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math/0003192
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Let MATH be the quadratic form determined by the Hessian at the origin. Denote the eigenvalues of MATH by MATH and note that each MATH has nonnegative real part. REF : change coordinates to make MATH exactly equal to the quadratic form MATH. Indeed since MATH, and the Hessian is nondegenerate, there is a locally smooth change of variables MATH such that MATH and the Jacobian at the origin is REF. REF : normalize by MATH. For any quadratic form MATH there is a linear change of variables MATH such that MATH. The change of variables matrix MATH satisfies MATH, the symmetric matrix representing MATH. Changing variables to MATH introduces an integrating factor of MATH which is a square root of MATH since MATH is just the Hessian. Let MATH be the region of integration over which MATH varies when MATH varies over an appropriately small neighborhood of MATH. CASE: Expand MATH into monomials. The function MATH has now become MATH, where MATH and the sign of the square root will be chosen later. We may expand MATH into monomials, using the same argument as in the proof of REF to show the remainder term can be made MATH for any MATH. It remains to evaluate the integral over the region of integration, MATH of MATH when MATH is a monomial. REF : move the region of integration to the real MATH-space. Let MATH be the projection of MATH onto MATH by setting the imaginary part to zero. We claim that changing the region of integration from MATH to MATH alters the integral by an amount rapidly decreasing in MATH. To show this, let MATH be the region MATH. The boundary of MATH (as a manifold) is composed of MATH (with opposite signs) together with MATH. For any MATH-form MATH, MATH. When MATH is a holomorphic MATH-form, we see that MATH vanishes (being the sum of MATH terms) so that MATH . We know that MATH is bounded away from REF on MATH, and its minimal value on MATH lies on MATH, hence the integral over MATH decays exponentially. REF : evaluate the integral. Factoring MATH into one-dimensional integrals and plugging into REF yields an asymptotic expansion whose leading term (when MATH) is equal to MATH. When MATH is the function MATH, then the positive square root is taken. The choice of square root must be continuous in the analytic topology on functions having nondegenerate Hessians and having eigenvalues with positive real parts, and the only such choice is the product of the principal square roots of the eigenvalues of the Hessian.
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math/0003192
|
Passing to MATH if necessary, we may assume MATH. Setting MATH for all but one index MATH, we cannot obtain the zero function (by minimality), and so some term in the expansion around MATH is a pure power of MATH, and we denote the minimal degree such term by MATH. If MATH is not a homogeneous point, then there is some MATH for which some monomial has total degree lower than MATH. Assume without loss of generality that MATH. The function MATH then has a minimal degree pure MATH term MATH, MATH, and some term MATH with MATH. In other words, the NAME Polygon of MATH around MATH has a support line passing through MATH with slope MATH in lowest terms, and MATH. It is well known that we may describe the solutions MATH of the equation MATH as follows. Write MATH for the polynomial collecting all the terms on this support line. Then for each MATH root of unity, MATH, and each root MATH of MATH, there is a solution MATH as MATH. A proof may be found in CITE. Varying MATH over the set MATH, we see that the solutions MATH must sometimes be in this set as well. For those MATH, the points MATH will be in MATH, violating minimality of MATH. By contradiction, we have shown that no monomial in the expansion around MATH has lower total degree than any pure power term, hence MATH is minimal.
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math/0003192
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Let MATH be any point on the boundary of MATH. For MATH and MATH the power series for MATH is convergent at MATH. As MATH and MATH therefore, MATH is finite and increasing. On the other hand, the power series for MATH is not absolutely convergent on MATH, since we know MATH to have some singularity on this torus. Hence MATH as MATH. Since MATH is meromorphic, it must have a pole at MATH, hence MATH and is a minimal point of MATH. As MATH varies over the boundary of MATH, we let MATH denote the curve traced out by this minimal point. Pick any MATH. The convex set MATH has horizontal and vertical support hyperplanes (by non-entirety of MATH and MATH), and therefore has a support hyperplane normal to MATH; let MATH be a point of intersection of this support plane with MATH. We have just seen that MATH is a minimal point of MATH. If MATH is a smooth point of MATH then MATH: either MATH is finitely minimal, in which REF applies, or it is toral, in which case the toral version of this theorem from CITE applies. Assume now that MATH is not a smooth point. By REF , MATH is a homogeneous point, and since MATH, MATH is a multiple point. REF then shows that MATH in this case as well. This finishes the proof.
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math/0003193
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We show that REF implies REF . To do this, we first prove that if MATH is another splitting of REF as above, with associated linear operator MATH on MATH, then for any MATH, MATH for all MATH. Indeed, from REF we have MATH for all MATH. Taking MATH and using the fact that MATH is primitive, REF gives MATH . But MATH is the class of a pure harmonic form, so the left-hand side of REF is the harmonic form associated to an element which is in the image of MATH, and hence is killed by MATH. We now claim there exists a splitting MATH such that for any MATH and MATH we have MATH . Indeed, if we let MATH be the linear transformation such that MATH, then MATH and it is an exercise to check that the space of transformations MATH is equal to the set of operators MATH on MATH satisfying MATH for all MATH and every MATH. Suppose REF holds and choose MATH satisfying REF . Let us choose a primitive basis for MATH. Applying MATH, we get half of a basis for MATH. By REF , we may apply MATH to the basis elements in MATH for each MATH to obtain another basis for MATH, which we view (via MATH) as the other half of our basis for MATH. Let MATH be one of the basis elements, and let MATH. By our conditions on MATH, a subset of our basis for MATH consists of MATH, the iterates MATH of MATH applied to MATH, the primitive element MATH satisfying MATH, and the iterates of MATH applied to MATH: MATH . The action of MATH is to send each element in REF to the element on its right, except that MATH is sent to MATH, and MATH to MATH. We construct MATH explicitly: define MATH . Then MATH (defined this way for every basis element MATH) satisfies the condition of REF . REF follows from REF by standard representation theory of MATH, as in CITE. To show that REF implies REF , note that by REF the condition in REF is independent of choice of splitting. If MATH is a nonzero primitive element and if we take MATH to be a splitting which satisfies REF , then MATH and REF fails.
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math/0003193
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Let MATH be a nonzero element of the kernel of MATH; the notation MATH refers to the direct sum decomposition REF , with respect to some splitting. We claim that MATH must be in MATH. Indeed, MATH for some MATH and MATH implies MATH since MATH vanishes. Also, by the classical hard NAME theorem, MATH. Now, if MATH denotes the arithmetic intersection pairing, then we have MATH . Hence, assuming MATH, we have MATH for every nonzero MATH. Moreover, if MATH is primitive, then the pairing of MATH with MATH has the required sign.
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math/0003193
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We shall see that REF implies REF except when MATH is exponentially large compared to MATH. For large MATH, the NAME polynomials are close approximations of classical orthogonal polynomials, in this case the NAME polynomials, and we know how to bound these. By NAME 's inequality, REF gives MATH . So, REF holds whenever MATH . Since MATH is concave increasing, the average value of MATH, MATH, MATH is at least MATH. As MATH, REF holds whenever MATH . To analyze the case where MATH is exponentially large compared to MATH, it is convenient to introduce the change of variable MATH and the rescaling MATH . Let MATH denote the MATH . NAME polynomial. It is known CITE that MATH where the constant in the error term depends on both MATH and MATH. For our purposes, we demonstrate REF Let MATH and MATH be positive integers such that MATH. Then MATH for all MATH with MATH. CASE: We have MATH whenever MATH and MATH. We have the following recurrences (CITE; for MATH this is classical) MATH with initial data MATH . Subtracting REF from REF leads to a recurrence in MATH. Then REF follows by induction on MATH, using the known bound MATH for all MATH and all MATH with MATH. REF is a corollary of REF . We have MATH when MATH or MATH. For MATH we have MATH. When MATH, we see from REF that MATH is a increasing linear function in MATH, attaining minimum when MATH, giving MATH, and maximum when MATH, giving MATH. For MATH the NAME identity CITE CITE gives MATH so the inequality is clear. When MATH or MATH, MATH is a quadratic or cubic polynomial, and it is a calculus exercise to check that MATH for every integer MATH with MATH. In fact, the integrality condition on MATH is required only when MATH, MATH. CASE: We have MATH for MATH, MATH and MATH. CASE: For MATH, we have MATH in REF whenever MATH. CASE: Assume MATH and MATH. Then MATH. The indicated bound on NAME polynomials is evident for MATH, and for larger MATH it follows from the inequality MATH . One obtains REF by using the transformed differential equation for MATH CITE; this is indicated in CITE. The proofs of REF are routine; the inequality MATH may be used for the latter. To complete the proof of REF , assume that REF fails, so that MATH. If MATH then MATH and REF follows from REF (note that the inequality in the proposition is strict unless MATH or MATH). When MATH we combine REF with REF to deduce REF . Indeed, MATH is bounded by MATH for every MATH, and by MATH over the middle half of the summation range. By pairing terms MATH with MATH and using the fact that MATH is monotone increasing for MATH, we obtain REF .
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math/0003195
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Expanding MATH as a geometric series and using unique factorization in MATH, one sees that the coefficient of MATH in the reciprocal of the left hand side is MATH times the number of monic polynomials of degree MATH, hence REF. Comparing with the reciprocal of the right hand side completes the proof.
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math/0003195
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Write the NAME expansion MATH. Then observe that MATH.
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math/0003195
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Recall that MATH is cyclic precisely when its characteristic polynomial and minimal polynomials are equal. From Subsection REF, these polynomials are equal when all MATH have at most one part. In the cycle index for MATH set MATH if MATH has at most REF part and MATH otherwise. It follows that MATH . By REF this equation can be rewritten as MATH . Recall that a product MATH converges absolutely if the series MATH converges. Thus using the crude bound MATH is analytic in a disc of radius greater than MATH. REF implies that MATH . Applying REF (with MATH, MATH and then MATH) gives that MATH .
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math/0003195
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The proof of REF shows that MATH . A matrix is separable if and only if all MATH have size MATH or MATH. Hence MATH . The result follows.
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math/0003195
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Taking coefficients of MATH on both sides of REF gives the relation MATH . Since MATH for all MATH, it follows that MATH as desired.
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math/0003195
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MATH is clearly non-negative when MATH and MATH. CITE established an equation which is equivalent to the sought identity MATH . As some effort is required to see this equivalence, we derive the identity directly using NAME 's line of reasoning. First observe that unipotent elements of MATH corresponding to nilpotent MATH matrices (subtract the identity matrix), and that the number of nilpotent MATH matrices is MATH by the NAME theorem CITE. The number of unipotent elements in MATH can be evaluated in another way using the cycle index of the general linear groups. Namely set MATH if MATH and set MATH otherwise. One concludes that MATH . Now multiply both sides by MATH, sum in MATH, and apply NAME 's identity MATH .
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math/0003195
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Recall the cycle index factorization MATH . Setting all MATH equal to MATH and using REF shows that MATH . Taking reciprocals and multiplying by the cycle index factorization shows that MATH . This proves the first assertion of the theorem. For the second assertion, use REF from Subsection REF.
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math/0003195
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From the above formula for NAME polynomials, it is clear that the only surviving term in the specialization MATH is the term when MATH is the identity. The rest is a simple combinatorial verification. (Alternatively, one could use ``principal specialization" formulas for NAME polynomials on page REF).
