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math/0006174 | By the definition of MATH, there is a unique representative up to isomorphism for the NAME class of MATH and it is both flat and regular. Hence the map MATH is an isomorphism. Then the tautological bundle MATH constructed above identifies an analytic neighborhood of MATH with the local semi-universal deformation space of MATH. Since MATH, it acts trivially on MATH and thus the local semi-universal deformation of MATH is in fact locally universal. |
math/0006174 | Choose a component MATH of the fiber product MATH, and let MATH be the induced map. It suffices to show that MATH in the obvious notation. Since both sides are algebraic, it suffices to show the following: let MATH be an irreducible curve such that MATH. Then MATH. Now choose a line bundle MATH of degree zero on MATH such that, if MATH is the canonical section of MATH, then MATH is a finite set of points. Let MATH be a point such that MATH. Since MATH, it follows from REF that, in an analytic neighborhood MATH of MATH, the bundle MATH is pulled back via MATH from MATH. It follows that the line bundle MATH and its canonical section are also pulled back via MATH. Thus MATH as claimed. |
math/0006174 | This follows from CITE, Corollary to REF , p. REF or CITE. |
math/0006174 | The groups MATH and MATH act on MATH by conjugation, and the action of MATH factors through the projection to MATH. Thus, the sheaves MATH and MATH are identified as well. |
math/0006174 | Let MATH and let MATH correspond to the identity in MATH. For a MATH-algebra MATH, if MATH, let MATH. Then there exists a unique homomorphism MATH corresponding to MATH, so that MATH is a MATH-algebra. Now MATH is the image of MATH under MATH, and the element MATH defines a unique homomorphism MATH. Thus MATH defines a unique homomorphism MATH. Conversely, suppose given a homomorphism MATH . Then MATH induces a homomorphism MATH and thus an element MATH induced by MATH, and the homomorphism MATH then gives an element MATH mapping to MATH. Clearly these are inverse constructions. It follows that MATH is represented by MATH. |
math/0006174 | We claim that the functor MATH is representable by an affine space. The proof is by induction on the length of the filtration MATH. If this length is zero, then MATH and there is nothing to prove. Suppose that the claim has been verified for every group and filtration of length less than MATH, and let MATH be a filtration of MATH length exactly MATH satisfying the hypotheses of REF . If MATH is the first term in MATH, then the filtration MATH of MATH has length MATH. By induction, the functor MATH corresponding to MATH is representable. Moreover, there is an obvious morphism of functors MATH. Now suppose that MATH is a MATH-algebra and that we are given an object MATH of MATH, in other words a pair MATH, where MATH is a principal MATH-bundle over MATH and MATH is an isomorphism from the principal MATH-bundle over MATH induced by MATH to MATH. We define the functor MATH on MATH-algebras MATH as follows: MATH consists of isomorphism classes of pairs MATH, such that MATH is a principal MATH-bundle over MATH and MATH is an isomorphism from the principal MATH-bundle over MATH induced by MATH to the pullback MATH of MATH to MATH, where MATH is the morphism corresponding to the homomorphism MATH. There is a natural map MATH. First we claim: There exists a a principal MATH-bundle MATH over MATH lifting the principal MATH-bundle MATH over MATH. In other words, for all MATH, MATH. By REF , the obstruction to finding such a lift lives in the group MATH. By REF , we can identify MATH with the sheaf MATH, which is the pullback via MATH of the vector bundle MATH on MATH. Now since MATH is affine, MATH . Since MATH by hypothesis, we can lift MATH to a bundle MATH. Once we know that there exists one lift MATH as in the claim, it follows from REF that the set of all such pairs MATH is classified by MATH. Next we claim that the map MATH is injective, and thus identifies MATH with the fiber MATH. To see this, it follows from the general formalism of nonabelian cohomology that there is a transitive action of MATH on the fibers of the map from MATH to MATH, which identifies the fibers with the coset space MATH. In our case, the unipotent group MATH is filtered, with successive quotients contained in MATH. Hence MATH is injective. Using REF and the fact that MATH is affine, MATH . If MATH is a basis for MATH, with dual basis MATH, this says that MATH. Thus MATH is representable by an affine space over MATH. Applying REF and induction, the functor MATH is then representable by an affine space, in other words there exists an affine space MATH and an object MATH such that the couple MATH represents MATH. The fact that the functor MATH defined above is representable implies in particular that there is a universal MATH-bundle over MATH, where MATH is the affine space representing MATH. A formal reduction also shows that obvious extension of MATH to a functor from schemes of finite type over MATH to sets is also representable by MATH. Lastly we must analyze the action of MATH on MATH. Suppose that MATH is proper and that MATH is a linear algebraic group. Then MATH is also a linear algebraic group, and the natural set-theoretic action of MATH on MATH is an algebraic action. Moreover, this action lifts to an action on MATH. Let MATH be the functor from MATH-algebras to groups corresponding to MATH: for a MATH-algebra MATH, MATH is the group of automorphisms of the pullback of MATH to MATH. Since MATH is proper and MATH is affine, MATH is representable by a linear algebraic group scheme MATH over MATH. To see this, first assume that MATH. If MATH is the vector bundle corresponding to MATH, MATH is an affine open subset of the affine space MATH and we claim that this linear algebraic group represents the associated functor. Indeed, an automorphism of MATH is the same thing as a section of MATH, in other words a MATH-valued point MATH of MATH, such that the determinant of MATH is an invertible element of MATH, and this is the same thing as a morphism from MATH to the NAME open subset of MATH consisting of elements with nonzero determinant. In general, choose an embedding of MATH as a closed subgroup of MATH for some MATH, defined by polynomials MATH. It is then straightforward to verify that MATH is representable by a closed subgroup of the corresponding group scheme for MATH. If MATH is the functor associated to MATH, there is an obvious morphism of functors MATH: if the points of MATH correspond to pairs MATH, where MATH is an isomorphism from the principal MATH-bundle over MATH induced by MATH to MATH, then the automorphisms of MATH act by composition with MATH. Since MATH and MATH are representable by MATH and by the affine coordinate ring of MATH respectively, there is a corresponding morphism MATH, which is easily checked to give an action. It again follows formally by representability that this action lifts to an action on MATH. This concludes the proof of REF . |
math/0006186 | The ideal MATH is the sheaf of sections of the sub-bundle MATH. By its construction, this bundle is MATH-invariant and its fiber over MATH is the subspace MATH. This latter subspace is a (minimal) MATH-invariant subspace of MATH where MATH is the stabilizer in MATH of MATH. Since MATH is a symmetric space, it follows that the sub-bundle MATH is parallel with respect to the NAME connection of the MATH-invariant metric associated to the NAME form MATH on MATH. Equivalently, MATH is a subspace of MATH. Since the NAME connection is torsion-free, the exterior derivative MATH on MATH-forms is just MATH where MATH is the bundle map induced by wedge product. The differential closure and holomorphicity of MATH now follow immediately. |
math/0006186 | Fix MATH with MATH and let MATH denote the symmetric group on MATH. Recall from CITE (whose notation I will follow) that one can associate to MATH an irreducible, unitary representation MATH. For example, MATH corresponds to the trivial representation of MATH, while MATH corresponds to the alternating representation, that is, MATH. Let MATH be the corresponding character. Define MATH for MATH. According to CITE, the formula MATH holds for any MATH. Using manipulations similar to those in REF can be rewritten in the form MATH where, for MATH and MATH, I have set MATH so that MATH is a MATH-form on MATH with values in MATH. Since MATH is a submersion, REF shows that MATH is indeed positive. |
math/0006186 | Fix any positive definite Hermitian inner product on MATH and define the corresponding invariant forms MATH on MATH. If MATH, then, since it is represented by MATH, which is positive by REF , and since MATH is NAME, the Generalized NAME Inequality of REF implies that MATH must vanish identically. |
math/0006186 | By REF and the positivity of MATH, it follows that, when MATH and MATH, the form MATH vanishes on the NAME variety MATH. As has already been seen, at each smooth point MATH, the tangent plane MATH is of type MATH and so must belong to MATH. Conversely, every subspace MATH of type MATH is tangent to the NAME cell MATH for some flag MATH on MATH. Since MATH is dense in the smooth locus of MATH, it follows that MATH must belong to MATH. |
math/0006186 | It follows from REF , together with the discussion in CITE of NAME 's construction of the of NAME functors (especially REF ), that, when the MATH-form MATH is written locally in the form MATH for some local MATH-forms MATH, these latter forms must be a local basis of the subspace MATH. The statements of the lemma follow immediately from this and REF . The representation-theoretic details are left to the reader. |
math/0006186 | It has already been noted that MATH for some MATH if and only if MATH for all MATH with MATH and MATH. Since, by REF , MATH is positive, the equation MATH holds if and only if MATH vanishes on MATH. In turn, by REF , this holds if and only if MATH is an integral manifold of MATH for all MATH with MATH and MATH, as claimed. |
math/0006186 | REF follows from REF together with REF and the fact that MATH. For REF , one direction is easy: If MATH is an integral element of MATH, then, by the first statement, MATH is an integral element of MATH for all MATH with MATH. Conversely, suppose that MATH has dimension MATH and is an integral element of MATH for all MATH with MATH and MATH. Then REF implies that the form MATH vanishes on MATH. Since MATH pulls back to MATH to be a strictly positive MATH-form, the Generalized NAME REF implies that MATH must vanish on MATH as well, that is, that MATH is an integral element of MATH, as desired. Finally, REF now follows from REF . |
math/0006186 | Set MATH and suppose that MATH is such that MATH and MATH. By REF , any subspace MATH of type MATH is an integral element of MATH. Since the dimension of such a MATH is MATH, it follows that MATH has integral elements of dimension MATH. It remains to show that MATH has no integral elements of dimension MATH. By the defining property of MATH, in the equation MATH all of the coefficients MATH with MATH are positive. It follows that if MATH were an integral element of MATH of dimension MATH, then MATH would be an integral element of MATH for all MATH with MATH. Since all such MATH satisfy MATH, REF , would then imply that MATH was an integral element of MATH, which is absurd, since MATH has no positive dimensional integral elements. |
math/0006186 | Fix MATH and consider any basis MATH of MATH with the property that MATH is spanned by MATH. Let MATH be the dual basis of MATH. The identification MATH can be written in the form MATH where MATH are a basis for the MATH-forms on MATH. In terms of these MATH-forms, the MATH-forms MATH with MATH generate MATH on MATH (see REF). A subspace MATH of dimension MATH is defined by a set of MATH independent linear relations among the MATH. Let MATH denote the restriction of MATH to MATH, so that exactly MATH of the MATH are linearly independent. The hypothesis that MATH be an integral element of MATH is then just that MATH so I assume these quadratic relations from now on. The MATH and (hence) the MATH depend on the choice of MATH. Choose the basis MATH so that the maximum number, say MATH, of MATH are linearly independent. (That is, so that the first `column' of MATH contains the maximal number of linearly independent MATH-forms.) Note that MATH satisfies MATH. By making an allowable basis change, I can assume that MATH are linearly independent and that MATH for MATH. Setting MATH, MATH, and MATH in REF yields MATH. Thus, it follows that MATH. All of the forms MATH must be multiples of MATH, since, otherwise, a new allowable basis MATH could be found that would result in at least two independent forms among the corresponding MATH, contradicting the maximality of MATH, which is equal to MATH. Since MATH, there must be MATH forms among MATH that are linearly independent modulo MATH. By making a basis change that fixes MATH, I can assume that MATH are linearly independent, but that MATH when MATH. Since there cannot be two linearly independent forms among MATH for any MATH, it follows that MATH for MATH and MATH, but it has already been shown that MATH for MATH and MATH. Since MATH for MATH, it follows that MATH for MATH and MATH. Finally, when MATH satisfies MATH, the same argument that showed that MATH is a multiple of MATH when MATH shows that MATH is also a multiple of MATH when MATH. Of course, this implies that MATH when MATH. The result of all this vanishing is that MATH . Since MATH is the identity map, MATH is just inclusion. In particular, MATH is a subspace of MATH where MATH, as desired. For the converse, just note that, when MATH, it follows that MATH when MATH. Since the left hand side of REF clearly vanishes when MATH, it follows that all of these expressions must vanish on MATH. Thus, MATH is an integral element of MATH. |
math/0006186 | Apply REF and the complementarity principle REF . |
math/0006186 | Apply REF and the duality principle REF . |
math/0006186 | Apply REF and the duality principle REF . |
math/0006186 | I will maintain the basic notation established during the proof of REF , especially the identification MATH, which will be used implicitly throughout the proof. Now, the ideal MATH is generated by the MATH-forms MATH . Note that MATH is skewsymmetric in its upper indices and symmetric in its lower indices. As in the proof of REF , let MATH be an integral element of MATH of dimension MATH and let MATH be the restriction of MATH to MATH. Then exactly MATH of the MATH are linearly independent and they satisfy the cubic relations MATH where MATH and MATH. Before embarking on the classification, I first verify that each of the four types of subspaces listed in the lemma are indeed integral elements of MATH. If MATH is of the first type, then it is possible to choose the basis MATH so that MATH for all MATH. In other words MATH is zero unless MATH or MATH. Since the expression on the right hand side of REF vanishes identically unless MATH, MATH, and MATH are distinct, it follows immediately that these expressions all vanish on MATH, that is, that MATH is an integral element of MATH. If MATH is of the second type, then it is possible to choose the basis MATH so that all of the MATH with MATH are multiples of a single MATH-form, say MATH. Again, since the expression on the right hand side of REF vanishes identically unless MATH, MATH, and MATH are distinct, it follows that every potentially nonzero term in any of these expressions contains a wedge product of two forms that are multiples of MATH, and hence must vanish. Thus, all of these expressions vanish on MATH, so that MATH is indeed an integral element of MATH. If MATH is of the third type, then it is possible to choose the basis MATH so that MATH unless MATH and so that MATH, while MATH. It is now not difficult to verify directly that all of the expressions on the right hand side of REF vanish. If MATH is of the fourth type, then it has dimension MATH, so any MATH-form on MATH is trivially zero. Hence, all of the MATH-dimensional subspaces MATH are integral elements of MATH. Now, suppose that MATH is an integral element. There is nothing to prove unless MATH is at least MATH, so assume this. I am going to show that MATH necessarily lies in an integral element of one of the first three types. Since no integral element of one of these types lies in an integral element of a different type, it will then follow that they are all maximal. As before, choose the basis MATH so as to have the maximum number MATH of linearly independent MATH in the first `column' and make a basis change so that MATH are linearly independent while MATH for MATH. Then the argument made in the course of REF shows that all of the forms MATH for MATH must be linear combinations of MATH (or else the maximality of MATH would be contradicted). Now, setting MATH, MATH, and MATH and MATH in REF yields MATH. Thus, MATH. First, suppose that MATH. Since MATH is a multiple of MATH when MATH, there must be at least MATH linearly independent forms in the first `row' of MATH, that is, among MATH. Write MATH for MATH and consider the MATH-dimensional subspace MATH of MATH spanned by the MATH elements MATH and the element MATH (note that the sum only contains terms with MATH). Then MATH contains MATH. If the rank of MATH is greater than MATH, then MATH, so MATH is an integral element of MATH of the second kind. If the rank of MATH is less than or equal to MATH, then MATH is a subspace of MATH where MATH is any MATH-dimensional subspace that contains MATH and the range of MATH. Thus, MATH lies in an integral element of the first kind. Either way, the assumption that MATH implies that MATH is a subspace of an integral element of one of the kinds listed in the lemma. Next, suppose that MATH, so that MATH, but MATH for MATH. Then MATH for all MATH. Since MATH, there must be at least one MATH-form in MATH that is nonzero modulo MATH. By making a basis change in MATH and in MATH, I can assume that MATH. Suppose, first, that it is possible to make such a basis change so that the four MATH-forms MATH are linearly independent. Since any three elements in any column of MATH must be linearly dependent, it follows that MATH for all MATH. However, it has already been shown that MATH for all MATH and the linear independence of MATH then implies that MATH for all MATH. Once this has been established, the same argument that showed that MATH for all MATH can be applied to the second column of MATH to conclude that MATH for all MATH. Combining these two congruences yields that MATH for all MATH. In other words, MATH is a subspace of the span of MATH, that is, MATH where MATH is the MATH-plane spanned by MATH and MATH. Thus, MATH lies inside an integral element of the first type. Suppose, then, that for any choice of basis, MATH are linearly dependent. By making a basis change in MATH, I can assume that the linear dependence is that MATH, that is, that MATH for some MATH. By subtracting MATH times the first column from the second column, I can assume that MATH, so MATH for some MATH. On the other hand, adding MATH times the first column to the second column and wedging together the first (that is, top), second, and MATH-th entries of the result gives MATH which must vanish for all values of MATH. The MATH-coefficient is MATH, which is already known to vanish. The MATH-coefficient is MATH. Since this must vanish as well, it follows that MATH for MATH, so that there exist numbers MATH for MATH so that MATH. First, suppose that MATH. Then the vanishing of the constant coefficient of the above expression yields MATH, which, combined with MATH implies that MATH for MATH. Now, since the top two entries of the second column of MATH are linearly independent, the same argument as was applied to the first column applies to the second and, indeed, to any linear combination of the first and second. In particular, it now follows that, for all MATH, MATH for any MATH and MATH. Using the fact that MATH is nonzero and separating the terms out by MATH-degree then leads to the conclusion that MATH for all MATH and all MATH. In other words, the only nonzero entries of MATH are in the first two rows. Thus, MATH is a subspace of MATH where MATH is the MATH-plane spanned by MATH and MATH. Thus, MATH lies inside an integral element of the first type. Thus, suppose, from now on, that MATH, that is, that MATH. If MATH for MATH, then all of the entries in the first