a Hamiltonian MATH-space MATH. Rescaling by MATH, we obtain a family of Hamiltonian spaces MATH, together with their exponentials. Let MATH be the corresponding fusion forms. We claim that MATH give the required isotopy of symplectic forms. Indeed, each MATH is symplectic, and for MATH, MATH showing that MATH. |
math/0005047 | Decomposing into irreducible factors we may assume MATH is simple. For any simple group other than MATH with MATH even, the center MATH is a cyclic group and the claim follows by choosing a lift of the generator. It remains to consider the case MATH even which has center MATH. We use the usual presentation CITE of the root system of MATH as the set of vectors MATH in MATH, where MATH are the standard basis vectors of MATH. The basic inner product on MATH is the standard inner product on MATH, and will be used to identify MATH. We choose simple roots MATH, and MATH. Then MATH is the highest root, and the fundamental alcove MATH is the polytope defined by MATH and MATH. The following four vertices of MATH exponentiate to the central elements of MATH: MATH . To describe the homomorphism REF, consider the exceptional element MATH . We have MATH where MATH, and MATH is the NAME group element, MATH . Similarly MATH with MATH (since MATH is even) and MATH given as MATH . We construct lifts MATH of MATH in two stages. First we lift to commuting elements MATH given in terms of their action on MATH as follows, MATH . We claim that any two lifts MATH to MATH still commute. The transformations MATH restrict to rotations on the MATH-dimensional subspace MATH and on its orthogonal complement. The restrictions of MATH to the subspace MATH lift to commuting transformations of MATH since MATH is the identity. The restrictions to MATH lift to commuting transformations of MATH since MATH is two-dimensional and MATH is abelian. |
math/0005052 | Suppose MATH is REF-avoiding. Choose a reduced expression for MATH for which a pair of MATH's is as close together as possible for some MATH. These two copies of MATH must be separated by at least one of MATH, otherwise our expression would not be reduced. But then our reduced expression looks like MATH where MATH. If MATH and MATH, then MATH has a reduced expression MATH. Such a MATH is not REF-avoiding, which is a contradiction. So either MATH or MATH must contain MATH. For the reverse implication, suppose that every two copies of the same generator MATH in some reduced expression for MATH are separated by both a MATH and a MATH. It is REF, that any two reduced expressions for MATH can be obtained from each other by a sequence of moves of the following two types: MATH . But, under our hypothesis, we are never able to apply a MATH move for such a MATH. So all reduced expressions for MATH must be obtainable by a sequence of MATH moves. Hence, MATH is REF-avoiding. |
math/0005052 | Let MATH be the smallest index for which MATH. Such a MATH must exist by our stipulation that MATH. Consider the sequence MATH. Since MATH is reduced, MATH. Hence, MATH. We now investigate the differences MATH for MATH. There are four possibilities (note that in each case, MATH): CASE: MATH, MATH. Then MATH, MATH. So MATH. CASE: MATH, MATH. Then MATH, MATH. So MATH. CASE: MATH, MATH. Then MATH, MATH. So MATH. CASE: MATH, MATH. Then MATH, MATH. So MATH. So, the only cases we need to consider are the second and the fourth. From this it follows that for each MATH, MATH . The conclusion of the lemma follows by induction upon setting MATH. |
math/0005052 | A picture is given in REF The claim follows immediately from Lateral Convexity by applying REF to the pairs MATH and MATH. Proof of REF . First consider the case where MATH for MATH. This is illustrated in REF By REF, MATH is in MATH. Since MATH, MATH contains the indicated hexagon by REF. Alternatively, we can have MATH for MATH. This is illustrated in REF Recall that the MATH are assumed to be distinct. So, starting at MATH, MATH must move down to the right at least twice (to cross MATH and MATH), and move down to the left at least once (to cross MATH). Hence, the lowest of the three crossings MATH must occur in MATH. By REF, MATH must therefore contain a hexagon. Proof of REF . By REF, in order to avoid a hexagon in MATH, we need at least one of MATH to be a distance of exactly MATH from MATH. Suppose first that both MATH and MATH. Then we are in the situation of REF Note that if MATH then MATH and we can appeal to REF. So we can consider only the case where there is a crossing at MATH. If MATH is either MATH or MATH, then it still needs to cross a string currently to its right (either MATH or MATH, respectively). This can only happen in MATH. The only alternative is that MATH. But then MATH cannot cross until MATH. Either way, MATH. Arguing analogously with MATH, we see that MATH. So MATH contains a hexagon. Now suppose that only one of MATH is a distance of MATH away from MATH. Without loss of generality, we