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hep-th/0010227 | In the case MATH the parity behaviour of the vacuum state can be obtained from an indirect argument. We know that the ground states in absence of potential term are holomorphic sections of the line bundle MATH with NAME class MATH (see Ref. CITE for a review and references therein). Any holomorphic section of MATH can have only MATH single nodes. For MATH only MATH - even states can have only one single node at MATH, whereas MATH - odd states have at least three-different nodal points. Thus, the vacuum state is MATH - even and has a node at the crossing of the two circles with holonomy MATH (the corresponding NAME lattice has one single vortex). The same state is parity odd with respect to MATH, MATH and MATH. (The property also holds for MATH). Those results can be explicitly checked from the exact analytic solutions MATH . However, the symmetry arguments already introduced in the case of the planar rotor allows us to generalize this result for more general potentials which makes it extremely useful especially for non-exactly solvable cases. The vanishing of MATH for any MATH implies that MATH for any energy level. If the ground state is degenerate there are at least two states MATH and MATH which are even and odd, respectively, with respect to MATH - parity. MATH vanishes at MATH and MATH at MATH, MATH and MATH. We know that the kinetic term contribution is minimized in a state with a single node at MATH. Since the potential term is reflection symmetric it has the same behaviour near the four points. Thus, the kinetic and potential energies are minimized on states with a unique node at MATH instead of three nodes at MATH, MATH and MATH. From NAME 's variational argument it is obvious that MATH and MATH cannot have the same energy. The existence of such non-trivial splitting implies that the vacuum state MATH is unique and, thus, has a unique node at MATH and is even with respect to MATH - parity and odd with respect to MATH, MATH and MATH parities. |
math-ph/0010005 | Let MATH be a minimizing sequence for MATH under the normalization condition MATH, with corresponding densities MATH. Using the one-dimensional NAME inequality CITE we have MATH . Moreover (compare REF ), MATH . Since the potential energy is relatively bounded with respect to the kinetic energy (compare REF), the right hand sides of REF are uniformly bounded. Hence the sequence MATH is bounded in MATH, and MATH is bounded in MATH. By the "diagonal sequence trick", there is a subsequence, again denoted by MATH, such that MATH converges to some MATH for each MATH, weakly in MATH for some MATH and pointwise almost everywhere. By NAME 's lemma MATH . Observe now that MATH for some MATH and for all MATH. Moreover, MATH, so MATH for all MATH. By the weak convergence we can conclude that MATH for each MATH. Moreover, by the dominated convergence theorem, we get MATH . Using the "diagonal sequence trick" once more, we can use the trace class property of the MATH's to conclude that there exists a subsequence of MATH and a MATH such that MATH in weak operator sense, for each MATH. It follows from weak convergence that MATH. Using NAME 's lemma twice, we have MATH . By the same argument MATH . Let now MATH denote the density of MATH. It remains to show that MATH for each MATH. From weak convergence it follows that MATH weakly on the dense set MATH. Since the operators are bounded by MATH, we see that REF holds weakly in MATH. With MATH considered as a multiplication operator, it is easy to see that MATH is a compact operator (it is even NAME). Thus it can be approximated in norm by finite rank operators. Using REF we can therefore conclude that MATH that is, MATH in the sense of distributions. Since we already know that MATH converges to MATH pointwise almost everywhere, we conclude that MATH for each MATH. We have thus shown that there exists a MATH with MATH and MATH. |
math-ph/0010005 | Observe that MATH where we set MATH. Using the positive definiteness of the NAME kernel and the fact that MATH implies MATH for all MATH, we immediately get the desired result. |
math-ph/0010005 | We proceed essentially as in CITE. For any MATH . In particular, for all MATH, MATH . Now if there exists a MATH with MATH and MATH we can choose MATH small enough to conclude that MATH which contradicts the fact that MATH minimizes MATH. |
math-ph/0010005 | For fixed MATH, MATH and MATH define MATH as above. Let MATH be the infimum of MATH under the constraint MATH. For MATH let MATH be the corresponding minimizers. We have, with MATH, MATH . Dividing by MATH and taking the limit MATH followed by MATH, we conclude that MATH . From REF we infer that MATH for all MATH, with equality for MATH. Therefore we get the inverse inequalities in REF , so equality holds. The assertions of the theorem follow from MATH where MATH is the largest integer MATH, and MATH is the MATH'th element of the set MATH in increasing order. |
math-ph/0010005 | With MATH the two-dimensional Laplacian, one easily computes that MATH for MATH. Multiplying REF with MATH and integrating over MATH, we therefore get, for any density matrix MATH (recall REF ), MATH where we used partial integration for the second step, and the fact that MATH for MATH. To treat the last term in REF , note that the function MATH has its maximum at MATH, because otherwise one could lower the energy by shifting the eigenvectors MATH. Therefore the second derivative of REF at MATH is negative. Setting MATH we see that the last term in REF is negative, so we can conclude that MATH . By the variational principle REF holds. The proof of REF is analogous, considering also the contribution from MATH in REF . |
math-ph/0010005 | This follows immediately from REF , noting that MATH for all MATH. |
math-ph/0010005 | After an appropriate scaling, this is a direct consequence of REF , using the estimates MATH and MATH . |
math-ph/0010005 | The lower bound is quite easy, using the results of CITE. As shown in REF, we have MATH where we have used the scaling properties of MATH. It is shown in CITE that the right hand side of REF divided by MATH converges to MATH. For the upper bound we assume MATH for the moment. We use as trial density matrices MATH where MATH is given in REF and MATH is defined in REF . The kinetic energy is easily computed to be MATH . For the attraction term we use REF to estimate MATH for some constant MATH depending on MATH. For the repulsion term we first estimate MATH which follows from monotonicity of MATH in MATH and the fact that MATH. Therefore we have MATH where we set MATH, and the function MATH is given by MATH . We now claim that MATH as MATH. Since MATH we have, for any MATH, MATH . Now MATH as MATH, which proves our claim. And since MATH for large MATH, this finishes the proof of REF , in the case where MATH is an integer. The proof for MATH is analogous, using REF with the density matrix corresponding to MATH multiplied by MATH as trial density matrices. |
math-ph/0010005 | The convergence of the densities in REF follows from the convergence of the energies in a standard way by considering perturbations of the external potential (compare for example, CITE). Moreover, since the convergence in REF is uniform in MATH, REF holds for any function MATH, so we can conclude that REF holds uniformly in MATH, too. |
math-ph/0010005 | We first treat the case MATH. From REF and the fact that MATH is convex in MATH, we get MATH . Suppose that MATH is not the ground state energy of some MATH. Then MATH, because the second lowest eigenvalue of MATH is equal to the ground state energy of the three-dimensional operator MATH . This follows because MATH is reflexion symmetric, so the eigenvector corresponding to the second lowest eigenvalue, MATH, has a node at MATH. Therefore MATH is an eigenvector of REF , and because it does not change sign, it must be a ground state. Since MATH, the ground state energy of REF is greater than MATH. So MATH would go to zero as MATH, in contradiction to REF . Therefore there exists a constant MATH such that MATH implies the assertion to the theorem. This constant can be chosen independent of MATH, because the limit REF is uniform in MATH (by the same argument as in the proof of REF ). Now assume that MATH for some MATH. From REF we infer that for large enough MATH there is some MATH such that MATH where MATH. We now will show that for MATH large enough MATH has at most one eigenvalue, for all MATH. By the same argument as above, we need to show that the three-dimensional operator REF has no eigenvalues. Using MATH and (compare the next section) MATH we can use REF to estimate MATH . Therefore, for MATH large enough, MATH . By means of the NAME bound CITE we can estimate the number of negative eigenvalues of REF as MATH which is less than REF for MATH large enough. |