two columns of MATH beyond the first two rows are zero. In particular, if I were to add MATH times the first column to the second, I would have a new second column whose only nonzero entries were the top MATH and the second entry MATH. It would then follow that MATH for all MATH and for all MATH. Separating out the powers of MATH in this expression, it would then follow that MATH so that MATH for all MATH. Thus, write MATH for MATH. If all of the MATH vanish, then, again, MATH is a subspace of MATH where MATH is the MATH-plane spanned by MATH and MATH, so again, MATH lies inside an integral element of the first type. If not all of the MATH vanish, then there is some integer MATH that is the rank of the MATH-MATH matrix MATH. By making a basis change in MATH and MATH, I can assume that MATH for MATH and that MATH otherwise when MATH, so I do this. I want to show that MATH for MATH. Suppose I can do this, say, MATH for all MATH. Then MATH will be a subspace of the MATH-dimensional integral element of MATH that is spanned by MATH and the element MATH, that is, an integral element of the second type of the lemma, and this subcase will be completed. To prove this claim, first suppose that MATH, consider the `column' obtained by first adding MATH times the first column of MATH and MATH times the second column of MATH to the MATH-th column of MATH, and then wedging together the first (that is, top), second, and MATH-th entries. This must vanish, so MATH . If this vanishes for all MATH and MATH, then MATH so, by the linear independence of MATH, it follows that MATH, that is, that MATH when MATH. Next, suppose that MATH. Consider the `column' obtained by first adding MATH times the first column of MATH and MATH times the second column of MATH and then the third column to the MATH-th column of MATH, and then wedging together the first (that is, top), second, and third entries. This must vanish, so MATH . Again, since this vanishes for all MATH and MATH, MATH so, by the linear independence of MATH, it follows that MATH, that is, that MATH when MATH. Thus, the desired claim is established. The only subcase left to treat now is when not all of the MATH vanish, so assume this. By making a basis change in MATH, I can assume that MATH, but that MATH for MATH (and MATH, of course). The argument applied to the first column that showed that all of the forms MATH with MATH must be linear combinations of MATH can now be applied to the second column. The result is that all of the forms MATH with MATH or MATH must be linear combinations of MATH. Explicitly, there are constants MATH, MATH, MATH when MATH so that, when MATH, MATH . If MATH for all MATH, then MATH lies in the span of the elements MATH and the element MATH . Consequently, MATH lies in an integral element of the second kind listed in the lemma. Thus, the subcase that remains to be treated is when not all of the MATH and MATH vanish, so assume this. (Note, by the way, that this subcase can only occur if MATH.) By subtracting from the MATH-th column MATH times the first column and MATH times the second column (which is effected by an appropriate basis change in MATH), I can actually assume that MATH, so I do this. I claim that MATH for all MATH. To see this, note that, adding to the MATH-th column MATH times the first column and MATH times the second column and then wedging together the top three entries of the resulting column gives MATH . This must vanish for all values of MATH and MATH. Expanding this out and taking the MATH coefficient yields MATH, so MATH as claimed. Now, by making a basis change in MATH, I can assume that MATH for MATH while MATH, so I do this. Now, I claim that MATH for MATH. This has already been established for MATH and MATH. If there were some MATH for which MATH, then the second, third, and MATH-th entries of the third column would be linearly independent, contrary to hypothesis. Thus, MATH for MATH. Now, if MATH, for some MATH and MATH, then adding the MATH-th column of MATH to the third column will produce a column with three linearly independent entries. Thus, MATH for MATH and all MATH, as claimed. The entries of MATH that remain to be understood are MATH (the remainder of the first row). Since MATH, there are constants MATH so that MATH. Adding MATH times the first column and MATH times the second column to the third column and wedging the top three entries yields MATH . Since this must vanish for all MATH and MATH, this gives MATH. Thus MATH. For MATH, adding the MATH-th column to the third column has the effect of replacing MATH by MATH in the upper left hand MATH-MATH minor. The above argument can then be repeated to conclude that MATH as well. Now, exchanging the first and third rows and then multiplying the top row by MATH yields a MATH whose upper left hand MATH-MATH minor is of the form MATH while all of the other entries of MATH vanish. Thus MATH has dimension MATH and has the third type listed in the lemma. Finally, it has been shown that every integral element of MATH of dimension at least MATH lies in either a (MATH-dimensional) integral element of the first type, a (MATH-dimensional) integral element of the second type, or a (MATH-dimensional) integral element of the third type. It only remains to observe that none of the integral elements of the second type lie in an integral element of the first type, and none of the integral elements of the third type lie in an integral element of either of the first two types. Thus, the first three types listed in the statement of the lemma are each maximal. The only integral elements not accounted for are the maximal ones of dimension at most MATH. Since every MATH-dimensional subspace is an integral element, the ones that do not lie in a subspace of any of the first three types must be maximal. The classification is now complete. |
math/0006186 | Apply REF and the complementarity principle REF . |
math/0006186 | Combine REF . |
math/0006186 | Again, the notation established in the previous proof-analyses will be maintained. The first difference is that the ideal MATH is generated by MATH-forms of the form MATH where MATH satisfies the relations MATH . (Essentially, MATH is the general element of MATH.) Note that MATH satisfies MATH . As in the proofs of REF , let MATH be an integral element of MATH of dimension MATH and let MATH be the restriction of MATH to MATH. Then exactly MATH of the MATH are linearly independent and they satisfy the cubic relations MATH for all MATH that satisfy the relations REF. Before going on to the classification, it is a good idea to verify that the subspaces listed in the statement of the lemma are indeed integral elements of MATH. If MATH for some line MATH, then it is possible to choose the basis MATH so that MATH is spanned by MATH. In this case, MATH for all MATH. Since the relations REF imply that MATH for all MATH, it follows that the right hand side of REF must vanish identically for all MATH satisfying REF. Thus, MATH is an integral element of MATH. If MATH for some line MATH, then it is possible to choose the basis MATH so that MATH is spanned by MATH. In this case, MATH for all MATH. Thus, the right hand side of REF vanishes unless MATH. However, the remaining expression MATH vanishes because MATH. Thus MATH is an integral element of MATH. Now, on to the classification. Fix MATH, and MATH satisfying MATH. Let MATH satisfy MATH while MATH and suppose further that MATH except in these three cases. Then MATH satisfies REF. The relation REF specializes in this case to MATH . Now, setting MATH and MATH, this relation reduces to the simple relation MATH . Thus, REF holds whenever MATH. The relation REF must hold on MATH and, moreover, because the condition of being an integral element of MATH is unaffected by the choice of basis MATH, it follows that these relations among triples of matrix entries in any MATH-MATH minor of MATH must continue to hold when MATH is pre- or post-multiplied by any matrices. This device will be very helpful in what follows. Now suppose that MATH is an integral element of MATH of dimension MATH. (Unless MATH, there is nothing to prove.) Suppose that the basis MATH has been chosen so as to have the maximum number MATH of linearly independent forms in the first column of MATH and that, moreover, it has been arranged that MATH for MATH. Then all of REF-forms MATH with MATH must be linear combinations of MATH (otherwise, the maximality of MATH would be contradicted). Suppose, first, that MATH. Then, for MATH and any MATH and MATH satisfying MATH, the relation MATH shows that MATH is a linear combination of MATH and MATH, so it follows that MATH is actually a basis for the MATH-forms on MATH. In other words, MATH. Thus, choosing any MATH, MATH, and MATH satisfying MATH and MATH satisfying MATH, the relations MATH and the independence of MATH imply that MATH. In other words, there are constants MATH so that MATH when MATH. Making a basis change in MATH (which has the effect of post-multiplying MATH by an invertible MATH-MATH matrix), I can assume that MATH for MATH. I claim that, for this choice of basis, MATH whenever MATH. To see this, fix any MATH and MATH satisfying MATH. Add the MATH-th row of MATH to the first row, resulting in a new matrix MATH. Choose a MATH with MATH and wedge together the MATH, MATH, and MATH entries of MATH, obtaining MATH . However MATH are linearly independent, so MATH. Thus, MATH for MATH, as desired. Now, if any MATH with MATH and MATH were nonzero, I could add the row that it appears on to, say, the top row, and get a new MATH that still satisfies all of the hypotheses so far but has a nonzero entry on the top row after the first column. Since I have just shown that this is impossible, it follows that MATH for all MATH. Of course, this implies that MATH, so that MATH lies in an integral element of the second kind. Now suppose, instead, that MATH. Then, by the first part of the argument, there must be MATH-forms among the MATH that are linearly independent modulo MATH. By a change of basis in MATH, I can assume that MATH are linearly independent and that MATH for MATH. Recall that, by hypothesis, MATH. Now, I claim that MATH for all MATH. To see this, first note that, when MATH, the relation MATH implies that MATH is a linear combination of MATH for MATH and MATH. However, the maximality of MATH has already shown that MATH. Thus, MATH when MATH. If MATH were nonzero for some MATH, then adding the MATH-th column of MATH to the second column would produced a MATH that still satisfied the MATH relations, but had a nonzero entry in the second column other than the top entry. It has just been shown, though, that this is impossible. Thus, MATH whenever MATH. Of course, this implies that MATH, so that MATH lies in an integral element of the first kind. Thus, the argument has shown that any integral element of MATH of dimension MATH or more lies in an integral element of one of the first two types listed in the statement of the lemma, as desired. |
math/0006186 | It is immediate that, for any MATH, the submanifold MATH is an integral manifold of MATH. Suppose that MATH satisfies the stated hypotheses and let MATH denote the smooth part of MATH, which is connected since MATH is irreducible CITE. By hypothesis, MATH is an integral manifold of MATH, that is, its tangent planes are integral elements of MATH. Since MATH, REF implies that for every MATH there is a MATH-plane MATH and a line MATH so that MATH . Now consider the set MATH consisting of the set of pairs MATH so that MATH and CASE: MATH spans MATH; CASE: MATH spans the annihilator of MATH; and CASE: MATH spans MATH. Then MATH is a holomorphic MATH-bundle over MATH where MATH is the parabolic subgroup that stabilizes MATH, MATH, and MATH, that is, MATH . From now on, all computations will take place on MATH or subbundles of MATH. Since MATH and MATH are connected, MATH is connected as well. Consider the structure equations MATH . By REF and the definition of MATH, it follows that MATH for all pairs MATH satisfying either MATH and MATH or MATH and MATH. On the other hand, MATH. Consequently, taking MATH and MATH satisfying MATH and computing exterior derivatives via the structure equations yields MATH . Since MATH and MATH, this implies MATH when MATH. Since MATH for all pairs MATH satisfying MATH, it follows that MATH . Thus, the span of the MATH-valued functions MATH is locally constant on MATH. Since MATH is connected, this span is constant. Let MATH be this span. By construction MATH contains MATH for all MATH. Thus, MATH lies in MATH. Since MATH is dense in MATH, it follows that MATH itself lies in MATH, as claimed. |
math/0006186 | Since MATH, it follows that MATH vanishes on MATH, the smooth part of MATH, for all MATH with MATH and MATH. Consider the positive MATH-form MATH. (This is where the hypothesis MATH is used.) By NAME 's REF MATH . Since MATH, it follows that MATH, so every term on the right hand side of the above equation vanishes on MATH. Thus MATH vanishes on MATH as well. Since MATH defines a NAME form on MATH, the generalized NAME REF implies that MATH must vanish on MATH. In particular, MATH is an integral manifold of MATH. Since MATH is irreducible and of dimension MATH, REF applies. The statements about degree now follow immediately. |
math/0006186 | Since MATH is generated by its sections, there exists a MATH and a surjective holomorphic bundle map MATH. Let MATH be the rank of the kernel bundle MATH. The mapping MATH defined by MATH then has the property that MATH. Moreover, MATH. Since MATH and MATH is a positive MATH-form, it follows that MATH is a positive MATH-form on MATH that represents MATH. Since MATH is compact and NAME, the hypothesis MATH implies that the representing positive form MATH must also be zero by REF . Equivalently, MATH is an integral variety of MATH. Since MATH is connected, MATH is irreducible. Now there are three cases: If the dimension of MATH is equal to MATH, then MATH is constant and MATH is trivial. This falls into both of the two cases allowed by the proposition. If the dimension of MATH is equal to MATH, then MATH is an algebraic curve MATH. Replace MATH by its normalization if necessary and define MATH to be the pullback to MATH of the bundle MATH. Then MATH. If the dimension of MATH is greater than MATH, then REF implies that there exists a MATH-plane MATH so that MATH. Let MATH be a subspace of dimension MATH that is a complement to MATH. The bundle MATH restricted to MATH splits as a sum MATH where MATH is a line bundle and MATH is trivial. Explicitly, for MATH, MATH and MATH. Setting MATH and MATH yields the desired splitting. |
math/0006186 | Apply complementarity to the proof of REF . |
math/0006186 | Apply complementarity to the proof of REF . |
math/0006186 | Let MATH be the space of global sections of MATH, a vector space of dimension MATH. Let MATH be the evaluation mapping, which, by assumption, is surjective, so that MATH. The kernel MATH is then a subbundle of rank MATH and can be used to define a mapping MATH that satisfies MATH. Consequently, MATH . Thus, the inequality MATH follows directly from REF . Moreover, if equality holds, MATH must be an integral variety of MATH. As in the proof of REF , there are now three cases: If MATH is a single point, then MATH is trivial. If MATH is a curve, let MATH be its normalization and let MATH be the pullback of MATH to MATH. If MATH has dimension greater than MATH, then REF implies that there is a MATH-plane MATH so that MATH. However, this implies that MATH is a subset of MATH. In other words, MATH consists of the global sections of MATH that vanish at all points of MATH. Of course, this implies that MATH, that is, that MATH, so that MATH, which is what needed to be shown. |
math/0006186 | Let MATH be an algebraic variety of codimension MATH and degree MATH. It is immediate that MATH is of codimension MATH. Moreover, a simple local calculation shows that, at its smooth points, its tangent spaces are integral elements of MATH (compare REF ). Thus, MATH vanishes on MATH, which implies that MATH for some MATH. Since MATH when MATH, it follows easily that MATH in all cases. Now, for the converse statement, it clearly suffices to prove the characterization when MATH is irreducible, so assume this. Thus suppose that MATH satisfies the stated hypotheses and is irreducible. Let MATH denote the smooth part of MATH, which is connected, since MATH is irreducible. Since MATH, this smooth part MATH must be an integral manifold of MATH, that is, its tangent planes must be integral elements of MATH. Thus, by REF , for every MATH there exists a line MATH and a codimension MATH subspace MATH so that MATH . Consider the set MATH consisting of the set of pairs MATH so that CASE: MATH spans MATH; CASE: MATH spans MATH; and CASE: MATH spans MATH. Then MATH is a holomorphic MATH-bundle over MATH where MATH is the parabolic subgroup that stabilizes MATH, MATH, and MATH. From now on, all computations will take place on MATH or subbundles of MATH. Of course, since MATH is connected and since MATH is connected, it follows that MATH is connected as well. Consider the structure equations MATH . By REF and the definition of MATH, it follows that MATH and, moreover, that these two relations are the only linear relations among the MATH forms MATH with MATH. Note that these latter forms generate the module of forms that are semibasic for the fibration MATH. Computing exterior derivatives via the structure equations yields MATH . Reducing these equations modulo MATH yields MATH . Since MATH are linearly independent modulo MATH, these equations imply that MATH . Thus, there exist functions MATH on MATH so that MATH. The structure equations now imply that MATH . Consequently, the map MATH is a holomorphic map of constant rank MATH. Moreover, since the forms MATH are MATH-semibasic, and MATH has connected fibers, it follows that there is a well-defined holomorphic map MATH of constant rank MATH that satisfies MATH, that is, MATH. By dimension count, the fibers of MATH have dimension MATH. Moreover, by construction, for each line MATH, the fiber MATH is embedded as a submanifold of the sub-Grassmannian MATH, which also has dimension MATH. It follows that MATH is an open subset of MATH. Since MATH is algebraic, it follows that MATH must actually contain MATH for all MATH. Let MATH be the set of lines MATH for which MATH lies in MATH. Then MATH is evidently a variety that has at least one component MATH of dimension MATH. Since, in particular, MATH must contain all of the MATH-planes that meet MATH, it follows that the dimension of MATH cannot be more than MATH (otherwise, it would impose at most one condition for a MATH to meet MATH). By connectedness, the image MATH must lie in a component of MATH, say MATH that has the maximum possible dimension, namely MATH. Let MATH denote the set of MATH-planes MATH satisfying MATH. Then, since MATH has dimension MATH, it follows that MATH has the same dimension as MATH. Since MATH is irreducible, MATH. Now, if MATH had any other component MATH of dimension MATH, then the corresponding MATH would also satisfy MATH and, consequently, MATH. However, this would imply that every MATH that meets MATH must also meet MATH and vice versa. But if MATH, this is absurd. Thus, MATH is unique. Finally, the equation MATH and the converse follow by the NAME calculus and calculation, respectively. |
