paper stringlengths 9 16 | proof stringlengths 0 131k |
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gr-qc/9910104 | As the discussion above shows, all independent functions are given by MATH and the functions MATH for all different values of MATH, MATH. In all cases MATH can be decomposed as MATH, MATH. The expansion REF then yields MATH . Induction over MATH then shows that all independent antisymmetric functions are given by MATH . This set of functions can be simplified if we can generate the functions MATH and MATH. This can indeed be achieved by using the sequence MATH for MATH odd, and analogously for even MATH. In the NAME measure this sequence converges to MATH (the norm of MATH is independent of MATH), whereas for even MATH we can obtain MATH. Thus all antisymmetric functions are generated by MATH with MATH. |
gr-qc/9910104 | According to REF all antisymmetric functions on MATH can be generated by the functions MATH for MATH. With the preceding equations we see that this set of functions is equivalent to the set MATH. That all functions contained in the set of generating functions are orthonormal has already been shown above. |
hep-th/9910037 | One has to verify that the map is consistent with REF ; this is done in REF. The verification that MATH is immediate if one uses MATH and MATH. Recall that we do not consider spinorial representations here, so MATH in REF . In the more ``down - to - earth" version with MATH, the calculation is more complicated, and sketched in REF. The bar is compatible with MATH, using the fact that MATH; note that for even MATH, MATH commutes with MATH, because MATH is invariant under the automorphism MATH. To see the last relation, observe that MATH is also an invariant tensor, hence it is proportonal to MATH, and in fact it is equal to MATH (consider for example, the highest weight state, where MATH). By REF , this implies that MATH and therefore MATH, so that MATH is equal to MATH. This is known explicitly for MATH. Hence REF follows from MATH, since MATH is a singlet. Therefore the bar is well - defined on MATH. |
hep-th/9910037 | REF follows from MATH . REF follows immediately from REF , and hermiticity is seen easily using REF : MATH . It only remains to prove that it is positive definite. For this, it is enough to check that MATH for all MATH, using the invariance of MATH and unitarity of MATH for MATH. This follows either by a continuity argument for MATH on the unit circle between REF and MATH using REF , or by a direct calculation, which is done in REF. There, the bound MATH will seen explicitly (the reason is that for larger MATH, MATH becomes indecomposable). |
hep-th/9910037 | Using REF , one can see that for MATH, MATH where MATH is the weight of MATH and MATH; this is because the term MATH in REF does not contribute since MATH, and the ``large" generators vanish as well, by the more explicit formulas in CITE. On MATH, one has MATH (no sum) by with MATH as in REF . Hence MATH, where MATH is the energy operator as defined in REF. Therefore MATH . Now MATH . Here we used MATH, MATH, and the fact that MATH for MATH, which is not hard to see. |
hep-th/9910153 | In order to reveal hidden dependencies between the NAME coefficients, we make use of the commutation relation MATH which implies that MATH where MATH is the lowest non-vanishing MATH-order of MATH, MATH . REF is equivalent to the consistency relations MATH . Note that the superscripts MATH denote the order in MATH, not in MATH. In order to solve REF we define MATH or inverted MATH . REF determines all MATH in terms of the MATH, MATH such that MATH may be rewritten as MATH . Comparison with REF shows that all breaking terms are NAME variations and can be absorbed into MATH by choosing MATH . Thus we have MATH that is, MATH . Now the lowest non-vanishing order of MATH is no longer MATH but MATH, and in this way MATH can be pushed higher and higher until MATH=REF to all orders in MATH. Putting MATH, the NAME identity REF becomes identical to the NAME identity on old minimal curved superspace CITE. Comparison of coefficients yields MATH . Correspondingly, we have MATH, MATH. REF together with REF yields the same value for MATH which is also obtained by requiring R invariance. Thus to zeroth order in MATH, the model is equivalent to the R invariant theory of CITE. The coefficients of the first order terms in MATH are obtained from REF as MATH . The linear terms will be treated separately in REF. |
hep-th/9910253 | CASE: The evenness of the fermionic MATH-operator is a direct consequence of the compatibility REF . CASE: Using the commutation relations REF and the projection REF for the matrices MATH as well as the compatibility REF and the NAME REF , the matrix elements on both sides of REF can be reduced to MATH . CASE: The proof is similar to the proof of REF . Using REF both sides of REF reduce to MATH . CASE: MATH . CASE: Using REF we obtain MATH and the assertion follows from REF . |
hep-th/9910253 | Let us first prove the case MATH. Then MATH and MATH. The last equation holds, since by hypothesis, MATH is unitary. For the case MATH we start from the NAME REF , MATH . By iterated use of REF we obtain MATH for MATH. Let us introduce the truncated cyclic permutations MATH, MATH, as above. MATH induces a cyclic shift on the MATH-tuple MATH and leaves the MATH-tuple MATH invariant. Using REF , it follows that MATH . Hence, MATH . Since MATH and MATH, the latter equation reduces the proof of REF for MATH to the case MATH, which was proved above. |
hep-th/9910253 | MATH . Here we used the regularity in the first equation. In the second equation we reversed the order of factors and introduced a product of NAME deltas. In the third equation we used the identity MATH which follows from REF . In the fourth equation we used that MATH and the fact that MATH is even. In the fifth equation we iterated the two previous steps of our calculation. Finally in the sixth equation REF entered. |
hep-th/9910253 | The lemma follows from REF to REF . |
hep-th/9910253 | Using REF , the corollary to REF we obtain MATH . It follows that MATH . Finally, by specifying MATH, we arrive at REF . |
math-ph/9910002 | This is easily proved by induction on the number of corners, starting from the case of a rectangle. |
math-ph/9910002 | Note first that MATH is connected. Let MATH be harmonic on MATH. Given the value of the imaginary part MATH at one vertex MATH, the value MATH for any other vertex MATH in MATH is uniquely determined as follows. Take a path in MATH from MATH to MATH. Each edge of the path crosses an edge of MATH. One of the NAME equations ( REF or REF ) at the crossing point determines the difference in values of MATH at the endpoints of this edge. The value MATH is obtained by summing this difference along the path. The harmonicity of MATH implies that the value MATH obtained is independent of the path chosen. |
math-ph/9910002 | Since we already have MATH, it suffices to show that MATH is real when MATH, pure imaginary when MATH and zero in the remaining cases. If we order the vertices of MATH in such a way that all the MATH are first, then MATH then MATH and then MATH, then the matrix MATH in this basis has the form MATH where MATH are real matrices. The conjugate of the above matrix by the matrix MATH is real. Hence the inverse of MATH has the same form as MATH. This completes the proof. |
math-ph/9910002 | The first two properties in both cases follow from MATH . This equation is valid at every vertex of MATH except the exposed vertices (which do not have MATH neighbors). The third property in each case follows by definition, since MATH is the harmonic conjuate of MATH and MATH is the harmonic conjugate of MATH. |
math-ph/9910002 | There is the following relation between MATH and the NAME 's function for the plane. The real part of MATH is the unique function on MATH satisfying MATH and tending to MATH at infinity (see REF , and recall that MATH is the limit of MATH on square regions centered at the origin). Now the classical NAME 's function MATH on MATH satisfies MATH and for any fixed MATH, MATH as MATH (see REF ). As a consequence we have MATH where on the right we used coordinates on MATH which has index MATH in MATH. Using REF we have MATH where we used MATH. A similar argument holds for the imaginary part. |
math-ph/9910002 | Let MATH be equal to MATH except in a MATH-neighborhood of the MATH, and such that MATH is flat and horizontal or vertical in a MATH-neighborhood of the MATH. We will first prove the theorem for MATH for any fixed MATH. We will do only the case MATH. The case MATH is identical using the imaginary part of MATH rather than the real part of MATH below. Let MATH be the NAME 's function on MATH (recall the construction of MATH from REF), that is, the function which satisfies MATH and MATH when MATH. The function MATH, considered as a function of MATH, is a linear combination of the NAME 's functions MATH and MATH for MATH since it is harmonic off of these vertices. In fact since MATH for some constants MATH, we have MATH . By REF below, the rescaled NAME 's function MATH (considered as a function of MATH) converges away from MATH to a continuous harmonic function with a logarithmic singularity at MATH and boundary values MATH. (This is the place where we need MATH rather than MATH.) Similarly by REF the difference MATH converges. It remains to show that the coefficients MATH in REF converge as MATH. Note that if MATH is simply connected then MATH and we are done. For general MATH, the right hand side of REF automatically satisfies REF defining the coupling function, but the NAME 's functions MATH do not in general have single-valued harmonic conjugate. It is necessary to choose the MATH so that the harmonic conjugate of the right-hand side of REF is single-valued. We show that in fact the MATH are uniquely determined by this property. We will use the language of electrical networks, see for example, CITE. Consider the graph MATH to be a resistor network with resistances MATH on each edge. The function MATH is the potential at MATH when one unit of current flows into the network at MATH and the boundary MATH is held at potential MATH. The MATH must be chosen so that, when currents MATH flow into the network at MATH, and current MATH flows into the network at MATH, and the boundary is held at potential MATH, then the net amount of current exiting each boundary component MATH is zero. For, the harmonic conjugate is the integral of the current flow: the integral of the current crossing a closed curve surrounding MATH is