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math/0003195
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The first assertion is clear. The second assertion is well known in the theory of partitions, but we argue probabilistically. Multiply both sides by MATH. Then note from the first assertion that the left hand side is the MATH chance of having a partition of size MATH. Now use the second equation in the proof of REF in Subsection REF.
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math/0003195
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From the formula for MATH, the conditioned measure for MATH is the same as for MATH. Now let MATH be the number of times that coin MATH comes up heads in the NAME Tableau Algorithm. Letting MATH denote conditioning, it suffices to show that MATH . In fact (for reasons to be explained later) we compute a bit more, namely the conditional probability that MATH given that MATH. By definition this conditional probability is the ratio MATH . The numerator and denominator are computed using REF as follows: MATH . MATH . Thus MATH and the result follows.
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math/0003195
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Suppose we know that MATH is a probability measure and that MATH . Then the MATH probability of choosing a partition with MATH for all MATH is MATH . Thus it is enough to prove the (surprising) assertion that MATH for all MATH. One calculates that MATH . Similarly, observe that MATH . Thus the ratio of these two expressions is MATH as desired. Note that the transition probabilities must sum to REF because MATH for any measure MATH on partitions. Thus to complete the proof, it must be shown that MATH is a probability measure and that MATH . Since MATH it follows that MATH . From this recursion and the fact that MATH, one solves for MATH inductively, finding that MATH . NAME 's identity (page REF) gives that MATH, so that MATH is a probability measure.
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math/0003195
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This is immediate from REF .
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math/0003198
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For all MATH, we have MATH, hence MATH is a dual basis for MATH as a right MATH-module.
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math/0003198
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Define MATH as follows: for MATH, let MATH be given by MATH . Given MATH, put MATH . We invite the reader to verify that MATH and MATH are well-defined and that they are inverses of each other.
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math/0003198
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Let MATH be a finite dual basis of MATH as a right MATH-module. Then for all MATH and MATH, MATH . Define MATH by MATH, with MATH for all MATH. To show that MATH is a left MATH-linear and right MATH-linear map, take any MATH, MATH and compute MATH . Conversely, for MATH define MATH . Then for all MATH that is, MATH. Finally, MATH and MATH are inverses of each other since MATH .
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math/0003198
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The result is a translation of REF in terms of MATH and MATH, using REF (for REF) and REF (for REF). We prove one implication of REF. Assume that MATH is a NAME pair. From REF , we know that MATH is finitely generated and projective. Let MATH and MATH be as in REF of REF, and take MATH, MATH. For all MATH and MATH, we have MATH and MATH .
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math/0003198
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Consider MATH and MATH. Due to the naturality of MATH and REF there is a commutative diagram MATH . Write MATH and MATH. Then it follows that MATH . We have seen before that MATH. It is easy to prove that MATH - the left MATH-action is induced by the multiplication in MATH - and MATH is a morphism in MATH. Thus MATH and MATH are left and right MATH-linear, and MATH proving REF. To prove REF, we first observe that MATH, the left MATH-coaction is induced by comultiplication in MATH in the first factor. Also MATH is a morphism in MATH, and we conclude that MATH is left and right MATH-colinear. Take MATH, and put MATH . Writing down the condition that MATH is left MATH-colinear, and then applying MATH to the second factor, we find that MATH . Since MATH is also right MATH-colinear, MATH and, applying MATH to the second factor, we find MATH and REF follows from REF. This proves that there is a well-defined map MATH. To show that the map MATH defined by REF is well-defined, take MATH, MATH, and let MATH be given by REF. It needs to be shown that MATH, that is, MATH is right MATH-linear and right MATH-colinear, and that MATH is a natural transformation. The right MATH-linearity follows from REF, and the right MATH-colinearity from REF. Given any morphism MATH in MATH, one easily checks that for all MATH and MATH that is, MATH is natural. The verification that MATH and MATH are inverses of each other is left to the reader.
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math/0003198
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We leave the details to the reader; the proof relies on the fact that MATH is left and right MATH-linear.
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math/0003198
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This follows immediately from REF .
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math/0003198
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Suppose that MATH is a NAME pair. Then there exist MATH and MATH such that REF hold. Let MATH, and MATH. Then REF can be rewritten as MATH for all MATH. Taking MATH, MATH, one obtains REF. For all MATH and MATH, one has MATH and REF can be written as MATH for all MATH and MATH. Taking MATH and MATH, one obtains MATH . Applying MATH to the second factor, one finds REF. Conversely, suppose that MATH and MATH satisfy REF implies REF, and REF implies REF. Let MATH, MATH. Then REF hold, and MATH is a NAME pair.
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math/0003198
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Let MATH and MATH be as in REF. Then for all MATH, MATH . Write MATH and for all MATH consider the map MATH . Then for all MATH and MATH, MATH so MATH is a finite dual basis for MATH as a left MATH-module.
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math/0003198
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CASE: With notation as in REF , let MATH be the MATH-module generated by the MATH. Then for all MATH, MATH . Since MATH is faithfully flat, it follows that MATH, hence MATH is finitely generated. CASE: From descent theory: if a MATH-module becomes finitely generated and projective after a faithfully flat commutative base extension, then it is itself finitely generated and projective. CASE: Follows immediately from REF: since MATH is a field, MATH is faithfully flat as a MATH-module, and MATH is projective as a MATH-module. CASE: Follows immediately from REF.
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math/0003198
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We first show that MATH. For all MATH and MATH, we have MATH proving that MATH is left MATH-linear. It is also right MATH-linear because MATH . Notice that the dual basis for MATH satisfies the following equality (the proof is left to the reader): MATH . Using this equality one computes MATH . This proves that MATH is right MATH-colinear. Conversely, given MATH, first one needs to show that MATH. It is now more convenient to work with MATH rather than MATH. For MATH, MATH, REF can be rewritten as MATH . Take any MATH, MATH and compute MATH . This proves that MATH satisfies REF. Before proving REF, we look at the right MATH-coaction MATH on MATH. Write MATH . Using REF, we find, for all MATH, MATH . This means that for all MATH . This can be used to show that MATH satisfies REF. Explicitly, MATH . It remains to be shown that MATH and MATH are inverses of each other. First take MATH. Then for all MATH, MATH . Finally, for MATH, MATH and MATH: MATH .
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math/0003198
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We have to show that MATH is left and right MATH-linear and right MATH-colinear. For all MATH and MATH, MATH proving that MATH is right MATH-linear. The proof of left MATH-linearity goes as follows: MATH . Next one needs to show that MATH is right MATH-colinear. Using REF, one finds MATH . Conversely, let MATH and put MATH. Using REF, we see that MATH, for all MATH, hence MATH, and MATH. Take MATH. Then MATH . Finally, take MATH, and write MATH. MATH and MATH are right MATH-comodules and left MATH-modules. Since MATH is right MATH-linear, right MATH-colinear and left MATH-linear, MATH and it follows that MATH, as required.
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math/0003198
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MATH. Let MATH and MATH be as in REF. Then MATH and MATH are morphisms in MATH, and MATH . The fact that MATH and MATH are right MATH-linear and left MATH-linear implies that MATH. Similarly, for all MATH, MATH . Since MATH and MATH are left MATH-linear, MATH. MATH. Obvious, since MATH and MATH are in MATH. MATH. Let MATH be the connecting isomorphism, and put MATH, MATH. Applying REF, one finds MATH . Evaluating this equality at MATH, one obtains REF. For all MATH, MATH . Applying MATH to the second factor, one finds REF implies that MATH is a NAME pair.
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math/0003198
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Details are left to the reader. Given MATH, for MATH, the natural map MATH is MATH .
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math/0003198
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We show that MATH is well-defined, leaving other details to the reader. Consider a commutative diagram MATH . The map MATH is left and right MATH-colinear. Write MATH. Then MATH . Applying MATH to the second factor, one finds MATH . The right MATH-colinearity of MATH implies that MATH and hence proves REF. To prove REF note that MATH is left and right MATH-linear, hence MATH .
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math/0003198
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We only prove REF. If MATH is a NAME pair, then there exist MATH and MATH such that REF hold. Take MATH and MATH corresponding to MATH and MATH and write down REF applied to MATH with MATH, MATH . Taking MATH, MATH, and applying MATH to the first factor, one obtains REF. Conversely, if MATH and MATH satisfy REF, then REF is satisfied for all MATH, and REF follows since MATH and MATH are right MATH-linear. Now write down REF applied to MATH, MATH . Take MATH, MATH, and apply MATH to the first factor. This gives REF. Conversely, if MATH and MATH satisfy REF, then application of REF to the second and third factors in MATH, and then MATH to the second factor shows that REF holds for all MATH. Finally note that REF is equivalent to REF.
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math/0003198
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Observe first that MATH is a left-left entwining structure. This means that REF hold, but with MATH and MATH replaced by MATH and MATH. In particular, MATH . This can be seen as follows: rewrite REF as commutative diagrams, reverse the arrows, and replace MATH by MATH. Then we have REF in diagram form. Now fix MATH such that MATH. Then for all MATH, MATH . Write MATH. For MATH, define MATH by MATH . Then MATH is a finite dual basis of MATH as a MATH-module.
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math/0003198
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We first prove that MATH is well-defined. CASE: MATH is right MATH-linear: for all MATH, MATH and MATH, we have MATH . CASE: MATH is right MATH-colinear: for all MATH and MATH, we have MATH . CASE: MATH is left MATH-colinear: for all MATH and MATH, we have MATH . The proof that MATH satisfies REF is left to the reader. The maps MATH and MATH are inverses of each other since MATH . At the last step, we used that for all MATH and MATH, MATH .
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math/0003198
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We first show that MATH is well-defined. Take MATH, and let MATH. CASE: MATH is right MATH-linear since for all MATH and MATH, MATH . CASE: MATH is right MATH-colinear since for all MATH and MATH, MATH . CASE: MATH is left MATH-colinear since for all MATH and MATH, MATH . Conversely, given MATH, we have to show that MATH satisfies REF. Take any MATH and write MATH. Since MATH is right and left MATH-colinear, we have MATH . Therefore we can compute MATH . Thus REF follows. Finally, we show that MATH and MATH are inverses of each other. MATH . We know that MATH is right MATH-linear. Hence suffices it to show that MATH for all MATH. From REF, we compute MATH . Now write MATH and compute MATH .
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math/0003198
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We only prove MATH in REF. First we show that MATH is a left inverse of MATH. Since MATH is right MATH-linear, it suffices to show that MATH . To show that MATH is a right inverse of MATH we use that MATH is right MATH-colinear and conclude that it suffices to show that for all MATH and MATH, MATH . Both sides of the equation are in MATH, so the proof is completed if we show that both sides are equal when evaluated at an arbitrary MATH. Observe that MATH hence MATH as required.
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math/0003208
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The steady state REF with MATH has general solution MATH, where we have used that MATH has an extremum at MATH. Hence the period is MATH, and for all small MATH we see that the perturbed function MATH is another positive periodic steady state solution. Thus MATH is not asymptotically stable in MATH with respect to even perturbations.