math/0006186 | Let MATH be an algebraic variety of dimension MATH and degree MATH. A simple local calculation verifies that MATH as defined in the proposition has codimension MATH in MATH and that its tangent space at a smooth point is of type MATH. Consequently (compare REF ), it follows that MATH for some MATH. Since MATH if and only if MATH is a linear MATH and since, in this case, MATH is a NAME cycle of type MATH, it follows that MATH in this case. It now follows easily by a degree argument that MATH in all cases. Now, to prove, the converse, it clearly suffices to treat the case in which MATH is irreducible, so assume this. Thus, suppose that MATH satisfies the stated hypotheses and let MATH denote the smooth part of MATH, which is connected, since MATH is irreducible. Since MATH, this smooth part MATH must be an integral manifold of MATH, that is, its tangent planes must be integral elements of MATH. Thus, by REF , for every MATH there exists a REF-plane MATH and a hyperplane MATH so that MATH . Consider the set MATH consisting of the set of pairs MATH so that CASE: MATH spans MATH; CASE: MATH spans MATH; and CASE: MATH spans MATH. Then MATH is a holomorphic MATH-bundle over MATH where MATH is the parabolic subgroup that stabilizes MATH, MATH, and MATH. From now on, all computations will take place on MATH or subbundles of MATH. Of course, since MATH is connected and since MATH is connected, it follows that MATH is connected as well. Consider the structure equations MATH . By REF and the definition of MATH, it follows that MATH and, moreover, that these two relations are the only linear relations among the MATH forms MATH with MATH. Note that these latter forms generate the module of forms that are semibasic for the fibration MATH. Computing exterior derivatives via the structure equations yields MATH . Reducing these equations modulo MATH yields MATH . Since MATH are linearly independent modulo MATH, these equations imply that MATH . Thus, there exist functions MATH on MATH so that MATH. The structure equations now imply that for all MATH, MATH . Consequently, the map MATH is a holomorphic map of constant rank MATH. Moreover, since the forms MATH are MATH-semibasic, and MATH has connected fibers, it follows that there is a well-defined holomorphic map MATH of constant rank MATH that satisfies MATH, where MATH. In particular, MATH. The same sort of argument as was made in REF now shows that there is an irreducible variety MATH of dimension MATH such that MATH is the closure of MATH and, moreover, that MATH consists exactly of the union of the MATH for MATH. Details are left to the reader. Finally, the equation MATH and the converse follow by the NAME calculus and calculation, respectively. |
math/0006186 | By REF , every integral element of MATH of dimension at least MATH is either an integral element of MATH or of MATH. Moreover, these two ideals have no integral elements of dimension MATH or more in common. Thus, any irreducible integral variety of MATH is either an integral variety of MATH or of MATH. Now apply either REF or REF , as appropriate. |
math/0006186 | In each case, the homological assumption implies that MATH vanishes on MATH and, hence, that MATH is an integral variety of MATH. Now apply REF . |
math/0006186 | Let MATH be the space of global sections of MATH, a vector space of dimension MATH. Let MATH be the evaluation mapping, which, by assumption, is surjective, so that MATH. The kernel MATH is then a subbundle of rank MATH and can be used to define a mapping MATH that satisfies MATH. Consequently, MATH . Thus, the inequality MATH follows directly from REF . If equality holds, MATH must be an integral variety of MATH. There are now five cases: If MATH is a single point, then MATH is trivial. If MATH is a curve, let MATH be its normalization and let MATH be the pullback of MATH to MATH. If MATH is a (possibly singular) surface MATH, let MATH be the pullback of MATH to MATH. If MATH has dimension greater than MATH, then REF implies that MATH is either an integral variety of MATH, in which case MATH, so that REF applies, or else of MATH, in which case MATH, so that REF applies. |
math/0006186 | Verifying that each of the types listed is indeed an integral manifold of MATH is relatively straightforward. Simple calculations via the moving frame show that the tangent spaces to these subvarieties at their smooth points are integral elements of MATH. Alternatively, the calculations below will provide a direct proof. Thus, let MATH be an irreducible integral variety of MATH. If MATH, there is nothing to prove, so assume that MATH. Let MATH be the smooth part of MATH. Then for each MATH, the subspace MATH is an integral element of MATH of dimension at least MATH. For MATH, let MATH be the smallest subspace for which MATH is contained in MATH. The function MATH defined by MATH is equal to its maximum value, say MATH, on a subset MATH that is open and dense in MATH and connected. (For any MATH, the set of MATH for which MATH is easily seen to be an analytic subvariety of MATH.) Suppose, first, that MATH. Then MATH is an integral element of MATH for all MATH, so MATH and, hence, MATH and MATH are integral varieties of MATH. By REF , there exists a MATH so that MATH. Choosing MATH, it follows that MATH. Thus, MATH lies in an integral manifold of the first category listed in the proposition. Suppose, second, that MATH. In this case, I claim that, for each MATH, the space MATH lies in a unique maximal integral element, namely MATH. To see this, first note that MATH is a maximal integral element of the first type. Now, since it has dimension at least MATH, MATH does not lie in an integral element of the fourth type listed in REF . Also, it cannot lie in an integral element of the third type, because these integral elements have dimension MATH, which would force MATH to equal a maximal integral element of the third type, but these integral elements do not lie in any integral element of the first type. Finally, suppose MATH were to lie in an integral element of the second type, say MATH, where MATH is a line and MATH has the property that MATH has rank at least MATH. Now, is easy to see that the only subspaces of MATH that have dimension at least MATH and that lie in a subspace of the form MATH where MATH has dimension at most MATH are the subspaces of MATH. Consequently, if MATH were to lie in MATH, then it would follow that MATH lies in MATH. But this would violate the assumption that MATH. Thus, the claim has been established. Because of the evident uniqueness of MATH for MATH, the family of vector spaces MATH is a holomorphic subbundle of MATH of rank MATH. Now I need to introduce another invariant. For MATH, say that a subspace MATH is free if the composition MATH is surjective. For each MATH, the set of non-free subspaces of MATH of dimension MATH is an algebraic subset of MATH. Let MATH be the dimension of the largest free subspace of MATH and let MATH be the maximum of MATH for MATH. Let MATH be subset consisting of those MATH for which MATH. Since the complement of MATH in MATH is evidently a proper analytic subvariety of MATH, it follows that MATH is open and dense in MATH and is connected. Now, I claim that MATH. This follows by elementary linear algebra from the assumptions MATH and MATH, so I will leave this to the reader. Suppose first that MATH and let MATH be the set of pairs MATH that satisfy the conditions CASE: MATH span MATH, CASE: MATH span a free subspace of MATH, and CASE: MATH spans MATH. Then MATH is connected, as follows from the connectedness of MATH. Consider the usual structure equations: MATH . The conditions defining MATH imply that MATH whenever MATH and MATH. Moreover, the freeness assumption implies that the MATH entries of the matrix MATH are linearly independent on MATH. Now, when MATH and MATH, the structure equations combined with the stated vanishing of forms give MATH . This implies, because of the stated linear independence, that MATH for each MATH. Since MATH by hypothesis, the stated linear independence implies that MATH, for all MATH. In turn, this implies that MATH . In other words, the MATH-plane spanned by MATH is locally constant on MATH. Since MATH is connected, this implies that there is a fixed MATH that is spanned by MATH at all points of MATH. By construction, this implies that MATH contains MATH for all MATH. Of course, this implies that MATH lies in MATH, and hence that MATH lies in MATH. Thus, suppose instead that MATH. For each MATH, the set of lines MATH that are free is the complement in MATH of an algebraic subset and is therefore open, dense, and connected. By hypothesis, for any MATH-plane MATH, the induced mapping MATH is not surjective. I claim that, for MATH outside a closed algebraic set in MATH, the mapping MATH has rank MATH. Certainly, the set of MATH for which the rank of MATH is at most MATH is an algebraic subvariety of MATH, so it suffices to show that it is not everything. However, this follows by linear algebra since MATH and MATH. I leave details to the reader. Say that a MATH is semi-free if the rank of MATH is MATH. When MATH is semi-free, the annihilator of MATH is a line in MATH. There are two subcases now to consider. The first is when the tensor rank of a generator of this line is generically equal to MATH. The second is when the tensor rank of a generator of this line is equal to MATH everywhere. Consider the first subcase and let MATH be the open, dense, connected subset consisting of those MATH for which there exist MATH so that the rank of a generator of the annihilator of MATH in MATH is equal to MATH. Let MATH be the set of pairs MATH that satisfy the conditions CASE: MATH span MATH, CASE: MATH spans MATH. CASE: MATH span a semi-free plane MATH, and, moreover, the annihilator of MATH is spanned by MATH. Then MATH is connected, as follows from the connectedness of MATH. Consider the usual structure equations: MATH . The conditions that define MATH give MATH whenever MATH and MATH. Moreover, the semi-freeness and the assumption about the annihilator imply that the four MATH-forms MATH satisfy exactly one linear relation, MATH, and are otherwise linearly independent on MATH. Just as in the case MATH, when MATH or MATH and MATH, the structure equations combined with the stated vanishing of forms give MATH . This implies, because of the stated linear independence, that MATH when MATH or MATH. Now, however, because the span of MATH intersects the span of MATH in the multiples of MATH, these congruences only imply that MATH for all MATH. Setting MATH for MATH and MATH and MATH and substituting this back into the relation MATH for MATH and MATH shows that MATH for MATH and MATH and MATH. Thus, MATH for MATH and MATH with these ranges, just as in the MATH case. In turn, this implies that MATH . In other words, the MATH-plane spanned by MATH is locally constant on MATH. Since MATH is connected, this implies that there is a fixed MATH that is spanned by MATH at all points of MATH. By construction, this implies that MATH contains MATH for all MATH. Of course, this implies that MATH lies in MATH, and hence that MATH lies in MATH. This finishes the first subcase of MATH. Consider the second subcase, in which the rank of the annihilator of MATH in MATH is equal to MATH for all semi-free MATH and MATH. Let MATH be the set of pairs MATH that satisfy the conditions CASE: MATH span MATH, CASE: MATH spans MATH. CASE: MATH span a semi-free plane MATH, and, moreover, the annihilator of MATH is spanned by MATH. Then MATH is connected, as follows from the connectedness of MATH. Consider the usual structure equations: MATH . The conditions defining MATH imply that MATH whenever MATH and MATH. Moreover, the semi-freeness and the assumption about the annihilator imply that the three MATH-forms MATH are linearly independent, while MATH. When MATH, the structure equations combined with the stated vanishing of forms give MATH so MATH for some functions MATH. The structure equations then give MATH which implies, first, that MATH for all MATH and then that there must exist functions MATH so that MATH. If all of the MATH vanish identically, then MATH when MATH, which implies, as before, that MATH . In other words, the MATH-plane spanned by MATH is locally constant on MATH. Since MATH is connected, this implies that there is a fixed MATH that is spanned by MATH at all points of MATH. By construction, this implies that MATH contains MATH for all MATH. Of course, this implies that MATH lies in MATH, and hence that MATH lies in MATH. Suppose, instead, that the MATH do not vanish identically. The set MATH on which at least one of the MATH is nonzero is the complement of an analytic subvariety of MATH and hence is open and dense in MATH and connected. Its image MATH is easily seen to be the complement of a proper analytic subvariety in MATH, so MATH is open and dense in MATH and is connected also. Now, I claim that for MATH, there exist functions MATH on MATH so that MATH. To see this, differentiate the relation MATH for MATH and MATH, which now yields MATH . Since not all of the MATH vanish, this implies the desired relations. Now that this has been established, the fact that there must be at least MATH one-forms among MATH that are independent modulo MATH shows that MATH and that MATH spans the unique line MATH with the property that MATH meets MATH in a subspace of dimension MATH. Write MATH where MATH has dimension MATH. Consider the subset MATH consisting of the MATH for which MATH is spanned by the MATH for which MATH. The above arguments show that MATH is a MATH-bundle where MATH and so is connected. Consequently, there are well-defined mappings MATH and MATH with the property that, for all MATH and MATH, the span of MATH is MATH and the span of MATH is MATH. In addition to the relations already found, the relations MATH hold for MATH while the MATH one-forms MATH are linearly independent and generate the semibasic forms for the map MATH. When MATH, the structure equations and the stated and derived vanishing (including MATH) give MATH . Since not all of the MATH vanish, it follows that MATH on MATH. Thus MATH is foliated by hypersurfaces that are the leaves of MATH. Since MATH is semibasic for MATH and since the fibers of this submersion are connected, this foliation pushes down to define a codimension MATH foliation MATH of MATH. The tangent space to the MATH-leaf through MATH is simply the unique rank MATH subspace of MATH of dimension MATH, namely MATH. Now, differentiating the relation MATH yields MATH so MATH for some function MATH on MATH. On the other hand, applying the structure equations to expand the relation REF yields the relation MATH which implies that MATH. Thus, MATH for some function MATH on MATH. Thus, setting MATH for notational consistency, the identities derived so far imply that, for MATH, MATH . Consequently, the differential of the mapping MATH has rank equal to MATH everywhere and each fiber is a union of leaves of the foliation MATH. In other words, if MATH, then MATH intersects MATH in a subvariety of dimension MATH. Now suppose that MATH is algebraic (as well as irreducible). Then MATH is the complement of a proper algebraic subvariety MATH and it is not difficult to see that MATH is the restriction of a rational map of MATH into MATH whose indeterminacy locus is contained in MATH. Since the rank of the differential of MATH is equal to MATH on MATH, an elementary argument shows that there is an irreducible algebraic curve MATH so that the graph of MATH over MATH is contained in MATH. In particular, the closure of this graph is contained in MATH. Now consider the subvariety MATH that is the union of the MATH for MATH. At its smooth points, the tangent spaces to MATH are integral elements of MATH, so MATH is an integral variety of MATH. Now, for MATH, the intersection MATH has dimension MATH and contains MATH, so it follows that MATH itself is contained in MATH. Since MATH is NAME dense in MATH, it follows that MATH is contained in MATH as well, so the case MATH is finally completed. Third, suppose that MATH. Each tangent space MATH for MATH must then lie in a maximal integral element of MATH of either the second or third types listed in REF . Now, none of the vectors in a maximal integral element of the third type is of tensor rank one when regarded as an element of MATH, while the vectors of tensor rank one in a maximal integral element of the second type form a canonical subspace of codimension MATH. Since the dimension of MATH is at least MATH, it follows that MATH lies in a unique integral element of exactly one of these two types, depending on whether or not MATH contains any vectors of tensor rank one. Consequently, there are two possibilities: Either there is an open, dense, connected subset of MATH consisting of those MATH for which MATH lies in a maximal integral element of the second type, or else, there is an open dense, connected subset of MATH consisting of those MATH for which MATH is an integral element of the third type. I will now treat these two cases in turn. Thus, suppose first that MATH is an open, dense, connected subset of MATH with the property that MATH lies in an integral element of the second type for all MATH. In particular, there exists a line MATH so that MATH has dimension MATH for all MATH. As before, let MATH be the subspace of dimension MATH so that MATH. The uniqueness of this line MATH implies that the family of lines MATH is a holomorphic line subbundle of MATH, while the family of subspaces MATH is a holomorphic subbundle of MATH. For each MATH, let MATH be the subspace that satisfies MATH. The rank of the differential of MATH is at least MATH everywhere since the kernel of MATH lies in MATH. Suppose first that that MATH is a holomorphic map whose differential has rank equal to MATH everywhere on MATH. Then, again, just as in the concluding subsubcase of the MATH argument, MATH is foliated in codimension MATH by leaves of the form MATH for MATH. Since MATH is an irreducible algebraic variety, it is not difficult to show that MATH is a rational map from MATH to MATH and, again, it follows, just as in the previous argument, that there is an irreducible algebraic curve MATH with the property that MATH contains the graph of MATH. In particular, MATH is a subvariety of MATH and therefore belongs to the second category of the proposition. Thus, suppose that the rank of MATH is sometimes greater than MATH. I am going to show that this implies that MATH and then that MATH necessarily belongs to the third category of the proposition. Let MATH be the NAME open subset on which the rank of MATH reaches its maximum and let MATH denote the set of pairs MATH that satisfy CASE: MATH spans MATH, CASE: MATH spans MATH, CASE: MATH spans MATH. Then MATH is a MATH-bundle over MATH where MATH, so MATH is connected. Consider the structure equations as usual. By the construction of MATH, the forms MATH with MATH and MATH together with the forms MATH with MATH are pairwise linearly dependent. Choose a MATH-form MATH (which will be unique up to multiples) so that there exist MATH and MATH so that MATH when MATH and MATH while MATH when MATH. By construction, the forms MATH are a basis for the MATH-semibasic MATH-forms on MATH. Moreover, because MATH is an integral element of the second type in REF , the rank of the MATH-MATH matrix MATH is at least equal to MATH everywhere. Now, I claim that MATH cannot be integrable. Indeed, suppose that MATH. Then, taking MATH and MATH, expanding out the structure equation MATH, using the congruences MATH when MATH and MATH, and reducing modulo MATH yields the relation MATH . Since MATH, these congruences, together with the linear independence of the MATH-forms MATH, imply that MATH. Thus, set MATH. Then the structure equations so far imply the relations MATH for MATH and this implies that MATH has rank MATH everywhere, contrary to hypothesis. Thus, MATH is not integrable, as claimed. Next, I claim that MATH for MATH and MATH. This follows because when the pair MATH satisfy these restrictions, expanding the structure equation MATH, using the congruences MATH when MATH and MATH, and reducing modulo MATH yields MATH, which implies MATH. Expanding the structure equation for MATH when MATH and MATH and reducing modulo MATH yields MATH . Since the matrix MATH has rank at least MATH everywhere, it follows that MATH must be decomposable modulo MATH. Moreover, since MATH for MATH, it follows that MATH for at least two distinct values of MATH. If MATH were to have rank equal to MATH at any point, then it would have rank MATH on a dense open set and this would force MATH to hold for at least three distinct values of MATH, which would, in turn, force MATH to hold, contradicting the nonintegrability of MATH. Thus, MATH has rank equal to MATH at all points. Because MATH has constant rank equal to MATH, it is now possible to define a sub-bundle of MATH on which MATH is normalized to some particular normal form. The choice of this normal form is not important for the structure of the argument, but a judicious choice (made with the desired end result in mind, I must confess) that simplifies the notation is to normalize so that MATH and so that MATH for all pairs MATH satisfying MATH and MATH except MATH and MATH. The subset MATH on which this holds is easily seen to be a MATH-subbundle over MATH with a connected structure group MATH that will be made explicit later on in the argument when it will be useful to do so. I will now use MATH to stand for MATH. Now, I claim that MATH. Suppose, instead that