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math/9908014 | Unlike in the case MATH, the NAME variety MATH can in general no more given explicitly in the form MATH. The spectrum is still the intersection of this complex variety with MATH, the parameter MATH labeling NAME waves. The intersection of the NAME variety with MATH is a set in the complex plane which is contained in a union of at most MATH closed curves. Let MATH denote the determinant function on the algebra of complex MATH matrices. We write MATH, also in the noncommutative case. With MATH, we have MATH where the mixed term MATH is a polynomial in MATH which is of degree MATH in MATH and of degree MATH in MATH. |
math/9908014 | Small changes are needed to the known proofs CITE for the same statements for selfadjoint operators MATH, where MATH. Write MATH for a point on MATH. CASE: It is enough to prove the statement for periodic, (nonergodic) transformations MATH. Proof. Every MATH can be approximated by periodic transformations MATH using NAME 's lemma CITE (which implies that there exist measurable sets MATH whose NAME measure on MATH goes to MATH for MATH so that MATH for MATH). The NAME formula is an identity for subharmonic functions MATH, parameterized by MATH. Let MATH denote the NAME measure of a subharmonic function MATH it satisfies MATH in the sense of distributions, where MATH is the Laplacian. It follows from the NAME lemma and the dominated convergence theorem that if MATH in the uniform topology for measure preserving transformations, then MATH weakly. Also MATH weakly holds because MATH is a weak limit of measures MATH, where MATH are the roots of MATH and MATH is continuous from MATH with the uniform topology to the space of measures on MATH with the weak topology. By assumption, we have MATH. Therefore MATH which is MATH. From this, MATH follows by NAME 's lemma CITE. CASE: Next, it is sufficient to assume that MATH is outside the spectrum of a periodic operator MATH (which is now contained in a finite union of one-dimensional curves REF ) as well as that MATH is outside the spectrum of MATH, where MATH is the NAME matrix on MATH with zero NAME boundary condition at MATH. The spectrum of MATH is a finite set of points. Proof. This is the NAME subharmonicity argument CITE. Taking balls MATH around a point MATH in the spectrum, the result holds for the over MATH averaged NAME exponent. Passing to the limit MATH is possible because of subharmonicity. CASE: If MATH is an eigenvalue of the MATH matrix MATH, (where the MATH matrix MATH is written as a MATH matrix of MATH matrices), then each root MATH of MATH gives rise to an eigenvector MATH of a truncation MATH of MATH, where MATH are put to zero for MATH as well as for MATH. The measures MATH converge for MATH weakly to the density of states MATH which is supported on finitely many curves. This is the NAME lemma CITE which holds in general for random finite-difference operators which do not need to be selfadjoint. CASE: The fact that MATH converges to MATH for almost all MATH in the limit MATH is seen in the same way as in the Appendix of CITE: If MATH are the matrix-valued NAME functions, where MATH and MATH, then MATH is the arithmetic mean MATH of the two largest NAME exponents of MATH. Furthermore, the maximal NAME exponent averaged over MATH is MATH . Because MATH is not an eigenvalue of MATH, there are two MATH-matrices MATH such that the functions MATH satisfy MATH. Here MATH is an other solution of MATH satisfying a NAME boundary condition at MATH. The matrix MATH is invertible because else there would exists MATH with MATH and MATH would be in MATH and contradict the fact that MATH was also chosen away from the spectrum of MATH. (In the selfadjoint case, this would already have been taken care of by choosing MATH.) Writing MATH and using MATH, one concludes MATH . |
math/9908014 | With the norm MATH, where MATH are the basis vectors, there is an explicit expression for an upper bound of the integral MATH. Following CITE, define MATH where the square-root takes the solution with positive imaginary part. For matrices MATH one has MATH. Furthermore, with the analytic function MATH where MATH, one has using the NAME formula MATH . Therefore MATH and MATH. For MATH, one has MATH. The function MATH, having a positive derivative, is strictly monotone. |
math/9908014 | We use NAME 's determinant theorem, which tells that for a finite matrix MATH, one has MATH, where MATH are the rows of the matrix MATH. (The relation follows from the fact that if MATH, then MATH, where MATH. The determinant theorem is equivalent to MATH, a relation which can easily be seen geometrically (that is, CITE). ) In the random case, where the MATH are vector valued random variables on MATH with NAME measure, and MATH is a finite dimensional MATH approximation of MATH and where MATH converges to to the NAME determinant MATH, we obtain from NAME 's determinant inequality and NAME 's ergodic theorem MATH . In the case of random NAME operators MATH, where MATH, this gives MATH . |
math/9908014 | If MATH and MATH is small enough, the cocycle maps the constant cone bundle MATH strictly into itself. (One could write the complex matrix cocycle as a real matrix cocycle on MATH for which the cone MATH in MATH is strictly invariant. This is equivalent to the fact that one can find a basis and MATH such that MATH. By (CITE), one can deal directly with the complex case. ) CITE is written in the case of continuous cone bundles and homeomorphisms, but REF extends this to the measure theoretic case. |
math/9908014 | The NAME derivatives of the function MATH can be computed (see REF) and satisfy MATH, where MATH is a constant independent of MATH. Actually, one only needs the first derivative: NAME 's theorem which says that if a function MATH is analytic in a disc MATH of radius MATH, then MATH contains a disc of radius MATH, where MATH is a constant not depending on anything (see CITE p. REF). |
math/9908014 | CASE: The operators MATH can be embedded in a single NAME algebra without changing the determinant. Proof. MATH is in the crossed product MATH of MATH with the MATH-action MATH and MATH is in the crossed product MATH of MATH constructed with the MATH-action MATH. The operator MATH is in the crossed product MATH of MATH, constructed with the MATH-action MATH, where MATH on MATH. One can embed the two crossed products MATH and MATH in MATH. The density of states of MATH exists in the same way as it exists for MATH and MATH. Moreover, the density of states of MATH (rsp. MATH) in MATH is the same as the density of states of MATH in MATH (rsp. in MATH). CASE: For the verification of the NAME formula, see REF. CASE: We can assume without loss of generality that MATH and MATH are invertible. Proof. By ergodic decomposition (that is, CITE), one can assume MATH to be ergodic. Approximate MATH by periodic transformations MATH using NAME 's lemma CITE: the measure of MATH is then smaller than MATH and MATH. By ergodic decomposition, the determinant is in such a case an integral of determinants of type MATH factors, in which case MATH as well as the product MATH are periodic operators which have their spectrum contained in a finite union of real smooth curves in the complex plane. In this case, there are arbitrarily small complex numbers MATH such that MATH are invertible allowing to apply the determinant formula. Now, MATH as well as MATH are continuous. MATH are invertible and MATH, MATH, and MATH for MATH. CASE: The product formula. Proof. We can now invoke the determinant theory for finite type NAME algebras CITE see REF which tells that MATH for invertible operators MATH in a type MATH NAME algebra. |
math/9908014 | Apply REF . |
math/9908014 | CASE: Assume first that MATH is analytic in MATH and that it has no roots in MATH. Clearly, MATH . Because MATH is analytic, the NAME equations MATH hold. With MATH, we obtain MATH . We have therefore, using MATH, MATH . CASE: In general, we partition MATH into MATH small sectors MATH and apply the formula to small sectors MATH which do not intersect MATH. The assumption on MATH and MATH assures that for all the integrals, the contribution of the sectors which intersect MATH can be neglected in the limit MATH. |
math/9908014 | For any general analytic parameterization MATH, the NAME exponent MATH of MATH is a a subharmonic function of MATH. The NAME decomposition theorem (see that is, CITE) implies the abstract NAME formula MATH, where MATH is harmonic. The name 'abstract NAME formula' is from CITE. If MATH is harmonic for large MATH and grows like MATH for MATH then MATH. The NAME measure MATH of the subharmonic function MATH satisfies MATH in the sense of distributions, where MATH is the Laplacian on distributions. The measure MATH coincides in the periodic case with the density of states because in this case the NAME exponent is zero exactly on the spectrum which implies that the density of states is the equilibrium measure on the spectrum. Consider the NAME measures MATH of the subharmonic functions MATH and let MATH be the NAME measure of the NAME exponent MATH. Because MATH converges by the multiplicative ergodic theorem to MATH for almost all MATH, the convergence holds also in the sense of distributions. Then also the NAME measures MATH converge to MATH as distributions for MATH. But then, because smooth functions are dense in all continuous functions, MATH converges weakly as measures to MATH. In the energy parameterization REF , the density of states of MATH is defined as the measure MATH in the parameter plane MATH satisfying MATH for every continuous function MATH. For MATH outside the spectrum of MATH, the sequence MATH converges for MATH uniformly to MATH. Because MATH where the NAME measure MATH is a finite point measure located on the point spectrum of MATH defined in an earlier section, the formula holds also after integration over MATH and MATH is the density of states of the random operator MATH. The functions MATH, where MATH is a dense set in the complement of the spectrum of MATH, span a dense set in MATH. For such functions, we have MATH by the NAME lemma. We know therefore MATH for all MATH so that MATH is the density of states. |
math/9908014 | The proof is the same as in the real case CITE: assume MATH is contained in a ball of radius MATH and assume that MATH is any ball of radius MATH around MATH. With MATH, we have MATH . Therefore, using in the first inequality of the following identities also that MATH on MATH and MATH, we have MATH . For MATH with MATH, and because MATH for MATH so that MATH which is the claim with MATH. |
math/9908014 | Given a point MATH with MATH. The set MATH is an open set. We have MATH for MATH. Write MATH . We have MATH and MATH where MATH. |
math/9908014 | CASE: MATH is nonnegative in the disc MATH and MATH. NAME 's inequality gives MATH so that MATH and MATH. CASE: The function MATH is positive in MATH. Let MATH be the analytic function in MATH which has MATH as its real part and which vanishes at MATH. By the NAME criterium CITE, the analytic function MATH is univalent in MATH because it has a derivative which has a positive real part in MATH. The function MATH is in the class MATH of functions which are analytic and univalent in the unit disc and satisfy MATH. By the distortion theorem for maps in the class MATH, we have MATH for MATH. Assuming MATH and applying this for MATH, we get for MATH . In other words, if MATH, then MATH which leads to MATH and implies the claim. |
math/9908021 | The result may be read off from the following estimate: MATH the estimate being valid for all MATH. Since MATH, MATH is dense in MATH, so we have MATH for all MATH, and this last property is equivalent to MATH being a pure shift on MATH. |
math/9908021 | The result follows from a substitution of the vectors in REF into the positivity requirement REF , and computing out the answer for the two cases of reflection MATH, that is, MATH and MATH. We refer to REF for more details, and additional comments on applications to interpolation theory. |
math/9908021 | The proof is rather long and will be broken up into its three REF , and REF . REF asserts the existence of a selfadjoint realization of the given operator MATH, while REF is uniqueness up to unitary equivalence. REF is an explicit construction which takes place in a certain reproducing kernel NAME space. The following observation gives a more concrete understanding of REF of REF . Let MATH be a period-MATH unitary operator in a NAME space MATH, and let MATH be the respective eigenspaces corresponding to eigenvalues MATH of MATH. If MATH is the projection onto MATH, then MATH. |