math-ph/0010005 | Using the definition of MATH it can be shown (compare CITE) that MATH for some coefficients MATH that fulfill MATH. Since MATH for all MATH (CITEb) we get MATH where we used the fact that MATH if MATH (CITEg). Moreover, using convexity of MATH (CITEi), we arrive at MATH . REF follows if we can show that MATH . This is of course trivial if MATH or MATH equals zero. If MATH we use MATH (CITEa) to estimate MATH which finishes the proof. |
math-ph/0010005 | Let MATH denote the eigenvectors of MATH, that is, MATH . Multiplying REF with MATH and integrating we get MATH where we used that MATH. Since MATH is convex and MATH as MATH we have MATH. Using this and partial integration we can estimate the last term in REF by MATH . Summing over all MATH and MATH (and, according to REF , multiplying the factor corresponding to the largest MATH by MATH), we arrive at MATH . Note that the last term in REF is equal to MATH. To treat the first term in REF we use symmetry and REF to get MATH . Inserting this into REF and dividing by MATH we arrive at MATH . The lower bound on MATH is quite easy. We just have to show that MATH has a bound state if MATH. Using MATH with MATH as a trial vector we compute MATH . Since MATH will be negative for small enough MATH, if MATH. |
math-ph/0010013 | The assumptions of REF imply those of CITE. |
math-ph/0010013 | See for example CITE. |
math-ph/0010013 | See REF. |
math-ph/0010013 | We may assume MATH, because the general case MATH follows therefrom. Let MATH be a monotone increasing sequence, MATH if MATH, of positive simple functions approximating MATH. More precisely, these functions are assumed to be of the form MATH with suitable constants MATH and bounded NAME sets MATH which are pairwise disjoint for each fixed MATH. Using REF and the MATH-homogeneity of the random potential (see REF ) one verifies that REF is valid for all simple functions. Thanks to the convergence MATH as MATH in MATH this implies MATH where we are using the abbreviation MATH. Hence there exists some sequence MATH of natural numbers such that MATH for all MATH by NAME 's inequality and REF . Thanks to monotonicity the right-hand side of the estimate MATH converges pointwise for all MATH as MATH. Since MATH by REF , the monotone- and dominated-convergence theorems imply that the right-hand side (and hence the l.h.s.) of REF converges in fact to zero for MATH-almost all MATH. In other words, the subsequence MATH is NAME in MATH for MATH-almost all MATH. Since the space MATH is complete, this sequence converges with respect to the NAME norm MATH to some MATH. Let MATH denote a multiplication operator associated with MATH. The above convergence and MATH for all MATH imply that MATH is the unique continuous extension of MATH from MATH to the whole NAME space MATH. Denoting this extension also by MATH, we thus have MATH for MATH-almost all MATH. We therefore get MATH . For the second equality we used the monotone-convergence theorem. Note that MATH is monotone increasing since MATH. |
math-ph/0010013 | See REF. |
math-ph/0010013 | If MATH for MATH-almost all MATH, then MATH using REF . Conversely, for every MATH, normalized in the sense MATH, there exists a bounded open cube MATH compatible with MATH such that MATH and therefore MATH . Taking the probabilistic expectation on both sides and using REF we arrive at the sandwiching estimate MATH by the assumption MATH. Since the magnetic translations MATH with MATH or MATH are total, the proof of CITE shows that MATH for MATH-almost all MATH. |
math-ph/0010013 | NAME 's inequality CITE, a subsequent shift in the MATH-integration, and an enlargement of its domain show that MATH where the cube MATH is the arithmetic difference of the unit cubes MATH and MATH. Using NAME 's inequality again, we thus arrive at MATH . Since MATH is contained in the cube centered at the origin and consisting of MATH unit cubes, the proof is complete. |
math-ph/0010013 | We first define the following one-parameter family MATH of smooth NAME probability densities on MATH which approximates the NAME measure MATH on MATH supported at MATH as MATH. Moreover, let MATH denote the convolution of MATH and MATH. The fundamental theorem of calculus yields MATH and hence MATH . The supremum on the right-hand side does not exceed MATH and hence converges to zero as MATH for all MATH. On the other hand, the supremum may be estimated by MATH such that the dominated-convergence theorem is applicable and one has MATH uniformly on MATH. We now claim that there exists some MATH, depending on MATH, with MATH such that MATH for all MATH. To prove this, we pick a compact subset MATH of MATH such that MATH and MATH. Since MATH converges uniformly to MATH as MATH, the bound REF is valid for all MATH. For any other MATH REF follows from the estimate MATH and an explicit computation. REF may then be employed to show MATH . The same holds true with MATH and MATH taking the place of MATH and MATH, respectively. We then estimate MATH . Here we have used the triangle inequality and NAME 's theorem in the integrals MATH and MATH. The integral on the right-hand side of REF tends to zero as MATH by the dominated-convergence theorem. It is applicable since the estimate MATH shows that the integrand in REF is bounded on MATH. Moreover, since MATH as MATH, this bound may be chosen independent of MATH. To complete the proof, we note that the other terms in REF stay finite as MATH and can be made arbitrarily small as MATH. |
math-ph/0010013 | REF : REF follows from REF and the finiteness MATH. Moreover, for every MATH one has MATH by partial integration. Vague convergence of MATH to MATH is now a consequence of the dominated-convergence theorem. It is applicable since REF implies the existence of a locally bounded function dominating all but finitely many of the non-decreasing functions MATH. CASE: For every MATH we define the following continuous ``indicator function" MATH of the half-line MATH. Moreover, we let MATH denote the MATH-continuous NAME measure with density MATH, that is, MATH for all MATH, and the NAME measures MATH are defined accordingly. From REF it follows that MATH for some MATH. Hence the vague convergence MATH as MATH and CITE imply that MATH for all MATH. Since MATH is a NAME measure, this implies the finiteness of MATH for all MATH. The sequence of total masses of MATH converges to the total mass of the limiting measure MATH. More precisely, defining the function MATH for each MATH, it follows that MATH . Here the first term on the right-hand side of the first equality tends to zero using REF and MATH. The second equality is a consequence of the vague convergence of MATH to MATH as MATH. The third equality follows from the monotone-convergence theorem. Hence CITE implies that MATH converges weakly to MATH as MATH, not only vaguely. We recall from CITE that weak convergence of the latter sequence requires that MATH tends to MATH as MATH for every bounded continuous function MATH. The claimed convergence REF of the corresponding distribution functions is therefore reduced to the content of CITE. |
math-ph/0010013 | To show that MATH is a positive NAME measure on MATH, it suffices that MATH for any compact energy interval MATH, MATH, MATH. This follows from the elementary inequality MATH, the spectral theorem applied to MATH and the functional calculus. REF below and REF ensure that the right-hand side of REF is indeed finite. To prove REF we employ an approximation argument with bounded truncated random potentials given by MATH . We denote by MATH, with MATH or MATH, the approximate finite-volume density-of-states measure associated with MATH, see REF . Moreover, MATH defines the approximate (infinite-volume) density-of-states measure. It is a positive NAME measure on MATH, see REF , and independent of the bounded open cube MATH due to MATH-homogeneity. In case MATH we let MATH and estimate as follows MATH . We first consider the limit MATH. In this limit, the third difference on the right-hand side of REF vanishes for all MATH and all MATH by REF below. Next we consider the limit MATH, in which the second difference vanishes for all MATH by REF . In the latter limit, the first difference vanishes by REF . This proves the claimed vague convergence of MATH to MATH as MATH for all MATH, hence for MATH-almost all MATH. In case MATH we estimate MATH . As MATH the first term on the right-hand side converges to zero for MATH-almost all MATH and the same is true for the second term thanks to REF below. |
math-ph/0010013 | Since we have already established the vague convergence of the density-of-states measures in REF , it remains to verify relation REF for the corresponding random distribution functions MATH for MATH-almost all MATH. To this end, we employ the elementary inequality MATH valid for all MATH, MATH with MATH defined in REF . Choosing MATH and MATH there, we get MATH for all MATH. Here, the second inequality results from REF choosing MATH there. Dividing REF by the volume MATH and using the NAME ergodic theorem in the formulation CITE we get MATH for MATH-almost all MATH and all MATH. Since the right-hand side of REF converges to zero as MATH, relation REF is fulfilled. The existence of the distribution function MATH of the limiting measure MATH for all MATH as well as the claimed convergence are thus warranted by REF . |
