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math/9910036 | Assume that MATH is MATH but neither MATH nor MATH . By applying NAME twists and in light of the analysis carried out in CITE, one may assume that MATH has character MATH on MATH with signs taken together. We will only address representations with character MATH as the arguments for MATH are practically identical. Up to conjugation, we may choose MATH . The group MATH consists of: REF tetrahedral matrices in the previous lemma, the four matrices MATH (signs in the second row determined by those in the first), and REF matrices obtained by permuting two non-zero terms REF and two zero terms REF in the first row of the matrix. One such example is: MATH . Let MATH with MATH be such that MATH is not contained in the isomorphic copy of MATH generated by MATH . Note that MATH and MATH . If each of MATH and MATH are MATH, then the argument produced in the previous lemma applies. Suppose that MATH is MATH and MATH is MATH . This leads to the system of equations: MATH and MATH . This system has non-real solutions. A similar argument holds in the event that MATH is MATH and MATH is MATH . We indicate how to proceed in the case where MATH is MATH and MATH is MATH. Since MATH is MATH, MATH and MATH must take on the values MATH or MATH . The case where both are zero reduces to MATH. We explicitly work out the case MATH and MATH is MATH . This leads to the equations MATH and MATH . This system has complex and real solutions, with real solution MATH which implies that MATH a contradiction. Similar arguments hold in the remaining REF cases obtained by assigning the values MATH to MATH and MATH as well as REF cases where MATH is MATH and MATH is MATH. All cases yield systems with only non-real solutions, or matrices MATH . This leaves cases where MATH and MATH are either MATH or MATH . In these cases, MATH, MATH, and MATH must take on the values MATH or MATH . We will work out the case in which MATH and MATH . The solutions to the associated system are: MATH with signs taken together. However, the trace of MATH is MATH . So the handle MATH is neither MATH nor MATH. The remaining REF cases are obtained by varying the values of MATH and MATH in MATH . The arguments proceed similarly to those exemplified above and lead to: CASE: matrices MATH inside MATH (a contradiction), CASE: a handle MATH that is neither MATH nor MATH CASE: one or more of MATH is not real (a contradiction). |
math/9910036 | Assume that MATH is MATH . By applying NAME twists, by the fact that MATH is in the center of MATH and by the previous lemmas, we may assume one of the following REF cases: CASE: MATH and MATH . CASE: MATH and MATH . CASE: MATH and MATH . CASE: By the uniqueness of coordinates, let MATH . The group MATH consists of: REF tetrahedral matrices (see above lemma), REF matrices MATH REF matrices MATH REF matrices MATH (signs in the second row determined by those in the first), and REF matrices obtained by multiplying those listed above by: MATH . Let MATH with MATH be such that MATH is not contained in the isomorphic copy of MATH generated by MATH . Note that MATH and MATH . If each of MATH and MATH are MATH, then the argument produced in REF applies. Suppose that MATH is MATH and MATH is MATH . This leads to the equations MATH and MATH . The solutions are MATH. However, MATH multiplied on the right by MATH is one of the matrices listed in MATH so MATH . A similar argument holds in the case where MATH is MATH and MATH is MATH . We indicate how to proceed in the cases where MATH is MATH and MATH is MATH. Since MATH is in MATH, MATH and MATH must take on the values MATH or MATH . The case where both are zero reduces to a MATH case. We will explicitly work out the case MATH and MATH . This leads to the system: MATH and MATH . These equations have no real solutions. Similar arguments hold in the other REF cases and REF cases where MATH is MATH and MATH is MATH. Note that some of these cases yield matrices MATH that are in MATH . This leaves cases where MATH and MATH are either MATH or MATH . In these cases, MATH, MATH, and MATH must take on the values MATH or MATH . First, we explicitly work out a special MATH case in which MATH and MATH are all zero. The solutions are MATH and MATH (signs taken together). However, the matrix MATH is in MATH a contradiction. We also explicitly work out the case MATH and MATH . The three resulting equations yield MATH . Thus, since MATH we see that MATH which violates MATH . The remaining REF cases are obtained by varying the values of MATH and MATH in MATH . The arguments proceed similarly to those exemplified above and lead to: CASE: matrices MATH inside MATH (a contradiction), CASE: a handle MATH that is not MATH nor MATH nor MATH, CASE: one or more of MATH is not real (a contradiction). CASE: By the uniqueness of coordinates, let MATH . Then MATH yields REF matrices in REF . A study of each of the MATH and MATH cases follows as in the previous case. As usual, each case leads to: CASE: matrices MATH inside MATH (a contradiction), CASE: a handle MATH that is not MATH nor MATH nor MATH, CASE: one or more of MATH is not real (a contradiction). CASE: By the uniqueness of coordinates, let MATH . Then MATH yields REF matrices in REF . A study of each of the MATH and MATH cases follows as in REF . As usual, each case leads to: CASE: matrices MATH inside MATH (a contradiction), CASE: a handle MATH that is not MATH nor MATH nor MATH, CASE: one or more of MATH is not real (a contradiction). |
math/9910038 | The minimum principle associated with the first MATH invariant eigenvalue problem, MATH states that MATH for all MATH invariant functions MATH with MATH. Equality holds if and only if MATH is an eigenfunction for MATH. Since MATH consists of constant functions and MATH, we see that MATH is an admissible solution of REF and therefore MATH . Equality holds if and only if MATH is the first MATH invariant eigenfunction. In this case, upon substitution of MATH into REF we obtain the equivalent equation MATH. Recalling that MATH and MATH must satisfy certain boundary conditions forces MATH and yields the unique solution MATH. In other words, MATH. |
math/9910038 | From REF MATH and MATH. Each of the sequences MATH is positive and strictly increasing so by truncating the above series after MATH terms and then replacing each term with the smallest one we obtain MATH . This produces the desired inequalities. |
math/9910038 | By REF , since MATH, then MATH. Upon integrating by parts we have MATH so that MATH . So there exists MATH with MATH, thus, by REF , precluding the possibility of an isometric embedding. |
math/9910038 | To prove this theorem we will find a MATH invariant metric of area MATH with this property. By REF, MATH is even if and only if MATH is not a MATH invariant eigenvalue, that is, if and only if MATH for any MATH. It is now clear that the first four multiplicities are even if and only if MATH, and, by REF, this will occur if our metric satisfies MATH. Using a variational principle, as in CITE, for the operator MATH, we obtain the upper bound: MATH such that MATH. Comparing this upper bound with the lower bound on MATH provided by REF ., the proof of this theorem may now be reduced to finding a function MATH and a suitable test function MATH such that MATH . We claim that MATH and MATH will satisfy REF . It is not difficult to see that MATH for this choice of MATH. So the right hand side of REF is greater than REF. Calculating the left hand side of REF for this choice of MATH and MATH yields: MATH where the first integral in brackets has been approximated in the obvious way. Since the left hand side is less than REF and the right hand side is greater than REF, the proof is finished. |
math/9910038 | Without loss of generality, we may assume the area of the metric is MATH. As seen in the proof of REF , the first four eigenvalues have even multiplicity if and only if MATH. This result will then follow from REF as long as we can prove that MATH. This is most easily accomplished by contradiction. Assume MATH so that MATH. Now each MATH for MATH must satisfy MATH for some MATH and MATH. However, by REF , MATH so if MATH it must be the case that MATH. By REF MATH so there are only three (possibly) distinct eigenvalues with these properties and their values coincide with MATH, MATH, and MATH. There are, therefore, at most three distinct values for the four distinct eigenvalues MATH for MATH, but this contradicts the pigeonhole principle. |
math/9910038 | If MATH satisfies the hypothesis for MATH then the statement of this result is simply REF as can be seen from REF . Hence there exists a point MATH such that MATH. If MATH satisfies the hypothesis for MATH, then as is well known, we must consider all cases with MATH. If MATH or MATH, then, again, MATH and the proof is the same as the previous case. If MATH then MATH (see CITE, p. REF) so that MATH and constant, and the statement of this result is simply, as we already know, that one of the first four, REF or REF-form, eigenvalues has odd multiplicity (in fact they all have odd multiplicity). Finally, if MATH then MATH. If, in this case, the hypothesis holds for MATH only then MATH is, topologically, the sphere and thus the statement of this theorem reduces to REF . |
math/9910039 | It is clear that the convex hull belongs to the set described above. It thus suffices to show that the only extreme points of the above set are the permutations of MATH. Let MATH be an extreme point; by symmetry we may assume that MATH. If MATH for some MATH, then we may modify MATH by a small multiple of MATH in either direction without leaving the set, contradicting the extremality of MATH. Thus MATH for all MATH, so that MATH, as desired. |