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math/0003208
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The proof depends on a number of elementary differential equations and inequalities that we derived in Section REF, and the reader may wish to skim those sections before proceeding. If MATH then MATH, as one sees directly from the formula in REF , using that MATH and MATH. So we assume MATH from now on, and MATH. By definition, MATH . First assume MATH; then MATH where we have used CITE and the identity MATH (valid for MATH). Differentiating the function MATH (which is inspired by REF ) with respect to MATH, we find from REF that MATH . Substituting MATH from CITE yields MATH . For MATH we obtain exactly the same REF for MATH, as follows: MATH using that MATH when MATH, by CITE. The last equation is REF for MATH. REF simplifies to MATH using that MATH by REF . Hence MATH . Also MATH by REF together with the facts that MATH and MATH as MATH (see CITE) and that MATH uniformly as MATH, by CITE. REF now follows from REF and CITE (which shows MATH when MATH or MATH). For example, if MATH on MATH then MATH on MATH; since MATH we deduce MATH on MATH, and so MATH for MATH.
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math/0003210
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Let MATH be a MATH-admissible tuple and MATH. Set MATH, MATH, and MATH, MATH. Using REF , one deduces that MATH; MATH with MATH; MATH and MATH for MATH. Therefore MATH. Assume now that MATH and MATH. Choose integers MATH as follows. Set MATH. Assume that MATH is chosen. If there is no MATH such that MATH, we set MATH and stop the process. Otherwise we choose for MATH minimal MATH with MATH. As MATH, there exists MATH with MATH (observe that in this case MATH). We choose minimal such MATH for MATH. Set MATH. Since MATH, using REF and analyzing the MATH-adic expansions of MATH and MATH, one can conclude that the tuple MATH is MATH-admissible and MATH.
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math/0003210
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This follows from REF .
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math/0003210
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Applying REF , and REF, we conclude that MATH is a composition factor of MATH and for each MATH-admissible tuple MATH the set MATH, so MATH.
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math/0003210
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One can assume that MATH. MATH . Let MATH. Since MATH, we have MATH. This implies that MATH. MATH . Let MATH. Then MATH. Let MATH. Assume that MATH. Since MATH, we have MATH, which is impossible. Therefore MATH, so MATH and MATH.
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math/0003210
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One can rewrite the formula in REF as follows. MATH where MATH . Recall that by convention MATH for all MATH, and MATH is the trivial one-dimensional MATH-module. So REF holds for MATH. Assume now that MATH and REF is valid for all MATH. We shall prove it for MATH. Then the theorem will follow by induction. It follows from CITE that MATH has a filtration by NAME modules for MATH. Then the classical branching rules for characteristic REF imply MATH where MATH, MATH, and MATH . On the other hand, by REF , MATH . Since MATH and the branching rules for MATH with MATH are assumed to satisfy REF , one can determine the branching of MATH. Therefore it suffices to check that the right part of REF is equal to MATH where MATH is the right part of REF . The latter sum can be rewritten as follows: MATH where MATH . We have to show that MATH for all MATH. Note that MATH. We proceed by steps. CASE: At most one summand in REF is nonzero. In particular, MATH. Assume that MATH and MATH with MATH and MATH. Since MATH, we have MATH. As MATH, by REF , MATH divides MATH, MATH, and MATH. Hence MATH, which yields a contradiction. Now assume that MATH and MATH with MATH. As above, we get that MATH divides MATH, MATH, and MATH. Let MATH. Then by REF , MATH and MATH. Since MATH, we have MATH, so MATH. Consider the following cases. CASE: MATH and MATH. Then MATH and MATH, so MATH, which is impossible. CASE: MATH. Then MATH. Since MATH, we have MATH, so MATH. This implies that MATH and yields a contradiction. CASE: MATH. Then MATH. Therefore MATH, so as above, MATH. CASE: MATH if and only if MATH (equivalently, MATH). Assume that MATH. By REF , this is equivalent to the following: there exist MATH with MATH such that MATH and MATH. Hence MATH and MATH. By REF , MATH divides MATH and MATH. Note that MATH. Therefore MATH and MATH, so MATH, as required. Assume now that MATH. Let MATH. Then MATH and MATH. Since both MATH and MATH are contained in MATH, we have MATH. Moreover, we have MATH. Since MATH, we get MATH. Hence MATH. Set MATH. Then MATH, so MATH and MATH. It remains to observe that MATH, so MATH and MATH. CASE: If MATH, then MATH, that is, MATH or MATH. Let MATH. First assume that MATH (see REF ), that is, MATH. Then by REF , MATH. One easily checks that if MATH, then MATH, and if MATH, then MATH, as required. Assume now that there exist MATH with MATH such that MATH and MATH. By REF , MATH and MATH. If MATH, then by REF , MATH. Hence we can assume that MATH. By REF , MATH and MATH, so MATH. Assume that MATH. Then MATH and MATH. If MATH, we have MATH, so MATH, as required. CASE: If MATH, then MATH. If MATH, then MATH, so assume that MATH. We have either MATH, or MATH. One needs to show that either MATH (that is, MATH), or there exist MATH with MATH such that MATH, MATH, and MATH (that is, MATH and MATH). First assume that MATH. Let MATH. We have MATH. If MATH, then MATH, MATH, and MATH, as required. If MATH, one gets MATH, as desired. Assume now that MATH. Consider the following cases. CASE: MATH. If MATH, then MATH. Assume that MATH. Set MATH, MATH. Then MATH and MATH, as required. CASE: MATH. Then MATH, so MATH. CASE: MATH. Let MATH. Since MATH, we have MATH, so MATH. As MATH, the integer MATH. If MATH, then MATH, so MATH. Assume now that MATH. This implies MATH. If MATH, then MATH. Therefore one can suppose that MATH. Then MATH, MATH, and MATH, as required.
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math/0003210
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By REF , MATH . It follows from REF that the factors MATH come from MATH and the factors MATH, and MATH if nonzero come from MATH. Note that MATH so MATH for all MATH. Therefore by REF , MATH . Similarly, we get MATH for MATH (and for MATH if MATH and MATH). Hence MATH . (Here the symbol MATH is extended to the zero module in the natural way.) Since MATH is selfdual, MATH is selfdual also. Let MATH be the socle series of MATH. Recall that MATH has a filtration by quotients of MATH and MATH. As the factor MATH has multiplicity REF and MATH is selfdual, REF implies that MATH is a factor of MATH with MATH, so MATH. If MATH, then MATH is the composition length of MATH and the theorem follows from REF . Assume that MATH. As above, by the selfduality of MATH and REF , MATH is a factor of MATH with MATH. Assume that MATH. Then MATH, so MATH is uniserial, which contradicts the selfduality of MATH. Hence MATH and the theorem follows from REF .
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math/0003210
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CASE: Assume that MATH. Choose maximal MATH such that MATH. Then MATH. Take minimal MATH such that MATH and set MATH. Then MATH and MATH. One has MATH. Moreover, if MATH, then MATH. Therefore REF implies that MATH is a composition factor of MATH for MATH, and MATH. As MATH, we get a contradiction. CASE: In view of REF , it suffices to verify that MATH is an inductive system. By REF , we need only to check that if MATH and MATH, then MATH if MATH, and MATH if MATH and MATH. But this is clear since MATH in the first case and MATH in the second one.
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math/0003210
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REF yield that MATH, MATH, MATH, and MATH, MATH, MATH, are inductive systems for MATH. Let MATH be an inductive system of fundamental representations. It is clear that either for every MATH there exist MATH and MATH such that MATH, MATH, and MATH, or MATH for some MATH and MATH. In the first case we claim that MATH. Indeed, fix MATH and MATH, MATH. Then one can choose MATH and MATH such that MATH, MATH, and MATH. Since MATH is an inductive system, REF implies that MATH and MATH. Hence MATH. Next, suppose that MATH. Choose minimal MATH and MATH with this property assuming that MATH if MATH and MATH if MATH. (Observe that for all MATH and MATH, MATH for MATH large enough.) We shall prove that MATH and MATH with MATH (in particular, MATH for MATH and MATH for MATH). First let MATH. We claim that MATH and MATH. As MATH and MATH and MATH are inductive systems, MATH for infinitely many integers MATH. So there exists MATH such that MATH for infinitely many MATH. Choose maximal such MATH. REF yields that MATH for all MATH and MATH. Now REF and the choice of MATH imply that MATH and MATH is an inductive system. It remains to show that MATH. Suppose this is not the case. As MATH, there exist MATH and MATH such that MATH, MATH, and MATH. Since MATH, MATH, and MATH are inductive systems, this implies that for every MATH there exists MATH with MATH which contradicts the choice of MATH. Hence MATH and MATH. Now we show that MATH if MATH. As MATH, for some MATH we have MATH. Since MATH, MATH, and MATH for MATH are inductive systems, this forces MATH for all MATH. Now REF yields that MATH, as desired.
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math/0003210
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Since the restrictions of the fundamental representations of a semisimple algebraic group over an algebraically closed field to relevant NAME groups over arbitrary subfields remain irreducible and can be realized over these subfields, only REF (or REF), REF, and REF require some analysis. Let MATH be the MATH-module MATH or MATH. First assume that MATH is algebraically closed. Set MATH. Let MATH be the NAME algebra of MATH. For a root MATH of MATH and MATH denote by MATH and MATH the root elements in MATH and MATH associated with MATH. It is well known that MATH for MATH and long MATH (see, for instance, CITE). For MATH set MATH and MATH. It is clear that MATH. It suffices to show that each MATH-submodule MATH is a MATH-submodule. Obviously, MATH and MATH for all long roots MATH, MATH, and MATH. But this forces MATH for all MATH. However, using the commutator relations for the NAME groups of type MATH (see, for instance, CITE), one can deduce that the subgroup generated by all MATH with MATH, MATH, and long MATH coincides with MATH. (Here, in fact, it suffices to make computations within subgroups of type MATH and show that our subgroup contains all short root subgroups). Hence MATH is a MATH-module, as desired. Now let MATH be arbitrary. For a finite dimensional MATH-module MATH set MATH and denote the socle of MATH by MATH. Since MATH for any MATH-module MATH, we have MATH if all composition factors of MATH are absolutely irreducible. The same holds for other members of the socle series of MATH. If MATH, then MATH and MATH are multiplicity-free and their submodules are completely determined by the sets of composition factors. Therefore the arguments on socles allow us to conclude that each submodule of MATH has the form MATH for some submodule MATH. Let MATH with MATH. Then the socle of MATH contains a submodule MATH. Since MATH and MATH are selfdual and MATH has only two composition factors isomorphic to MATH, there exists a submodule MATH of MATH such that MATH and MATH is the unique submodule in MATH with MATH. Now one can see that the socle series of MATH is determined by that of MATH and is described by REF .
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math/0003212
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Choose a basis MATH for MATH . Let MATH denote the bilinear form defining the NAME bracket in MATH with respect to this basis. Let MATH denote the MATH-variety of upper triangular matrices of order MATH. Hence the MATH-points of MATH are just the upper triangular matrices of order MATH with coefficients in MATH. We have a morphism MATH which takes the matrix MATH to the subspace spanned by MATH . Let MATH denote the inverse image of MATH under this map. Since there is a finite partition of MATH into locally closed subvarieties over which MATH induces morphisms which are locally trivial fibrations for the NAME topology, it suffices to show that MATH is a constructible subset of MATH. CASE: a matrix MATH in MATH defines a subspace of codimension MATH in MATH if and only if the diagonal MATH contains exactly MATH zero entries. CASE: a matrix MATH in MATH defines a subalgebra of MATH if and only if for each MATH there exist MATH such that MATH where MATH denotes the MATH-th row of MATH CASE: a matrix MATH in MATH defines an ideal of MATH if and only if for each MATH there exist MATH such that MATH where MATH denotes the MATH-tuple with REF in the MATH-th entry and zeros elsewhere. The two conditions on a matrix MATH in REF define MATH as a constructible subset of MATH.