math/9908021 | The main idea in the proof is in CITE, but we include a sketch. This will also give us a chance for introducing some terminology which will be needed later anyway. Suppose MATH is a closed subspace which satisfies REF . For MATH we have MATH, where MATH and MATH. But MATH for all MATH by REF , and if we define MATH, then MATH is well-defined and contractive from MATH to MATH. The reasoning shows that the converse argument is also valid, so the lemma follows except for the assertion that MATH must be automatically closed. Let MATH be a sequence of vectors in MATH such that MATH. Then by REF , MATH . So the limit MATH exists in MATH, and MATH . Since MATH is assumed closed in MATH, we get MATH, and MATH. This shows that MATH is closed, and the proof is completed. |
math/9908021 | CASE: Let the operator MATH be given as in the statement of the theorem. Let MATH be the invariant subspace with projection MATH, and let MATH be the reflection. It is assumed to satisfy REF - REF . In view of REF , we have MATH where MATH denotes the given inner product from MATH. (Note that MATH is not invariant under MATH, so the vector MATH is typically not in MATH if MATH is.) Applying the NAME - NAME inequality, we get MATH . The idea is to get a new NAME space MATH from the form MATH, that is, that this form should be the new inner product. So we must form the quotient space MATH where MATH . In view of REF , we get MATH . Since MATH we conclude that MATH passes to the quotient MATH and defines there a symmetric operator. When MATH is completed in the new norm MATH, MATH the induced operator becomes selfadjoint in this NAME space MATH . The induced operator will be denoted MATH, and we will now show that it satisfies REF - REF , starting with REF , that is, showing first that MATH is a bounded operator in the NAME space MATH. The argument for boundedness is essentially in CITE, but we include it here for the convenience of the reader. Let MATH, and use recursion on REF as follows: MATH . We have MATH the spectral radius, and MATH if MATH. We have therefore proved the estimate MATH for MATH, and it follows that the induced operator MATH on MATH satisfies REF , as claimed. Since we already showed that MATH is selfadjoint, we conclude that MATH has bounded spectrum inside the interval MATH . If MATH on MATH is unitary, this is the interval MATH. If MATH, MATH, is a group of operators, then MATH, MATH, is a semigroup of selfadjoint operators, and so MATH for all MATH, and the spectrum of MATH is therefore positive in that case, and we get the representation MATH for some (generally unbounded) selfadjoint operator MATH in MATH. |
math/9908021 | For a given operator MATH which has a pair MATH defining a reflection symmetry, we showed in REF that there is a system MATH with a selfadjoint operator MATH in MATH, and an intertwining operator MATH, which satisfy REF - REF in the statement of the theorem. We now prove that this system is unique up to unitary equivalence. So suppose there are two systems MATH, MATH, both satisfying REF - REF and with the two ``extension" operators MATH and MATH both selfadjoint and bounded. We will now show that there is then a unitary isomorphism MATH which defines the equivalence, that is, which satisfies REF in the theorem. We will make REF into a definition, setting MATH for MATH. Since both MATH and MATH satisfy REF , we conclude that MATH or, stated equivalently, MATH where MATH is defined in REF . Hence, REF makes a good definition of a linear operator MATH mapping a dense subspace in MATH into one in MATH. But REF for MATH and MATH implies that MATH is also isometric, indeed MATH . Hence MATH is a unitary isomorphism of MATH onto MATH. Using now REF for the two systems, we get MATH . Since MATH has dense range, we get the desired intertwining REF as claimed in the theorem. |
math/9908021 | The assertion in REF is that there are examples where the induction MATH has intertwining operator MATH with zero kernel, or equivalently, MATH. We already mentioned this in REF - REF, and in fact this is a one-parameter semigroup of isometries MATH, MATH. In fact, it arises as the restriction to an invariant subspace of a unitary one-parameter group. It is a representation MATH, MATH, of the multiplicative group MATH, or equivalently, via MATH, a representation of the additive group MATH. We get as a corollary of REF that MATH is equivalent to the group of translations on MATH for some infinite-dimensional NAME space MATH as described in REF - REF in the conclusion of REF above. Now recall the NAME space MATH and its subspace MATH from REF. When MATH, MATH is defined by the norm MATH from REF and the subspace MATH is the completion of MATH in the MATH-norm. We may pick some MATH, and consider the isometry MATH of MATH. From REF we see that MATH also depends on MATH. The new inner product is MATH (defined for MATH), and depends on MATH as well. It is worked out explicitly in REF . It follows from REF that MATH is defined from the integral kernel MATH. The corresponding operator MATH is a special case of the NAME - NAME intertwining operator, see CITE. (See also CITE and CITE.) This operator MATH is defined more generally and also in MATH. Then the integral kernel is MATH, and MATH. If MATH is the positive NAME operator in MATH, that is, MATH, then it is shown in CITE that MATH, and the NAME transform of MATH is MATH . Hence up to a constant, the norm MATH of REF may be rewritten as MATH and the inner product MATH as MATH where MATH is the usual NAME transform suitably extended to MATH, using NAME 's singular integrals. Intuitively, MATH consists of functions on MATH which arise as MATH for some MATH in MATH. This also introduces a degree of ``non-locality" into the theory, and the functions in MATH cannot be viewed as locally integrable, although MATH for each MATH, MATH, contains MATH as a dense subspace. In fact, REF , for the norm in MATH, makes precise in which sense elements of MATH are ``fractional" derivatives of locally integrable functions on MATH, and that there are elements of MATH (and of MATH) which are not locally integrable. On the other hand, vectors in MATH are not too singular: for example the NAME function MATH is not in MATH. To see this, pick some approximate identity MATH, say MATH, MATH, MATH, and set MATH; then a calculation shows that MATH for some positive constant MATH. Hence MATH is not in MATH, and then of course also not in the subspace MATH. Nonetheless, if we pass to the new norm MATH of REF , then from REF we get MATH . Hence the limit MATH defines a bounded linear functional on MATH, relative to the norm MATH on that NAME space. From the NAME lemma, and the definition of MATH, we conclude that MATH is in MATH. The same argument shows that the distributions MATH given by MATH for MATH, are also in MATH. In fact, the norm computes out as MATH . In the next lemma we provide the detailed proof of the fact that the iterated derivatives MATH of the NAME delta function are all in the completion of MATH relative to the ``new" norm of the NAME space MATH. But recall that MATH, or its derivatives, are not in MATH. For the NAME mass and its derivatives, we have MATH, MATH. The restriction on MATH is, as before, MATH. First note that if MATH, then MATH restricts to a MATH-function on MATH. By this we mean that there is a MATH-function MATH on MATH such that MATH holds for all MATH in MATH. Hence, if MATH is a distribution with compact support in MATH, then MATH is well-defined. The same argument shows that MATH is well-defined, and that MATH is also MATH up to the endpoints in the closed interval MATH. Hence, the distribution MATH may be applied again, and we get the expression MATH . Moreover, if MATH, then MATH is well-defined in the distribution sense, and MATH where MATH is REF . Hence for each MATH, we must show the following implication: MATH . The interpretation of the brackets MATH is in the sense of distributions as noted. In particular, MATH where MATH is really the compactly supported distribution MATH evaluated at the monomial MATH. Recall, it is assumed that the distribution MATH is supported in MATH. Now pick MATH such that MATH, and MATH, and let MATH, for MATH. We prove next that MATH where both sides are understood in the sense of distributions. But we also have MATH for all MATH, by the assumption in REF . To complete the proof we will then only need to check that MATH . Explicitly, MATH and this last expression can be estimated directly: If MATH, there is a constant MATH REF such that the MATH term in REF is estimated by MATH. In particular, we have the desired estimate REF . The left-hand side of REF may therefore be estimated by MATH. Since MATH for all MATH and all MATH, by assumption, see REF , we will then have MATH, which is the claim. It remains to check that the limit (as MATH) in REF is as stated. The argument is much as the previous one, so we will merely sketch the details for the case of MATH: Since MATH is an distribution with support in MATH, we need to check that MATH and MATH for all MATH; and both of these limits can be verified by calculus. Indeed the left-hand side in REF is of the order MATH which is differentiable in MATH, for every MATH. The corresponding expression in REF is MATH, MATH. Since the distribution is of compact support (in MATH) we also have, for some MATH, the estimate MATH for all MATH. Applying this to the functions MATH REF in the left-hand side of REF , we finally arrive at the desired conclusion REF . This completes the proof of the lemma. Hence if MATH, MATH, we get the inner product MATH is well-defined. A calculation yields MATH . However, if MATH is not locally integrable, then the right-hand side in REF must be understood as a singular integral, see, for example, CITE. Recall that MATH is obtained as the completion of MATH relative to the norm MATH of REF . If MATH is in MATH, then the NAME transform MATH of REF clearly has an entire analytic extension, that is, it extends to complex values of MATH as an entire analytic function with exponential growth factor MATH, MATH. We wish to show that this also holds for MATH. Note if MATH, it has finite MATH-norm, and MATH or rather MATH. Since MATH can be rather singular, the claim requires a proof. We have MATH, and the NAME measures MATH, for MATH, MATH, are in MATH. Hence MATH. But a calculation yields, for MATH, MATH . Let MATH, and multiply by MATH, to get MATH and so MATH. We conclude that MATH is supported in the interval if MATH is in MATH. This localizes the computation of MATH but still interpreted as a singular integral. Since MATH, and MATH, there is a sequence MATH such that MATH. Then of course also MATH . But MATH by REF . It follows that there is a subsequence MATH such that MATH converges pointwise almost everywhere on MATH. We wish to use NAME 's theorem CITE to conclude that the NAME transform MATH of MATH also has an entire analytic extension. To do this we need only check that MATH, MATH, is an equicontinuous family. Now pick MATH, and consider MATH . Let MATH, and pick MATH such that MATH on MATH. Continuing the calculation, we get MATH and MATH . But we have from REF that MATH, and the second term is independent of MATH, and it can be estimated in terms of MATH by calculus. This shows that the entire functions MATH do form an equicontinuous family. Since MATH is convergent MATH as noted, we conclude that the entire functions MATH converge uniformly for MATH in compact subsets of MATH, and that the limit function is also entire analytic. But by the argument above, this limit is an extension of MATH, for MATH. From REF , we have MATH . Since MATH, MATH, and the left-hand side vanishes for all MATH. Hence all the derivatives MATH vanish at MATH. Since MATH is analytic, it must vanish identically. Finally use REF to conclude that MATH as an element of MATH. This completes the proof of REF , and therefore the proof of the theorem. |
math/9908021 | While it is possible to give a direct proof along the lines of the last two pages in REF, we will derive the result here as a direct corollary to REF , that is, the uniqueness up to unitary equivalence. Given MATH, we already established the system MATH in REF . We wish to show that there is a second system MATH which also satisfies REF - REF . The MATH-dependence of MATH will be suppressed in the proof for simplicity. For MATH we take the transformation defined in REF above, and we get MATH by the following formula: MATH for MATH, and MATH. To see that MATH in REF is selfadjoint in MATH, we compute the inner products as follows: MATH . Since the kernel functions MATH are dense in MATH by construction, we conclude that MATH is indeed selfadjoint in MATH when MATH and MATH. We now show that MATH in REF is contractive. For MATH, we have MATH which shows that MATH is contractive as claimed. But we also proved that MATH for all MATH. Hence MATH where MATH denotes the projection of MATH onto MATH. Hence REF in the statement of REF is also satisfied. We leave the verification of MATH from REF to the reader. The conclusion of REF is now immediate from REF . |