math-ph/0010013 | See CITE where the appropriate NAME formula for the infinite-volume and the NAME semigroup is employed; see also CITE. |
math-ph/0010013 | Thanks to REF below and REF , the integrals MATH and (analogously) MATH are finite for all MATH and MATH-almost all MATH such that REF yields MATH . Here the quantity MATH, which depends on MATH and MATH, was introduced in REF and vanishes for MATH. We further estimate the first term with the help of REF choosing MATH there. The upper limit MATH of the first term after dividing by the volume MATH is then seen to be finite by the NAME ergodic theorem CITE, MATH for MATH-almost all MATH. The second term in REF is bounded with the help of REF where we again choose MATH. This bound together with the same ergodic theorem yields MATH for MATH-almost all MATH. In the limit MATH, the right-hand side and hence the l.h.s. of REF vanishes for MATH-almost all MATH thanks to REF . This completes the proof since the first term on the l.h.s. of REF may be made arbitrarily small as MATH. |
math-ph/0010013 | Thanks to REF , the integrals MATH and (analogously) MATH are finite for all MATH. Moreover, MATH for all MATH by REF again. This implies vague convergence by REF . |
math-ph/0010013 | The proof consists of an approximation argument. To this end, we recall REF of the truncated random potential MATH. Since MATH is bounded and MATH enjoys REF , we may apply CITE (or CITE together with REF ) which gives MATH for all MATH, all MATH and all MATH. Using the triangle inequality we estimate MATH . The proof is then completed with the help of REF . |
math-ph/0010013 | CASE: Let MATH if MATH and zero otherwise stand for the signum function of a complex-valued function MATH. The polar decompositions MATH and MATH together with NAME 's inequality CITE and CITE show that MATH . Let MATH denote an orthonormal eigenbasis associated with MATH and MATH the eigenvalue corresponding to MATH. Then the trace in REF may be calculated in this eigenbasis and estimated as follows MATH . The inequality is a consequence of the uniform boundedness MATH for all MATH which follows from the explicitly known expressions for MATH, see CITE. CASE: Using the integral represention of powers of resolvents, we get MATH . The claimed bound hence follows from the estimates MATH which are obtained by NAME bracketing CITE and the explicitly known CITE spectrum of MATH. |
math-ph/0010013 | CASE: The claim follows from the chain of REF . Here the first inequality is a consequence of the polar decomposition MATH. The second one is a special case of CITE. For the third one we used the diamagnetic inequality CITE in the version MATH for any MATH, together with CITE. The fourth inequality eventually follows from REF . CASE: We repeatedly use NAME 's inequality CITE for NAME norms MATH . The proof is completed using REF of the present lemma. |
math-ph/0010013 | The proof is split into the following three parts: CASE: Proof of REF for MATH, CASE: Proof of REF for MATH, CASE: Approximation argument for the validity of REF . Throughout the proof we use the abbreviations MATH in agreement with REF . As to REF. Let MATH. We may then apply the (second) resolvent equation CITE MATH . By the triangle inequality for the NAME MATH-norm MATH, NAME 's inequality and the standard estimate MATH, involving the usual (uniform) operator norm MATH, the resolvent equation yields MATH . Using REF we thus have MATH because MATH. The proof is finished by the upper bound in REF . As to REF. We start from the resolvent equation for powers of resolvents MATH see CITE. Moreover, by the standard iteration of the resolvent REF we have MATH where MATH if MATH and MATH if MATH, and the second sum extends over all subsets MATH with MATH elements. Using REF and a suitable analogue with MATH replaced by MATH in REF , the trace norm of the l.h.s. of REF is seen to be bounded from above by a sum of finitely many terms of the form MATH with MATH, MATH. Each of these terms involves multiplication operators MATH suitably chosen from the set MATH where exactly one is equal to MATH. The estimate REF is again NAME 's inequality. The proof is then finished with the help of REF , because MATH appears only once and the other three operators in the above set are all bounded by MATH since MATH. As to REF. We approximate MATH by MATH defined through MATH with MATH and MATH, MATH. Consequently, we let MATH. Monotone (decreasing) convergence for forms CITE yields the strong convergence MATH for all MATH and all MATH. On the other hand, monotone (increasing) convergence for forms CITE yields MATH for all MATH. We therefore have MATH where we used the non-commutative version of NAME 's lemma CITE (see also CITE) twice. Similarly, MATH by the strong resolvent convergences REF , its analogue with MATH replaced by MATH and CITE. Applying REF of the present proof to the pre-limit expressions in REF completes the proof of REF . |
math-ph/0010013 | Throughout, we assume that MATH is MATH-ergodic. The proof is split into three parts. CASE: Proof of REF , CASE: Proof of REF with MATH replaced by MATH with MATH arbitrary, CASE: Approximation argument for the validity of REF with MATH replaced by MATH and of REF . We use the abbreviations MATH for the resolvents of MATH and MATH. As to REF. We write MATH, where MATH is the complex conjugate of MATH. Suitably iterating the (second) resolvent equation CITE MATH we obtain the analogue of REF for MATH. Using this equation and its adjoint, we are confronted with estimating finitely many terms of the form MATH . Here MATH, MATH and MATH denotes some product of MATH factors each of which either being MATH or its adjoint. Moreover, MATH respectively MATH are random potentials suitably chosen from the set MATH. The estimate in REF is just NAME 's inequality for the trace norm and for the expectation. Thanks to CITE we may use the estimate MATH inside the expectation. We therefore obtain the inequality MATH and analogously for the other factor, involving MATH instead of MATH. The second inequality in REF is a consequence of CITE and the diamagnetic inequality CITE which upon iteration gives MATH for all MATH. To complete the proof we use the iterated NAME inequality as in CITE. Taking there MATH, MATH, MATH for MATH and MATH, MATH, we in fact obtain MATH with MATH see also CITE. Since MATH for all MATH the proof is complete. As to REF. We let MATH, MATH. The resolvent equation for powers of resolvents gives MATH see also CITE, with MATH if MATH and MATH otherwise. Using the resolvent REF and its adjoint, we may accumulate in total MATH resolvents MATH, analogously to what was done to obtain REF , such that we are confronted with estimating finitely many terms of the form MATH . Here MATH, MATH and MATH is some product of MATH factors each of which either being MATH, MATH or one of their adjoints. Moreover, MATH are random potentials suitably chosen from the set MATH and exactly one of these is equal to MATH. We now copy the steps between REF and take MATH, MATH, MATH and MATH for MATH in CITE to obtain the bound MATH for MATH times the first expectation on the right-hand side of REF . The second expectation is treated similarly. Since exactly one of the MATH is equal to MATH and all others in the above set may be bounded by MATH, the proof is complete. As to REF. Since MATH as MATH for all MATH and MATH is a common core for all MATH and MATH for MATH-almost all MATH by REF , CITE implies that MATH for all MATH and MATH-almost all MATH. REF of the present proof together with REF shows that MATH . Analogous reasoning as in the proof of REF yields the existence of some sequence MATH of natural numbers such that MATH for MATH-almost all MATH. In other words, the subsequence MATH is NAME in MATH for MATH-almost all MATH. Thanks to completeness of MATH and the strong convergence REF , we have the convergence MATH in MATH for MATH-almost all MATH. The latter implies MATH by NAME 's lemma. Since REF holds for all MATH, the proof of REF with MATH replaced by MATH is complete. For a proof of REF we proceed analogously using REF of the present proof and again NAME 's lemma. |
math-ph/0010013 | We consider the product of the probability spaces MATH and MATH. The latter corresponds to a uniform distribution on the open unit cube MATH. On this enlarged space we define the random potential MATH . It is MATH-ergodic by construction CITE and enjoys REF . The latter assertion is proven by tracing the claimed properties of MATH back to the respective properties of MATH. It remains to prove that the validity of REF for MATH implies the one for MATH. For this purpose, we note that the integral transform REF of the (infinite-volume) density-of-states measure corresponding to MATH obeys MATH . Here the first equality results from REF and the definitions of MATH and the cube MATH together with NAME 's theorem. To obtain the second equality we have used the fact that the trace does not depend on MATH after performing the MATH-integration. This follows from MATH-homogeneity as well as from the fact that one may ``re-arrange" MATH in the form of MATH by MATH-translations since MATH is compatible with the lattice. Moreover, one computes MATH using REF and NAME 's theorem. This completes the proof of REF with MATH replaced by MATH. The other parts of REF in the MATH-ergodic case are proven similarly. |