MATH if and only if the harmonic conjugate is single-valued around that curve. We claim that given any MATH real numbers MATH such that MATH, there exists a unique choice of reals MATH such that, when currents MATH flow into the network at MATH, and the boundary is held at potential MATH, the net current flow out of each boundary component MATH is MATH. This will then determine the MATH, because letting MATH be the current flow out of the boundaries from the function MATH (we mean, when MATH unit of current flows in at MATH and MATH flows out at MATH), we must choose the unique MATH to exactly cancel this flow. To prove the claim, note that the map MATH which gives the outgoing currents MATH (and therefore MATH as well) as a function of MATH is linear (this is the principle of superposition). It suffices to show that the determinant of MATH is nonzero. However on each column of the matrix of MATH (in the basis MATH and MATH) the diagonal entry is the only negative entry: MATH induces a positive net current flow out of each boundary component except the component MATH which contains MATH, since MATH is a positive harmonic function. Furthermore the diagonal entry in MATH is larger than the absolute value of the sum of the other entries in that column, since a nonzero amount of current flows out of MATH, that is, MATH (and the total inflowing current equals the total outflowing current). This implies that MATH (see REF below). Now as MATH tends to MATH, the rescaled NAME 's function MATH converges REF . This implies that the entries of the matrix of MATH converge: the pointwise convergence of a sequence of harmonic functions implies convergence of their derivatives (even in the discrete case), due to NAME 's formula: the derivative at a point is determined by integrating the values of the function on a neighborhood of that point against (the derivative of) the NAME kernel. By integrating the derivative we get convergence of the net current flow out of each boundary. Furthermore the amount of current out of MATH due to MATH is bounded from below. This implies that MATH is bounded away from MATH REF . Since the difference in NAME 's functions MATH also converges REF , the net current out of MATH from MATH converges. Therefore the MATH converge as well. We conclude that MATH converges. The MATH-convergence of MATH implies convergence of its derivatives and so by integrating we get local convergence of MATH as well. By uniqueness of the harmonic conjugate (up to an additive constant) we have that MATH converges (the constant is determined by the fact that it is zero at MATH). In conclusion when MATH, MATH converges to an analytic function (of MATH) with all the properties of the function MATH. Furthermore the proof shows that there is a unique function with these properties. When MATH then MATH converges to MATH which is also unique. When MATH, the main contribution to MATH is from MATH; the unrescaled NAME 's functions MATH contribute at most MATH. Since MATH (see the proof of REF ), we conclude that MATH. This gives the ``local" term in the statement. The above holds for MATH for any MATH. It remains to see that when MATH the functions MATH on MATH converge to MATH on MATH. This follows from REF below, and the fact that the NAME map from MATH to MATH converges (if appropriately normalized) to the identity mapping. Therefore the result holds for MATH as long as MATH sufficiently slowly. |
math-ph/9910002 | We use the notation of the previous proof. If MATH, then by REF , the function MATH is already MATH for MATH near the boundary of MATH except at the point MATH. Therefore the MATH will all tend to MATH as well. The result follows if we define MATH. On the other hand if MATH, then by REF , the function MATH has two poles (each of residue MATH) within MATH of MATH. The remainder of the proof is similar to that of the previous theorem. |
math-ph/9910002 | We already know that MATH are analytic in the second variable. Going back to the coupling function, for a fixed black vertex MATH not adjacent to MATH we have MATH . If MATH and MATH this gives in the limit (using REF ) MATH and if MATH and MATH this gives MATH . These can be combined into a single complex equation MATH . Similarly if MATH this gives MATH . Summing these gives MATH and taking their difference and conjugating gives MATH. This proves the first two statements. As a function of MATH, the function MATH has all the properties of MATH except that the residue at MATH is MATH. Similarly the function MATH has all the properties of MATH except that the residue at MATH is MATH. So letting MATH be the real and imaginary parts of MATH we have that MATH has residue MATH at MATH, and all the other properties of MATH, and so must equal MATH since MATH is unique. A similar argument shows that MATH . The equations for MATH and MATH follow. |
math-ph/9910002 | Gaussian elimination using rows preserves this property: if for each MATH we multiply the first row by MATH and subtract it from the MATH-th row, the first column of the new matrix is all MATH except for the first entry MATH, and the remaining MATH submatrix still has the property in the statement. For example the first column of the submatrix is MATH and MATH . |
math-ph/9910002 | Let MATH. From REF , on the plane MATH we have MATH . The function MATH is harmonic (as a function of MATH) on all of MATH (including MATH) and has bounded boundary values, since MATH is MATH on the boundary of MATH and MATH is zero there. Let MATH be the continuous harmonic function which has boundary values equal to the boundary values of the limit MATH . Since these boundary values are continuous in the limit, MATH exists and is unique. Note that the boundary values of MATH are within MATH of the limiting values REF . Restrict MATH to a function on the vertices of MATH. The discrete Laplacian of MATH at a vertex MATH is: MATH and when MATH is small we can approximate this using the NAME expansion of the smooth function MATH, yielding MATH . Therefore MATH has discrete Laplacian which is MATH on MATH, and the boundary values are MATH. A standard argument now shows that MATH is close to MATH: the function MATH has discrete Laplacian which is a constant; choose constants MATH sufficiently large so that MATH and MATH on MATH. By the maximum principle for superharmonic functions, these functions must take their maximum value on the boundary of the domain MATH. Since MATH on the boundary of MATH, we conclude that MATH . Therefore MATH converges to the function MATH which has boundary values MATH and a single ``pole" of residue MATH at MATH. This is MATH times the MATH-derivative of the continuous NAME 's function. |
math-ph/9910002 | Reflect MATH across the boundary edge near MATH (the edge consisting of vertices in MATH) to get a graph MATH. Glue MATH and MATH along their common edge in a MATH-neighborhood of MATH. A harmonic function MATH on MATH which is zero on the boundary extends to a harmonic function on this glued graph by setting MATH when MATH is the reflection of MATH. In other words the NAME 's function MATH on MATH is the difference of two NAME 's functions on MATH; one centered at MATH and one centered at MATH. On the glued graph MATH, the vertices MATH are at distance at least MATH from the boundary MATH, but only distance MATH from each other. The argument of REF can then be applied in this case, replacing MATH by MATH. |
math-ph/9910002 | This follows from REF , induction on MATH and the fact that MATH . |
math-ph/9910002 | Letting MATH denote the ``size" of the moment, it suffices to show that a moment of size MATH is smaller than MATH for a constant MATH. Let MATH be the paths of integration in REF . We can choose the MATH so that no two are closer than MATH for some constant MATH; indeed, we can choose the paths so that the distance between MATH and MATH is at least MATH. Since MATH and MATH are MATH, in the determinant in REF the MATH-entry is at most MATH in absolute value. The determinant of a matrix is bounded by the product of the MATH-norms of its rows, and each row of the determinant in REF has MATH-norm bounded by MATH for another constant MATH. Therefore the sum of the integrals in REF is bounded by MATH for a constant MATH. This completes the proof. |
math-ph/9910011 | Define MATH, for MATH, and MATH for MATH. Then, since MATH is a hereditary cone, for MATH with MATH one has both MATH, and thus MATH is evident from MATH. For MATH with MATH at least one of MATH must not be in MATH. Hence, MATH and MATH by definition of MATH. Thus additivity holds in any case. That MATH is also positive homogeneous is clear. Finally, remind that, according to polar decomposition MATH for MATH, one has MATH and MATH, with the partial isometry MATH. Since MATH is a two-sided ideal of MATH, from this one infers MATH if, and only if, MATH. From this in view of the definition and since MATH is invariant on MATH also invariance of MATH on MATH follows. |
math-ph/9910011 | The sequence MATH is increasing, for each MATH. Thus MATH exists in the extended sense. Especially, from REF in the limit then additivity of MATH follows, whereas from REF homogeneity and invariance can be seen. Thus, MATH is a trace (see REF). By construction MATH for each compact positive MATH of finite rank. Thus MATH is non-trivial. However, since MATH is infinite dimensional, MATH will occur for some positive compact operators. To see that MATH is semifinite requires to prove that for MATH with MATH there existed a sequence MATH with MATH and MATH such that MATH. Note that by definition of MATH, MATH implies that MATH cannot be of finite rank. Hence, MATH can be written as MATH, with infinitely many mutually orthogonal one-dimensional orthoprojections MATH and all MATH. Clearly, for each MATH the operators MATH are of finite rank and obey MATH. Also, owing to MATH for MATH, one has MATH, and therefore MATH follows. Thus MATH is semifinite. Suppose MATH is a non-trivial trace. Thus MATH, for some positive compact MATH. Suppose MATH for some MATH of finite rank. According to additivity and homogeneity of MATH there has to exist a one-dimensional subprojection MATH of a spectral orthoprojection of MATH with MATH. The same arguments for MATH ensure that MATH, for some one-dimensional subprojection MATH of some spectral projection of MATH. But since MATH and MATH, with MATH, by invariance of MATH one has MATH. Hence MATH, and MATH for each one-dimensional orthoprojection MATH. Put MATH. Then MATH, and thus MATH for each positive operator MATH of finite rank. Finally, if MATH is not of finite rank, let MATH be the above approximating sequence of MATH by finite rank operators MATH. Also in such case MATH follows. Hence, in view of the above relation over the operators of finite rank, and since MATH holds, MATH. |