math/9910039 | For the first statement it suffices to prove that all tuples satisfying REF, and REF are contained in the convex hull of MATH. This in turn follows immediately from REF and the observation that MATH contains all tuples which have MATH elements equal to MATH, MATH elements equal to MATH, and the remaining element equal to MATH. The second statement follows immediately from the observation that the right hand side of REF is greater than or equal to one with strict inequality in case MATH. To see the third statement, assume by symmetry that MATH has bad index MATH. Define MATH if MATH and MATH. Then REF shows that MATH, so we can enlarge the entries of MATH so that MATH and MATH for MATH. We can write MATH where MATH is some tuple with MATH. Then a similar application of REF as before implies that MATH is in the convex hull of those elements MATH for which MATH. This implies the third statement of REF . |
math/9910039 | Let MATH be a good tuple. By symmetry we can assume that MATH. Then we have MATH and MATH for all MATH. By REF we find MATH such that MATH where MATH is an admissible tuple in MATH with bad index MATH. We can arrange that MATH. For MATH let MATH be the best constant such that MATH for all sets MATH of finite measure with MATH and functions MATH. Let MATH be the supremum of all MATH. By the a priori smoothness and support assumptions on MATH, MATH is finite and the point is to prove that it is bounded. By splitting MATH appropriately and using restricted type MATH from REF we obtain MATH . On the other hand, if MATH then we can use REF to find an index MATH such that MATH . By splitting MATH appropriately and using restricted type MATH from REF we obtain MATH . Since MATH is negative, we can iterate the previous inequality to obtain for sufficiently large MATH: MATH . Combining this with REF gives MATH which proves boundedness of MATH. |
math/9910039 | By symmetry we can assume that MATH for MATH and MATH for MATH for a certain MATH. REF implies MATH for all MATH in a small neighborhood of MATH satisfying MATH . Fix functions MATH. Then NAME interpolation as in CITE implies MATH for all functions MATH, MATH. |
math/9910039 | Fix MATH; by symmetry we may assume that MATH has bad index MATH. By REF we find MATH such that MATH where MATH is an admissible tuple in MATH with bad index MATH. We have MATH and by the last statement of REF we can assume that MATH is negative for all MATH. For MATH let MATH be the best constant such that for all sets MATH of finite measure with MATH there is a major subset MATH of MATH such that MATH for all MATH. Let MATH be the supremum of all MATH. Using restricted type MATH from REF we obtain MATH . On the other hand let MATH then we can find an index MATH such that MATH and we can use restricted type MATH from REF to conclude MATH . Since MATH is negative, we can iterate the previous inequality to obtain for sufficiently large MATH: MATH . Together with REF this proves bondedness of MATH. |
math/9910039 | We assume for simplicity that MATH for all MATH. If this was not the case, we could freeze the function MATH and the exponent MATH whenever MATH and run the argument on the remaining functions only, as done in the proof of REF . Let MATH be functions such that MATH for MATH. We have to show that MATH . We may assume that the MATH are non-negative. By a measure-preserving rearrangement, we may assume that the MATH are supported on the half-line MATH and are monotone non-increasing on this half-line. Let MATH denote the function MATH. We can expand the desired estimate as MATH . Since MATH, we have the elementary inequality MATH so it suffices to show that MATH . By symmetry we may restrict the summation to the region MATH . Fix MATH. Let MATH be arbitrary, and consider the set MATH . Let MATH be an admissible tuple close to MATH; we may thus assume MATH has bad index MATH. Since MATH is of restricted type MATH, and MATH, we may thus find a major subset MATH of MATH such that MATH . By definition of MATH, we thus have MATH . Solving for MATH, and optimizing in MATH, one obtains MATH for some MATH, where MATH. By symmetry one may obtain the same bound when MATH is replaced by MATH . Integrating this over all MATH, one then obtains MATH . To prove REF, it thus suffices to show MATH . Write MATH. For fixed MATH and MATH there are at most MATH choices of MATH. Fixing MATH, and then applying NAME 's inequality using REF, we can estimate the left-hand side of REF by MATH . The MATH sum is convergent, and the expression inside the product is essentially MATH. The claim is thus proved. |
math/9910039 | Suppose first that MATH intersected MATH. Then there exists MATH such that MATH for all MATH. This implies that MATH for all MATH, which implies REF. Now suppose that MATH was such that MATH was disjoint from MATH, but MATH intersected MATH. By arguing as before we see that MATH for all MATH, and MATH, which again implies REF. |
math/9910039 | By translation and scale invariance we may make MATH the unit cube MATH. We may write MATH, where MATH is supported on MATH and satisfies the same bounds REF as MATH; in other words, MATH is a bump function on MATH. Since MATH it suffices to show the estimate MATH . From NAME 's theorem and REF one sees that MATH where MATH and the kernel MATH satisfies the estimate MATH for arbitrarily large MATH. In particular, we have MATH and the claim follows. |
math/9910039 | Since MATH is non-degenerate, we can write it as a graph MATH where MATH is a linear map from MATH to MATH. Let MATH, MATH denote the centers of MATH and MATH respectively. From the definition of MATH we have MATH and MATH and similarly for MATH. Since MATH contains MATH, we have MATH . Combining this with REF we see that MATH which implies that MATH for all MATH as desired. Now suppose MATH. By subtracting REF for MATH and MATH we thus have MATH which implies that MATH . On the other hand, from REF we have MATH . If MATH is sufficiently large, this guarantees that there exist MATH such that MATH which combined with the previous observations gives MATH and MATH as desired. |
math/9910039 | We first deal with the contribution of those multi-tiles MATH such that MATH. From REF there are only MATH of these multi-tiles, and the contribution can be handled by the estimate MATH . Now let us consider those multi-tiles for which MATH. From REF there exist MATH, MATH such that MATH for MATH; by pigeonholing we may make MATH, MATH independent of MATH. If one then uses REF for all MATH, one reduces to showing that MATH . But this follows from NAME. |
math/9910039 | The expression in the norm is a variant of a NAME square function. Thus, we shall use NAME techniques to prove this estimate. By frequency translation invariance we may assume that MATH contains the origin. Let us first assume that MATH is supported outside of MATH. From REF we have MATH . Applying this estimate, we obtain MATH and the claim follows from NAME. It thus remains to show that MATH for all MATH. Fix MATH. Perform a NAME decomposition at level MATH where MATH, the MATH are intervals such that MATH and the MATH are supported on MATH and satisfy MATH and MATH. To control the contribution of MATH, it suffices from NAME to verify the MATH bound MATH . The left-hand side of this is MATH . However, from NAME and the definition of MATH we have MATH . Thus we may bound REF by MATH where MATH ranges over the set MATH. But the desired bound of MATH then follows from NAME and the lacunary nature of the MATH. To deal with the MATH, it suffices from the triangle inequality, NAME, and to show that MATH for all MATH. In fact we prove the stronger MATH . Fix MATH. We may restrict the summation to those MATH such that MATH. From REF we have MATH in particular, from the hypothesis MATH we have MATH . Also, by playing off the moment condition on MATH against the smoothness of MATH, we have MATH . Combining all these estimates, we obtain MATH . Inserting this into REF we obtain the result. |
math/9910039 | Let MATH be elements in the unit ball of MATH, and let MATH denote the MATH-linear operator MATH and let MATH be the associated MATH-form as in REF. We need to show that MATH . By the reductions in REF it suffices to show that MATH is of restricted type MATH for all exponent MATH-tuples MATH such that MATH for all indices MATH which are not equal to the bad index MATH of MATH, and MATH . Fix MATH; by symmetry we may assume that MATH has bad index MATH. Let MATH be sets of finite measure. We have to find a major subset MATH of MATH such that MATH for all MATH. By scaling we may take MATH. We choose MATH to be the set defined by REF, with MATH replaced by MATH throughout. Since REF is sub-additive in MATH, and the unit ball of MATH is the convex hull of the plane waves, we may assume that each MATH is a plane wave MATH for some constants MATH. By modulating the MATH suitably, and translating the symbol MATH by a direction in MATH, one may set MATH. The functions MATH are now completely harmless, and the claim follows from REF with MATH replaced by MATH. |
math/9910043 | It is clear. |
math/9910043 | It is a direct calculation. |
math/9910043 | It is clear that the complex MATH is equal to the NAME complex MATH where MATH considered as a complex of MATH-bimodules, and MATH is considered as MATH-bimodule with zero action. On the other hand, MATH is the bar-complex, it is quasi-isomorphic to MATH for any algebra MATH with unit as complex of bimodules, where MATH is equipped with zero action. Then, MATH. The last complex is dual to the bar-complex. |
math/9910043 | The proof is analogous to the proof of theorem in REF, but now the bar-complex MATH quasi-isomorphic to MATH as a complex of MATH-bimodules, when MATH is equipped with zero structure of MATH-bimodule. |
math/9910045 | Let MATH denote the Cartesian product of MATH copies of the positive integers. Define an additive weight-function MATH by MATH where the sum is over all MATH . For each MATH, define the subset MATH of MATH by MATH . The NAME Principle states that MATH . We remark that the term on the right-hand side of REF arising from the subset MATH is MATH by the usual convention for intersection over an empty set. Next, note that the left-hand side of REF is simply MATH. Finally, observe that all terms on the right-hand side of REF have depth strictly less than MATH - except when MATH, which gives MATH . This latter observation completes the proof of REF . |