math-ph/0010016 | Recall that MATH is the solution of MATH with MATH . Thus MATH is analytic in MATH with at worst algebraic singularities at MATH and MATH, that is, the boundaries of the stability interval. But, as was determined before, both MATH have identical analytic properties, and by the definition of MATH and MATH, MATH so we are done. |
math-ph/0010016 | Suppose that MATH for all MATH, and hence all MATH by analyticity. We know then that for every MATH and MATH, the NAME solution MATH is MATH (for all MATH); that is, MATH is exponentially decaying in this region of the upper half plane. Thus MATH is the NAME solution for the perturbed equation MATH. We may therefore calculate the NAME MATH-function (see for example, CITE), MATH, for MATH on the half line MATH, MATH where the latter is the MATH-function of MATH on MATH. As the MATH-functions are analytic in the entire upper half plane, we conclude that MATH for all MATH. Thus, by the recent results of CITE and CITE, which provide generalizations (applicable in our setting) to the classical results of CITE, this implies MATH is identically zero. |
math-ph/0010016 | Since MATH, we have that MATH, MATH, and MATH from MATH and MATH. Thus MATH is invertible. For the rest of the proof we will drop the fixed parameter MATH. First note that expressing the solutions MATH and MATH in terms of the NAME solutions MATH yields MATH . Clearly, MATH and using that MATH are the NAME solutions we have MATH . Hence we see MATH . Let MATH then it is easily checked that MATH with MATH, MATH and MATH as defined in MATH, MATH. Thus MATH follows from MATH. |
math-ph/0010016 | CASE: By (MATH), MATH uniformly in MATH and MATH. Thus by dominated convergence, MATH, if MATH. CASE: We have that MATH where the last inequality follows as MATH for all MATH by REF and (MATH). |
math-ph/0010016 | CASE: We first show that MATH weakly as MATH. Note that in general if MATH weakly and MATH weakly, then MATH weakly (for the definition of the convolution, see REF). By MATH we have that MATH weakly and an application of NAME implies that every subsequence of MATH contains a weakly convergent subsequence. Using this, the convergence of the convolutions, and the uniqueness of the invariant measures, we see that MATH. Now, continuity follows from (MATH), that is, noting that MATH, and the estimate (MATH), see REF. CASE: This follows as in the proof of (MATH), with REF replaced by corresponding estimates for MATH, which again follow from REF , see REF. |
math-ph/0010016 | One calculates that MATH . The result follows by letting MATH and noting both REF of the above Corollary. |
math-ph/0010016 | REF confirms that both assumptions of REF hold uniformly with respect to MATH. That this implies the above estimates, again uniformly with respect to MATH, follows from observing that the proofs of REF yield uniform results under uniform assumptions. |
math-ph/0010016 | REF follows as in V. REF using that MATH . REF follows as in REF using REF as done therein. |
math-ph/0010016 | This lemma can be proved in exactly the same way REF is proved in CITE. For the reader's convenience, we sketch the argument briefly. By our choice of the interval MATH and the results from REF, in particular REF , we have MATH. Using the inequality MATH and NAME 's inequality, one shows as in CITE that for every MATH and every MATH, we have MATH for some finite constants MATH, MATH. Hence, by REF , we have for some MATH, uniformly in MATH and MATH in the unit sphere, MATH for some MATH, provided MATH is small enough. Iterating REF as in CITE yields MATH for all MATH and MATH, for some MATH and MATH, where MATH is the largest integer less or equal MATH. |
math-ph/0010016 | We will follow the same strategy as NAME in their proof of CITE, that is, we will use NAME continuity of the integrated density of states to derive the estimate REF. The only difficulty that arises is that the cutoff of eigenfunctions as performed by NAME in the discrete case may produce elements outside the domain of the local Hamiltonian. We therefore use a smooth cutoff procedure and show that the argument still goes through. Note first that it suffices to prove REF for small MATH. It follows from REF that there are constants MATH and MATH such that for every MATH, MATH . Now fix MATH, MATH, and MATH. Let, for MATH, MATH be the interval MATH and denote by MATH the operator MATH restricted to MATH with NAME boundary conditions. Let MATH be the event MATH has an eigenvalue MATH such that the corresponding normalized eigenfunction MATH satisfies MATH. Let MATH. Clearly, MATH is independent of MATH and equals the left hand side of REF. Fix some MATH and let MATH be the operator MATH restricted to MATH with NAME boundary conditions at MATH and MATH. Let MATH be distinct and such that the event MATH occurs. For each such MATH, we shall construct a normalized function MATH in the domain of MATH which is supported on MATH (hence MATH form an orthonormal set) such that for every MATH, MATH where MATH is a constant which only depends on the single site potential MATH and the single site distribution MATH. By disjointness of supports, we also have MATH . By REF imply that the number of eigenvalues of MATH (counted with multiplicity) in MATH is bounded from below by MATH. In other words, MATH . Thus, if MATH is small enough (namely, such that MATH or, equivalently, MATH), we have MATH where the intermediate steps hold true for almost every MATH. It remains to construct MATH with the desired properties. Fix some MATH and consider the function MATH which is defined on MATH and vanishes at the boundary points. In principle we would like to extend MATH by zero on MATH. However, this function will in general not belong to the domain of MATH, so we could not even evaluate MATH applied to this function. Instead we will use a smooth extension of MATH to MATH. Fix once and for all a smooth function MATH which obeys MATH, MATH for MATH, and MATH for MATH. Let MATH and MATH and define MATH for MATH by MATH . Then MATH clearly belongs to the domain of MATH and it has norm bounded by MATH. We want to estimate MATH. Now MATH is an eigenfunction corresponding to the eigenvalue MATH, so we write MATH. To estimate MATH, we consider MATH for MATH. Hence, we have MATH where the constant MATH depends on the single site potential, the single site distribution, and the function MATH. Here we have used REF . Let us define MATH. By construction and REF , we have, for MATH sufficiently small, MATH and hence REF holds true with a suitable MATH which depends only on the single site potential, the single site distribution, and the function MATH. Moreover, by construction MATH form an orthonormal set and obey REF. This concludes the proof of the lemma. |
math-ph/0010016 | We closely follow the proof of REF and make the necessary modifications. Let MATH be as above and for each odd MATH, we set MATH with some MATH to be chosen later. For every MATH and MATH, we define the events MATH . Let MATH with MATH and MATH from REF . Then MATH and REF immediately implies MATH provided MATH is large enough. Together with REF yields for MATH large enough, MATH with MATH independent of MATH and MATH. Finally, using REF, for MATH large enough, one proves similarly to CITE MATH for some suitable MATH if MATH is chosen small enough. The assertion now follows from REF - REF. |
math-ph/0010016 | This follows from REF. Note that in the case MATH the assumptions required there are equivalent to MATH for some MATH and that MATH is non-compact and strongly irreducible. In particular, non-compactness is equivalent to the contractivity required in CITE; see CITE. Integrability of MATH with respect to MATH for all MATH follows from REF and boundedness of the distribution of MATH. |
math-ph/0010016 | This follows from REF, whose assumptions, that is, MATH strongly irreducible and MATH for some MATH, are satisfied. |
math-ph/0010016 | We closely follow the proof of REF. Define MATH by MATH . Then MATH . Define MATH by MATH . For MATH, choose representatives MATH and MATH such that MATH and the angle between MATH and MATH is at most MATH. Then, by REF, MATH recall that MATH. This implies, using notation from REF, MATH . The definition of MATH shows MATH . Noting that MATH, we may use REF above to conclude that for MATH, there exists MATH and MATH such that for MATH and all MATH, MATH where in the last step, MATH is assumed to be sufficiently small, that is, such that MATH, and MATH. Next choose the unit vector MATH. We therefore have MATH for all MATH. It then follows from REF that there exist MATH and MATH such that MATH . Inserting REF into REF completes the proof of REF with MATH. We have assumed that MATH is sufficiently small, but the result extends to large MATH with unchanged MATH. |
