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cond-mat/0002219 | If a set of bosonic fields with four-species has the same commutation relation as REF , the kernel MATH whose dual fields are replaced with those fields is equivalent to the original kernel. We put c-numbers MATH and MATH into the dual fields. Under a simultaneous replacement MATH and MATH the commutation relation REF is invariant and the kernel MATH is then changed into MATH. It thus follows that MATH . Note that the kernel MATH contains the dual fields MATH. In order to produce the same vacuum expectation values in both sides, a dual field MATH must be equal to MATH. |
cond-mat/0002219 | By REF we have MATH where MATH. Due to the orthogonality MATH and MATH, MATH it follows that MATH . Using the relation MATH we obtain the lemma. |
cs/0002008 | Suppose, by way of contradiction, this were not true. Let MATH be a state of MATH such that a deadlock MATH is reachable from MATH, but not by an initial nontrivial motion which is trivial in MATH. Let MATH be a behaviour of MATH with initial state MATH and reaching the deadlock. Write MATH where MATH and MATH. We claim some MATH is a nontrivial motion of MATH. If not, then MATH. Since MATH is not a deadlock state of MATH, there is some motion with source MATH, say MATH labelled MATH. Since MATH is not looking at MATH in state MATH, it must be that MATH is trivial. Hence we can extend MATH with the motion MATH in MATH and the trivial motion in MATH, and thus MATH did not reach a deadlock state, contrary to choice of MATH. Let MATH be minimal such that MATH is a nontrivial motion in MATH. Note that MATH is trivial for MATH, MATH, MATH. Thus MATH . The triviality of MATH for MATH also implies that the action performed by MATH on the boundary MATH is trivial. The action performed by MATH on the boundary MATH is also trivial, since MATH is not looking at MATH in state MATH. Thus, since MATH synchronizes with MATH, we have that the action performed by MATH on the boundary MATH is also trivial, for MATH, MATH, MATH. It is now evident that MATH is a behaviour of MATH with initial state MATH that leads to the specified deadlock MATH, and that has first motion trivial in MATH. Hence the desired contradiction. |
cs/0002008 | Using the algebra of designs, organize the system as a composite of the given subsystem and its complement: MATH . Now apply REF to the composite of the evaluation of the two subsystems. |
cs/0002012 | Let MATH be the number of mismatches between MATH and MATH at the positions in MATH. Let MATH . First, we prove the following claim. For any MATH such that MATH, where MATH is the constant in the algorithm closestString, there are indices MATH such that for any MATH. MATH . Consider indices MATH such that MATH. Then for any MATH and MATH, we have MATH where REF is from the fact that MATH and Equality REF is from the fact that MATH. MATH . Consider MATH such that MATH. Then among the positions where MATH mismatches MATH, for at least one of the two strings, say, MATH, the number of mismatches between MATH and MATH is at least MATH. Thus, among the positions where MATH matches MATH, the number of mismatches between MATH and MATH is at most MATH. Therefore, MATH. So, MATH . Thus, at least one of MATH, MATH, MATH, MATH, MATH is less than or equal to MATH. MATH . If MATH, them from REF , there must be a MATH such that MATH. From REF , MATH . Hence, there are at most MATH bits in MATH where MATH differs from MATH while agrees with MATH. The lemma is proved. MATH . |
cs/0002012 | CASE: Let MATH. By REF , MATH-where the last inequality is because MATH and that MATH is increasing for MATH. It is easy to verify that for MATH, MATH . Therefore, REF is proved. CASE: Let MATH. By REF is proved. MATH . |
cs/0002012 | Let MATH. Then, for any two strings MATH and MATH of length MATH, we have MATH. Thus for any MATH, MATH . Therefore, the following optimization problem MATH has a solution with cost MATH. Suppose that the optimization problem has an optimal solution MATH such that MATH. Then MATH . Now we solve REF approximately. Similar to CITE, we use a REF variable MATH to indicate whether MATH. Denote MATH if MATH and MATH if MATH. Then REF can be rewritten as a REF optimization problem as follows: MATH . Solve REF by linear programming to get a fractional solution MATH with cost MATH. Clearly MATH. Independently for each MATH, with probability MATH, set MATH and MATH for any MATH. Then we get a solution MATH for REF optimization problem, hence a solution MATH for REF . It is easy to see that MATH takes MATH or MATH randomly and independently for different MATH's. Thus MATH is a sum of MATH independent REF random variables, and MATH . Therefore, for any fixed MATH, by REF , MATH . Considering all sequences, we have MATH . If MATH, then, MATH. Thus we obtain a randomized algorithm to find a solution for REF with cost at most MATH with probability at least MATH. The above randomized algorithm can be derandomized by standard method of conditional probabilities CITE. If MATH, MATH is a polynomial of MATH. So, we can enumerate all strings in MATH to find an optimal solution for REF . Thus, in both cases, we can obtain a solution MATH for the optimization REF with cost at most MATH in polynomial time. Since MATH, MATH for a constant MATH. Let MATH and MATH. From REF , MATH where the last inequality is from REF . This proves the lemma. MATH . |
cs/0002012 | Given an instance of Closest String, suppose MATH is an optimal solution and the optimal cost is MATH, that is, MATH for all MATH. Let MATH be defined as REF of Algorithm closestString. Since for every position in MATH, at least one of the MATH strings MATH conflict the optimal center string MATH, so we have MATH. As far as MATH is a constant, REF can be done in polynomial time by REF . Obviously the other steps of Algorithm closestString runs in polynomial time, with MATH as a constant. If MATH, then by the definition of MATH, it is easy to see that the algorithm finds a solution with cost at most MATH in REF . If MATH, them from REF , the algorithm finds a solution with cost at most MATH. This proves the theorem. MATH . |
cs/0002012 | Obviously, the size of MATH in REF is at most MATH. REF takes MATH time. Other steps take less than that time. Thus, the total time required is MATH, which is polynomial in term of input size for any constant MATH. From REF , the performance ratio of the algorithm is MATH. MATH . |
cs/0002012 | Let MATH be an optimal center string and MATH be the length-MATH substring of MATH that is the closest to MATH. Let MATH. Let MATH be any small positive number and MATH be any fixed integer. Let MATH. If MATH, then clearly we can find a solution MATH within ratio MATH in REF . So, we assume that MATH from now on. By REF , Algorithm closestSubstring picks a group of MATH in REF at some point such that REF For any MATH, MATH . Obviously, the algorithm takes MATH as MATH for at some point in REF . Let MATH and MATH satisfy REF . Let MATH be defined as in REF . Let MATH be a string such that MATH and MATH. Then we claim: CASE: With high probability, MATH for all MATH. For convenience, for any position multiset MATH, we denote MATH for any two strings MATH and MATH. Let MATH. Consider any length MATH substring MATH of MATH satisfying MATH . It is easy to see that MATH implies either MATH or MATH. Thus, we have the following inequality: MATH . It is easy to see that MATH is the sum of MATH independent random REF variables MATH, where MATH indicates a mismatch between MATH and MATH at the MATH-th position in MATH. Let MATH. Obviously, MATH. Therefore, by REF , MATH where the last inequality is due to the setting MATH in REF. Similarly, using REF we have MATH . Combining REF , we know that for any MATH that satisfies REF , MATH . For any fixed MATH, there are less than MATH substrings MATH that satisfies REF . Thus, from REF and the definition of MATH, MATH . Summing up all MATH, we know that with probability at least MATH, MATH for all MATH. MATH . From REF , MATH . Combining with REF and MATH, we get MATH . By the definition of MATH, the optimization problem defined by REF has a solution MATH such that MATH. We can solve the optimization problem within error MATH by the method in the proof of REF . Let MATH be the solution of the optimization problem. Then by REF , for any MATH, MATH . Let MATH be defined in REF , then by REF , MATH . It is easy to see that the algorithm runs in polynomial time for any fixed positive MATH and MATH. For any MATH, by properly setting MATH and MATH such that MATH, with high probability, the algorithm outputs in polynomial time a solution MATH such that MATH is no more than MATH for every MATH, where MATH is a substring of MATH. The algorithm can be derandomized by standard methods CITE. MATH . |