math/9908021 | While the details are essentially in REF, we sketch REF . CASE: Given REF , and defining MATH and MATH by REF - REF , we saw that MATH is closed, and that, by REF , MATH is well-defined and contractive. CASE: Given REF , the subspace MATH in MATH, defined in REF , is positive. Indeed, if MATH, MATH, then MATH since MATH is assumed contractive. We also easily check that MATH in REF is closed when REF holds, that is, MATH is closed, and the operator MATH in REF is contractive. |
math/9908021 | Let MATH be a positive subspace, and let MATH be the corresponding contraction with closed domain MATH, see REF . We saw that then MATH; and, if MATH then MATH . It follows that the assignment MATH then passes to respective quotients MATH where MATH is defined in REF . If MATH is the corresponding operator MATH induced by MATH, then MATH is isometric relative to the two new norms, and it passes to the respective completions MATH . From REF - REF , we read off REF for the contraction MATH. Using again REF , we conclude that MATH satisfies REF . Conversely, if MATH and MATH are constructed from MATH and MATH, respectively, then, if we set MATH, MATH, then MATH is isometric, and extends naturally to a unitary isomorphism of MATH onto MATH. |
math/9908021 | The proof is based on REF above. Since MATH is given by REF at the outset, the two subspaces MATH are then determined from REF , applied to MATH. Let MATH, and set MATH. Then MATH, where MATH . Let MATH, MATH, be given, and define MATH by MATH . Then it follows from MATH that MATH is a contractive operator with domain MATH and mapping into MATH. The corresponding positive subspace, see REF , is that which is given by REF . The space MATH is maximally positive. A positive subspace MATH satisfying MATH would correspond to a contractive operator MATH mapping MATH into MATH and extending MATH, in the sense that the graph of MATH contains that of MATH. But then MATH and therefore MATH by the uniqueness part in REF . This proves that MATH is maximally positive in MATH. The contractive property for the operator MATH in REF follows from the two assumptions on MATH, that is, MATH, and MATH. Indeed, if MATH, then MATH . This proves that the operator MATH in REF is indeed well-defined and contractive. We then conclude from REF that the corresponding positive subspace MATH is the graph of MATH. An application of REF from REF then finally yields REF as claimed. If it were the case that MATH REF were invariant under the shift MATH of REF , then from NAME 's theorem, there would be a unitary function MATH such that MATH . (Recall MATH is said to be unitary if the corresponding multiplication operator MATH on MATH is unitary.) But identity in REF for some unitary MATH is possible only if the factor MATH in REF vanishes identically on MATH, and it follows therefore that MATH can only be shift-invariant if MATH. In this case, MATH reduces to the special case which we studied in REF. In that case, the contraction MATH from REF reduces to MATH, and MATH . Hence MATH is one-dimensional. Since MATH has zero constant term, the selfadjoint operator MATH on MATH, induced from MATH, is zero, and the proof is completed. |
math/9908021 | Since MATH we see that REF holds if and only if MATH, which is the conclusion. |
math/9908021 | It follows from REF that REF on the sequence MATH is equivalent to MATH being dissipative. Hence, since MATH, the operator MATH has a dense domain MATH, and the closure of MATH is well-defined. We will work with the closure, and refer to MATH as the closed operator. Notice that if MATH is defined from a sequence MATH, then the adjoint operator MATH is defined from the sequence MATH; and so, by REF , both are dissipative. In particular, MATH for all MATH. To prove REF , suppose MATH. Then MATH, and MATH. Since then MATH, this contradicts REF , unless MATH. Hence MATH is dense in MATH. But it is also closed since MATH is closed and dissipative. |
math/9908021 | We begin with a lemma. Let MATH, MATH, be a sequence such that all the sums MATH satisfy MATH for sequences MATH which are eventually zero. Let MATH be a positive NAME measure on MATH with finite moments MATH . Let MATH be the (possibly unbounded) NAME operator with symbol sequence MATH. CASE: Then the following are equivalent: CASE: MATH, CASE: MATH for some MATH, CASE: MATH for all MATH, and CASE: MATH. CASE: If one, and therefore all, the conditions hold, then MATH . CASE: The conditions are satisfied if MATH . But REF is more restrictive than REF - REF . We view MATH as an operator on MATH, and note that, if MATH, then MATH . Equivalently, setting MATH, MATH . The equivalence of REF - REF is immediate from this. Indeed, if MATH, then MATH. So this decides REF ; and REF also follows. Hence for REF , it is enough to show that REF follows from REF . Let MATH be a sequence which is eventually zero. Then MATH and the integral on the right is finite by REF . It follows that the sequence MATH defines a bounded linear functional on MATH, and so it is in MATH by NAME 's theorem. Equivalently, MATH defines an element of MATH, and so REF holds, and in fact MATH is densely defined as an operator on MATH. We now continue with the proof of REF . Let MATH be given, and assume it has the properties stated in the theorem. Then from REF , we know that there is a selfadjoint version MATH in a NAME space MATH. With the data from REF , we also know that the pair MATH is unique up to unitary equivalence. Since the spectral radius of MATH in the present theorem is clearly one, we get, from REF , that MATH. Suppose for the moment that MATH is realized as multiplication by MATH on MATH. Then the spectrum of MATH must be contained in MATH, and so the support of MATH must be contained in MATH. We saw in REF that MATH must have the desired form REF for some dissipative operator MATH with dense domain MATH in MATH. Since MATH is mapped into itself by MATH, we get the commutation identity REF . Writing out the positivity REF for MATH, MATH, MATH, we get MATH . But this means that the NAME moment problem is solvable for the sequence MATH. If the solution is represented as in REF , then it follows that MATH is represented as multiplication by MATH on MATH, and we saw (using REF ) that this forces MATH to be supported in the interval MATH. Since MATH, it is known from the theory of moments that MATH is unique from MATH. We include the argument for why MATH is indeed multiplication by MATH on MATH. Returning to REF , we note that MATH is determined from the identity MATH for MATH, MATH; and we have: MATH . Consider finite sums MATH and MATH, and the corresponding restrictions to MATH. Using MATH and MATH, we get MATH . This concludes the proof of existence. |
math/9908021 | Let MATH be a finite positive NAME measure on MATH which is supported in MATH, and assume that MATH is in MATH. We wish to reconstruct MATH such that MATH is a closed dissipative operator with dense domain in MATH. Note that if MATH has been found, then MATH . It follows that if MATH for some MATH, then MATH and therefore the corresponding norm-completion MATH only depends on the sequence MATH, that is, from REF , MATH. Equivalently, we may assume without loss of generality that the sequence MATH is real-valued. Now set MATH and let MATH be the corresponding positive NAME operator. For domain MATH, take the functions MATH which derive from corresponding MATH as MATH . Recall MATH is the unilateral shift, and therefore MATH or, in function form, MATH . We then set MATH and note that MATH is a NAME operator, which is closed with dense domain MATH and given by REF . Moreover, MATH has the desired properties, with MATH and restricting MATH to MATH. Moreover, for MATH, MATH is the function MATH given in REF above. |
math/9908021 | It follows from REF that MATH . So for some MATH, MATH, and we may view MATH as the restriction to MATH of the corresponding function MATH defined and analytic in MATH. If MATH has accumulation points in MATH, and MATH, then by REF , MATH vanishes on a subset of MATH of full measure. This subset must also have accumulation points, and since MATH is analytic in MATH, it must vanish identically. To prove the converse, suppose MATH contains only isolated points. Then MATH must have the form MATH where MATH, and MATH and MATH. Recall MATH is finite, and supported in MATH. Then pick MATH, MATH, such that MATH, for example MATH. Then MATH. |
math/9908031 | It is enough to exclude the reflection which sends MATH, and MATH. We must show that in every MATH which is invariant under MATH and under MATH, and for every MATH such that REF holds, and MATH and MATH, we must have MATH REF the trivial zero-dimensional space. Recalling the new norm in MATH, MATH, and MATH we conclude that the induced transformations MATH and MATH on MATH satisfy: CASE: MATH is a unitary one-parameter group on MATH and its spectrum is a subset of MATH; CASE: MATH is a contraction semigroup on MATH satisfying MATH; and CASE: MATH for all MATH, and MATH. Combining REF - REF , and using a theorem of CITE, we may assume that the selfadjoint generator MATH of MATH REF and MATH have the representation MATH relative to MATH for closed subspaces MATH. In this representation it is possible to check all the candidates for MATH with the stated properties; see also REF below. The presence of a nontrivial MATH leads to REF and the conclusion that MATH. But, of the nontrivial candidates for MATH, none can satisfy the added axioms MATH and MATH . We postpone further details to a future paper on semidirect products. |
math/9908031 | For MATH, MATH, we have MATH . Since MATH, and MATH, the result follows. By this we get MATH which shows that MATH, whence MATH passes to a bounded operator on the quotient MATH and then also on MATH satisfying the estimate stated in REF . If both the operators in REF leave MATH invariant, so does MATH and the operator induced by MATH is MATH as stated. |
math/9908031 | The basic idea is contained in CITE. |
math/9908031 | If MATH, then MATH, and MATH is unitary on MATH. Thus MATH proving that MATH is selfadjoint in the MATH-inner product. Since MATH is unitary on MATH and the contractivity property follows. |
math/9908031 | The last statement follows by the spectral theorem. By construction MATH is a semigroup of selfadjoint contractive operators on MATH. The existence of the operators MATH as stated then follows from a general result in operator theory; see, for example, CITE or CITE. |
math/9908031 | This follows as the spectrum of MATH and MATH is contained in MATH. |
math/9908031 | See REF . |
math/9908031 | REF follow by the NAME theorem and REF , as the resulting representation of MATH is obviously continuous. CASE: Let MATH be a MATH-invariant subspace in MATH. Then MATH is MATH invariant. Let MATH, MATH and MATH. Define MATH by MATH . Then MATH is holomorphic in MATH, and MATH for every (real) MATH. Thus MATH is identically zero. In particular MATH for every MATH. Thus MATH . As MATH it follows that MATH. By continuity we get MATH. Thus MATH is reducible as a MATH-module. The other direction follows in exactly the same way. |
math/9908031 | This is obvious from the spectral theorem. |
math/9908031 | Let MATH. Then, as MATH is invariant by construction, we conclude that MATH . By differentiation at MATH, it follows that MATH. This shows that MATH is an ideal in MATH. The last part follows as MATH is generating (in MATH). |
math/9908031 | By CITE (see also CITE), MATH exists if MATH. But MATH. Hence MATH. |
math/9908031 | If MATH is of NAME type then MATH and MATH, MATH, MATH. Thus MATH. We also have (see CITE) MATH . From this the theorem follows. |
math/9908031 | The first claim follows as MATH and MATH act by MATH on MATH, and thus map MATH onto MATH, and MATH onto MATH. The second follows as we can realize MATH by conjugation by an element in MATH. |
math/9908031 | We may assume that MATH is big enough such that the integral defining MATH converges. The general statement follows then by analytic continuation. We have MATH . Now MATH. By REF we get MATH . The second statement follows in the same way. Let MATH be the subspace of MATH consisting of functions MATH with support in MATH, and with MATH compact. |
math/9908031 | REF is a simple calculation. For REF see CITE and CITE. |
math/9908031 | Let MATH and MATH. Then MATH only if MATH. Thus MATH. |
math/9908031 | CASE: Let MATH and MATH be as above. Then MATH . This proves REF . CASE: This follows from REF and the following calculation: MATH where we have used that MATH . CASE: By REF we get: MATH . Thus MATH passes to a contractive operator on MATH. That MATH follows from REF . |