math-ph/0010016 | From REF we get for all MATH, MATH with probability at least MATH. This yields the assertion. |
math-ph/0010016 | Let MATH be the solutions of MATH with NAME boundary conditions at MATH, that is, MATH, MATH. Then the NAME 's function MATH (that is, the kernel of MATH) is given by MATH where the Wronskian MATH is constant in MATH. Setting MATH, we get MATH where MATH denotes the transfer matrix from MATH to MATH. By stationarity we can use REF to conclude MATH . Note that MATH if and only if MATH. Thus the event in REF implies MATH. Let MATH and MATH. Then MATH by REF . Also MATH . Again by REF , MATH and by stationarity and REF (note MATH), MATH . Combining REF - REF and using MATH, we get MATH for MATH sufficiently large. In a completely analogous way the same estimate is found if MATH, MATH. From this it can now be seen easily that for every MATH, there exist MATH and MATH such that for MATH and MATH, we have MATH and MATH with probability at least MATH. The theorem now follows by NAME 's test. |
math-ph/0010016 | With probability MATH we have that both the event in REF and the complementary event in REF hold. Thus, by assumption we have that for every MATH, we have MATH and moreover, by the resolvent equation, MATH . Thus for these MATH's, the cube MATH is MATH-good. |
math-ph/0010016 | Fix an arbitrary compact interval MATH. It follows from REF that we have both a NAME estimate and a fixed energy initial length scale estimate for every MATH. REF shows that these two results imply a variable-energy initial length scale estimate for a ball MATH of explicit radius around MATH . The variable-energy multiscale analysis as presented, for example, in CITE then establishes variable-energy resolvent decay estimates on a sequence MATH of length scales for energies in MATH. These estimates, together with the existence of polynomially bounded eigenfunctions for spectrally almost every energy, yield pure point spectrum in MATH with exponentially decaying eigenfunctions for almost every MATH; see, for example, CITE for details. Thus, we have exponential localization in MATH for almost every MATH. Finally, since, by general principles CITE, the set MATH carries almost surely no spectral measure, we have exponential localization in MATH for almost every MATH and hence REF . |
math-ph/0010016 | It essentially follows from CITE that the variable-energy resolvent decay estimates, as given by the output of the the variable-energy multiscale analysis, imply strong dynamical localization in the sense of REF . For the curious reader we briefly sketch the argument, referring him to CITE for necessary notation. Given a compact interval MATH, a compact set MATH, and MATH, we first let MATH. Next we choose MATH large enough so that, for every MATH, REF imply both MATH, MATH, and MATH of CITE with parameters sufficient to cover the desired MATH. Having this length scale fixed, we decompose the interval MATH into a finite disjoint union of intervals MATH, each of them having length bounded by MATH. We split the projection MATH in REF into the finite sum MATH and treat each term separately. For every MATH, we can apply REF, with initial length scale MATH, since all the other conditions (for example, (INDY), (GRI), (WEYL), (EDI), REF ) are known to hold for the concrete operators MATH under consideration CITE. After establishing REF, with MATH replaced by MATH, for every MATH, we get REF and hence REF . |
math-ph/0010016 | For MATH one has MATH that is, MATH, which implies the lemma. |
math-ph/0010016 | Without restriction let MATH. The solutions MATH and MATH satisfy MATH and thus MATH . NAME 's lemma, for example, CITE, yields MATH . By REF we have for all MATH that MATH . Inserting this into REF yields the result. |
math-ph/0010016 | By REF there are constants MATH only depending on MATH such that for all MATH . With MATH and MATH we get MATH . It now follows from elementary geometric considerations, for example, CITE, that MATH contains an interval of length MATH on which MATH. This yields REF . |
math-ph/0010016 | See REF , and REF. |
math/0010019 | Assume then that MATH is bounded. To check the continuity, let MATH be a net weakly convergent to MATH. Then for all MATH we have MATH as MATH is dense in MATH and MATH is bounded by assumption, it follows that MATH weakly. It remains to show that MATH is bounded if MATH is norm closed. We shall show that MATH is closable, thus MATH will be bounded by the closed graph theorem. Let MATH be a sequence in MATH and MATH be such that MATH and MATH in norm. As MATH is a bounded set, by the just proved weak continuity of MATH we have that MATH weakly, thus MATH. |
math/0010019 | Assuming in MATH that MATH is contained in the ball of radius MATH, we shall show that the same is true for MATH, that is, MATH holds. Let MATH. By NAME density theorem CITE, there exists a net of operators MATH strongly convergent to MATH. Since MATH, we may assume, possibly restricting to a subnet, that MATH weakly converges to MATH, MATH. Now take MATH. We have MATH . Since MATH is self-adjoint, this means that MATH and MATH, that is, MATH. Now MATH and MATH follows by REF as MATH is norm closed. |
math/0010019 | If MATH is pure, then MATH, hence, if the boundedness condition holds, MATH is bounded. As MATH, it follows that MATH is bounded, thus MATH is semibounded. The converse is obvious. |
math/0010019 | Immediate by REF . |
math/0010019 | Clearly MATH for all MATH. If MATH is MATH-holomorphic on MATH and MATH, then MATH is boundary value of a function in MATH, thus by REF MATH. If moreover the bound REF holds, then MATH thus MATH is MATH-bounded. Conversely, if MATH is MATH-bounded, then MATH and the same computation done above yields, by REF , that MATH for all MATH and MATH so the the bound REF holds with MATH. Since by REF MATH is MATH-holomorphic on MATH iff MATH, the rest follows by REF . |
math/0010019 | Immediate by REF . |
math/0010019 | As the map MATH satisfies MATH, REF holds with MATH (see REF), where MATH namely MATH . Assuming first that MATH is also separating, that is, MATH ,if MATH and MATH we then have MATH . Since MATH implements automorphims of MATH and MATH, by the modular theory MATH and MATH commute, thus there exists a strongly dense subalgebra MATH of MATH such that MATH is a core for every continuous function of MATH or of MATH. Taking MATH, the above equation gives MATH thus MATH as MATH is a core for both MATH and MATH. In general, if MATH, we may consider the the reduced NAME algebra MATH on MATH. Since MATH is separating for MATH and MATH commutes with MATH, the above shows that MATH . Thus, by REF , we have for all MATH that entails MATH . Combining REF we get the desired REF . |
math/0010019 | By REF applied to the MATH-fold tensor product, we have MATH. By the modular theory MATH and MATH commute, thus a simple application of the NAME theorem implies MATH for all MATH, thus MATH. Taking the limit as MATH we obtain MATH . As MATH and MATH, the above inequality also entails that MATH . Taking logarithms, as MATH and MATH commute, these inequalities imply respectively MATH and MATH, where MATH are the spectral projections of MATH corresponding to the positive/negative half-line. The projection MATH onto the kernel of MATH clearly commutes with MATH and MATH, thus denoting by MATH the closure of MATH, MATH is a bounded positive linear contraction, MATH, commuting both with MATH and MATH. As MATH and MATH and MATH, we also have MATH. |
math/0010019 | Let MATH be the NAME operator and MATH be the spectral projection of MATH corresponding to the interval MATH. Then MATH commutes with the (real, unbounded) projection MATH onto MATH (see REF ) because it commutes with MATH: indeed MATH commutes both with MATH and with MATH (being a real even function of MATH). Denoting by MATH the space of elements of MATH whose spectrum under the modular group MATH . Ad-MATH lies in MATH and by MATH the real subspace of selfadjoint elements of MATH, we have MATH . Let MATH denote the NAME space MATH equipped with the MATH-graph scalar product MATH and notice that any MATH subset of MATH is contained in MATH and its closure MATH in MATH coincides with its closure in MATH (because the restriction of MATH to MATH is bounded). Notice also that, as a linear operator of MATH, MATH is bounded, indeed MATH is the (real) orthogonal projection of MATH onto MATH. Therefore, on the NAME space MATH, we have by REF that MATH . As the REF holds for all MATH, it then holds for all MATH. Given MATH the sequence of vectors MATH belongs to MATH and MATH therefore MATH . |