MATH. Then, MATH while MATH. Differentiating these two equations and reducing modulo MATH then yields MATH . Of course, this implies that there exist functions MATH and MATH so that MATH and MATH. Substituting these relations into the equations MATH for MATH and MATH, then yields MATH . Since MATH are linearly independent by hypothesis, this is impossible unless MATH, but this vanishing would make MATH integrable. Thus, MATH, as claimed. Since MATH for MATH, differentiating these equations and reducing modulo MATH yields MATH from which it follows that MATH for all MATH. Thus, there exist MATH for MATH so that MATH for all MATH. Differentiating the relations MATH and reducing modulo MATH implies MATH so there exist MATH, MATH, MATH, and MATH so that MATH and MATH. Substituting this into the derivatives of the equations MATH and MATH and reducing modulo MATH yields MATH so it follows that MATH. The structure equations now imply MATH . Since the rank of MATH is greater than MATH, it follows that MATH cannot vanish and, hence, that the rank of MATH is identically equal to MATH. To save writing, introduce the abbreviations MATH and MATH. Thus, for example, MATH. Moreover, MATH and MATH. I am now going to reduce to the case MATH. (If MATH already, these next two paragraphs are unnecessary.) Fix MATH. Differentiating the identities MATH and using the structure equations yields MATH that is, MATH and MATH. On the other hand, differentiating the equation MATH and reducing modulo MATH yields MATH . In view of the congruences for MATH and MATH, this gives MATH. Consequently, MATH for all MATH. Moreover, there exist MATH and MATH so that MATH and MATH for MATH. However, I now claim that MATH. This follows since, if I now differentiate the relations MATH and MATH and reduce modulo MATH, the result is MATH from which the claim follows, since MATH. This vanishing, in turn, now implies the congruences MATH . In other words, the MATH-plane MATH is locally constant on MATH and hence, by connectedness constant. It follows that MATH lies inside MATH for some MATH. Thus, it clearly suffices to take MATH for the rest of the argument, so I will do so. I am now going to reduce further to the case MATH. (I remind the reader that MATH in what follows. If MATH already, then these next two paragraphs are unnecessary.) This argument is essentially the `dual' of the argument just given. Differentiating the relations MATH for MATH and using the structure equations yields MATH so that MATH and MATH. Now differentiating the relation MATH, using the structure equations, and reducing modulo MATH yields MATH . Using the known congruences, this yields MATH which, of course, implies that MATH. Now that MATH, it follows that MATH and MATH for some functions MATH and MATH. Again, differentiating these relations and then reducing modulo MATH implies the relations MATH which, of course, implies that MATH for all MATH. This vanishing, in turn, now implies the congruences MATH . In other words, the MATH-plane spanned by MATH is locally, and, hence, globally constant on MATH. Let MATH be this constant MATH-plane. Then MATH and, hence, MATH are contained in MATH. Thus, as claimed, it suffices to take MATH for the rest of the argument, so I will do so. At this point, I am going to switch over to the standard language of the moving frame and assume that the reader is familiar with it. (The reader who is not might consult CITE. Of course, such a reader probably could not have followed the argument to this point anyway.) It is a good idea to take stock of the problem. At this moment, MATH is an algebraic variety of dimension MATH that contains a NAME subset MATH over which there exists a `moving frame' MATH satisfying the condition that MATH is spanned by MATH and MATH as well as the structure equations MATH where MATH are linearly independent and MATH. My goal now is to show that the existence of such a frame field implies that MATH is an open subset of the isotropic NAME associated to some nondegenerate quadratic form on MATH. This will have to be done in a series of steps. First, to simplify the argument, note that it suffices to prove this in the case where MATH is a non-singular, connected, and simply connected MATH-dimensional complex submanifold in MATH that possesses such a frame field, so assume that MATH has these properties. Replacing MATH by MATH yields a new frame for which the corresponding MATH and MATH are zero. Thus, without loss of generality, I can assume that MATH. Using the simple-connectivity of MATH, write MATH for some function MATH on MATH. Replacing the given frame MATH by MATH yields a new unimodular frame for which the corresponding MATH is equal to MATH. Thus, again, without loss of generality, I can further assume that MATH. Thus, I will say that a frame field MATH is MATH-adapted to MATH if the map MATH is the inclusion MATH and, moreover, MATH satisfies MATH for some independent MATH-forms MATH on MATH. By standard methods in the theory of moving frames, one sees that the MATH-adapted frame fields over MATH are the sections of a principal MATH-bundle MATH where MATH is a MATH-dimensional NAME subgroup whose NAME algebra MATH is the set of matrices of the form MATH . The group MATH is not connected because of the usual complication caused by the fact that MATH has a nontrivial, finite center. Instead, MATH is equal to the product of its identity component (which is determined by the NAME algebra MATH) and the elements of the form MATH where MATH. Thus, MATH has five components. It follows either that MATH is connected or else that it has MATH components. Following the standard method of the moving frame, for any MATH-adapted frame field MATH, let MATH, where MATH and MATH. By hypothesis, the relations MATH hold. (I will continue to use the abbreviations MATH, MATH, and MATH.) Taking the exterior derivatives of these relations, applying the structure equations, and then collecting terms and applying NAME 's Lemma shows that there exist functions MATH so that MATH . In particular MATH. Since MATH, applying the structure equation for MATH and using the above relations yields MATH which implies MATH. In particular MATH, so going back to its structure equation yields MATH so it follows that MATH also. Now, computing how the MATH vary under a change of MATH-adapted frame (a detail that can be safely left to the reader), one sees that by adding the appropriate multiples of MATH and MATH to MATH and MATH, one can construct a MATH-adapted frame for which MATH. I will say that a MATH-adapted frame that satisfies this additional property is MATH-adapted. Again, the usual methods show that the MATH-adapted frame fields are the sections of a principal MATH-bundle MATH where MATH is the MATH-dimensional NAME subgroup whose NAME algebra MATH is the space of matrices of the form MATH and that is generated by its identity component together with the elements of the form MATH where MATH. (Thus, MATH, like MATH, has five components.) For a MATH-adapted coframe field, in addition to the relations REF, there are now relations MATH . Taking the exterior derivatives of these six relations, applying the structure equations, and then collecting terms and applying NAME 's Lemma (keeping in mind that MATH) implies the further relations MATH . The relations REF, and REF combine to imply that the matrix MATH satisfies MATH, where MATH . That is, MATH takes values in the subspace MATH that is the NAME algebra of the group MATH of matrices MATH that satisfy MATH and MATH. Of course, MATH is isomorphic to MATH. Since the MATH are linearly independent except for the relations REF, and REF, it follows that the projection MATH immerses each component of MATH into a single left coset of MATH. Moreover, each component of MATH is open in such a coset. Since the MATH that forms the center of MATH does not lie in MATH but does lie in MATH, it follows that the image of MATH actually maps into five distinct left cosets of MATH and so must consist of five components instead of one. In particular, the inverse image of one of these components is a component MATH that is a MATH-bundle over MATH. I now restrict all forms and functions to this component MATH. Since MATH is an open immersion into a single left coset of MATH, it now follows that there exists a (unique) non-degenerate inner product MATH on MATH with the property that MATH. Thus, the MATH-plane MATH is MATH-isotropic for all MATH. Since MATH and the MATH-isotropic NAME are both MATH-dimensional submanifolds of MATH, it follows that MATH is an open subset of the MATH-isotropic NAME, as was to be proved. As already mentioned, this implies that any MATH-dimensional irreducible algebraic variety MATH that contains a NAME subset MATH that supports a MATH-adapted frame field must actually be the MATH-isotropic NAME for some non-degenerate inner product MATH on MATH. Thus, at last, this subcase is finished; such varieties fall into the third category of the proposition. Finally, all that remains is to address the last subcase, that of an irreducible MATH-dimensional subvariety MATH that contains a NAME subset MATH that is smooth and whose tangent spaces are integral elements of the third type listed in REF . My goal is to prove that such a MATH is necessarily a component of the variety listed in the fourth category of the proposition. The first task is to reduce to the case MATH. This will be reminiscent of the previous argument's reduction to MATH. Let MATH consist of the pairs MATH that satisfy the following conditions: CASE: MATH spans MATH, and CASE: the tangent space MATH is spanned by the three vectors MATH . (The indexing is a little unfortunate, but, for consistency with the conventions I have used so far, it is unavoidable.) Now, MATH is a submersion and is a principal MATH-bundle where MATH is a closed subgroup of MATH of codimension MATH. I will not need the full definition of MATH right now, so I postpone this. As usual, the structure equations hold on MATH. By construction, MATH if either MATH or MATH while the matrix MATH is skew-symmetric. Moreover, introducing MATH-forms MATH, MATH, and MATH by the equations MATH one has MATH. Suppose MATH and fix MATH. Differentiating the equations MATH and using the structure equations yields the relations MATH . The linear independence of MATH then implies that MATH. This vanishing implies the relations MATH . In other words, the MATH-plane spanned by MATH is locally constant. Set MATH for some fixed MATH. Since MATH is connected, it follows that MATH (and hence MATH) must lie in MATH. Thus, it suffices to analyze the case MATH, so I will assume this from now on. Suppose MATH (remember that MATH now) and fix MATH. Differentiating the equations MATH and using the structure equations yields the relations MATH . Again, the linear independence of MATH then implies that MATH. This vanishing implies the relations MATH . In other words, the MATH-plane spanned by MATH is locally constant. Set MATH for some fixed MATH. Since MATH is connected, it follows that MATH (and hence MATH) must lie in MATH. Thus, it suffices to analyze the case MATH, so I will assume this from now on. At this point, it is worthwhile to make the group MATH explicit. The usual calculation shows that this is a NAME subgroup of matrices whose NAME algebra is the space MATH of matrices of the form MATH where MATH and MATH are arbitrary MATH-MATH matrices. The group MATH is not connected, but is generated by its identity component and the center of MATH, a cyclic group of order MATH that consists of the matrices of the form MATH where MATH. Consequently, MATH actually has MATH components (the MATH-subgroup MATH already lies in the identity component of MATH). From this point on, the argument is much like the argument for the integral manifolds of the third category, so I will just indicate the steps without explicitly writing out the details. The first step is to note that the following six relations hold on MATH: MATH . Differentiating these relations, applying the structure equations, and applying NAME 's Lemma shows that there exist functions MATH on MATH so that the relations MATH hold for any MATH and where MATH is an even permutation of MATH. (It is important to remember that MATH, since this relation figures into these calculations.) The six equations MATH define a principal MATH subbundle MATH where MATH is the subgroup whose NAME algebra MATH consists of the matrices of the form MATH where MATH and MATH are MATH-MATH matrices with MATH. Again, MATH is generated by its identity component and the (finite) center of MATH and so has three components. On MATH, in addition to the six REF , the nine equations MATH also hold for all MATH. Differentiating these relations, applying the structure equations, and applying NAME 's Lemma shows that MATH . In view of the relations REF, and REF, it follows that MATH satisfies MATH where MATH . The MATH are otherwise linearly independent, so the map MATH immerses each component of MATH as an open subset of a left coset of MATH a subgroup isomorphic to MATH. Since MATH does not contain the full center of MATH while MATH does, it follows that the image MATH lies in three distinct left cosets of MATH and that MATH must therefore consist of three distinct components. Let MATH be one of these three components and restrict all forms and functions to MATH henceforth. As in the argument for the third case, it follows that there exists a nondegenerate quadratic form MATH on MATH with the property that MATH lies in the MATH-dimensional submanifold of MATH-isotropic MATH-planes in MATH. Thus MATH (and hence MATH) must be an open subset of one of the two components of this isotropic NAME. Since MATH was assumed to be irreducible and algebraic, it follows that MATH must actually be one of these components, that is, it must belong to the fourth category of the proposition. At last, the proof of the proposition is complete. |
math/0006186 | Combine complementarity and REF . |
math/0006186 | Any variety MATH of pure dimension MATH that satisfies MATH is necessarily an integral variety of MATH. REF implies that any such irreducible variety must, for dimension reasons, fall into the first category listed there. Thus, if MATH is irreducible, then MATH for some MATH. Since the ray generated by MATH is extremal, this implies the first rigidity statement of the theorem. Similarly if MATH has MATH (that is, the length of MATH is MATH), then any variety MATH of pure dimension MATH that satisfies MATH is an integral variety of MATH. REF implies that any such irreducible variety of dimension MATH must fall into the first category listed there. Thus, if MATH is irreducible, then MATH for some MATH. Since the ray generated by MATH is extremal, this implies the second rigidity statement of the theorem. |
math/0006186 | The proof is very similar to the proofs of REF , so I will only sketch the argument. Such a MATH, is, of course, an integral variety of both MATH and MATH. Conversely, any MATH-dimensional integral variety of both of these ideals is homologous to MATH for some MATH. REF describes the three-dimensional integral elements of MATH, pointing out that they form five distinct orbits under MATH, which, in REF , are denoted MATH, MATH, MATH, MATH and MATH. Let MATH be the complement of the singular locus of MATH. Each of the tangent spaces to MATH lies in one of the five orbits and there is one of these five orbits for which the set MATH consisting of the points whose tangent spaces lie in that orbit is a non-empty NAME open set in MATH. The argument now breaks into five cases. If the tangent spaces at the points of MATH lie in MATH, then one can apply a moving frame argument to show that MATH must fall into the first category of REF . Conversely, any subvariety MATH that falls into this category is an integral of MATH, so it must be homologous to some multiple of MATH. If the tangent spaces at the points of MATH lie in MATH, then the final part of the proof of REF shows that MATH falls into the last category of REF . Conversely, since the tangent spaces to MATH are integral elements of MATH, it must be homologous to a multiple of MATH. If the tangent spaces at the points of MATH lie in MATH, then arguments similar to those of REF show that MATH falls into either the fourth category or the sixth category of REF . Conversely, varieties in these two categories have their tangent spaces at generic points of type MATH or MATH, so they are integrals of MATH. If the tangent spaces at the points of MATH lie in MATH, then arguments similar to those of REF show that MATH falls into either the third category or the fifth category of REF . Conversely, varieties in these two categories have their tangent spaces at generic points of type MATH or MATH, so they are integrals of MATH. Finally, if the tangent spaces at the points of MATH lie in MATH, then arguments similar to those of REF show that MATH falls into one of the first four categories of REF . Conversely, varieties in these four categories have their tangent spaces at generic points fall into one of MATH, MATH, MATH, MATH, or MATH, so they are integrals of MATH. |
math/0006186 | The inequality MATH is, of course, immediate from REF . Moreover, by REF , if MATH, then MATH is an integral variety of MATH, where MATH. Now apply REF and interpret each of the possible cases. Only two cases require any comment: One of these is the second category of integral varieties of REF , that is, the case in which MATH has dimension at least MATH and lies in an integral variety of the form MATH for some curve MATH. In such a case, examination of the proof-analysis for REF shows that the rational mapping MATH that was defined in this case has the property that MATH is actually well-defined globally on MATH. Then, for every MATH, MATH is a hyperplane in MATH. Since MATH is canonically isomorphic to MATH, the line bundle MATH is then defined by MATH. The quotient MATH is canonically isomorphic to MATH, but this latter space depends only on the point MATH, so the quotient bundle MATH is necessarily pulled back from MATH, as claimed. The other is the third and fourth category of integral varieties of REF : Suppose first, that MATH has dimension MATH and is an integral variety of the form MATH for some MATH and MATH with MATH and MATH is a nondegenerate inner product on MATH. By the very definition of MATH, the subspace MATH must be zero and MATH must have dimension MATH. By the discussion in the second half of REF , it follows that, after identifying MATH with MATH via the embedding MATH, the bundle MATH can be written in the form MATH where MATH. Last, suppose that MATH has dimension MATH and is a component of MATH for some MATH and MATH with MATH and MATH is a nondegenerate inner product on MATH. By the very definition of MATH, the subspace MATH must be zero and MATH must have dimension MATH. By the discussion in the first half of REF , it follows that, after identifying MATH with MATH via the embedding MATH, the bundle MATH can be written in the form MATH where MATH. |
math/0006186 | I will first show that if MATH is not isotropic then one can construct forms in each of MATH that do not vanish on MATH. To begin, suppose that the inner product is nondegenerate on MATH. Then there exists an oriented orthonormal basis MATH of MATH, with dual basis MATH of MATH, so that MATH is spanned by the vectors MATH. Then neither of the forms MATH vanishes on MATH. If MATH has nullity MATH, then one can choose the basis MATH as above so that MATH is spanned by MATH and, again, both of MATH are nonzero on MATH. Now, suppose that MATH is isotropic and has positive chirality. Then there is an oriented orthonormal basis MATH of MATH so that MATH is spanned by the vectors MATH . Straightforward computation now shows that, when MATH is any even permutation of MATH, the MATH-form MATH vanishes on MATH. Since such MATH-forms span MATH, all the forms in MATH vanish on all of the isotropic planes of positive chirality. Since MATH is the direct sum of the spaces MATH, not all of the forms in MATH can vanish on MATH. By applying an orientation reversing isometry, it follows that not all of the elements of MATH vanish on any given MATH-plane of negative chirality. This proves the first statement in the lemma. The proof of the second statement is similar. |