assume that this point is MATH. We argue depending on whether or not MATH where MATH. Assume first that MATH. We are in the situation of REF Since MATH, MATH for some MATH. Hence, in order to avoid a hexagon, we must have MATH cross as shown. But then it is easily seen that the crossing MATH must occur in MATH. This ensures that MATH contains the indicated hexagon. If MATH, then we are in the situation of REF Since MATH must go left once below MATH (to cross MATH) and MATH must go right once (to cross MATH), we see that the lowest of the crossings MATH must occur in MATH. If MATH, then by REF, MATH contains a hexagon. If MATH, then we need the additional fact that MATH to ensure that MATH. But this follows from the assumption that MATH is not at a distance of MATH from MATH. |
math/0005052 | Assume that MATH is not a forest - that is, MATH contains a cycle. We will assume that MATH is REF-avoiding and show that if MATH contains a cycle then MATH contains a hexagon. Note that since MATH is REF-avoiding, REF (Lateral Convexity) holds. Let MATH be a minimal subset such that the subgraph MATH of MATH spanned by MATH is a cycle. Hence, for each MATH, MATH has degree at least REF. Choose MATH as large as possible such that MATH is on the line MATH for some MATH. Now choose MATH to be minimal among such MATH. By choice of MATH, MATH must be shared and we must have MATH for some MATH. So our heap looks like REF In the discussion that follows, ``shared" should be interpreted in the context of MATH. Since MATH is minimal, either MATH for some MATH, or MATH is shared. In the first case, MATH must be in MATH by Lateral Convexity. Consider the second case - where MATH is shared. By REF, MATH. So MATH for some MATH. So in both cases, we have the following fact which we state for reference. If MATH, then MATH. Two other simple facts we state for reference are the following. By Lateral Convexity, any point encountered by a string that still needs to cross below that point must be in the heap (after pushing together connected components). For example, if MATH, then MATH must be in the heap. Recall that MATH is defined as right critical zero of the left critical zero of MATH (see REF). If MATH does not contain a hexagon, then the point MATH must lie along the boundary of MATH. We now show that, regardless of the characteristics of MATH (that is, values of MATH, MATH, and whether or not MATH), MATH must contain a hexagon. Suppose that MATH. By REF, the only way this can happen is if the other string encountering MATH is MATH. Since MATH is minimal, we then need either MATH or MATH shared. Consider REF. Suppose MATH is shared. By choice of MATH on the line MATH, this implies that MATH. But then by REF, MATH. Then by Lateral Convexity, MATH. The alternative is that MATH is shared. Again, this implies that MATH. Since MATH by REF , MATH contains a hexagon. So we can assume that MATH. But by choice of MATH, MATH must be shared. This implies that MATH for some MATH. We now argue that MATH must contain a hexagon according to the position of MATH relative to MATH. CASE: MATH for MATH. There are three cases to consider. REF depicts the first. Here, MATH and MATH both encounter MATH. Since MATH is minimal, either MATH or MATH must be shared. By choice of our line MATH and the fact that MATH, we see that, in fact, MATH must be shared. But then MATH. Since MATH, MATH contains the indicated hexagon. The second alternative is that MATH but MATH does not encounter MATH along any of the nodes between MATH. This is depicted in REF If MATH, then MATH must cross in MATH. If MATH , then MATH must cross in MATH. In either case, MATH must contain the indicated hexagon. The third possibility is that REF . In fact, this is the only possibility for MATH when MATH. Here we see that the path of MATH must be as shown in order to avoid MATH. But then MATH cannot cross until MATH. So we have the indicated hexagon. CASE: MATH for some MATH. The situation is depicted in REF For both MATH and MATH to cross outside of MATH, we need MATH to cross in MATH. This is shown in REF We mention three additional assertions we have made in REF First, MATH must cross MATH as shown in REF in order to avoid having MATH contain a hexagon. Second, MATH by REF . Third, since MATH must be shared, MATH as shown. So, by Lateral Convexity, MATH contains the hexagon indicated in REF (It is possible that MATH or MATH, but this does not change our conclusion.) |
math/0005052 | Partition MATH where MATH consists of all masks in MATH ending in MATH for MATH. There are natural bijections MATH and MATH given by MATH. So, to prove the lemma, we need only compare MATH to MATH. If MATH, then MATH. In this case, if MATH (MATH), then MATH, so MATH. Alternatively, if MATH (MATH), then MATH and MATH. This accounts for the first term in REF. Since MATH, proof of the second term in REF reduces to the above case. |