math/9910045 | By REF of MATH, MATH as claimed. |
math/9910045 | By REF of MATH, MATH . |
math/9910045 | From REF of the NAME convolution, we have MATH . Now multiply both sides by MATH and apply REF of MATH-duality. |
math/9910045 | Let MATH be a positive integer, and let MATH be a nonnegative integer. We have MATH . By duality, MATH . Letting MATH and forming the generating function, it follows that MATH . Expanding MATH in powers of MATH and integrating term by term yields MATH . Since MATH, we may ignore the terms in REF which are independent of MATH. Thus formally, but with the divergences coming only from the terms independent of MATH and hence harmless, MATH where we have used the abbreviations in REF . It is now a routine matter to extract the coefficient of MATH to complete the proof. |
math/9910045 | Write the left-hand side as an iterated integral as in REF : MATH . Now let MATH at each level of integration. This sends MATH to MATH and, by partial fractions, MATH . The change of variable gives MATH . Now carefully expand the noncommutative product and reinterpret each resulting iterated integral as a MATH-function to complete the proof. |
math/9910045 | Let MATH . Now let us use duality, and then we let MATH at each level of integration. We get MATH . Now let us carefully expand the noncommutative product. We get MATH where the sum is over all sign choices MATH, MATH, MATH, and where by MATH we mean the cardinality of the set MATH. Let us now interpret the iterated integrals as MATH-functions. In this case, they are all unit NAME MATH-sums, as we defined in REF . Thus, MATH where, as usual, MATH. Now if MATH of the MATH equal MATH, then MATH of them equal MATH. Hence, MATH . Finally, MATH is the same as the product over all the signs MATH, and this latter observation completes the proof of REF . |
math/9910045 | Let MATH . Again, let us make the change of variable MATH at each level. Then MATH . Again, let us carefully expand the noncommutative product. We get MATH where this time, the sum is over all MATH with MATH. Note that each MATH in the integrand contributes MATH to the sign and MATH to the depth. Since MATH it follows that MATH is a sum of MATH-values with all MATH coefficients. That is, MATH where the sum is over all vectors MATH and such that MATH . In other words, the sum is over all MATH independent positive integer compositions (in the technical sense of combinatorics) of the numbers MATH, MATH. |
math/9910045 | If MATH and MATH denote the left-hand and the right-hand sides of REF , respectively, then by elementary manipulations (under the assumption MATH) we can show that MATH. The easy observation MATH then completes the proof. |
math/9910045 | Using notation of REF let us observe that MATH . Plugging in MATH and applying REF now completes the proof. |
math/9910045 | Both sides of the putative identity start MATH and are annihilated by the differential operator MATH . Once discovered, this can be checked in NAME or Maple. |
math/9910045 | NAME 's MATH summation theorem gives MATH . Hence, setting MATH in the generating function REF , we have MATH . |
math/9910049 | We show that the functions MATH can be expressed in terms of the flag coordinates and the MATH's. The proof is by induction on MATH, using the total ordering MATH . The key property of the ordering MATH is that for each MATH, there exists an edge MATH such that MATH, MATH, MATH. Using the formulas of REF, we can express the sections MATH, MATH, and MATH entirely in terms of the flag coordinates (and the basis MATH). Since MATH determines the line MATH, MATH determines the plane MATH, and MATH determines the MATH-plane MATH, we can express all of the corresponding functions MATH, MATH, and MATH in terms of the flag coordinates. For the case MATH, since MATH (by REF ) and MATH determines MATH, we can express the functions MATH in terms of MATH and the flag coordinates. For MATH, MATH and MATH determines MATH, so we can express the functions MATH in terms of MATH and the flag coordinates. Expressions for the remaining functions are obtained similarly. |
math/9910049 | The vanishing of the linear polynomials follows from the definition of the edge coordinates. The quadric relations follow from this definition and REF . The cubic (respectively, quartic) relations can be obtained from the quadric relations by eliminating coordinates corresponding to edges in MATH and MATH (respectively, MATH). It follows that MATH is a subvariety of the variety defined by the given polynomials. To see that MATH coincides with this variety, it suffices to show that they are both irreducible and have the same dimension. Since MATH is the closure of the connected MATH-dimensional nonsingular variety MATH, it is a MATH-dimensional irreducible variety. Let MATH be the variety defined by the vanishing of the linear and quadric polynomials. A simple computation using NAMEREF CITE shows that MATH has three irreducible components, two of which correspond to the ideals MATH . The remaining component MATH is MATH-dimensional. Since MATH contains points where all MATH are nonzero, we have MATH. |
math/9910049 | Let MATH with MATH. Then the section MATH, (respectively, MATH) is defined, at each point MATH, to be the linear span of the nonzero vector MATH (respectively, MATH). With MATH as in REF, it follows from the formulas in REF that MATH and MATH are nonzero on MATH. Hence the image MATH is contained in MATH. Taking suitable products of these sections, we then have MATH. Using the various bundle trivializations above, one can show that the section MATH is given by MATH and MATH. By REF , we have MATH . Applying MATH to this equation, and using MATH, we have MATH. Applying MATH to this equation, and using MATH, we have MATH. Thus MATH is given by MATH. |
math/9910049 | Since the flag coordinates on MATH are pulled back from the coordinates on the flag variety, the composition MATH is equivariant. An explicit calculation using the sections of REF and the definition of the edge coordinates shows that the composition MATH is equivariant. And finally, since for each MATH, the group MATH acts via the same character on MATH, for all MATH, the induced action on each MATH will be trivial. It follows that MATH embeds equivariantly into MATH; hence, so does its closure. To see that the embedding is an isomorphism, we let MATH. Since MATH is the closure of MATH in MATH, MATH is dense in MATH. It follows from the description of the MATH-action that the unipotent subgroup of MATH acts freely and transitively on MATH, and that the diagonal subgroup of MATH acts fiberwise on the product of the complements of the zero sections in MATH. Thus MATH acts with dense orbit on each fiber of MATH. Since each MATH is nonzero on the image of MATH in MATH, the image of MATH intersects this MATH-orbit for every fiber of MATH. It follows that the image of MATH is dense in MATH, so the image of its closure MATH coincides with MATH. |
math/9910049 | For any fiber, we can choose MATH whose MATH-orbit is open in that fiber. The description of the MATH-action in REF therefore implies that for any MATH, there will be some MATH such that MATH on the image of MATH in MATH. But MATH implies MATH, where MATH. Therefore MATH is split, and the result follows since MATH is constant on MATH-orbits. |
math/9910049 | We can find a subspace MATH such that MATH, and such that MATH. Then for MATH we can take any one-parameter subgroup that scales in MATH with a negative weight and scales in MATH with a positive weight. |
math/9910049 | Let MATH be the set of split points in MATH. This set is open, and thus MATH is in its closure. But since the MATH-orbit of any point in MATH lies in MATH, the entire fiber MATH must also be in the closure of MATH. Letting MATH be a split point in this fiber completes the proof of the first statement. For the second statement, a look at the MATH-action shows that passing to a point in MATH that is in an orbit closure either preserves MATH or drops it down to MATH. |
math/9910049 | We use the lemmas above to collapse the configuration associated to MATH so that only two subspaces of each dimension remain. We use implicitly that in passing to a point in the closure of a MATH-orbit, the number of planes in any given dimension cannot increase. We begin with the subspaces of dimension MATH. By assumption, MATH. If MATH, then we can use REF to find MATH such that MATH. The orbit MATH must lie in MATH, and since MATH acts transitively on our charts that cover MATH (see REF), we can find a MATH-translate MATH of MATH such that MATH. Since the functions MATH are constant on MATH-orbits, we have MATH. REF implies that we can find a split point MATH with MATH, and such that MATH is MATH and is MATH. Since any point in MATH is also in MATH, we can repeat this procedure until we find a split point MATH with MATH. Now we induct on MATH to produce points MATH and MATH. The key point is that we can apply the lemmas to reduce MATH while preserving MATH for MATH. Since the final point MATH is minimally split, this completes the proof. |
math/9910049 | The quadric relations in REF imply that related triangles will have the same shape. Using this fact and the linear relations of REF it is then easy to verify that the only combinations of zero values for such a pair are those shown in REF . |
math/9910049 | Let MATH be minimally split, and let MATH be is its image in MATH. Then MATH for MATH, which implies MATH. Up to symmetry a minimally split point must have its subspaces partitioned as follows: CASE: The lines must collapse together as MATH or MATH. (The notation MATH means that the two distinct lines are the image of MATH and MATH lines, where MATH.) CASE: The MATH-planes must collapse together as MATH, MATH, or MATH. CASE: The MATH-planes, like the lines, must collapse together as MATH or MATH. With these facts and REF in hand, computing MATH becomes a combinatorial exercise. We leave the pleasure of this computation to the reader. |
