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math-ph/0106024 | To prove the limits REF , start by noticing that MATH (Compare REF for proof.) Suppose MATH. If MATH is even and MATH, then by REF MATH . By the lemma we know the summand converges to MATH for each MATH, as MATH. But we may also bound the summand by a summable sequence. Namely, for any MATH, MATH and MATH for any MATH and MATH. So the summand is bounded (uniformly in MATH) by MATH, which is summable. Thus by DCT the limit exists, that is, MATH . Hence MATH . Since MATH (compare REF keeping in mind that here we rescaled the bi-infinite kink states so that MATH), we have the desired result when MATH is even. The other limit is proved just the same, except using odd integers and the translation MATH. |
math-ph/0106024 | The fact that MATH and MATH are annihilated by MATH follows trivially from the fact that MATH and MATH are annihilated by each pairwise interaction MATH. So, in fact these states are frustration-free ground states. Next, MATH by REF . We observe that each MATH is an orthogonal projection. Moreover MATH commutes with MATH for every MATH and MATH. So MATH . But MATH, which proves REF . |
math-ph/0106024 | First, MATH because MATH and MATH. It is also clear that MATH, and MATH, in agreement with REF . Because MATH and MATH are eigenvectors of the self-adjoint operator MATH, all that remains is to check that REF holds on MATH. But this is true by REF , since MATH and MATH on MATH. |
math-ph/0106024 | Define MATH. Define MATH an infinite matrix such that MATH. Let MATH be an orthonormal family in any NAME space, and let MATH. Then MATH and MATH. For simplicity let MATH, and let MATH. We consider MATH. Then we calculate MATH . Since MATH, this shows that MATH is bounded and MATH is invertible. Under the invertibility condition, it is true that MATH is also invertible on MATH, and considering this as its domain, MATH. If we let MATH and MATH operate on proper superspaces of MATH and MATH, then they will be identically zero on the orthogonal complements. But it is still true that MATH . In particular, if we let MATH be the orthogonal projection onto MATH, then MATH . This proves REF . To prove the second part, let MATH be a state in MATH. Let MATH. Then MATH . We calculate MATH . Breaking the sum into two pieces yields, for any MATH, MATH . So, using REF , we have MATH for any nonzero MATH. This proves REF . |
math-ph/0106024 | Partition MATH into MATH intervals MATH each of length MATH. If MATH then MATH . By REF , MATH . So MATH . In other words, MATH for some MATH. Since MATH, MATH. Let MATH, then REF holds. Note that for any orthogonal projection MATH and any operator MATH we have the decomposition MATH . If MATH is nonnegative, then MATH is as well. Hence MATH . On the other hand, it is obvious that MATH which implies MATH for any nonzero MATH. Moreover, MATH . In our particular case, where MATH and MATH, REF imply MATH . All that remains is to calculate MATH. Notice that MATH and that MATH commutes with MATH for all MATH except MATH and MATH. (We define MATH.) Straightforward computations yield MATH and MATH where MATH and MATH. It is easy to deduce that MATH is zero unless MATH. (MATH has a tensor factor MATH and MATH has a tensor factor MATH, which implies MATH is zero unless MATH. The term MATH is treated similarly.) Another straightforward computation yields MATH and MATH . So MATH where MATH. In particular MATH, so that MATH and MATH . Thus MATH, which along with REF proves REF . |
math-ph/0106024 | Since MATH, it is clear that MATH . So REF implies REF . To prove REF notice that for any operator MATH, any orthogonal projection MATH, and any nonnegative operator MATH, MATH . So, for any nonzero MATH, MATH . Setting MATH, MATH and MATH we have MATH . Since MATH the corollary is proved. |
math-ph/0106024 | We first prove that MATH . It is easy to see that, just as for the droplets on an interval, MATH where MATH. In fact, using the same tools as in REF, we can calculate exactly, for MATH, MATH . It is verifiable that this satisfies the bounds above. The other expectations MATH are similar. Applying REF , proves REF . Now we prove that, considering MATH acting on the invariant subspace MATH, MATH where MATH. To do this, we use REF . There exists a MATH and MATH such that, if MATH then for any MATH with MATH, REF guarantess the existence of a ``subinterval" MATH satisfying MATH, MATH, and MATH . We can take MATH and MATH. By ``subinterval", we mean that there exists an interval MATH, such that MATH. Without loss of generality, we assume MATH. Next, MATH and MATH. We estimate MATH, first. Of course, MATH, and since MATH, we see that MATH. Then using REF , MATH where MATH. But MATH where by MATH, we mean MATH, and by MATH, we mean MATH. We omit the calculations here. So MATH . Symmetrically, MATH where MATH denotes the spin-flip. But MATH. Also, MATH. So, for any MATH, MATH . REF together imply the corollary. |
math-ph/0106024 | The proof that MATH is essentially the same as in REF. One fact we should check is that for each MATH, MATH. We observe that MATH . But as before, MATH . An obvious fact is MATH . Taking MATH, yields the desired result. We have the usual orthogonality estimates MATH . In fact, the estimate of MATH follows by REF , taking the limit that MATH, and the other estimates are consequences. Applying REF proves REF . For the second part, suppose MATH. Then MATH . Furthermore MATH, and MATH by virtue of the fact that MATH, the total number of down spins in the state MATH, is finite. Essentially the same fact is restated as MATH, where MATH . Let us define MATH where MATH is the droplet state subspace for the finite chain. By REF , MATH . Since MATH in the norm-topology, as MATH, all we need to check is that MATH converges weakly to MATH. It helps to break up MATH into two pieces, MATH and MATH. Define MATH . Note that for any sequence MATH such that MATH, we have MATH . The reason is that MATH because the the left and right interfaces of the droplet in MATH are a distance at least MATH from the left and right endpoints of the interval MATH, and the probability of finding an overturned spin decays MATH-exponentially with the distance from the inteface. For the same reason, for any fixed MATH, MATH. These two facts imply that MATH converges weakly to MATH. Now MATH converges weakly to zero, because every state in MATH has over half its downspins concentrated in the annulus MATH, and the inner radius tend to infinity. This means that MATH, as claimed. Thus, taking the appropriate limits, MATH which finishes the proof of the theorem. |
math-ph/0106024 | For the NAME domain MATH of MATH, we observe that MATH . But the centroid of MATH is REF. Thus MATH . The lemma follows by decomposing MATH into the MATH affine copies of MATH, one for each site, and adding the inequalities obtained from REF . |
math-ph/0106024 | This can be obtained as follows. Let consider the grand canonical probability MATH with MATH where MATH is the i-th one dimensional stick that we are decomposing our volume in, and where MATH is the grand-canonical partition function. Clearly, we have MATH . Define MATH and we have MATH . The idea now is to make use of the local central limit theorem for the probability distribution of the occupation number in the i-th stick (see REF .). Let MATH. For any integer MATH, consider, the probability MATH . Due to the factorization property of MATH, the MATH's are independent identically distributed random variables. For centered i.i.d. random variables MATH with variance MATH, the local central limit theorem guarantees that the random variable MATH is close to a Gaussian in the sense that the quantity MATH fulfills the bounds MATH where MATH is the constant MATH . By applying REF to the centered quantity MATH, we obtain the following bounds on the ratio of probabilities: MATH where MATH . In terms of the non-centered variables MATH we have MATH where MATH is the average number of particles of a REFD stick MATH, in the grand canonical ensemble with chemical potential MATH. From this and REF , we obtain MATH . Note that in case MATH is chosen so that MATH or MATH then we can replace MATH by MATH, with the result that MATH may be replaced by MATH as well. Also, from REF , we have MATH . Using REF (and observing that MATH), we have MATH and MATH . Changing to base MATH then leads to REF. By the derivation of REF, we have the bounds on the variance for the number of particles in a REFD stick: MATH . In conjunction with the remark about replacing MATH by MATH, this gives REF . |
math-ph/0106024 | (Proof of Corollary) It follows from the monotonicity of MATH proved in REF, that the equation MATH always has a unique solution for MATH. Then, REF follows immediately from REF , once we observe that MATH as MATH in the sense prescribed in the corollary. For REF , take MATH, with base MATH, and MATH such that MATH where MATH denotes the largest integer MATH. Then, MATH solving REF , is easily seen to converge to MATH, and REF follows from REF . |
math-ph/0106024 | Let MATH be determined by REF , and define MATH as follows: MATH where MATH. We will obtain the equivalence of ensembles by combining two facts. The first is that MATH is approximately equal to MATH, and the second is an estimate showing that MATH . But first, let us recall the definitions of the expectation of an observable MATH: MATH . Since MATH is an observable localized in MATH, we note that MATH. Moreover, we may decompose the grand canonical state into a superposition of canonical states: MATH . Since MATH commutes with MATH, it does not have off-diagonal matrix elements between these canonical states with all different values of the total spin. Therefore, MATH . Note also, that since we have a decomposition MATH and using the previously described properties, we have MATH . This differs from the definition of MATH only by the final factor, which is a ratio of partition functions hence amenable to our activity bounds. In fact, we have MATH where MATH, which equals MATH for our choice of MATH. Thus we obtain MATH where MATH . Now we use the activity bounds REF , but replacing MATH by its actual value, MATH. We arrive at the bounds MATH where MATH . Therefore, MATH. We now use the triangle inequality and the fact that the exponent is negative to obtain: MATH so that MATH . Similarly, MATH . We will use the NAME inequality to control the expectation term in REF . Specifically, for any MATH, MATH . In REF we show that MATH. One choice for MATH is MATH. This gives the bound MATH . The leading order term in the bound is MATH for fixed MATH, strictly between REF. Also, let MATH which is greater than both MATH and MATH. Then MATH. In particular, MATH, which is to say that MATH. Then, using the triangle inequality, we have MATH . So, defining MATH, the theorem is proved. |