gr-qc/0002016 | Suppose the MATH- and MATH-scalars are related as in REF , that is, MATH. Then the above calculations prove that REF are equivalent. Since REF is equivalent to the vanishing of MATH and since REF is equivalent to MATH being a MATH-potential of a NAME potential of the NAME spinor, the theorem follows. |
gr-qc/0002034 | From REF , MATH if MATH so MATH cannot cross MATH from below. Since MATH for small MATH, the lemma follows. |
gr-qc/0002034 | If MATH for MATH, then MATH exists (because MATH and MATH) so MATH exists as well. We will show that the orbit can be continued beyond MATH provided that MATH. Since MATH, the only obstruction to extending the solution is the possibility that MATH,MATH, or MATH might be unbounded. To see that MATH is bounded we note that MATH. Choose MATH such that MATH for MATH; then MATH and integrating from MATH gives that MATH and hence, by the NAME inequality, MATH. Thus, MATH is bounded. This implies by REF that MATH is also bounded so both MATH and MATH are bounded. Now, REF says that MATH is bounded so MATH is bounded. |
gr-qc/0002034 | A simple calculation yields MATH . REF follows immediately from REF . To prove REF note that MATH if MATH, so MATH cannot cross zero from above. |
gr-qc/0002034 | CASE: We note that MATH, so MATH if MATH. CASE: If MATH and MATH, then MATH by REF . Next, we note that MATH if MATH and MATH. If MATH then MATH but MATH since MATH when MATH by REF . CASE: The function MATH changes sign at each zero for which MATH. From REF , MATH changes sign at most once. Thus, for MATH, MATH and at MATH, the first zero of MATH, if MATH then by REF MATH for all MATH. If MATH then MATH changes sign at MATH, and hence, at MATH, the next zero of MATH, MATH and hence MATH for all MATH. For MATH, MATH, MATH near MATH and if MATH then MATH, hence, MATH for all MATH. |
gr-qc/0002034 | The proof is standard so we just outline it. We introduce new variables MATH, MATH, MATH, and rewrite REF as the first order system MATH . We will use the sup norm throughout this discussion: MATH means the MATH. Consider the space MATH of quadruples of functions MATH where MATH, and MATH and each of the four functions is in MATH, the space of continuous functions defined on the interval MATH with the sup norm. MATH is a complete metric space if we take as metric the maximum of the four components. We define a map MATH by MATH where MATH . One verifies easily that MATH does in fact take MATH to MATH and that MATH is a contracting map if MATH is sufficiently small, and that a fixed point of MATH is a solution to our equations. The proof that the solution depends continuously on MATH is also routine. |
gr-qc/0002034 | Let MATH. Then, REF become MATH with the behavior at the origin MATH . For MATH (decoupling of gravity) REF with REF have constant flat-spactime solutions MATH, MATH. Inserting these solutions into REF gives the NAME equation MATH whose unique solution satisfying REF is MATH . This solution has infinite rotation as MATH. If MATH then MATH so for MATH close to MATH, say MATH, we have MATH because solutions of REF are continuous in MATH and MATH. This concludes the proof of REF . |
gr-qc/0002034 | Note that MATH so integrating gives MATH. Therefore, MATH for MATH. To show that MATH assume MATH. Then MATH so if MATH we have MATH or MATH so MATH is bounded. Since MATH, MATH is bounded below. Thus, by integrating one concludes that MATH. But MATH and MATH so MATH. This contradicts MATH so we must have MATH. |
gr-qc/0002034 | In order to prove that MATH becomes positive at some point MATH, we will show that MATH. By REF we have MATH, so we must show that MATH. The proof of this fact is divided into two cases: CASE: MATH, and REF MATH. Before considering these cases we list some useful properties of the function MATH. We have: CASE: MATH; CASE: if MATH, then MATH for all MATH; CASE: MATH if MATH; REF is a calculation. For REF note that MATH so MATH cannot cross zero from above. For REF we have MATH so integrating gives MATH and hence, MATH. We return now to the proof that MATH. We first consider REF MATH. A calculation shows that MATH, hence MATH if MATH. Since MATH, this implies that MATH for some MATH and therefore MATH for MATH. Note that MATH is monotone decreasing because MATH. Thus MATH and therefore MATH . |
gr-qc/0002034 | Note that MATH. Let MATH be the first zero of MATH, that is, MATH and MATH. If MATH then MATH. Thus MATH can have a zero for MATH only if MATH. Then, from REF , MATH for all MATH. Define a function MATH. A calculation gives MATH so by integrating we get MATH. On the other hand we have MATH, so MATH; contradiction. |
gr-qc/0002034 | If MATH is bounded above then there is an integer MATH such that MATH for all MATH but MATH for some MATH and hence, by REF for all MATH, MATH. We next show that there is a MATH such that for all MATH, MATH (that is, the orbit is ultimately in QREF or QREF). Note that, by REF the orbit must satisfy either MATH or MATH, that is the orbit must lie in QREF or QREF if MATH is odd and in QREF or QREF if MATH is even. We must rule out the possibility that the orbit is in QREF or QREF. Assume that the orbit lies in QREF or QREF for all MATH. Then MATH for all MATH, so MATH for all MATH. From REF we have MATH so MATH and hence MATH goes to zero in finite MATH. This contradiction concludes the proof. |
gr-qc/0002034 | Suppose that MATH for all MATH. We claim that this implies MATH. To see this, suppose that MATH for some MATH. Let MATH which exists because MATH. Note that MATH for all MATH. Choose a MATH such that MATH. If MATH for some MATH, then by REF MATH for MATH, where the last but one inequality follows from the fact that MATH is monotone increasing. Integrating this inequality from MATH to MATH say, we get MATH; contradiction. Thus, MATH and hence MATH. Since MATH exists, MATH also exists and is finite. Next, from REF we know that the MATH-orbit is ultimately in QREF or in QREF. For concreteness we consider the case of REF (the proof of the QREF-case is identical), that is MATH and MATH for sufficiently large MATH. Then, from REF , MATH exists so MATH exists as well (where MATH might be infinite; the point is that MATH). Now, by NAME 's rule, MATH. But REF says MATH which implies MATH, a contradiction. |
gr-qc/0002034 | From the previous lemma we know that there exists a MATH such that MATH for MATH. Let MATH for MATH. A calculation shows that MATH so MATH if MATH. Multiplying REF by MATH we obtain MATH . The right hand side is positive for MATH so MATH is negative and increasing, hence it has a finite non-positive limit. This implies that MATH is integrable. Similarly, multiplying REF by MATH we obtain MATH . The right hand side is positive for MATH so MATH is negative and increasing, hence it has a finite non-positive limit. This implies that MATH is integrable (recall that MATH is monotone decreasing). The integrability of MATH and MATH implies via REF that MATH exists. This concludes the proof that MATH. Having MATH we can strengthen REF by showing that MATH exists. To see this choose a MATH such that MATH. Then MATH, hence MATH. Since MATH is monotone decreasing, we have MATH for MATH and thus MATH. Hence, MATH. Since MATH, MATH exists and MATH. Now we have all we need to derive the asymptotics of MATH. Let MATH. Then MATH, where MATH. Let MATH for MATH and assume that MATH is sufficiently large so that MATH. If MATH, then clearly MATH becomes eventually positive which contradicts that the orbit is eventually in QREF or QREF. If MATH, then MATH - this is impossible because then by NAME 's rule MATH. Therefore MATH must be sandwiched in the interval MATH. Since MATH is arbitrarily large and MATH, we conclude that MATH. The asymptotics of MATH given in REF follows immediately from this. Finally, inserting the derived leading asymptotic behavior of MATH and MATH into REF , we obtain MATH, from which the asymptotics of MATH follows trivially. |
gr-qc/0002034 | Suppose that MATH for all MATH, so MATH for all MATH. We have from REF that MATH. Integrating this inequality from some MATH to some MATH, we obtain MATH which implies (by the NAME inequality) that MATH is bounded. Next, MATH, MATH, implies that MATH; moreover, by REF MATH, hence MATH is not integrable near MATH. Since MATH is bounded, this shows that MATH is not integrable near MATH. But from REF , MATH, so MATH is not integrable near MATH, which contradicts the fact that MATH is a bounded function. |
gr-qc/0002034 | CASE: Suppose the MATH-orbit crashes in QREF or QREF. By REF , MATH for some MATH with MATH; hence, for MATH sufficiently near MATH we have MATH with MATH. By REF , MATH for all MATH. CASE: This case is much more difficult and will require several auxiliary results. It follows from REF that nearby orbits have rotation MATH; we must prove a much more difficult result, namely that nearby orbits have rotation MATH. |