math/9910049 | We apply the Jacobian condition for nonsingularity in an affine neighborhood of each point of MATH. Let MATH. Fix an affine neighborhood of MATH in MATH as follows. For each MATH, choose one MATH with MATH, and set this coordinate equal to MATH. The remaining variables MATH form a system of local parameters at MATH in MATH. Let MATH be the MATH-vector space of differentials of MATH at the point MATH. Since MATH is a threefold, we have MATH, with equality only if MATH is nonsingular at MATH. Furthermore, MATH where MATH is the subspace generated by the differentials (evaluated at MATH) of all functions vanishing on MATH. To study this quotient, we will use the following combinatorial rules for computing with differentials in MATH. These follow immediately from the equations in REF ; we omit the simple proof. Suppose that MATH vanishes on MATH. CASE: If MATH and MATH, then MATH. CASE: If MATH and MATH, then MATH. Using these rules, we add data for MATH to MATH as in REF . The MATH means that the differential of the variable corresponding to the edge is MATH, and the two MATH s indicate that the two differentials coincide in MATH. First we verify nonsingularity at the isolated points of MATH. Since the computations for the various points are all very similar, we explain the case MATH in detail and will leave the others to the reader. We fix the affine neighborhood of a point of type MATH by assigning the value MATH to exactly one thin edge in each of the MATH components in REF . Since the differentials of the linear polynomials are in MATH, the differential corresponding to any thin edge is a linear combination of differentials corresponding to bold edges. Thus, MATH is generated by MATH, where MATH is a bold edge of the component MATH. Consider the hypersimplex connected components of REF . Applying REF , we find three independent differentials MATH, MATH, and MATH in these components; the other differentials in these components are MATH. Now consider the other connected components of REF . Using REF we see that the remaining differentials are either MATH or are equal to MATH, MATH, or MATH. The result is summarized in REF . Hence MATH is MATH-dimensional, and all the points of type MATH are nonsingular points of MATH. Finally consider the family of special points MATH in REF . In contrast to the isolated case, to verify nonsingularity we have to use the cubic and quartic polynomials of REF . Let MATH be a point in a subvariety of type MATH. As before we construct an affine neighborhood of MATH choosing a thin edge in each connected component of REF and setting it to MATH. At the point MATH, the linear relations imply that all of the thin edges except those in the four thin triangles will also have value one. To complete the graph MATH to represent the point MATH, we apply the quadratic relations to find MATH such that the values are as in REF (up to the choice of the edges with values MATH). Hence, a priori, this is a MATH-dimensional component of MATH. The quartic relations from REF , however, imply that the two parameters MATH and MATH satisfy a linear relation. For the choice of parameters in REF , for example, this relation is MATH . Thus this component of MATH is in fact a curve. We now complete the proof of the theorem. As in the isolated case, the differentials on all thin edges, except for those in the four thin triangles, can be expressed as linear combinations of the differentials on bold edges. Moreover the differentials on edges of the thin triangles can be expressed as linear combinations of MATH and MATH. Thus, using REF , we can find MATH differentials that span MATH REF : MATH . Note that the span of these is at most MATH-dimensional, because of the relations MATH induced by the differentials of the linear relations, and the relation MATH induced by the linear relation between MATH and MATH. To finish, we claim that the spans of MATH and MATH are each MATH-dimensional. Indeed, a quadric relation implies that the front face of the octahedron in REF has the same shape as the corresponding face in REF , which implies MATH . This relation, and a similar one involving MATH and MATH, shows that MATH. This completes the proof of the main theorem. |
math/9910053 | One considers MATH which, via the natural inclusion to MATH is the inverse image MATH (here we abuse of notation, still denoting by MATH the map coming from the filtration), and via the map coming from the natural inclusion MATH and the identification with the first term of the filtration on MATH is the inverse image MATH. |
math/9910054 | If MATH be a small enough ball around the MATH-th singularity of MATH for MATH then MATH is transverse to each MATH and also to MATH. By compactness, there exists MATH so that MATH is also transverse to each of these spheres for MATH. Let MATH and MATH. Then MATH is a submersion of a compact manifold with boundary, so by NAME 's theorem (see, for example, CITE) it is a locally trivial fibration. Since it is a fibration over a disk it is a trivial fibration, so MATH. This gives the embedding MATH used in the definition of NAME fibers at infinity. We can extend MATH to the collars outside the regular boundary components of MATH (by definition of regular boundary components) to get MATH compatible with the map MATH and with the trivial structure of MATH outside regular boundary components of MATH. Let MATH. Then it is clear that MATH is diffeomorphic to MATH so we will identify MATH with this subset MATH of MATH. The closure of MATH consists of MATH pieces, of which the last MATH are ``NAME disk" neighborhoods of the singularities of MATH. Gluing these back in to MATH gives an embedding MATH, the closure of whose complement consists of pieces MATH attached at the boundary components MATH of MATH for MATH. If we show each MATH is homeomorphic to a half-open collar on MATH, then adding MATH to MATH has the same effect up to homeomorphism as removing MATH, so the proof is complete. To see MATH is a collar we can use a standard vector-field argument. Since MATH is transverse to large spheres about MATH, we can find a vector-field MATH in a neighborhood of any point of MATH outside MATH so that MATH is tangent to fibers of MATH and has radially outward component of magnitude MATH. Gluing these local MATH's by a partition of unity, we can find a vector-field MATH which is defined on all MATH, is zero off a neighborhood of MATH, is tangent to fibers of MATH, and has radially outward component of magnitude MATH on MATH and of magnitude at most MATH elsewhere. We can also assume MATH is non-zero on the part MATH of MATH. Let MATH be the inward radial vector-field MATH on MATH. Again, by gluing local choices by a partition of unity, we can find a vector-field MATH on MATH whose image under MATH is MATH, which is tangent on the part MATH of MATH, and which has globally bounded magnitude. The sum MATH is then a vector-field on MATH whose flow-lines all lead in backwards time to MATH and intersect MATH transversally, and whose forward flow lines continue for infinite time. Integrating the vector-field from MATH thus gives a homeomorphism of MATH with MATH, completing the proof. |
math/9910054 | In the previous proof we identified MATH with a subset of MATH in such a way that MATH differs from MATH by closing some MATH boundary components by disks and adding collars to some other boundary components. It follows that the homology of MATH and MATH differ only in degree MATH. Since MATH vanishes if MATH is not in the range MATH, the lemma follows from REF . |
math/9910054 | The above lemma implies that MATH by an isomorphism that fits in a commutative diagram: MATH . We identify MATH with a subset of MATH as in the proof of REF . Thus MATH maps MATH to the disk MATH. Moreover, MATH is the union of the outer shell MATH, where MATH, and an inner core MATH isomorphic to MATH. Express the disk MATH as the union of two half-disks MATH and MATH by cutting along a diameter. Let MATH and MATH be the parts of MATH that lie over MATH and MATH and put MATH. We have MATH with MATH . Thus MATH has MATH as a deformation retract and the pair MATH has its intersection with MATH (isomorphic to MATH) as a deformation retract. In particular, we see that each of the following maps induces an isomorphism in homology, since they are, respectively, a homotopy equivalence, an excision map, and a homotopy equivalence: MATH . Since MATH we get a homology isomorphism MATH . The NAME theorem thus gives MATH . Summarizing, we have an isomorphism MATH . The composition MATH is the variation map (up to sign). Indeed, for the isomorphism MATH, a relative cycle MATH is taken to a relative cycle MATH, mapped by MATH (we are identifying MATH with MATH). We are interested in the boundary of this cycle as a cycle for MATH. When we retract MATH to MATH, the subset of MATH given by MATH maps to MATH by the identity on one component and by MATH on the other. The resulting cycle in MATH thus represents MATH. REF now follows because MATH by excision. |
math/9910054 | The first part of REF just says that the image MATH is contained in MATH, which is part of REF (it is also proved in REF). The second part of REF is immediate from the above proof of REF . |