math/0010019 | As is well known, MATH, where MATH is the NAME 's operator associated with MATH and MATH. With MATH, let MATH the operator given by MATH, MATH, and define analogously MATH with respect to MATH. Then MATH and MATH and MATH, thus MATH and MATH. |
math/0010019 | Since the kernel of MATH is invariant under MATH and MATH, one can decompose MATH as a direct sum of two components, one corresponding to the kernel of MATH, and one to its orthogonal complement, thus REF can be proved for each component separately. Since the inequality is obviously satisfied on the kernel of MATH, we may just suppose the kernel of MATH to be trivial. Now we give an explicit description of the vectors of MATH (compare CITE) which allows an immediate verification of REF . Let us chse choose a selfadjoint antiunitary MATH commuting with MATH and MATH, and set MATH, so that MATH. Then denote with MATH the real vector space of MATH-invariant vectors in the spectral subspace MATH and by MATH the maps MATH, MATH, where the operator MATH is defined by MATH, MATH. Since MATH maps the spectral space MATH onto the spectral space MATH, both MATH and MATH are isometries, and a simple calculation shows that their ranges are real-orthogonal. Moreover, decomposing MATH as MATH, one can show that any solution of the equation MATH can be written as a sum MATH, namely the map MATH is an isometric isomorphism of real NAME spaces. Moreover, for any MATH, MATH is purely imaginary, therefore MATH namely the inequality should be checked on MATH and MATH separately. Finally, MATH since MATH and MATH are commuting positive operators on MATH, and the same holds on the range of MATH. |
math/0010019 | Set MATH as in REF . Note first that MATH is equal to MATH where MATH is the NAME operator. As MATH commutes both with MATH and MATH, the same is true for MATH and thus MATH commutes with MATH. It follows that MATH . We then have for all MATH . Clearly the same is true if we replace the NAME algebra MATH by MATH (infinite tensor product with respect to the constant sequence of vectors MATH), the vector MATH by MATH and MATH by MATH. The permutation shift then acts in a strongly cluster fashion on the latter system. By the NAME REF either MATH or this latter system satisfies the KMS condition at some inverse temperature MATH. Clearly the KMS condition then holds true also for the original system. To prove the last assertion we have to show that, given MATH, MATH is not completely MATH-bounded with respect to MATH and MATH. Indeed, if this were not the case, we would have MATH, by REF , which is not possible if MATH unless MATH in which case also MATH. |
math/0010019 | MATH . If MATH satisfies the KMS condition at inverse temperature MATH then it is MATH-holomorphic with constant MATH. It is also immediate by REF that MATH is not completely MATH-holomorphic if MATH. For the same reason it is not completely MATH-holomorphic if MATH. If MATH is a ground state, then MATH is obviously completely MATH-holomorphic for all MATH. MATH . By considering the GNS representation of MATH, it is sufficient to show that REF holds true without assuming that MATH is separating, but allowing MATH to be a ground state. Indeed let MATH be the projection onto MATH. Clearly MATH is MATH-bounded on MATH with respect to MATH and MATH. Hence Ad-MATH implements the rescaled modular group of MATH. Thus MATH, for some MATH, where MATH is the modular operator of MATH acting on MATH. Moreover, by REF , we have MATH. In particular MATH, thus MATH. Thus MATH, where MATH is positive. By the completely MATH-holomorphic assumption, the same relation holds true by replacing the system with its tensor product by itself. This is possibly only in two cases: either MATH, thus MATH and MATH is a ground state, or MATH, namely MATH is cyclic and separating and MATH. |
math/0010019 | Let MATH be the family of projections associated with MATH by the spectral theorem, namely MATH. Then MATH if and only if MATH, that is, MATH, where MATH is the finite NAME spectral measure associated with MATH. By the next NAME (with a change of sign of MATH) this holds iff MATH is the boundary value of a function holomorphic in MATH and continuous in MATH therefore MATH. If MATH holds, then MATH extends to a function in MATH, namely MATH is bounded in the strip MATH, because MATH for all MATH. If MATH, the function MATH can be checked to belong to MATH by standard methods, thus MATH. Last assertion follows by the spectral theorem. |
math/0010019 | If MATH belongs to MATH, then also MATH belongs to MATH for all MATH and MATH defines a function in the strip MATH, that can be easily seen to belong to MATH. Conversely suppose MATH to be the boundary value of a function holomorphic in MATH and continuous in MATH. Decompose MATH as MATH, where the first term is supported in the positive axis, the second in the negative axis. We have MATH for any positive MATH, hence MATH is holomorphic in the upper half plane. Therefore we may restrict to the case where MATH is supported in the positive axis. In this case MATH is holomorphic in the lower half plane and in the strip MATH, and is continuous on the real line both from above and from below, therefore MATH extends to a holomorphic function on MATH, continuous on the boundary. Set MATH. Then MATH is analytic in MATH, continuous on MATH, and for MATH is given by MATH. If MATH, the dominated convergence theorems entails MATH, hence by monotone convergence we obtain MATH. The analyticity implies that for MATH where the last equality follows by monotone convergence. Again by monotone convergence and the continuity of MATH on the boundary we get MATH, namely MATH. |
math/0010019 | By replacing MATH with MATH, we may assume that MATH is non-negative. Let MATH be a vector orthogonal to MATH. We have to show that MATH. If MATH is a function in the NAME space MATH, we have MATH for all MATH, where MATH the NAME anti-transform of MATH. If MATH is a bounded NAME function with compact support, we may choose a sequence of smooth functions MATH with compact support such that MATH weakly, thus REF holds for such a MATH. If MATH is a bounded NAME function with compact support, we may write MATH, therefore MATH . We can then choose a sequence MATH of such functions such that MATH strongly. It follows that MATH for all MATH, hence MATH because MATH is dense. |
math/0010019 | Note first that, as a consequence of the commutation relations REF , we have MATH for all MATH CITE. By the the criterion given in REF below, it will suffice to construct a core MATH for MATH such that MATH for all MATH. The set MATH is contained in the domain of MATH, and clearly MATH . If moreover MATH is Ad-MATH-invariant, then MATH . We may thus apply REF below to conclude that MATH is a core for MATH. On the other hand, if MATH is MATH-bounded, then also MATH is MATH-bounded by REF , thus we are in the previous case as MATH is Ad-MATH-invariant. |
math/0010019 | Since MATH implements automorphisms of MATH, it commutes with the modular operator MATH associated with MATH, thus MATH is a one-parameter group of unitaries. We denote by MATH its self-adjoint generator. By the modular theory MATH also commutes with the modular conjugation MATH associated with MATH, thus MATH. We then have, for all MATH, MATH . We shall now show that the above bound holds for all MATH in the unit ball of the MATH-algebra MATH. By NAME theorem CITE, MATH has the same commutation relations REF as MATH with MATH, thus MATH commutes with MATH. Therefore if MATH namely the MATH-boundedness property with respect to MATH holds for MATH. Now, by the following REF , MATH is irreducible on MATH, thus MATH is semi-bounded by REF . As MATH, MATH is indeed a bounded operator. We now follow an argument in CITE. By the NAME derivation theorem (compare CITE), there exists a selfadjoint element MATH, indeed a minimal positive one, such that MATH and indeed MATH by the canonicity of the minimal positive choice for MATH. Therefore MATH, MATH, and this implies MATH CITE, thus MATH by the uniqueness of the MATH-invariant vector. As MATH is separating, MATH, thus MATH as MATH is minimal. It follows that MATH for all MATH. |
math/0010019 | Let MATH commute with MATH and MATH. Then MATH and MATH for some MATH. As MATH is separating for MATH, then MATH and this entails the thesis. |
math/0010019 | Let MATH be the weak closure of MATH. Clearly MATH, hence, by REF, see also CITE, MATH. By the mean ergodic theorem, the one dimensional projection MATH onto MATH belongs to NAME algebra generated by MATH, hence to MATH. Then MATH by REF . |
math/0010019 | Set MATH and apply REF . |