gr-qc/0002034 | We set MATH and then compute that MATH . Note that MATH and all terms on the right side of REF are negative except for the last two. If MATH, we combine the term MATH with MATH; clearly, MATH. Next, we combine the term MATH with MATH to get MATH where MATH; the maximum value of this expression occurs when MATH and that value is MATH by REF . Hence, if MATH, then MATH. Thus, MATH implies that MATH; consequently, MATH. Since MATH, and MATH is bounded and MATH, MATH, hence MATH. |
gr-qc/0002034 | The assumption on MATH tells us that the orbit lies in QREF or QREF for MATH near MATH. For simplicity of exposition we only discuss the case of REF, that is, MATH. In particular, MATH is a monotone function and hence has a limit at MATH. Thus, MATH is continuous; in particular, if we suppose that MATH, then MATH for MATH near MATH. Since MATH, we get from REF that MATH for MATH near MATH. We conclude that MATH is bounded above, hence MATH and MATH. Since MATH, the right hand side of REF is positive and hence MATH is bounded and since MATH is bounded we conclude that MATH is bounded; thus MATH. Then, from REF , MATH, we see that MATH near MATH so there is no crash. This is a contradiction so we conclude that MATH and hence MATH. |
gr-qc/0002034 | If the MATH-orbit crashes at MATH with rotation MATH, then there is a MATH such that MATH. Let MATH. By REF , if MATH is sufficiently close to MATH, MATH (along the MATH-orbit) is bounded in QREF by MATH; MATH is a uniform bound on MATH in QREF for all MATH sufficiently near MATH. Next, choose MATH such that MATH and such that MATH, and MATH; this is possible by REF . Then, for MATH sufficiently near MATH we have MATH and MATH. We shall find a MATH that works for all MATH, that is, it satisfies MATH for all MATH sufficiently near MATH and MATH. So let MATH satisfy: CASE: MATH (along the MATH-orbit) is bounded by MATH, CASE: MATH, CASE: MATH, and REF MATH. If MATH for all MATH and all MATH near MATH we are done - let MATH. Otherwise, we define MATH, etc. by MATH, where MATH, and MATH are the largest values of MATH with that property. For MATH we have from REF MATH since MATH so MATH or MATH. We now integrate the above from MATH to MATH to get MATH . Now, MATH, so MATH. Using the uniform bound on MATH in the interval MATH, we have MATH for some MATH. But MATH and hence we may take MATH. |
gr-qc/0002034 | From REF we have MATH. Moreover, since MATH and MATH, so if MATH then MATH. Since MATH, we have MATH if MATH. Thus, if MATH, and MATH in the interval MATH; if MATH, then MATH for all MATH in the interval MATH because MATH cannot cross that value from above. |
gr-qc/0002034 | To show that MATH goes to MATH, we note that MATH for all MATH near MATH and MATH by REF . Hence, MATH. By REF , MATH, hence MATH for MATH. Thus, MATH. Let MATH be chosen so that MATH. Then, if MATH is sufficiently close to MATH, MATH so MATH. For such MATH we then have MATH and MATH because MATH; thus, MATH. |
gr-qc/0002034 | Suppose that the MATH-orbit crashes at MATH with MATH for all MATH and MATH for MATH near MATH. For MATH near MATH there is a MATH with MATH by REF . Since MATH, MATH, that is, MATH, so the MATH-orbit crashes, MATH, or exits QREF to REF (or QREF to REF), MATH, never to return. In either case, the MATH-orbit has rotation MATH. |
gr-qc/0002034 | Let MATH for all MATH for which the a-orbit is defined. Note that MATH and MATH by REF and hence, MATH. Also note that MATH is a lower bound for MATH by REF ; hence, MATH has a greatest lower bound MATH. We will show that the MATH-orbit is a globally regular solution and MATH for large MATH. We first show that MATH, that is, MATH is the smallest element in MATH. If MATH for some MATH then MATH for all MATH near MATH so MATH for these MATH's and this contradicts the fact that MATH is the greatest lower bound of MATH. Thus, MATH. In particular, the MATH-orbit has bounded rotation. Next we show that the MATH-orbit does not crash. Recall from REF that if the MATH-orbit crashes at MATH then MATH for MATH near MATH. If the MATH-orbit crashes in QREF or QREF, that is, if MATH for MATH near MATH then MATH and MATH for all MATH near MATH which implies by REF that the MATH-orbit must have MATH for all MATH. Thus, MATH for all MATH near MATH and this contradicts the fact that MATH is the greatest lower bound of MATH. Similarly, if the MATH-orbit crashes in QREF or QREF, that is, at some MATH with MATH, then by the crash lemma MATH for all MATH in the domain of definition of the MATH-orbit for all MATH near MATH and this contradicts the fact that MATH is the greatest lower bound of MATH. Thus, the MATH-orbit is defined for all MATH and hence is a globally regular solution by REF . Also, by REF , the MATH-orbit is in QREF or QREF for large MATH. It remains to prove that MATH for large MATH. Suppose that MATH for large MATH. By REF we have that MATH for large MATH and hence, MATH for all MATH near MATH. Then, by REF the MATH-orbit must have MATH for all MATH and thus MATH, and this contradicts the fact that MATH is the greatest lower bound for MATH. This completes the proof of REF . |
hep-th/0002169 | From REF the left hand side is equal to MATH . By performing the sum over MATH we obtain MATH . However, the last expression is equal to the right hand side of REF. |
hep-th/0002169 | This is a direct consequence of REF. |
hep-th/0002169 | If MATH, we find that MATH . As for the case MATH, we have MATH where we used MATH. Thus REF shows that MATH . This completes the proof of REF. |
hep-th/0002169 | We see that MATH by REF . Then REF follows from REF . |
hep-th/0002169 | See CITE. |
hep-th/0002169 | REF is an easy consequence of REF while REF follows from REF , MATH and NAME. |
hep-th/0002169 | Recall that MATH since MATH for the inclusion MATH. |
hep-th/0002169 | If MATH the only relevant part for the calculation of MATH is the term MATH in REF. Then use the series expansion REF for MATH. |
hep-th/0002169 | Let MATH be a point of MATH. By He CITE, the NAME tangent space of MATH at MATH is given by MATH the obstruction of infinitesimal liftings belong to the kernel of the composition of homomorphisms MATH and MATH where MATH is the hypercohomology associated to the complex MATH. Moreover there is an exact sequence MATH where MATH. Then the NAME dual of MATH is the composition of homomorphisms MATH . So we shall prove that MATH. Let MATH be the universal extension, that is, the extension class corresponds to the identity element in MATH . We set MATH. Since MATH by REF, it is sufficient to prove that CASE: MATH is injective, CASE: MATH. MATH . Proof of REF : Since there is an exact sequence MATH it is sufficient to prove that MATH. We note that MATH . Since MATH is a subspace of MATH, MATH. Hence REF holds. MATH . Proof of REF : It follows from CITE that MATH, that is, MATH. Hence MATH by NAME duality. By the stability of MATH, we also have MATH. By the exact sequence MATH. Since MATH fits in an exact sequence MATH and MATH, we have MATH. Thus REF holds. |
hep-th/0002169 | For a coherent system MATH belonging to MATH, our assumptions and CITE imply that CASE: MATH is surjective in codimension REF (and hence MATH) and MATH is a (slope) stable sheaf, or CASE: MATH is injective and MATH is a (slope) stable sheaf according as REF MATH or REF MATH. For the second case, MATH is also generically surjective. There is an exact sequence MATH . Since MATH is generically surjective, MATH is injective. Hence we obtain MATH. Since MATH is torsion-free or of pure dimension REF implies that MATH. Since MATH is a free module, MATH for all MATH. Thus we obtain MATH. We set MATH. We shall prove that MATH is a (slope) stable sheaf of MATH. We first compute MATH: In the NAME group MATH, we have MATH . For MATH, we set MATH. Then we get MATH . Hence we see that MATH. We next show that MATH is (slope) stable: By using the diagram MATH we have an exact sequence MATH . Hence MATH is torsion-free or of pure dimension REF according as MATH or MATH. If MATH, then MATH is a (slope) stable vector bundle. Hence MATH is also stable, which implies that MATH is also (slope) stable. Thus MATH is an element of MATH. If MATH, then MATH, and hence MATH. Since MATH is irreducible and reduced, MATH is a stable sheaf. Therefore MATH also belongs to MATH. Hence we obtain a map MATH . We shall prove that this map is holomorphic. For this purpose, we consider a family MATH of coherent systems parametrized by a scheme MATH such that MATH is flat over MATH and MATH is a vector bundle of rank MATH on MATH. Let MATH be a surjective homomorphism from a locally-free sheaf MATH to MATH. We set MATH. Since MATH, MATH is torsion-free or a coherent sheaf of pure dimension REF implies that MATH is a locally-free sheaf. We consider a homomorphism MATH sending MATH to MATH, where we regard MATH and MATH as subsheaves of MATH. Then we obtain a morphism of complex which is quasi-isomorphic: MATH . Since the construction of MATH is compatible with base change and MATH is generically surjective, MATH is injective, where MATH is the dual of MATH. Hence MATH is flat over MATH and MATH. Let MATH be the homomorphism induced by the natural inclusion MATH. Then MATH is a family of coherent systems. Therefore MATH is a holomorphic map. In the same way, we can construct a holomorphic map MATH. Then MATH is the inverse of MATH. Indeed, by using the diagram MATH we obtain the following diagram MATH . Then we can easily show that MATH is identified with MATH. Thus MATH. MATH also follows from the same argument. |