math-ph/0106024 | We begin by calculating how a two-site hamiltonian MATH acts on the perturbed state. We consider the decomposition of our lattice into the relevant bond MATH and everything else MATH. Thus MATH where MATH is the unit vector from REF on the pair MATH, and MATH . Here MATH is as would be defined by REF , but with MATH replaced by MATH and MATH replaced by MATH. For example MATH. But MATH is orthogonal to MATH and MATH, since MATH lies in the sector of total spin REF. And MATH . Now it is straightforward to see MATH where we have defined MATH . Then we may write MATH . Actually, MATH depends on MATH only through MATH. So from here on, we'll write it as MATH, and observe the following: MATH where MATH. Let us estimate the term MATH. We have an inequality MATH (which is actually an equality in the limit MATH for our ansatz). Also, MATH . In fact, using the inequality MATH one may conclude that the error in REF is bounded by MATH. Incorporating this estimate into the inequality of REF , we have MATH . Finally, as MATH, the sum over each MATH becomes increasingly well-approximated by the integral over MATH, we is proved in REF immediately following this proof. The lemma gives us a bound MATH where MATH is the Laplacian and MATH is the maximum radius for the NAME domain. (Note that by its definition, as the trace of the second-derivative tensor, the Laplacian enjoys the bounds MATH which may be combined with error term in REF .) Combining REF gives us the theorem, modulo the term MATH, for which we derive the necessary in REF . |
math-ph/0106024 | For each MATH, MATH . This clearly leads to the bound MATH . From this, the lemma follows easily. |
math-ph/0106024 | Recall MATH . The ratio of partition functions in the equation above is clear: It is the probability of finding one particle shared by the sites of MATH, and MATH particles shared by the sites of MATH, conditioned on finding MATH total particles on MATH. We consider the operator MATH . Then MATH and MATH where MATH, where MATH, and MATH is a bond in the stick containing the origin, which we denote by MATH. The restriction of the state in MATH with MATH spins down is of the form MATH where MATH is any observable commuting with MATH, as is, for example, MATH, and the MATH are non-negative numbers summing up to one. We will now derive an upper bound for MATH, that is independent of the coefficients MATH. We start from MATH where MATH denotes the probability in the ground state with MATH spins down for a one-dimensional system on MATH, the sites of which we label by MATH. Each term in the Right-hand side of REF can be estimate as follows. MATH REF gives the following bounds MATH . Combining these inequalities and summing over MATH yields MATH for all MATH. Together with REF this concludes the proof. |
math-ph/0106024 | The proof of REF is a direct consequence of the equivalence of ensembles because, since MATH is a unitary operator, MATH. Let us now consider the proof of REF . We wish to bound the denominator from below; that is, to demonstrate that MATH is not too small. This is tantamount to showing that MATH is not too close to REF. Furthermore, we know this quantity lies between REF. We estimate the actual canonical average with the grand canonical average, and take the logarithm in order to exploit the factorization properties of the grand canonical ensemble. First, we note MATH . Recall the definition MATH. This allows us a more convenient form in place of REF MATH . We partition the product over planes and estimate the logarithm, thus: MATH . We may approximate MATH by MATH, with an error no larger than MATH which is the same as MATH. In this case MATH . We may approximate the sum over MATH with an integral such that the error is bounded by MATH. We may bound the sum MATH from below by its largest term (since all the terms are positive). The largest term occurs for that integer MATH which is closest to MATH. Thus, defining MATH, we see MATH . Using these bounds, we may continue the estimate of MATH. We arrive at MATH . Since MATH and since MATH, we have REF . |
math-ph/0106024 | The periodicity of MATH follows directly from its definition. To prove REF , define MATH for MATH as MATH . Then, MATH . And clearly the remainder term MATH satisfies REF . For the bounds, we first restrict ourselves to MATH. For MATH, we note that REF implies MATH . Then we use REF in combination with this bound to also get the upper bound for MATH. MATH . Due to the peridicity REF , the upper and lower bound are automatically extended to all real MATH. The special values stated in REF are straightforward from REF . |
math-ph/0106024 | We observe that for MATH, MATH where for the square root we use the principle branch (see REF ). This is well-defined, that is, in a single connected, simply connected domain not intersecting the branch cut, since MATH. From this it follows that MATH . We also have MATH . So, for MATH, and MATH, and any MATH, MATH . So, to estimate MATH, we have to estimate two things: the integral MATH and the prefactor MATH. We estimate the integral first, since this will lead us to a particular choice of MATH, from which it will be easy to extract MATH. The point to estimating the integral is to show that the largest constribution comes from a very small neighborhood of MATH, roughly of length MATH. For this reason, for any MATH such that MATH, we break the integral up as follows MATH . We estimate the large MATH-integral first. For MATH, MATH . Hence MATH . Hence, MATH . We define MATH . Then MATH . Hence MATH . This means that an exponentially small portion of the integral comes from the large MATH range. However, we must choose MATH, otherwise our estimate is order REF. Next, we consider small MATH. First we estimate MATH for MATH. Suppose, first, MATH, MATH and MATH. Then MATH and MATH . Of course, for MATH, MATH so MATH . Thus, MATH . So MATH . We will see that for the exponential, the proper scaling of MATH comes from the definition MATH . Then the arument of cosine will be divergent as MATH, unless MATH, that is, MATH. This is the condition for a stationary phase, because it is the condition that makes the phase, that is, the argument of cosine, become MATH as MATH. With this choice of MATH, MATH . So MATH . By symmetry, this same condition is true for any signs of MATH and MATH, as long as MATH and MATH. Implementing the scaling we mentioned before, MATH . Then, for MATH, MATH . This equation will be useful to us, shortly. Next we derive upper bounds for the exponential term. For MATH, MATH . Thus MATH . This means MATH and MATH . Hence MATH . Then, incorporating REF , MATH . Now we come to lower bounds for the exponential term. For MATH, MATH . Since MATH we have MATH . Thus, MATH which implies, using the scaling for MATH, MATH . Then, employing REF , MATH where MATH . Combining this with REF , gives MATH . Finally, we come to the prefactor MATH. We observe that MATH . Then we see MATH . On the other hand, MATH . Thus, MATH . Puting REF together yields REF . |
math-ph/0106024 | . MATH the last equation following by REF . Defining MATH, we have MATH . |
math-ph/0106024 | We observe that MATH where MATH is the obvious isomorphism, induced by the shift MATH, MATH. This means MATH . Also, by its definition, MATH . We view these two sums as expectations on a discrete probability space. Specifically, define two probability measures MATH, MATH on the discrete space MATH given by MATH . For any random variable MATH on MATH, any subset MATH, and any finite (signed) measure MATH on MATH, let MATH be the expectation value, that is . MATH . Then MATH . For a fixed MATH, we decompose MATH into two subsets MATH . From REF , we know that MATH is small for large MATH, and in particular it is decreasing like MATH to the quadratic function MATH. This is a faster decay rate than the exponential factor in the grand canonical expectation. Thus, if MATH is large enough then MATH is quite small. Specifically, MATH . We define MATH. Next, we observe that for any bounded random variable MATH on MATH, MATH where for any signed measure, MATH, the absolute value of MATH is the nonnegative measure MATH . Then by the triangle inequality MATH. So, MATH where as usual MATH is defined when MATH by MATH . By the definition of MATH, MATH, and therefore MATH. We define MATH . On the other hand, we also know MATH. So MATH . Thus, we obtain MATH so that MATH . In particular, this implies MATH since for any MATH, MATH . We already have a bound for MATH, all we need to do now is bound MATH. This is where the activity bounds come in. We observe MATH . Now, by REF , MATH where MATH . Similarly, MATH where MATH . Also, from REF , we have MATH while simply, from the definition of MATH, we have MATH . Puting it all together, we have MATH . Taking the supremum and infemum over MATH, yields the stated bound for MATH. |
math-ph/0106028 | The projection MATH belongs to MATH and MATH is equivalent to the GNS representation of MATH. If the central support of MATH in MATH is equal to MATH, then the homomorphism (induction) MATH is an isomorphism that gives the quasiequivalence between MATH and MATH. Conversely, quasiequivalent representations have equivalent amplifications. |
math-ph/0106028 | Let MATH be the operator on MATH corresponding to the quasifree state MATH via REF , and let MATH be REF-particle NAME space, as defined in REF . In addition, let us define the following closed, real linear subspaces of MATH: MATH . We note that MATH and it is not difficult to see that MATH if and only if MATH is a projector. By REF , we can write MATH . By an argument due to CITE and NAME, we know that REF holds if and only if MATH, where ``prime" denotes the symplectic complement of a set in MATH, defined as MATH . Let now MATH. We must show that MATH. Now, since MATH, we have that MATH, which means that MATH for all MATH. Moreover, since MATH, it is not difficult to see that MATH for all MATH. Altogether, this means that MATH . Consequently, we know that MATH . Because the boundary of MATH is smooth, it follows easily that the space MATH is dense in MATH in the norm induced by the scalar product REF. This implies that MATH or equivalently that MATH. Let us now introduce an auxiliary quasifree state MATH whose corresponding operator MATH is a projection, so that MATH holds for the corresponding REF-particle space. Let us write MATH, where MATH with MATH. Then, since MATH it follows that the limit MATH exists in MATH. If MATH, with MATH an arbitrary element of MATH with MATH, then REF implies that MATH where MATH. It follows from this that MATH for all MATH, and therefore that MATH. But, by REF, this also implies that MATH, which proves the proposition. |