math/9910054 | We first recall from CITE (see also CITE) how the NAME linking form for the regular link at infinity can be defined on MATH. Let MATH be a large disk in MATH which contains all MATH for which MATH is either singular or fails to be NAME at infinity (in the sense of CITE, see also CITE; there are finitely many such MATH and they include all irregular values of MATH). Then there is a radius MATH so that for any MATH the boundary of the disk MATH intersects all fibers MATH with MATH transversally. Then MATH is homeomorphic to MATH. The embedding of MATH as MATH with MATH gives a NAME surface for the regular link at infinity. Let MATH be a neighboring copy of MATH, obtained by replacing MATH by a nearby point MATH of MATH. If MATH is a cycle for homology MATH, let MATH be a copy of the cycle in MATH. The NAME form is the form MATH where MATH is linking number in MATH. It can be computed (up to a sign which depends on conventions; following CITE and CITE the sign is MATH) by letting MATH and MATH bound chains MATH and MATH in MATH and taking the intersection number MATH. We now choose our base point MATH for which MATH is our ``standard" regular fiber to be the above point MATH, so we have paths MATH (as chosen before REF ) from MATH to the irregular values MATH. We can assume these paths run in the disk MATH. Suppose that we have a homology class MATH in the image of the map MATH. Here, we will consider, for the moment, MATH to be a regular fiber MATH with MATH, so MATH is on MATH. We write the cycle MATH as MATH with MATH a relative cycle in MATH. By transporting MATH around the circle MATH we obtain a map of MATH to MATH. The boundary of this chain MATH consists of the union of MATH, MATH, and MATH. The part MATH bounds a mapping of MATH mapping to MATH, so we can glue this to MATH to get a chain MATH in MATH with boundary MATH representing MATH mapping to MATH. If we want MATH be a cycle for homology of our standard fiber MATH we glue onto the above MATH a copy of MATH mapping into MATH. We call the resulting chain MATH. Note that MATH lies completely in the ``shell" MATH. We can also construct MATH in a smaller shell obtained by replacing MATH by MATH and removing a thin collar from MATH. We denote the version of MATH constructed this way by MATH. Suppose now the two homology classes MATH are in the image of the map MATH, where MATH is now our standard regular fiber. We can assume they both lie in MATH, since MATH retracts to MATH. We can then make MATH bound a cycle MATH as above. We can also make MATH bound a cycle MATH constructed as above but using a path MATH running parallel to MATH to a point on MATH. This path runs through a point MATH next to MATH on MATH. The chains MATH and MATH intersect only in the fiber MATH and the intersection number MATH is, up to sign, the intersection number in MATH of MATH and MATH. (See REF .) With standard sign conventions, the sign is in fact MATH (this is most easily checked by using the standard formulae MATH and MATH relating NAME form MATH, intersection form MATH and monodromy MATH, since for a knot the relationship to be proved is MATH). This proves the claim of REF about the diagonal blocks of the NAME form. The claim about vanishing of appropriate off-diagonal blocks is the same as the corresponding proof in CITE, as suggested by REF . |
math/9910054 | Let MATH be a REF-codimensional submanifold (with boundary) of the sphere MATH and suppose MATH is fibered over the circle MATH with fiber MATH. Then we can define a NAME form MATH and homological monodromy MATH on the homology of MATH as for fibered links and the obvious geometric relation MATH can be written in matrix form as MATH (this is a well-known relation in the case of fibered links, see, for example, CITE). If we apply this in the situation of REF it gives the equation MATH . An inductive argument, which we omit, shows that this equation is equivalent to the collection of equations of the theorem. |
math/9910054 | Let MATH. That MATH is a union of solid tori was proved in CITE. The argument (due to NAME) is that for MATH the intersection MATH is transverse and the result is a union of disks by the maximum modulus principle, since it is equivalent to the set MATH. On the other hand, MATH by the argument that identifies MATH with collars on the MATH's, so MATH is a union of solid tori. The same argument applies to show the sets MATH are unions of solid tori for MATH. But the components of these sets are homeomorphic to the MATH's. |
math/9910055 | The argument is essentially taken from NAME: see CITE. We check that MATH is perpendicular to MATH, for any MATH : MATH where MATH on MATH and then REF were used. |
math/9910056 | MATH is the order of a subgroup of MATH, and MATH is a finite group of order at most MATH. |
math/9910056 | We compute MATH so MATH. Conversely, if MATH, write MATH with MATH and MATH, and note that the polynomial MATH has degree at most MATH and span MATH; write it as MATH with MATH and MATH of degree less than MATH. It is equal, in MATH, to MATH where the two summands don't overlap, and therefore cannot equal MATH. If MATH, it is clear that MATH, and if MATH the same holds by symmetry. |
math/9910056 | We compute MATH so MATH. The argument in the proof of the previous proposition shows that no smaller value satisfies this equation. |
math/9910056 | In MATH, we may consider a subset MATH . It is a vector subspace of dimension MATH, as the MATH are linearly independent for MATH. Elements of MATH are polynomials and therefore can be composed, an operation we denote by MATH. This operation is internal to MATH, and endows MATH with a MATH-algebra structure: MATH as soon as all the monomials of MATH have degree a power of MATH. Moreover, MATH is Abelian (on the basis MATH we have MATH), and MATH through the natural map MATH extended by linearity. Indeed MATH and for any MATH where MATH is a polynomial divisible by MATH. It follows that any polynomial representing MATH in MATH maps to a multiple (for MATH) of MATH, which in turn represents MATH in MATH. Now the evolution MATH is mapped in MATH to MATH; thus for all MATH such that MATH in MATH, one has MATH in MATH, and MATH in MATH. |
math/9910056 | MATH is irreducible and primitive if and only if MATH is a field and MATH is generated by MATH; this is equivalent to MATH, its maximal possible value. We have MATH the first and third equalities following from Conjecture REF and the second from induction. |
math/9910056 | It suffices to show that MATH; if MATH is even, then MATH, while if MATH is odd, then MATH and MATH . |
math/9910056 | Let us denote by MATH the MATH-th power of the NAME automorphism. We must show that MATH in MATH, and MATH in MATH; but in MATH we have MATH of order MATH, and in MATH we have MATH of order MATH. |
math/9910056 | The factors MATH of MATH are irreducible, but the MATH have at least two factors, one of them being MATH. According to the proposition, MATH . If we distribute the MATH direct sums over the tensor products, we obtain an expression of MATH as a direct sum of MATH algebras. Among these is MATH; all the MATH others are summands of MATH. Among these others are the MATH. |
math/9910056 | By REF , MATH factors as claimed. By induction, it suffices to consider a factorization MATH, with MATH and MATH coprime, and to show that in that case MATH . As MATH and MATH are coprime, apply NAME 's theorem to decompose the identity MATH, for MATH and MATH polynomials. Apply the ``hat" operator: MATH . We may now define the two mutually inverse maps MATH . |
math/9910056 | MATH is semisimple and commutative whenever both MATH and MATH are, so MATH is a direct sum of fields. Let MATH and MATH be the NAME automorphisms of MATH and MATH: then the NAME automorphism of MATH is MATH, so is of order exactly MATH, and all subfields of MATH are of degree at most MATH. On the other hand, MATH splits as a sum of fields each containing MATH and MATH (see CITE). |
math/9910056 | A reformulation of REF is that there is a group homomorphism MATH, mapping MATH to MATH. Now by REF is of order MATH in MATH, so MATH in MATH, and MATH splits as a direct sum of fields of degree dividing MATH. A generator of MATH maps to an automorphism of MATH, whose MATH-th power is MATH. It must act by permutation and automorphisms on a set of MATH subalgebras of MATH, who are then all isomorphic; call them MATH. |
math/9910058 | For REF , see REF , and CITE. For REF , see REF . See also REF. |
math/9910058 | The isomorphism of the intermediate NAME was proved in the previous paragraph. Let MATH. By CITE or CITE, the global NAME Theorem holds for smooth REF-dimensional cubics, so there exists the unique cubic threefold MATH such that MATH as p.p.a.v. Let MATH and MATH be the unique cubics associated to MATH and MATH. Since MATH, then MATH. Let now MATH and MATH be associated to the same cubic threefold MATH, and let MATH. Then by the above MATH. Let MATH, MATH be related by, say, a NAME - Takeuchi birational isomorphism. By REF , MATH contains a smooth elliptiic quintic curve and admits a birational isomorphism of NAME - NAME type with some cubic MATH. Then, as above, MATH by Global NAME, and MATH, MATH are associated to each other by the definition of the NAME - NAME birational isomorphism. Conversely, if we start from the hypothesis that MATH, MATH are related by a NAME - NAME birational isomorphism, then the existence of a NAME - Takeuchi one from MATH to some cubic MATH is affirmed by REF . Hence, again by Global NAME, MATH and we are done. |
math/9910058 | Let MATH be a normal elliptic quintic. By REF , there exists a MATH together with a birational isomorphism MATH. REF implies that MATH and MATH are associated to each other. By REF , we have MATH, where MATH denotes the NAME variety in MATH parametrizing the lines MATH contained in MATH. It is standard that MATH is the scheme of zeros of a section of the dualized universal rank REF vector bundle MATH on MATH. Hence MATH is the scheme of zeros of a section of MATH. Hence MATH can be obtained by NAME 's construction from MATH, and by uniqueness, MATH. By REF, MATH, so, to prove the assertion about global sections, it is enough to show the injectivity of the natural map MATH. The latter is obvious, because the quartic da NAME is not contained in a hyperplane. Thus we have MATH. For the identification of MATH with MATH, it is sufficient to show that MATH is defined by the sections of MATH in the image of the map MATH and that MATH is an isomorphism. This is proved in the next lemmas. The uniqueness modulo MATH is proved in REF . |