hep-th/0002169 | Let MATH be an element of MATH. Since MATH for MATH, we obtain MATH . Since MATH is a stable sheaf of positive degree, NAME duality and REF imply that MATH . By using the canonical exact sequence MATH we see that MATH . |
hep-th/0002169 | Let MATH be a family of coherent systems parametrized by a scheme MATH such that MATH for all MATH, where MATH is a line bundle on MATH. Replacing MATH by MATH, we may assume that MATH. We consider a locally-free resolution REF MATH . Then MATH is injective and it defines an effective NAME divisor MATH on MATH. MATH is called the scheme-theoretic support of MATH and MATH is a MATH-module. Thus we can regard MATH as a sheaf on MATH and we get a homomorphism MATH. Since the construction of MATH is compatible with the base change, MATH is flat over MATH and MATH for all MATH. Thus we get a morphism MATH. Conversely, for a flat family of NAME divisors MATH and a family of coherent systems MATH, MATH gives a family of coherent systems on MATH, where we regard MATH as a sheaf on MATH. Hence we have a morphism MATH. Clearly MATH. Since every member MATH is irreducible and reduced, set-theoretically MATH. In particular, MATH is isomorphic to the reduced subscheme MATH of MATH. Therefore it is sufficient to prove that MATH induces an injective homomorphism MATH of NAME tangent spaces for all MATH. Let MATH be a coherent system corresponding to a point MATH. Assume that MATH for a tangent vector MATH. Let MATH be a family of coherent systems corresponding to MATH, where MATH is a flat family of NAME divisors over MATH. We claim that MATH. Then MATH, which implies that MATH. MATH . Proof of the claim: Our assumption implies that MATH. In particular, MATH. Since MATH is generated by one element on MATH, by the construction of MATH, we get MATH . Since first order deformations of MATH are classified by MATH and the map MATH is injective, it follows from REF that MATH. This completes the proof of REF. |
hep-th/0002169 | That the assertions hold is guaranteed by CITE unless MATH and MATH is not ample. Hence we may assume that MATH and MATH satisfies MATH. The following argument is very similar to the last part of the proof of CITE. Let MATH be an ample line bundle in REF. Replacing MATH by MATH, MATH, we may assume that the evaluation map MATH is surjective for all MATH. By CITE, MATH is a stable sheaf. Then the correspondence MATH gives an immersion. Since MATH is irreducible (indeed deformation equivalent to MATH), MATH is an isomorphism. Therefore MATH is also deformation equivalent to MATH. |
hep-th/0002169 | See CITE. |
hep-th/0002169 | We may assume that MATH is smooth by applying REF successively. Since MATH is a projective morphism, the NAME spectral sequence for MATH degenerates. Moreover we obtain MATH, and hence MATH for MATH. Since MATH is the NAME dual of MATH, we obtain our claim. |
hep-th/0002169 | By REF , MATH is isomorphic to MATH. Hence we shall prove that MATH is smooth. Let MATH be an element of MATH. Then condition (MATH-REF) implies that MATH is injective and MATH is a rank-REF torsion-free sheaf when restricted to its support MATH. In order to prove the smoothness of MATH at MATH, it is sufficient to prove that MATH. Since MATH and MATH is simple, we obtain our claim. |
hep-th/0002169 | It is sufficient to prove the deformation equivalence of MATH. By the connectedness of the moduli space of polarized MATH surfaces, there is a family of polarized MATH surfaces MATH such that MATH is irreducible and there are two points MATH which satisfy MATH. Then there is a family of moduli spaces of coherent systems MATH such that MATH and MATH is a projective morphism. Assume that every member of MATH is irreducible and reduced for a point MATH. Let MATH be a point of MATH. By the proof of REF , MATH is injective. By a standard argument, the obstruction of infinitesimal lifting lives in MATH. Let MATH be the relative cohomology class of MATH. Since MATH is smooth (indeed isomorphic), the injectivity of MATH implies that infinitesimal deformations of MATH are unobstructed. Hence MATH is a smooth morphism at MATH. In particular, MATH is a proper closed subset of MATH. Since MATH and MATH is smooth, we obtain our claim. |
hep-th/0002169 | Using REF the left hand side can be rewritten as MATH . Then the integrand of each factor becomes MATH where we used REF. By performing the integrals using REF we thus obtain MATH which can be cast in the desired form thanks to the identity MATH . |
hep-th/0002169 | Using REF we see that the left hand side is equal to MATH . |
hep-th/0002169 | The calculation is similar to that in the proof of REF . The left hand side is equal to MATH . By performing the Gaussian integrals we obtain that MATH . This readily leads to the desired result. |
hep-th/0002193 | Obvious. |
hep-th/0002193 | The proposition holds for the exterior algebra, and therefore also for the algebra MATH since the latter is a ``small" deformation of the exterior algebra. |
hep-th/0002193 | For MATH this is a well-known fact about the symmetric algebra MATH. Since the augmented NAME complex is exact for MATH it is also exact for small MATH and consequently for all MATH. |
hep-th/0002193 | We have to check that MATH vanishes. Note that MATH and that the differentials MATH and MATH restricted to MATH coincide with the multiplication maps MATH and MATH respectively. Thus we have the following commutative diagram: MATH . Now the Lemma follows because MATH (associativity) obviously annihilates MATH. |
hep-th/0002193 | The MATH - bigraded component of MATH is equal to MATH hence the MATH - bigraded component of the bicomplex MATH vanishes for MATH or MATH. Thus the MATH - bigraded component of the bicomplex MATH is bounded. Therefore both spectral sequences of the bicomplex MATH converge to the cohomology of the total complex MATH. The first term of the first spectral sequence reads MATH . Hence the spectral sequence degenerates in the first term, and we have MATH . On the other hand, the first term of the second spectral sequence reads MATH . Hence the spectral sequence degenerates in the second term, and we have MATH . Therefore MATH and we have an exact sequence MATH . Looking at MATH - bigraded component of this sequence we see that MATH . Thus MATH. |
hep-th/0002193 | This follows immediately from the definition of MATH and MATH. |
hep-th/0002193 | By ampleness a sheaf MATH can be covered by a finite sum of sheaves MATH. Now the statement follows from the Proposition, because MATH for all MATH. |
hep-th/0002193 | The group MATH coincides with MATH. Let MATH be the maximal integer (it exists because the global dimension is finite) such that for some MATH there exists arbitrarily large MATH such that MATH. Assume that MATH. Choose an epimorphism MATH. Let MATH denote its kernel. Then for MATH we have MATH hence MATH implies MATH. This contradicts the assumption of the maximality of MATH. |
hep-th/0002193 | To prove this we only need to show that multiplication by MATH is a monomorphism. If MATH is a bundle, it can be embedded into a direct sum MATH because by ampleness the dual bundle MATH is covered by a direct sum of line bundles. Now, since the morphism MATH is mono for any MATH the same is true for the morphism MATH. |
hep-th/0002193 | We have the following exact sequence in the category MATH: MATH . Since MATH we have MATH. Assume that MATH has a nontrivial section. Then MATH has a nontrivial section too. For the same reason MATH has a nontrivial section, and so on. Thus for any MATH the bundle MATH has a nontrivial section. By REF we have isomorphisms: MATH . On the other hand, by REF the last group is trivial for MATH. Hence MATH for all MATH and consequently MATH. Further, assume that MATH is nontrivial. Since MATH for all MATH we have from the exact sequence REF with MATH that MATH is nontrivial too for all MATH. But this contradicts REF . Therefore MATH. This completes the proof. |
hep-th/0002193 | Let MATH be a finitely generated graded MATH-module such that MATH. Consider an exact sequence: MATH . Since MATH the module MATH is finite dimensional. This implies that for MATH the map MATH is surjective. Moreover, these maps are isomorphisms for MATH because all MATH are finite dimensional vector spaces. Let us identify all MATH for MATH with respect to these isomorphisms. Using the MATH-module structure on MATH, we obtain a representation of the NAME algebra MATH on the vector space MATH. But it is well known that the NAME algebra does not have finite dimensional representations. Thus MATH for all MATH and MATH is finite dimensional. Therefore MATH . |