math/9908031 | REF follows from REF as obviously MATH. CASE: This follows now from REF. CASE: By REF we know that MATH defines an isometric MATH-intertwining operator. Let MATH. Differentiation and the fact that MATH is holomorphic gives CASE: MATH. CASE: MATH. But those are exactly the relations that define MATH. The fact that MATH generates MATH implies that MATH induces a MATH-intertwining operator intertwining MATH and MATH. As both are also representations of MATH, it follows that this is an isometric MATH-map. In particular as it is an isometry by REF , it follows that the map MATH is an isomorphism. This proves the theorem. |
math/9908031 | Let MATH. Then we can choose MATH such that MATH. As MATH and MATH acts unitarily, it follows that we can assume that MATH. Let MATH be such that MATH. Then MATH. Hence we can assume that MATH. Let MATH and let MATH. We want to show that MATH. Choose MATH such that MATH. For MATH and MATH we have MATH. Thus MATH. Choose MATH such that MATH for every MATH. Then MATH for every MATH. It follows that for MATH: MATH . By REF we know that MATH is holomorphic on MATH. As MATH and MATH both have compact support it follows by REF that MATH is holomorphic on MATH. But MATH for MATH. Thus MATH for every MATH. In particular MATH for every MATH. By continuity MATH. Thus MATH. |
math/9908031 | This follows from REF |
math/9908031 | The details will only be sketched here, but the reader is referred to CITE and CITE for definitions and background literature. An important argument in the proof is the verification that, if a column vector of the form MATH is in MATH, then MATH must necessarily be zero in MATH. But using positivity, we have MATH . Using this on the vectors MATH and MATH, we get MATH . But, since MATH is also in MATH, we conclude that MATH, proving that MATH is the graph of an operator MATH as specified. The dissipativity of the operator MATH is just a restatement of REF . |
math/9908031 | Since MATH is a unitary isomorphism MATH we may make an identification and reduce the proof to the case where MATH and MATH is the identity operator. We then have MATH and if MATH, then MATH while, if MATH, then MATH . Using only the MATH part from (MATH), we conclude that MATH is invariant under MATH. If the projection MATH of MATH onto MATH is written as an operator matrix MATH with entries representing operators in MATH, and satisfying MATH then it follows that MATH which puts each of the four operators MATH in the commutant MATH from REF . Using REF , we then conclude that MATH is a dissipative operator with MATH as dense domain, and that MATH is the graph of this operator. Using REF , and REF, we finally conclude that MATH is a normal operator, that is it can be represented as a multiplication operator with dense domain MATH in MATH. |
math/9908031 | See, for example, CITE. |
math/9908031 | Immediate from the discussion, and the NAME theorem classifying the closed subspaces in MATH which are invariant under the multiplication operators, MATH, MATH. We refer to CITE, or CITE, for a review of the NAME theorem. |
math/9908031 | Suppose there are unitary functions MATH such that MATH. Then this would violate the NAME REF , and therefore REF . Using irreducibility of MATH and of MATH, we may reduce to considering the cases when one of the spaces MATH is MATH. By REF , we are then back to the case when MATH or MATH is the graph of a densely defined normal and dissipative operator MATH, or MATH, respectively. We will consider MATH only. The other case goes the same way. Since MATH it follows that MATH must anti-commute with the multiplication operator MATH on MATH. For deriving this, we used REF at this point. We also showed in REF that MATH must act as a multiplication operator on the NAME side. But the anti-commutativity is inconsistent with a known structure theorem in CITE, specifically REF in that paper. Hence there are unitary functions MATH in MATH such that MATH. But this possibility is inconsistent with positivity in the form MATH (see REF ) if MATH. To see this, note that MATH is invariant under the unitary operators REF for MATH. The argument from REF , now applied to MATH, shows that the two subspaces MATH and MATH are both invariant under the whole group MATH. But the commutant of this is MATH-dimensional: the only projections in the commutant are represented as one of the following, MATH relative to the decomposition MATH of MATH. The above analysis of the anti-commutator rules out the cases MATH and MATH, and if MATH, we are left with the cases MATH and MATH. Recall, generally MATH, as a starting point for the analysis. A final application of the NAME theorem (as in CITE; see also CITE) to REF then shows that there must be a pair of unitary functions MATH in MATH such that MATH where MATH are the two NAME spaces given by having MATH supported in MATH, respectively, MATH. The argument is now completed by noting that REF is inconsistent with the positivity of MATH in REF ; that is, we clearly do not have MATH semidefinite, for all MATH and all MATH. This concludes the proof of the Corollary. |
math/9908031 | The group-law in the NAME group yields the following commutator rule: MATH for all MATH. We now apply MATH to this, and evaluate on a general vector MATH: abbreviating MATH for MATH, and MATH for MATH, we get MATH valid for all MATH, MATH. Note, in REF , we are assuming that MATH takes on some specific value MATH on the one-dimensional center. Since MATH is unitarily equivalent to MATH by assumption (see REF ), we conclude that MATH . The argument really only needs that the two representations MATH define different characters on the center. (Clearly MATH since MATH.) Multiplying through first with MATH, and integrating the resulting term MATH in the MATH-variable, we get MATH. The last conclusion is just using that MATH is a closed subspace. But we can do the same with the term MATH and we arrive at MATH. Finally letting MATH, and using strong continuity, we get MATH and MATH both in MATH. Recalling that MATH are general vectors in MATH, we conclude that MATH, and therefore MATH. Since the converse inclusion is obvious, we arrive at REF with MATH. |
math/9908034 | See classification of pencils of finite-dimensional linear operators, say in CITE or CITE. |
math/9908034 | Consider the projectivization MATH of the vector space of matrices. Let MATH be the projection of the set of matrices of rank MATH to MATH, denote projections of MATH, MATH on MATH by MATH, MATH. Then MATH is a closed subset, and the line through MATH, MATH does not intersect MATH. Due to compactness of MATH, nearby lines do not intersect MATH as well. |
math/9908034 | Indeed, multiplication of MATH by a constant changes the identification of MATH with MATH by the same constant. For a diffeomorphism MATH and a NAME structure MATH on MATH denote by MATH the NAME structure MATH transferred to MATH via MATH, MATH. Apply this definition in the case when MATH is multiplication by MATH in MATH. Now the only thing to prove is that MATH if MATH is the canonical NAME structure on MATH, and MATH is an arbitrary manifold. One can check it momentarily in local coordinates on MATH (as in REF ). |
math/9908034 | Since for a NAME web MATH, one can take MATH, and MATH, thus the NAME structure MATH is defined on the whole total space of MATH for any MATH. Fix MATH. Since a NAME structure is a bivector field, one can associate to MATH an element MATH of MATH. We know that this element depends analytically on MATH and is homogeneous of degree REF in MATH. However, any mapping of MATH to a vector space which is of homogeneity degree REF is linear, which finishes the proof. |
math/9908034 | The only case which remains to be proven is MATH. If a NAME web MATH is real-analytic, then one can consider the complex-analytic continuation, and one returns to the case MATH. Thus in real-analytic REF implies the theorem. Consider now MATH-case. First, note that the theorem follows from some ``abstract nonsense" remarks. Recall that one of the contributions of algebraic geometry to differential geometry is the understanding of the importance of considering ``formal" objects as tools for investigation of ``geometric" objects. In particular, given a MATH-manifold, one can consider ``a MATH-jet of the complex-analytic neighborhood" (or a ``formal neighborhood") of this manifold. This formal neighborhood is canonically defined, and carries many properties of complex-analytic manifolds. Readers familiar with the language we used above may immediately recognize that all the objects needed for the proof of REF (foliations, tensors, cotangent bundles, NAME structures) make sense in settings of ``formal geometry", thus one can finish the proof in MATH-case in the same way we did it in real-analytic case (when we had a no-nonsense complex neighborhood instead of a formal one). For other readers we provide a stripped-down version of the proof below. This version will not use ``hard" notions of formal geometry (such as jets of manifolds etc), but will use only notions of (finite-order) jets of sections of ``real" bundles on ``real" manifolds. This simplified version of the proof goes until the end of this section. Consider a vector bundle MATH over MATH, fix a point MATH. To describe a vector subbundle MATH of MATH of rank MATH is the same as to describe a section of the bundle MATH (the fibers of this bundle are NAME of fibers of MATH). Recall that a MATH-jet near MATH of a section MATH of a bundle is the collection of NAME coefficients of MATH of the order MATH or less. If a geometric object can be described locally by a section of some bundle, one can define a MATH-jet of this object near MATH as being a MATH-jet of a section of this bundle. Thus one can define a MATH-jet of a vector subbundle MATH of MATH near MATH. MATH . In particular, one can define a MATH-jet MATH of a preweb. Consider two prewebs MATH and MATH which have the same MATH-jet near MATH. Then the subbundles MATH and MATH have the same MATH-jet near MATH. Thus there is a canonically defined MATH-jet of an isomorphism MATH between vector bundles MATH and MATH. This leads to a MATH-jet of an isomorphism between the total spaces of these bundles. On the other hand, given a MATH-jet of an isomorphism MATH between two manifolds MATH and MATH, and a tensor field MATH on MATH, one can define a MATH-jet of the tensor field MATH on MATH. Suppose that both prewebs MATH and MATH are in fact webs. As we know, total spaces of both bundles MATH and MATH carry natural NAME structures, and one can consider NAME structures as bivector (thus tensor) fields MATH and MATH. Thus given a MATH-jet MATH of an isomorphism between these manifolds, one can consider MATH, which is a MATH-jet of a NAME structure on MATH. Obviously, MATH coincides with the MATH-jet of MATH. This immediately implies Consider two NAME webs MATH and MATH on MATH which have the same MATH-jet near MATH. Then there is a canonically defined MATH-jet of an isomorphism MATH between MATH and MATH, and this isomorphism identifies MATH-jets of NAME structures MATH (on MATH) and MATH (on MATH) near fibers of these bundles over MATH. Thus In the conditions of REF suppose that MATH, and the web MATH is in fact real-analytic. Then the tensor field MATH on the fiber of MATH over MATH depends linearly on MATH. In particular, the theorem follows from the case MATH of Fix MATH. Given a NAME web MATH on an open subset MATH, MATH, one can find a real-analytic NAME web MATH on MATH, MATH, such that MATH-jets of MATH and MATH at REF coincide. In fact, in the case MATH we propose the following amplification: Given a NAME web MATH on an open subset MATH, MATH, one can find MATH, MATH, and a diffeomorphism MATH, MATH, such that MATH-jet of MATH (the transfer of MATH via MATH) coincides with MATH-jet of an appropriate translation-invariant NAME web MATH. Not only this conjecture implies the case MATH of Conjecture REF, but it would also directly imply the theorem we are proving. However, since these conjectures are not settled yet, one needs an alternative way to prove the theorem. We are going to do this without the extension-properties for webs which are mentioned above by introducing integrability conditions on MATH-jets of prewebs. Since jets are just collections of numbers, thus are not sensible to a change of the base field, we will be able to consider a MATH-jet of a MATH-web as a MATH-jet of a complex-analytic web, thus we will be able to extend MATH to MATH, then apply the arguments of REF . MATH . MATH . Due to REF , if a MATH-jet near MATH of preweb MATH on MATH is NAME, then the preweb MATH is NAME in a neighborhood of MATH. Thus the definition of a NAME MATH-jet-web above is compatible with taking a MATH-jet of a NAME web. Consider a MATH-jet-web MATH. It is a MATH-jet of a preweb, denote this preweb by MATH. Given MATH, the MATH-jet of MATH, MATH, is naturally identified with the MATH-jet of the cotangent