math/9910058 | Let MATH, MATH be dual bases of MATH respectively, and MATH the corresponding bases of MATH, MATH. Identify MATH in the source of MATH with a REF-form MATH. Then MATH can be given by the formula MATH, where MATH, and MATH stands for the contraction of tensors. Notice that MATH sends REF-forms of rank REF, respectively, REF to bivectors of rank REF, respectively, REF. Hence MATH is not defined on MATH and contracts MATH into MATH. In fact, the NAME of a REF-form MATH of rank REF are exactly the NAME coordinates of MATH, which implies MATH. In order to iterate MATH, we have to identify its source MATH with its target MATH. We do it in using the above bases: MATH. Let MATH. Then each matrix element MATH is a polynomial of degree REF in MATH, vanishing on MATH. Hence it is divisible by the equation of MATH, which is the cubic Pfaffian MATH. We can write MATH, where MATH are some linear forms in MATH. Testing them on a collection of simple matrices with only one variable matrix element, we find the answer: MATH. Hence MATH is a birational involution. |
math/9910058 | The restriction of MATH to the lines in MATH is written out in CITE on pages REF (for a non-jumping line of MATH, REF ) and REF (for a jumping line). These formulas imply the assertion; in fact, the quadric surface has rank REF for a non-jumping line, and rank REF for a jumping one. |
math/9910058 | Let MATH be the natural desingularization of MATH parametrizing pairs MATH, where MATH is a skew-symmetric MATH matrix and MATH is a line in the projectivized kernel of MATH. We have MATH, where MATH is the tautological rank REF vector bundle on MATH. MATH has two natural projections MATH and MATH. The classifying map of MATH is just MATH. MATH is isomorphic over the alternating forms of rank REF, so MATH. MATH is at least bijective on MATH. In fact, it is easy to see that the fibers of MATH can only be linear subspaces of MATH. Indeed, the fiber of MATH is nothing but the family of matrices MATH whose kernel contains a fixed plane, hence it is a linear subspace MATH of MATH, and the fibers of MATH are MATH. As MATH does not contain planes, the only possible fibers are points or lines. By the previous lemma, they can be only points, so MATH is injective. |
math/9910058 | Let MATH be the coordinate functions on MATH, dual to the basis MATH. The MATH can be considered as sections of MATH. Then MATH can be considered either as an element MATH of MATH, or as a section MATH of MATH. For a point MATH, the NAME coordinates of the corresponding plane MATH are MATH for a non zero bivector MATH. By construction, this is the same as MATH. This proves the assertion. |
math/9910058 | The representations MATH, MATH define two isomorphisms MATH, MATH. Identifying MATH, define MATH. Assume that MATH. Then the two REF-dimensional cubics MATH and MATH are isomorphic by virtue of the map MATH. By construction, we have MATH for any MATH. Hence MATH and MATH represent two cross-sections of the map MATH defined in the proof of REF over their common image MATH, and MATH is a morphism over MATH. These cross-sections do not meet the indeterminacy locus MATH of MATH, because it is at the same time the singular locus of MATH and both REF-dimensional cubics are nonsingular. The fibers of MATH being linear subspaces of MATH, the generic element of a linear pencil MATH represents also a cross-section of MATH that does not meet MATH. So there is a one-dimensional family of representations of MATH as a linear section of the Pfaffian cubic which are not equivalent under the action of MATH but induce the same vector bundle MATH. This family joins MATH and MATH and its base is an open subset of MATH. This cannot happen for generic MATH, because both the family of vector bundles MATH and that of representations of MATH as a linear section of MATH are REF dimensional for generic MATH REF . |
math/9910058 | It is sufficient to verify this property for a special MATH. Take NAME 's cubic MATH . NAME REF gives the representation of this cubic as the Pfaffian of the following matrix: MATH . Its quadratic NAME are given by MATH where MATH, MATH is a permutation of MATH, and MATH is nothing but its sign. A direct computation shows that REF quadratic NAME are linearly independent. |
math/9910058 | By REF , the curve MATH represented by the generic point of MATH is a (smooth) normal elliptic quintic. Let MATH be the associated vector bundle, represented by the point MATH. Choose any representation of MATH as a linear section of the Pfaffian cubic MATH as in REF , so that MATH. The projective space MATH is naturally identified with MATH. This follows from the proof of REF . Indeed, the curves MATH represented by points of MATH are exactly the zero loci of the sections of MATH, and the latter are induced by linear forms on MATH via the natural surjection MATH. The zero loci of these sections are of the form MATH, where MATH runs over all the hyperplanes in MATH. Let MATH be a line in MATH. By REF , the quadratic pencil of lines with base MATH sweeps out a quadric surface MATH of rank REF or REF. Let MATH be the linear span of MATH in MATH. Then MATH is a component of MATH if and only if MATH. Such hyperplanes MATH form the pencil MATH in MATH. Obviously, the pencil MATH contains exactly all the curves, represented by points of MATH and having MATH as an irreducible component. |
math/9910058 | As concerns the numerical values for the MATH, the proof goes exactly as that of REF with only one modification: the authors used there the property of a normal elliptic quintic MATH, proved in REF. Here we should verify directly this property for our curve MATH. This is an easy exercise: one can use the identifications of the normal bundles of MATH and the three natural exact sequences MATH where MATH, MATH is the sky-scraper sheaf with the only nonzero stalks at MATH equal to MATH, MATH, and MATH denotes NAME 's MATH of a singularity; we have MATH for nodal curves. The values of MATH in REF imply the stated properties of the moduli space, because MATH is self-dual, and so MATH. For the remaining assertions of REF , we will apply REF . It states that if the elementary transformations of the normal bundles to MATH satisfy MATH, then MATH is strongly smoothable. In fact, by REF and MATH we see that MATH with MATH. By CITE, MATH or MATH, so MATH or MATH. For MATH, MATH may be one of the sheaves MATH, MATH, or MATH. So, the hypotheses of the theorem may be not verified only if MATH. In interchanging the roles of MATH and assuming that MATH, we can see that MATH or MATH. The second case is impossible, because MATH and the tangent directions of MATH at the points MATH are not coplanar, so the centers of the elementary transformation MATH, corresponding to the directions of MATH at MATH, do not lie on the same section of MATH. Thus, in both cases the theorem can be applied, and we conclude that the natural maps MATH are surjective. Hence the discriminant divisor MATH has locally analytically two nonsingular branches with tangent spaces MATH, each unfolding only one of the two singular points of MATH, and their transversal intersection MATH parametrizes the deformations preserving the two singular points. To conclude the proof, remark that the linear normality and MATH are preserved under small deformations. |
math/9910058 | We have to show that MATH and that the natural map MATH remains surjective when restricted to MATH. It suffices to do this only for one special cubic threefold MATH and for one special MATH, because both conditions are open. So, choose a curve MATH of type MATH in MATH, then a cubic MATH passing through MATH. Take, for example, the closures of the following affine curves: MATH . The family of quadrics passing through MATH is REF-dimensional with generators MATH . The cubic hypersurface in MATH with equation MATH is nonsingular for generic linear forms MATH, so we can choose MATH to be of this form. We verified, in using the NAME program CITE, that the choice MATH yields a nonsingular MATH such that MATH. Look at the following commutative diagram with exact rows and columns, where the first row is the restriction of REF to the subsheaf of the sections of MATH vanishing along MATH. MATH . It allows to identify the tangent space MATH with the image of MATH in MATH. So, we have to show that the derivative MATH is surjective. Using the basis REF of MATH, we easily verify that this is the case (in fact, MATH generate MATH). |
math/9910058 | According to CITE, the family of rational normal quartics in a nonsingular cubic threefold MATH has dimension REF, and is irreducible for generic MATH. By REF , each rational normal quartic MATH has exactly REF chords MATH in MATH, so the family MATH of pairs MATH is equidimensional of dimension REF. It suffices to verify that one of the components of MATH, say MATH, meets MATH at some point MATH with local dimension MATH for one special cubic threefold MATH, for one special MATH and for at least one MATH. But this was done in the previous lemma. Indeed, the fact that MATH can be smoothed inside MATH implies that MATH, hence MATH for some MATH. The assertion for general MATH follows by the relativization over the family of cubic threefolds and the standard count of dimensions. |
math/9910058 | We know already that the family of pairs MATH is REF-dimensional. Now we are to show that the second components MATH of these pairs move in a dense open subset in the NAME surface MATH of MATH. This can be done by an infinitesimal argument: let MATH be such a pair. Then a small neighborhood of MATH in MATH contains a smooth subvariety MATH of codimension REF with tangent space MATH, which parametrizes the curves MATH of the same type. It suffices to show that the natural projection of MATH to MATH is surjective. The exact triples REF with MATH together with the observation that all the MATH's vanish imply that the natural map MATH is surjective and that it restricts to a surjective map between the kernels of the respective maps to MATH: MATH . We want to see that it will remain surjective even if we shrink its source to MATH. Let MATH be as above. Then the triple REF determines an isomorphism MATH in such a way that MATH is the image of the restriction map MATH. Hence we can identify MATH with MATH. Look at the exact triple MATH . By REF , MATH. By REF , we have MATH for generic MATH. As in the proof of REF , the fact that the centers of the elementary transformation MATH of MATH do not lie on the same section of MATH implies that MATH and that the map MATH is surjective. As MATH, the last exact triple gives the surjectivity of MATH. Restricting the map to the kernels of the natural surjections onto MATH, we obtain the result. |
