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math-ph/9911013 | Given MATH, there exists a finite collection of non-overlapping congruent cubes MATH for which (MATH)- REF hold (for MATH) with MATH . Note that MATH depends on MATH and hence so does the lower bound MATH of MATH. Let MATH and set MATH for MATH. Then, MATH where MATH . We have that MATH and MATH . It then follows that MATH . From REF with MATH (which, according to the remark after REF , is allowed since MATH) MATH and by REF MATH . Let MATH. On using REF , we have that MATH . It follows from REF that MATH . The cases MATH follow similarly from REF for MATH in CITE. The reverse inequality is obtained by choosing MATH and repeating the argument. |
math-ph/9911013 | We first note that MATH is properly defined as a form sum by REF . Also, from CITE for MATH, MATH and hence, for any MATH and MATH . Suppose that MATH, MATH, MATH satisfy (MATH)- REF. Then REF follows as before, and so do REF , with MATH replaced by MATH. The remaining term on the right-hand side of REF is estimated by REF . Since MATH, we have, with MATH, MATH where MATH as MATH, uniformly in MATH and MATH, by REF . It follows as for REF that MATH . For a general MATH we proceed in a similar way to NAME in CITE, using NAME 's estimate REF . Let MATH and choose non-overlapping cubes MATH, MATH, with sides parallel to the co-ordinate planes, and a piecewise constant MATH which is constant in each MATH, is zero outside MATH and MATH . Note that REF . and REF. imply that MATH, MATH and MATH . We prove the result for MATH, the proof for MATH being similar. Set MATH. Then, we have the NAME inequality MATH for any MATH. By REF MATH . Also, from REF , MATH and from REF we have MATH . Hence, MATH . Let MATH, MATH in that order to get MATH . The proof of the upper bound is similar. We use MATH and argue as before. |
math-ph/9911013 | The operator MATH is properly defined as a self-adjoint operator on the domain of MATH since it is a bounded perturbation of MATH. The point to note is that MATH is compactly supported in MATH. On choosing MATH in REF to contain this support, and the piecewise constant function MATH to be such that MATH in MATH, we have for MATH that MATH for all MATH. The proof follows easily from REF . |
math-ph/9911013 | We prove the result for MATH, the other cases being similar. As in the proof of REF , given MATH choose a piecewise constant function MATH taking constant values in non-overlapping congruent cubes MATH, MATH, and such that MATH in which MATH, MATH outside MATH. Thus, REF hold for MATH, and, indeed, for any function taking constant values in each MATH and zero outside MATH. Let MATH and set MATH where MATH, MATH, and MATH. Then, for MATH, MATH . With MATH and MATH, we have MATH . Since MATH, MATH . Hence, MATH . Also, MATH which implies that for any MATH . On substituting REF , we have that MATH where MATH. Whence, on choosing MATH, MATH with MATH and MATH . Since MATH is piecewise constant and vanishes outside MATH, the analysis leading to REF holds with MATH and MATH, and this yields the estimate MATH where MATH, MATH is arbitrary and MATH. Also, by REF MATH with MATH. It follows that MATH A similar argument yields the reverse inequality (compare REF ) MATH . Note that MATH for MATH, and also MATH, and is zero for MATH. Therefore, we have MATH from REF . Hence, as MATH, MATH . Finally, by REF , MATH . It follows from REF , and REF that, with MATH, MATH . The reverse inequality is proved similarly, with MATH in REF . The proof is complete. |
math-ph/9911013 | The only difference with the proof of REF is in the way the error terms involving MATH are estimated to give the analogue of REF . These are now dependent on REF rather than on the magnetic CLR inequality. We again prove it for MATH. From MATH, it readily follows that, for any MATH, MATH and, for any MATH, on choosing MATH, MATH . Consequently, MATH and, from REF , with MATH, MATH as in the proof of REF , where MATH as MATH, uniformly in MATH and MATH. This gives REF as in REF . From REF we have, MATH where, with MATH, MATH . The MATH term is handled as in the proof of REF and the first part above. The MATH term on MATH is again estimated by REF . |
math-ph/9911013 | We have that MATH where MATH is the NAME operator in two dimensions, namely MATH with MATH and MATH the magnetic NAME operator in MATH. The eigenvalues of MATH with NAME boundary conditions on MATH are MATH, MATH. It follows that the eigenvalues of MATH are of the form MATH, where MATH are the eigenvalues of MATH on MATH (with NAME boundary conditions). Hence, for any MATH, MATH . Since MATH, we have MATH. Hence, for any MATH . Replacing MATH by something smaller seems rather wasteful, but will be seen to provide the correct scaling for MATH by a suitable choice of MATH. It is reminiscent of a technique (running energy-scale renormalization) used in a paper of CITE. On substituting in REF , this yields MATH where the sum is over all MATH such that MATH . To estimate MATH we use the result of NAME and NAME CITE with (in their notation) MATH . Then, MATH as it generates a positivity preserving contractive semigroup with the MATH-boundedness (ultracontractivity) property having kernel MATH, where, on the diagonal, MATH see REF, for the terminology. The operator MATH is dominated by MATH in the sense that MATH that is, MATH in the language of REF. It follows from CITE with the choice MATH for some MATH, that MATH for MATH and MATH . Now choose MATH, MATH. Then MATH and MATH as MATH. Furthermore, on integration by parts, MATH implying that MATH . It now follows from REF , and REF that MATH and REF is proved. |