math-ph/9910011 | Let MATH be an o.n.s. with MATH, for all MATH, and be MATH the orthoprojection with MATH. Then, by positivity of MATH one has MATH, and therefore and in view of REF one gets MATH. According to REF then MATH follows. On the other hand, if MATH is the orthoprojection onto MATH, according to REF for each MATH certainly MATH. From this in view of REF once more again MATH is seen. Taking together this with the above estimate provides that equality has to occur. |
math-ph/9910011 | If MATH is fulfilled, then MATH, for each one-dimensional orthoprojection MATH. Owing to MATH (see REF ) then MATH follows. Suppose MATH. Then, MATH, and if a decomposition MATH with singular MATH exists, then MATH, for some (and thus any) one-dimensional orthoprojection MATH, and the following two alternatives have to be dealt with : firstly, if MATH is vanishing on all positive operators of finite rank, MATH is singular, and MATH and MATH have to be chosen (see above). Secondly, if MATH does not vanish on all positive operators of finite rank, according to REF there exists unique MATH with MATH, for each MATH, with equality occuring on any operator of finite rank. Hence, in defining MATH, for each MATH with MATH, and MATH else, we get a positive map MATH which does not vanish identically on the positive compact operators, but which is vanishing on all positive operators of finite rank. From the previous and since both MATH and MATH are traces, also additivity, positive homogeneity and invariance of MATH at once follow. Hence, MATH is a singular trace, which is easily seen to obey MATH, with MATH. |
math-ph/9910011 | Note that, in contrast to the spectral characterization of positive compact operators, the spectral theorem in case of a non-compact MATH with MATH provides that MATH has to be fulfilled, for some non-zero MATH and orthoprojection MATH with MATH (for one MATH at least the corresponding spectral eigenprojection MATH has to meet the requirement). But then, due to normclosedness of the compact operators, and since for each positive MATH one has MATH (uniform closure), such type of estimate has to exist in each case of a non-compact positive operator MATH. On the other hand, if MATH is fulfilled, for some non-zero MATH and infinite dimensional orthoprojection MATH, in view of this relation the equivalence of MATH with the unit operator MATH will imply MATH to be invertible, for the partial isometry MATH achieving MATH, MATH. Thus, owing to the non-triviality of the ideal MATH, MATH has to hold. Due to two-sidedness of MATH the latter requires that also MATH was non-compact. |
math-ph/9910011 | For MATH the assertion is trivial. For non-zero MATH and positive MATH with MATH also MATH must be compact since otherwise the criterion of REF were applicable to MATH with resulting in a contradiction to the assumed compactness of MATH, by the same criterion. |
math-ph/9910011 | Let MATH be a sequence of orthoprojections with MATH, for each MATH. Then, for each MATH, MATH. Note that MATH holds. Since also MATH is a non-trivial trace, by REF there is unique MATH with MATH, for each MATH. Hence, by normality of MATH and since MATH is normal, MATH follows, for each MATH. |
math-ph/9910011 | In view of REF and since MATH is a two-sided ideal, the validity of the first assertion follows as an immediate consequence of REF together with the fact that for each operator MATH of finite rank MATH is a null-sequence and thus is bounded. Finally, owing to REF for MATH and MATH with MATH one infers MATH, and then MATH according to REF implies also MATH. |
math-ph/9910011 | Let us consider the sequence MATH of partial averages MATH, MATH. Since these all are unital positive linear maps, by MATH-weak compactness of the closed unit ball in MATH the sequence of partial averages then must have a MATH-weak cluster point MATH which has to be a unital positive linear map, too. Since then MATH has to be fulfilled for some appropriately chosen subnet MATH, the inclusion MATH gets evident. Since MATH and MATH hold, for each MATH, and since owing to normality of MATH for each MATH also MATH is fulfilled, by argueing with the mentioned subnet one infers that MATH. From this MATH and MATH follow, for each MATH. Thus in view of the above MATH follows. Hence, MATH is a projection of norm one (conditional expectation) projecting onto the fixpoint algebra of MATH and which satisfies REF . To see REF , note first that owing to MATH for MATH odd, and MATH for MATH even, one certainly has MATH, for each MATH. Hence, the action of the MATH-th average MATH to the orthoprojection MATH can be estimated as MATH (here MATH means the integer part), and thus for all MATH one has MATH. From this and MATH then especially MATH follows, for each MATH. Hence MATH, for each MATH. Since for each MATH with MATH one has MATH, from the previous together with positivity of MATH also MATH follows. By linearity of MATH and since MATH is the linear span of MATH this remains true for each MATH. But then, for MATH with MATH by continuity of MATH and in view of REF one infers MATH, which is equivalent with REF . |
math-ph/9910011 | Let MATH be constructed as in REF . By positivity and unitality of MATH, for each MATH also MATH is a state. In view of REF - REF this state then obviously satisfies REF - REF . That MATH is MATH-compact and convex is evident from the linear nature of REF - REF . Finally, in accordance with REF one has MATH, for each MATH and all MATH. Now, let MATH be any orthoprojection with MATH. Then, MATH, and owing to MATH there has to exist MATH with MATH. Thus MATH and MATH. In view of the above from the latter by positivity of MATH then MATH follows. Hence, each MATH is singular. |
math-ph/9910011 | Since MATH is the positive cone of a two-sided ideal of compact operators, for MATH and MATH we have that MATH, and these are compact operators again. Hence, in view of REF from REF both MATH and MATH follow, which in line with REF means that MATH is positive homogeneous and invariant. It remains to be shown that MATH is additive. First note that according to the left-hand side estimate of REF within MATH one has MATH. Hence, by positivity and linearity of MATH, REF yields MATH . Now, to each compact operator MATH let MATH, that is, MATH arises from MATH by application of the one-step left-shift. Also, on MATH let a linear map MATH be defined by MATH, for all MATH, at MATH. One then has MATH . Note that MATH whenever MATH. We are going to estimate MATH for MATH. Since both MATH and MATH are monotoneously increasing, in view of REF for each MATH the follwing estimates at once can be seen to hold, for all MATH : MATH . On the other hand, we also have MATH . From these two estimates we infer that MATH for MATH is a null-sequence in MATH, that is, MATH is fulfilled. According to the choice of MATH and in accordance with REF we thus have the following to hold : MATH . Let us come back to our above MATH. Having in mind the definitions of the positive linear operators MATH and MATH as well as the meanings of MATH and MATH, it is easily inferred that from the right-hand side estimate in REF when divided by MATH, and considered for all MATH, the estimate MATH can be followed to hold in MATH. By positivity and linearity of MATH from this then MATH follows. Now, in view of REF one has MATH, whereas from REF one concludes that MATH. These facts together with REF fit together into the estimate MATH, which in view of REF says that MATH has to be valid. The latter and REF then make that the desired additivity MATH holds. |
math-ph/9910011 | The validity of REF follows since the traces in question all are obtained as extensions of the maps given in REF, which satisfy REF and which have range MATH (and not merely MATH). Since each state MATH obeys REF, in view of the previous and REF also REF follows. Finally, for each MATH the sequence MATH is a null-sequence, and therefore especially MATH, and as a special case of REF then MATH follows. Hence, MATH is a singular trace. |
math-ph/9910011 | For compact operator MATH suppose MATH to be fulfilled. Then, in case of MATH, for MATH with MATH, let MATH be chosen such that MATH . From this for each MATH we get MATH . Since MATH is a strictly monotone decreasing function and the sequence of the singular values is decreasingly ordered, with the help of MATH which holds for MATH the above estimate REF implies MATH . From this for all MATH is obtained. Considering these estimates for MATH then yields MATH . Note that in case of MATH by positivity of all MATH instead of the previous one finds MATH, for any MATH. Thus, since MATH can be chosen arbitrarily small, in either case MATH follows. |
math-ph/9910011 | In case of MATH one has MATH as well as MATH, by triviality. From the latter MATH follows. Hence, for each operator MATH of finite rank the assertion is true, with MATH. Suppose now that MATH is not of finite rank, MATH. In view of REF , upon possibly considering instead of MATH a scaling MATH by a suitably chosen real MATH, without loss of generality it suffices if the assertion for MATH with MATH can be shown. In line with this assume such MATH. By the spectral theorem there exists a spectral representation of MATH as an operator NAME MATH, with projection-valued measure MATH derived from a left-continuous spectral family MATH, that is, a family of orthoprojections obeying MATH, for MATH, MATH, for MATH and MATH, for each MATH, with MATH. By convention, for MATH, then MATH and MATH, and so on accordingly, where for example, MATH stands for the greatest lower bound MATH. By means of some functional calculus and owing to normality of the trace MATH it is easily inferred that REF can be represented as an ordinary NAME integral : MATH with the monotone increasing function MATH given by MATH . But then, if the assumptions on MATH are fulfilled with MATH, all conditions for an application of NAME 's theorem CITE, are given (we refer to the formulation in CITE). In line with this the conclusion is that asymptotically MATH has to be fulfilled as MATH tends to infinity. Since MATH is a compact operator, in view of the properties of the spectral resolution MATH together with normality of MATH REF provides a right-continuous, integral-valued step function which is constant between inverses of neighbouring spectral values of MATH. Especially, in case of MATH with MATH one infers that MATH for all MATH with MATH yields MATH. A moments reflection then shows that with respect to each term of the ordered sequence MATH of all subscripts where the value of MATH jumps the relation REF in view of MATH and by continuity of the parameter MATH in particular also implies both MATH and MATH to be fulfilled. But then, since MATH holds for MATH, also MATH can be obtained from these limit relations. Thus under the condition of the hypothesis also MATH in case of MATH and which is not of finite rank. In accordance with our preliminary remarks the assertion then has to be true, in either case under the mentioned hypothesis. |