math/9910058 | One can easily relativize the constructions of MATH, etc. over a small analytic (or étale) connected open set MATH in the parameter space MATH of REF cubics, over which all the cubics MATH are nonsingular. We have to restrict ourselves to a ``small" open set, because we need a local section of the family MATH in order to define the maps MATH. The fibers MATH are equidimensional and nonsingular of dimensions MATH respectively. Moreover, it is easy to see that a normal elliptic quintic MATH in a special fiber MATH can be deformed to the neighboring fibers MATH. Indeed, one can embed the pencil MATH into the linear system of hyperplane sections of a REF-dimensional cubic MATH and show that the local dimension of the NAME scheme of MATH at MATH is REF, which implies that MATH deforms to all the nearby and hence to all the nonsingular hypeplane sections of MATH. Hence the families MATH are irreducible, flat of relative dimensions REF, respectively, REF over MATH, and the degree of MATH is constant over MATH. If there is a reducible fiber MATH, then the degree sums up over its irreducible components, so it has to be strictly greater than REF. But we know, that MATH is REF over the generic fiber, hence all the fibers are irreducible and MATH for all MATH. |
math/9910058 | As we have already mentioned in the proof of REF , the family MATH of rational normal quartics in MATH is irreducible of dimension REF. The nonsingularity of MATH follows from the evaluation of the normal bundle of a rational normal quartic in the proof of REF . We saw also that MATH is dominant, so the generic fiber is equidimensional of dimension REF and we have to prove its irreducibility. Let MATH be the quasi-finite covering of MATH parametrizing the irreducible components of the fibers of MATH over points of MATH. Let MATH be generic, and MATH the fiber of MATH. By REF , for a generic line MATH, we can represent MATH as MATH for a rational normal quartic MATH having MATH as one of its chords. Let MATH be the rational map sending MATH to the component of MATH containing MATH. Let MATH. REF implies that MATH is dominant. Hence it is generically finite. Then MATH is also generically finite, and we have for their degrees MATH. Let us show that MATH. Let MATH be two distinct points in a generic fiber of MATH. By REF , MATH contains only one pencil of curves of type MATH, where MATH is a fixed chord of MATH, and MATH is a rational normal quartic meeting MATH in REF points. But REF imply that both MATH and MATH contain such a pencil. This is a contradiction. Hence MATH. Now, choose a generic rational normal quartic MATH in MATH. We are going to show that MATH is birational to some MATH, associated to MATH, and hence birational to MATH itself. Namely, take the MATH obtained by the NAME - Takeuchi transformation MATH from MATH with center MATH. Let MATH be the indeterminacy point of MATH. The pair MATH is determined by MATH uniquely up to isomorphism, because MATH is the image of MATH under the map defined by the linear system MATH and MATH is the image of the unique divisor of the linear system MATH. By REF , a generic MATH defines an inverse map of NAME - Takeuchi type from MATH to the same cubic MATH. As MATH is generic, it has no biregular automorphisms, and hence this map defines a rational normal quartic MATH in MATH. We obtain the rational map MATH, MATH, whose image contains MATH. As MATH, the whole image MATH is contracted to a point by the NAME - NAME map. Hence, to show that MATH is birationally equivalent to MATH, it suffices to see that MATH is generically injective. This follows from the following two facts: first, the pair MATH is determined by MATH uniqueley up to isomorphism, and second, a generic MATH has no biregular automorphisms. If there were two points MATH giving the same MATH, then there would exist an automorphism of MATH sending MATH to MATH, and hence MATH. Another proof of the generic injectivity of MATH is given in REF . We did not find an appropriate reference for the second fact, so we prove it in the next lemma. |
math/9910058 | As MATH is embedded in MATH by the anticanonical system, any biregular automorphism MATH of MATH is induced by a linear automorphism of MATH. Hence it sends conics to conics, and thus defines an automorphism MATH of the NAME surface MATH, parametrizing conics on MATH. In CITE, the authors prove that the NAME scheme MATH parametrizing pairs of points on the KREF surface MATH of degree REF in MATH is isomorphic to the NAME REF-fold MATH parametrizing lines on MATH, where MATH is REF-dimensional linear section of the Pfaffian cubic in MATH associated to MATH. The same argument shows that MATH, where MATH is the cubic REF-fold associated to MATH, and MATH is the NAME surface parametrizing lines on MATH. Hence MATH induces an automorphism MATH of MATH. Let MATH be the induced linear automorphism of MATH, and MATH its differential at the origin. By CITE, the projectivized tangent cone of the theta divisor of MATH at MATH is isomorphic to MATH, so MATH induces an automorphism of MATH. MATH being generic, MATH is also generic, so MATH. Hence MATH. By the Tangent Theorem for MATH CITE, MATH is identified with the restriction of the universal rank REF quotient bundle MATH on MATH, and all the global sections of MATH are induced by linear forms MATH on MATH via the natural map MATH. Hence the fact that MATH acts trivially on MATH implies that MATH permutes the lines MATH lying in one hyperplane section of MATH. For general MATH, there are REF lines MATH, and in taking two hypeplane sections MATH which have only one common line, we conclude that MATH fixes the generic point of MATH. Hence MATH is the identity. This implies that every conic on MATH is transformed by MATH into itself. By REF , we have REF different conics MATH passing through the generic point MATH, which are transforms of REF chords of MATH in MATH. Two different conics MATH cannot meet at a point MATH, different from MATH. Indeed, their proper transforms in MATH (we are using the notations of REF ) are the results MATH of the floppings of two distinct chords MATH of MATH. Two distinct chords of MATH are disjoint, because otherwise REF points MATH would be coplanar, which would contradict the linear normality of MATH. Hence MATH are disjoint. They meet the exceptional divisor MATH of MATH at one point each, hence MATH. As MATH and MATH, this implies MATH. This ends the proof. |
math/9910058 | Let MATH be a rational normal quartic. Then there exists a unique MATH-orbit MATH transforming MATH to the normal form MATH . There is one particularly simple rational normal scroll MATH containing MATH: MATH . Geometrically, MATH is the union of lines which join the corresponding points of the line MATH and of the conic MATH, MATH. Conversely, any rational normal scroll can be obtained in this way from a pair MATH whose linear span is the whole MATH. Remark that MATH is the only correspondence from MATH to MATH such that the resulting scroll contains MATH. Now, it is easy to describe all the scrolls containing MATH: they are obtained from MATH by the action of MATH. Each non-identical transformation from MATH leaves invariant MATH, but moves both MATH and MATH, and hence moves MATH. |
math/9910058 | NAME constructs in CITE for any elliptic quintic curve MATH on MATH a rank two vector bundle MATH such that MATH, MATH, MATH, and proves that the map from MATH to MATH given by the sections of MATH and composed with the NAME embedding is the standard embedding of MATH into MATH. Hence MATH is isomorphic to the restriction of the universal rank REF quotient bundle on MATH (in particular, it has no moduli), and the zero loci of its sections are precisely the sections of MATH by the NAME varieties MATH over all hyperplanes MATH. These zero loci are l. c. i. of pure dimension REF. Indeed, assume the contrary. Assume that MATH has a component of dimension MATH. Anyway, MATH, hence if MATH, then MATH has a divisor of degree MATH. This contradicts the fact that MATH has index REF and NAME number REF. One cannot have MATH, because otherwise MATH would be reducible. Hence MATH, and it is l. c. i. of pure dimension REF as the zero locus of a section of a rank REF vector bundle. All the zero loci MATH of sections of MATH form a component MATH of the NAME scheme of MATH isomorphic to MATH. The curves MATH from MATH passing through MATH are the sections of MATH by the NAME varieties MATH for all MATH containing REF-plane MATH represented by the point MATH, and hence form a linear subspace MATH in MATH. Now, let us prove the last assertion. Let MATH be generic. We have MATH, therefore the map MATH, given by the linear system MATH, sends it to a curve MATH of degree MATH. So, the image is a quintic of genus REF. Let MATH. The inverse MATH being given by the linear system MATH, we have for the degree of MATH: MATH, hence MATH, that is, MATH is a simple point of MATH. Thus the generic MATH is transformed into a smooth elliptic quintic MATH passing through MATH. By the above, such curves form a MATH in the NAME scheme, and this ends the proof. |
math/9910058 | For the construction of MATH and for the fact that MATH, see REF. According to REF , there exists a REF-dimensional family of varieties MATH, associated to a fixed generic cubic REF-fold MATH, which is birationally parametrized by the set MATH of isomorphism classes of vector bundle MATH obtained by NAME 's construction starting from normal elliptic quintics MATH. By REF , MATH is an open subset of MATH. Hence all the assertions follow from REF . |