math/0006186 | Let MATH be the subgroup of MATH consisting of the matrices MATH that satisfy MATH . Then MATH acts on MATH and induces a transitive action on MATH. Let MATH denote the inclusion and write MATH where MATH are regarded as (holomorphic) mappings. Then MATH is a holomorphic principal fiber bundle over MATH. Moreover, the map MATH makes MATH into a holomorphic fiber bundle over MATH, that is, the set of isotropic MATH-planes in MATH. In accordance with the moving frame, write the structure equations as MATH where MATH and MATH. (These relations follow in the usual way from the exterior derivative of REF.) Moreover, the structure equation MATH holds since MATH. Now suppose that MATH is a MATH-dimensional complex submanifold with the property that all of its tangent planes are isotropic and let MATH denote the set of pairs MATH that satisfy CASE: MATH; and CASE: The projectivized isotropic MATH-plane MATH is tangent to MATH at MATH. Then MATH is a holomorphic fiber bundle over MATH. We now consider the functions and forms on MATH to be pulled back to MATH in the usual way of the moving frame. Then, by construction, MATH that is, MATH for MATH, while the fact that MATH is a submersion implies that MATH. When MATH, the structure equations imply MATH . Since MATH, NAME 's Lemma implies that there exist functions MATH so that MATH. However, since MATH, it follows that MATH, that is, MATH. The structure equations thus imply that MATH is locally constant, that is, that, locally, MATH is tangent to the projectivization of a fixed isotropic MATH-plane in MATH. Since MATH is connected, smooth, and holomorphic, it follows that MATH must be everywhere tangent to this linear MATH, as desired. |
math/0006186 | Each of MATH is a positive MATH-form and the above analysis shows that MATH . Since MATH, it follows that MATH and MATH generate MATH. Thus, any MATH-dimensional subvariety MATH that satisfies MATH must be a union of irreducible subvarieties whose homology classes are multiples of MATH. Thus, I may assume that MATH is irreducible and satisfies MATH. In particular, MATH vanishes on MATH. By REF , it follows that, on the smooth locus of MATH, each of its tangent spaces is an isotropic MATH-plane. By REF , it follows that MATH must contain an isotropic MATH in MATH. Since MATH is irreducible, it follows that MATH must itself be such a plane. In particular, MATH. Since MATH is homologous to MATH, it must have positive chirality. The argument when MATH is essentially the same. |
math/0006186 | Let MATH be the identity map. For any basis MATH of MATH, write MATH, where MATH are REF-forms on MATH. Note that MATH is a basis for the dual space. Now, MATH occurs as a constituent of MATH, but MATH does not. Consequently, the MATH-forms of the form MATH when MATH, MATH, and MATH are symmetric in their indices, must lie in MATH. Taking, as a particular example, MATH and all other MATH, MATH and MATH equal to zero yields MATH, so the span of the MATH is nontrivial. Since this span is invariant under MATH and since MATH is irreducible, this span must be all of this subspace. Thus, MATH is the span of the MATH-forms of the form REF. Now suppose that MATH is a subspace of dimension MATH on which all of the forms in MATH vanish. My goal is to show that there is a unique line MATH so that MATH and, conversely, that all of these MATH-forms vanish on MATH for any line MATH. The converse assertion is easy, so let me do this first. Since all of the conditions are invariant under the action of MATH, it suffices to prove this for the case that MATH. In this case, MATH has dimension MATH and is defined by the equations MATH when MATH, so suppose that all of these MATH-forms have been set to zero. Then for MATH of the form REF, all of the terms vanish unless exactly one entry of each of the pairs MATH, MATH, and MATH is equal to MATH. Moreover, the entries not equal to MATH in these pairs must all be distinct. Clearly, it suffices to treat the case where all of these entries are drawn from the set MATH. There are eight possible ways to assign the value of MATH to one element of each of the pairs MATH, MATH, and MATH. In six of those ways, two of these indices will enter the same MATH, MATH, or MATH coefficient and the corresponding sub-sum will vanish. For example, when MATH, the part of the sum in REF corresponding to this choice is the sub-sum MATH, which vanishes, since MATH is symmetric and MATH is skewsymmetric in the pair MATH. The two exceptional configurations are MATH and MATH, but these two sub-sums cancel: MATH . Thus, all the forms in MATH vanish on MATH, as desired. Now suppose that MATH is an integral element of MATH and that MATH. Then MATH will be defined by some set of linear relations among REF-forms MATH. (By hypothesis, however, at least three of the MATH are linearly independent on MATH.) My goal is to show that one can choose the basis MATH so that these relations include MATH for MATH. Suppose that the basis MATH has been chosen so that the maximum number, say MATH, of the forms MATH are linearly independent on MATH. Clearly, MATH. By making a change of basis in MATH, I can assume that MATH on MATH but that MATH on MATH for MATH. I claim that the maximality property implies that MATH whenever MATH. To see this, let MATH be parameters and consider the basis MATH defined by MATH for MATH. Then MATH . Suppose that there exist MATH and MATH so that MATH. Then set MATH for MATH and MATH and consider the expansion MATH . Clearly, there will be a nonempty open set of values for MATH for which the left hand side of this equation will be nonzero, thus contradicting the maximality of MATH. It follows immediately that MATH. Now, so far, no use has been made of the hypothesis that MATH be an integral element of MATH. Its first use is to show that MATH. This follows because, as has already been noted, one of the elements of MATH is MATH. Since MATH, it follows that MATH is a linear combination of MATH and MATH. If it were true that MATH, then all of the forms MATH with MATH would also be linear combinations of MATH and MATH, so there could not be three linearly independent MATH-forms among the MATH. Thus, MATH, as claimed. Now, the same argument that showed that MATH is in MATH shows that MATH is in MATH for all MATH. In particular, it follows that MATH for all MATH. Combined with the previous argument, showing that MATH when either MATH or MATH is greater than MATH, this shows that MATH must actually be a basis for REF-forms on MATH, that is, MATH. Now, the fact that MATH on MATH for MATH combined with the skew-symmetry MATH implies that there exist unique numbers MATH for MATH so that MATH . Moreover, for any distinct MATH satisfying MATH, REF shows that the form MATH lies in MATH, so the fact that this vanishes on MATH implies that MATH. It follows that there are constants MATH so that MATH for all MATH. Consequently, MATH for MATH. Thus, by replacing MATH by MATH, I get a new basis in which MATH holds on MATH for MATH, so I assume this from now on. Next, taking MATH satisfying MATH, the above form simplifies on MATH to MATH. Since this must vanish, it follows that MATH for all such triples. Since MATH, this implies MATH whenever MATH. Thus, set MATH for some quantities MATH. Now, observe that the sum MATH is in MATH, and take MATH and MATH with MATH not equal to either MATH or MATH. (This last is possible since MATH.) The last two terms in the sum vanish since MATH and the first two terms simplify to MATH. Since this must vanish, it follows that MATH for some constants MATH when MATH. Thus, MATH. It follows that, by replacing MATH by MATH, I can arrange that MATH whenever MATH, so assume this from now on. Finally, go back to the above sum and assume MATH. Then all of the terms except the first are zero. The vanishing of the first term MATH implies that MATH is a linear combination of MATH and MATH for any distinct pair of indices MATH and MATH satisfying MATH. Since MATH, this forces MATH. Thus MATH satisfies the relations MATH for MATH and for MATH and MATH. It follows that MATH is a basis for MATH, as desired. Now, I will compute the integral elements of MATH. My goal is to show that any integral element MATH of dimension MATH or more is actually of the form MATH for some (unique) MATH-dimensional subspace MATH. First, I must describe a set of generators of MATH. Now, it is easy to calculate that MATH occurs as a constituent of MATH, but that MATH does not. Consequently, every MATH-form of the form MATH when MATH is skewsymmetric in its indices and MATH is symmetric in its indices, must actually lie in MATH. Taking, as particular examples, MATH and letting MATH be zero unless MATH while MATH, gives MATH, so it follows that the MATH span a nontrivial subspace of MATH. Since this subspace is evidently invariant under MATH and since MATH is irreducible, it follows that the forms MATH of the form REF must span MATH. Now suppose that MATH is an integral element of MATH whose dimension is at least MATH. Just as in the first part of the argument, choose a basis MATH of MATH so that the maximum number, say, MATH, of MATH are linearly independent on MATH and so that MATH for MATH. The argument given in the first half of the proof shows that MATH, but the fact that MATH lies in MATH implies that MATH. Thus, MATH. As before, the fact that MATH is maximal implies that MATH whenever MATH. In particular, all REF-forms on MATH must be linear combinations of MATH, MATH, and MATH. Consequently, MATH, but since MATH by hypothesis, MATH. Moreover, since there must be at least three linearly independent forms on MATH, it follows that MATH. Now, the same argument as showed that there cannot be more than two independent forms among the MATH shows that there cannot be more than two independent forms among the MATH or the MATH. In particular, for MATH, the MATH-form MATH must be a linear combination of MATH and MATH. However, I have already shown that it must also be a linear combination of MATH and MATH. Consequently MATH must simply be a multiple of MATH, say MATH for MATH. Similarly, MATH must simply be a multiple of MATH, say MATH for MATH. Now, replacing MATH by MATH produces a new basis for which MATH for MATH, so assume that this has been done. Now, for each MATH, consider MATH as in REF where MATH while MATH unless MATH and MATH while MATH unless MATH. The result is MATH . Since MATH on MATH, the vanishing of MATH on MATH forces MATH to be a linear combination of MATH and MATH. Since MATH, this forces MATH, that is, MATH vanishes on MATH. If MATH, it has now been demonstrated that, for any integral MATH-dimensional integral element of MATH, there is a basis MATH so that MATH is defined by the equations MATH when MATH. If MATH assume MATH and consider MATH where MATH while MATH unless MATH and MATH while MATH unless MATH. Then, on MATH, MATH. Now, permuting MATH in this construction shows that MATH . This implies that MATH, on MATH, as desired. Finally, if there is a basis MATH so that MATH is defined by MATH when MATH, then it is clear that MATH vanishes on MATH for all MATH and MATH, so that MATH is, indeed, an integral element of MATH. |
math/0006186 | The first task (which will be needed in the next proposition as well), is to establish the equations of the moving frame for submanifolds of MATH. Define MATH be the subgroup of MATH consisting of the matrices MATH that satisfy MATH . Also, let MATH denote the set of matrices MATH that satisfy MATH and MATH. Evidently, MATH is an orbit of MATH acting on MATH on the right. I will regard MATH as a matrix-valued function and denote its columns as MATH where MATH are regarded as (holomorphic) mappings. Define MATH so that MATH is a surjective submersion MATH. The fibers of MATH are the orbits of the parabolic subgroup MATH consisting of elements of the form MATH . Thus, MATH is a principal right MATH-bundle over MATH. In accordance with the usual moving frame conventions, write the structure equations as MATH where MATH but the components of MATH, MATH, and MATH are otherwise linearly independent. The relations REF follow in the usual way from the exterior derivative of REF. The structure equation MATH holds since MATH. These expand to MATH . Now suppose that MATH is an irreducible integral variety of MATH of dimension MATH, and let MATH denote its smooth locus, which is an embedded submanifold of MATH. For every MATH, the tangent space MATH is an integral element of MATH of dimension MATH. By REF , it follows that, for every MATH, there exists a MATH so that CASE: MATH is spanned by MATH, and CASE: MATH is spanned by MATH. Let MATH denote the set of such MATH as MATH ranges over MATH. Then MATH is a principal MATH-bundle over MATH, where MATH is the subgroup consisting of the matrices of the form REF with MATH in MATH. By construction, the forms MATH are linearly independent on MATH and span the MATH-semibasic MATH-forms, while MATH for MATH and MATH when both MATH and MATH are bigger than MATH. This paragraph of the argument is necessary only if MATH, so suppose this is so for the moment. Choose a pair MATH satisfying MATH and differentiate the relation MATH. By the structure equations, this is MATH . Since MATH, and since MATH are linearly independent, MATH for MATH. Now choose a pair MATH with MATH and differentiate MATH. The structure equations give that MATH . Equivalently, MATH . Wedging this relation with MATH gives MATH for all MATH. Again, because MATH and because MATH are linearly independent, it follows that MATH for MATH. In particular, there exist functions MATH on MATH so that MATH. Substituting this back into REF and again using the linear independence of MATH and MATH, it follows that MATH for all MATH. Again, since MATH, this implies that MATH for MATH. In other words, MATH for MATH. Since the previous paragraph showed that MATH for all MATH, this combines to give that MATH for all MATH. This vanishing together with the fact that MATH for all MATH yield the congruences MATH . In other words, the mapping MATH defined by MATH is constant. Let MATH be this constant MATH-plane. By construction MATH lies in MATH, so it follows that MATH, and, hence, MATH lie in MATH, as desired. That MATH really is an integral variety of MATH follows immediately from the proof of the first part. |
math/0006186 | Recall the moving frame notation and constructions from the first part of the proof of REF . Suppose now that MATH is an irreducible MATH-dimensional integral variety of of MATH and let MATH be its smooth locus, which is connected since MATH is irreducible. By REF , for every MATH, there exists a MATH so that CASE: MATH is spanned by MATH, and CASE: The tangent space MATH is spanned by MATH. Let MATH denote the set of such MATH as MATH ranges over MATH. Then MATH is a principal right MATH-bundle over MATH where MATH is the subgroup consisting of the matrices of the form REF with MATH in MATH. Since MATH and MATH are each connected, it follows that MATH is also connected. By construction, the MATH-forms MATH are linearly independent on MATH and span the MATH-semibasic MATH-forms, while MATH if either MATH or MATH is greater than MATH. Let MATH be fixed and differentiate the identities MATH using the structure equations. The result is equations of the form MATH . By the linear independence of MATH, it follows that MATH. This vanishing for all MATH implies MATH that is, the MATH-plane MATH is locally constant on MATH. Since MATH is connected, this map must be constant. Thus, let MATH be the isotropic plane so that MATH. By construction, MATH for all MATH, so it follows that MATH and, hence, MATH are subsets of MATH, as desired. |
math/0006186 | First of all, it follows by either CITE or REF and the general results of NAME mentioned above that MATH. Let MATH be the MATH-invariant NAME form on MATH whose cohomology class is a generator of MATH. By REF, there is a sum of the form MATH where MATH and MATH and MATH and MATH are positive MATH-invariant forms dual to the generalized NAME cycles MATH and MATH of complex dimension MATH whose cohomology classes generate MATH. It follows that MATH is an irreducible MATH-dimensional integral variety of MATH and, so, by REF , must be of the form MATH for some isotropic MATH plane MATH. Thus MATH is homologous to MATH. It also follows that MATH is an irreducible MATH-dimensional integral variety of MATH, and so, by REF , must lie in MATH for some unique MATH. Since MATH must be a generator of MATH, it follows easily that it must be homologous to MATH, where MATH is a linearly embedded projective MATH-space. Thus, MATH is homologous to MATH. Finally, if MATH is irreducible and satisfies MATH, then the integral of MATH over MATH must be zero, so MATH must vanish on MATH. Thus MATH is an integral manifold of MATH and REF applies. The argument when MATH is similar. |
math/0006186 | Using arguments that should, by now, be familiar, one sees that the homology class of a codimension MATH irreducible cycle MATH is some multiple of MATH if and only if the form MATH vanishes on MATH, which is the same as saying that, at any smooth point MATH, the normal space to MATH in MATH is an integral element of MATH. By REF , it follows that, for every MATH, there is a MATH so that CASE: MATH is spanned by MATH, and CASE: MATH is spanned by the MATH, where MATH and MATH. The set of all such MATH as MATH ranges over MATH is a principal right MATH-bundle MATH, where MATH is the subgroup consisting of the matrices of the form REF with MATH in MATH. Since MATH and MATH are each connected, it follows that MATH is also connected. By construction, the MATH-forms MATH with MATH and MATH are linearly independent and span the MATH-semibasic forms on MATH while the forms MATH, MATH, and MATH are all identically zero. To save writing, adopt the conventions that MATH while MATH. Differentiating the relations MATH and applying the structure equations then yields the relations MATH (summation on MATH). Judicious use of NAME 's Lemma, together with the linear independence of the MATH, implies that there exist functions MATH on MATH so that MATH . After computing how the functions MATH vary on the fibers of MATH, one sees that the equations MATH define a principal right MATH-bundle MATH over MATH where MATH is the connected subgroup of matrices whose NAME algebra consists of the matrices of the form MATH . The relations MATH hold on MATH. Differentiating these relations and applying NAME 's Lemma shows that the relations MATH must also hold. These identities combine to show that, on MATH, MATH . In other words, the map MATH is locally constant. Since MATH is connected, this map must be globally constant. Let MATH be its constant value. Note that MATH lies in MATH if MATH is odd and in MATH if MATH is even. By construction MATH for all MATH. It follows that MATH, and, hence, MATH lie in MATH, as desired. Moreover, examining the argument given shows that MATH is indeed an integral manifold of MATH and has codimension MATH in MATH. |
math/0006186 | Using arguments that should, by now, be familiar, one sees that the homology class of a codimension MATH irreducible cycle MATH is some multiple of MATH if and only if the form MATH vanishes on MATH, which is the same as saying that, at any smooth point MATH, the normal space to MATH in MATH is an integral element of MATH. By REF , it follows that, for every MATH, there is a MATH so that CASE: MATH is spanned by MATH, and CASE: MATH is spanned by the MATH, where MATH and either MATH or MATH. The set of all such MATH as MATH ranges over MATH is a principal right MATH-bundle MATH, where MATH is the subgroup consisting of the matrices of the form REF with MATH in MATH. Since MATH and MATH are each connected, it follows that MATH is also connected. By construction, the MATH-forms MATH with MATH and and either MATH or MATH are linearly independent and span the MATH-semibasic forms on MATH while the forms MATH, MATH, and MATH are all identically zero. To save writing, adopt the conventions that MATH while MATH. Differentiating the relations MATH and applying the structure equations then yields relations of the form MATH for MATH, MATH, and MATH. Judicious use of NAME 's Lemma, together with the stated linear independence of the entries of MATH, implies that there exist functions MATH, MATH, and MATH on MATH so that MATH . In particular, it follows from the structure equations that MATH . Consequently the mapping MATH is constant on the fibers of MATH and its differential has rank MATH everywhere. Thus, MATH where MATH is a holomorphic map whose differential has rank MATH everywhere. When MATH is an algebraic variety, it is not hard to argue that MATH is a rational map and then that the closure of MATH in MATH is an algebraic variety MATH of dimension MATH. At this point, it is evident that MATH. Details will be left to the reader, along with the verification that the degree MATH of MATH satisfies MATH. |