math/0010019 | In this proof we drop the subscript MATH on the operators associated with MATH. MATH: As MATH, then MATH for all MATH, which immediately implies MATH. MATH is obvious. MATH: We only need to show the spectrum condition. By NAME covariance it is sufficient to show that the positivity of the generator of a one-parameter group MATH of light-like translations associated with MATH; this satisfies MATH, MATH, and the commutation relations REF with MATH. Thus the positivity property follows by the criterion in REF . MATH: Since MATH implements automorphisms of MATH, it commutes with MATH, that is, MATH is a one-parameter group of unitaries. Denote by MATH its self-adjoint generator and note that, since by the NAME theorem MATH commutes with the modular conjugation MATH of MATH, we have for all MATH, MATH namely MATH is MATH-bounded with respect to MATH and MATH. Moreover, by NAME commutation relations CITE and by NAME theorem, MATH commutes with all translations. Therefore, denoting by MATH the translation unitary group, MATH namely MATH for all MATH with MATH. By REF MATH is a MATH-subalgebra of MATH, which is also translation invariant. By REF MATH is thus irreducible, thus MATH is semi-bounded by REF . The rest now follows as in the proof of REF . It remains to show the last assertion. Let's then assume that MATH intersects the edge of MATH in a half line. By REF the boundedness of MATH implies that also MATH is bounded. Let MATH be the one-parameter unitary group of translations along the edge of MATH. Clearly MATH commutes with MATH, therefore if MATH then MATH namely the MATH-boundedness condition hold for MATH. As MATH is a dense MATH-algebra of MATH, by REF MATH, thus we obtain all the properties in the statement by the above proof. |
math/0010019 | MATH: By the split property the set MATH is compact for MATH CITE and metrically nuclear for MATH CITE. By the KMS property MATH for all MATH and MATH (see CITE); we omit the suffix MATH on MATH and MATH. MATH: Assume first that the underlying NAME MATH space is separable. Let MATH and choose a sequence MATH norm dense in MATH; the function MATH is bounded and holomorphic in the open strip MATH. Indeed, by the uniform boundedness assumption, MATH for some constant MATH independent of MATH and MATH. As MATH the limit MATH exists except for MATH in a set MATH of NAME measure zero. Choose MATH; as MATH, the weak limit MATH exists, thus also the weak limit MATH exists. By the spectral theorem, this implies MATH and MATH, namely MATH is MATH-bounded with respect to MATH and MATH and this entails MATH by the previous theorem. If MATH is non-separable, it is sufficient to apply the above argument to the separable NAME subspace generated by MATH as MATH varies in the complex continuous functions on MATH vanishing at infinity. Then MATH follows because the MATH-metrical nuclearity of MATH implies the split property for MATH CITE. |
math/0010019 | The proof is similar to the one of REF . We only notice that in this case condition MATH refers to a bounded interval MATH. This is possible because MATH and, by scaling the interval, this gives MATH for any MATH, thus the boundedness condition holds for MATH. |
math/0010019 | MATH. The inclusion MATH is half-sisd modular, therefore the translations (with positive generator) can be constructed as in CITE, compare also PropositionREFA. REF. MATH. First we extend the net MATH to all intervals in MATH setting MATH, MATH. Clearly MATH and MATH act as translations and dilations on the net, therefore REF applies, hence the generator of MATH is positive. Conformal invariance then follows by CITE. |
math/0010020 | That MATH acts faithfully on MATH is well-known and easy to prove. If MATH are represented by sections MATH, MATH, then there is a fiberwise translation in the part of MATH that is smooth over MATH which sends MATH to MATH. As recalled above, this translation extends as an automorphism MATH of MATH. Then MATH fixes the difference of any two sections, so it certainly acts as the identity in MATH. If MATH and MATH meet a reducible fiber MATH in the same component, then this component is special. So MATH fixes every irreducible component of MATH. The rest of the argument is now straightforward or follows from the above mentioned property of NAME fibrations. |
math/0010020 | Let MATH. So for every MATH we have MATH. It follows that for every reducible fiber MATH, MATH preserves the root subsystem MATH of MATH generated by MATH. So MATH normalizes the associated affine NAME group MATH. Choose a section MATH. Then MATH sends its class MATH to an element of the form MATH with MATH, where MATH is the class of a section MATH. There is a unique MATH that sends MATH to MATH. We show that MATH is in MATH. It is clear that MATH is the identity on the orthogonal complement of MATH and fixes MATH. Also, for every reducible fiber MATH, MATH normalizes MATH and its image in MATH is induced by a translation. Since MATH, it follows that this image is trivial: MATH acts in the span of MATH as an element of MATH. This is true for all reducible fibers and hence MATH. |
math/0010020 | Let MATH be the locus of pairs of distinct points of MATH with the same image in MATH. For the first assertion it is enough to show that MATH is of dimension MATH. A point of MATH for which MATH are mutually distinct can be represented by a triple MATH with MATH so that MATH is the divisor defined by MATH. Notice that the vector MATH is unique up to a scalar factor. An element of MATH is representable by a quadruple MATH in MATH with MATH. This identity can also be written as MATH. If the righthand side is nonzero, then it is factored by the lefthand side into two forms of degree six. The family of such factorizations (with fixed nonzero righthand side) is of dimension one. Since MATH lies in a projective space of dimension MATH, it follows that MATH. To prove the second assertion we consider the vector bundle MATH over MATH. Denote the projections on its summands by MATH repectively. So for MATH as above, the expression MATH defines a homomorphism MATH. Its zero set in the associated projectivized bundle MATH is a NAME curve over MATH with modular function MATH. If MATH and MATH are linearly independent, then minimal resolution of its singularities gives an elliptic surface for which the first summand of MATH defines a section. This surface is rational. If MATH and MATH have no nontrivial common zero, then MATH has degree MATH and MATH for MATH. NAME 's theory (see for example REF and Subsection REF below) implies that this elliptic surface is unique up to MATH-isomorphism. |
math/0010020 | In view of REF we must show that the MATH-orbit space of MATH is rational. Generically MATH is fibered in lines over the product of projective spaces MATH. Let MATH be the locus where MATH acts freely. Then MATH is open-dense in MATH, and the orbit space MATH is a rational curve. So if MATH denotes the preimage of MATH in MATH, then MATH is a morphism to a rational curve whose generic fiber has the structure of a fibration of lines over a projective space. This implies that MATH is rational. |
math/0010020 | We begin with proving the first part of REF . Suppose that MATH has a singular point MATH over MATH. Denote the closures of the connected components of MATH by MATH and MATH. Then on MATH we have a NAME fiber of type MATH for some MATH. The NAME characteristic of such a fiber is MATH and hence the degree of MATH on MATH is MATH. To see that the birational map from MATH to MATH is in fact a morphism, we consider the closure MATH of the diagonal embedding of MATH in MATH. Since MATH is normal it suffices to prove that the projection MATH is a bijection. Or equivalently, that any curve germ in MATH is the image of one in MATH. Moreover, we want this lift to be as prescribed by the proposition. This can be checked in a straightforward manner. |
math/0010020 | The first statement follows in a straightforward manner from the fact that MATH, our computation of MATH, and the MATH-linearity of MATH. The second follows from the first if we bear in mind the Formulae REF for MATH. |
math/0010020 | From the definitions we find that MATH and MATH. We know a priori that MATH and MATH are covering transformations, hence it is enough to show that these elements act on MATH as respectively, MATH and MATH. This is verified in a straightforward manner using REF . |
math/0010020 | Choose an affine equation for MATH as before. First note that MATH lies in MATH. At a point of multiplicity MATH, a local equation of MATH is MATH. A straightforward calculation shows that the pull-back of MATH under normalization has in each the preimage of this singularity a zero of order MATH for MATH. Any other element of MATH is of the form MATH and as in the smooth case we find that MATH cannot have any poles, hence must be constant. |
math/0010020 | We use our fixed two dimensional vector space MATH equipped with a generator MATH of MATH. Given a semistable MATH, regard MATH as a homogeneous function on MATH. Then MATH defines a degree MATH covering of MATH. It is an affine surface with good MATH-action (so that MATH has weight MATH) whose curve at infinity is a MATH-curve MATH as above. Then MATH is a MATH-invariant rational form whose residue at infinity, MATH, is a nonzero element of MATH. So MATH is the residue of MATH. Think of MATH as the linear form on the line MATH in MATH spanned by MATH which takes the value MATH on MATH. The MATH-orbit of such a linear form defines an element of the complement of the zero section of MATH and vice versa. Since the constructions are MATH-equivariant, we thus get a map from the complement of the zero section of MATH to the moduli space in question. It is easy to see this this extends to an isomorphism of MATH to the moduli space. |