math/9910063 | Choose a local trivialization of MATH : MATH . This reduces the problem to the computaion of MATH. In using the resolution MATH we arrive at the conclusion that MATH for MATH, MATH for MATH, and MATH for MATH. Taking into accout the functorial behavior of the last isomorphism with respect to the choice of bases in MATH, we obtain the wanted formulas with MATH instead of MATH, which is the same by self-duality. |
math/9910063 | REF follow from CITE. REF is verified for MATH in the proof of REF , CITE and is obviously extended to MATH by considering the exact triple MATH with MATH. MATH is not contained in a hyperplane by projective normality, so MATH. As MATH equals MATH for MATH and is zero for all MATH, we obtain the wanted assertion. REF for MATH is nothing but CITE. By NAME duality (on MATH), it remains to verify REF only for MATH. This follows from REF with MATH, the exact sequence REF , twisted by MATH, and from MATH for MATH. |
math/9910063 | This follows immediately from REF . |
math/9910063 | The first isomorphism is proved in CITE. For the second one, look at the exact sequences MATH obtained by tensoring REF with MATH. REF follows from REF . REF give sufficiently many values of MATH, which is a cubic polynomial in MATH by NAME - NAME - NAME, that one can deduce that MATH. Thus, MATH. We have also MATH, hence MATH, and this allows to determine the remaining NAME numbers from REF , thus proving MATH. Now, use again REF to show that MATH for MATH. First, show that MATH. By REF , MATH. Hence MATH (we used the fact that MATH is self-dual). Consider the natural inclusion MATH and apply CITE : MATH for any invertible MATH of degree REF. By NAME duality, MATH. Hence MATH, and all the more MATH. Hence MATH. Next, MATH for MATH by the same argument as the one applied in the proof of REF . Now REF gives the wanted assertion. |
math/9910063 | As in the proof of REF, we have the exact sequence MATH where MATH is a rank two vector bundle with MATH and MATH. It is either the direct sum of two line bundles MATH with MATH, or fits into the non-split exact sequence MATH where MATH is a theta characteristic on MATH, and MATH a point of MATH such that MATH. This implies that MATH. By REF , the map MATH is surjective. This map factors obviously through MATH, hence the map MATH is also surjectve. By NAME, MATH, and by REF , MATH, and MATH. By CITE, MATH, hence, the exact sequence MATH yields MATH. By projective normality of MATH, MATH, hence MATH. Consider the following commutative diagram with exact rows and columns: MATH . The left column gives MATH, and the first row MATH. The lower row gives MATH, MATH. The middle row gives MATH, MATH, which implies also MATH. Now, look at another commutative diagram in which all the triples are exact: MATH . From the middle row, we see that MATH is surjective on global sections. We can think of elements of MATH as of sections from MATH vanishing on MATH. Thus, a section MATH is in MATH if and only if MATH. Hence MATH and MATH and MATH by REF . |
math/9910063 | REF and the last lemma imply the result. |
math/9910063 | CASE: For the smoothness, remark that MATH, and there is an equality if and only if MATH is smooth at MATH. But it is indeed an equality, because MATH contains explicitly a REF-dimensional family of pairwise non-isomorphic sheaves: we have REF parameters specifying the support MATH of the sheaf, and the moduli space MATH with fixed support MATH is smooth of dimension REF by CITE. The openness of MATH in the moduli space of sheaves on MATH follows exactly as in REF in using an appropriate relativization of NAME 's construction. The existence of the sheaves MATH in REF and of the bilinear form in REF follow from REF and from REF. The existence of the MATH is obtained by an obvious modification of the arguments of CITE. The skew symmetry follows from the smoothness: the NAME square MATH of an element MATH is exactly the first obstruction map MATH, and it is zero by smoothness of MATH at MATH. By general properties of spectral sequences of composite functors with a ring structure, the NAME coupling induces on the bi-graded object MATH a natural ring structure MATH . The isotropy of MATH follows from the fact that MATH REF . |
math/9910063 | To see that these are indeed the same, we will express both of them in a local basis of MATH. The latter is just the multiplication of coefficients: MATH . For the former, notice that MATH contains the trivial subbundle MATH as a direct summand. By the above considerations, MATH. Hence every section from MATH can be written locally as MATH. The self-duality of MATH being given by a trivialization of MATH, we can assume that MATH is the trivializing section, and then the dual basis of MATH is MATH, where MATH. Hence the contraction of middle terms MATH giving the NAME coupling is the identity MATH on the basic sections. This implies that the NAME coupling is given by multiplication of coefficients, MATH. |
math/9910063 | We are to verify, that for any linear form MATH, the map MATH is surjective. By REF , the last map is identified with MATH . Using REF , as it is and twisted by MATH, and REF , we can continue the sequence of identifications: MATH . Only the isomorphisms involving MATH are not obvious. Those involving MATH are established in the proof of REF . We will verify now the isomorphisms involving MATH. To this end, write down REF twisted by MATH: MATH . The left column yields for the vector MATH the values MATH. Similarly the left column of REF gives MATH. Now, the middle row of REF gives MATH. Further, MATH, and by CITE, MATH. We proved in the previous section that MATH, and so much the more MATH. Hence, from the last column of REF we can deduce that MATH. We see already that MATH. Now, let us verify that the connecting homomorphism MATH of the first row is indeed an isomorphism. Look at the second column of REF . By CITE, MATH. We saw in the previous section that MATH, and so much the more MATH. Hence MATH. Hence all the connecting homomorphisms of the first row are isomorphisms. Now, coming back to REF , we see that the vertical arrow on the right hand side is obviously surjective, and we are done. |
math/9910063 | By REF , for any point MATH, one has a canonical isomorphism MATH. It is enough to show that there is a canonical isomorphism MATH . To this end, fix an equation MATH of the cubic MATH in MATH and take any section MATH vanishing on MATH; this section MATH defines an isomorphism MATH such that the substitution MATH induces MATH. Next, by REF there is a canonical isomorphism MATH . Now take a general section MATH such that MATH is a smooth projectively normal elliptic quintic. By computations of REF, MATH where MATH is the span of MATH, or of MATH, in MATH. The section MATH defines an exact sequence MATH . Twisting it by MATH and using the canonical isomorphism MATH, we obtain the exact triple MATH. Thus we have a composed isomorphism MATH which is proportional, by the above, to MATH and to MATH . Next, similar to REF we have a morphism MATH inducing a map MATH which gives, in composition with MATH, the isomorphism MATH . The exact triples MATH and MATH give together with REF and NAME duality on MATH a chain of canonical isomorphisms: MATH . Next, MATH, composed with MATH and with the canonical isomorphisms MATH, gives the following isomorphism: MATH . Composing it with REF , we obtain the isomorphism MATH . Now, composing REF , we get the desired isomorphism MATH in REF with MATH that is MATH does not depend on the choice of scalar multiples of MATH and MATH. Thus, to show that MATH is a canonical isomorphism, it is enough to check that MATH does not depend on the choice of the point MATH . The last assertion is clear, since, by construction, MATH is defined for any point MATH , thus giving a morphism of trivial bundles MATH. We have MATH, hence MATH is constant as a function on the projective REF-space MATH. The tangent distribution of the fibers of MATH is the vector bundle MATH, and its isotropy was proved in REF . This implies that the fibers of MATH are Lagrangian. |
math/9910063 | The first two isomorphisms follow from the definitions of the intermediate Jacobian and of the NAME variety, as well as the identification MATH mentioned above. The third one is a relative version of the tangent bundle formula for the intermediate Jacobian of a cubic REF-fold. To prove it, write down the relative conormal bundle sequence of MATH over MATH, taken to the third exteriour power and twisted by MATH: MATH . Applying MATH and using the NAME formulae, we obtain the canonical isomorphism MATH . We have MATH, MATH and MATH, so both sheaves in REF are canonically isomorphic to MATH. It is also well known that the incidence divisor MATH is identified with the projectivized cotangent bundle of MATH: MATH, hence there are canonical isomorphisms MATH. |
math/9910063 | The natural inclusions MATH give the following exact sequences: MATH . We have MATH. The NAME sequence MATH implies MATH. We find successively the determinants: MATH, MATH, MATH, MATH, MATH. Dualizing REF and tensoring by MATH, we obtain the exact triple MATH which gives the canonical isomorphism MATH . As MATH and by relaive NAME duality, we obtain the canonical isomorphism MATH . The natural identification MATH induces the canonical isomorphism MATH, which implies the assertion of the lemma. |
math/9910063 | One can represent MATH as the complete intersection MATH, where MATH is the universal NAME variety MATH over MATH. It is the scheme of zeros of MATH, where MATH is the equation of MATH. As MATH, we have proved the first statement. The second one follows immediately from the NAME - Welters computation of the differential of the NAME - NAME map (see REF, or REF below). |
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