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math/0003212
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We just have to add the constructible condition: CASE: a matrix MATH in MATH defines a MATH-submodule of MATH if and only if for each MATH there exist MATH such that MATH .
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math/0003212
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We want to put a condition on a subspace of MATH that it is a subalgebra of codimension MATH in MATH . The trouble is now that MATH is not necessarily an ideal in MATH . However we use the fact that MATH . Let MATH be a basis for MATH with MATH a basis for MATH, and MATH be a basis for MATH with MATH a basis for MATH . Let MATH be the bilinear map defining the NAME bracket from MATH with respect to these choices of basis. There is still a morphism MATH which takes a matrix MATH, MATH, to the subspace spanned by MATH . REF a matrix MATH in MATH defines a subspace of codimension MATH in MATH if and only if the diagonal MATH contains exactly MATH zero entries and the diagonal MATH contains MATH non zero entries and MATH. REF the subspace MATH defines a subalgebra of MATH if and only if, for MATH and for all MATH there exist MATH such that MATH where there are MATH zeros. The point is that we are going to be guaranteed MATH so we just need to check that MATH is a subalgebra modulo MATH which is finite dimensional and therefore leads to a constructible subset.
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math/0003212
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Let MATH be a finite set of generators. Then, for any fixed MATH, there exists some MATH and MATH such that MATH and MATH . So we can carry out the same analysis essentially as the previous lemma. Let MATH be a basis for MATH with MATH a basis for MATH, and MATH be a basis for MATH with MATH a basis for MATH . Let MATH define the action of MATH taking MATH with respect to these choices of basis. Then REF the subspace MATH defines an ideal of MATH if and only if it is a subalgebra of MATH and for MATH and MATH for all MATH there exist MATH such that MATH where there are MATH zeros.
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math/0003212
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Again we just have to add the constructible condition: REF a matrix MATH in MATH defines a MATH-submodule if and only if, for each MATH, there exist MATH such that MATH .
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math/0003212
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Suppose MATH is a closed ideal of codimension MATH in MATH . Let us show that each ideal MATH of codimension MATH in MATH contains some fixed term of the filtration. We can assume MATH. Let us consider MATH where MATH is the lower central series of MATH which might be distinct from the closure of the lower central series with respect to the filtration MATH. The ideal MATH is closed and hence contains some term of the filtration MATH . There must be some MATH such that MATH else the ideal would have codimension greater than MATH . Hence MATH . So MATH . Now MATH since MATH is an ideal. Hence MATH and therefore MATH . Continuing this analysis shows that MATH for all MATH hence MATH . But MATH for some MATH since MATH is a closed subalgebra of finite codimension. This implies that MATH. Since MATH is closed, MATH. Similarly, if MATH is a finite extension of MATH, every closed ideal of codimension MATH in MATH contains MATH, whence MATH is a stable class of subalgebras.
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math/0003212
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This follows from the fact that a MATH-submodule MATH of codimension MATH in MATH must contain MATH. In REF we apply this to MATH which must have codimension bounded by MATH. Recall that the case MATH either is empty or else MATH is abelian and MATH .
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math/0003212
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Let MATH be a closed subalgebra of MATH of codimension MATH . Suppose MATH is the decomposition of the underlying simple NAME algebra. Then MATH where MATH is central, MATH is a derivation and MATH is an automorphism of MATH which has order MATH (MATH is assumed to have primitive MATH-th roots of unity) and MATH . We can restrict our attention to showing that a subalgebra MATH of codimension MATH in MATH must contain some fixed term MATH of the filtration on MATH depending only on MATH . For convenience we drop the subscript MATH and since we are only considering one twisted algebra at a time we put MATH . The assumption that MATH has no infinite abelian sections means that MATH is not abelian and hence MATH is perfect. This implies in turn that MATH . Let MATH . The codimension of MATH is then equal to the sum of the codimensions of MATH in MATH . Therefore MATH for at most MATH values of MATH . Therefore consider MATH there must be some value of MATH with MATH such that both MATH and MATH . Hence MATH for all MATH . Thus MATH .
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math/0003212
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This depends on the fact that any derivation MATH with MATH can be realised as a linear multiple of the NAME product MATH where MATH and MATH. For the NAME algebras MATH and MATH the result follows, since we can then realise the derivation in the grading of weight MATH as a NAME product of elements of weight MATH and MATH for all values of MATH . We shall have to work harder to get the corresponding result for the other NAME algebras MATH or MATH, since we need to realise elements of weight MATH as sums of commutators of elements of weight MATH and MATH which still stabilize the form defining the corresponding NAME algebra. For MATH let us show why, for MATH where MATH, that MATH can be realised as a linear combination of commutators of elements of weight MATH and MATH for MATH values of MATH . We are going to use the following identity which can easily be checked. Suppose that MATH then MATH . Since MATH we know that there is some MATH . CASE: Suppose firstly that MATH . Then choose MATH and MATH with MATH, MATH and MATH and MATH. For MATH, we get MATH with MATH and MATH of weight MATH and MATH of weight MATH . CASE: If MATH then choose some MATH and define MATH and MATH with MATH, MATH and MATH and MATH. Form MATH where note that MATH . To deal with MATH we can use the analysis of REF since the weight is concentrated still at MATH but now we are considering a derivation MATH with MATH . Hence choose MATH then MATH, and MATH, MATH and MATH where MATH . Note that the weight at MATH of MATH is now MATH . Hence we have MATH with MATH and MATH have weight MATH and MATH and MATH have weight MATH . This proves then that MATH is well-covered. For the Hamiltonian algebra, we can use the following identity (see REF) MATH . Consider now realising MATH of weight MATH . So there exists some MATH such that MATH . We then put MATH and MATH with MATH and hence can express MATH as a commutator of elements of the Hamiltonian algebra of weight MATH and MATH for MATH. Hence MATH is well-covered. For the contact algebra MATH, we shall use the following identity (see REF): MATH . This algebra is slightly trickier than the previous algebras. We shall need to consider realising MATH of weight MATH as a linear combination of commutators of weight MATH and MATH for MATH different values of MATH . Recall that in this algebra MATH has weight two whilst the other variables have weight one. There exists some MATH with MATH . CASE: Suppose that MATH . We prove by induction on MATH that we can express MATH as a linear combination of commutators of weight MATH and MATH for MATH . Put MATH and MATH where MATH and MATH . Then MATH . Suppose that MATH . Then we are done. We then suppose by induction that we have proved the claim for MATH . Then the identity REF shows how to express MATH as a linear combination of commutators as desired since the first expression on the right hand side of REF can be dealt with by the inductive hypothesis. CASE: Suppose now that MATH . We show how to shift the weight from MATH to the other variables such that eventually we are in a position to invoke REF . We use the following identity MATH where MATH . Note that we shall let MATH for MATH and MATH . Since MATH this will mean that MATH is always non-zero. The choice of MATH means that we can keep on applying the above identity to realise the expressions MATH as commutators MATH with an error term which has the values of MATH increasing for MATH and MATH decreasing whilst still ensuring that MATH so that we can choose MATH for MATH . Eventually the error terms will have some MATH with MATH and we can apply REF to finish the realisation.
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math/0003212
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Let MATH and MATH be the free generators. If MATH is any NAME word in MATH and MATH we define the length MATH to be the number of terms in MATH . We take a NAME set MATH as a basis for MATH (see CITE II. REF) which is defined as follows: REF if MATH and MATH and MATH then REF MATH and REF an element MATH of length MATH belongs to MATH if and only if it is of the form MATH with MATH in MATH, MATH and MATH and MATH where the ordering is defined by the ordering on the index set. For example the construction provided by REF II. REF starts with the following basis for each layer MATH: MATH . Let MATH be the dimension of MATH . Then we claim that the additive subspace of MATH generated by the following basis elements is actually a subalgebra: MATH . So this is a vector subspace of codimension REF which skips the basis elements MATH and MATH where MATH appears MATH times. The NAME basis has the property that each word MATH has a unique decreasing factorization MATH with MATH (see REF). We prove by induction on the length of MATH for MATH and MATH that MATH . If MATH has length MATH then MATH and by REF MATH is in MATH . The property of unique factorization implies it is not the element MATH . Suppose we have proved that MATH for all words MATH of length less that MATH and take MATH of length MATH . There is a unique decreasing factorization MATH with MATH . Using the NAME identity we can rewrite MATH . Now we can use our induction hypothesis applied to MATH and MATH .
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math/0003212
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We may assume MATH is contained in MATH. By REF applied to MATH and MATH, MATH since MATH. We deduce that MATH . The result follows since MATH and MATH.
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math/0003212
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By taking a cover of the NAME MATH by open NAME cells corresponding to different choices of bases of the lattice MATH, one deduces the result from the following elementary REF .
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math/0003212
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Let MATH be the subalgebras of codimension MATH. Then MATH and MATH .
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math/0003212
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The first line REF follows from REF and the remark that MATH if MATH is a subalgebra of MATH . Let MATH . Let MATH be in MATH and write MATH for MATH. We have MATH if and only if, for every MATH, there exist MATH such that MATH, for some finite field extension MATH of MATH. Here MATH is the bilinear mapping MATH corresponding to the product in MATH. Let MATH denote the matrix whose rows are MATH. We then have MATH . Then MATH if and only if, for every MATH, one can solve the matrix equation MATH for MATH, with MATH a finite field extension of MATH. Let MATH denote the adjoint matrix, we can then rewrite REF as MATH . Let MATH denote the MATH-th entry of the MATH-tuple MATH. Then the set MATH has the following description: MATH . The set MATH is therefore MATH-semi-algebraic. Let MATH denote the conjunction of REF . Since MATH is a simple function this implies that MATH is a semi-algebraic set and in particular is constructible. As we promised in REF this provides an alternative way to show the constructibility of this class of subalgebras. We prove that MATH is MATH-semi-algebraic. We have MATH if and only if, for every MATH, there exist MATH, with MATH some finite field extension of MATH, such that MATH . Let MATH denote the MATH-th entry of the MATH-tuple MATH. Then an argument similar to the above implies that the set MATH has the following description: MATH . Hence MATH is MATH-semi-algebraic. Indeed it is defined by MATH, the definable condition which is the conjunction of REF follows from REF and the last statement is clear, since when MATH is of the form MATH, the above definable conditions are all semi-algebraic.
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math/0003212
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This follows from the fact that MATH. Hence the codimension of MATH is given as usual by MATH . The codimension of MATH in MATH on the other hand is given by MATH.
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math/0003212
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Everything has already been proven, except for REF , which follows from REF .