hep-th/0002193 | The existence of such a monad was proved above. The uniqueness follows from REF . The equality of dimensions of MATH and MATH follows immediately from the exact sequence REF . |
hep-th/0002193 | Let MATH be a bundle on MATH trivial on the line MATH. We showed above that any such bundle comes from a monad unique up to an isomorphism. Conversely, suppose we have a monad MATH with MATH such that its restriction to the line MATH is a monad with the cohomology MATH. Then the cohomology of this monad is a bundle on MATH which belongs to MATH. Indeed, the cohomologies of the dual complex MATH coincide with MATH and MATH. Hence, to prove that MATH is a bundle, it is sufficient to show that the dual complex is a monad too, that is, that the map MATH is an epimorphism. The restriction of the dual complex to MATH is a monad which is dual to the restriction of the monad REF to MATH. Hence the restriction of MATH on MATH is an epimorphism. Then, by REF , MATH is an epimorphism as well. Thus to describe the moduli space MATH we have to decsribe the space of all monads REF modulo isomorphisms preserving trivialization on MATH. Consider a monad MATH with MATH and MATH. Denote by MATH its cohomology bundle. The maps MATH and MATH can be regarded as elements of MATH and MATH respectively, where MATH is the vector space spanned by MATH. The maps MATH and MATH can be written as MATH where MATH and MATH are constant linear maps. Let us restrict the monad to the line MATH. The monadic condition MATH implies now: MATH . Moreover, since the restriction of MATH to MATH is trivial, the composition MATH is an isomorphism (see REF ). We can choose bases for MATH so that MATH (the identity matrix) and MATH . Using the equations MATH and MATH we can write: MATH . Now the monadic condition MATH can be written as: MATH . Therefore we obtain the following matrix equation: MATH . Note that the last MATH basis vectors of MATH give us a trivialization of the restriction of MATH to the line MATH. It is easy to check that any isomorphism of a monad which preserves trivialization on MATH and the choice of the bases of MATH made above has the form MATH . This proves the theorem. |
hep-th/0002193 | Obvious. |
hep-th/0002193 | By the preceding theorem, it is sufficient to show that the NAME transform of every multiplier is a convolute, and vice versa. The former fact is proved in REF, ch. CASE: Let us prove the converse. By REF , every convolute has the form MATH for some MATH and rapidly decreasing continuous functions MATH. Let MATH be the NAME transform of MATH . Since the integrals MATH are absolutely convergent, the functions MATH are MATH functions. Furthermore, the NAME transform of MATH is equal to MATH (see REF), hence MATH is also a MATH function. Finally, since by the preceding theorem the NAME transform of any element of MATH is again an element of MATH and all its derivatives are polynomially bounded. Hence MATH is a multiplier. |
hep-th/0002193 | We will prove that the MATH - product of two convolutes of MATH is well-defined, and is again a convolute of MATH. The rest is obvious. It is sufficient to consider the case when MATH . Then, integrating by parts, we may rewrite the MATH - product in the following form: MATH . Derivatives acting on the exponential bring down powers of MATH so the integral can be rewritten as MATH where MATH is a homogeneous polynomial of degree MATH. We now use the NAME rule repeatedly to rewrite the expression above as MATH where MATH is a homogeneous polynomial of degree MATH. Because both MATH and MATH are rapidly decreasing, the integral converges absolutely and defines a MATH function of MATH which is rapidly decreasing. Hence the MATH - product of MATH and MATH has the form MATH where the functions MATH are continuous and rapidly decreasing. It follows that the space of convolutes is closed under the MATH-product. |
hep-th/0002193 | The first statement is an immediate consequence of REF . The second statement follows from a simple computation. |
math-ph/0002003 | CASE: The time evolution of MATH is unitary and thus MATH. The stated analyticity is an immediate consequence of the elementary properties of the NAME transform . The asymptotic behavior follows then from the NAME lemma. CASE: We have in MATH, MATH . For MATH we push the integration contour through the upper half plane. At the two poles in the upper half plane MATH equals MATH and MATH respectively, so that MATH where MATH is the root of MATH in the upper half plane. Thus MATH with the branch satisfying MATH as MATH in MATH. As MATH varies in MATH, MATH belongs to the lower half plane MATH and then MATH varies in the fourth quadrant. For MATH we have MATH and relation REF follows. |
math-ph/0002003 | To get a contradiction, assume MATH is a MATH solution of REF . Multiplying REF by MATH, and summing with respect to MATH from MATH to MATH we get MATH . For MATH the imaginary part of MATH is nonpositive, by REF , and is strictly negative for MATH large enough. Thus if for some such MATH, MATH is nonzero then the last sum in REF has a strictly negative imaginary part, which is impossible since the left side is real. If on the other hand MATH is zero when MATH is large negative, then solving REF for MATH in terms of the MATH it would follow inductively that MATH, contradicting the assumption. |
math-ph/0002003 | CASE: We look at MATH, the case of MATH being similar. Dropping the MATH superscript we have from REF MATH . To find the analytic properties of the solution MATH it is convenient to regard REF as a contractive equation in the space MATH of sequences MATH in the norm MATH. Let MATH be large. The map MATH defined by MATH depends analytically on MATH and is NAME continuous of exponent at least MATH if MATH. In addition, if MATH we have for sufficiently large MATH . Similarly, MATH for sufficiently large MATH which shows that MATH is contractive in the unit ball in MATH. Thus, REF has a unique solution in MATH, which depends analytically on MATH and is NAME continuous of exponent at least MATH if MATH. This also implies the convergence of REF . Note that given MATH and MATH both compact, MATH can be chosen the same for all MATH and MATH. CASE: From REF it is seen that MATH for large MATH. Thus, we may write, for large MATH, MATH . The estimates REF now follow by a straightforward calculation. Since MATH, the estimates follow from REF and the NAME summation formula. CASE: As before, we only need to look at MATH. We take two compact sets MATH and MATH, and choose MATH as in the note at the end of the proof of REF . Taking the log in REF , the infinite sums are absolutely convergent. By standard measure theory, MATH has the same analyticity properties in the interior of MATH and NAME continuity in MATH as those of MATH, when MATH. Now, REF easily implies that the same is true for MATH as well. If MATH and MATH are solutions of REF then MATH and thus MATH . Thus, if MATH for some MATH then MATH and MATH. The smoothness properties follow from the proof of REF . |
math-ph/0002003 | By analyticity and continuity MATH does not vanish for any MATH and MATH. By REF the function MATH defined through MATH, where MATH and MATH has the required properties. Since no solution of the homogeneous equation is bounded on MATH, MATH is the unique solution with the desired properties. |
math-ph/0002003 | Since MATH is analytic in MATH, continuous and nonzero in MATH, MATH is bounded below in compact sets in MATH. Then, the smoothness properties of MATH derive easily from those of MATH on which we concentrate now. CASE: For MATH we write, using REF , MATH . The last term in parenthesis can be rewritten, using also REF , as MATH . Thus we see that MATH is continuous as MATH and MATH [compare REF ], if MATH. A very similar calculation shows the continuity of MATH if MATH. CASE: By REF , MATH is continuous as MATH with MATH or MATH. Now, REF written in the form MATH shows that MATH is NAME continuous as MATH if MATH thus for all MATH. The value of MATH is easily calculated using REF . |
math-ph/0002003 | Indeed, MATH . |
math-ph/0002003 | Consider MATH endowed with the norm MATH, where MATH. If MATH is continuous and MATH, a straightforward calculation shows that MATH where the last relation follows from the NAME lemma. The integral REF can be written as MATH . Since MATH is locally in MATH and bounded for large MATH it is clear that for large enough MATH, and for any MATH, REF is contractive if MATH. |