math/0010020 | This is clear for the group of covering transformations. Any such automorphism that is not a covering transformation must permute the ramification points nontrivially. It is easy to see that such an automorphism acts nontrivially on MATH. |
math/0010020 | Let MATH be a stable effective degree MATH divisor in MATH (so all multiplicities MATH). Given a neighborhood MATH of MATH in the space of effective degree MATH divisors, denote by MATH the divisors that are reduced. Then MATH defines a locally constant sheaf of MATH-modules. If MATH has multiplicities MATH (so that MATH), and MATH is sufficiently small, then the local monodromy group is isomorphic to subgroup of MATH. Since the ranks MATH are all MATH, the latter is finite by Subsection REF, and hence so is the monodromy group. The assertions of the lemma are a formal consequence of this fact. |
math/0010020 | Since MATH is MATH-equivariant, it is enough to prove that MATH is an isomorphism. To this end, one first shows that MATH is a local isomorphism in codimension one (this is based on simple type of local NAME theorem) and has discrete fibers. This implies that MATH has no ramification. So MATH is a local isomorphism every where. We wish to show that MATH is proper; the simple connectivity of MATH will then imply that MATH is an isomorphism. This will follow if we prove that MATH is proper. In other words, we want to show that MATH extends continuously to the one-point compactifications of its domain and range. Let MATH be a strictly semistable divisor of degree MATH on MATH. So MATH has a point of multiplicity MATH. Let MATH be a small oriented circle around this point. Then the preimage of MATH in MATH consists of MATH disjoint circles. If MATH is one of these, then MATH is by REF the residue of a differential with a simple pole and hence nonzero. The cycle MATH subsists under small deformations of MATH and for MATH in a neighborhood of MATH the corresponding integral MATH is then analytic in MATH and nowhere zero. If MATH is reduced, then MATH defines an isotropic element of MATH. On the other hand, by REF , MATH tends to MATH, as MATH approaches MATH. So the same is true for the expression MATH . It now follows from our explicit description of the NAME topology in REF that the image of MATH under MATH tends to the cusp of MATH, as MATH tends to MATH. This proves that MATH is proper. So both MATH and MATH are isomorphisms. Since MATH and MATH are normal one point compactifications of MATH and MATH respectively, the continuous extension MATH is in fact an isomorphism. |
math/0010020 | The group MATH may be identified with the orbifold fundamental group of MATH. Via the orbifold isomorphism MATH, we then get a MATH-covering. This covering factorizes over a covering of MATH with the kernel of MATH as covering group. Since MATH is trivial in MATH, a simple loop around a deleted hyperplane has monodromy of order three, and so the covering over MATH extends as an unramified covering over the smooth part of MATH: we now have a connected unramified covering over MATH. Since MATH is simply connected, this covering must be trivial. We conclude that MATH is injective. From this it follows that MATH is injective as well. |
math/0010020 | It is clear that the projection MATH is a MATH-covering. There is no ramification outside the discriminant divisor MATH since MATH is there locally liftable to a morphism to MATH. The remaining statements follow easily. |
math/0010020 | The first assertion follows from the fact that MATH is MATH-equivariant and of degree MATH and the second from the observation that MATH. The last clause requires more work. In view of the connectedness of MATH, it is enough to prove that assertion for one particular rational elliptic surface. We take the case studied in REF, where MATH is the set of MATH-th roots of unity and MATH is the curve with MATH-action. As noted in REF , MATH is the discriminant divisor of an elliptic surface, but we will exhibit such a fibration more directly. Consider the action of the (order REF) subgroup MATH generated by MATH. The orbit space MATH is a MATH-covering of MATH. If we identify the latter with MATH by means of the affine coordinate MATH, then we see that MATH has total ramification over MATH, a fiber with two points over MATH and a fiber with three points over MATH. These properties imply that MATH has genus one and more than that, namely that MATH is MATH-equivariantly isomorphic to the NAME curve MATH. The NAME curve supports a MATH-equivariant elliptic fibration. This pulls back to a MATH-equivariant elliptic fibration over MATH and that in turn descends to an elliptic fibration on MATH. We therefore denote the resulting MATH-morphism MATH. The induced map on the first cohomology MATH is MATH-equivariant. We identify the MATH-module MATH with the algebra MATH defined in REF. It is clear that the image of MATH is the MATH-submodule spanned by MATH . The image MATH of this element in MATH is easily calculated to be of the form MATH, with MATH . We claim that MATH is a MATH-vector: this is a straightforward computation or one invokes REF and the fact that the Hermitian form is multiplied by MATH. So MATH is a MATH-vector. |
math/0010020 | Consider the case when the closed fiber represents a general point of MATH, MATH, MATH or MATH. The image of such a point in MATH is a semistable orbit of a degree MATH divisor on MATH of type MATH, MATH, MATH, MATH respectively. So its image under the period isomorphism is going to be perpendicular to a (primitive) sublattice MATH of MATH of type MATH, MATH, MATH, MATH respectively. In the last two cases, the central component of MATH is in MATH and so the morphism MATH will be zero. This implies that in these cases MATH contains MATH. This shows that in terms of the notation of REF MATH is of type MATH in in the MATH-case and of type MATH and in the MATH-case. So we then find a point of MATH and MATH respectively. We now show that for MATH we cannot end up with a point of MATH. Since we have a period isomorphism, it then will follow that we must get a point of MATH and that in the remaining case MATH we get a point of MATH. We note that in the MATH-case, the lattice MATH is accounted for by MATH, where MATH is the irreducible component of MATH that lies over MATH. Since the map MATH is constant on MATH, it follows that MATH. It follows that MATH is of type MATH. A priori this lattice might be imprimitive, but it certainly does not contain a lattice of type MATH. We know that both MATH and MATH map to MATH. So they will map to distinct cusps of MATH. Hence is enough to show that MATH maps to MATH: then MATH must necessarily map to the other cusp MATH. A similar argument as used for MATH shows that a generic point of MATH is mapped to cusp of MATH that is perpendicular to a sublattice MATH isomorphic to MATH. Then MATH is a primitive isotropic line whose image in MATH is the cusp in question. A primitive isotropic line of type MATH is not perpendicular to a lattice of type MATH, whereas one of type MATH is. So MATH maps to a cusp of type MATH. |
math/0010020 | Let MATH be the maximal integer for which MATH is nonempty. So MATH is an isomorphism, but MATH is not. So MATH is locally the intersection of MATH members of MATH in general position. From this it follows that the blowup of MATH factorizes over MATH. The pull-back of MATH to MATH is up to a twist with a principal ideal equal to MATH. The lemma now follows with induction. |
math/0010020 | Consider the set MATH of pairs of MATH-vectors MATH in MATH with MATH. The mod MATH reduction of a pair MATH is pair of vectors MATH in MATH with symplectic product MATH. The number of such pairs of vectors is MATH. The MATH-vectors mapping to MATH are the elements of the MATH-orbit of MATH and likewise for MATH. So the preimage of MATH in MATH is the MATH-orbit of the pair MATH. Hence MATH has MATH elements. The image of the map MATH consists of MATH-vectors, hence is the set of all MATH-vectors, since it is MATH-invariant. As there are MATH-vectors, we see that each MATH-vector occurs precisely three times. If MATH, then MATH and MATH are two other elements of MATH with the same sum. So there are no more elements in MATH with that property. Hence the span of MATH and MATH only depends on MATH. All the assertions of the lemma now have been proved. |
math/0010020 | Let MATH be a MATH-vector in MATH with MATH. It follows from the preceding that MATH is MATH-equivalent to MATH, with MATH and MATH a vector mentioned in one of the cases MATH. In these cases the exponent MATH is determined by the inner product of MATH with MATH. The last part of the corollary is straightforward. |
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