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math/0003212
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We need to prove that the image of a MATH-closed subgroup MATH under the map MATH is a MATH-subalgebra. The proof of REF can be applied in this setting to prove that MATH . But since MATH is MATH-closed and MATH we have MATH . Hence MATH that is, MATH is additively closed. Once we have this then REF implies that MATH is also closed under the NAME bracket by applying the inverse of the NAME series. We have implicitly used the identity MATH above when MATH. This is true in fact for all values of MATH . This follows from the fact that there exist polynomials MATH such that MATH . Since MATH is true for all MATH and MATH we get that the following is a polynomial identity for all MATH and MATH . Hence this must be a formal identity of polynomials which is then true for all values of MATH . Conversely, if MATH is a MATH-subalgebra of MATH then MATH is well-defined on MATH being a finite polynomial map. By using the NAME series one obtains directly that MATH is a subgroup and the identity MATH again implies that MATH is a MATH-closed subgroup taking MATH and using the identity MATH . Since MATH and MATH are bijective maps, we have a one-to-one correspondence as detailed in the statement of the Theorem. The correspondence preserves codimension since this is defined by lengths of chains of MATH-closed subgroups or MATH-subalgebras.
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math/0003212
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The proof works just the same as the one in CITE for the explicit formula for MATH-adic cone integrals, using REF for performing the change of variable, compare the proof of REF.
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math/0003212
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This follows from REF and the fact that if MATH for almost all primes MATH then MATH .
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math/0003213
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Let MATH be a general line of MATH (in particular, MATH is the only component of MATH containing MATH), and let MATH be a plane containing MATH. Then MATH splits as a union MATH where MATH is a plane curve of degree MATH. So MATH has length MATH and it is formed by points that are singular for MATH, hence either tangency points of MATH to MATH or singular points of MATH. We will prove that, if MATH is general among the planes containing MATH, then exactly MATH of these points are singular for MATH. To this end, let us consider the family (possibily reducible) of planes MATH; its dimension is MATH. Claim. The general plane through MATH cannot be tangent to MATH in more that two points. Proof of the Claim We have to prove that MATH does not possess a MATH-dimensional family of MATH-tangent planes, with MATH . Assume by contradiction that MATH possesses such a family MATH . Let MATH be a general point of MATH, MATH. The projection MATH, centered at MATH, is a covering of degree MATH, with branch locus a surface MATH contained in MATH. There is a MATH-dimensional subfamily MATH of MATH formed by the planes passing through MATH: they project to lines MATH-tangent the surface MATH. Then MATH satisfies the assumptions of the following lemma: Let MATH be a reduced surface and assume that there exists an irreducible subvariety MATH with MATH whose general point represents a line in MATH which is tangent to MATH at MATH distinct points. Then MATH and MATH is a plane parametrizing the lines contained in a fixed plane MATH which is tangent to MATH along a curve. Therefore there exists a plane MATH tangent to MATH along a curve of degree MATH. But MATH is the projection of a MATH-space MATH passing through MATH, which must contain the planes of MATH. So these planes are MATH-tangent also to MATH, which is a surface of MATH: this means that all planes tangent to MATH are MATH-tangent. Since MATH is not a plane, this is a contradiction. Therefore, we have at least MATH singular points of MATH on MATH . Assume there are MATH . Let MATH be a hyperplane containing MATH . Let us denote by MATH the NAME cycle in MATH parametrizing lines contained in MATH . Then, for general MATH the intersection MATH is proper, namely it is purely MATH-dimensional. In fact, if infinitely many lines of MATH were contained in MATH then MATH a contradiction. Moreover, since we assume that MATH is generically reduced, both MATH and MATH are smooth at MATH . We will show, now, that if MATH contains MATH singular points of MATH then MATH and MATH do not intersect transversally at MATH and this will yield a contradiction. In fact, MATH acts transitively on MATH and we can use CITE because we have assumed that our base field MATH has characteristic zero. Before we start, let us recall briefly for the reader convenience some basic facts about MATH . Let MATH be the MATH-dimensional linear subspace corresponding to MATH that is, MATH . Then MATH can be identified with MATH hence for a non zero MATH we have MATH or MATH . In both cases we can associate to MATH in a canonical way a double structure on MATH . When MATH this structure is obtained by doubling MATH on the plane MATH hence it has arithmetic genus zero CITE. When MATH the doubling of MATH is on a smooth quadric inside MATH and the arithmetic genus is MATH . In both cases we have MATH and MATH . To prove the non transversality of MATH and MATH at MATH it is harmless to assume that MATH is not tangent to MATH at any smooth point of MATH . Therefore, the singularities of the surface MATH on the line MATH are exactly those points which are already singular for MATH . To fix ideas, let MATH be defined by the equations MATH defined by MATH and MATH defined in MATH by MATH . Then, the restriction to MATH of the NAME map of MATH is given analytically as follows: MATH . We can regard the MATH's as polynomials of degree MATH in the coordinates of MATH on MATH . Since we assume MATH has MATH singular points on MATH the four polynomials MATH have a common factor of degree MATH . Therefore, if we clean up this common factor, the above map can be represented analytically by polynomials of degree MATH . Therefore, the double structure on MATH defined by MATH has arithmetic genus MATH and it arises from a non zero vector MATH . Now, for every MATH which is a smooth point for MATH we have MATH and in particular we have MATH . This means that MATH is also a tangent vector to the NAME scheme MATH of the lines on MATH (see CITE, pp. REF), that is, MATH. Since we assume that MATH is the only component of MATH containing MATH and MATH is reduced at MATH by the usual criterion for multiplicity one, we conclude that MATH and MATH are not transversal at MATH and the proof is complete (for general facts about intersections multiplicities the reader is referred to CITE).
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math/0003213
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of REF The lines in MATH which are tangent to MATH are parametrized by a ruled threefold MATH: any line on MATH corresponds to the pencil of lines in MATH which are tangent to MATH at a fixed smooth point. Then MATH . MATH is a surface: otherwise, a general point MATH would be contained in infinitely many lines of MATH, therefore, every tangent line to a general plane section MATH of MATH would be MATH-tangent to MATH with MATH a contradiction. Let MATH be a line corresponding to a smooth point of MATH then MATH is tangent to MATH at least at points MATH. Since a general point of MATH represents a line which is tangent to MATH at a unique point, MATH has three branches at MATH . We denote by MATH the tangent spaces to these branches at MATH that is, MATH is contained in the tangent cone to MATH at MATH . We have MATH . The intersection of this plane with MATH is the union of two lines. Then, a direct, cumbersome computation proves that these lines inside MATH represent respectively the pencil of lines in MATH through MATH and the pencil of lines in MATH through MATH . Claim. For a general point MATH we have MATH . It is sufficient to show that MATH contains three distinct lines. From MATH we get that the two lines of MATH are contained also in MATH . If we translate all this into equations, an easy computation shows that MATH . By symmetry we get MATH . Therefore, the three distinct lines in MATH which correspond to the pencils in MATH of centres respectively MATH are contained in MATH, and the claim is proved. By continuity, all the tangent planes MATH belong to one and the same system of planes on MATH . Therefore, the tangent planes at two general points of MATH meet, and either MATH is a NAME surface, or its linear span MATH is a MATH . The first case is impossible because the tangent planes to a NAME surface fulfill a cubic hypersurface in MATH whereas MATH is a quadric. On the other hand, the quadric hypersurface MATH in MATH contains planes, hence it is singular. Therefore, the hyperplane MATH is tangent to MATH at some point MATH and all the lines of MATH meet the fixed line MATH in MATH . Were the lines of MATH not lying on a unique plane through MATH then any plane MATH through MATH would contain infinitely many lines MATH-tangent to the plane section MATH of MATH a contradiction.
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math/0003213
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Let MATH be a line such that MATH and MATH . Let MATH be the NAME variety parametrizing the lines in MATH which intersect MATH . The only singular point of MATH is MATH . In fact, it is easily seen that MATH is the intersection of MATH with the (projectivized) tangent space to MATH at MATH . In particular, the points of MATH different from MATH are exactly the tangent vectors to MATH at MATH which are of rank MATH . Then, by using the facts on tangent vectors to NAME briefly recalled in the proof of REF , it is easily seen that MATH is the affine cone inside MATH over a MATH . It is clear that MATH and MATH intersect properly at MATH . We claim that MATH is reduced at MATH if and only if MATH is reduced at MATH. Assume that MATH is reduced at MATH . Let MATH be the local ring of MATH at MATH and let MATH and MATH denote respectively the ideals of MATH and MATH in MATH. Then the NAME ring MATH is reduced, that is, it is a field, and we want prove that MATH is reduced. The NAME locus of MATH is certainly open and non empty. So, by genericity, we can assume that MATH is NAME. We have MATH . But MATH is generated by a regular sequence of length MATH since MATH is smooth at MATH . Therefore, the same is true for MATH being MATH a NAME ring. But MATH is a field, hence MATH is the maximal ideal of MATH. It follows that this last ring is a regular local ring. Assume, conversely, that MATH is reduced at MATH. Then MATH is reduced at MATH bevause MATH is general and because of NAME 's criterion of transversality of the generic translate, already used in the proof of REF . Denote by MATH the NAME cycle in MATH parametrizing the lines in MATH through MATH . A moment's thought shows that the local rings at MATH of MATH and MATH are the same. Then we are reduced to compute the ideal of MATH inside MATH . To do this, we replace the parametric representation of a general line MATH containing MATH namely MATH and MATH (MATH where MATH varies in the base field MATH) in all the equations MATH for MATH . From MATH we get simply MATH . Then, since the MATH are homogeneous polynomials, the other generators for the ideal of MATH at MATH are the MATH . An obvious change of variables completes the proof.
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math/0003213
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Assume by contradiction that there exists a plane MATH such that MATH for every MATH . We perform some local computations and we use the same notations as in REF . So, let MATH be an affine chart in MATH with coordinates MATH . Assume that the origin is a general point MATH of MATH and that MATH is defined by MATH . Let MATH and MATH be defined respectively by MATH and MATH . Let MATH be the equation of MATH in this chart. We write also MATH where the MATH's are homogeneous polynomials in MATH . Since the line MATH is represented in MATH by the point MATH we have MATH where the MATH are homogeneous polynomials of degree MATH or zero. Now, if we move the origin of our system of coordinates to the point MATH by a change of coordinates of type MATH and MATH for MATH and MATH (hence MATH), then in the new system of coordinates MATH is defined by the equation MATH where MATH . Now, since MATH for every MATH the above equation shows that, necessarily the linear term of MATH belongs to the ideal MATH for every MATH . Therefore, the linear forms MATH are in the ideal MATH for every MATH . But in this case the curves in MATH defined by MATH are either singular at MATH, or with tangent line MATH at MATH . This contradicts REF , and the proof is complete.
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math/0003213
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First of all, the dimension of MATH must be at least MATH otherwise MATH would contain a MATH-dimensional family of planes, hence a MATH-dimensional family of lines, a contradiction. So assume by contradiction that MATH . But then along each fibre of the NAME map there is even a fixed tangent hyperplane, contradicting REF .
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math/0003213
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Assume the contrary. Then, when MATH moves on MATH to MATH the plane MATH moves to a limit plane MATH . The intersection MATH is a curve which has the line MATH as a "double component" ; in particular, this curve is singular along MATH . Then MATH for every MATH . In fact, if MATH then MATH would be smooth at MATH contradiction.
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math/0003213
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By REF there are two tangency points of MATH to MATH on MATH and two on MATH. The point MATH is singular for MATH, but it cannot be singular for MATH because, otherwise, letting MATH and MATH vary, every point of MATH would be singular. So MATH is a tangency point of MATH to MATH. Hence, MATH is tangent to MATH at exactly three points lying on MATH or MATH.