math-ph/9910011 | see CITE, p. CASE: The principal symbol of the NAME operator is MATH (NAME multiplication on spinors), thus MATH is a first order (elliptic) MATH-DO. Multiplication with MATH is a zero order operator, therefore MATH is a MATH-DO of order MATH. Its principal symbol is MATH, where MATH is the identity map of the fibre MATH of MATH. This principal symbol reduces on the cosphere bundle MATH to MATH. Thus, REF gives MATH . The area MATH of the unit sphere MATH leads to the right factor MATH. |
math-ph/9910011 | We refer to CITE, p. REF. |
math-ph/9910022 | Both results follow from the second-order resolvent identity REF , which yields: MATH . For the proof of the first claim, we take MATH and MATH. Then, the first term of REF is zero because MATH decouples MATH and MATH and the second term is zero because MATH decouples MATH and MATH. Thus MATH . It follows that for any MATH . (note that for MATH: MATH.) In estimating the terms on the right hand side of REF let us consider first the conditional expectation of the central factors, MATH. Only these factors depend on the values of the potential at MATH and MATH, and therefore they can be replaced by their conditional expectation MATH. As will be proven in the appendix, under the regularity condition MATH these are uniformly bounded REF : MATH . (The proof involves a reduction to a two-dimensional problem via the NAME formula, and a two-dimensional NAME estimate.) Once the central factor in each expectation on the right hand side of REF is replaced by the above bound, what remains there are two independent random variables which are MATH and MATH . The expectation now factorizes, and the resulting expression yields the first claim of the Lemma. For the second claim, we take MATH. Once again the first term of REF is zero because MATH decouples MATH and MATH. However, the second term is non-zero, and we obtain MATH . At this point we may not use the previous argument, since in the last expectation MATH affects each of the first two factors and MATH affects each of the last two factors. However, the dependence of each of these factors on the potentials is of a particularly simple form: they are ratios of two functions (determinants) which are separately linear in each potential variable. Using the decoupling hypotheses, that is, the regularity conditions MATH and MATH, the expectation may be bounded by the product of expectations. Specifically, we prove in REF that: MATH . Once again, we are left with a product of two independent random variables, MATH and MATH. The factorization of the remaining expectation yields the second claim of the Lemma, REF . |
math-ph/9910022 | Starting from the first order resolvent identity, REF , and taking expectation values of its matrix elements, we find: MATH where MATH, and MATH. It suffices, therefore, to show that in the last term the factor MATH may be replaced (for an upper bound) by the constant MATH. This follows through a decoupling argument which we present in the Appendix - see REF . |
math-ph/9910022 | Assume that for some MATH and a finite region MATH the smallness REF holds. By REF and translation invariance, we learn that for any region MATH and any MATH with MATH: MATH where MATH of REF , and MATH is the translate of MATH by MATH. By REF , each of the terms in the sum is bounded by MATH. Since the sum is normalized by the prefactor MATH, REF permits to improve that bound for MATH by the factor MATH. Furthermore, the inequality may be iterated a number of times, each iteration resulting in an additional factor of MATH. One should take note of the fact that the iterations bring in NAME functions corresponding to modified domains. It is for this reason that the initial input assumption was required to hold for modified geometries, that is, not just for MATH but also for all its subsets. REF can be iterated as long as the resulting sequences REF do not get closer to MATH than the distance MATH. Thus: MATH with MATH. |
math-ph/9910022 | Our first goal is to show that under REF there is MATH such that for all pairs MATH with MATH and MATH, MATH with non-negative weights satisfying: MATH . We shall use this inequality along with its conjugate: MATH where MATH satisfy the suitable analog of the normalization REF . It is important that - unlike in REF , the functions which appear on the right hand side of REF are computed in the same domain as those on the left hand side. The first step is by REF , which yields MATH whenever MATH and MATH, with MATH specified in REF . Next, we apply REF , to bound MATH in terms of a sum of quantities of the form MATH with MATH. The result is initially expressed as a sum over bonds: MATH where, using translation invariance, MATH . Collecting terms, and pulling out normalizing factors, one may cast REF in the form REF with MATH . The smallness REF is nothing other than the assumption that MATH. The above argument proves REF . By the transposition, or time-reflection, symmetry of MATH REF also REF holds. (Such symmetry of MATH is not essential for our analysis: it suffices to assume that the smallness condition REF holds along with its transpose.) We proceed in the proof by iterating REF . However an adaptation is needed in the argument which was used in the proof of REF since the iteration can be carried out only as long as the two points (the arguments of the resolvent) stay at distance MATH not only from each other but also from the boundary MATH. The relevant observation is that for every pair of sites MATH there is a pair of integers MATH such that: CASE: MATH , CASE: the ball of radius MATH centered at MATH and the ball of radius MATH centered at MATH form a pair of disjoint subsets of MATH. For the desired bound on MATH, we shall iterate REF MATH times from the left, and REF MATH times from the right. Similar to REF , we obtain: MATH with MATH. |
math-ph/9910022 | Suppose that REF holds with some MATH and MATH. We need to show that also in finite systems the NAME function is sufficiently small between an interior point and the boundary. To bound the finite volume function in terms of the infinite volume one, we may use REF , by which MATH for any finite region MATH containing the origin. We need to show that for MATH with MATH large enough MATH . After applying REF to the terms on the left side of REF we find that the number of summands involved and their prefactors grow only polynomially in MATH, whereas under our assumption the relevant factors MATH are exponentially small in MATH. Hence REF is satisfied for MATH large enough. |
math-ph/9910022 | One may read the claim as saying that the function MATH lies in the space MATH of functions for which the following norm is finite: MATH . We shall deduce this claim after arriving first at a bound formulated within the larger space of bounded functions MATH. Let MATH be the linear operator with the kernel MATH. Within MATH the operator acts as a contraction, since its norm there is MATH (using REF ). It is convenient to paraphrase the assumption on MATH in the following form, which holds for all MATH: MATH with MATH the ``indicator function" of MATH. Iterating this relation MATH times, one obtains a bound in the form of a finite geometric series with a ``remainder" which is uniformly bounded by MATH. As MATH the reminder vanishes, since MATH, and one is left with a bound in the form of a convergent series: MATH . We now note that for a finite region MATH, the function MATH lies in the ``weighted-MATH space" MATH. The norm of MATH as an operator within MATH is easily seen to obey: MATH . The expression on the right hand side is convex in MATH, and by the temperedness assumption (the analog of REF ) it is finite for small enough MATH. Since convexity implies continuity, using REF we conclude that there is some MATH for which MATH . With this choice of MATH we conclude: MATH . |
math-ph/9910022 | To establish the claimed bound REF fix MATH, and let MATH. We note that for each MATH there is a unique element of the symmetry group, MATH, such that MATH. Starting from the kernel MATH which appears in REF , let us define a shift-invariant kernel MATH by: MATH . Due to the shift invariance of the distribution of MATH, REF implies that the function MATH is sub-harmonic, in the sense of REF , with respect to the kernel MATH, which satisfies REF and is tempered . Thus, a direct application of REF yields now the claimed bound REF . |
math-ph/9910022 | The situation to be discussed now is different from that encountered in the last proof in that now for each MATH the basic sub-harmonicity bound can be assumed only for points which are not too close to the boundary MATH. The claim made for the special case MATH is covered by the above analysis. However, the second claim, that is, REF , requires a somewhat different argument. The argument we shall use shadows the proof of REF , replacing there the weighted-MATH estimate by its weighted-MATH version. The starting observation is that MATH has the sub-mean property with respect to averages over either MATH or MATH - provided the point is at distance at least MATH from the other and from the boundary MATH. (In allowing the averaging procedure to occur from either side, we rely on the fact that the smallness condition holds for both the kernel MATH and its conjugate, or equivalently the fact that the smallness condition is assumed to hold for both MATH and MATH .) To cast the situation in terms reminiscent of the proof of REF , let us consider the function MATH as defined over the space of pairs, MATH, equipped with the distance function MATH . For MATH not in the set MATH, with MATH, we have the basic sub-mean estimate: MATH with MATH where MATH is given by REF . By repeating the arguments seen there we find that MATH obeys the analog of REF - formulated within the space MATH, with the set MATH replaced by MATH, and the operator MATH replaced by MATH defined by the kernel MATH. Unlike in the previous case, we have no fixed bound on the size of the set MATH. Thus we shall not use here the weighted-MATH estimate. However, we may reuse the argument applying it to weighted-MATH norm of MATH, which is defined as: MATH . The conclusion is that there is some MATH at which MATH. Equivalently: MATH as claimed in REF . |