math/0006186 | This proof is very similar in spirit and structure to the proof of REF , so, to save space, I will not go into as much detail here as I did there. Instead, I will limit my discussion to the outline, except where some essentially new or different idea is needed. Let MATH be the identity map. For any basis MATH of MATH, write MATH, where MATH are REF-forms on MATH. Note that MATH is a basis for the dual space. Now, MATH occurs as a constituent of MATH, but MATH does not. Consequently, the MATH-forms of the form MATH when MATH and MATH are symmetric in their indices, must lie in the subspace MATH of MATH. Taking, as a particular example, MATH and all other MATH, MATH equal to zero yields MATH, so the span of the MATH is nontrivial. Since this span is invariant under MATH and since MATH is irreducible, this span must be all of this subspace. Thus, MATH is the span of the MATH-forms of the form REF. Now, to begin, it must be checked that for every line MATH, all of the forms MATH vanish on the MATH-dimensional subspace MATH. By equivariance, it suffices to check this for the line MATH, which is defined by the equations MATH for MATH. Thus, the claim is equivalent to the claim that the forms MATH all lie in the ideal generated by the MATH-forms MATH with MATH. By obvious reductions (keeping in mind that MATH), it suffices to check this for the cases where MATH but MATH otherwise and MATH but MATH otherwise. In this case, there are MATH terms in the sum REF, and each term either vanishes identically, cancels in combination with one or two other terms, or else is a multiple of some MATH with MATH. Thus, the claim is established. Now suppose that MATH is a subspace of dimension MATH on which all of the forms in MATH vanish. The goal is to show that there is a unique line MATH so that MATH. Let MATH be the restriction to MATH of MATH. Now, MATH will be defined by some set of linear relations among the MATH-forms MATH. (By hypothesis, at least three of the MATH are linearly independent on MATH.) I need to show that one can choose the basis MATH so that these relations include MATH for MATH. Suppose that the basis MATH has been chosen so that the maximum number, say MATH, of the forms MATH are linearly independent. (This maximum independence will hold on a NAME open set of bases MATH.) Clearly, MATH. It is not difficult to show that, by making a change of basis in MATH, I can assume that MATH but that MATH for MATH. Then, by the same sort of analysis as was done in the proof of REF , one can show that maximality implies that MATH . Since the MATH must span MATH, the relations REF imply that MATH. So far, no use has been made of the assumption that the forms in MATH vanish on MATH, that is, that MATH for all MATH and MATH symmetric in their indices. To make any further progress, these relations will have to be used. Since, as has already been seen, the relations REF include MATH, it follows that MATH. Now, replacing MATH by MATH in this relation gives MATH, so, in particular, MATH . On the other hand, polarizing the identity MATH gives MATH for all MATH. Since MATH, taking any pair MATH with MATH and setting MATH in the above relation yields MATH . Wedging this relation with MATH gives MATH whenever MATH. Since MATH by hypothesis, it follows that MATH . It follows from REF, and REF that MATH is a basis for MATH. Consequently, MATH. Moreover, analysis of REF, shows that there must exist MATH so that MATH . It follows that, by replacing MATH by MATH, I can arrange that MATH, so assume this. Now, in the same way that REF was derived, one can derive the relations MATH and MATH . The relations REF imply that there exist MATH for MATH so that MATH when MATH, while REF implies MATH when MATH. In particular, it follows that MATH . Thus, the hyperplane MATH is of the form MATH where the line MATH is spanned by the vector MATH and MATH is spanned by MATH. The analysis so far shows that every MATH of dimension MATH on which all the elements of MATH vanish contains a hyperplane MATH of the form MATH where MATH and MATH are subspaces of MATH of dimensions MATH and MATH, respectively, that are independent, that is, MATH. To finish the characterization of these integral elements, it suffices to show that, for any basis MATH, every integral element of MATH that contains the MATH-plane spanned by MATH and MATH must be itself be a subspace of the MATH-plane MATH (which has already been shown to be an integral element). To see this, consider, for every MATH, the MATH-forms MATH which manifestly belong to MATH. When MATH, MATH while MATH . In particular, if MATH is to span an integral element of MATH, then MATH when MATH, that is, MATH must lie in the span of MATH, that is, MATH, which is what needed to be shown. I now turn to the analysis of the integral elements of the ideal MATH. First, MATH occurs as a constituent of MATH, but MATH does not. Consequently, the MATH-forms of the form MATH when MATH is symmetric in its indices and MATH is skewsymmetric in its indices, must lie in the subspace MATH of MATH. Taking, as a particular example, MATH with all other MATH equal to zero, and MATH but MATH unless MATH yields MATH. Thus, the span of the MATH is nontrivial. Since this span is invariant under MATH and since MATH is irreducible, this span must be all of this subspace. Thus, MATH is the span of the MATH-forms of the form REF. Now suppose that MATH is a subspace of dimension MATH on which all of the forms in MATH vanish. The goal is to show that there is a MATH-plane MATH so that MATH. Let MATH be the restriction to MATH of MATH. Now, MATH will be defined by some set of linear relations among the MATH-forms MATH. (By hypothesis, at least three of the MATH are linearly independent on MATH.) I need to show that one can choose the basis MATH so that these relations include MATH when MATH. Suppose that the basis MATH has been chosen so that the maximum number, say MATH, of the forms MATH are linearly independent. (This maximum independence will hold on a NAME open set of bases MATH.) Clearly, MATH. It is not difficult to show that, by making a change of basis in MATH, I can assume that MATH but that MATH for MATH. Then, by the same sort of analysis as was done in the proof of REF , one can show that maximality of MATH implies that MATH . Since the MATH must span MATH, the relations REF imply that MATH. So far, no use has been made of the assumption that the forms in MATH vanish on MATH, that is, that MATH for all MATH symmetric in its indices and MATH skewsymmetric in its indices. To make any further progress, these relations will have to be used. Since, as has already been seen, the relations REF include MATH, it follows that MATH, that is, MATH. Moreover, since MATH, the relations REF imply that MATH must span MATH. Since MATH, by hypothesis, it follows that MATH and that MATH must be a basis for MATH. Now, fix MATH and let MATH with MATH unless MATH. Letting MATH be the general symmetric symbol with MATH unless MATH, substituting this into the relations REF, and using the fact that MATH yields the relations MATH which implies MATH. Now, fix MATH. Letting MATH be the symmetric symbol that satisfies MATH but MATH unless MATH is a permutation of MATH gives the relation MATH. Letting MATH be the symmetric symbol that satisfies MATH but MATH unless MATH is a permutation of MATH gives the relation MATH. Letting MATH be the symmetric symbol that satisfies MATH but MATH unless MATH is a permutation of MATH gives the relation MATH. These three identities imply that MATH when MATH, which is what remained to be shown. |
math/0006186 | The first task (which will be needed in the next proposition as well), is to establish the equations of the moving frame for submanifolds of MATH. Define MATH be the subgroup of MATH consisting of the matrices MATH that satisfy MATH . I will regard MATH as a matrix-valued function and denote its columns as MATH where MATH are regarded as (holomorphic) mappings. Define MATH so that MATH is a surjective submersion MATH. The fibers of MATH are the orbits of the parabolic subgroup MATH consisting of elements of the form MATH . Thus, MATH is a principal right MATH-bundle over MATH. In accordance with the usual moving frame conventions, write the structure equations as MATH where MATH but the components of MATH, MATH, and MATH are otherwise linearly independent. The relations REF follow in the usual way from the exterior derivative of REF. The structure equation MATH holds since MATH. These expand to MATH . Now suppose that MATH is an irreducible integral variety of MATH of dimension MATH, and let MATH denote its smooth locus, which is an embedded submanifold of MATH. For every MATH, the tangent space MATH is an integral element of MATH of dimension MATH. By REF , there is a NAME subset MATH (which may be empty) that consists of the elements MATH such that MATH where MATH and MATH are subspaces of MATH of dimensions MATH and MATH, respectively, and MATH. (See REF .) The proof has to be broken up into two cases now, depending on whether or not MATH is empty. (Note that MATH cannot be empty unless MATH.) The first case is that MATH is not empty, so assume this. Note that MATH is connected since MATH is irreducible. Now, for every MATH, there exists a MATH so that CASE: MATH is spanned by MATH, and CASE: MATH is spanned by MATH. Let MATH denote the set of such MATH as MATH ranges over MATH. Then MATH is a principal MATH-bundle over MATH, where MATH is a subgroup of the matrices of the form REF with MATH in MATH. The reader can write out the exact conditions defining MATH and verify that it is connected. By construction, the forms MATH are linearly independent on MATH and span the MATH-semibasic MATH-forms, while MATH when either MATH or MATH and MATH are both greater than MATH. This paragraph of the argument is necessary only if MATH, so suppose this is so for the moment. Choose a pair MATH satisfying MATH and differentiate the relation MATH. By the structure equations, this is MATH . Since MATH, and since MATH are linearly independent, MATH for MATH. Now choose a pair MATH with MATH and differentiate MATH. The structure equations give that MATH . Equivalently, MATH . This relation implies that there exists a function MATH on MATH so that MATH for MATH. Computing how the function MATH varies on the fibers of MATH (a standard computation in the technique of the moving frame) shows that the equation MATH defines a principal right MATH-bundle MATH over MATH where MATH is a certain connected NAME subgroup of codimension MATH. Since the previous paragraph showed that MATH for all MATH, it now follows that the identities MATH for all MATH hold on MATH. This vanishing together with the fact that MATH for all MATH yield the congruences MATH . In other words, the mapping MATH defined by MATH is locally constant and hence, by connectedness, globally constant. Let MATH be MATH-plane that is the image of MATH. Of course, it now follows that MATH is the MATH-plane MATH. Thus, MATH lies in MATH. Of course, this implies that MATH itself lies in MATH as well, as desired. Now, consider the second case, in which MATH. Then for every MATH, the tangent space MATH is of the form MATH where MATH and MATH are subspaces of MATH of dimensions MATH and MATH, respectively, and MATH. (Again, see REF .) Note that MATH is connected since MATH is irreducible. For every MATH, there exists a MATH so that CASE: MATH is spanned by MATH, and CASE: MATH is spanned by MATH. Let MATH denote the set of such MATH as MATH ranges over MATH. Then MATH is a principal MATH-bundle over MATH, where MATH is the group consisting of the matrices of the form REF with MATH in MATH. By construction, the forms MATH are linearly independent on MATH and span the MATH-semibasic MATH-forms, while MATH when either MATH or MATH and MATH are both greater than MATH. This paragraph of the argument is necessary only if MATH, so suppose this is so for the moment. Choose a pair MATH satisfying MATH and differentiate the relation MATH. By the structure equations, this is MATH . Since MATH, and since MATH are linearly independent, MATH for MATH. Now choose a pair MATH with MATH and differentiate MATH. The structure equations give MATH . Equivalently, MATH . This relation implies that there exists a function MATH on MATH so that MATH for MATH. Computing how the function MATH varies on the fibers of MATH (a standard computation in the technique of the moving frame) shows that the equation MATH defines a principal right MATH-bundle MATH over MATH where MATH is a certain connected NAME subgroup of codimension MATH. Since the previous paragraph showed that MATH for all MATH, it now follows that the identities MATH for all MATH hold on MATH. This vanishing together with the fact that MATH for all MATH yield the congruences MATH . In other words, the mapping MATH defined by MATH is locally constant and hence, by connectedness, globally constant. Let MATH be MATH-plane that is the image of MATH. Of course, it now follows that MATH is the MATH-plane MATH. Thus, MATH lies in MATH. Of course, this implies that MATH itself lies in MATH as well, as desired. That MATH really is an integral variety of MATH follows immediately from the argument in the second case (with MATH). |
math/0006186 | Recall the moving frame notation and constructions from the first part of the proof of REF . Suppose now that MATH is an irreducible MATH-dimensional integral variety of of MATH and let MATH be its smooth locus, which is connected since MATH is irreducible. By REF , for every MATH, there exists a MATH so that CASE: MATH is spanned by MATH, and CASE: The tangent space MATH is spanned by MATH. Let MATH denote the set of such MATH as MATH ranges over MATH. Then MATH is a principal right MATH-bundle over MATH where MATH is the subgroup consisting of the matrices of the form REF with MATH in MATH. Since MATH and MATH are each connected, it follows that MATH is also connected. By construction, the MATH-forms MATH are linearly independent on MATH and span the MATH-semibasic MATH-forms, while MATH if either MATH or MATH is greater than MATH. Let MATH be fixed and differentiate the identities MATH using the structure equations. The result is equations of the form MATH . By the linear independence of MATH, it follows that MATH. This vanishing for all MATH implies MATH that is, the MATH-plane MATH is locally constant on MATH. Since MATH is connected, this map must be constant. Thus, let MATH be the isotropic plane so that MATH. By construction, MATH for all MATH, so it follows that MATH and, hence, MATH are subsets of MATH, as desired. |
math/0006186 | First of all, it follows by either CITE or REF and the general results of NAME mentioned above that MATH. Let MATH be the MATH-invariant NAME form on MATH whose cohomology class is a generator of MATH. By REF, there is a sum of the form MATH where MATH and MATH and MATH and MATH are positive MATH-invariant forms that are dual to the generalized NAME cycles MATH and MATH of complex dimension MATH whose cohomology classes generate MATH. It follows that MATH is an irreducible MATH-dimensional integral of MATH and so, by REF , must be of the form MATH for some isotropic MATH plane MATH. Thus MATH is homologous to MATH. It also follows that MATH is an irreducible MATH-dimensional integral of MATH, and so, by REF , must lie in MATH for some MATH. When MATH, the dimension of MATH is MATH, so MATH must be equal to MATH, which is MATH in this case. The definition of MATH now makes it clear that MATH represents MATH when MATH as well. Finally, if MATH is irreducible and satisfies MATH, then the integral of MATH over MATH must be zero, so MATH must vanish on MATH. Thus MATH is an integral manifold of MATH and REF applies. The argument when MATH is similar. |
math/0006186 | The analysis is very similar to that for the proof of REF , so I will leave the details to the reader. Only one aspect of the proof requires comment: Since there are two types of integral elements of MATH, there are, correspondingly, two types of integral elements of MATH. This forces a subdivision into two cases like that of REF , but this offers no essential new difficulty. |
math/0006186 | Follow the pattern of the proof of REF . |
math/0006191 | A splitting of Type REF is given as follows. Note that MATH is the mapping cylinder of a (single) NAME twist on the torus. Thus, if we begin with the rational elliptic surface MATH let MATH denote the image of a fishtail fiber, and let MATH be a disk around MATH containing no other singular points for MATH, then MATH splits MATH into a pair of elliptic fibrations MATH and MATH over disks. Thus we can realize MATH as a union of fiber sums MATH and MATH joined along MATH, where MATH is the fiber sum of three copies of MATH. Neither side is negative-definite: both sides contain a torus of square zero and a sphere (constructed from vanishing cycles) which meets this torus in a single, positive point. Since every simply-connected elliptic surface with MATH can be obtained from MATH by fiber sums with MATH, logarithmic transformations, and blow-ups (see CITE, or CITE) the result follows. A splitting of Type REF is realized by finding an elliptic surface MATH over MATH which contains MATH singular values for the elliptic fibration whose holonomy is a NAME twist along a given curve in the fiber. In fact, it is REF (see CITE, also REF) that if MATH is a nodal elliptic surface without multiple fibers and MATH singular fibers, then we can think of the monodromy representation around MATH of the singular fibers, of which we select MATH, as being a REF NAME twist around a fixed non-separating curve in the fiber, and the monodromy around the remaining MATH as being a NAME twist around another curve. Let MATH be a disk in MATH which contains only the MATH distinguished singular points and no others. Now, it is easy to see that MATH, which separates the elliptic surface. Forming fiber sums with rational elliptic surfaces on both sides as before, we get a decomposition of the elliptic surface MATH along MATH into two pieces with MATH. |
math/0006191 | First, we reduce to the case where MATH is absent (that is, zero-dimensional). This is done in two steps, first establishing an inclusion MATH where both are thought of as subsets of MATH, and then seeing that the map MATH is surjective. To see Inclusion REF we describe the geometric representatives for the generators of the cohomology ring MATH . Given a point MATH and a line MATH in the fiber of the spinor bundle over MATH, the class MATH is NAME dual to the locus MATH of pairs MATH with MATH. Moreover, given a curve MATH, the corresponding one-dimensional cohomology class determined by the homotopy type of the map MATH given by measuring the holonomy of MATH (relative to some fixed reference connection MATH) around MATH; that is, it is NAME dual to the preimage MATH of a regular value of MATH. This cohomology class is denoted MATH. (Note that geometric representatives cohomology classes in the configuration spaces of four-manifolds are constructed in an analogous manner.) Now, fix a curve MATH and consider the one-parameter family of maps MATH indexed by MATH defined by measuring the holonomy of MATH around the curve MATH. Since the configurations in MATH converge exponentially to a stationary solution (see CITE), MATH extends continuously to MATH. Now, MATH represents MATH of the one-dimensional class MATH, while MATH represents the restriction (to the moduli space) of the one-dimensional class MATH, where MATH. A similar discusion applies to the two-dimensional class to show that MATH (now we use the connection MATH to identify the fiber ``at infinity" with the fiber at some point inside MATH). This completes the verification of Inclusion REF. NAME of MATH follows from classical properties of the cohomology of symmetric products MATH (see CITE), according to which the cohomology ring is generated by ``symmetrizations" of the cohomology of MATH. It is then a straightforward verification (which is spelled out in REF) using the geometric interpretations of the cohomology classes given above to see that that MATH corresponds to the symmetrization of MATH, while MATH corresponds to the symmetrization of the point MATH on MATH (where we think of MATH as the NAME dual of MATH for some choice of line MATH). Thus, it remains to prove that MATH pairs trivially with classes in MATH . We can think of the cohomological pairing MATH as counting the (signed) number of points to the NAME equations, which satisfy constraints in the compact subset MATH; that is, if MATH, where MATH are curves in MATH, and MATH are generic points in MATH, and MATH are generic lines MATH, then we have gemetric representatives MATH and MATH for these cohomology classes, so that MATH where MATH. In fact, if we consider the solutions MATH which satisfy these same constraints, then we get a manifold of dimension one with ends corresponding to MATH. Thus, counting boundary points with sign, we see that MATH . |