math-ph/0106028 | Let MATH be the unitary operator on MATH implementing the automorphism MATH, MATH. By the remark following REF , we know that the MATH-invariant (that is, even) part of of MATH is trivial. The central support of MATH is invariant under MATH, hence it is MATH. The result now follows from REF . |
math-ph/0106028 | As explained above, we only need to prove the quasiequivalence of the representations MATH and MATH, where MATH is a region of the form MATH. Let MATH and MATH be the self-adjoint operators on MATH associated to the the quasifree states MATH and MATH via REF . The partial states MATH and MATH are then again quasifree states, corresponding to the operators MATH and MATH, where MATH is the projector onto the closed subspace MATH of MATH, given by MATH with MATH the characteristic function of the set MATH. By REF and CITE, the partial states MATH and MATH are guaranteed to be quasiequivalent if MATH where MATH is the NAME norm on MATH, see Sec. CASE: In order to estimate the left side of REF, we recall the the NAME inequality, which says that MATH for any two positive bounded operators MATH and MATH on a NAME space, where MATH denotes the trace-norm, see Sec. CASE: Taking MATH for MATH and MATH for MATH in that inequality, we find that REF follows if we can show that MATH . We now define a bidistribution MATH on MATH associated with the operator MATH by putting MATH. This bidistribution can be expressed in terms of the two-point functions MATH and MATH as follows: By the NAME property combined with a microlocal argument (see for example, CITE), the latter possess a well-defined pull-back to the submanifold MATH, denoted by MATH respectively MATH, where MATH is the embedding map. Using REF , we compute that MATH where MATH is the normal of MATH and where we have used the identity MATH (see CITE) in the last line, with MATH the dual of the restriction map MATH. It is not difficult to see from this that MATH . Since MATH and MATH are by assumption NAME states, we know that MATH is smooth on MATH (see the first remark following REF ). REF therefore implies that MATH can be identified with a smooth section in MATH, with compact support in MATH. Let MATH be a smooth partition of unity covering MATH with the property that the support of each MATH is contained in a coordinate patch. Then we can write MATH, and each MATH can be identified a matrix of smooth, compactly supported functions in MATH via a local trivialization of MATH over the patch corresponding to the pair MATH. As it is well known, a matrix of operators is in the trace class if each matrix entry is in the trace class. Thus, in order to show that MATH is in the trace class, it suffices to show that the matrix elements of each MATH is in the trace class. In fact any smooth, compactly supported integral kernel MATH in MATH is in the trace class, as one can see for example, as follows: We can write MATH where MATH. The multiplication operator by MATH is in the NAME class for MATH sufficiently large since MATH then, and the integral kernel MATH is in MATH for any MATH, and therefore also in the NAME class. Hence, MATH is the product of two NAME operators, and hence in the trace class. |
math-ph/0106028 | The partial state MATH is a quasifree state corresponding to MATH in the notation of the proof of REF . Hence, the algebra MATH is a factor by the results of CITE. By local quasiequivalence, also MATH is a factor. The type of this factor can be obtained by again invoking local quasiequivalence. In fact, the local v. NAME algebras are type MATH factors in a pure, quasifree NAME state by CITE, therefore they are MATH factors for all quasifree NAME states. |
math-ph/0106028 | The first part of REF follows immediately from the definition. In order to show the norm estimate in REF , we notice that REF implies that MATH where MATH. We therefore get the inequality MATH and hence MATH. REF can be demonstrated by adapting a method developed in CITE, we briefly sketch the argument. Let MATH be a function which is equal to one on MATH and let MATH be the corresponding multiplication operator on MATH, defined by MATH. Then one can show that the operators MATH are in the NAME class for all MATH. A proof of this is given in the Appendix of ref. CITE for the case of NAME space; the main steps for a proof of this statement in the case when MATH is not flat are given in the Appendix of this paper. Now it trivially follows from REF of MATH that one can write MATH . Since the operator MATH is bounded for any MATH, this shows that MATH can be written as the product of an arbitrary number of NAME - NAME operators. This implies that MATH for all MATH and hence MATH for all MATH, thus proving REF. The proof of REF is outlined in the Appendix. REF is equivalent to the statement that the range of the map MATH is dense in MATH, which in turn is equivalent to the statement that the NAME state on MATH given by MATH has the ``NAME. That this property holds in the situation under consideration has been shown in CITE. |
math-ph/0106028 | REF is just REF . The estimate for MATH follows from this and REF . In order to prove the last statement of the proposition, we note that the nuclearity index in the rescaled spacetime with Killing field MATH is equal to the nuclearity index in the unscaled spacetime at inverse temperature MATH and mass MATH. Since MATH does not change under these rescalings, this means that the nuclearity index for the rescaled spacetime is given estimated by the same expression as for the unscaled spacetime, but with MATH replaced by MATH. |
math-ph/0106028 | Our proof follows the chain arguments given by CITE in the context of a linear NAME field, thereby using the results obtained in the previous section and some results obtained in CITE. Let us first consider the special case when MATH is ultrastatic and when MATH is the ground state. In that case we can define a map MATH as in REF , and we know by REF that this map is nuclear with a nuclearity index estimated by the bound given in REF . It can be shown (see CITE) that such a bound entails the split-property, thus proving our theorem in the case when MATH is the ground state in an ultrastatic spacetime. In order to generalize this statement from an ultrastatic spacetime to an arbitrary globally hyperbolic spacetime (with spin-structure), one now makes use of the following facts: CASE: The ground state on an ultrastatic spacetime is of NAME form. CASE: More generally, if MATH is a spacetime which is ultrastatic in an open neighborhood of some NAME surface MATH, then the state obtained from the ``ground state construction on MATH" is NAME throughout MATH. REF is a corollary of the results obtained in CITE. If the NAME surface MATH is compact, then one can obtain the following alternative, somewhat more direct proof of REF using a construction of Junker CITE (and using ideas of CITE to make Junker's construction applicable to the NAME case). We first write the projector MATH on the positive spectral subspace of MATH as MATH. The operators MATH and MATH can be defined for every NAME surface MATH perpendicular to the timelike Killing vector MATH, and can thereby act on smooth spinor fields MATH on MATH. This makes it possible rewrite the two-point function REF of the ground state on an ultrastatic spacetime as MATH where in the third line we have used that MATH for all smooth spinor fields MATH on MATH, by the NAME equation. The key observation is that in an ultrastatic spacetime, we have that MATH where MATH is the NAME scalar. The operator MATH is the self adjoint extension on MATH of the elliptic differential operator MATH, where MATH is the induced (negative definite) Riemannian metric on MATH. Therefore, by standard results about powers of positive elliptic pseudo differential operators on compact manifolds (see for example, CITE), MATH is a pseudo differential operator of order REF on MATH with principal symbol MATH, where MATH is the identity map in the fibers of MATH, and MATH is a pseudo differential operator of order MATH. REF therefore provides a presentation of the two-point function MATH to which REF is applicable, and we conclude by that theorem that MATH, thus showing that MATH is of NAME form. The above arguments do not apply as stated to the case of non-compact MATH, since it is not guaranteed then that MATH is a pseudo differential operator. This can presumably be shown provided that the spatial metric MATH satisfies suitable fall-off conditions at infinity, but would require a considerable effort. REF follows immediately from REF , together with the second remark following REF . It is known that if the split property holds for a state MATH, then it also holds for any other state that is locally quasiequivalent to MATH. Therefore, since the ground state on an ultrastatic spacetime is NAME by REF of the above lemma, and since all NAME states are locally quasiequivalent by REF , we conclude that the split-property holds for any quasifree NAME state on an ultrastatic spacetime. More generally, by the same argument and REF of the above lemma, it follows that if a spacetime has a NAME surface MATH with an ultrastatic neighborhood, then the split property holds for any quasifree NAME state and any pair of concentric double cones MATH and MATH whose bases are in MATH. In order to further generalize this to arbitrary quasifree NAME states on an arbitrary globally hyperbolic spacetime, we now employ a deformation argument identical to the one given in CITE. For this, one notes that one can construct, besides the original, arbitrary globally hyperbolic spacetime MATH (with spin structure), a deformed spacetime MATH with spin structure, with the following properties: CASE: MATH possesses a neighborhood MATH of some NAME surface MATH which is isometric to some neighborhood MATH containing a NAME surface MATH in the deformed spacetime MATH. We chose the spin structure of MATH so that its restriction to MATH coincides (via the isometry) with the restriction of spin structure of MATH to MATH. CASE: The spacetime MATH is ultrastatic in a neighborhood of a NAME surface MATH. CASE: If MATH are sets as in the hypothesis of the theorem, then there exist open subsets MATH with compact closure such that MATH and MATH where the sets MATH correspond to MATH and MATH corresponds to MATH via the isometry postulated in REF. Now let MATH be an arbitrary quasifree NAME state on MATH for the original spacetime, and let MATH be the uniquely determined quasifree NAME state for the deformed spacetime, whose two-point function coincides with that of MATH in MATH via the isometry identifying the neighborhoods MATH and MATH described in REF . The trick is now to use the split-property for the deformed spacetime, MATH . (which is already known, by REF ) and REF above and by what we have said so far to prove the split-property in the undeformed spacetime. By REF above, we have that MATH therefore we have by REF that MATH . Since MATH agrees with MATH on MATH, this therefore shows the split-property on the undeformed spacetime. |