bundle of the corresponding foliation MATH, in other words, one can construct a MATH-jet of an identification MATH of MATH with MATH. Recall that a MATH-jet of a diffeomorphism acts on MATH-jets of tensor fields. MATH carries a natural NAME structure, thus it carries the tensor field which describes the bracket of the NAME structure. Using the identification MATH mentioned above, one obtains a MATH-jet of a tensor field MATH on MATH. Obviously, one can consider this MATH-jet as living on MATH. It does not depend on the choice of MATH. If MATH, and MATH, then one obtains a MATH-jet of a tensor field MATH for any MATH, and the same arguments as in the proof of REF show that Let MATH, consider a NAME MATH-jet-web MATH near MATH. The family of tensors fields MATH on the total space of MATH is well defined on the fiber over MATH of the projection MATH. This family depends linearly on MATH, MATH. The only thing to prove is that MATH depends smoothly on MATH. The definition of MATH depended on the foliation MATH, and a priori we have no conditions on how MATH depends on MATH. It is enough to show that the MATH-jet of MATH depends smoothly on MATH. Any foliation MATH on a manifold MATH can be locally described by REF; here MATH is a local coordinate system, MATH, MATH, MATH, and MATH. If we know the subbundle MATH, then we know that MATH here MATH is a matrix function of MATH variables which can be deduced from equations of MATH in MATH. Given MATH, one can find the unique solution of a part of REF with initial conditions MATH: Require that MATH is given by REF if MATH. Moreover, given a MATH-jet of MATH, this uniquely determines MATH-jet of MATH. This implies that if there is a family of foliations MATH and the subbundle MATH depends smoothly on MATH, then MATH depends smoothly on MATH. In turn, this implies that MATH depends smoothly on MATH. To finish the proof of the theorem, it is enough to show that given a NAME MATH-jet-web over MATH, one can consider it as a NAME MATH-jet-web over MATH. Recall that a jet of a preweb is just a collection of NAME coefficients of a section of a bundle, thus a collection of numbers. Any collection of real numbers can be considered as a collection of complex numbers, thus a MATH-jet of a preweb over MATH can be considered as MATH-jet of a preweb over MATH. Since the condition of being NAME does not change when we change field of scalars, the only thing we need to prove is the integrability condition for complex MATH. Recall NAME integrability condition: Consider a vector subbundle MATH of MATH. Suppose that for any small open subset MATH and any section MATH of MATH over MATH its NAME differential MATH (which is a MATH-form on MATH) can be written as MATH with MATH being sections of MATH and MATH being arbitrary differential forms on MATH. Then MATH is integrable. By constructing jet-solutions of ordinary differential equations, one can easily prove the following jet-analogue of REF : Consider a MATH-jet of a vector subbundle MATH of MATH near MATH. Suppose that for any open subset MATH and for any MATH-jet MATH of a section of MATH near MATH its NAME differential MATH (which is a MATH-jet of a MATH-form on MATH) can be written as MATH with MATH being MATH-jets of sections of MATH and MATH being arbitrary differential forms on MATH. Then MATH is MATH-integrable. In fact one can do more. Given a section MATH of MATH, consider the image of MATH under projection MATH. This defines a mapping MATH from sections of MATH to sections of MATH. A priori it is a differential operator of order REF, but in fact it has order REF, thus is a linear mapping between bundles. Indeed, if MATH, then one can easily check that MATH. Thus MATH defines a section of the vector bundle MATH, call this section the torsion of MATH. Given a MATH-jet of a vector subbundle MATH, MATH is defined as MATH-jet of a section of MATH. If a vector subbundle MATH depends smoothly on a parameter MATH, then MATH depends smoothly on MATH (that is, it is a smooth section of the vector bundle MATH over MATH). Suppose MATH. Apply the construction of MATH above to the vector subbundle MATH of a NAME preweb MATH in the case MATH. Here MATH, thus MATH is a section of MATH over MATH. Restrict this section to MATH, MATH. We obtain a vector bundle over MATH, and a regular section MATH of this vector bundle. There may be only two different cases: either MATH, or MATH vanishes at a finite number of points of MATH. Consider a NAME MATH-jet-web MATH near MATH (and MATH). The complexification of this MATH-jet-web is a MATH-jet MATH of a NAME preweb with MATH. Then MATH is a NAME MATH-jet-web. Arbitrarily extend MATH to a preweb MATH in neighborhood of REF in MATH, and arbitrarily extend MATH to a preweb MATH in a neighborhood of REF in MATH. Consider the torsion MATH of the subbundles MATH restricted to MATH, and the torsion MATH of the subbundles MATH restricted to MATH. Obviously, MATH is a restriction of MATH from MATH to MATH. On the other hand, MATH vanishes, since MATH-jet MATH of MATH is a MATH-jet-web. Thus MATH vanishes on MATH, thus at infinitely many points of MATH, thus MATH. This implies that the conditions of REF are satisfied, which finishes the proof. This finishes the proof of REF . |
math/9908034 | This follows immediately from the explicit form of a NAME block. |
math/9908034 | Since MATH is isotropic with respect to MATH, MATH induces a natural pairing of MATH with MATH, or a mapping MATH. Similarly, MATH induces a mapping MATH. Consider the relation MATH in MATH. Looking on the explicit form of a NAME block of a pair of skew-symmetric bilinear forms, one can easily recognize in the relation MATH a direct sum of NAME blocks. |
math/9908034 | This follows immediately from the following Consider a vector space MATH and numbers MATH. Consider the product MATH of NAME MATH and the subset MATH consisting of MATH-tuples of subspaces such that the linear span of such a MATH-tuple has dimension MATH. Consider the mapping MATH here MATH is a vector subspace of MATH of dimension MATH. Then the mapping MATH is smooth. Let MATH, MATH be a small neighborhood of MATH in MATH. For MATH denote by MATH the MATH-th component of MATH, MATH. Choose a basis MATH in MATH, MATH, MATH, which depends regularly on MATH. Pick up a basis MATH out of vectors MATH, MATH, MATH (each MATH has a form MATH). Then in an open subset MATH of MATH the vectors MATH remain linearly independent, thus on MATH they span MATH. This implies that MATH depends regularly on MATH. This finishes the proof of REF . |
math/9908034 | By REF action subspaces from a subbundle of MATH. Consider MATH and the NAME structure MATH on MATH. This NAME structure has corank MATH, consider its symplectic foliation MATH. Then the kernel of the bilinear form MATH in MATH is the normal space to the leaf of MATH through MATH. Now the statement follows from the following lemmas: Consider a pair of skew-symmetric bilinear forms in MATH which has no NAME blocks. Suppose that dimensions of all the NAME components of MATH are MATH. If MATH, MATH, is a finite subset of MATH with MATH elements, then MATH. Direct corollary of the explicit form of a NAME block. Consider vector subbundles MATH of MATH, such that MATH. Suppose that MATH coincides with the normal bundle to a foliation MATH on MATH, MATH. Then there is a foliation MATH on MATH such that MATH coincides with the normal bundle to MATH. Leaves of MATH can be described as intersections of leaves of foliations MATH. This follows immediately for the NAME integrability criterion REF . This finishes the proof of REF . |
math/9908034 | Indeed, consider MATH and the projection MATH of MATH to MATH. The (co)differential MATH of the mapping of projection identifies the vector space MATH with the action subspace at MATH. However, by REF the action subspace at MATH is equipped with a NAME linear relation, thus MATH is equipped with such a relation as well. To show that MATH has a structure of a NAME preweb, it is enough to show that this relation in MATH does not depend on the choice of the point MATH over MATH. Let MATH, MATH be two point of MATH over MATH. Let MATH, MATH be the action subspaces in MATH and MATH. Both MATH and MATH are identified with MATH, thus one with the other. For any MATH the identification between MATH and MATH sends MATH to MATH. Let MATH. Consider the NAME structure MATH on MATH. Then MATH is the normal bundle to the symplectic foliation MATH for this NAME structure. Consider the leaf MATH of this foliation which passes through MATH. This leaf contains the leaf of the action foliation MATH through MATH, thus MATH is a preimage of a submanifold MATH. Thus MATH is the image of the normal space to MATH under the (co)differential of the projection mapping. By the same reason MATH is in MATH, and MATH is also the image the same normal space. Thus the identification of MATH and MATH via projection to MATH indeed sends MATH to MATH. Consider vector spaces MATH and MATH with NAME linear relations MATH and MATH in MATH and MATH correspondingly. Consider an isomorphism MATH. If MATH sends MATH to MATH for any MATH, then MATH sends MATH to MATH. This lemma is equivalent to REF proven in REF. We had shown that the base MATH of the action foliation has a canonically defined structure of a NAME preweb. Denote by MATH the relation in MATH, MATH. To show that this preweb is in fact a web, it is enough to describe the MATH-integrating foliation MATH of MATH. Fix MATH, consider the NAME structure MATH on MATH. Since this is a NAME structure of constant rank, symplectic leaves form a foliation on MATH. This foliation depends only on MATH, denote this foliation MATH. The normal space to this foliation at MATH is the kernel of the bilinear form MATH in MATH. In particular, the normal bundle to MATH is contained in the action bundle of the bihamiltonian structure. Thus MATH is a subfoliation of MATH. In particular, MATH induces a ``quotient" foliation MATH on the base MATH of the foliation MATH. Now one can immediately see that the normal space to the foliation MATH at MATH coincides with MATH, thus MATH is the MATH-integrating foliation of the preweb on MATH. Thus the preweb structure on MATH is indeed a NAME web. |
math/9908034 | Given a bisurjective relation MATH in a vector space MATH, suppose that MATH. Define the lifting MATH of MATH into MATH as the collection of vectors MATH. One can easily check that MATH. Now suppose that MATH is NAME. Let MATH, MATH, and consider MATH. It is sufficient to show that these vector subspaces span MATH. Indeed, assume that this is true. Then the statement of the amplification is obvious if MATH is NAME. In general, in the conditions of the amplification let MATH; here MATH is a sum of NAME blocks of MATH, MATH is a sum of NAME blocks of MATH. Let MATH be restriction of MATH to MATH. We know that vector subspaces MATH span MATH, that MATH are subspaces of MATH, and that MATH is spanned by some subspaces of MATH, thus MATH. If MATH, MATH are the numbers of NAME blocks in MATH and MATH, then MATH and MATH. Hence MATH. On the other hand, MATH, MATH. Thus MATH. Hence MATH, thus MATH. What remains to prove is that that vector subspaces MATH span MATH if MATH is NAME. Decomposing into a direct sum, we can restrict our attention to the case when MATH is one NAME block of dimension MATH or less. By decreasing MATH we may assume that MATH. Then MATH, MATH for any MATH. Thus it is enough to show that MATH, MATH, are linearly independent. Since all the NAME blocks of dimension MATH are isomorphic, it is enough to do this for one particular NAME block of dimension MATH. Consider a MATH-dimensional vector space MATH with basis MATH, MATH (as in REF ). Let MATH be the symmetric power MATH. Consider MATH as a subspace of the vector space of polynomials on MATH, then two partial derivatives MATH, MATH define two mappings MATH. Let MATH consists of pairs MATH such that MATH. One can momentarily check that MATH is a NAME block in MATH. Obviously, MATH is spanned by MATH. Let MATH, introduce a mapping MATH given by MATH. One can easily check that MATH is spanned by the vector MATH. Thus to show that MATH are linearly independent, it is enough to show that for MATH non-proportional linear functions MATH on MATH the polynomials MATH are linearly independent. In turn, this is an obvious corollary of non-vanishing of NAME determinant. |
math/9908034 | The first statement follows from the definition of the NAME bracket MATH on MATH. If MATH, the second statement is a direct corollary of the first one. If MATH, the second statement follows from the following Consider a pair of skew-symmetric bilinear forms MATH and MATH in a vector space MATH. Then MATH is constant for MATH inside an open subset of MATH, call this common value MATH. Suppose that for any finite subset MATH there is MATH such the vector subspaces MATH, MATH, span a vector subspace of MATH of dimension MATH. Then the pair MATH, MATH is NAME. Lemma follows immediately from the classification of REF . The remaining statements of the proposition follow immediately from the first two statements. |