math/0005052 | CASE: Assume MATH is REF-hexagon-avoiding. We need to show that the MATH are the NAME polynomials. Now, every MATH has three critical zeros. Furthermore, by REF, no point is a critical zero for REF distinct defects. So the number of edges in MATH equals the number of shared critical zeros. Hence, MATH . Now, by REF, MATH is a forest with MATH vertices. Hence, MATH has at most MATH edges (see, for example, CITE). Hence, MATH . So by REF, MATH. Therefore the inequality MATH holds. Now apply REF, from which it follows that MATH for all MATH. CASE: We shall prove (not REF) MATH (not REF). Assume MATH is not REF-avoiding. We can find a reduced expression for MATH of the form MATH with MATH. Set MATH . Then MATH and MATH. By REF, MATH. So MATH does not satisfy the properties of the MATH listed in REF. Now assume MATH is REF-avoiding but not hexagon avoiding. Then we can write MATH where MATH as in REF and MATH. Set MATH . The mask MATH is depicted graphically in REF. Then MATH and MATH. By REF, MATH. So MATH does not satisfy the properties of the MATH listed in REF: We first appeal to a result of NAME and NAME relating the intersection NAME polynomial of the NAME variety MATH to the NAME polynomials MATH CITE: MATH . Now, we are assuming that MATH for all MATH. So we need only show that MATH . We proceed by induction, the result being obvious for MATH. Choose a MATH such that MATH. From CITE, we know that: MATH . Using REF, along with REF, we can write MATH . If MATH, then MATH, so this becomes MATH . The last line is by the induction hypothesis. REF: CITE proves that for any NAME group MATH, we can always find a subset MATH such that MATH for all MATH. (More generally, he shows that such a MATH exists when the coefficients of MATH are already known to be non-negative. Due to their interpretation in terms of dimensions of intersection cohomology groups, this is known for any NAME group.) Hence, for such a MATH, we have the following string of equalities: MATH . Setting MATH, we find that MATH. But then MATH. So MATH for all MATH. CASE: This follows from CITE. CASE: This is the content of CITE. CASE: This is a standard result on small resolutions. See, for example, CITE. CASE: Recall that MATH denotes the NAME resolution of MATH (corresponding to some reduced expression MATH). By CITE, MATH . We are assuming that MATH. By NAME duality, we know that MATH. Combining REF with this isomorphism yields MATH as desired. This completes the proof of REF. |
math/0005052 | As a permutation, MATH . This is easily seen to be REF-hexagon-avoiding. So by REF, MATH. The claim is true for MATH. The proof is by induction. The situation of the general case is illustrated in REF for some MATH. Let MATH. In REF, no new defect is introduced by MATH, so MATH. In REF, we have MATH. The claim follows by the induction hypothesis. |
math/0005052 | We only sketch the proof. We see that MATH is clearly REF-hexagon-avoiding, so by REF, MATH. The idea is to use recursion on MATH. From REF, it is easy to see that MATH. Similar recurrences can be found for MATH where MATH. Solving these recurrences for MATH yields REF. |
math/0005052 | It has been proved by CITE that for MATH and MATH, MATH is smooth on the NAME cell MATH if and only if MATH. By REF, MATH for every MATH. So to show that MATH is singular, we need only show that MATH contains a mask of positive defect. Let MATH correspond to MATH. Since every defect must have two critical zeros (in addition to the defect itself), MATH. Lateral Convexity tells us that if MATH for some other MATH, then MATH. So for MATH, if MATH is singular at MATH, MATH is maximally singular. Now, the conditions in REF imply that MATH. By REF, this implies that MATH. So MATH. The only fact that remains to be checked is that if MATH is a singular point of MATH, then MATH for some MATH. So pick some MATH with MATH. Choose MATH and suppose MATH and MATH. Now define a mask MATH by setting MATH . Using the characterization of NAME order in terms of subexpressions (see, for example, CITE), it is easily checked that MATH. Since MATH is in MATH, we are done. |
math/0005054 | By translating the sets MATH if necessary, we may assume that each MATH contains the origin. By hypothesis, there exists a packing of MATH into each MATH, so we may choose MATH for each MATH. The set MATH is closed and bounded, hence compact, and so by REF the space MATH is also compact. Since the sets MATH all contain the origin, the space MATH is contained in MATH by REF ; we know by REF that MATH is a closed set, and so it is itself compact by REF . Therefore by REF , the sequence MATH of points in MATH has a convergent subsequence. By replacing the sequence MATH by this subsequence, we may assume that the MATH converge to some element MATH. It remains to show that this element MATH in fact represents a packing of MATH into MATH. For each MATH, the sequence MATH is contained (except for at most the first MATH terms) in MATH. Since this set is closed by REF , we see by REF that the limit MATH is itself an element of MATH. Because this is true for all MATH, REF implies MATH which establishes the lemma. |