math-ph/9910022 | By REF , to establish exponential decay at the energy MATH it suffices to show that for each MATH . Because the off diagonal elements are tempered we have the following bounds MATH for some MATH, and all MATH. Under the assumption REF : MATH . For this bound the sum was split according to MATH, and in the first case we used the uniform upper bound MATH. It is now easy to see that with an appropriate choice of MATH and MATH REF implies the claimed bound REF - for the given energy MATH. The extension to an interval of energies around MATH then follows from the continuity of the fractional moments of finite volume NAME functions. To show the sufficiency of the second condition, we first use REF to bound finite volume NAME functions in terms of the corresponding infinite volume funtions MATH . Splitting the sum as in REF , we get MATH . The combination of REF with REF , yields the claim - for the given energy. Again, the existence of an open interval of energies in which the condition is met is implied by the continuity of the finite-volume expectation values. |
math-ph/9910022 | Assume that REF is satisfied for all MATH. We shall show that this implies that for any power MATH with the constant MATH uniform in MATH. The stated conclusion then follows by an application of REF (and the uniform bounds seen in its proof). We shall deal separately with large and small MATH, splitting the two regimes at MATH. The case MATH is covered by the general bound of CITE, which states that: MATH for any MATH such that MATH . To estimate the resolvent for MATH, we shall use the fact that the function MATH is subharmonic in the upper half plane, and continuous at the boundary. The subharmonicity is a general consequence of the analyticity of the resolvent in MATH, and the continuity is implied through the continuity of the distribution of the potential. MATH serves as a convenient cutoff, which may be removed after the bounds are derived (since MATH in the strong resolvent sense). Let MATH be the triangular region in the upper half plane in the form of an equilateral triangle based on the real interval MATH with the side angles equal to MATH - determined by the condition MATH . The NAME representation of harmonic functions yields, for MATH, MATH where MATH is a certain probability measure on MATH. We now rely on the fact that this probability measure satisfies MATH . (This is easily understood upon the unfolding of MATH by the map MATH applied from either of the base corners of MATH, that is, from MATH, and a comparison with the NAME kernel in the upper half plane.) For MATH the integrand satisfies the exponential bound REF . Along the rest of the boundary of MATH we use the NAME bound REF . Putting it all together we get MATH . The claimed REF follows by simple integration, and the relation REF . |
math-ph/9910022 | We first prove REF and then indicate how the proof can be modified to show REF . Fix an energy MATH. For MATH, define MATH and let MATH . We will show that for suitable MATH, MATH and MATH, if MATH then the input REF : MATH is satisfied for all energies MATH. Exponential localization in the corresponding interval (and strip, with MATH) follows then by REF (and REF ). First we must show how to estimate MATH in terms of MATH. This is achieved by considering separately the contributions from the ``good set": MATH and its complement, the ``bad set": MATH. On the ``good set", MATH, the energy MATH is at a small yet significant distance REF from the spectrum of MATH. In this situation, we use the CITE bound, by which: MATH . The above estimate does not apply on the ``bad set". However, using the NAME inequality, we find that the net contribution to the expectation is small because MATH is small. The two estimates are combined in the following bound: MATH where MATH is any number greater than MATH for which the distribution of the potential is still MATH-regular (that is, MATH). The required bound, REF , is satisfied once one chooses MATH small enough so that MATH, and MATH large enough so that for MATH . Finally let us remark on how this argument can be adapted to prove REF . We simply define the good and bad sets differently: MATH and MATH , and then proceed as in the proof of REF using NAME 's inequality to estimate the contributions from MATH. It is easy to see that for large MATH, REF implies that the input for REF is satisfied. |
math-ph/9910022 | It is convenient to derive the result through the analysis of the finite volume operators obtained by restricting MATH to finite regions, MATH. It is generally understood that for each MATH and each increasing sequence of finite regions MATH which contain MATH and whose union is MATH, the associated spectral measures, MATH, converge in the vague topology to MATH. Thus, by REF, for any MATH: MATH. The upshot is that it suffices to prove the following statement regarding finite volume operators. Under the assumptions of REF there exist MATH (which depend only on the regularity assumptions for MATH) such that for any finite region MATH, any MATH, any MATH, and any MATH: MATH . Following is a summary of the proof of this assertion. Let us fix a finite region MATH and a pair of sites MATH. For simplicity of notation, we will suppress the region MATH and denote the restricted operator by MATH and the associated spectral measure by MATH . Since MATH is finite dimensional, MATH is a weighted sum of NAME measures supported on the eigenvalues of MATH. Integrals with respect to this measure are discrete sums. The argument of ref. CITE makes an essential use of the following representation of this measure. Let MATH, and let MATH be any other value in MATH. Denote MATH, with MATH the operator with the potential at MATH changed to MATH. Then, MATH . In what follows, we will take MATH to be a random variable independent of MATH and identically distributed. In this REF holds almost surely. A special case of REF is the formula (which was the basis for the important ``Kotani-argument"CITE) for the spectral measure at MATH . The above is a probability measure. Another normalizing condition is: MATH (which typically holds as equality). The reason for REF is that by the general structure of the spectral measures, MATH, with MATH satisfying MATH, where MATH is the projection onto the cyclic subspace for MATH which contains MATH. Let us first present the necessary estimates for the case that MATH is of finite NAME measure. Using the bound REF , and the NAME inequality, MATH where MATH is a small number to be specified later. By a further application of the NAME inequality, followed by the NAME inequality we obtain MATH where MATH is fixed by the equation MATH. Finally we evaluate: MATH where MATH is a uniform upper bound for MATH. These estimates can be combined to provide a bound of the form REF for MATH a finite interval, which was the case considered in ref. CITE. We shall now improve the argument, to obtain a statement which covers the case that the localized spectral regime is unbounded. Since we do not wish our final estimate to depend on the NAME measure of MATH, we seek a way of introducing an integrable weight MATH, so that the final bound involves the integral of MATH in place of MATH. This may be accomplished with the following inequality: MATH where MATH and MATH is any continuous function which is bounded and bounded away from zero. To prove REF , write MATH, and apply the NAME inequality followed by MATH . It is convenient to choose MATH, since MATH where MATH is a bounded random variable which depends only on the off-diagonal part of MATH. Upon taking expectations followed by a further application of the NAME inequality this leads to MATH where MATH. We estimate the two factors on the right hand side of this inequality separately. The first factor can be controlled by choosing MATH so that MATH . The exponents MATH, MATH, MATH, MATH are all specified once we choose MATH. Specifically, MATH, MATH, and MATH. Note that MATH. To estimate the second factor, we note that MATH is a sub-probability measure and MATH, so by the NAME inequality, MATH . Estimating the right hand side with the argument outlined above for MATH with finite NAME measure, we find that MATH which is uniformly bounded provided we choose MATH such that MATH. This is possible since MATH which can be made as large as we like. Thus, for any finite volume MATH can be bounded by a constant multiple of MATH raised to a certain power. Which multiple and which power depend only on the MATH-moments of the potential and the uniform bound on the conditional distributions MATH. By the vague convergence argument outlined at the start of the proof, this proves the theorem. |
math-ph/9910022 | Let us first consider MATH. For such energies REF is a consequence of a NAME type estimate on REF-dimensional subspace spanned by MATH. The key is to determine the correct expression for the dependence of MATH on MATH and MATH. Such an expression is given by the `NAME formula': MATH where MATH is a MATH matrix whose entries do not depend on MATH or MATH. In fact, MATH where MATH denotes the operator obtained from MATH by setting MATH and MATH equal to zero. The regularity condition MATH implies a NAME type estimate: MATH where MATH is any finite number such that for every MATH, MATH, and MATH . The desired bound REF follows easily from REF . (The factor, MATH, on the right hand side of REF arises as the square of the ``volume" of the region MATH. In the case MATH, we could replace this factor by MATH.) Although the NAME REF is true when MATH is replaced by any MATH, the resulting matrix MATH may not be normal if MATH. (The resolvent, MATH, is normal. However, given an orthogonal projection, MATH, the operator MATH may not be normal!) Yet, the NAME estimate REF holds only when MATH is a normal matrix. At first, this seems to be an obstacle to the extension of REF to all values of MATH. However, once the inequality is known for real values of MATH, it follows for all MATH from analytic properties of the resolvent. Specifically, the function MATH is sub-harmonic in the upper and lower half planes and decays as MATH. Hence, MATH is dominated by the convolution of its boundary values with a NAME kernel: MATH . By NAME 's theorem and REF for MATH, REF is seen to hold for all MATH. |