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math/0003213
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Assume by contradiction that MATH contains the lines MATH. Then the residual curve of MATH in MATH splits as MATH. Hence, by REF , there is a new tangency point on MATH, against REF .
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math/0003213
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We assume by contradiction that MATH is a curve. Then every point of MATH belongs to infinitely many lines of MATH. The curve MATH is not a line because every line of MATH meets MATH in MATH points, and MATH . If MATH is general, from MATH it follows that through MATH there are two secant lines of MATH say MATH and MATH . By NAME 's lemma the tangent space to MATH must be constant along MATH and also along MATH . Therefore, the plane spanned by MATH and MATH is (contained in) a fibre of the NAME map, hence it is contained in MATH . So, through a general point of MATH there is a plane on MATH contradiction. This proves that MATH is a surface. To prove the assertion on the degree, we consider a general plane of MATH. If it intersects properly MATH then this intersection contains at least MATH points, and the claim follows. If the intersection is not proper, then MATH contains a family of plane curves of dimension MATH hence it is a plane. Let MATH be a hyperplane containing MATH then MATH splits as the union of MATH with a surface MATH . If MATH is general, there are at least two lines on MATH passing through MATH . Each of them meets MATH hence is contained in MATH and therefore in MATH . This shows that MATH is a union of smooth quadrics.
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math/0003213
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The first assertion of MATH is clear. To prove the second, it is enough to observe that exactly MATH lines of MATH, different from MATH, pass through a general point of MATH, and that these lines are separated by the blow-up of MATH along MATH. The assumption of MATH means that, for every MATH, the lines of MATH intersecting MATH intersect also infinitely many other lines of the family, so MATH is doubly ruled, hence it is a smooth quadric, or a finite union of smooth quadrics. In the second case, the algebraic family described by the surfaces MATH has dimension two, so this case is excluded.
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math/0003213
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To evaluate MATH we choose the lines MATH and MATH so that they intersects at a point MATH smooth for MATH . Since MATH by REF we can also assume that MATH and MATH are the only lines of MATH contained in the plane MATH, so that the lines intersecting both MATH and MATH are those passing through MATH. The conclusion follows from REF .
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math/0003213
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Note first that MATH is equal to the degree of the curve, intersection of MATH with a hyperplane MATH. We can assume MATH then MATH splits in the union of MATH with MATH other lines meeting MATH. Indeed, if MATH and MATH, there exists a line passing through MATH and meeting MATH, which is necessarily contained in MATH. Moreover, MATH and MATH meet along MATH with intersection multiplicity MATH REF . Therefore MATH. To compute MATH, the number of lines meeting MATH and contained in a MATH-space MATH, we can assume that MATH is tangent to MATH at a point MATH of MATH. In this case MATH contains the MATH lines through MATH different from MATH. To control the other MATH lines, we use the following degeneration argument. Since MATH is tangent to MATH at MATH the intersection multiplicity of MATH and MATH at MATH is MATH (this will be proved in REF ). According to the so called"dynamical interpretation of the multiplicity of intersection", in any hyperplane MATH "close" to MATH (if we are working over MATH this means: in a suitable neighbourhood of MATH for the Euclidean topology of MATH) there are two distinct lines MATH which both have MATH as limit position when MATH specializes to MATH . Note that the lines MATH and MATH are skew, because otherwise MATH which becomes MATH when MATH specializes to MATH a contradiction with REF . Therefore, we can choose a family of MATH-spaces MATH, parametrized by a smooth curve MATH, such that MATH and, for general MATH, MATH is generated by two skew lines MATH and MATH, having both MATH as limit position for MATH. The lines in MATH meeting MATH come from lines in MATH meeting either MATH or MATH. In other words, the intersections MATH and MATH both move to MATH. Therefore to preserve the degree of these intersections, the remaining lines intersecting MATH have to come from the MATH lines of MATH meeting both MATH and MATH. Note that, if MATH is one of these "remaining" lines, then the multiplicity of MATH in MATH is MATH . In fact, otherwise, MATH would be tangent to MATH at some point of MATH but MATH is already tangent to MATH at MATH and MATH . We can conclude by the previous proposition that MATH.
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math/0003213
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It was already remarked in the Introduction that for the degree MATH of MATH we have MATH. Let MATH be a general point of MATH and fix a system of affine coordinates MATH such that MATH. Let MATH be the equation of MATH . As usual, we assume that MATH is defined by MATH and, moreover, we write MATH for MATH . The polynomials MATH define (if not zero) curves in the plane MATH . In particular, MATH is a conic MATH whose points represent tangent lines to MATH having at MATH a contact of order MATH, and MATH is a cubic MATH the points of MATH represent the tangent lines to MATH having at MATH a contact of order MATH, and so on. Clearly the points of MATH corresponding to lines contained in MATH are exactly those of MATH . We have MATH at any general point of MATH because, otherwise MATH would be a hyperplane of MATH . On the other hand, since MATH at any general point of MATH we have also that MATH is not a multiple of MATH REF . In particular, we have MATH and MATH is not contained in MATH . Now, MATH so MATH is an irreducible conic (see CITE or CITE), and we are done.
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math/0003213
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The first part of the statement was already shown in the proof of REF . Moreover, in the same proof we saw that, if MATH is not tangent to MATH at some point of MATH then MATH and MATH are transversal inside MATH . In fact, the GCD of the polynomials MATH in MATH has degree exactly MATH . Hence the double structure on MATH they define has arithmetic genus MATH and does not represent any vector in MATH . Therefore MATH and the intersection is transversal. So we have proved that MATH if and only if MATH is not tangent to MATH at any point of MATH . Hence, we assume now that MATH is tangent to MATH at some point of MATH . To show that MATH we perform some local computations. Let MATH be a system of homogeneous coordinates in MATH such that the line MATH is defined by the equations MATH . Let MATH where MATH and MATH is defined by MATH . Let MATH be the related NAME coordinates. So MATH has coordinates MATH hence MATH . We will restrict, from now on, to work in the affine chart MATH of MATH given by MATH coordinates in this chart are MATH and MATH. The equations of MATH inside MATH are MATH. Then the general point of a line MATH is MATH. In a suitable system of coordinates, the equation of MATH is of the form: MATH where MATH are forms of degree MATH and MATH respectively. Here we have used the condition MATH. Moreover the homogeneous part of degree MATH in MATH of MATH can be normalized in this way because there is no fixed tangent plane to MATH along MATH CITE. From MATH it follows that the coefficient of MATH in the polynomial MATH is not zero, and we can set MATH . The point MATH is MATH in the affine chart MATH, and if we dehomogeneize MATH with respect to MATH we get MATH where MATH. The condition MATH implies that MATH is identically zero as a polynomial in MATH. If we set MATH, then we can compute MATH from REF , and we get: MATH where MATH are the constant terms of the polynomials MATH and MATH respectively. Note that MATH are some of the equations of MATH . By setting MATH we define inside the four dimensional affine space MATH a curve which is smooth at MATH the point in MATH which represents MATH . The direction of the tangent line to this curve at MATH is given by the vector MATH . Assume by contradiction that MATH . Then this vector annihilates MATH, and MATH . It follows that MATH divides MATH, hence MATH does not give any contribution to MATH. Therefore, the reduction modulo MATH (the equation of MATH in MATH) of the polynomial MATH is MATH . This is the equation of the conic MATH embedded in MATH. But, since the dual variety of MATH has dimension MATH, by REF this conic should be smooth (CITE, CITE).
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math/0003213
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We analyze in detail the case MATH. A similar proof can be given if MATH. For a different proof of this last case, see CITE. Let us recall that MATH, so given MATH general and skew, there are three lines MATH meeting both MATH and MATH. The lines MATH are pairwise skew, otherwise MATH would fail to be skew. Since MATH there exists a third line in MATH, besides MATH and MATH, meeting both MATH and MATH . The same conclusion holds for the pairs MATH and MATH . If MATH and MATH are general lines of MATH, then the three lines of MATH constructed above starting from the pairs MATH, MATH and MATH are distinct. Assume the contrary. Then there exists a unique line MATH, different from both MATH and MATH, which meets MATH. Note that all the six lines MATH, MATH, MATH, MATH, MATH, MATH are contained in the linear span of MATH and MATH. We consider now a family of pairs of lines MATH on MATH, parametrized by a smooth quasi - projective curve MATH, such that MATH and MATH are disjoint for a general MATH, while for MATH the lines MATH and MATH meet at a point MATH, general on MATH. Therefore for MATH general MATH is a MATH: we get a family of MATH-spaces whose limit position MATH is the tangent space MATH. We can assume that the plane of MATH and MATH does not contain other lines of MATH (because MATH). For general MATH, we have three lines MATH, MATH, MATH, meeting MATH and MATH, and a third line MATH, meeting MATH, MATH and MATH, which exists by assumption. For MATH, the lines MATH, MATH, MATH still meet MATH and MATH, and MATH meets MATH, MATH and MATH. Hence MATH, MATH, MATH pass through MATH. By REF , MATH cannot coincide with MATH, MATH or MATH, therefore by the assumption MATH and MATH, either MATH or MATH. Assume MATH . By REF , the intersection multiplicity of MATH and MATH is two at each of the five points corresponding to the lines MATH, MATH, MATH, MATH, MATH, therefore, by the dynamical interpretation of the intersection multiplicity, there exist four more lines in MATH moving to MATH, MATH, MATH, MATH respectively. Let MATH be a line of MATH, having MATH as limit position: by REF MATH. Let us assume that MATH. In this case, from MATH, it follows that there exist six lines in MATH, three of them meeting both MATH and MATH, three meeting both MATH and MATH. The limit position of each of these six lines passes through MATH: but in this way we get too many lines passing through MATH in MATH, contradicting the "multiplicity two " statement of REF . Therefore MATH meets either MATH (or, symmetrically, MATH) or MATH (or, symmetrically, MATH or MATH). CASE: MATH. In this case MATH, otherwise we would have four lines meeting both MATH and MATH. Also MATH (and analogously MATH and MATH), otherwise the three lines MATH, MATH and MATH would be coplanar. Therefore there exist three lines meeting MATH and MATH, two more lines meeting MATH and MATH, two meeting MATH and MATH, two meeting MATH and MATH: summing up, we get nine new lines. We get again a contradiction with REF , because we have found MATH lines tending to lines of MATH passing through MATH. We conclude that MATH. CASE: MATH. So, being MATH, MATH. In this case, we can construct four new lines, two meeting MATH and MATH and two meeting MATH and MATH. Summing up we have MATH lines moving to lines of MATH passing through MATH: this contradiction proves the Claim. Hence, given MATH and MATH, general lines on MATH, there exist lines MATH, MATH and MATH meeting both of them, and two by two distinct lines MATH, MATH, MATH meeting MATH and MATH, MATH and MATH, MATH and MATH respectively. Moreover: MATH for MATH; MATH, MATH. Using the assumption MATH, we get the existence of six more lines: MATH meeting MATH and MATH, MATH meeting MATH and MATH; MATH meeting MATH and MATH, MATH meeting MATH and MATH; MATH meeting MATH and MATH, MATH meeting MATH and MATH. Altogether there is a configuration of MATH lines obtained from MATH and MATH. The first observation is that the MATH's tend to lines through MATH, but MATH tends neither to MATH nor to MATH, because MATH meets MATH and MATH. Therefore there are three possibilities, that we examine separately: CASE: MATH; in this case the lines tending to MATH are only MATH and MATH. Now we consider MATH: there are two subcases: CASE: MATH: hence MATH. Since MATH meets both MATH and MATH, then it moves either to MATH or to MATH; similarly MATH, which meets both MATH and MATH, moves either to MATH or to MATH, and also MATH does the same. This contradicts REF . CASE: MATH: then we consider MATH, which moves either to MATH or to MATH. If MATH: then MATH, which meets MATH and MATH, goes to MATH; MATH, which meets MATH and MATH, goes to MATH; MATH which meets MATH and MATH could go to MATH or to MATH or to MATH: but all three cases are excluded by REF again. If MATH, the conclusion is similar. CASE: MATH; this case is analogous to REF . CASE: MATH. We consider MATH: since it meets MATH and MATH, it goes to MATH, or to MATH, or to MATH. The last two possibilities are excluded as in REF for MATH, so MATH and finally MATH. By considering the limit positions of MATH, MATH, MATH, we find that also in this case the " multiplicity two" statement of REF is violated. This concludes the proof.