math/9908034 | Consider Hamiltonian mappings MATH of NAME structures on MATH. Restrict these mappings to the action subspace MATH. By REF the action subspace is isotropic with respect to both pairings in MATH, thus the elements MATH, MATH, are orthogonal to MATH. By definition, MATH coincides with the tangent space to the fiber MATH of the action foliation which passes through MATH. In particular, there are two mappings MATH. Note that these mappings form a pencil which corresponds to the linear relation in MATH (as in REF and in the proof of REF ). Since the statement of the proposition is local on MATH, one can suppose that the action foliation is a fibration. Let MATH be the projection to the base of foliation, and MATH. Then MATH induces a canonical isomorphism MATH. Now MATH is a NAME web, thus MATH carries a NAME relation, and this relation is compatible with the relation on MATH with respect to the isomorphism above. On the other hand, let MATH be the fiber of MATH over MATH. Then the linear relation in MATH can be described both by a pencil MATH and by a pencil MATH. REF gives a canonical isomorphism between MATH and MATH, MATH. As a corollary there is a canonical identification of tangent spaces to MATH at different points (since they all project to the same point of MATH). This induces a flat connection MATH on the tangent bundle to MATH. What remains is to show that this connection on MATH is a locally-affine structure. Pick up any MATH, for example, MATH. Consider the corresponding NAME structure MATH of the pencil. By REF , MATH for any functions MATH, MATH on MATH. This implies that fibers of MATH are involutive submanifolds of MATH with respect to MATH. On the other hand, if the rank of bihamiltonian structure on MATH is MATH, then symplectic leaves of MATH have dimension MATH, and MATH has dimension MATH. Since MATH has half the dimension of the symplectic leaf, and is involutive, it is Lagrangian. Thus MATH is a Lagrangian foliation with respect to MATH. Now REF imply that the leaves of MATH are equipped with canonical locally-affine structures. Thus it is enough to prove that the connection MATH on MATH constructed above coincides with the connection of this locally-affine structure. Recall that the connection MATH on MATH is constructed basing on the mappings MATH, in other words, on the isomorphisms MATH for different MATH with MATH. However, MATH coincides with MATH; here MATH is the fiber of integrating foliation MATH which passes through MATH. Let MATH, recall that MATH is a symplectic leaf of the NAME structure MATH on MATH. The foliation MATH allows a restriction MATH to MATH, this restriction is a Lagrangian foliation on a symplectic manifold. Both the identification MATH and the identification of MATH with MATH in REF are constructed as restrictions of Hamiltonian mappings. Setting MATH, MATH, and replacing MATH with MATH shows that the connection MATH coincides with the connection of the locally-affine structure of REF. This finishes the proof. |
math/9908034 | MATH is a vector bundle over MATH. Let MATH be the multiplication by MATH in this bundle. To show that MATH is an anti-involution, note that MATH is isomorphic to the NAME structure on MATH (here MATH is the integrating foliation on MATH), and this isomorphism is a mapping of vector bundles. Thus it is enough to show that multiplication by MATH is an anti-involution on the cotangent bundle to a foliation. In turn, it is enough to prove this for the cotangent bundle to a manifold, which can be checked immediately in local coordinates. |
math/9908034 | Given a locally-affine structure MATH and a point MATH, the exponential mapping of the connection on MATH gives a canonical identification of a neighborhood of MATH in MATH and a neighborhood of REF in MATH. Similarly, given a bundle MATH with fibers having a locally-affine structure, and a section MATH of this bundle, one obtains a canonical identification of a neighborhood of MATH with a neighborhood of MATH-section in an appropriate vector bundle MATH. Obviously, fibers of MATH are vertical tangent space of MATH at points of MATH. In conditions of the proposition MATH. In particular, any submanifold MATH which is transversal to leaves of MATH, and such that MATH, gives a natural identification MATH of a neighborhood of MATH with a neighborhood of MATH-section of MATH (one can identify MATH with the local base MATH). We need to show that if MATH is a cross-section, then this identification is compatible with bihamiltonian structures on MATH and on MATH. It is enough to show that MATH is compatible with MATH (the same argument will be applicable to MATH). Consider a symplectic leaf MATH for MATH. It is a preimage of a submanifold MATH. The restriction foliation MATH is well-defined and is a Lagrangian foliation. Moreover, MATH is Lagrangian. Consider the identification of a neighborhood of MATH in MATH with a neighborhood of MATH-section of MATH (see REF ). Since MATH is Lagrangian with respect to MATH, this identification is compatible with the NAME structure MATH. Composing this identification with MATH, one obtains an identification MATH of MATH with a submanifold of MATH. Obviously, MATH sends a fiber of MATH to a fiber of MATH, and MATH-section of MATH into a MATH-section of MATH. Due to REF , MATH is compatible with affine structures on fibers of MATH and of MATH, and since it sends REF to REF, it is a linear mapping of vector bundles MATH. However, given MATH, MATH is canonically isomorphic to MATH (as in REF), and by REF MATH is a leaf of the foliation MATH. Thus there is a canonical isomorphism MATH which is compatible with the NAME structure MATH on MATH. It is enough to show that MATH. But this is a direct corollary of the proof of REF . |
math/9908034 | This proposition follows immediately from the following Consider a symplectic manifold MATH with a Lagrangian foliation MATH and an anti-involution MATH which sends each leaf of MATH into itself. Then fixed points of MATH form a Lagrangian submanifold of MATH which is transversal to leaves of MATH. It is enough to restrict our attention to a neighborhood of MATH, thus one may assume that MATH is an open subset of MATH, and MATH is the foliation on fibers of projection MATH. Since the locally-affine structure on fibers of a Lagrangian foliation does not change if one multiplies the NAME bracket by a number, the restriction of MATH on any leaf of MATH induces an affine transformation of this leaf (which is an open subset of MATH for an appropriate MATH). Recall that the set of fixed points of an involution MATH of MATH forms a manifold MATH, and if MATH, then MATH coincides with the invariant subspace of MATH. Consider a symplectic vector space MATH, and a Lagrangian subspace MATH. Consider a linear involution MATH such that MATH for any MATH. Suppose that MATH, and that MATH induces an identity mapping of MATH into itself. Then MATH is multiplication by MATH, and MATH is a complement to the invariant subspace of MATH in MATH, which is a Lagrangian subspace of MATH. Let MATH is the vector subspace of fixed points of MATH, MATH be the eigenspace of MATH with eigenvalue MATH. Obviously MATH, and due to the conditions on MATH both MATH and MATH are isotropic. Thus both MATH and MATH are Lagrangian. Similarly, MATH, MATH, MATH. Let MATH be any complementary subspace to MATH in MATH, MATH. Since the action of MATH in MATH is isomorphic to the action of MATH in MATH, we see that MATH, thus MATH. Since MATH is Lagrangian, MATH, which finishes the proof. Applying this lemma to MATH, MATH (here MATH is such that MATH, MATH is the leaf of MATH through MATH) finishes the proof of REF . This finishes the proof of REF . |
math/9908034 | Recall that MATH . Clearly, if MATH is antiholomorphic, and MATH is a complex-analytic subvariety, then MATH is a complex analytic subvariety. Indeed, if MATH is an equation of MATH, then MATH is an equation of MATH. Similarly, one can transfer a NAME structure and a bihamiltonian structure via an antiholomorphic bijection. If MATH is a complexification of a real-analytic manifold MATH, then MATH is equipped with an antiholomorphic involution MATH such that MATH. If MATH has a NAME or bihamiltonian structure, so has MATH, and the structures on MATH are invariant with respect to MATH. Consider the antiholomorphic involution MATH for MATH, MATH. Consider a cross-section MATH of the projection MATH which passes through MATH. Then MATH is also a cross-section which passes through MATH. Since fibers of MATH have an locally-affine structure, the section MATH is well defined on an appropriate neighborhood of MATH. Moreover, this section is a cross-section. Indeed, this immediately follows Consider a Lagrangian foliation MATH on a symplectic manifold MATH, and submanifolds MATH, MATH, MATH of dimension MATH which are transversal to MATH. Suppose that for any leaf MATH of MATH the intersection MATH, MATH, consists of one point MATH, and MATH is the midpoint of the segment MATH in the locally affine structure on MATH. If MATH and MATH are Lagrangian, so is MATH. Indeed, we may assume that MATH, leaves of MATH are fibers of MATH, MATH is MATH-section of MATH, MATH are graphs of sections MATH of MATH, and MATH. Consider a section MATH of MATH. The graph of MATH (which is a submanifold of MATH) is Lagrangian iff MATH. Application of this obvious lemma finishes the proof of REF . Now the section MATH of MATH is MATH-invariant, thus it is a complexification of the section MATH for MATH. Since the complexification of MATH is a cross-section, so is MATH. This finishes the proof of the proposition. |
math/9908034 | Indeed, the Hamiltonian vector field of any function with respect to a NAME bracket MATH preserves MATH. Thus the bracket MATH is preserved by MATH, thus by MATH. On the other hand, due to the condition on MATH, MATH is proportional to the Hamiltonian flow of MATH with respect to any bracket MATH (as far as MATH), thus MATH preserves the NAME structure MATH as well. By linearity, it preserves MATH too. |
math/9908034 | Due to REF it is enough to show that the span of biflow vectors in MATH coincides with MATH. Decrease MATH so that the action foliation becomes a fibration MATH with a base MATH. Let MATH. Consider the pencil MATH from the proof of REF . A vector MATH is a biflow vector iff MATH can be written as MATH with MATH, MATH, MATH, MATH. If MATH is a function on MATH, and MATH is the Hamiltonian flow of MATH with respect to the NAME bracket MATH, then the value of MATH at MATH coincides with MATH. Since locally any NAME function for MATH can be written as MATH for an appropriate MATH such that MATH, this implies the ``only if" part of the lemma. On the other hand, if MATH, then MATH is normal to the leaf of the integrating foliation MATH on MATH. Decreasing MATH, one can find a function MATH on MATH such that MATH is constant on leaves of MATH, and MATH. This implies the ``if" part of the lemma. Consider a NAME relation in a vector space MATH, and the associated pencil MATH. Let MATH be the span of vectors MATH of the form MATH with MATH and MATH, MATH, MATH. Then MATH. We may assume MATH, MATH. Since MATH, it is enough to show that vectors MATH such that MATH for an appropriate MATH span MATH. But this is a corollary of REF . In conditions of REF assume that the bihamiltonian structure on MATH is micro-Kronecker, and MATH is the action foliation. Then for any MATH points MATH and MATH are on the same leaf of MATH. This follows from the fact that biflow vectors are tangent to leaves of action foliation. This finishes the proof of REF . |
math/9908034 | This is an immediate corollary of REF . |
math/9908034 | Let MATH be the subvariety of irregular elements. Let MATH be the projection of MATH to MATH. The existence of a MATH-compatible element is equivalent to MATH. If MATH is regular, then any non-zero scalar multiple of MATH is also regular. Thus one can consider a closed subvariety MATH of irregular elements in the projectivization MATH of MATH. Given a strongly regular element MATH and a compatible element MATH, one obtains a line MATH in MATH, and MATH. Clearly, any nearby line MATH will also not intersect MATH. Thus the set of elements MATH which are compatible with MATH is NAME open. Thus the intersection of this set with MATH is non-empty. Since MATH, MATH has codimension REF or more, thus the same is true for MATH. On the other hand, if MATH has codimension REF or more, then the projection of MATH to MATH is not surjective; here MATH is arbitrary. This implies that any regular element of MATH is strongly regular. |