math-ph/0106028 | Since MATH is a positive operator in MATH, we can calculate its trace norm by (here and in the following we use the constant convention, meaning that different numerical constants, are denoted by the same symbol MATH): MATH . This proves the first inequality. For notational simplicity, we will only explicitly prove the second inequality for the case MATH, but our arguments can be generalized straightforwardly to deal with general values for MATH. We first introduce some notation. Let MATH be smooth functions of compact support in MATH such that MATH on the support of MATH, MATH on the support of MATH, MATH on the support of MATH, etc. Furthermore, let us set MATH so that MATH are partial differential operators of degree REF. Let MATH denote the norm on the NAME space MATH, and let MATH be an arbitrary smooth spinor field. Since MATH is elliptic, we obtain the estimate: MATH . If we move the operator MATH through MATH, then this can be further estimated by MATH where we have used that MATH in the last line, and where MATH is an arbitrary but fixed natural number. We may repeat the above chain of inequalities for the second expression in the last line, replacing MATH by MATH and MATH by MATH. If we iterate this procedure, we get MATH where neither of the above constants depend on MATH, but only on MATH and MATH. This inequality implies that MATH is a continuous map from MATH to MATH for every MATH. By duality, MATH is a continuous map from MATH to MATH for every MATH. Therefore, MATH is a continuous map MATH for every MATH and therefore has a smooth (and by definition compactly supported) integral kernel on MATH. From the dependence upon MATH of the constants in the above inequality, one gets furthermore that MATH . This estimate can now be used to bound the MATH-th NAME norm MATH of MATH by writing MATH as a product of an arbitrary number of NAME operators as in REF , MATH where the MATH is chosen sufficiently large. The multiplication operator by MATH is then NAME, and MATH using REF . Hence MATH for all MATH and MATH, where MATH depends only on MATH. The chain of inequalities leading to REF can be repeated for arbitrary MATH. It follows in particular that MATH (choosing MATH). This the claim of the lemma for MATH. |
math/0106001 | Let MATH be a PROP action. Then MATH defines a cyclic algebra structure on MATH. NAME, let MATH be a cyclic algebra. Pick any ribbon graph MATH and realize it as an NAME; call this NAME MATH. Give MATH the structure of a MATH-colored NAME by coloring all the edges of MATH with MATH and coloring any vertex MATH with MATH. Denote the MATH-colored NAME obtained this way by MATH. Two realizations of MATH as an NAME differ by a finite sequence of moves MREF, MREF. Since MATH is self-dual, the graphical calculus MATH is independent of the orientation on the edges, that is, it is invariant with respect to the move MREF. Moreover relations provided by REF give the invariance of MATH with respect to moves of type MREF: MATH . Therefore, MATH is well defined and is a PROP action. |
math/0106001 | An action MATH endows MATH with a symmetric algebra structure, as in the proof of REF. Conversely, assume we are given a cyclic algebra structure on MATH. Since the tensors MATH are symmetric, they are, in particular, cyclically invariant: so MATH is a cyclic algebra and an action MATH is defined. Since the tensors MATH are symmetric, this action factors through an action MATH. |
math/0106001 | By linearity of the MATH's and REF, MATH . If MATH is odd, MATH is zero, and the set of graphs considered in the statement is empty. If MATH is even, according to graphical calculus rules, MATH corresponds to the juxtaposition of MATH univalent vertices, MATH bivalent vertices, etc., up to MATH vertices of valence MATH. In this case, REF translates MATH into edges connecting these vertices in all possible ways. |
math/0106001 | Let MATH be the set of all ribbon graphs obtained by: CASE: juxtaposing MATH vertices of valence MATH, MATH vertices of valence MATH, etc., up to MATH vertices of valence MATH, and, CASE: connecting them in all possible ways by means of arcs. The constant MATH counts the number of occurrences of graphs isomorphic to MATH in the set MATH. The semi-direct product MATH acts on MATH as follows: the image of a graph MATH is obtained by permuting vertices of the same valence and rotating edges incident to each vertex. Since this action is transitive on isomorphism classes, MATH where MATH is the stabilizer of MATH under the action of MATH. |
math/0106001 | Expand in NAME series the left-hand side: MATH by REF . |
math/0106001 | Exponentiate MATH to find: MATH where each MATH is a connected graph. Now recall that juxtaposition defines a tensor product MATH in the category of graphs (compare REF) and that MATH is multiplicative with respect to this structure: MATH . Therefore, MATH . For a graph MATH having MATH connected components MATH. Let MATH be the set of all possible juxtapositions of MATH; all graphs in MATH are isomorphic to MATH. The semi-direct product MATH of MATH and MATH acts transitively on MATH; the stabilizer of any element is isomorphic to MATH. Therefore, MATH . So we reckon: MATH . |
math/0106001 | To evaluate MATH we need an explicit expression for the NAME element MATH of the cyclic algebra MATH. Since MATH is Hermitian positive definite, there exists an orthonormal basis MATH of MATH in which MATH for some MATH positive real numbers. Any choice of a like basis induces an identification of MATH with MATH, and, consequently, of MATH with the space MATH of MATH complex matrices. Let MATH be the canonical basis for MATH: MATH . A basis for MATH is given by matrices MATH . It is orthonormal with respect to the inner product MATH, whereas MATH that is, the matrix of MATH with respect to the basis MATH is MATH . So we get the following expression for the NAME element: MATH . Rewrite this identity as: MATH but, for MATH, MATH . So, MATH . According to standard graphical calculus rules, evaluation MATH is performed through the correspondence MATH . If we introduce the notation MATH then we can depict REF as MATH which turns MATH into a sum of ribbon graphs equipped with a number in MATH on each side of every edge, and operators MATH on each MATH-valent vertex. The map MATH is the restriction of a map MATH defined on MATH, namely, the trace of a MATH-fold product. We have MATH . Therefore, the only graphs that give non-zero contribution to the sum giving MATH are the ones whose boundary components have the same index on all edges - that is, we need only account for graphs equipped with a map MATH. An edge whose sides are indexed MATH brings in a factor MATH; combining this with REF, we can conclude the proof. |
math/0106008 | We first reduce the case of arbitrary MATH to the special case MATH. To this end let MATH be a positive function such that MATH, MATH, for all MATH. Multiplication by MATH induces isomorphisms MATH with inverse induced by MATH. Therefore, MATH . But for large MATH, MATH where MATH. Since MATH and MATH are of the same quality as MATH and MATH, respectively, and MATH and MATH, we can assume from the very beginning that MATH. The term MATH certainly behaves in the right way, since it is rapidly decreasing in MATH. Also the term MATH is good by the standard NAME theorem. The two remaining terms MATH and MATH we shall consider in the spaces MATH . If MATH is the group action from REF, then MATH for all MATH. Hence for an arbitrary operator MATH we have MATH . Now let MATH have a kernel MATH as described in REF (with MATH replaced by MATH and MATH). Then the operator-norm of MATH is the same as that of MATH, which has the kernel MATH. But this kernel is MATH in MATH, compare REF. To treat the last term we can pass to local coordinates, that is, we assume MATH and MATH . By a tensor product argument we can assume that MATH is independent of MATH. Conjugating with MATH, we have to show that MATH is uniformly bounded in MATH for large MATH. Since then MATH is bounded away from zero and infinity, a simple calculation shows that MATH uniformly in MATH, that is, MATH uniformly in MATH. Then the result follows, see the end of REF. |
math/0106008 | Let MATH denote the symbol in question. For shortness let us write MATH and MATH. Since the eigenvalues of MATH are proportional to MATH (uniformly for MATH) and do not lie in MATH, there exists a constant MATH such that MATH is a smooth function in MATH . Thus, if we choose MATH in such a way that MATH vanishes for MATH and MATH, then MATH is smooth on MATH for MATH. To verify that MATH is a symbol, it suffices to show that MATH uniformly in MATH and MATH. Since MATH is anisotropic homogeneous of order MATH for large MATH, estimate REF holds on MATH. It also holds for MATH and MATH sufficiently large, since then MATH is MATH due to the above described behavior of the eigenvalues. For MATH and MATH simultaneously small, estimate REF holds anyway. |