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math/0003213
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Since we assume that MATH is not a quadric bundle we have that the dimension of MATH is MATH by REF . Then, through a general point of MATH there are infinitely many curves MATH and, by REF we conclude MATH .
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math/0003213
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The algebraic system of dimension two MATH on the surface MATH is linear because there is exactly one curve of the system passing through two general points (MATH). Also the self - intersection is equal to MATH, therefore MATH is a homaloidal net of rational curves, which defines a birational map MATH from MATH to the plane, such that the curves of the net correspond to the lines of MATH. The degree of the curves MATH is MATH by REF . So the birational inverse of MATH is given by a linear system of plane curves of degree MATH. Hence we get immediately the weak bound MATH. Let MATH denote the number of lines of MATH contained in a MATH-plane: by NAME calculus, MATH. To evaluate MATH, we consider two general skew lines MATH, MATH on MATH, generating a MATH-space MATH. The lines MATH and MATH have a common secant line MATH. The set - theoretical intersection MATH is the union of MATH, MATH and two more lines MATH, MATH by REF . Similarly we get two new lines MATH, MATH in MATH. The line MATH (respectively, MATH) cannot meet both MATH and MATH because MATH, so there are two new lines in MATH . So we have found at least MATH lines in MATH, hence MATH. The assumption MATH together with MATH gives at once MATH. Let MATH be a general hyperplane section of MATH. If MATH, then it is well known (see for example the classical book of CITE) that under our assumptions one of the following happens: MATH is a ruled surface (in particular a cone) or a NAME surface or a NAME surface with a double irreducible conic. None of these surfaces is section of a threefold MATH with the required properties. In the first case MATH would have a family of lines of dimension MATH, in the second case MATH would be a cone, in the third case MATH (see CITE and CITE). Therefore the degree of MATH is exactly MATH . We can apply, now, REF which gives MATH since MATH . If MATH denotes the sectional genus of MATH (that is, the geometric genus of a general plane section of MATH) we deduce MATH . To exclude MATH we show that there exist planes containing three lines of MATH. Indeed let MATH be a general line of MATH. We fix in MATH a MATH-plane MATH not containing MATH, intersecting MATH at a point MATH. Let MATH be a hyperplane section of MATH. By REF , MATH has degree MATH, hence there exists a trisecant line MATH passing through MATH and meeting MATH again at two points MATH and MATH. Let MATH be the plane generated by MATH and MATH: it contains also the lines of MATH passing through MATH and MATH, so MATH contains three lines contained in MATH. Now we consider MATH. By REF in MATH there must be a "new" tangency point, hence MATH contains at least five points. Therefore MATH and MATH. If MATH, the curves intersection of MATH with its tangent planes have a new singular point, so they split. Then by the NAME - NAME theorem, MATH is ruled, a contradiction. So we have MATH and MATH is a projection of a linearly normal NAME surface MATH of MATH of the same degree MATH (see CITE), which is necessarily a linear section of MATH. This proves the theorem.
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math/0003213
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Let MATH be general and set MATH for simplicity. Let MATH denote a normalization of MATH . The proof of the proposition is based on the following two lemmas. The curve MATH is irreducible, hyperelliptic of genus MATH . Hence MATH can be embedded into MATH as a smooth quintic. Let MATH be the surface parametrizing the secant lines of MATH . Let MATH be a fixed general secant line of MATH we will denote by MATH and MATH the points of MATH . The family of all secant lines of MATH that intersect MATH has three irreducible components: the secant lines through MATH those through MATH and "the other ones". This last component is represented on MATH by an irreducible curve that we will denote by MATH . There exists a birational map MATH such that the image via MATH of every curve MATH is the curve MATH on MATH just introduced. If MATH are general, then MATH if and only if MATH . We will prove now REF assuming REF . Let MATH be a general point of MATH . There are four secant lines MATH of MATH through MATH and we can assume that MATH with MATH . By REF we have MATH for every MATH . The first possibility is that, for a general MATH the four lines MATH all lie in a plane MATH . By REF the family of such planes has dimension at most MATH and, therefore, the same plane MATH corresponds to infinitely many points of MATH . This implies that every plane MATH contains infinitely many lines of MATH hence MATH . Then MATH contains at least a MATH-dimensional family of planes: a contradiction. Therefore, for a general MATH the four lines MATH all contain one fixed point MATH and we get a rational map MATH by setting MATH . This map is dominant because MATH is birational, and it has degree MATH because MATH . Hence MATH is birational to MATH via MATH . Note that MATH is not regular at the points of MATH so MATH is defined by a linear system of surfaces MATH of degree MATH all containing MATH . Let MATH be the maximum integer such that these surfaces contain the MATH infinitesimal neighbourhood of MATH . So MATH where MATH is a plane divisor in MATH . We claim that MATH and MATH . The second part of the statement of REF makes clear that any secant line of MATH is transformed by MATH into a line of MATH . Therefore we must have MATH if we intersect one of the surfaces MATH with the unique quadric surface MATH containing MATH by NAME and NAME we get MATH hence MATH . If MATH we get MATH and the surfaces MATH contain the first infinitesimal neighbourhood of MATH . Let MATH denote the saturated ideal of MATH . Since MATH is arithmetically NAME, the saturated ideal of the first infinitesimal neighbourhood of MATH is MATH (CITEEF). Now, MATH can be minimally generated by one polynomial MATH of degree MATH (the equation of MATH) and two polynomials of degree MATH therefore, every homogeneous polynomial of degree MATH in MATH must contain MATH as a factor. So the case MATH is excluded. Hence, the linear system defining MATH is a system of cubic surfaces of MATH containing MATH with multiplicity MATH . The linear system of all such surfaces defines a rational map MATH whose image is a NAME threefold, complete intersection of two quadric hypersurfaces of MATH. This completes the proof of REF .
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math/0003213
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The proof is divided into several steps. CASE: There is a birational map MATH where MATH denotes the symmetric product of the curve MATH by itself. On MATH there is the algebraic system of curves MATH of dimension MATH . Since MATH, there are exactly MATH curves of the system containing two fixed general points on MATH moreover MATH. The map MATH is defined as follows: let MATH be a general line of MATH; let MATH be the two lines of MATH intersecting both MATH and MATH . The corresponding points on MATH actually lie on MATH . We set MATH it is easily seen that MATH is birational. Note that the map MATH depends on the choice of MATH . In particular, from MATH irreducible it follows that MATH is also irreducible. CASE: The characteristic series of the algebraic system MATH on the curve MATH is a complete MATH . Therefore also the algebraic system MATH is complete. From the fact that the dimension and the degree of the algebraic system MATH are both MATH, it follows at once that the characteristic series has degree MATH and dimension MATH that is, it is a MATH . Assume it is not complete; then MATH is necessarily a rational curve and the characteristic series generates a complete MATH. In this case MATH is a rational surface and we can embed MATH into the complete linear system MATH of dimension MATH. Let MATH be the linear span of MATH inside MATH . Let MATH be the linear system of those ruled surfaces on MATH which correspond to the curves of MATH . Fix a general point MATH of MATH and denote by MATH the subsystem of surfaces of MATH containing MATH: MATH contains MATH linearly independent surfaces, hence its dimension is at least MATH: a contradiction. CASE: Let MATH denote the geometric genus of MATH . Then MATH . By the previous step we already know that MATH assume MATH . Then, by the well known fact that the irregularity of MATH equals the (geometric) genus of MATH the irregularity of MATH is MATH . But the surface MATH, which parametrizes the curves of MATH is therefore fibered by a MATH-dimensional family of lines, each line representing a linear pencil of curves MATH from MATH it follows that every such pencil has MATH base points. This also means that on MATH we have a MATH-dimensional family of linear pencils of elliptic ruled surfaces MATH each pencil having exactly two base lines. We fix one of these pencils MATH and we let MATH and MATH denote the two base lines. Every surface of the pencil is of the type MATH with MATH intersecting both MATH and MATH . Set MATH . We claim that, for general MATH the lines MATH and MATH don't meet on MATH . Indeed, if MATH, then also the fourth line of MATH through MATH would be contained in MATH the base locus of the pencil: a contradiction. So MATH is a simple unisecant for MATH. Since MATH is irreducible, from MATH it follows that MATH . Then we have a contradiction because MATH has multiplicity MATH on MATH by REF . Therefore, MATH is hyperelliptic of geometric genus MATH. To complete the proof of REF it remains to show: CASE: The genus of MATH is MATH . In particular, MATH is embedded in MATH with degree MATH . Let MATH be a general point, and let MATH denote the lines through MATH different from MATH . Moreover, let MATH be such that MATH on MATH . Then MATH is a positive divisor on MATH of degree MATH . When MATH varies on MATH the divisors on MATH of type MATH are all linearly equivalent because they are parametrized by the rational variety MATH . We denote by MATH the pencil of such divisors. Since the two rational maps MATH defined respectively by MATH and MATH are clearly different, it is easily seen that MATH . Hence, by NAME 's theorem MATH is non special. Since MATH it follows then by NAME - NAME that MATH and that MATH . Then MATH is also very ample on MATH .
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math/0003213
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Since MATH and MATH there are MATH secant lines of MATH through a general point of MATH and MATH secant lines of MATH contained in a general plane of MATH . Therefore, the class of MATH in the NAME group MATH is MATH with traditional notations. It follows that the degree of MATH is MATH this means that there are MATH secant lines of MATH intersecting two general lines MATH and MATH in MATH . Assume, now, that MATH and MATH are chords of MATH and set MATH . To compute MATH we have just to compute the number of the spurious solutions among these MATH secant lines. Let MATH be the plane generated by MATH and MATH besides MATH the plane MATH intersects MATH at the points MATH . Therefore, we have the MATH secant lines MATH on MATH . By repeating this argument for the planes MATH we get MATH spurious secant lines, MATH of them have been counted twice. Hence, MATH . It follows easily that the index of MATH is also MATH .