math-ph/9910022 | Note that given MATH with MATH where MATH is any number with MATH. |
math-ph/9910022 | Let MATH and MATH be two functions of the appropriate form for the decoupling lemma. Then, with MATH where MATH indicates an independent variable distributed identically to MATH. Now, if MATH and MATH are functions of MATH variables of the given form, then at fixed values of MATH, they satisfy the MATH variable decoupling lemma, so MATH . For fixed values of MATH and MATH, MATH and MATH (as functions of MATH) are again of the correct form to apply the MATH variable decoupling lemma. Thus, MATH . |
math-ph/9910022 | For each MATH, we define MATH . Property MATH amounts to the statement that MATH . In fact, if we let MATH then by the NAME inequality, it suffices to show that MATH and MATH are uniformly bounded. However this is elementary since MATH and MATH are continuous functions which are easily shown to have finite limits at infinity. |
math-ph/9910025 | We first note that, in the special case where MATH by CITE the spectral decomposition of MATH is given by the `NAME formula', which is equivalent to applying the multiplier MATH to the signature MATH where MATH . Here, and in what follows of this proof, the NAME formula also requires the condition that MATH where MATH, and we will refer to a multiplier of this type as a simple multiplier. Now applying this simple multiplier to the signature MATH we have MATH . But the first sum, with MATH, is just the spectrum of MATH contained in the spectrum of MATH . We next note that a similar situation occurs when we apply a second simple multiplier MATH to the above spectral decomposition MATH. That is, the sums are grouped into those terms whose last entry is zero, and those terms whose last entry is non-zero; MATH . We then extend this idea to the general case where the spectral decomposition of MATH is given by applying the multiplier MATH to the signature MATH where MATH is a compound multiplier computed as the `CITE; MATH . Here the simple multipliers MATH are regarded as permutable operators and the determinant is expanded in the usual way, with MATH and MATH for MATH. From this last statement it is obvious that the compound multiplier MATH is equal to MATH since the MATH row used to compute the determinant corresponding to MATH-is just MATH. Now for notational convenience we set the simple multiplier MATH and using the usual formula for determinant (summing over MATH, the symmetric group on MATH symbols) we have MATH . So MATH . But the sum REF is just the spectrum of MATH appearing in the spectrum of MATH. Finally, the requirement that MATH guarantees that the application of a simple multiplier to a signature MATH extends the length of the signature by at most one, since MATH. Thus, since there are only MATH sums in REF , application of a compound multiplier corresponding to a signature of length MATH decomposes the tensor product into a spectrum of signatures of length at most MATH, proving that the spectrum stabilizes. |
math-ph/9910025 | From CITE we know that for sufficiently large MATH the multiplicity of MATH in MATH is equal to the multiplicity of the identity representation MATH in the augmented tensor product MATH . For each MATH let MATH denote the homomorphism sending the irreducible representation of MATH with signature MATH into the MATH-module MATH . Then MATH where the homomorphisms MATH are defined as in REF . Let MATH denote the signature of the representation of MATH as the inverse limit of irreducible representations of MATH with signature MATH. Then we can define a homomorphism MATH by MATH where in REF , MATH denotes the projection of MATH onto MATH. Note that MATH or MATH are just the trivial MATH or MATH modules MATH, and that the MATH-module MATH is considered as a MATH-submodule of the MATH-module MATH. As remarked in REF, the dimension of MATH the space of all homomorphisms intertwining MATH and MATH stabilizes as MATH gets large. But this dimension is just the multiplicity of MATH in MATH which, in turn is equal to the multiplicity of MATH in MATH. It follows that at the (inductive) limit we have MATH or equivalently the multiplicity of MATH in the tensor product MATH is equal to the multiplicity of MATH in the tensor product MATH . |
math-ph/9910025 | To first show that MATH in fact lies in MATH it is sufficient to show (by the NAME theorem) that if MATH then, as in REF MATH . But since MATH `lies ' in MATH it transforms covariantly with respect to the NAME subgroup defined in REF so we have MATH . We next show that the MATH transform under the representation MATH in the same manner as the MATH. Since MATH is invariant with respect to the action MATH of MATH we have MATH which can succinctly be written as MATH. We also have that MATH where the MATH are the MATH-functions for the representation MATH. Now for any MATH we can assume that MATH for some MATH, so that MATH. Hence MATH, and it follows from the definitions of the symbols involved that MATH . Thus we seek to show that MATH . By the preceding remarks and the definition of MATH we then have MATH . Finally we have MATH which is REF . |
math-ph/9910034 | The substitution MATH in the left integral in REF yields MATH . Using the inverse MATH of MATH and the fact that MATH has regular MATH-decay REF , one shows that for every MATH there exists MATH such that MATH for all MATH. Since MATH, that is, MATH, MATH, the bounds REF imply MATH . The claimed result now follows by interchanging limit and integration by applying the dominated-convergence theorem. In order to show that this theorem is indeed applicable we have to distinguish the cases MATH and MATH. In the first case, we may use the decomposition REF together with REF to further estimate REF and construct an upper bound on MATH which is independent of MATH and has an integrable long-distance decay. In the second case, we use REF instead of REF . |
math-ph/9910034 | As in the proof of REF we construct asymptotically coinciding upper and lower bounds on MATH. For the upper bound we pick MATH and split the convolution integral into two integrals with domains of integration inside and outside a disk with radius MATH centered about the origin and estimate the two parts separately as follows MATH for sufficiently large MATH, since MATH has a regular MATH-decay and MATH is strictly increasing. Moreover, estimating the Gaussian MATH on the domain of integration yields MATH . Since MATH has sub-Gaussian decay it follows that MATH for all MATH such that the first term dominates the asymptotics of MATH. Employing the facts that MATH is strictly increasing and in MATH, we therefore arrive at MATH . For a lower bound we may proceed similarly MATH for sufficiently large MATH, which gives MATH . This completes the proof since MATH may be picked arbitrarily small. |
math-ph/9910034 | We will only give an outline since the proof copies exactly the strategy of CITE. First note that the theorem is immediate if MATH as MATH since MATH. We will therefore assume throughout the rest MATH REF as MATH and, moreover, MATH without loss of generality. Using REF below one shows that for every MATH provided that REF holds, respectively. To complete the proof we note that REF below gives MATH again supposing that REF holds, respectively. |
math-ph/9910034 | Since MATH for all MATH (compare CITE) the inequality follows by choosing MATH. |
math-ph/9910034 | We put MATH and split the domain of integration in REF into the three parts MATH, MATH, and MATH. The integral over the first part is estimated according to MATH due to monotonicity of MATH. We now employ REF or REF directly to show that MATH for sufficiently small MATH which implies the following upper bound for the integral over the second part MATH . Here we used REF with MATH and MATH which yields MATH for sufficiently small MATH, since MATH and MATH. Finally, MATH for sufficiently small MATH, for which MATH and MATH as MATH. |
math/9910001 | The necessity is obvious since a MATH-vector bundle isomorphism MATH restricts to a MATH-vector bundle isomorphism MATH, and the sufficiency follows from the fact that MATH is functorial. |
math/9910001 | It is easy to check that the map MATH sending MATH to MATH gives the inverse of the map in the lemma. |
math/9910001 | Let MATH and MATH. We know that the fibers MATH and MATH of MATH at MATH and MATH are isomorphic as MATH-modules. On the other hand, since MATH for MATH, the map MATH becomes a MATH-equivariant isomorphism if we consider a MATH-action on MATH given through an automorphism of MATH given by MATH. This implies the proposition. |
math/9910001 | We shall prove the latter statement in the proposition. In this proof, we do not need the assumption that MATH is isomorphic to a subgroup of MATH. Suppose that MATH has MATH-extensions MATH and MATH. Then MATH is a MATH-module and one-dimensional by NAME 's lemma because MATH is irreducible. One can easily check that the map MATH sending MATH to MATH is a MATH-linear isomorphism. Therefore, any MATH-extension of MATH is the tensor product of MATH and a one-dimensional MATH-module viewed as a MATH-module. Moreover, since MATH as MATH-modules, different one-dimensional MATH-modules produce different MATH-extensions of MATH. This proves the latter statement. It is well-known and easy to see that there are two MATH- or MATH-modules of dimension one for MATH odd, and four MATH-modules of dimension one for MATH even. This proves the statements on the number of MATH-extensions in REF . It remains to see that MATH has a MATH-extension in REF . REF is well-known if MATH is finite cyclic, see for instance CITE. The book CITE treats only finite groups (so that MATH is finite), but some direct proofs work even if MATH is infinite. The reader can find one of the direct proofs in CITE. We suspect that REF is known even when MATH is isomorphic to MATH, but there is no literature as far as we know. Since the proof we found is rather long and has independent interest, we will give it in CITE. REF is also well-known, see CITE or CITE for an elementary proof. |