math-ph/0002014 | Let us define MATH for all MATH by REF, and let us extend MATH to all MATH by setting MATH when MATH. To prove REF Note that MATH, which implies that MATH is subharmonic (we use MATH and MATH, by REF ). Set MATH with MATH and small. Obviously, MATH is subharmonic on the open set MATH because MATH is harmonic there. Clearly, MATH as MATH and MATH. Suppose that REF is false at some radius MATH and that MATH. In the annulus MATH, MATH has its maximum on the boundary, that is, either at MATH or at MATH (since MATH is subharmonic in MATH ). By choosing MATH sufficiently small and positive we can have that MATH and this contradicts the fact that the maximum (which is at least MATH ) is on the boundary. REF is proved by noting (by subharmonicity again) that the maximum of MATH in MATH occurs on the boundary, that is, MATH for any MATH. REF is proved by studying the function MATH. Since MATH and MATH are continuous, the falsity of REF implies the existence some open subset, MATH on which MATH. On MATH we have that MATH is subharmonic (because MATH). Hence, its maximum occurs on the boundary, but MATH there. This contradicts MATH on MATH. |
math-ph/0002014 | In polar coordinates, MATH, one has MATH. Therefore, it suffices to prove that for each angle MATH, and with MATH denoted simply by MATH, MATH where MATH denotes the distance of the origin to the boundary of MATH along the ray MATH. If MATH then REF is trivial because the right side is zero while the left side is evidently nonnegative. (Here, MATH is used.) If MATH for some given value of MATH, consider the disc MATH centered at the origin in MATH and of radius MATH. Our function MATH defines a spherically symmetric function, MATH on MATH, and REF is equivalent to MATH . Now choose some MATH and note that the left side of REF is not smaller than the same quantity with MATH replaced by the smaller disc MATH. (Again, MATH is used.) According to REF , and linearity in MATH, this integral over MATH is at least MATH. Hence, for every MATH, MATH . The proof is completed by noting that MATH, by multiplying both sides of REF by MATH and, finally, integrating with respect to MATH from MATH to MATH. |
math-ph/0002021 | By REF is continuous, hence we can find a seminorm MATH so that MATH for all MATH. Thus there are positive constants MATH and MATH so that MATH holds for all MATH. In the first equality, the invariance of MATH was used, and in the second, REF was applied. Thus, for any NAME MATH, and any MATH, MATH in MATH, one obtains that the following function of MATH and MATH, MATH depends smoothly on MATH and satisfies the estimate MATH with suitable constants MATH, MATH. Hence, this function satisfies the assumptions of REF. Application of the said Lemma entails the following: Suppose that for some open neighbourhood MATH of MATH we can find some MATH with MATH and MATH for all MATH. Then this implies that the analogous relation holds with MATH replaced by MATH for any MATH when simultaneously MATH is replaced by some slightly smaller neighbourhood MATH of MATH. Consequently, relation REF - with MATH, MATH - entails that MATH is absent from MATH. CASE: Some notation needs to be introduced before we can proceed. For MATH, we define MATH . Then we will next establish MATH . To this end, let MATH be such that MATH, and pick some MATH and an open neighbourhood MATH of MATH so that MATH for all MATH. Now pick two functions MATH such that their NAME MATH have the property MATH. Define MATH by MATH. Then observe that one can find MATH such that the functions MATH vanish for all MATH and all MATH. Consequently, also the functions MATH vanish for all MATH and all MATH. Denoting the NAME of MATH by MATH, one obtains for all MATH, MATH: MATH for all testing-families MATH, MATH. Invariance of MATH under MATH was used in passing from the second equality to the last. In view of REF . above, this shows REF. CASE: In a further step we will argue that MATH . So let again MATH and MATH as above, and MATH as in REF with NAME MATH. Let MATH have MATH. Then there is an open neighbourhood MATH of MATH and a MATH so that MATH for all MATH. The support of MATH is contained in the support of MATH, and there is clearly some MATH such that MATH for all MATH as soon as MATH. By the characterization of a ground state given in REF, and using also the MATH-invariance of a ground state, one therefore obtains MATH for all MATH. Relation REF is thereby proved. CASE: Now we turn to the case MATH . Consider a MATH with MATH and pick some MATH and an open neighbourhood MATH of MATH so that MATH for all MATH. Choose again MATH and MATH as above and define correspondingly MATH and MATH as in REF, respectively. Denote again their NAME by MATH and MATH. Note that MATH and MATH are in MATH for all MATH and all MATH, so their NAME are entire analytic. Moreover, a standard estimate shows that MATH . One calculates MATH and now the NAME REF yields for all MATH, MATH for suitable MATH, where REF have been used. Making also use of the MATH-invariance of MATH one finds, with suitable MATH, MATH for all MATH. This establishes REF Combining now REF , one can see that for each MATH there holds MATH . Since the set on the right-hand side obviously has no proper conic subset in MATH, one concludes that REF holds true. CASE: As MATH is KMS at MATH, this means that it is a trace: MATH. Since MATH is also MATH-invariant, we have MATH . The set on the right hand side has precisely two proper closed conic subsets MATH . These two sets are disjoint, MATH, and we have MATH. Hence, since MATH is a trace, one can argue exactly as in CITE to conclude that either MATH or MATH imply MATH. This establishes REF. |
math-ph/0002021 | Select a local trivialization of MATH over MATH. With respect to it, there are smooth MATH-valued functions MATH of MATH such that MATH . Now suppose that MATH is not in MATH, so that MATH isn't contained in any of the MATH. Then, making use of the fact that the wavefront set of a scalar distribution may be characterized in terms of the decay properties of its localized NAME in any coordinate chart (compare CITE) in combination with REF, one obtains immediately the relation REF. Conversely, assume that REF holds. Since MATH is in MATH for each MATH and depends smoothly on MATH, we can find a smooth family MATH of functions of MATH so that MATH, MATH. Since REF holds, one may apply REF to the effect that for some open neighbourhood MATH of MATH and for all MATH where only the MATH-th entry is non-vanishing, one has MATH . Then one concludes from REF that MATH isn't contained in MATH for each MATH. |
math-ph/0002021 | Let MATH be any point in MATH. Then there is a coordinate chart MATH around MATH so that, for small MATH, MATH holds on a neighbourhood of MATH, where MATH . In such a coordinate system, we can also define ``spatial" translations MATH for MATH in a sufficiently small neighbourhood MATH of the origin in MATH. Let MATH be any smooth family of local morphisms around MATH covering MATH, where MATH (on a sufficiently small neighbourhood of MATH). Now let MATH be another point, and choose in an analogous manner as for MATH a coordinate system MATH, and MATH and MATH. CASE: In a further step we shall now establish the relation MATH . Since we have MATH by REF , this then allows us to conclude that MATH and we observe that thereby the possibility MATH with MATH or MATH is excluded, because that would entail both MATH and MATH. For proving REF it is in view of REF and according to our choice of the coordinate systems MATH, MATH and corresponding actions MATH and MATH sufficient to demonstrate that the following holds: There is a function MATH with MATH, and for each MATH with MATH or MATH there is an open neighbourhood MATH so that MATH as MATH holds for all MATH. (The notation MATH should be obvious.) However, making use of REF of REF, for proving REF it is actually enough to show that there are MATH and MATH as above so that MATH as MATH holds for all MATH and all MATH. In order now to exploit the strict passivity of MATH via REF , we define the set MATH of testing families with respect to the NAME algebra MATH in the same manner as we have defined the set MATH of testing families for the algebra MATH in REF . In other words, a MATH-valued family MATH is a member of MATH, for arbitrary MATH, whenever for each continuous seminorm MATH on MATH there is some MATH so that MATH . Now if MATH is in MATH, then MATH defined by MATH is easily seen to be a testing family in MATH. The same of course holds when taking any MATH and defining MATH accordingly. Since MATH, it follows from REF that, with respect to the time-translation group MATH, MATH . And this means that there is some MATH with MATH, and for each MATH with MATH or MATH an open neighbourhood MATH in MATH so that MATH as MATH for all MATH and MATH in MATH. When MATH relates to MATH as in REF, and if their primed counterparts are likewise related, then for sufficiently small MATH and MATH, MATH one has MATH for small enough MATH. Whence, upon taking MATH and MATH, where MATH is in MATH with MATH, and with MATH and MATH having sufficiently small supports, it is now easy to see that REF entails the required