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math/0003213
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Let us remark first of all that the curves MATH and MATH are birational. Indeed let MATH . If MATH and MATH then the plane MATH intersects MATH at the points MATH . We get a birational map MATH by setting MATH . We fix now a general secant line MATH of MATH . Starting from the just constructed map MATH, we can also construct, in a canonical way, a map MATH which is again birational. In the first step of the proof of REF we have constructed a birational map MATH . Since MATH and MATH are birational, we get also a map MATH . Finally, the algebraic system MATH-allows us to construct a birational map MATH as follows. Let MATH be a general pair of secant lines of MATH and assume that each of them intersects MATH . By REF we have MATH one of these intersections is MATH the other one is, by definition, MATH . If we compose MATH, MATH and MATH we get the desired map MATH . It remains to show that MATH. Consider a curve MATH such that MATH intersects MATH . It is mapped by MATH to the curve on MATH formed by all the pairs of elements of MATH containing MATH . Therefore, MATH sends MATH to the curve on MATH formed by all the pairs of elements of MATH containing MATH, and clearly MATH maps this last curve to MATH .
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math/0003213
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If MATH is a fixed line of MATH, then the lines of MATH meeting it generate the rational ruled surface MATH having MATH as simple unisecant. Hence MATH results to be a rational surface. Similarly for MATH. There are two possibilities regarding the algebraic system MATH, whose dimension is two (because MATH is not a quadric bundle): either it is already linear, or it can be embedded in a larger linear system of curves in MATH, which corresponds to a linear system of rational ruled surfaces on MATH. We will prove now that the second case can be excluded. To this end, we reformulate the problem in a slightly different way. We consider the rational map MATH associated to the complete linear system MATH. The map MATH sends a point MATH to the subsystem formed by the ruled surfaces passing through MATH. From MATH, it follows that MATH contracts the lines of MATH, which are therefore the fibres of MATH. Hence MATH is a surface MATH of degree MATH. By an argument similar to that of REF , we have that MATH. The inverse images of the hyperplane sections of MATH are the surfaces of MATH, so MATH is a surface with rational hyperplane sections. We replace now MATH with a general projection in MATH, so we can apply REF - NAME and we get only three possibilities: CASE: MATH: in this case the considered algebraic system is already linear and MATH; CASE: MATH is a scroll and MATH; CASE: MATH is a NAME surface, projection of a NAME surface, with MATH. We have to prove that only the first case happens. Assume by contradiction that MATH is like in MATH or MATH . Note that any section of MATH with a tangent plane is reducible. If MATH is a scroll, such a section is the union of a line MATH with a plane curve MATH of degree MATH. Let MATH be the arithmetic genus of MATH. The following relation expresses the arithmetic genus of a reducible plane section of MATH: MATH, so MATH, MATH and MATH is a quadric. Moreover MATH, so a general ruled surface in the linear system MATH is a scroll of type MATH or MATH. The case MATH is excluded because every surface of the system should have a unisecant line and our threefold MATH contains a family of lines of dimension exactly MATH. So a general scroll of the system should be of type MATH, hence contain a MATH-dimensional family of conics. In this case MATH contains a MATH-dimensional family of conics, and a general hyperplane section MATH of its contains a MATH-dimensional family of conics. By the usual argument, MATH is a quadric or a cubic scroll or a NAME surface: all three possibilities are easily excluded. We assume now that MATH is a projection of a NAME surface. In this case MATH, so a general ruled surface in the linear system MATH is a scroll of type MATH or MATH. The reducible plane sections of MATH are unions of conics and correspond to reducible ruled surfaces on MATH, unions of two scrolls of degree three. Necessarily they are both of type MATH so each of them contains a family of conics of dimension MATH: we conclude as in the previous case. So we have proved that for both systems of lines MATH, hence MATH. Also the curves in the NAME MATH corresponding to these ruled surfaces have degree MATH. So the surface MATH ( for MATH) contains a linear system of dimension two of rational cubics, with self - intersection one: it defines a birational map from MATH to MATH, whose inverse map is defined by a linear system of plane cubic curves. Hence MATH and MATH has rational or elliptic hyperplane sections. Moreover there is a natural birational map between plane sections of MATH and some hyperplane sections of MATH. Precisely, let MATH be the singular hyperplane section of MATH given by lines meeting a plane MATH: then MATH represents lines of MATH passing through the points of MATH. Since there is only one line of MATH through a general point of MATH, we get the required birational map between MATH and MATH. We conclude that also the plane sections of MATH are rational or elliptic curves. In particular a general hyperplane section of MATH is a surface of MATH with the same property. The case of rational sections can be excluded using the NAME - NAME theorem as in REF . So a hyperplane section of MATH is a NAME surface and MATH is a (projection of) a NAME threefold. Looking at the list of NAME threefolds we get the proposition.
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math/0003213
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For every pair of indices MATH and general MATH, the surface MATH is a smooth quadric and it is clear that the linear systems MATH and MATH coincide: we call it MATH. We want to study the intersection of two quadrics belonging to two families of the form MATH and MATH, MATH. Let us remark first that, if MATH, MATH are two general coplanar lines in MATH, MATH respectively, then two cases are possible: either the plane MATH does contain a line of MATH, or it does not. In the first case MATH is a cubic REF . So if MATH and MATH, MATH, then MATH. This immediately implies that MATH. Let us consider now MATH: it can be written also as MATH for a fixed MATH and MATH varying in a ruling of the second quadric. Now MATH certainly meets all the quadrics of MATH and is not contained in any of them, so there exists a MATH such that MATH and MATH meet at a point MATH. Let MATH be the line of MATH through MATH . Then: MATH so we fall in the previous case. We conclude that two general quadrics of these families meet along a line of the family having the common index. As a consequence, we have that through a general point MATH of MATH there pass one quadric of the family MATH and one line of MATH. Now, we embed the MATH containing MATH as a subspace of a MATH and call MATH the image of the NAME embedding MATH . If MATH is a fixed general quadric of the family MATH, by acting on MATH with an element of the projective linear group, we can assume that MATH as well. Let MATH be a linear subspace of dimension MATH in "general position" with respect to MATH that is, MATH is a curve. Let MATH and MATH denote the three families of lines on MATH to fix ideas, assume that MATH contains lines of the families MATH on MATH . We define a rational map MATH as follows. Let MATH be general; then, the line MATH, such that MATH, intersects MATH at a single point MATH . Let MATH be the line (on MATH) containing MATH . Set MATH. It is clear that MATH is birational. Moreover, by considering the case of a hyperplane through MATH we see that MATH takes hyperplane sections of MATH to hyperplane sections of MATH . There are suitable MATH's in MATH let us call MATH one of them, such that the restriction MATH of the projection MATH is birational. The inverse map MATH is defined by a linear system MATH where the MATH's are three lines, pairwise skew. Since MATH takes hyperplane sections of MATH to hyperplane sections of MATH the birational map MATH is defined by a linear subsystem of MATH that is, MATH is a projection of MATH and the proof is complete.
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math/0003216
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Given MATH, we may write MATH, where MATH and MATH, for some constant MATH depending on MATH. Then MATH and MATH by the NAME Embedding Theorem, with MATH the norm of the embedding MATH, MATH by the diamagnetic inequality (see CITE). The lemma follows from this.
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math/0003216
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The proof is similar to that in CITE. The following formal argument for deriving REF is instructive, and will be helpful for obtaining the analogous result in MATH for MATH. The set of Hamiltonian quaternions MATH is the unitary MATH-algebra generated by the symbols MATH with the relations MATH . Multiplication is associative but obviously not commutative. If we identify a magnetic field MATH and a magnetic potential MATH with purely imaginary quaternionic fields on MATH then the equation MATH where MATH, is equivalent to MATH . We can solve the equation MATH by the convolution of MATH with the NAME 's function of MATH. Since MATH then MATH is the NAME 's function for MATH if MATH is the NAME 's function MATH for MATH. The identity REF is exactly this convolution of MATH with MATH.
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math/0003216
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CASE: Let MATH. Then MATH . Thus REF follows by continuity, and this implies REF once REF is established. CASE: Let MATH and MATH. Then MATH whence MATH by the diamagnetic inequality. Thus REF is established, and so REF .
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math/0003216
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Let MATH. Then MATH . Hence, MATH if and only if MATH with MATH. Moreover, for any MATH whence MATH . The result follows since MATH.
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math/0003216
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This is quite standard, but we give the short proof for completeness. We show that MATH is compact. Let MATH be a sequence which converges weakly to zero in MATH, and hence in MATH by REF . Then, in particular MATH, say. Given MATH, set MATH where MATH with support MATH and MATH say, and MATH. Then MATH . The first term on the right-hand side tends to zero as MATH by the NAME Theorem. Consequently MATH is compact and hence so is MATH. REF follows from REF .
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math/0003216
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For MATH and MATH where MATH . From REF and since MATH and MATH we have MATH for some constant MATH, and so MATH satisfies MATH and extends to an operator in MATH, the space of bounded linear operators on MATH. Thus, there exists a neighbourhood MATH of MATH such that for MATH in MATH. It follows that MATH and MATH note that MATH for any MATH. For MATH . Hence MATH, and MATH, are bounded on MATH. We may therefore write MATH where MATH and the series lies in MATH for MATH. The preceeding argument implies that with MATH, MATH has an extension in MATH. It follows from REF that MATH, and this yields the lemma.
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math/0003216
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The kernel of MATH is finite-dimensional, and we have the orthogonal decomposition MATH . With respect to this decomposition, we can represent MATH as MATH where MATH and MATH . We are required to prove that, for any MATH, there exists a neighbourhood MATH such that MATH for all MATH. We can write MATH where MATH, MATH are bounded self-adjoint operators and MATH is bounded. As MATH, we know from REF that MATH, MATH and MATH in norm. Choose a neighbourhood MATH of MATH such that MATH for MATH, where MATH is the constant in REF . Then MATH is invertible for all MATH. The operator MATH where MATH is the identity, is a bounded injection on MATH, and we have MATH . It follows that MATH whence the lemma.
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math/0003216
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We already know from REF that MATH in MATH. It is therefore sufficient to prove the theorem for MATH, where MATH. Define MATH . The theorem will follow if we prove that MATH, in view of the compactness of MATH. We shall prove that MATH is both open and closed. Since MATH, we know that MATH. It is clear from REF that MATH is open. To prove that it is closed, let MATH be a sequence in MATH and MATH; we may assume that MATH. In the notation of the proof of REF , set MATH . Then, from REF MATH and MATH . If we can prove that MATH for all MATH in some deleted neighbourhood MATH of MATH, it will follow from REF that MATH, as required. Since MATH, then any minor MATH of MATH of order greater than MATH must vanish when MATH. Hence, since MATH is analytic, MATH in some neighbourhood MATH of MATH, and so MATH in MATH. By REF there exist a minor of MATH of order MATH which does not vanish on some subsequence of MATH, and hence can have a zero only at MATH within some neighbourhood MATH of MATH. Consequently, MATH for MATH, and, the theorem is proved.
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math/0003216
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Let MATH in REF be denoted by MATH and set MATH. We shall prove that MATH is an open subset of MATH; the theorem will then follow from REF since the density of REF is a consequence of REF . For MATH, let MATH be magnetic fields which satisfy MATH. Then, if MATH are the associated vector potentials given in REF , MATH for some MATH. It follows as in the proof of REF that, with MATH and MATH, MATH where, for MATH, MATH and MATH on using REF and NAME 's inequality. Moreover, MATH satisfies MATH and MATH for MATH sufficiently small. Also, as MATH, MATH in MATH. It follows that, as MATH, MATH in MATH, and that, as in the proof of REF , the map MATH is upper semi-continuous. The set REF is therefore open and the theorem is proved.
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