math/9910001 | The necessity follows from REF , so we prove the sufficiency. Let MATH be a MATH-module with MATH-invariant character. We distinguish three cases according to MATH. CASE: The case where MATH is infinite, that is, MATH or MATH. In this case the MATH-action on MATH is transitive, and the isotropy subgroup MATH of a point in MATH is MATH when MATH, and contains MATH as an index two subgroup when MATH. Therefore MATH has a MATH-extension by REF since the character of MATH is MATH-invariant (in particular, MATH-invariant) and MATH is trivial or of order two. This together with REF implies the existence of a MATH-vector bundle over MATH with MATH as the fiber MATH-module. CASE: The case where MATH. In this case MATH. Set MATH and consider the product MATH-vector bundle MATH. Choose a MATH-submodule isomorphic to MATH in the fiber of MATH at MATH, and identify it with MATH. The MATH-invariance of the character of MATH implies that MATH. Viewing MATH and MATH as MATH-invariant subspaces of MATH, one can connect MATH and MATH through a continuous family of MATH-invariant subspaces along the arc of MATH joining MATH and MATH, in other words, for each MATH in the arc one can find a MATH-invariant subspace in the fiber of MATH at MATH so that the family of MATH-invariant subspaces varies continuously on the points MATH. This is always possible because the set of such MATH-invariant subspaces of MATH is homeomorphic to a product of NAME manifolds which is arcwise connected. Translating the family of MATH-invariant subspaces by the action of MATH repeatedly yields the desired MATH-subbundle of MATH. CASE: The case where MATH. By REF , there exist MATH- and MATH-extensions MATH and MATH of MATH respectively. Set MATH and consider the product MATH-vector bundle MATH. Then MATH and MATH are contained as MATH- and MATH-submodules in the fibers of MATH at MATH and MATH, respectively. Since MATH, it is possible to connect MATH and MATH through a continuous family of MATH-invariant subspaces along the arc of MATH joining MATH and MATH as we did in REF above. We translate it using the action of MATH and then using the action of MATH repeatedly to obtain the desired MATH-subbundle of MATH. |
math/9910001 | Since MATH is injective, MATH acts freely on MATH; so taking orbit spaces by MATH gives an isomorphism MATH . In fact, the inverse is given by pulling back elements in MATH by the quotient map from MATH to MATH. Because of this isomorphism, it suffices to study the semi-group structure on MATH. Note that MATH is again a circle or a point, and that the pullback of a trivial bundle is again trivial. CASE: The case where MATH. In this case MATH, that is, MATH is the trivial group, so the semi-group MATH is generated by one element, that is the trivial line bundle, as is well known. This implies REF in the lemma. CASE: The case where MATH. In this case MATH is of order two and MATH is a point. Therefore, MATH is generated by two elements of dimension one. This implies REF in the lemma. CASE: The case where MATH for some MATH. In this case MATH is again a circle, MATH is of order two, and the action of MATH on MATH is a reflection. In the sequel it suffices to treat the case where MATH. But this case is already studied in CITE. (NAME treats real bundles but the same argument works for complex bundles.) The result in CITE says that MATH-vector bundles over MATH are distinguished by the fiber MATH-modules over the fixed points MATH, and that any pair of MATH-modules of the same dimension is realized as the fiber MATH-modules at MATH of a MATH-vector bundle over MATH. Since MATH is of order two, there are two one-dimensional MATH-modules (one is the trivial one MATH and the other is the nontrivial one MATH) and that any MATH-module is a direct sum of them. Therefore, there are four inequivalent MATH-line bundles MATH, where MATH (MATH and MATH stand for MATH or MATH) denotes the MATH-line bundle with the fiber MATH-modules MATH at MATH and MATH at MATH, and the structure of MATH is as stated in REF . For the reader's convenience, we shall give the argument in CITE briefly. First we observe that any pair of one-dimensional MATH-modules can be realized as the fiber MATH-modules at MATH of a MATH-line bundle over MATH. In fact, the trivial (real) line bundle and the (real) NAME line bundle have respectively two different MATH-liftings to the total space, and each MATH-lifting of the trivial (real) line bundle has the same fiber MATH-modules at MATH while that of the (real) NAME line bundle has different fiber MATH-modules at MATH. We consider complexification of them. Then any pair of MATH-modules of dimension MATH can be realized as the fiber MATH-modules at MATH by taking the NAME sum of suitable MATH number of those complexified MATH-line bundles. On the other hand, the same technique used in the proof of REF shows that any MATH-vector bundle over MATH decomposes into the NAME sum of the above MATH-line bundles. |
math/9910001 | The necessity part is obvious, so we prove the sufficiency. We note that if there exists an equivariant isomorphism MATH, then it must satisfy the equivariance condition MATH for any MATH where MATH. By the assumption we have a MATH-linear isomorphism MATH (and a MATH-linear isomorphism MATH when MATH). In the following we will define MATH for all MATH using the above equivariance condition to get an equivariant isomorphism MATH. We consider three cases according to the images of MATH by MATH. CASE: The case where MATH or MATH. In this case the MATH-action on MATH is transitive, so for any MATH we define MATH with MATH such that MATH. The well-definedness follows from the MATH-equivariance of MATH. This gives the desired equivariant isomorphism MATH. CASE: The case where MATH. Let MATH and define MATH by MATH. The map MATH is also a MATH-equivariant isomorphism. We connect MATH and MATH along the arc of MATH joining MATH and MATH, in other words, we find a MATH-equivariant linear isomorphism MATH for each MATH in the arc of MATH so that MATH is continuous at those MATH. (This is always possible because the set of MATH-linear isomorphisms between MATH and MATH is arcwise connected, in fact, homeomorphic to a product of MATH's.) Now we define MATH for any MATH using the equivariance condition MATH. This gives the desired isomorphism MATH. CASE: The case where MATH. Note that the equivariance condition of MATH is MATH . We connect MATH and MATH along the arc joining MATH and MATH to obtain MATH for MATH in the arc. Then using the equivariance condition MATH, we define MATH for MATH in the arc joining MATH and MATH. Thus we have defined MATH for MATH in the arc joining MATH and MATH. We then define MATH for all MATH using the equivariance condition MATH. This gives the desired isomorphism. |
math/9910001 | It is obvious that MATH is a homomorphism and that the characters of all representations in MATH (when MATH) and MATH (when MATH) are MATH-invariant. The surjectivity follows from REF (when MATH) and the first remark following it in REF (when MATH or MATH), and the injectivity follows from REF . |
math/9910001 | CASE: The necessity is trivial and the sufficiency follows from REF . CASE: Let MATH and MATH be two MATH-extensions of MATH and suppose that the product bundles MATH and MATH are isomorphic. Then MATH (and MATH if MATH). It is easy to see from REF that each MATH-extension of MATH is distinguished by its restriction to MATH (and MATH if MATH). Therefore, MATH and MATH are isomorphic as MATH-modules, proving REF . |
math/9910001 | This follows from REF . |
math/9910001 | The corollary follows from REF . |
math/9910001 | If MATH, then MATH for any MATH. Therefore the corollary follows from REF . |
math/9910001 | Let MATH be an extension of MATH to MATH. Since (the character of) MATH is MATH-invariant (in particular, MATH-invariant) and MATH is an index two subgroup of MATH, such an extension exists by REF . CASE: It is elementary to see that MATH and MATH for MATH. Since MATH is MATH-invariant we have MATH for MATH. By an inductive application of the above identity, we have MATH . It follows that MATH . CASE: The necessity is obvious, so we shall prove the sufficiency. Let MATH be a MATH-th root of MATH. Then MATH by the identity in REF above, so there is an integer MATH (determined module MATH) such that MATH where MATH. The equality MATH is equivalent to MATH being even. We define MATH. Then MATH. Therefore, to see that the extended MATH is a MATH-extension of MATH, it only remains to check that MATH. (Remember that MATH.) The left hand side at the identity is equal to MATH while the right hand side is equal to MATH which agrees with MATH because MATH by the choice of MATH above. |
math/9910001 | This follows from REF , and the observation done in REF . |
math/9910001 | REF at the end of REF implies that MATH is contained in the image of MATH, so we prove the converse. Choose an element MATH in MATH. Then the fibers of MATH and MATH at the base point MATH are isomorphic as MATH-modules. In particular, MATH and MATH have the same fiber MATH-module and thus MATH is isomorphic to MATH. Hence, one can express the image of MATH, that is, MATH in MATH as MATH where MATH and MATH are irreducible MATH-submodules of MATH and MATH, respectively, such that MATH. Here we note that an irreducible MATH-module MATH is uniquely determined by MATH if it is reducible, because MATH in this case, see CITE. Therefore, if MATH is reducible, then MATH in MATH. On the other hand, if MATH is irreducible, then both MATH and MATH are MATH-extensions of MATH. Thus MATH is isomorphic to MATH or MATH by the last statement of REF . It follows that MATH is either zero or MATH in MATH. Therefore, the image of MATH is contained in the ideal MATH. |
math/9910002 | As MATH we have MATH, so that MATH. Since MATH and MATH the canonical bundle MATH of MATH is trivial, and this implies that MATH. Thus MATH. But we are given that MATH. Hence MATH for MATH, and the proposition follows from REF . |
math/9910002 | Let MATH be a point in MATH. Then by REF we can choose complex coordinates MATH near MATH such that MATH and MATH are given by REF at MATH, with MATH. Define real coordinates MATH on MATH near MATH such that MATH. Then from REF we see that MATH, MATH and MATH are given at MATH by MATH where MATH. It follows from this equation that MATH coincides with REF-form MATH defined in REF . As this holds for all MATH, we see that MATH is a MATH-structure on MATH, in the sense of REF . Now MATH, where MATH is the NAME connection of MATH, and so MATH. But MATH is the torsion of MATH, so that MATH is torsion-free, as we want. |