math/0106008 | By conjugation with MATH we can assume that MATH (compare the proof of REF ). If we split the integral into three terms according to the decomposition of MATH in REF, the integral over MATH can be estimated in the desired way, since MATH is bounded on MATH. By symmetry, MATH and MATH can be treated in the same way. So we shall assume for the rest of the proof that MATH (for notational convenience we replace MATH by REF). Also for convenience we suppress the MATH-variables from the notation. We shall frequently make use of the fact that, substituting MATH, we have MATH . According to REF , MATH for MATH is an integral operator (with respect to the scalar product in MATH) with kernel MATH where, for some MATH, MATH (the fact that MATH is classical will not play a role for the following calculations). Then MATH is an integral operator with kernel MATH . Writing MATH with the characteristic function MATH of MATH, the proposition will be true, if we can show that in any of the four cases MATH the associated integral operators are bounded in MATH, uniformly in MATH for some MATH. REF are equivalent by symmetry (that is, passing to the adjoint). The proofs of all REF rely on the following NAME inequalities: MATH for any non-negative function MATH on MATH and MATH (compare REF , page REF). To begin with REF we use the fact that, for some fixed MATH, MATH uniformly in MATH and MATH, to obtain MATH . Since MATH is negative, the factor MATH is uniformly bounded by REF for MATH. If MATH the factor MATH can be estimated from above by a constant uniformly in MATH. Since furthermore the kernel function MATH belongs to MATH and thus induces a continuous operator in MATH, it remains to consider the kernel MATH. Because this kernel is symmetric in MATH and MATH, indeed it suffices to treat MATH . If MATH denotes the associated integral operator, then MATH by NAME 's REF . This finishes REF . For REF observe that MATH for any MATH uniformly in MATH and MATH. Then MATH . This expression equals zero if MATH and for MATH we can estimate MATH with kernel functions MATH and MATH given by MATH . In order to check the uniform boundedness of the integral operator MATH associated with MATH, MATH, on MATH we observe that MATH with MATH. The boundedness of MATH is equivalent to the boundedness of MATH on MATH. To show it, we employ NAME 's lemma: if MATH is sufficiently large, then MATH and MATH . To handle MATH first observe that we can drop the factor MATH, since this is uniformly bounded by REF in MATH and MATH, and if MATH and MATH then MATH for MATH (for MATH anyway MATH). Thus we can assume that MATH . But then NAME 's REF shows that the integral operator associated with the kernel MATH is continuous in MATH with operator norm bounded by MATH. This finishes REF . For the final REF we use that MATH for any MATH uniformly in MATH and MATH. Then MATH . The factor in front is obviously uniformly bounded in MATH for MATH sufficiently large. Since MATH is negative, MATH . All these kernel functions belong to MATH for sufficiently large MATH and thus induce continuous operators in MATH. Hence it remains to investigate MATH and by symmetry even MATH . Again NAME 's REF shows that the associated operator is MATH-continuous. |
math/0106008 | Without loss of generality, we can set MATH. We have to show that MATH is uniformly bounded for MATH, MATH, MATH and MATH. The totally characteristic derivatives in MATH can be handled very simply, observing that MATH, MATH and both symbols MATH and MATH are of the same type as MATH. Since the derivatives with respect to MATH, MATH and MATH can be taken under the integral sign, it suffices to assume MATH and to show that MATH uniformly in MATH, MATH, MATH and MATH. By hypothesis, we have MATH and on MATH we can estimate MATH from above by MATH. The transformation MATH yields MATH where MATH means the path MATH. Since the support of MATH is compact, we may assume without loss of generality that MATH. Then we obtain the estimate MATH and the statement follows, since MATH is uniformly bounded in MATH. |
math/0106008 | Without loss of generality let MATH. We shall also suppress MATH from the notation and instead assume that MATH. We can also assume MATH confined to a compact subset of MATH. By REF , MATH where MATH denotes a local symbol of MATH as defined in REF . In view of REF , we therefore may assume that MATH where we have used the substitution MATH. We have to estimate this expression as in REF . The factor MATH behaves correctly, since MATH is uniformly bounded in MATH and MATH. For MATH we have MATH with suitable constants MATH and MATH. Thus for large enough MATH we have MATH and the spectrum of MATH is located to the right of the path MATH. By NAME 's theorem we can then replace the path MATH by MATH, and obtain for large MATH . Then we can estimate (as in REF ) MATH with a polynomial MATH. However, since we can replace MATH by MATH for some MATH (as noted in the comments on REF ), this yields the uniform symbol estimates of MATH for large MATH. For small MATH, we now shall show that MATH where we have set MATH and MATH is the path given in the following picture (with MATH to be chosen appropriately): In fact, the difference of both sides from REF equals MATH . Since, for small MATH, the spectrum of MATH is contained in some ball of finite radius, MATH is holomorphic in MATH for MATH, if MATH is chosen large enough. Thus MATH for all MATH, by NAME 's theorem. For any fixed MATH and MATH the integrand in REF is MATH for MATH and, on the radial part of MATH, the integrand is MATH. Hence, MATH, and REF holds. To estimate the right-hand side of REF, we split the integral into four parts, which we briefly analyze separately. First of all, observe that MATH can be estimated from above by MATH on the whole path. This and the fact that MATH are enough to get the desired estimates for the terms obtained integrating along the two arcs MATH and MATH, since they can be treated with essentially the same technique we used to prove REF . The term obtained integrating along MATH is MATH . The derivatives with respect to MATH, MATH and MATH can be taken under the integral sign, so that we could again start with a symbol MATH and prove that, for any MATH, MATH uniformly in MATH. This is true for MATH, as one can easily check. For MATH we get MATH and this also satisfies the desired estimate. In fact, the first term is of the same kind as MATH, while for the second it suffices to use the definition of MATH and the fact that MATH. The result for arbitrary MATH can be proved by induction and, obviously, the contribution obtained integrating along MATH behaves in a completely similar way. This yields the desired symbol estimates for small MATH and finishes the proof. |
math/0106008 | Let us set MATH. The rescaled symbol of MATH is MATH where MATH refers to the metric MATH on MATH. Hence MATH satisfies the ellipticity REF for any MATH in question. The conormal symbol of MATH, compare REF, is MATH . If MATH are the eigenvalues of MATH, then MATH is not bijective if and only if MATH . Note that, in particular, MATH is invertible for all MATH with MATH, and thus by REF for all MATH with MATH. This shows that MATH is elliptic with respect to MATH in the sense of REF and has only one closed extension MATH by REF . The model cone operator is MATH that is, MATH is the Laplacian on MATH with respect to the metric MATH. As before, MATH has a unique closed extension MATH . Since MATH is symmetric and non-negative, MATH is self-adjoint and MATH. Let us show that MATH . By REF (in the version for operators in the cone algebra MATH on MATH, which is introduced in REF), the spectrum of MATH is independent of MATH. Thus we can set MATH and write MATH. We can assume MATH, by passing to the adjoint (that is, MATH and MATH). Then MATH, since MATH in view of MATH. The fact that MATH is invertible for MATH implies that MATH we shall give the argument below. As a consequence, we have for the adjoint MATH hence MATH is bijective for MATH. In order to see REF set MATH. The invertibility of the conormal symbol implies that MATH is elliptic with respect to MATH. Moreover, the minimal and maximal extensions of MATH, considered as unbounded operators in MATH, coincide and their domain is MATH. In particular, MATH is a subset of the maximal domain, thus it is included in MATH. Iterating this process, we see that MATH for all MATH. Choosing MATH large enough we get REF. |
math/0106008 | In REF , it is shown that then MATH, where the right-hand side denotes the space of bounded continuous functions on MATH, multiplied by MATH. Thus MATH where the first constant MATH is the maximum of the norm of the total derivative of MATH, as MATH varies over the bounded set of all values of MATH, MATH. |
math/0106008 | The proof is similar to that of REF . For the convenience of the reader we provide the details. Choose a cut-off function MATH and MATH with MATH. Then MATH. We first focus on the analysis near the boundary, that is, we show that MATH is NAME continuous on bounded sets. We abbreviate MATH and MATH, MATH, for MATH. Since MATH and MATH have their support near MATH, we have according to REF MATH . Here, we chose MATH, and we employed NAME 's inequality. Next we use the embedding MATH, valid for MATH, compare CITEEF . Since we assumed that MATH we deduce that MATH . The second estimate results from the continuity of MATH; for the third we used that MATH. In the same way we estimate MATH and finally MATH . Next set MATH and MATH and note that the norm of MATH coincides with that of MATH on their supports. Then the estimate MATH plus the fact that the norms of MATH, MATH and MATH can be estimated by the norms of MATH, MATH, and MATH in MATH completes the argument. |
math/0106008 | We have MATH for any MATH and MATH. Choosing MATH close to MATH and MATH large, we have MATH . We can then pick MATH with MATH and MATH and apply REF . |
math/0106009 | This follows from the standard trace argument. |
math/0106009 | It is sufficient to prove this over MATH. In that case every MATH is simple by REF . Then by REF it follows that MATH is semistable (in fact stable) for MATH. |
math/0106009 | Again it is sufficient to do this over MATH. First we note that if MATH then by REF MATH and in particular MATH acts freely on MATH. By REF below MATH is smooth at MATH. Thus the restriction of MATH to MATH is smooth. It follows that the induced map MATH is also smooth. Since MATH acts freely on MATH we deduce that MATH is also smooth. This then yields that MATH is surjective on tangent spaces and hence smooth. |