math/9908034 | By REF , the set MATH of compatible with MATH elements of MATH is NAME open (thus dense). Show that MATH (or MATH in the case MATH) satisfies the conditions of the proposition. One may assume that MATH. We need to show that for MATH the symplectic leaf through MATH of MATH, MATH, has codimension MATH. For MATH the normal space to this leaf coincides with MATH, thus regularity of MATH implies the statement. Thus we may assume MATH. The NAME structure MATH is a MATH-translation of MATH. If MATH, then MATH is regular. Thus it is enough consider MATH. But the normal space to the leaf of symplectic foliation through MATH is MATH, which finishes the proof. |
math/9908034 | Indeed CITE, irregular elements of a semisimple NAME algebra form a NAME closed subvariety of codimension REF. This implies that the same statement for reductive algebras, thus the proposition. |
math/9908034 | Consider an admissible anti-involution MATH such that MATH. Let MATH be the NAME open subset of MATH where MATH is micro-Kronecker. Since MATH is an anti-involution of MATH, MATH is an anti-involution of the NAME structure MATH on MATH. Since MATH preserves MATH, and MATH is the derivative of MATH with respect to translations in the direction of MATH, MATH is an anti-involution of MATH as well. Thus MATH is MATH-invariant, and MATH is an anti-involution of a micro-Kronecker bihamiltonian structure. Note that the same arguments as in the proof of REF show that MATH. Let MATH. This is a non-empty NAME open subset of MATH. Let MATH. The principal step is to show that the defect of MATH at MATH is REF. Let MATH be a neighborhood of MATH in MATH such that the action foliation MATH of the bihamiltonian structure becomes a fibration MATH. It is enough to show that for any function MATH on MATH the function MATH on MATH is preserved by MATH. In turn, it is enough to do the same for a large enough collection of functions MATH on MATH. Take as such collection functions MATH which are constant on fibers of the integrating foliation MATH on MATH, for each one of MATH (for a sufficiently large MATH). We may suppose MATH. Obviously, MATH is constant on fibers of MATH iff MATH is constant on fibers of the symplectic foliation MATH of the NAME structure MATH. Thus it is enough to show that MATH preserves such functions. Since MATH is the result of translation of MATH by MATH, it is enough to show this for MATH and MATH, MATH, taken instead of MATH. However, MATH is in MATH for MATH in an open subset of MATH, thus we can restrict our attention to a given MATH and MATH. Since MATH, any MATH-orbit which passes near MATH intersects MATH. Since MATH sends a MATH-orbit to a MATH-orbit, this implies that MATH preserves any MATH-orbit which passes near MATH. Decrease MATH so that MATH-orbit of any MATH intersects MATH. However, symplectic leaves of MATH coincide with MATH-orbits in MATH. Thus on a neighborhood of MATH any function which is constant on symplectic leaves of MATH is MATH-invariant. This implies that the defect of MATH near MATH is indeed REF, and we are in conditions of REF . This implies the theorem for MATH being the union of fibers of the action foliation which intersect MATH. |
math/9908034 | Identify MATH with MATH using the Killing form. Due to REF , it is enough to consider the case MATH. MATH . The NAME anti-involution of a semisimple NAME algebra is admissible. Identification of MATH with MATH allows one to consider MATH instead of MATH. Any anti-involution sends a MATH-orbit to a MATH-orbit. Since MATH, and an orbit of a regular element of MATH is transversal to MATH, MATH satisfies the second condition of REF . Thus to prove the lemma it is enough to show that irregular elements in MATH form a subvariety of codimension REF or more. (Note that this statement is not true if one substitutes MATH instead of MATH) Due to homogeneity of the set of regular elements, a translation to algebraic geometry in a projective space show that it is enough to prove this statement for an arbitrary vector subspace in MATH taken instead of MATH (see REF ). Recall that CITE: There are numbers MATH, MATH, MATH, MATH, such that for the elements MATH, MATH, MATH the vector subspace MATH is a NAME subalgebra isomorphic to MATH. The adjoint action of MATH on MATH is a direct sum of MATH odd-dimensional irreducible representations of MATH. Since the action of any non-zero element of MATH in an odd-dimensional irreducible representation has MATH-dimensional null-space, this shows that the stabilizer of any non-zero point of MATH has dimension MATH. But MATH, thus all the non-zero elements of MATH are regular. Moreover, conjugating MATH with elements of the NAME group MATH of MATH, one may assume that MATH, MATH. Thus the subspace MATH is MATH-invariant, let MATH be the MATH-dimensional invariant subspace MATH of MATH. Since MATH intersects irregular elements on MATH, which is a subvariety of codimension REF, irregular elements in MATH form a submanifold of codimension REF or more. This finishes the proof of REF implies that any regular MATH which is on a MATH-orbit of MATH is admissible. But any regular semisimple element of MATH is conjugate to an element of MATH, it is on a MATH-orbit of a MATH-invariant element! Translating from MATH to MATH, any regular semisimple MATH is admissible. Hence there is an open subset MATH on which Conjecture REF holds. Let us show that MATH can be taken NAME open, thus dense. Let MATH be the NAME open subset where foliation MATH makes sense. Recall that MATH is the union of leaves of MATH which intersect MATH. Show that MATH contains a NAME open subset. There is a NAME open subset MATH of MATH such that MATH is MATH-invariant, and MATH-invariant polynomials distinguish MATH-orbits in MATH. On the other hand, locally the fibers of the action foliation MATH are intersections of a finite number of MATH-shifted MATH-orbits, MATH, MATH. Taking a large enough finite collection MATH, MATH, of invariant polynomials on MATH, we see that fibers of MATH coincide with connected components of level sets of MATH, MATH, MATH; here MATH. Let MATH; here MATH is the parallel translation of MATH by MATH. Obviously, MATH is NAME open and is a union of fibers of foliation MATH. Let MATH be the polynomial mapping of MATH to MATH, MATH, with components MATH. Let MATH, clearly MATH contains a NAME open subset MATH of MATH. Decreasing MATH, one may assume that MATH is a submersion to MATH. Since MATH is constant on leaves of MATH, MATH. Since MATH contains an open subset of MATH, MATH contains an open subset of MATH, thus a NAME open subset of MATH. Thus one can assume that MATH. Let MATH, MATH. We obtain the following mappings: MATH; here MATH is NAME open in MATH, MATH is NAME closed in MATH, MATH is (a restriction of) a polynomial mapping which is a submersion onto MATH, and MATH is surjective. Let MATH be the union of connected components of fibers of MATH which intersect MATH. Since MATH, it is enough to show that MATH coincides with MATH (which is a union of fibers of MATH which intersect MATH). Since MATH is a submersion, the number of connected components of fibers of MATH can only jump up during specialization. This implies that MATH is open in MATH. Decreasing MATH, we may assume that the number of connected components of the fiber of MATH over MATH does not depend on MATH. This implies that MATH is closed in MATH. Since MATH is connected (as a NAME open subset of a vector space), MATH. This finishes the proof of REF . |
math/9908034 | As the proof of REF shows, it is enough to show that Consider a reductive NAME algebra MATH and its NAME anti-involution MATH. Then any element MATH of MATH is on MATH-orbit of MATH-invariant element. First of all, the statement for reductive algebras follows momentarily from the case of semisimple NAME algebras. For semisimple elements the statement is obvious (since MATH is MATH-invariant). Consider the case when MATH is nilpotent. One may assume that MATH. If MATH, then any nilpotent element is MATH-conjugate to MATH which is symmetric, thus MATH-invariant. Thus to prove the proposition for the case of nilpotent MATH is enough to show For any nilpotent MATH with a reductive NAME algebra MATH there is a subalgebra MATH such that MATH, MATH, MATH is the NAME anti-involution of MATH, and MATH is conjugate to MATH. Here MATH is the NAME anti-involution of MATH. It is enough to prove this for a semisimple MATH. The classification of nilpotent elements up to conjugation is well-known (CITE, or it can be deduced from CITE): For any nilpotent element MATH, MATH, there is a reductive subalgebra MATH of MATH with MATH and a NAME set of generators MATH, MATH, MATH, MATH, MATH, of MATH such that MATH. Here MATH. The NAME subalgebra of MATH is a NAME subalgebra of MATH, thus after conjugation one can assume that MATH generate MATH. Then MATH for any MATH. Let MATH be coefficients in relations MATH, MATH. Given MATH and MATH, these relations determine MATH, MATH uniquely up to proportionality. This implies that MATH is proportional to MATH. Thus MATH, moreover, after a rescaling MATH may be supposed to be the NAME anti-involution of MATH. Substituting MATH instead of MATH, it follows that it is enough to prove the statement of REF for MATH. In this REF implies that MATH is MATH-conjugate to the element MATH of REF . Moreover, doing another MATH-conjugation one can ensure that the vector space MATH of REF is MATH-invariant. Now the proposition is proven for semisimple and for nilpotent elements MATH. For an arbitrary MATH, there is unique representation MATH as a sum of commuting semisimple and nilpotent elements. Doing conjugation, we may assume MATH. Let MATH. Obviously, MATH, MATH is a reductive NAME algebra, MATH, and MATH is the NAME anti-involution of MATH. Since MATH, we know that MATH is MATH-conjugate to an element of MATH; here MATH is the MATH-group of MATH. Since MATH is MATH-invariant and MATH-invariant, the conjugation by MATH above sends MATH to MATH. This finishes the proof of the proposition. This finishes the proof of the amplification. |
math/9908034 | Due to REF , it is enough to show that on a dense open subset of MATH the NAME web which is a (local) base of the action foliation is flat. This was proven in CITE (in less generality); here we reproduce a more general form of these arguments: Consider a NAME algebra MATH, MATH, and MATH-invariant polynomials MATH on MATH. Let MATH consists of points MATH such that MATH are linearly independent. Suppose that MATH is MATH-regular, MATH, MATH, and that MATH is regular. Consider the NAME structure on the local base MATH of the action foliation MATH of the bihamiltonian structure on MATH associated to MATH. This NAME structure is flat on an open subset. Argue as in the end of the proof of REF . On a NAME open subset MATH of MATH the polynomials MATH, MATH, locally distinguish MATH-orbits (thus symplectic leaves of MATH) on MATH . Thus on a NAME open subset MATH of MATH the polynomials MATH, MATH, MATH, locally distinguish leaves of MATH (here MATH). Associate to a given MATH the coefficients MATH, MATH, MATH, of polynomials MATH in MATH. This is a polynomial mapping MATH, MATH. We conclude that on MATH connected components of fibers of MATH coincide with leaves of MATH. But the leaves of MATH have codimension MATH, thus in the conditions of the lemma the mapping MATH is a submersion, and leaves of MATH are connected components of fibers of MATH. Thus on MATH the manifold MATH may be considered as a base MATH of the action foliation. Describe the structure of NAME web on MATH. Fix MATH. The symplectic leaves of MATH are MATH-translations of MATH-orbits, thus on MATH they coincide with level sets of MATH. Thus the projections of these leaves to MATH may be described by REF, MATH; here MATH are coordinates on MATH. Thus fibers of integrating foliations MATH on MATH are parallel planes in MATH, hence the NAME web structure on MATH is translation-invariant. To finish the proof of the theorem, it is enough to recall CITE that reductive NAME algebras satisfy the lemma. |
math/9908034 | It is enough to show that a micro-Kronecker translation-invariant bihamiltonian structure can be represented as a product of structures of REF . This a direct corollary of REF . |
math/9908034 | The only thing to prove is that MATH is MATH-regular. Constant mappings give an inclusion MATH, and regular elements of MATH go to regular elements of MATH. This shows that MATH intersects irregular elements of MATH over a subset of codimension REF or more, thus MATH is MATH-regular. |
math/9908034 | Start with MATH . Due to REF , if MATH, MATH, is a sequence of different elements of MATH, then MATH. Thus it is enough to show that the MATH-filtration in MATH does not depend on the choice of decomposition into NAME blocks. Now REF taken together imply that MATH. Thus MATH does not depend on the choice of decomposition. |