math/9910002 | Let MATH be a NAME metric on MATH. Then MATH is also a NAME metric on MATH, and so MATH is a MATH-invariant NAME metric on MATH. Let MATH be the NAME class of MATH. Then MATH is MATH-invariant, regarded as an equivalence class of metrics on MATH. By REF we know that MATH and MATH, and that MATH is simply-connected, where MATH is the singular set of MATH. Thus by REF , the NAME class MATH contains a unique metric MATH such that MATH is a NAME - NAME orbifold. As MATH is MATH-invariant we see that MATH is MATH-invariant, by uniqueness of MATH. REF shows that there exists a holomorphic volume form MATH on MATH. Since MATH is antiholomorphic, it is easy to show that MATH, for some MATH. Define MATH. Then MATH is a holomorphic volume form for MATH, and MATH, as we want. Let MATH be as in REF . Then MATH, where MATH is the NAME form of MATH. As MATH and MATH we have MATH, and MATH as MATH. Thus MATH and MATH are both MATH-invariant. |
math/9910002 | Since MATH and MATH form a NAME - NAME structure on MATH, there certainly exists an isomorphism MATH which identifies MATH with MATH and MATH with MATH. This MATH is unique up to the action of MATH on MATH. That is, if MATH then MATH also identifies MATH with MATH and MATH with MATH. Now MATH is complex antilinear, and so MATH identifies MATH with the map MATH given by MATH for some MATH complex matrix MATH. In fact MATH is only defined up to multiplication by a power of MATH. As MATH preserves MATH and takes MATH to MATH on MATH, it follows that MATH preserves MATH and takes MATH to MATH on MATH. These imply that MATH and MATH, and so MATH. Also, MATH as MATH, and this implies that MATH for MATH or REF. And because MATH fixes only MATH in MATH, the only fixed point of MATH in MATH is REF. So MATH lies in MATH and satisfies MATH. When we replace MATH by MATH for MATH, the matrix MATH is replaced by MATH. We wish to show that we can choose MATH such that the maps MATH of REF and MATH of REF coincide. That is, we must show that there exists MATH and MATH or REF such that MATH . Now MATH shows that MATH and MATH commute, and so MATH. Thus MATH is a real matrix, which implies that MATH or REF, and MATH. By studying the eigenvectors of MATH, one can prove that there exists MATH such that MATH is one of MATH . We exclude the first three possibilities because MATH fixes MATH in MATH, contradicting the fact that the only fixed point of MATH in MATH is REF. Putting MATH in the fourth case and MATH in the fifth, we see that REF holds. Thus MATH identifies MATH with MATH and satisfies all the conditions of the proposition, and the proof is complete. |
math/9910002 | The derivative of MATH at REF is the identity map on MATH. Thus the derivative of MATH at REF is MATH, and so MATH, since MATH identifies MATH and MATH. Therefore MATH at REF in MATH. As MATH is a REF-form on a subset of MATH, we can pull it back to MATH, and regard MATH as a REF-form on the ball MATH of radius MATH in MATH. Then MATH is a smooth MATH-invariant REF-form on MATH which vanishes at REF. But MATH contains MATH, and any REF-form invariant under this map MATH has zero first derivative at REF. Hence MATH vanishes to first order at REF in MATH, and so by NAME 's Theorem we can show that MATH and MATH on MATH. Now MATH and MATH are closed, so that MATH is closed, and as MATH is contractible we can write MATH for some smooth REF-form MATH on MATH. Since MATH vanishes to first order at REF we can easily arrange that MATH vanishes to second order at REF, and therefore MATH for MATH, using NAME 's Theorem as above. Thus there exists MATH such that MATH on MATH, for MATH and MATH. |
math/9910002 | Since MATH is simply-connected by REF and MATH acts freely on MATH, we see that the fundamental group of MATH is MATH. The natural inclusion of MATH in MATH induces a homomorphism from MATH to MATH, which is easily shown to be surjective. Also, as MATH is MATH or MATH we have MATH. Therefore, MATH is MATH if the generator of MATH projects to the nonzero element of MATH for all MATH, and MATH is trivial otherwise. But calculation shows that the generator of MATH is nonzero in MATH if and only if MATH. |
math/9910002 | Expanding REF we find that MATH in MATH. Since MATH, REF show that MATH . Combining these with the previous equation and using the facts that MATH and MATH is bounded independently of MATH, we soon prove REF . |
math/9910002 | Outside the overlaps MATH for MATH we either have MATH or MATH. In both cases MATH, and so MATH is zero outside the MATH. In MATH we apply REF with MATH and MATH, to get MATH . Combining this with REF gives MATH . Now each MATH is an annulus in MATH with inner radius MATH and outer radius MATH, and the metric MATH on MATH is close to the flat metric MATH on MATH. Therefore we can find MATH independent of MATH such that MATH. Hence MATH . Taking roots gives REF, with MATH. REF are elementary. The metric MATH is made by scaling MATH by a factor MATH. Thus MATH and MATH. We make MATH by gluing together the MATH on the patches MATH for MATH and MATH on MATH. It is clear that for small MATH, the dominant contributions to MATH and MATH come from MATH and MATH for some MATH, and these are proportional to MATH and MATH. This proves REF for some MATH, and the theorem is complete. |
math/9910002 | Let MATH be as in REF . Then REF gives a constant MATH. Choose MATH with MATH and MATH. Let MATH be the MATH-structure MATH on MATH, and MATH REF-form MATH. Then MATH by REF , and REF - REF imply REF - REF , as MATH. Therefore all the hypotheses of REF hold, and the theorem shows that there exists a torsion-free MATH-structure MATH on MATH. It remains to identify the holonomy group MATH of MATH. We can regard the MATH-orbifold MATH as the limit as MATH of the MATH-manifolds MATH. Because of this, it is not difficult to show that MATH. Now MATH, and thus MATH. If MATH then MATH is connected. But the only connected NAME subgroup of MATH containing MATH is MATH, so MATH. If MATH then MATH by REF . This forces MATH, and it is then easy to see that MATH. |
math/9910002 | Define MATH and MATH as in REF, where MATH. Then MATH is the set of REF points MATH in MATH with MATH, and so MATH. Now MATH is a REF-fold branched cover branched over MATH, so by REF we have MATH . Similarly, MATH is a REF-fold branched cover branched over MATH, so that MATH . And MATH is a REF-fold branched cover of MATH branched over MATH, giving MATH . Finally, MATH is a REF-fold branched cover of MATH branched over MATH, and so MATH as we want. |
math/9910002 | REF shows that MATH for MATH. Since MATH is REF for MATH even with MATH and REF otherwise, this shows that MATH and MATH, and so MATH as MATH. REF also gives MATH, so MATH is simply-connected. As the nonsingular set of MATH is simply-connected, we can strengthen this to show that MATH is simply-connected. The isomorphism MATH above identifies MATH with MATH, and so MATH for MATH. Hence MATH. |
math/9910002 | We first calculate the NAME numbers of MATH. As MATH fixes REF points in MATH, by properties of the NAME characteristic we find that MATH. But MATH by REF , so MATH. As MATH is the MATH-invariant part of MATH we see from REF that MATH and MATH. Also MATH is generated by MATH and MATH, so MATH acts as MATH on MATH, and MATH. Thus MATH, MATH and MATH, giving MATH. REF then gives the NAME numbers of MATH, and REF gives MATH. REF shows that there exist torsion-free MATH-structures MATH on MATH, with MATH as MATH is simply-connected. By REF the moduli space of metrics on MATH with holonomy MATH is a smooth manifold of dimension MATH. |
math/9910002 | By REF is the blow-up of MATH at MATH. Each blow-up fixes MATH and MATH and adds REF to MATH and MATH. So the NAME numbers of MATH follow from REF . As MATH is simply-connected and MATH, we see that MATH is simply-connected and MATH. |
math/9910002 | As MATH fixes REF point in MATH we have MATH, so MATH by the previous proposition. Since MATH is the MATH-invariant part of MATH we have MATH and MATH. Now MATH, and MATH is generated by MATH and the cohomology classes NAME dual to the two exceptional divisors MATH introduced by blowing up MATH and MATH. But MATH swaps MATH and MATH, so MATH swaps the corresponding classes in MATH, and MATH by definition. Therefore MATH, where MATH acts as REF on MATH and MATH on MATH. Hence MATH, and MATH. Thus MATH, MATH and MATH, giving MATH. REF then gives the NAME numbers of MATH, and REF gives MATH. REF shows that there exist torsion-free MATH-structures MATH on MATH, with MATH as MATH is simply-connected. By REF the moduli space of metrics on MATH with holonomy MATH is a smooth manifold of dimension MATH. |
math/9910002 | Calculating the NAME numbers of MATH in the usual way gives MATH . As MATH is modelled on MATH at each point of MATH, the resolution MATH is modelled on MATH. Since MATH, the NAME numbers of MATH satisfy MATH . But MATH has genus REF, and so its NAME numbers are MATH and MATH. Combining this with REF gives the NAME numbers of MATH. The last part follows as in REF . |
math/9910002 | As in REF , we find MATH and MATH. Thus MATH . Using REF we find that MATH has MATH and MATH, and it soon follows that MATH also has MATH and MATH. Since MATH we see that MATH. Now MATH is the blow-up of MATH along MATH, so that each point of MATH is replaced by a copy of MATH. It can be shown that the NAME numbers of MATH satisfy MATH . But MATH can be thought of as an octic in MATH, and by the usual method we find that the NAME numbers of MATH are MATH, MATH, MATH, MATH and MATH. Combining these with REF and the NAME numbers of MATH above gives the NAME numbers of MATH. The last part follows as usual. |
math/9910003 | It is straightforward by REF . |
math/9910003 | We can verify the relations REF case-by-case, by a direct substitution of REF; for details, see CITE. |
math/9910003 | Combining REF, the unitarity REF, and an equality MATH, we obtain REF for generic MATH and thus for all MATH. The form REF is due to the fact that MATH implies the exchange condition CITE, MATH for some MATH. |
math/9910003 | It is sufficient to check it for the generators MATH. By REF , we see that MATH, for MATH by REF, and for MATH by REF noting that the unitarity REF vanishes. Hence MATH for all MATH. |
math/9910003 | First notice that MATH consists of nonaffine MATH-matrices, MATH for MATH, because MATH is minuscule. Substituting the MATH-matrices REF into MATH and expanding them, we see that every term includes a translation operator of the form MATH, where MATH . Let us show that MATH implies MATH. Suppose MATH and MATH, then we have MATH. From REF, MATH while MATH by a direct calculation, which leads to a contradiction. This implies that the term including MATH, MATH appears if and only if MATH. The coefficient of this term can be easily calculated, MATH . The MATH-invariance of the operator MATH yields the form REF. |
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