math/0005054 | For any MATH and any pair MATH, MATH of elements of MATH, we have MATH . We certainly have MATH by REF of the metric MATH. On the other hand, all entries of the matrix MATH are also at most MATH in absolute value, while the entries of the vector MATH are at most MATH in absolute value. Therefore each entry of MATH is bounded by MATH in absolute value, and so REF becomes the upper bound MATH (we have made no effort to obtain a strong constant in the inequality). Now if MATH is an element of MATH such that MATH lies in the open set MATH, then there exists some positive number MATH such that MATH. If we set MATH, then for any MATH such that MATH, the upper bound REF tells us that MATH and therefore MATH as desired. |
math/0005054 | Since MATH and MATH are open sets that are not disjoint, we can choose a point MATH and a positive number MATH such that MATH. Using REF we may set MATH and MATH, so that MATH and MATH; we also set MATH . Then for MATH or REF, for any MATH such that MATH the upper bound REF tells us that MATH so that MATH by REF . In particular, this shows that MATH is an element of MATH, which is therefore nonempty as desired. |
math/0005056 | For such a choice of MATH, the operator MATH descends to give a homogeneous differential operator on MATH. The symbol of the NAME operator is NAME multiplication, which is invertible, so the operator is elliptic. We may therefore apply REF to compute its index, and by REF we have MATH using the homogeneous NAME REF in the final line to decompose the virtual representation MATH into irreducible representations of MATH. Finally, we have MATH if and only if MATH for MATH. |
math/0005056 | As in the proof of REF , we use the homogeneous form REF of the NAME theorem to decompose the kernel of MATH as the direct sum MATH over the finite dimensional NAME operators MATH acting on the spaces MATH where all but finitely many of these kernels vanish since the operator MATH is NAME. Ignoring the signs of the half-spin representations, REF for the kernel of MATH implies that the kernel of MATH is MATH if MATH for some MATH and MATH otherwise. |
math/0005057 | To compute the central charge of the spin representation MATH, we extend it to obtain the spin representation associated to the entire NAME algebra MATH. Since the construction of spin representations is multiplicative, we have MATH . These two spin representations have the same central charge since they differ only by the finite dimensional factor MATH. However, the extended spin representation MATH admits an action of the full NAME algebra MATH, and in fact, MATH is the direct sum of MATH copies of an irreducible positive energy representation of MATH. Examining the structure of this representation, the first three energy levels of MATH are as follows: MATH . Letting MATH denote the highest root of MATH, and MATH the central charge of MATH, the highest weights of MATH and MATH are then MATH and MATH respectively, while the weight MATH is not present in MATH. The weights MATH and MATH thus form a complete string of weights for the root MATH, and so they must be related to each other by the affine NAME element MATH, the reflection through the hyperplane orthogonal to MATH. By REF , the difference of these weights is MATH, so we obtain MATH where MATH and MATH in the basic inner product since MATH is the highest root. The central charge of the spin representation is thus MATH. The quadratic NAME operator of a NAME algebra does not depend on the choice of orthonormal basis, and it commutes with the action of the NAME algebra. It therefore acts by a constant times the identity on each irreducible representation. On the irreducible representation of highest weight MATH, the value of the NAME operator is MATH. In particular, if MATH is simple, then the adjoint representation is irreducible, and taking MATH to be the highest root of MATH, which satisfies MATH in the basic inner product, we again obtain the value MATH as desired. |
math/0005057 | We first note that the construction of the spin representation is multiplicative, provided that the underlying vector spaces are even dimensional. In our case, the positive and negative energy subspaces pair off, while for the zero modes, the maximal rank condition implies that the complement of MATH in MATH and the complement of MATH in MATH are even dimensional, so we have MATH . Applying the NAME character REF to the left side of REF , and factoring the NAME denominator using REF , we obtain MATH . Recall that the affine NAME group acts simply transitively on the NAME alcoves. Due to the MATH shift, the weight MATH lies in the interior of the fundamental NAME alcove for MATH, and thus for any MATH, the weight MATH likewise lies in the interior of some NAME alcove. Furthermore, the NAME alcoves for MATH are completely contained inside the NAME alcoves for MATH, and so there exists a unique element MATH such that MATH lies in the interior of the fundamental NAME alcove for MATH. Shifting back by MATH, we see that the weight MATH is anti-dominant for MATH. Putting MATH, we can write MATH, and more generally we have MATH. Using this decomposition to rewrite the numerator on the right side of REF , we have MATH where the second line follows by applying the NAME character REF for MATH. This proves the character form of the identity REF . |