math/0106009 | We consider the projection map MATH . According to CITE the image of MATH consists of indecomposable representations. Since MATH is indivisible, representations of dimension vector MATH are absolutely indecomposable if and only if they are indecomposable. Thus the image of MATH consists of absolutely indecomposable representations. Let MATH denote the constructible subset of absolutely indecomposable representations in the affine space MATH. It is also shown in CITE that the elements of MATH lift to MATH. More precisely the inverse image of MATH can be identified with MATH. Starting from a variant of the NAME formula we compute MATH where we have used that MATH is the symmetrization of the NAME form on MATH. Since MATH the inequalities defining genericity also hold in MATH. Hence we will assume this. By REF our choice of MATH insures that MATH contains only simple representations. Thus if MATH then MATH and hence MATH has trivial stabilizer in MATH. Using CITE we obtain MATH . |
math/0106009 | Let MATH, MATH be the restrictions of MATH to MATH and MATH. Using the previous proposition and the fact that MATH commutes with base change we find for MATH: MATH on MATH. We may consider MATH and MATH as the MATH-modules given by MATH and MATH respectively. Since the NAME action is determined by the action of MATH CITE this proves what we want. |
math/0106009 | We use NAME 's generic base change result for direct images CITE. This result was only stated for torsion sheaves, but the corresponding result for MATH-adic sheaves is an easy consequence. Since MATH is of finite type there are only a finite number of MATH for which MATH is non-zero. So we may treat each MATH separately. Put MATH. Let MATH be the structure map and let MATH be the map given by adjointness. Let MATH be the kernel and cokernel of MATH. By CITE MATH and hence MATH will be constructible over an open subset MATH. Below we show that MATH is an isomorphism. By CITE we have MATH. Hence MATH. From the fact that MATH and MATH are constructible it follows that MATH and MATH are constructible subsets of MATH whose image in MATH does not contain the generic point. Hence we find MATH for a suitable open MATH. Now we prove our claim that MATH is an isomorphism. To do this we replace the etale topology on MATH with the analytic topology. Then the claim follows from the comparison theorem REF below and the fact that MATH is connected. |
math/0106009 | Let MATH. We will use the hyper-Kähler structure on MATH which was introduced by CITE. For the benefit of the reader we recall the basic facts. First we define a Riemannian metric on MATH via the trace form: MATH where MATH is the conjugate transpose to MATH. Let MATH be the quaternions. We define an action of MATH on MATH via MATH . It is clear that with respect to this quaternionic structure the metric REF is hyper-Kähler. Let MATH be the kernel of the reduced trace map on MATH. If MATH then there is an associated real symplectic form on MATH defined by MATH. Let us write MATH and MATH where MATH is the maximal compact subgroup of MATH given by the product of unitary groups MATH. The hyper-Kähler structure on MATH is clearly MATH-invariant and it is a standard fact that the symplectic form MATH has an associated moment map MATH given by MATH for MATH, MATH. Below we will write MATH for MATH. The three moment maps MATH, MATH, MATH may be combined into a so-called hyper-Kähler moment map MATH . From the explicit description of MATH we deduce for MATH: MATH where MATH is the conjugate of MATH in MATH. From REF we deduce that REF is MATH-invariant if we let MATH act on MATH by MATH. For this action MATH is a homogeneous space and hence if we choose MATH and a contractible subset MATH containing MATH then there is a continuous map MATH which is a section (above MATH) for the map MATH. Choose a MATH-invariant MATH and let MATH, MATH. Then MATH defines a trivialization of MATH. Thus we have proved that above MATH, MATH is a trivial bundle. Moreover this trivialization is clearly MATH-equivariant. Put MATH. This is a complex MATH-invariant symplectic form on MATH and it is easy to see that the associated moment map MATH is given by MATH where we have extended MATH to linear maps MATH. A straightforward computation shows that MATH where we have identified MATH, MATH with their duals via the trace form MATH (the minus sign makes the form positive definite on MATH). From the description MATH we obtain: MATH which yields MATH . From the fact that MATH is contractible we deduce as explained above that MATH is trivial above MATH. Since on the inverse image of MATH, MATH and MATH are basically the same we deduce that MATH is a trivial family in a way that is compatible with the MATH-action. We now use this to construct the following commutative diagram of continuous maps: MATH . Here MATH is obtained from the trivialization of MATH we have constructed above (recall that MATH) and MATH are obtained from the inclusion MATH CITE. To prove the lemma it is now sufficient to show that the vertical maps on the left are homeomorphisms. This is true by construction for MATH. We claim that it is also true for MATH. It suffices to consider MATH since MATH is obtained from MATH by restricting to a fiber. By CITE MATH is a bijection. Hence it suffices to show that MATH is proper. Clearly MATH is the restriction to MATH of the first map in the following diagram MATH . By REF below the composition of these two maps is proper. It follows that the first map is also proper. This finishes the proof. |
math/0106009 | By the existence and uniqueness of the NAME REF. follows from REF. Hence we only have to prove REF. Let us call a subset MATH of MATH good if it has the following properties. CASE: The elements of MATH have constant NAME. CASE: MATH is constructible. CASE: MATH has a dense intersection with a unique irreducible component of MATH. By induction it is clearly sufficient to prove the following claim: Claim. Let MATH be semistable and let MATH be good. Assume that MATH for MATH arbitrary. Define MATH as the set of all MATH which contain a semistable subrepresentation MATH such that MATH, MATH. Then MATH is good. The only non-obvious property to prove is that MATH has a dense intersection with a unique irreducible component of MATH. So this is what we do below. Let MATH be the semi-stable locus of MATH and let MATH be the set of REF MATH with MATH, MATH, MATH, MATH, MATH such that MATH is exact. It is easy to see that MATH is a constructible subset of MATH. Due to the uniqueness of the NAME the non-empty fibres of the projection map MATH are isomorphic to MATH and hence they have dimension MATH. There is another projection map MATH. According to CITE its fibers have dimension MATH and the proof also shows that these fibers are irreducible and locally closed. According to CITE we also have MATH and furthermore according to REF we have MATH. Substituting we find that the fibers of MATH have dimension: MATH . According to REF below we find that MATH contains a dense irreducible locally closed subset MATH such that MATH. Furthermore the dimension of MATH is: MATH . Now we have for MATH: MATH A trite computation shows that MATH has dimension MATH and MATH has smaller dimension. Hence MATH. Since MATH is irreducible it follows that MATH is dense in some irreducible component MATH of MATH. This finishes the proof. |
math/0106009 | Left to the reader. |
math/0106009 | By the NAME we have MATH . In this formula we may collect the MATH's with equal slope. Let MATH be given by the righthand side of REF. Then we have MATH and by induction it follows MATH. This finishes the proof of REF. |
math/0106009 | Recall that MATH is such that MATH and MATH for all MATH. Now it is clear that NAME semistability for MATH is equivalent to slope semistability for MATH. Hence for this particular MATH we need to show that MATH. This follows immediately from REF. |
math/0106009 | It is clearly sufficient to show that the action of MATH on MATH has a unique eigenvalue MATH and that in addition MATH. Write MATH and MATH for MATH odd. Since MATH is pure the eigenvalues of MATH acting on MATH are given by MATH where MATH and MATH. From the hypotheses and the trace formula we obtain MATH where MATH. Dividing by MATH we find MATH . Using a NAME der Monde type argument we see that the limit on the righthand side only exists if MATH for all MATH. Subtracting the leading term in MATH from REF and repeating the same argument we ultimately find that MATH for all MATH. Since MATH for odd MATH we find that MATH for odd MATH. This finishes the proof. |
math/0106009 | Let MATH be the decomposition into connected components and for each MATH define MATH . According to CITE the MATH, MATH are smooth and the MATH are locally closed in MATH. Furthermore the limit map MATH is a NAME locally trivial affine fibration. Furthermore in CITE it is shown that there is a filtration MATH of MATH by closed subsets such that for every MATH, MATH is one of the MATH (this depends on MATH being quasi-projective). Looking at NAME open sets we find MATH . Thus MATH . By the NAME conjectures MATH is pure. Since MATH and MATH are smooth, REF below imply that MATH is smooth as well. Applying REF finishes the proof. |
math/0106009 | This follows by NAME duality. |
math/0106009 | This follows from the fact that in the long exact sequence MATH the connection maps must be zero by purity. |
math/0106009 | Recall that if MATH is a reductive algebraic group over an algebraically closed field MATH then a MATH-representation of countable dimension is said to have a good filtration if it has an ascending filtration by co-Weyl-modules MATH, or equivalently if MATH for all MATH and all MATH CITE. In particular MATH is exact on representations with a good filtration and the category of representations with good filtrations is stable under taking cokernels of surjective maps and extensions. It is a deep theorem CITE that the category of representations with a good filtration is stable under tensor product. Put MATH. If MATH is a MATH-module free of finite rank over MATH and if MATH (MATH as in the statement of the theorem) has a good filtration then it follows from exactness of MATH that MATH is the number of MATH's in a good filtration of MATH. This can be computed in terms of characters so we conclude MATH . By the universal coefficient theorem the canonical map MATH is a monomorphism and hence by REF it an isomorphism. If MATH is not necessarily of finite rank but has a filtration MATH such that each MATH is free of finite rank and MATH has a good filtration then it is easy to see that REF is still an isomorphism. Since the action of MATH is locally finite there exist a finitely generated MATH module MATH such that MATH is a quotient of MATH. By increasing MATH we may assume that MATH is free. If the characteristic of MATH is large with respect to MATH (in a suitable sense) then MATH is simple CITE. It follows that if MATH is large then the finite dimensional MATH-representation MATH has a good filtration. It then follows from CITE that MATH has a good filtration as well. From the proof it follows that this good filtration is compatible with the grading. Now we filter MATH by degree and we put the induced filtration on MATH. We choose a compatible filtration on MATH such that MATH is a finitely generated MATH-module (confusingly such a filtration is also called a good filtration!) CITE. Since MATH and MATH are finite over the noetherian ring MATH their MATH-torsion is supported on a finite set of primes. Hence by increasing MATH we may and we will assume that MATH and MATH are torsion free. Since MATH has finite global dimension it is easy to see that (at the cost of possibly increasing MATH) we may construct a graded resolution of MATH whose terms are of the form MATH with MATH a free MATH-representation of finite rank. Increasing MATH again if necessary we may assume that all MATH have a good filtration. Thus it follows that MATH will also have a good filtration compatible with the grading for all MATH. Thus MATH has vanishing cohomology. The rest of the theorem follows from the fact that REF is an isomorphism with MATH. |