math/9908034 | Suppose that dimensions of NAME blocks of MATH are equal to MATH. Let MATH has MATH element, let MATH, MATH. By Amplification REF the collection MATH uniquely determines MATH. Express possible decompositions of MATH into NAME blocks in terms of this collection. Suppose that MATH, MATH. By REF , MATH, denote by MATH, MATH, the projection of MATH on MATH according to this decomposition. As one can easily check, MATH, thus MATH, provided MATH and MATH is a NAME block with MATH. Hence MATH for an arbitrary MATH-isotypic MATH. We see that projections MATH identify all the MATH with MATH, thus one with another. Assume MATH. Due to the identifications above, a choice of a basis MATH in MATH induces bases MATH in each of the subspaces MATH, MATH, thus a basis in MATH. Let MATH is spanned by MATH, MATH. Consider a vector space MATH, MATH, and one-dimensional subspaces MATH, MATH, MATH. Suppose that each collection of MATH subspaces out of MATH spans the whole vector space. Then there is one and only one NAME relation MATH in MATH such that MATH, MATH, The ``only one" part follows from Amplification REF. On the other hand, if MATH is a NAME block in MATH, MATH, then there is (exactly one up to proportionally) linear mapping MATH from MATH to MATH such that MATH, MATH. Since MATH is invertible, putting MATH finishes the proof. Apply the lemma to MATH, MATH, MATH. By the construction of the basis MATH, MATH is one-dimensional, thus the conditions of the lemma apply. This gives a NAME linear relation MATH in MATH. Then MATH is a NAME linear relation in MATH with all the NAME blocks having dimension MATH, and MATH for MATH. By Amplification REF, MATH, thus MATH is the required decomposition of MATH into a direct sum of NAME blocks. |
math/9908034 | This follows directly from decomposability into NAME blocks. |
math/9908034 | Consider again the pencil MATH. It is clear that MATH as subspaces in MATH. If we prove that MATH (which coincides with MATH) does not intersect MATH, then one can split MATH and MATH into direct summands (in MATH and MATH correspondingly), which will prove the lemma. Suppose that MATH does intersect MATH. Consider an arbitrary MATH-isotypic block MATH in MATH. Conditions of the lemma imply that MATH. This implies that MATH intersects MATH, which is a contradiction. |
math/9908034 | By REF , one may suppose that MATH. Consider MATH-chains which form bases in MATH and MATH. Denote these chains in MATH by MATH, in MATH by MATH (here MATH enumerates chains, and MATH vectors inside a chain). By REF we may assume that MATH. Let MATH be the polynomial in MATH which corresponds to the MATH-chain MATH, similarly introduce MATH. Fix MATH. Obviously, MATH can be written as MATH, and the polynomial MATH corresponds to an appropriate MATH-chain MATH. Moreover, all the vectors MATH are in MATH. Decompose MATH into direct sum of NAME components, consider projections of MATH to these components. Clearly, for any such projection MATH the vectors MATH form a MATH-chain. Now the proposition follows from the following Consider a NAME block MATH in MATH and a MATH-chain MATH which is a basis of MATH. Let MATH be an arbitrary MATH-chain in MATH. Let MATH, MATH. Then there is a polynomial MATH such that MATH . Write MATH in the basis MATH and compare the coefficients using the definition of a MATH-chain. This finishes proof of the proposition. |
math/9908036 | I will use some basic notions from derived categories CITE for succinctness, but the argument can be rewritten without them. Let MATH. It is enough to show that the sequence MATH is part of a distinguished triangle. This can be checked at the stalks. If MATH, then the stalk of the sequence at MATH reduces to MATH which is certainly part of a distinguished triangle. If MATH, let MATH denote the fiber MATH. Then the stalk of the sequence decomposes as sum of sequences MATH and MATH which both extend to distinguished triangles. |
math/9908036 | The argument goes along the lines presented in CITE. REF and the exactness of MATH and MATH on the category of MATH-Hodge structures implies that MATH strictly preserves MATH and MATH. The same goes for MATH by REF. Thus the first direct and recursive filtrations for MATH and MATH on MATH coincide. Therefore MATH, when equipped with these filtrations, becomes a pure MATH-Hodge structure of weight MATH because it is a subquotient of MATH. Consequently MATH is a morphism of NAME structures of different weights and so it must vanish. As zero strictly preserves everything, the first direct and recursive filtrations on MATH agree, then by similar reasoning MATH. Continuing in this fashion proves REF . It follows from the previous paragraph that MATH and MATH induce weight MATH-Hodge structures on MATH. This implies REF . The proof of the last part is based on the fact that given a spectral sequence of finite dimensional vector spaces (lying in a translate of the first quadrant), there is an inequality MATH and equality holds if and only if MATH for all MATH. It suffices to prove the opposite inequality for MATH. Applying the above inequality for a different spectral sequence gives MATH . There is a commutative diagram MATH where the top isomorphisms follows from NAME 's lemma CITE, and the bottom from REF . As MATH carries a MATH-Hodge structure and MATH is exact for these, it follows that MATH . By REF , MATH and hence MATH as required. The argument for MATH is identical. |
math/9908036 | We verify that MATH satisfies the axioms of REF. MATH certainly contains a terminal object. Given two morphisms MATH with MATH, let MATH. Let MATH be the Boolean algebra generated by fibered products MATH with MATH. Then MATH is the fibered product of MATH and MATH in MATH. The next step is to show that every object possesses a resolution with respect to MATH. If MATH is a stratified variety, let MATH be the disjoint union of the irreducible components. We can pull MATH back to MATH to obtain a surjective proper morphism MATH. Since we can work with one component at a time, we may assume that MATH is irreducible. The proof will proceed by induction on the cardinality of MATH. If MATH has less than or equal to MATH elements, in other words if the stratification is trivial, then the existence of a resolution follows immediately from CITE or CITE. In general, let MATH be the set of atoms of MATH. Choose an element MATH, which is minimal with respect to MATH. Then MATH is necessarily closed. Let MATH be the Boolean subalgebra generated by the closures of elements of MATH. MATH is easily seen to be a stratification not containing MATH. Therefore, by induction, there is a resolution MATH. Let MATH be the preimage of MATH. Then there exists a proper surjective morphism MATH from a second nonsingular variety such that the preimage MATH of MATH is a divisor with normal crossings CITE. Then MATH is a resolution of MATH. It remains to check the last condition for denseness. Given a morphism MATH and a proper surjective morphism MATH, let MATH be a resolution of the fibered product. |
math/9908036 | Fix MATH. It suffices to construct a functorial retraction MATH to the map MATH, because then the augmented complex MATH carries a contracting homotopy defined by MATH . See CITE for details. We have MATH . If MATH then MATH, define MATH to act as the identity on each summand MATH with MATH, and zero on the other summands. |
math/9908036 | Let MATH be the full subcategory of MATH where the underlying variety is proper. The intersection MATH is dense in MATH. Let MATH denote the category of constant constructible sheaves on MATH. This determines an Abelian family of coefficients. Let MATH be an object of MATH, and let MATH be a constant constructible sheaf on it. There is no loss of generality in assuming that MATH is connected, so that MATH is the Boolean algebra MATH generated by the components of a divisor with normal crossings. In this case, the components of MATH are fairly easy to describe: MATH is a constant sheaf. Let MATH be the associated NAME complex. The differentials of MATH extend to give a differential graded NAME complex MATH. Then MATH is a NAME complex on MATH. The theorem now follows from REF . |
math/9908039 | Let MATH, MATH, MATH. We rewrite the table for MATH as The fact that several spin representations of small dimensions have natural realizations given by composition algebras can be found in CITE, from which all cases except for MATH can be deduced. The case of MATH follows from REF below, which gives a general relation between NAME algebras and composition algebras. MATH . Note that in the magic chart for MATH, the tangent space to points in the third row does not have an analogous interpretation. |
math/9908039 | By REF algebras, see CITE, we have to construct a map MATH such that MATH, as then there exists a unique extension to a map MATH. Consider MATH . A short calculation shows that MATH has the required property. Moreover, since MATH is generated by the products of even numbers of vectors in MATH, the diagram follows. |
math/9908039 | Note that MATH to obtain that the points of the second row are indeed MATH and the first are MATH as in the four by four case. For the third row, one uses the same argument as above, only note that the corresponding Plucker type mapping is of degree MATH. |
math/9908039 | It is sufficient to show that the vector space MATH is MATH. Suppose to the contrary that MATH is a proper subspace. Since each of the four "matrix" components of MATH is weighted differently for the cubic MATH, we see the subspace must be the sum of linear subspaces of each of the four components. In fact the two one-dimensional components must be present as they are in MATH. Moreover, the other two components are dual to one another so must be cut equally. So it is sufficient to consider the action on MATH. But the identity matrix is in an open orbit and so we obtain everything. |
math/9908040 | Since MATH and MATH are disjoint unions of MATH and MATH cells of dimension MATH and MATH, respectively, MATH and MATH . Note also that with respect to the NAME cellular decomposition of MATH, the subspace MATH is a cell subcomplex. The dimensions of cells in the complement of MATH run from MATH through MATH. Thus, MATH . Observe that for any pair MATH and integer MATH, either MATH or MATH is REF. Thus the spectral-sequence differentials MATH will all vanish for MATH, which implies the collapse of the spectral sequence at MATH. |
math/9908040 | Notice that for the space MATH, MATH, the group MATH is naturally isomorphic to MATH by NAME duality. The space MATH is the disjoint union of the strata MATH, MATH running over the set of MATH-trees MATH which have exactly MATH vertices. Each stratum MATH is naturally isomorphic to the product of spaces MATH over the set of vertices MATH of the tree MATH. Each space MATH is homotopy equivalent to the configuration space MATH of MATH little disks. Thus MATH where the superscript MATH means the component of degree MATH. The differential MATH takes the component MATH corresponding to a MATH-tree MATH to the sum of components MATH over all MATH-trees MATH such that the tree MATH may be obtained by contracting an interior edge of MATH, merging two adjacent vertices MATH and MATH on the tree MATH into a single vertex MATH of MATH. The matrix element MATH of the differential is induced (up to a sign, which is treated below) by the map MATH which is the dual of the corresponding operad structure map MATH . This map is induced on homology by the map MATH gluing the unit disk with a configuration of MATH little disks into the MATH-th little disk in a configuration of MATH little disks, MATH corresponding to the contracted edge in MATH, if we assume that the contracted edge is directed from MATH to MATH. The sign for MATH comes from the choice of orientation of the strata MATH. This orientation may be chosen by ordering all the edges of each MATH-tree MATH, except the root edge. The orientation on the stratum MATH is then given by ordering the MATH and the MATH coordinates of the points in MATH corresponding to the edges of MATH, according to the order of the edges, skipping for each vertex MATH the coordinates of the point corresponding to the first edge and the MATH coordinate of the point corresponding to the second edge - remember that MATH. Then the compatibility of orientations on MATH and MATH implies that MATH is MATH multiplied by the sign of the permutation from the ordered set of edges of MATH to the ordered set MATH, where MATH is the contracted edge in MATH. This description of MATH in terms of trees and the operad MATH of vector spaces means that MATH is the cobar construction of the operad MATH, just by the definition of CITE. |
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