math/0005057 | First, we show that the operators MATH and MATH are super-derivations with respect to the NAME multiplication REF . For MATH and MATH, we have MATH . Now, to prove the identities REF , we need only verify them for the generators MATH, but it follows from the definition of the NAME algebra that MATH and by applying REF and the identity MATH, we obtain MATH . To prove REF , we first verify that it holds when acting on a generator MATH: MATH . Finally we show that MATH is a super-derivation for NAME multiplication: MATH where we use the expansion MATH . |
math/0005057 | All of the commutation relations follow immediately from REF and the above discussion with the exception of that for MATH. For example, we derive MATH . To compute MATH, we note that the fundamental REF-form is closed, so we have MATH . Written in terms of an orthonormal basis MATH for MATH, the fundamental REF-form is MATH and so its norm is given by MATH which yields the desired anti-commutator MATH. |
math/0005057 | Since the NAME group acts by isometries, the weights MATH satisfy the identity MATH and it follows from REF that the NAME operator MATH vanishes on any MATH-invariant subspace of MATH transforming like MATH. To complete the proof, it remains to show that each of these representations occurs exactly once in the domain of the NAME operator and that no other MATH-representations appear in its kernel. We establish these facts in the following two lemmas (see also CITE). |
math/0005057 | The highest weight space of an irreducible representation of MATH is always one dimensional, so the weight MATH appears with multiplicity REF in MATH. Now consider the complex spin representation MATH associated to the orthogonal complement of a NAME subalgebra MATH in MATH. Given a positive root system for MATH, the character of this spin representation is MATH, and so the highest weight MATH of MATH appears with multiplicity REF. The highest weight of the tensor product MATH is then MATH, appearing with multiplicity REF, and likewise the weights MATH for MATH all have multiplicity REF in MATH. Choosing a common NAME subalgebra MATH, the spin representation factors as MATH. As we noted above, the weight MATH appears with multiplicity REF in the second factor MATH. It follows that the weights MATH for MATH can appear at most once in the tensor product MATH, as each such weight contributes one weight of the form MATH to the tensor product MATH. On the other hand, we see from the homogeneous NAME REF that the irreducible representations MATH for MATH appear at least once in the decomposition of the tensor product MATH. We therefore conclude that the representations MATH for MATH each occur exactly once in MATH. |
math/0005057 | If MATH is a weight of MATH, then MATH is a weight of the tensor product MATH. Since the NAME group acts simply transitively on the NAME chambers, there exists an element MATH such that MATH is dominant, where we recall that a weight MATH is dominant if and only if MATH for all positive roots MATH. Note that every weight of MATH can be obtained from its highest weight MATH by subtracting a sum of positive roots. Likewise, for the tensor product MATH, the difference MATH is a sum of positive roots, and it follows that MATH with equality holding only when MATH. As for the uniqueness of MATH, if MATH is dominant, then the weight MATH lies in the interior of the positive NAME chamber for MATH, and thus the weights MATH for MATH are distinct. |
math/0005057 | The bracket MATH is simply the definition of the NAME algebra, while the brackets MATH and MATH and MATH follow immediately from REF . By the NAME identity, for any MATH and MATH we have MATH which shows that MATH is a projective representation of MATH on MATH. In REF , we established that this spin representation has central charge MATH, which gives us the brackets MATH and MATH. To compute MATH, we write it as the sum MATH of homogeneous forms of degrees REF. (We shall see that MATH has no components of degrees REF or REF.) Since MATH for MATH is a derivation with respect to the backet, we have MATH . Taking one further interior contraction, we obtain MATH which is a constant. It follows that MATH has no components of degree higher than REF, and that MATH is REF-cocycle determining the central extension of MATH for the spin representation MATH. In fact, this REF-cocycle is a multiple of MATH, and we have MATH . Going back up one level, we see that MATH . Finally, the value of the constant MATH is the value of MATH acting on the minimum energy subspace MATH of the spin representation, since MATH vanishes there. However, all the terms in MATH vanish on MATH except the contribution from the constant loops, and thus MATH as we proved in REF . |