math/0006192 | Fix an initial metric MATH which is product-like along all MATH, and which agrees with MATH away from a tubular neighborhood of MATH. REF gives us a constant MATH so that for all MATH-tuples MATH with MATH, MATH is allowable. We can view the metric MATH as the result of inserting tubes with parameters MATH into a one-parameter (compact) family of metrics away from the MATH. Thus, REF gives us a number MATH with the property that if all MATH, then for all metrics in the path MATH, the theta divisor misses MATH. Hence, if MATH, then MATH is a MATH-allowable metric. |
math/0006192 | According to NAME 's theorem, for any generic MATH, the fiber MATH is a compact, canonically oriented, zero-dimensional manifold. We investigate the conditions necessary to show that the fiber misses the boundary of the domain of MATH. The set of points MATH for which the fiber MATH does not contain boundary points of the form MATH or MATH is an open set which contains MATH (since MATH is MATH-allowable for MATH). Let MATH be a connected neighborhood of MATH in this set. Moreover, the fiber MATH cannot contain points of the form MATH if the spaces MATH and MATH are disjoint; but MATH is equivalent to the condition that the image of MATH. This is a codimension MATH sub-torus MATH of MATH, so its complement is dense. Thus, for a dense set of perturbations MATH, the fiber MATH is a smooth submanifold which misses the boundary of the domain of MATH. Moreover, given two generic perturbations MATH, a generic path in MATH misses the locus MATH as well, since it has codimension MATH (this is what distinguishes the case where MATH from the case MATH). By NAME 's theorem, then, a generic such path induces a compact cobordism between MATH and MATH. By lifting to the covering space MATH, one can easily see that the cobordism respects the partitioning into MATH structures. Thus, MATH has the required property. Fix MATH and MATH. Since the space of MATH-allowable metrics is path connected and the space of metrics over MATH is simply-connected, we can connect MATH and MATH by a two-parameter family of metrics with MATH, MATH, MATH is MATH-allowable and MATH is MATH-allowable. This, together with a small generic perturbation, gives rise to a cobordism between MATH and MATH (which once again respects the partitioning into MATH structures). This completes the proof of the proposition. |
math/0006192 | To see this, we make use of the NAME duality map. NAME duality gives rise to a natural involution on MATH, with the property that MATH for any MATH and MATH. For any given metric MATH, this involution preserves the theta divisor (by NAME duality), and fixes the points associated to spin structures. The involution on MATH can be lifted to an involution of MATH which we can assume fixes MATH. The involution then carries MATH to MATH. Unfortunately, this involution is not defined on the symmetric product. Instead, it gives an involution on the set of points in the symmetric product which map injectively to the theta divisor, and more generally, it gives a relation. We write MATH if MATH. For any MATH, let MATH denote the moduli space MATH where MATH. For generic MATH, MATH, MATH, MATH is a smooth cobordism from MATH to MATH. This follows from the fact that the set of points in MATH which do not map injectively into the theta divisor is empty if MATH and has complex codimension MATH for generic MATH if MATH (see p. REF). As before, MATH can be partitioned according to MATH structures; let MATH denote the corresponding set. The same dimesnion counting also shows that for generic MATH, MATH, and MATH, MATH . This proves the lemma. |
math/0006192 | If MATH is not a torsion class, then MATH and MATH are disjoint, so the proof of REF applies. For the torsion MATH structure MATH (that is, the one corresponding to the spin structure), there are a priori two invariants, depending on the sign MATH for the perturbation MATH. But REF guarantees that MATH and since MATH comes from the spin structure, MATH, while MATH so the invariants for both perturbations are equal. |
math/0006192 | Note that the cylindrical-end metric MATH on MATH is conformal equivalent to the metric MATH on MATH inherited from MATH. Indeed, we can write MATH where MATH is a real-valued function, which agrees with the real coordinate projection on the cylindrical ends. The spinor bundles of the two manifolds can be (metrically) identified accordingly, with a change in the NAME action to reflect the conformal change. With respect to this change, (see CITE, bearing in mind that we are in two dimensions), the NAME operator over MATH can be written: MATH that is, multiplication by MATH induces a vector space isomorphism from the MATH-harmonic spinors to the MATH-harmonic ones. Moreover, a section of a bundle is in MATH for the cylinder iff MATH is in MATH for MATH. From this discussion, it follows that the MATH-harmonic spinors on MATH are identified with the space of harmonic spinors over MATH for which MATH lies in MATH (for MATH). The proposition follows from this, along with some considerations in the neighborhoods of the punctures. Consider MATH, with the trivial line bundle endowed with a connection MATH - this is the model of the punctured neighborhood of the MATH. Under the standard identification MATH, the function MATH (which is MATH over MATH) corresponds to the radial coordinate MATH on the disk. Moreover, multiplication by MATH induces an isomorphism from the space of (ordinary) holomorphic functions on MATH to the space of MATH-holomorphic functions. Under these correspondences, a holomorphic function MATH which vanishes to order MATH corresponds to a MATH-harmonic spinor on MATH iff MATH is bounded, that is, iff MATH. In the borderline case where MATH, the holomorphic functions on MATH which vanish to order MATH correspond to harmonic sections over the cylinder whose pointwise norm is bounded, and hence they lie in the space MATH (of the cylinder) extended by constants. The proposition follows. |
math/0006192 | Since the operator MATH is translationally invariant in the cylinder MATH, there is a single constant MATH which works for all the manifolds MATH, so that for any form MATH, MATH (see for instance CITE); thus, MATH . This together with the NAME lemmas shows that the forms MATH converge to zero in MATH over any compact set. But the cohomology class of any of the MATH is determined by its restriction to the (compact) subset MATH. |
math/0006192 | Fix a genus MATH . NAME decomposition of MATH. There is a ``stabilized" genus MATH . NAME decomposition of MATH, corresponding to the natural decomposition MATH that is, let MATH and consider the NAME decomposition MATH . We would like to show that the invariant MATH associated to the NAME decomposition MATH agrees with that associated to MATH, which we will denote MATH. Fix a metric on the torus MATH. We observe that one can find MATH-allowable metrics MATH on a NAME surface MATH with the property that for all sufficiently large MATH, MATH are MATH-allowable. To see this, let MATH denote the metric on MATH which is stretched out along MATH of the attaching circles of MATH. We show there is a MATH so that for all MATH, MATH is MATH-allowable. If this were not the case, we could stretch both tube-lengths simultaneously, and extract a subsequence of spinors, which would converge to a non-zero harmonic spinor either on the punctured MATH (with a cylindrical end attached) - which cannot exist in light of the holonomy constraint coming from MATH, see REF - or a harmonic spinor on the genus zero surface with MATH punctures obtained by degenerating the punctured version of MATH. This is ruled out by the holonomy constraints at infinity, as in the proof of REF (the holonomy around the curves corresponding to the attaching circles vanish as in the proof of that lemma; around the curve corresponding to the connected sum neck it vanishes since that curve bounds in MATH). Thus, for sufficiently large MATH, the metric MATH has the desired properties. In view of this observation, we can find a path of metrics MATH on MATH to calculate MATH, with the property that for all sufficiently long connected sum tubes, the family of metrics obtained by connecting MATH with a constant metric on the torus MATH can be used to calculate MATH. Choose a point MATH. The fiber of the map MATH (used in definition of MATH) over a generic point MATH misses the submanifold of divisors MATH which contain the point MATH. In other words, there is a compact region MATH so that MATH. Consider the one-parameter family of maps MATH used in defining the invariant MATH for the NAME decomposition MATH (using metrics with length parameter MATH). Note that we have an isomorphism MATH . Under this isomorphism, the origin corresponds to MATH, where MATH is a spin structure on the torus which bounds. Let MATH be the pre-image of MATH under the NAME map MATH . Since MATH admits no harmonic spinors, it follows that MATH is not the connected sum point. Given a sequence of points MATH with MATH using compactness on the MATH-side, we obtain a subsequence which converges to a divisor MATH and numbers MATH, so that MATH. It follows from our choice of MATH, that the divisor is actually supported in MATH. Moreover, looking on the MATH side, we see that the fiber points must converge to the divisor MATH. Thus, we see that for all MATH sufficiently large, the divisors in the fibers of MATH are contained in the range of the splicing map MATH where MATH is a compact set whose interior contains MATH, and MATH is some compact subset of the punctured torus (punctured at the connect sum point) which contains MATH. But applying REF , we see that the maps MATH obtained by mapping MATH (which we will denote MATH in a mild abuse of notation) converge in MATH to the map which sends MATH . (Note that MATH does not depend on MATH and MATH since we are fixing the metric on the torus side.) The preimage of MATH under this limiting map is the fiber MATH (the equality of the two sets follows from the fact that MATH), which is used to calculate MATH. Now the MATH convergence, identifies this fiber with the fiber MATH which is used to calculate MATH (this is how we chose the subsets MATH and MATH). Thus, MATH. Note that our sign conventions are compatible with stabilization, since if MATH, MATH are positively ordered for MATH and MATH, then MATH, MATH are positively ordered for MATH and MATH, since the boundary of the two-cell corresponding to MATH is MATH times the boundary of the one-cell corresponding to MATH. |
math/0006192 | Our hypothesis on MATH gives us a compact set MATH with the property that MATH. Similarly, our hypothesis on MATH gives a compact set MATH with the property that MATH. We let MATH, MATH be any pair of compact sets whose interior contains MATH and MATH. Consider pairs MATH over MATH which correspond to the intersection of the theta-divisor with MATH, and which are normalized so that the MATH norms over MATH of MATH is MATH. By local compactness, together with the fact that the tube admits no translationally invariant harmonic spinor, any such sequence of pairs MATH for tube-lengths MATH must admit a subsequence which converges in MATH to a MATH solution MATH and MATH on the two sides MATH and MATH, at least one of whose MATH norm is non-zero. By transfering back to MATH REF , our assumption on MATH ensures that MATH. By MATH convergence, then, the zeros of MATH must converge to the zeros of MATH over MATH. Without loss of generality, we might as well assume that all the MATH are of the form MATH for fixed MATH, MATH. Note that for each MATH, there is a unique MATH-holomorphic section MATH over MATH which, after transferring to MATH, admits an asymptotic expansion MATH (the growth here corresponds to the pole at MATH which we have introduced in our convention for the NAME map). Existence of the section follows from the fact that the MATH-fold NAME map has degree one (this is the ``NAME inversion theorem", see for instance p. REF). Uniqueness follows from the fact that a difference of two such would give a MATH section, showing that MATH actually lies in the theta divisor, which we assumed it could not. We show the restrictions of MATH to the MATH-side come close to approximating MATH or, more precisely, that its zeros converge to those of MATH. Rescale MATH so that over MATH, it has the form MATH . Consider the section MATH, viewed as a section of MATH (we can do this, as its support is contained in the support of MATH) . Note that MATH . Thus, MATH. Since MATH is NAME with index zero (it is a spin connection) and no kernel (it is not in the theta divisor), it has no cokernel, and we can conclude that MATH for some constant MATH independent of MATH. Since the restriction of MATH to MATH is MATH-holomorphic, elliptic regularity on this compact piece shows that the section MATH is bounded by some quantity of order MATH. Thus, the zeros of MATH in MATH converge to those of MATH. |
math/0006192 | We would like to apply a version of REF , with more than one neck (note that the proof works in this context as well). Let MATH be the theta divisor of MATH. It contains none of the connect sum points, of course, because it has degree zero. Moreover, the set MATH misses the theta divisor for MATH for MATH (the theta divisor of MATH consists of a single point where the holonomy around MATH is MATH). Hence, REF applies: for all sufficiently long necks, the theta divisor hits MATH in a region corresponding to the splicing map from REF . Thus, the composite of MATH with the splicing map is MATH close to the map MATH which is a product of MATH copies of the NAME map with the inclusion of the point (theta-divisor for the MATH). Since the NAME map in this case is a diffeomorphism, the points where the MATH-holonomy is trivial forms a smoothly embedded curve. In fact, it is easy to see that this curve is isotopic to MATH (see CITE and also CITE). Also, it is clear that post-composing with evaluation along MATH gives us map to MATH with MATH as a regular value, whose fiber is isotopic to MATH. It is easy to see that any other MATH close map must have MATH as a regular value, with an isotopic fiber. Thus, the corollary follows from REF . |
math/0006192 | The proof will rely on the fact that MATH has bounded variation along any one-parameter family of metrics. Specifically, let MATH be a one-parameter family of metrics, and fix a norm on MATH. Then, there is a constant MATH with the property that for any MATH, MATH, MATH . This follows immediately from the compactness of MATH, together with the fact that MATH is a MATH-equivariant map. Note that MATH is calculated by the number of points (counted with signs) in the zero-dimensional submanifold of MATH a space we denote by MATH. We construct a cobordism between this space and the points in the intersection stated in the lemma, as follows. We can assume without loss of generality that MATH is constant between MATH and MATH. Moreover, let MATH be a non-decreasing smooth function on MATH which is monotone increasing in the range MATH, with MATH and MATH. Consider the subspace of MATH (which agrees with MATH when MATH): MATH . We argue that for all sufficiently large MATH, MATH . Since MATH and MATH are not negative multiples of one another, we see that that as MATH, the distance between the point MATH and the ray MATH goes to infinity, and similarly the distance between the point MATH and the ray MATH goes to infinity. Fix MATH large enough that both distances are larger than the constant MATH from REF . This condition ensures that all points MATH have MATH and MATH. Monotonicity of MATH over MATH, and the choice of MATH then also ensures that the identification REF holds. Thus, REF is established once we construct a smooth cobordism between MATH and MATH. Consider the spaces obtained by connecting MATH to MATH by first allowing MATH to go from MATH to MATH (to connect MATH to MATH) and then allowing MATH to go from MATH to MATH (to connect MATH to MATH). Since MATH and MATH are allowable metrics, the MATH and MATH boundaries are excluded in this one-parameter family for all small MATH and MATH. Thus, we get a cobordism between MATH and MATH, provided that the MATH do not hit the MATH boundary, which is guaranteed if MATH . Taking MATH of both spaces, we get a pair of lines in MATH, which generically miss each other when MATH. The case where MATH requires a slightly closer investigation. In the first part of the cobordism, where we allow MATH to vary in MATH, there are still no MATH boundaries, as we can arrange for MATH (since, applying MATH, we have a point and a line in a two-space). Now, as MATH varies in the MATH, it is easy to see that the only possible MATH boundaries lie in the range where MATH, by our hypothesis on MATH, and hence they must lie in the set MATH since MATH is constant for MATH. Now, consider the intersection point MATH of the induced rays MATH and MATH. Note that as MATH is stretched out normal to the attaching disks, the image under MATH of the intersection intersection MATH converges to a discrete subset of the ray MATH, consisting of points separated by some distance MATH (which depends on the MATH and MATH). If MATH misses this discrete set, then if MATH is sufficiently stretched out, then all sufficiently small MATH and MATH have the property that MATH . If, on the other hand, MATH lies on the discrete set, then, given any sufficiently small MATH, if MATH is sufficiently stretched out, then all sufficiently small MATH and MATH have the property that MATH . Thus, in both cases, we have obtained the requisite cobordism. |
math/0006192 | Take a weak limit of connections which lie in MATH, as all the MATH curves are stretched. Under this limit, the surface degenerates into a collection of genus zero surfaces (with cylindrical ends), whose ends correspond to attaching circles for MATH or separating curves for MATH. Thus, the weak limit of connections in MATH induces a connection over these genus zero surfaces, whose holonomies around all its bounding circles is zero. But none of these support harmonic spinors according to REF , proving the lemma. |
math/0006192 | Fix a metric MATH on MATH as in REF . Let MATH denote the metric which is stretched by MATH along the MATH and MATH along the separating curves. The lemma guarantees that for all MATH, MATH is allowable. Now, for all sufficiently large MATH, REF gives the isotopy of MATH with the subset MATH, where MATH. |
math/0006192 | Let MATH be MATH-allowable metrics for MATH. Given MATH, we construct a natural element MATH closely related to the MATH defined in the beginning of this section. The MATH will be a translate of the following analogue of the MATH, which is assigned to a pair of lifts MATH and MATH (to MATH) of the manifolds MATH and MATH: MATH . To do define the MATH, we must assign a MATH structure to each intersection point of MATH with MATH. To this end, we assume that MATH and MATH are strongly allowable for some MATH. If MATH, then let MATH be a lift of MATH. There is a pair of lifts MATH and MATH of MATH and MATH which meet in MATH. Note that there are unique lifts MATH and MATH whose image under MATH lie in a MATH neighborhood of MATH and MATH respectively. Let MATH denote the MATH structure which corresponds to the difference between these two lifts (that is, if MATH and MATH are the two lifts, then the pair MATH and MATH are translates of MATH and MATH by a single cohomology class in MATH). By summing over all intersection points which correspond to a given MATH structure (with signs), we obtain the element MATH, which is clearly the translate by some MATH structure of the polynomial MATH defined above. It follows from REF that MATH is given by adding up the intersection number of certain lifts of MATH and MATH. Moreover, the intersection point MATH will contribute for each MATH in the MATH structure MATH (as those are the MATH structures for which MATH lies on the corresponding rays). This proves that MATH . Finally, the proposition is proved once we establish that the MATH is a translate of MATH, as defined in REF . To see this, recall that REF guarantees that the spaces MATH and MATH are isotopic, for suitable choices of allowable metrics MATH and MATH, to MATH and MATH respectively. Now the polynomial MATH, which is clearly a translate of MATH, depends on the submanifolds MATH and MATH only up to isotopy. Thus, the proposition follows. |