math/0106009 | First note that MATH is isomorphic to MATH for MATH. Therefore, it is enough to show that MATH is equal to MATH. Let MATH be the decomposition of the fixed point set MATH into connected components. Each MATH is a nonsingular projective variety. Moreover, MATH is contained in MATH. (We have used REF .) We consider the NAME decomposition of MATH with respect to the MATH-action: MATH where MATH. By REF , the right hand side coincides with the whole space MATH. It is known that the natural projection MATH is an affine fibration whose fiber is isomoprhic to the direct sum of positive weight space in the tangent space at MATH. Therefore, we have MATH where MATH is the dimension of the fiber. We also consider the NAME decomposition of MATH: MATH . Then MATH is also an affine fibration over the same base MATH. The tangent space of MATH (at a point in MATH) decompose into the sum of the tangent space of MATH (fiber direction) and MATH (base direction). Therefore, the dimension of the fiber is equal to MATH. Thus MATH . On the other hand, MATH . Therefore the conclusion follows. |
math/0106010 | Let MATH and MATH, then we clearly have MATH as algebras (where MATH). We identify the map MATH with the antipode map of MATH restricted to MATH. It can be easily deduced from the axioms of a weak NAME algebra that for all MATH we have MATH and that MATH is non-degenerate. Therefore, there exists an invertible element MATH such that MATH. Hence, MATH and MATH. Since MATH, we have : MATH . Since MATH is a separability element of MATH, we have MATH where MATH is such that MATH, that is, MATH for any irreducible representation MATH of MATH. The counit and antipode properties imply that MATH, since MATH . We compute : MATH . The comultiplication on bases is determined by the value of MATH, therefore MATH which completely defines the structure of MATH. One can check that the above operations indeed define a weak NAME algebra. Let MATH be the minimal weak NAME algebra defined by MATH and MATH be an isomorphism of weak NAME algebras. Then MATH . Since MATH preserves the regular trace of MATH and MATH we have MATH. |
math/0106010 | It is clear that no automorphism of MATH can map MATH to MATH. |
math/0106010 | The identities for counital maps follow from applying MATH to REF. That MATH follows from the uniqueness of the inverse element and antipode. |
math/0106010 | This is a straightforward dualization of REF is equivalent to MATH and REF is equivalent to MATH. |
math/0106010 | Without loss of generality we may assume that MATH, as every minimal weak NAME algebra is a direct sum of weak NAME algebras with this property. Let MATH be a group-like element in MATH, where MATH and MATH are linearly independent sets. We may assume that each MATH is invertible. Then MATH whence MATH by linear independence. Therefore, MATH and MATH. It is straightforward to check that this MATH satisfies the second equality of REF if and only if MATH. |
math/0106010 | A direct computation. |
math/0106010 | It follows from the properties of bases of a weak NAME algebra that MATH for all MATH. |
math/0106010 | By the previous remark, to every MATH there corresponds a cosemisimple subcoalgebra MATH of MATH, the coradical of MATH; and MATH if and only if MATH and MATH define the same coset. But there are only finitely many mutually non-equal cosemisimple subcoalgebras of MATH, so the quotient group is necessarily finite. |
math/0106010 | We will only prove the first statement since the proof of the second one is completely similar. We have MATH for all MATH, using REF and properties of the counit. Also, for all MATH we compute, using REF : MATH where MATH stands for the second copy of MATH. |
math/0106010 | If MATH has a structure of a right MATH-module via MATH, such that its restricted MATH-module is the regular MATH-module, then the functional MATH belongs to MATH by REF , since MATH for all MATH. Clearly, REF establish a bijective correspondence. Next, MATH and MATH are isomorphic if and only if there is an invertible element MATH such that MATH for all MATH. This is equivalent to MATH that is, MATH, where MATH. Hence, MATH in MATH. Conversely, if MATH for some MATH, then the map MATH, where MATH, is an isomorphism between MATH and MATH. In a similar way one can show that if MATH is a right MATH-module structure on MATH that restricts to the regular MATH-module, then MATH defines an element of MATH. This correspondence is also bijective and MATH-modules corresponding to MATH are isomorphic if and only if MATH for some MATH. |
math/0106010 | Since MATH restricts to the right regular MATH-module, every self-intertwiner of MATH is of the form MATH. The condition that MATH in MATH for all MATH is equivalent to MATH . This, in turn, is equivalent to the identity MATH . This means that MATH. Observe that MATH is an algebra isomorphism between MATH and MATH that maps MATH to MATH. Clearly, MATH is an algebra isomorphism. |
math/0106010 | The map MATH is a linear isomorphism between MATH and MATH. For all non-zero MATH we have MATH, whence for all MATH we have MATH if and only if MATH, that is, MATH is separating. It is cyclic since MATH. The proof of the statement regarding MATH is similar and uses that MATH for all MATH. |
math/0106010 | What we need to show is that the equality MATH for MATH and MATH implies that MATH. By REF there exists a unique well-defined linear map MATH such that MATH for all MATH. Clearly, MATH is an algebra anti-homomorphism and the composition MATH is an automorphism of MATH. Since MATH when MATH by REF , we have MATH, because any automorphism of a semisimple algebra preserves the regular trace. |
math/0106010 | Follows from REF . |
math/0106010 | Since both MATH and MATH are non-degenerate, we have MATH for some invertible MATH, and the action MATH of MATH on MATH is related to the action MATH of MATH on MATH by MATH for all MATH. This means that right MATH-modules MATH and MATH are isomorphic via MATH. Therefore, by REF , the corresponding group-like elements MATH and MATH define the same coset in MATH. |
math/0106010 | Take MATH and compute MATH . Here we used the definition of MATH and that MATH for all MATH. Thus, MATH, which proves the claim. |
math/0106010 | We compute, using the invariance REF of left integrals : MATH for all MATH. |
math/0106010 | First, we note that MATH, by the definition of MATH. On the other hand, MATH by REF . Since MATH is non-degenerate, we have MATH . Applying the target counital map to the both sides of the last identity we get MATH therefore MATH, that is, MATH. |
math/0106010 | The statements concerning MATH follow from REF in the the proof of REF and those concerning MATH follow by duality. |
math/0106010 | We establish these relations by a series of direct computations that use the invariant properties of integrals and REF : MATH for all MATH. |
math/0106010 | From REF we deduce MATH for all MATH, where MATH. Since the maps MATH and MATH are bijective, it follows that MATH . Combining these identities we get MATH . Replacing MATH by MATH we get the desired formula. |
math/0106010 | By the previous remark we may assume that MATH. Since any weak NAME algebra automorphism of MATH preserves the class of distinguished group-like elements, one can see that weak NAME algebra automorphisms MATH differ by a trivial automorphism, that is, MATH and MATH commute in MATH. Since both MATH and MATH have finite order modulo trivial group-like elements,the claim follows. |
math/0106010 | Under the given assumption, all trivial group-like elements are equal to the identity element MATH. |
math/0106010 | Follows from REF for MATH. |
math/0106010 | Fix a non-degenerate left integral MATH of MATH. Using the dual bases tensor REF we see that MATH implies that MATH. Let MATH be a group-like element of MATH such that MATH. Such a MATH exists by REF and depends on MATH. Using the definition of integrals we compute : MATH whence MATH . Let MATH and let MATH be such that MATH. For all MATH we have MATH and MATH. Due to this observation and non-degeneracy of MATH, we can rewrite REF as MATH . By REF , MATH is a cosemisimple subcoalgebra of MATH and MATH is a simple subcoalgebra of MATH (since the element MATH is a primitive idempotent in MATH). Recall from REF that MATH is a coalgebra automorphism of MATH which preserves MATH only if MATH is trivial. This is exactly the case here since the relation REF insures that MATH maps MATH to itself for all primitive idempotents MATH. Thus, one can choose MATH in such a way that MATH, that is, one can assume that MATH (in which case MATH is a two-sided integral). Then REF implies that MATH, therefore, MATH is a non-zero multiple of MATH. We conclude that MATH is an invertible central element of MATH. Hence, MATH is a normalized right integral. By NAME 's theorem, MATH is semisimple. |
math/0106010 | Note that MATH, where MATH is a primitive idempotent, is naturally identified with the dual space of MATH, where MATH, so that MATH. |
math/0106010 | Note that since MATH is semisimple, MATH is an inner algebra automorphism of MATH by CITE. For all MATH we have MATH where MATH is the minimal weak NAME quotient of MATH (viewed as a minimal ideal of MATH) and the sum is taken over the rest of minimal two-sided ideals of MATH (so that each MATH is a simple algebra). Since MATH is a primitive idempotent of MATH, it follows from REF that MATH for some MATH, so that MATH by the same reasoning as in CITE. Since MATH on MATH, we have MATH. Therefore, REF implies that MATH for all primitive idempotents MATH. |