math/0005057 | In order to simply our calculations, we first note the following identites: MATH . We now show that the commutator of MATH with an element MATH is MATH . It follows that the operator MATH commutes with the action of MATH, and therefore takes a constant value on each irreducible representation. Acting on the minimum energy subspace MATH of MATH, the only terms contributing to MATH are those coming from the constant loops, and thus this constant is MATH . The desired result then follows immediately. |
math/0005057 | Since MATH is MATH-equivariant, it respects the decompositions of MATH and MATH into their constant energy subspaces, and it can be written in the block diagonal form MATH, with MATH. If both MATH and MATH are of finite type, then each of the subspaces MATH and MATH is a finite dimensional MATH-module, and so the MATH-equivariant index of MATH is given by the MATH-valued formal power series MATH . Since a representation of the full loop group MATH is uniquely determined by its constant energy components, the MATH-equivariant index must therefore be the difference of the domain and the range, hence MATH. |
math/0005060 | The estimates in REF are immediate. The proof of REF is also an easy estimate, which can be found in CITE, for example. The arguments for REF are also quite standard. We leave the proof for the reader. Let us see that REF holds. If MATH and MATH are concentric, the identity MATH is a direct consequence of the definition. In case MATH and MATH are not concentric we have to make some calculations: MATH . So we must show that MATH . We set MATH . The integral MATH is easily estimated above by some constant MATH, since MATH for MATH. An analogous calculation yields MATH. For MATH we have MATH and we are done with REF . We leave the proof of REF for the reader too. |
math/0005060 | Let MATH be the biggest cube centered at MATH with side length MATH, MATH, such that MATH. Then, MATH. Otherwise, MATH and since MATH and MATH we get MATH which contradicts the choice of MATH, assuming MATH. Now we have MATH . Thus MATH . Let MATH be the smaller doubling cube of the form MATH, MATH. Then MATH. Also, MATH. Otherwise, MATH and MATH . This is not possible if we assume MATH. Now MATH satisfies the required properties, since it is doubling, it is contained in MATH, and MATH . |
math/0005060 | Suppose, for example, MATH. Then, MATH. Let MATH be the smallest doubling cube of the form MATH, MATH. We have MATH . Thus MATH . We also have MATH . Since MATH and MATH have comparable sizes, MATH, and so MATH . Therefore, MATH . By REF , the proposition follows. |
math/0005060 | Let MATH be an atomic block supported on some cube MATH, with MATH, where MATH are functions supported on cubes MATH such that MATH. We will show that MATH. First we will estimate the integral MATH . For MATH and MATH, since MATH, we have MATH . Thus MATH . Now we will show that MATH and we will be done. If MATH and MATH, then MATH . So MATH . For MATH and MATH, we have MATH . Therefore, MATH and REF follows. |
math/0005060 | The arguments are quite standard. For any MATH there exists some cube MATH which contains MATH, with MATH and such that MATH. Then by NAME 's Covering Theorem, there are points MATH such that MATH and so that the cubes MATH, MATH, form an almost disjoint family. Observe that REF cannot be applied to the cubes MATH (they are non centered), however we have applied it to the cubes MATH, which are non centered too, but fulfil the condition MATH . That is, the point MATH is ``far" from the boundary of MATH. Under this condition, NAME 's Covering Theorem also holds. Since, for each MATH, MATH and MATH, it is easlily seen that MATH. Then, MATH where MATH is some constant that will be fixed below. Now, we have MATH . The last inequality follows from MATH (notice that the cube MATH is MATH-doubling). Therefore, by REF (which also holds for cubes that are MATH-doubling instead of MATH-doubling, with constants MATH and MATH instead of MATH and MATH) we have MATH . So if we choose MATH, we get MATH . |
math/0005060 | Both statements are a straightforward consequence of REF , since MATH and MATH . |
math/0005060 | We can assume MATH. Let MATH be some fixed constant. If MATH, then MATH. So, if MATH is a cube centered at MATH with side length MATH, we have MATH. By REF we get MATH . Since MATH if we set MATH, we obtain MATH . Then for MATH small enough we have MATH . This implies MATH, which is not possible. |
math/0005060 | By the previous lemma, MATH and MATH. This gives MATH. |
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