relation REF, proving REF, whence REF is also established. CASE: Now we shall show the assumption that MATH is smooth at causal separation to imply that also MATH and hence, MATH itself is smooth at causal separation. The same conclusion can be drawn assuming instead that MATH is smooth at causal separation. We will present the proof only for the first mentioned case, the argument for the second being completely analogous. We define MATH as the set of pairs of causally separated points MATH. The restriction of MATH to MATH will be denoted by MATH. By assumption, MATH has empty wavefront set and therefore MATH. Since MATH iff MATH, the `flip' map MATH is a diffeomorphism of MATH. Then MATH is a morphism of MATH covering MATH. Thus one finds MATH implying MATH for all MATH. Noting that multiplication by constants different from zero doesn't change the wavefront set of a distribution, this entails, with REF MATH . Now it is easy to check that MATH for all MATH, and this implies MATH . However, since we already know from REF that MATH and MATH, we see that MATH. Combining this with REF yields MATH . And thus we conclude that MATH is smooth at causal separation. CASE: Now we will demonstrate that the wavefront set has the form REF for points MATH on the diagonal in MATH, by demonstrating that otherwise singularities for causally separated points would occur according to the propagation of singularities REF . To this end, let MATH be in MATH with MATH not parallel to MATH. In view of the observation made below REF that we must have MATH and MATH, we obtain from REF MATH. For any NAME surface of MATH, one can find MATH in MATH with MATH and MATH lying on that NAME surface because of the inextendibility of the bi-characteristics. Since MATH is not parallel to MATH, one can even choose that NAME surface so that MATH (if such a choice were not possible, the bi-characteristics through MATH with cotangent MATH and MATH would coincide). But this is in contradiction to the result of REF since MATH and MATH are causally separated. Hence, only MATH with MATH can be in MATH. Applying the constraint MATH found in REF gives MATH. Together with the other constraint MATH of REF we now see that if MATH is in MATH it must be in MATH. CASE: It will be shown next that MATH is smooth at points MATH in MATH which are causally related but not connected by any lightlike geodesic: Suppose MATH were in MATH with MATH as described. Using global hyperbolicity and the inextendibility of the bi-characteristics, we can then find MATH in MATH with MATH lying on the same NAME surface as MATH. As MATH cannot be equal to MATH by assumption, it must be causally separated from MATH, and so we have by REF a contradiction to REF. Thus, MATH must indeed be smooth at MATH. CASE: Finally, we consider the case of points MATH connected by at least one lightlike geodesic: Let MATH be in MATH. To begin with, we assume additionally that MATH is not co-tangential to any of the lightlike geodesics connecting MATH and MATH. As in REF we then find MATH in MATH with MATH and MATH lying on the same NAME surface and MATH, thus establishing a contradiction to REF. To cover the remaining case, let MATH be co-tangential to one of the lightlike geodesics connecting MATH and MATH. As a consequence, we find MATH with MATH. By REF, we have MATH, MATH, showing MATH to be in MATH. |
math-ph/0002027 | Since the eigenbasis MATH of MATH is an orthonormal basis for MATH, it suffices to show that for each MATH, MATH. But MATH . |
math-ph/0002027 | The proof is by induction on MATH. The formula clearly holds when MATH. For MATH, the determinant is a rational function of MATH with a double pole at MATH; we can write MATH . The coefficient MATH is zero since the determinant is even under the exchange of MATH and MATH (exchange the first two rows and exchange the first two columns). The coefficient MATH is the determinant of MATH, the matrix obtained from MATH by deleting the first two rows and columns. Therefore the right and left-hand sides of REF both represent rational functions (in each variable) with the same poles and residues; hence they differ by a constant. This constant is zero by homogeneity. |
math-ph/0002027 | When MATH is the upper half plane this follows by combining REF with REF and the calculation REF . For arbitrary MATH, REF shows that MATH (where MATH is the NAME 's function on MATH) by conformal invariance of the height moments and of MATH. This completes the proof. |
math-ph/0002038 | It follows by a standard averaging argument CITE that the set MATH of invariant probability measures supported on MATH is non-empty: for any probability measure MATH supported on MATH, the set of weak* limit points of the set of finite time averages MATH gives at least one non-trivial element of MATH. Since MATH consists of only finitely many orbits, any invariant measure in MATH is a finite sum of (ergodic) measures, each supported on just one orbit. The non-compact orbits MATH obviously cannot carry invariant probability measures; hence, at least one orbit must be compact. |
math-ph/0002038 | In the case of a Hamiltonian flow with NAME vector field MATH, this is a case of NAME 's theorem CITE. In fact, MATH where MATH. In the case of MATH actions, we can apply NAME 's theorem to any one parameter subgroup. |
math-ph/0002038 | By REF no such invariant foliation exists unless MATH, so we may assume this is the case. The NAME metric MATH is then a well-defined metric on MATH. We denote by MATH the associated homogeneous Hamiltonian (length squared of a covector). Since the sets MATH and MATH are the same, the latter carries a NAME foliation by tori which project regularly to MATH . Since the geodesic flow MATH of MATH on MATH coincides (up to a time re-parametrization) with MATH this foliation is invariant under MATH. Now let MATH be the dilation MATH . Then MATH intertwines the geodesic flows on these sphere bundles (up to constant time reparametrization). Since MATH is conformally symplectic it also carries the invariant NAME torus foliation of MATH to an invariant NAME torus foliation of MATH. It follows that MATH carries a NAME torus foliation invariant under the geodesic flow of the NAME metric. By Mane's theorem, the geodesic flow has no conjugate points and so by NAME 's theorem, (E - V)g must be flat. |
math-ph/0002038 | The assumption implies that MATH is flat for all MATH in the interval. Let MATH denote the curvature tensor of MATH . It is clearly a real analytic function of MATH. Since MATH in MATH, it must vanish identically. Therefore the NAME 's flow MATH on MATH has no conjugate points. By REF, it follows that MATH is flat and MATH is constant. |
math-ph/0002038 | CASE: MATH: By definition, MATH where MATH is a cutoff to MATH . Since MATH is a finite set for each MATH, there exists MATH such that MATH . We form the sequence MATH and then choose a sub-ladder MATH with a unique quantum limit. Then MATH . CASE: MATH: It is clear that for each ladder MATH we have MATH . Therefore the same holds after taking the supremum over MATH . |
math-ph/0002038 | Let MATH be a MATH-invariant neighbourhood of MATH. Let MATH be a sequence of regular points such that MATH. For each MATH and sufficiently large MATH, there exists at least one component MATH of MATH such that MATH. By REF , MATH is charged by an amount MATH. We now break up the discussion into two cases: CASE: All MATH-orbits of MATH are compact In this case, we just need a positive lower bound for the quotient MATH as MATH . A lower bound for the numerator is given by the minimal period of NAME 's theorem REF . Since all orbits (including the limit) are compact, the masses in the denominator have uniform upper bounds. Indeed, by REF the masses are the co-volumes of the period lattices of MATH . Since the period vectors generating the lattices are uniformly bounded as MATH, the volumes are also uniformly bounded above. Hence the denominator is bounded above, and therefore the quotient is bounded below by a positive constant. CASE: There exists a non-compact orbit in MATH . In this case, the denominator will tend to infinity, so we need a better lower bound on the numerator. We claim that there exists MATH such that MATH and MATH. To prove this, it suffices to find MATH such that MATH . The natural candidate is to choose MATH such that MATH . We now prove that this choice of MATH satisfies REF . We write MATH . The second term is bounded above by a constant MATH independent of MATH. The first term tends to infinity since MATH contains a non-compact orbit. Indeed, at least one vector of the period lattice of MATH must tend to infinity as MATH since the limit orbit is non-compact. It follows that the set of period lattices MATH is non-compact in the manifold of lattices of full rank of MATH. Now according to NAME 's theorem, any set MATH is compact. By NAME 's theorem CITE, the minimal period stays bounded below, so non-compactness of the lattices forces some volume MATH as MATH . It follows that when a non-compact orbit exists in MATH, then for each MATH, MATH . Then REF follows if we select MATH as in REF . We now complete the proof of REF . By the finite complexity condition, we have found MATH such that MATH. Further, for each MATH, there exists a ladder MATH which gives charge MATH to MATH. Let MATH denote the unique weak limit measure of the ladder. Then let MATH denote any weak* limit of the sequence MATH. It follows that MATH is an invariant probability measure supported on MATH . Indeed, its support must be contained in the set of limit points of the sequence of orbits MATH, hence in MATH . Since MATH is closed in the weak* topology (since it is a set of limit points), it follows further that MATH . Hence there exists a ladder MATH such that MATH and which charges MATH by an amount MATH . This proves the first part of the lemma. The second statement is an immediate consequence of REF : There must exist at least one compact singular orbit MATH . Since MATH is an invariant probabililty measure, it must be supported on union of the compact singular orbits, hence must charge at least one such orbit. |