math/0006192 | We adopt the notation and most of the argument for the proof of REF . The difference arises when one wishes to prove that the moduli spaces MATH and MATH are cobordant, that is, when one wishes to exclude the possible MATH boundary components. To do this, it is no longer possible to argue that MATH . Rather, to exclude MATH boundaries, we show that MATH (which suffices). To see this, begin by choosing generic MATH and MATH so that MATH is positive (we are free to do this according to REF , or just REF for integral homology three-spheres). If the intersection were non-empty, by applying MATH to both sides, we would get: MATH . Thus, MATH which is impossible, as it is a sum of four non-negative terms at least one of which (the first) is positive. |
math/0006192 | This is a direct consequence of REF . |
math/0006192 | REF shows that MATH is determined by MATH and the classes MATH and MATH. After a series of handleslides, we can arrange that all MATH and MATH are equal to one, fixed generator of MATH. It follows from REF that all MATH equal MATH up to translation. Since the NAME polynomial MATH (modulo multiplication by MATH) is the greatest common divisor of the MATH (compare REF ), it follows that MATH is a translate of the symmetrized NAME polynomial. But REF also shows that MATH is symmetric. |
math/0006192 | Since MATH, there is a unique MATH structure over MATH. Moreover, there is a complex-antilinear involution MATH of the spinor bundle which commutes with NAME multiplication (actually, this involution exists in much more general contexts, and can be thought of as the basis for the conjugation action on the set of MATH structures described in REF). It follows that if MATH is the connection on MATH coming from a spin structure MATH on MATH, MATH, then MATH . We can express any given closed path MATH as MATH, where MATH is a one-parameter family of closed one-forms which induces a closed path MATH in MATH; and the homotopy invariance of the spectral flow ensures that the spectral flow of the NAME operator around MATH depends only on the free homotopy class of MATH (in particular, it is independent of the spin structure). Now, the conjugation symmetry gives us that MATH but these two spectral flows have opposite signs: the path MATH is homotopic to the path MATH given the opposite orientation. Thus, the spectral flow around the MATH must vanish. |
math/0006193 | Morally this means that MATH-equivariant sheaves descend naturally to the quotient MATH. To check that the cohomology of complexes MATH and MATH are canonically isomorphic it is sufficient to consider infinitesimal gauge transformation MATH, MATH. Then the map MATH gives the needed chain isomorphism. |
math/0006193 | It is straightforward to check that MATH, MATH. In our case MATH-module is the complex REF with MATH-action MATH, MATH. The operators MATH for MATH define the lifting of this action to MATH-action. |
math/0006193 | MATH . |
math/0006193 | We need to check first that MATH for MATH. It is enough to prove this for MATH. Applying the standard commutation rules MATH one gets: MATH . Assume that MATH, or equivalently MATH . Applying the operator REF to MATH we get MATH on the other hand MATH . Let MATH be a solution to NAME equation describing a deformation of complex structure MATH. Let MATH be a set of complex coordinates with respect to the fixed initial complex structure MATH. The differentials of the deformed complex coordinates can be written as MATH, where MATH. Therefore MATH. Notice also that for MATH one has MATH. It follows that MATH for MATH. It is easy to see that there exist elements MATH, MATH, MATH, such that MATH, MATH, MATH and such that MATH give a basis in MATH. Then given a solution to NAME equation MATH over an NAME algebra MATH describing MATH one has MATH for sufficiently large MATH. Therefore MATH and MATH . |
math/0006193 | Let MATH describes a MATH-deformation MATH. Let MATH describes a family of elements of the varying semi-infinite subspaces. Then for any vector field MATH: MATH . |
math/0006193 | It follows from REF that MATH is transversal to MATH. One can show that MATH-module MATH is free (it follows, for example, from REF ). Therefore MATH. |
math/0006193 | It follows from REF that MATH . Therefore there exist MATH such that MATH . On the other hand, MATH . Therefore MATH . The coordinates MATH were chosen so that MATH is linear in MATH. Therefore in these coordinates MATH and MATH. |
math/0006194 | The orientation and partitioning into MATH structures are described in REF. When MATH, the genericity statement follows from REF , which is proved in REF of the present paper. Note that solutions with MATH correspond to points in MATH which map to MATH, which is a discrete set of points (since MATH is a homology three-sphere); so these are excluded by dimension counting. When MATH, it is easy to see that the moduli space is empty: in that case, the theta divisor consists of a single, isolated spin structure, which does not bound. |
math/0006194 | Let MATH be a generic allowable homotopy connecting MATH and MATH. Consider the moduli space MATH . According to the generic metrics result, REF , this space is an oriented one-manifold with boundary. Since MATH is an allowable homotopy, the only boundary components are MATH, MATH and MATH. Thus, counting boundaries, with sign and multiplicity, we get the result as stated. |
math/0006194 | The vanishing of the kernel follows from the vanishing of the kernel over the three pieces: MATH, MATH, and MATH respectively. Over the handlebodies the kernel vanishes for generic choices of MATH and MATH (this will be proved in REF ). Over the cylinder, it vanishes for generic paths MATH, according to REF . Moreover, the spectral flow statement follows from the splitting principle, the chambered property of MATH, and the chambered property of MATH, which in turn follows from REF together with REF . |
math/0006194 | The allowability of MATH and diffeomorphism statement for MATH were proved in REF; note that the hypothesis that MATH was not used in the proof of this fact. One can arrange for MATH and MATH to be generic simultaneously by varying the family MATH in a region MATH which does not contain the connected sum point. This statement is proved in REF . |
math/0006194 | The connection MATH naturally induces a connection on the spinor bundle of the boundary. More precisely, under the splitting MATH into the MATH-eigenspaces of NAME multiplication by MATH times the volume form of MATH, the connection MATH naturally induces a connection MATH on MATH. If MATH is compactly supported in MATH, then the induced connection MATH has normalized curvature form. The NAME operator in a neighborhood of MATH takes the form MATH . According to REF, this operator is NAME if the kernels of MATH and MATH are trivial. Now, if MATH is reducible, or even if it differs from a reducible by a compactly supported one-form, then MATH. Thus, the NAME condition is guaranteed if MATH is MATH-allowable. The genericity statement is an application of the NAME theorem (see CITE). Let MATH and MATH . With in NAME completions, MATH is a NAME manifold which is transversally cut out from MATH by the NAME equation: that is, if MATH were in its cokernel at MATH, then by varying the spinor component, we would see that MATH is MATH-harmonic. By varying the connection (indeed, in any open set), we see that MATH must vanish identically (by the unique continuation principle). It is easy to see that the projection map from MATH to MATH which forgets the spinor is NAME of index MATH. It follows then from the NAME theorem that for generic MATH, there are no harmonic spinors. |
math/0006194 | Fix some constant MATH, and fix a smooth, non-decreasing function MATH with MATH . Then, the hypersurface MATH inherits a metric from MATH. The region where MATH and MATH is diffeomorphic to half of the three-ball (its two boundaries, are the loci where MATH and MATH vanish respectively). Since the function MATH is constant for small values, the metric is easily seen to respect the corners. Moreover, the metric is easily seen to have all non-negative sectional curvatures (see for example CITE), as the hypersurface is locally described as a graph of MATH (after possibly renumbering the four variables), which is clearly a convex function. |
math/0006194 | Both statements are proved in CITE: the NAME paramatrix is constructed in the proof of REF , and the exponential decay estimate is NAME REF of that reference. |
math/0006194 | We can introduce coordinates on MATH with respect to which MATH takes the form MATH. For any real numbers MATH, MATH, let MATH. Clearly, MATH gives a one-parameter family of complex structures whose differential at MATH acts by MATH . The lemma follows. |
math/0006194 | We calculate the (connection form) difference between the NAME connection and MATH over MATH. Suppose for the moment that MATH. Let MATH be a (time dependent) moving coframe for MATH. There are functions MATH over MATH for which MATH where MATH is the connection matrix for the metric MATH over MATH. Thus, by the NAME formalism, the connection matrix with respect to the coframe MATH is given by MATH . For general MATH, MATH form a moving coframe, and the above calculation shows that the difference form MATH where MATH is a one-form whose MATH length is independent of MATH. In fact, by glancing at the connection matrix, we have that MATH for some endomorphism MATH which is independent of MATH. It follows that the square can be written MATH . The anticommutator term can be expressed in the local coframe: MATH . The two-form MATH is independent of MATH (only its NAME action depends on MATH); and the NAME action of a fixed form in MATH scales like MATH. |
math/0006194 | By REF , the spectral flow localizes to the region where MATH meets the MATH-theta divisor. By homotopy invariance of both quantities, then, it follows that the spectral flow must be some multiple of the above intersection number. The factor is then calculated in a model case, as in REF (note that in that proposition, the ``spectral flow" refers to real spectral flow, hence the difference in factors of MATH). |
math/0006194 | We recall from CITE that MATH, where the MATH structure MATH corresponds to the integer MATH under the identification MATH which sends the spin structure to MATH. |
math/0006194 | Every integral homology three-sphere can be obtained from MATH by a sequence of MATH surgeries. So, it follows that MATH is uniquely determined by its surgery formula and its value on MATH. It is easy to see that MATH: fix a genus one NAME decomposition of MATH, and let MATH be a constant family of metrics on the torus, which is clearly an allowable path, and indeed MATH for this path. Furthermore, the correction terms cancel each other by REF . Since NAME 's invariant MATH satisfies the same surgery formula (see CITE), and MATH as well, we get the result. |
math/0006194 | Smoothness follows from the generic metrics statement REF . There is no MATH boundary since the metric MATH is MATH-allowable. The MATH boundary lies in MATH, which maps near MATH, as above. The MATH boundary corresponds to the intersection of the theta divisor with MATH, which in turn maps to MATH. Indeed, this latter circle corresponds to the circle of reducibles MATH: the point MATH corresponds to the spin structure which extends over MATH, while the point MATH corresponds to the spin structure on MATH. |
math/0006194 | Fix a real number MATH with MATH. Fix curves in MATH: MATH, MATH, and MATH. Note that the restriction of MATH and MATH to the torus gives the spin structures corresponding to MATH and MATH respectively. Since the moduli space misses these spin structures, it follows that MATH and MATH are well-defined, and indeed by the definition of the MATH-invariant, we have that MATH and MATH; similarly, MATH . Consider the oriented subset MATH which does not contain MATH, and whose boundary is MATH. Then, by transversality, MATH . According to REF , and our choice of MATH, MATH consists of boundary components where MATH. In fact, REF shows that MATH so that counting points in REF , we obtain REF . |
math/0006194 | Recall that there is a natural cobordism MATH from MATH to MATH. This cobordism is obtained from MATH, and then attaching a two-handle with MATH framing. We have a natural inclusion MATH which maps MATH to the boundary of MATH (that is, taking MATH to the knot complement, as a subset of MATH, and MATH to the corresponding subset of MATH). Fix a metric on MATH whose restriction to the region MATH is a product metric, its restriction to the MATH boundary agrees with MATH, and its restriction to MATH agrees with MATH. Consider the MATH structure MATH on MATH whose first NAME class generates MATH. Endow MATH with a MATH connection in MATH whose restriction to MATH is a path of reducibles, and whose first NAME form is compactly supported away from the boundary (and hence it interpolates between MATH and MATH). By definition, MATH . The index is calculated by an excision principle. Let MATH. According to CITE, the NAME operator on both MATH and MATH is a NAME operator, since the NAME operator has no kernels on the ``corners" MATH and the various boundaries MATH (coupled to MATH and MATH respectively) and MATH (coupled to a sufficiently slowly-moving one-parameter family of connections on MATH which bound - see REF ). Thus, the index splits as MATH . The first term is the spectral flow MATH. To understand the second term, we replace the knot complement MATH by a much simpler knot-complement MATH, endowed with a family of connections whose holonomy around the MATH factor goes from MATH to MATH. Note that the manifold MATH is a cobordism from MATH to MATH: indeed, it is MATH punctured at two points. Moreover, the connection MATH extends over MATH, to give a connection MATH. Now, by the same splitting principle, MATH . The first term on the right hand side is a spectral flow for the NAME operator through flat connections on the manifold MATH, which has non-negative sectional curvatures; thus, the index vanishes. The second term on the right is, of course, the same as MATH; thus, we have that MATH . Finally, by the same reasoning which gave REF , we have that MATH which is a difference of the APS invariant for the three-sphere MATH. Now, according to REF (together with the fact that MATH for any MATH connections with traceless curvature), we have that MATH . Combining REF, we have established the lemma. |
math/0006194 | This follows from the surgery formula, and the calculation of MATH from CITE, according to which MATH . |
math/0006194 | Since any rational homology three-sphere MATH can be obtained from MATH by a sequence of surgeries on rational homology three-spheres, REF and the constants MATH determine the sum MATH (see REF ). A suitably normalized version of the NAME invariant satisfies a formula of the same shape, with constants MATH. We spell this out as follows. Let MATH, so that MATH and MATH, MATH. NAME 's surgery formula (see p. REF) says that: MATH . Here, MATH is the sum MATH where MATH is the symmetrized NAME polynomial of MATH, normalized so that MATH. (The symmetrization forces us to allow half-integer powers of MATH, if MATH is even.) We compare MATH with MATH. By comparing with the NAME torsion, one sees that MATH up to possible multiples of MATH and constants. Now, MATH are the coefficients in the NAME polynomial of MATH, normalized so that MATH. With these normalizations, then, it follows that MATH . Then, MATH . For the renormalized version MATH we then have: MATH . It follows that MATH. Thus, it remains to show that MATH . For MATH it follows from REF that MATH, so we have that MATH as claimed. We now argue that in fact MATH is determined by the surgery formula and the values of MATH. To this end, we find it convenient to make the following definitions. Choose three fiber circles in MATH, and let MATH denote the manifold obtained by performing MATH surgery on the MATH circle, with respect to the framing induced by the product structure. Similarly, let MATH denote the manifold obtained from surgeries along only two circles, and let MATH be the three-manifold obtained from MATH by deleting a tubular neighborhood of the third circle. Note that MATH is either a lens space or MATH. Note that MATH is obtained from MATH by a MATH surgery. Note also that MATH is a connected sum of lens spaces MATH. This is clear, for example, from the NAME calculus picture in REF . Suppose first that MATH and MATH are relatively prime. Then, since MATH, it follows that MATH is obtained from MATH by a sequence of surgeries of divisibility MATH. With this in place, we turn our attention to the cases where MATH. We can assume that MATH is even. Consider the manifold MATH. It has divisibility equal to MATH. It is easy to see that MATH is gotten from MATH by a MATH surgery. Both of these manifolds can be obtained from MATH by surgeries of divisibility MATH. (Note that the first manifold is MATH, and MATH has divisibility MATH, since MATH is odd.) For the general case, we use induction on MATH. Note that MATH, with equality iff MATH and MATH. Similarly, MATH, with equality iff MATH and MATH. Now it follows that MATH can be obtained from either MATH or MATH with surgeries of divisibility less than MATH, unless MATH and MATH. Since MATH, this would imply that MATH and MATH are not relatively prime. Now, since MATH, it follows from the above argument that MATH for all MATH. This finishes the proof of REF . |
math/0006194 | The MATH boundaries correspond to the intersection of the theta divisor with the MATH. Using the identification between the Jacobian and the MATH, MATH corresponds to those MATH representations of MATH which extend to representations of MATH. Since MATH bounds in MATH, these representations must take MATH to a MATH-torsion point. NAME with any element of MATH which maps to a generator of MATH, and taking the reducible representative on MATH, changes the holonomy around MATH by MATH: this shows that the number MATH appearing above is the level of the MATH structure as defined in the beginning of this section. |
math/0006194 | This follows immediately from the splitting principle for spectral flow, applied to the zero-surgery MATH. |
math/0006194 | Note first of all that MATH is homologous to MATH. Thus, we can consider the oriented MATH-chain MATH which does not contain MATH, and whose boundary is MATH. By elementary differential topology, MATH . These are to be viewed MATH-chains; that is, we have that MATH where MATH denotes the multiplicity of MATH at MATH times the degree of MATH to MATH. Note that MATH . Note that the multiplicity of MATH is constant on the open interval MATH and MATH, and we denote the constant by MATH and MATH respectively. Thus, we can write the final term in REF as: MATH thanks to REF . Moreover, it is a simple homological fact that MATH, so we can rewrite the above quantity as a sum of MATH where MATH is the spectral flow around the circle from REF . Substituting this back into REF , along REF, we obtain REF . Note that in the case where the level is MATH, the spectral flow terms might not be defined, since the path of reducibles fails to induce a NAME operator. This can be compensated by using spectral flow through nearly-reducible connections (that is, translating MATH, once again, by a small amount). |
math/0006194 | Since MATH can be thought of as the MATH index of the NAME operator for MATH, with ends attached, the above statement is an easy application of the splitting principle for the index, bearing in mind that the difference MATH is the index of the NAME operator on MATH, and topological terms which depend only on MATH, MATH, MATH, and MATH. |
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