math/0106010 | By REF , we have MATH for some MATH. Also, the bases are necessarily commutative, so that MATH. By REF we have MATH for all primitive idempotents MATH of MATH. For any dual pair MATH of non-degenerate integrals, the element MATH is the dual bases tensor for MATH, therefore, MATH . Let MATH, then MATH. Since the latter holds for all primitive idempotents MATH, we conclude that MATH is invertible. Therefore, MATH is a normalized left integral and, hence, MATH is semisimple by NAME 's theorem. |
math/0106010 | By REF , it suffices to show that MATH in MATH. From REF we see that MATH . Therefore, MATH, where the summation is taken over all irreducible representations of the semisimple algebra MATH. From the formulas for MATH and MATH and explicit REF we have : MATH whence we can compute MATH and, similarly, MATH so that MATH. |
math/0106016 | Since MATH is absolutely irreducible, so are MATH and MATH . Define MATH . The absolutely irreducible rational projective representations of MATH are precisely the restrictions to MATH of the MATH, so by REF , MATH where MATH and the MATH are field homomorphisms. Since MATH has no non-trivial Abelian quotients, by REF there is a unique lift of MATH to MATH, whence MATH . To determine MATH it then remains to study its action on the matrices MATH with MATH. The identity REF gives MATH whence MATH . In light of REF , this relation holds if we replace each MATH by MATH. It follows that MATH commutes with every MATH. As MATH runs through all elements of MATH, the matrices MATH generate MATH CITE. Since MATH is absolutely irreducible, NAME 's Lemma implies that MATH for some element MATH. Since MATH is a homomorphism, MATH is in fact a homomorphism of multiplicative groups. Thus MATH when restricted to MATH and to the subgroup of MATH generated by MATH for all the MATH. These two subgroups together generate MATH, so we are done. |
math/0106016 | Denote by MATH and MATH the eigenvalues in MATH of MATH. Write MATH for a fixed primitive MATH-th root of unity. Then MATH . The equality REF then becomes MATH . Since this holds for infinitely many distinct MATH, this becomes an equality of monic NAME polynomials in the variable MATH over the field MATH. Since the ring of (one variable) NAME polynomials over MATH has unique factorization, we are done. |
math/0106016 | Suppose MATH is a totally real field. We consider two cases. Case: MATH, so MATH. Since MATH is odd, CITE shows that the image of MATH contains MATH for all but finitely many MATH. Following the notation in REF , the restriction of the abstract homomorphism MATH to MATH then gives rise to a non-trivial homomorphism MATH for some finite extension MATH. Since MATH is compact, the argument in CITE shows that MATH is equivalent to a homomorphism with image in MATH. By REF , this is impossible for MATH sufficiently large. Case: MATH. Suppose MATH is an abstract, absolutely irreducible representation coming from MATH. Following the notation in REF - REF , that means MATH is the composition of some MATH with an abstract homomorphism MATH. Since MATH is absolutely irreducible, so does MATH. Apply REF and we are done. |
math/0106016 | Our goal is to understand the dependency of the parameters MATH, MATH, MATH and MATH on MATH. NAME shows that MATH is semisimple, and that MATH is the commutant of MATH in the End-MATH. The argument in CITE then shows that the NAME algebra of MATH is equal to MATH. Since MATH is equivalent to a representation whose image lies in MATH (compare CITE), it follows that MATH . In particular, there exists an infinite subset MATH such that, for a fixed MATH, MATH and with pairwise distinct traces. Since MATH is a strictly compatible family, that means MATH lies in MATH and is independent of MATH. Let MATH be two distinct finite places of MATH of residual characteristic MATH and MATH (MATH is allowed). Following the notation in REF , we have MATH . Since this holds for all MATH, by REF that means MATH and, up to reordering, MATH for all MATH. It remains to study the multiplicative homomorphism MATH. Since MATH is the MATH-adic cyclotomic character CITE, it is surjective onto MATH for all but finitely many MATH. Since MATH is dense in MATH, we can find an infinite subset MATH and a prime MATH such that MATH. Moreover, MATH is independent of MATH. From MATH for all MATH we then see that MATH for all MATH, whence MATH. |
math/0106016 | Every irreducible representation of MATH and MATH gives rise to an irreducible representation of MATH and MATH, respectively, so it suffices to examine the three latter class of groups. By NAME 's Lemma, any irreducible complex representation MATH takes the center of MATH to the center of MATH. Compose MATH with the canonical map MATH and we get an irreducible projective representation MATH. To bound the minimal degree of non-Abelian irreducible representations of MATH, it then suffices to bound the minimal degree of non-Abelian irreducible projective representations of MATH. This is done in CITE, and the lower bound we desire follows from the main theorem there. The same discussion applies to MATH as well. Finally, let MATH be an irreducible representation of MATH. Its restriction to MATH remains irreducible CITE, so MATH if MATH is non-Abelian since MATH is its own commutator subgroup. The Lemma for MATH is then reduced to the case of MATH. This completes the proof of the Lemma. |
math/0106016 | Suppose there exists an injective map MATH. We can assume that MATH is irreducible. Denote by MATH the kernel of the reduction map MATH. Every irreducible representation of a nilpotent group MATH is induced from a degree one character of some subgroup of MATH CITE, so the degree of every irreducible representation of the MATH-group MATH is either MATH or is divisible by MATH. But MATH, so for MATH every irreducible representation of MATH appearing in MATH has degree MATH. Thus MATH is Abelian. But MATH is non-Abelian if MATH, so MATH. REF covers the case MATH, so it remains to consider the case MATH. We first introduce some notation. Let MATH be a normal subgroup of a finite group MATH. For any character MATH and any element MATH, denote by MATH the conjugate of MATH by MATH: MATH . We now resume the proof of the Lemma. Take MATH, so MATH is Abelian. NAME 's theory CITE gives a decomposition MATH where MATH are integers, MATH, and MATH is an irreducible representation of MATH. Moreover, MATH permutes the MATH transitively via REF . We claim that this action on the MATH, when restricted to MATH, is trivial. To see this, note that for any MATH, MATH . This verifies the claim. As a result, MATH permutes the MATH transitively. The left side of REF has degree MATH, so MATH. Thus MATH permutes transitively a set of size MATH. We have two cases: CASE: MATH: Then MATH is the direct sum of copies of a fixed irreducible representation MATH of MATH. Since MATH is Abelian, that means MATH, whence MATH has cyclic image. This is impossible since MATH is non-cyclic and MATH is injective. CASE: MATH: Then MATH contains a proper subgroup of index MATH. That means MATH has a non-trivial permutation representation, and hence a non-trivial irreducible representation, of degree MATH. This is impossible if MATH is large enough, by REF (note that MATH has no non-trivial degree one character if MATH). This completes the proof of the Lemma for MATH. The same argument applies to MATH and MATH. Finally, if MATH contains a subgroup isomorphic to MATH or MATH, then it contains one isomorphic to MATH or MATH, so we are done. |
math/0106016 | For MATH and MATH see CITE. For MATH see CITE. |
math/0106016 | Again we give the argument for MATH. We can assume that MATH. Denote by MATH and MATH the maximal ideal of MATH and MATH, respectively. CASE: We can assume that MATH is not trivial. For MATH, denote by MATH the kernel of the reduction map MATH . Every quotient MATH is a finite Abelian MATH-group, whence the same holds for MATH. Every MATH has finite index in MATH. Every MATH is normal in MATH, so every MATH is normal in MATH. In light of REF , it suffices to show that none of the MATH is contained in the center of MATH. Suppose otherwise. Then MATH is infinite and injects into MATH, which is finite. This is a contradiction. The MATH form a basis of open neighborhoods of the identity element in MATH. REF then implies that MATH has finite index in MATH for every sufficiently small open set MATH containing the identity; and hence for all open sets MATH. That means MATH is continuous. Finally, suppose MATH is not contained in the subgroup of scalar matrices. In light of REF , for some MATH we then get an injective map of MATH or MATH into MATH. Restrict this map to the subgroup MATH or MATH and note that we have a (discontinuous) inclusion MATH, we get a contradiction, by REF . This completes the proof of REF . CASE: We claim that if such MATH exists, then we can find an injective map of abstract groups MATH . By the work of CITE it then follows that MATH. For every integer MATH, denote by MATH the kernel of the reduction map MATH. Every quotient MATH is a finite Abelian MATH-group, whence the same holds for the MATH. We claim that MATH . Suppose otherwise. Since MATH is a proper subgroup of MATH and since MATH, if REF is false then there exists a smallest integer MATH such that MATH; necessarily MATH. On the other hand, every MATH is normal in MATH. Since MATH, REF implies that MATH or MATH for some MATH. Then MATH or MATH, which is not a finite Abelian MATH-group. This is a contradiction, so REF must hold, and we get an injective map MATH . By REF , we get an injective map MATH for some MATH. Every element of MATH either is semisimple or contains a non-trivial NAME block (over MATH). In the first case the element has order prime to MATH. In the second case, if MATH then every non-semisimple element has order divisible by MATH but not by MATH. On the other hand, since MATH, both MATH and MATH contain elements of order MATH (coming from transvections). Since the maps in REF are injective, that means MATH. This completes the proof of REF . |
math/0106017 | Since MATH the corresponding arc MATH crosses at least two vertical lines MATH. Since MATH, in between such crossings MATH must cross at least one horizontal line MATH. |
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