math-ph/0002038 | Existence of a compact singular orbit contradicts the the uniform boundedness of eigenfunctions assumption. Indeed, it follows from REF that, for any MATH, there exist a compact, singular orbit MATH and MATH-normalized joint eigenfunctions MATH such that for some MATH . However, the estimate in REF cannot hold since by definition, compact singular orbits have dimension MATH. Therefore, there cannot exist singular levels of the moment map MATH. |
math-ph/0002038 | We fix an energy level MATH and consider eigenvalues of MATH lying in MATH for some fixed MATH . The eigenfunctions we consider are the joint eigenfunctions of MATH with joint eigenvalues MATH respectively, satisfying MATH for some MATH . We recall that MATH and MATH corresponds to the energy shell MATH of the classical Hamiltonian MATH corresponding to the quantum Hamiltonian MATH. By assumption, the eigenfunctions corresponding to these joint eigenvalues are uniformly bounded independently of MATH . By REF , it follows that all NAME torus orbits of MATH on MATH project regularly to the base. Indeed, the proof that the torus MATH projects regularly only involves trace formula and quantum limits over joint eigenvalues in the set MATH . Hence our assumption on uniform boundedness of the eigenfunctions of MATH with eigenvalues in the interval MATH is sufficient to obtain the result of REF for the tori on the energy shell MATH . Hence, by a simple covering space argument, we can without loss of generality assume that the base manifold is a torus. By REF , there are no singular levels of the moment map MATH . Hence MATH has a smooth NAME foliation invariant under MATH . By REF , we must have that MATH and the NAME metric MATH is flat. If we additionally assume that the sup norms are bounded indepedently of MATH and MATH in some interval MATH, then the NAME metrics MATH are flat for all MATH in this interval, and it follows by REF that MATH is flat and MATH is constant. |
math-ph/0002039 | The surjectivity of MATH guarantees that the normal bundle MATH is disjoint from the wave front set of MATH and hence MATH is well-defined (see CITE). Thus we can define MATH . To verify the equation of the lemma, it suffices by the continuity of MATH to consider the case where MATH. In this case, MATH, and hence MATH . Since MATH and MATH are intrinsic volume forms, it follows that MATH is independent of the choice of local frame MATH (and local coordinates). To show that MATH does not depend on the choice of connection on MATH, write MATH, MATH, MATH. Then if we consider the flat connection MATH, MATH, we have MATH . Hence, MATH so that MATH. |
math-ph/0002039 | The proof is by induction on MATH. From REF , MATH where the summation goes over all partitions MATH with at least two elements in the partition (that is, MATH). From REF , we have MATH where the sum is taken over all pairs MATH of partitions of the set MATH. If we use the inductive assumption that REF holds when the number of points is less than MATH and apply it to MATH in REF , then we obtain that MATH where the sum on the right is taken over all partitions MATH of the set MATH and all partitions MATH of the set MATH which are ``mutually disconnected" in the sense that there is a proper subset MATH of the set MATH that is simultaneously a union of some subsets MATH and a union of some subsets MATH. When we substitute REF into REF and take the difference on the right of REF , disconnected pairs MATH will be cancelled out and we will be left with mutually connected MATH. This proves the lemma. |
math-ph/0002039 | By the determinant formula, MATH where the sum goes over all MATH-tuples MATH of permutations of MATH. We claim that MATH where the summation on the right goes over the set of MATH-tuples MATH such that no proper subset of MATH is invariant under the group generated by the MATH. (Each such MATH corresponds to a connected graph consisting of edges beginning at MATH and ending at MATH, for all MATH.) Indeed, MATH where the summation on the right goes over all partitions MATH with MATH. Using this equation, we prove REF by induction (compare the proof of REF ). The estimate REF now follows from REF . |
math-ph/0002039 | We have the identity MATH . The proof of this identity is the same as that of REF . Indeed, the connected functions of MATH are equal to REF (except that REF-point connected function equals REF); hence REF follows. The identity REF can be rewritten as MATH where the summation on the right goes over all mutually connected pairs of partitions MATH with at least two elements in MATH (that is, MATH). Now the estimate REF follows by induction from REF and identity REF . |
math-ph/0002039 | We must show that MATH . Assume without loss of generality that MATH. Let MATH be an oriented connected graph with zero boundary as in the definition of MATH. Since MATH and MATH are connected by a chain of loops in MATH, we can choose disjoint sets of edges MATH such that MATH forms a path starting at MATH and ending at MATH, and MATH forms a path starting at MATH and ending at MATH. This means that there is a sequence MATH such that MATH, where MATH begins at MATH and ends at MATH. By removing any loops in MATH, we can assume that the MATH are distinct and thus MATH. A similar description holds for MATH. Let MATH. We note that MATH where the second inequality is by NAME. We then have MATH and hence MATH . The same inequality also holds for the product over the path MATH. Since each term of the product in REF is less than REF, we then have MATH . Taking the supremum over all graphs, we obtain REF . |
math-ph/0002045 | The existence of continuous sections can be proved applying NAME 's Lemma to constant sections in the (trivial) part of the vector bundle over one chart MATH, getting continuous sections of the whole vector bundle supported in one chart MATH over which the vector bundle is trivial. Any bundle homomorphism MATH maps sections in MATH to sections in MATH. If MATH is a bundle isomorphism, then MATH is a MATH-module isomorphism. We observe that MATH. These MATH-modules are free and finitely generated. Since MATH for a given vector bundle MATH, some vector bundle MATH and MATH by NAME 's theorem, MATH is projective and finitely generated. |
math-ph/0002045 | MATH: MATH . Assume MATH by NAME 's theorem. Let MATH be the fibrewise projection, MATH. Define MATH by MATH for MATH, MATH. Then MATH is a correctly defined surjective bundle homomorphism. Let MATH be the induced MATH-module map, a projection onto MATH. Note, that MATH for every MATH. Therefore, MATH by construction. MATH: MATH . Note, that MATH are the fibres of MATH. The family of projections MATH is continuous, and MATH becomes a locally trivial vector bundle. Thus, MATH. |
math-ph/0002045 | Because of the categorical equivalence by NAME and NAME it is sufficient to indicate the existence and the properties of MATH-valued inner products on finitely generated projective MATH- or MATH-modules. For MATH the MATH-valued inner product is defined as MATH. For direct summands MATH of MATH we reduce this MATH-valued inner product to elements of them. The relation between two MATH-valued inner products follows from an analogue of NAME 's representation theorem for MATH-linear bounded module maps from MATH into MATH. (Attention: This may fail for more general MATH-modules with MATH-valued inner products.) If MATH is a smooth manifold, then the MATH-valued inner product defined above maps elements with smooth entries to smooth functions on MATH. A perturbation of the MATH-valued inner product by a positive invertible operator MATH that preserves the range MATH of it or the restriction to a direct summand of MATH do not change this fact. |