math-ph/9911013 | As in CITE, let MATH, MATH, be non-overlapping squares obtained by translating MATH parallel to the coordinate axes to form a tesselation of MATH. Choose a square MATH such that MATH with MATH extended by periodicity from MATH to MATH. Let MATH and suppose, without loss of generality, that MATH for MATH. Set MATH and denote by MATH the torus with fundamental domain MATH, MATH. It is proved in CITE that if MATH . For the lower bound we need MATH in order to use REF . Let MATH for MATH, and set MATH, MATH. Then, MATH for MATH. Also, since MATH from REF ; note that MATH is of one sign since MATH is continuous and MATH. Therefore, MATH and MATH which implies that MATH where MATH depends on MATH and MATH. We now proceed as in REF . The analogue is MATH for MATH in the domain of MATH and a partition of unity MATH subordinate to the covering of MATH by MATH which satisfies REF with MATH and MATH replacing MATH. From this we conclude that MATH . From REF MATH and by REF MATH . It follows from REF that MATH . Since MATH we have from REF MATH which yields REF on recalling REF . Note that REF gives, for any MATH, MATH . A similar argument then gives REF . |
math-ph/9911013 | By REF MATH . From REF MATH and MATH . Therefore, MATH and REF follows since MATH is arbitrary. The proof of REF follows in a similar manner. |
math-ph/9911013 | We prove the result for MATH, the proof for MATH being similar. The last assertion in the theorem, concerning MATH, is proved in CITE. Let MATH, MATH, MATH be as in REF (MATH)-(MATH). As in REF, for MATH, we partition each MATH into a finite number of cubes MATH, MATH, such that for each pair MATH . Let MATH and MATH the complementary set. In view of REF we have that MATH for MATH, MATH. Since MATH, MATH and from REF MATH . Since MATH is arbitrary, then MATH for each MATH. For the reverse inequality we proceed as in the proof of REF . Let the interior of each MATH be denoted by MATH. Set MATH for some MATH. Construct a partition of unity MATH subordinate to the covering MATH of MATH: MATH for some MATH where MATH is the characteristic function for MATH. It follows that MATH and for every MATH . Let MATH and MATH. Then, MATH since MATH in MATH,and MATH for MATH. This implies that MATH . From REF MATH and from REF , MATH . From REF MATH . It now follows from REF , on allowing MATH, in that order, that MATH which together with REF implies MATH . Similarly, we have for any MATH and MATH . The proof for general MATH is similar to the corresponding part of the proof of REF, using the NAME inequalities in the same way as NAME does in CITE, but, instead of REF we now use the inequality derived by NAME in REF, namely, MATH . |
math-ph/9911013 | In the notation of the proof of REF , MATH where MATH. The expression MATH has a self-adjoint realization in MATH with essential spectrum MATH (see REF ), and it is readily shown that there are negative eigenvalues at the solutions of the equation MATH corresponding to the eigenfunctions MATH where MATH. If these negative eigenvalues are denoted by MATH, MATH, then MATH and MATH . We now proceed in a similar way to the proof of REF . We have that for any MATH . Now substitute MATH in REF. This gives, as in REF , MATH where MATH and MATH satisfies REF . Choose MATH . Then MATH and the result follows as before. |
math-ph/9911013 | The proof follows from REF just as REF follows from REF . As for REF , we show that (in the notation of (MATH)- REF) MATH where MATH, and MATH . The rest of the proof is similar to that of REF . |
math-ph/9911015 | From REF and the numerical inequality MATH, valid for MATH and MATH, we obtain a bound MATH with positive numbers MATH where MATH can be chosen arbitrarily small. Then the NAME Theorem yields the first statement. From REF we obtain MATH . Hence the operator inequalities MATH hold, and we have derived the second statement. |
math-ph/9911015 | Since the operator MATH has an absolutely continuous spectrum, the measure MATH is absolutely continuous with respect to the NAME measure MATH on MATH. Consequently, the measure MATH is absolutely continuous with respect to the NAME measure on any interval MATH with MATH. The identity MATH and the NAME Lemma therefore imply MATH. Given a number MATH REF yields the existence of a MATH such that MATH . From the inequality MATH we then obtain MATH for sufficiently large MATH. Since the number MATH can be arbitrarily large the integral REF diverges for MATH. |
math-ph/9911019 | Differentiating REF once yields MATH . Denoting MATH, REF can be rewritten as MATH . If we now take MATH, then due to the assumptions of the lemma, MATH and REF follows. MATH . |
math-ph/9911023 | For one parameter groups weak continuity implies strong continuity. It is clearly sufficient to show that the group is weakly continuous on MATH, viewed as a subspace of MATH, which is equivalent to MATH . Given MATH we have MATH where MATH . We fix a finite interval MATH and find that MATH is compact. For each MATH we therefore have MATH. This implies that MATH is bounded as a function on MATH, and therefore there exists a function MATH, such that MATH. Hence, we can perform the limit under the integral, and since MATH, weak continuity follows. The group MATH commutes by construction with MATH. It is now not difficult to check on the generators that the corresponding group of NAME automorphisms is strongly continuous. The covariance is clear by construction. |
math-ph/9911023 | Let MATH. Then for each MATH we have at least for some open neighbourhood MATH of REF: MATH . Since MATH is a NAME REF or a ground state (MATH), MATH is the boundary value of a function MATH, which is analytic on the strip MATH and bounded and continuous on MATH. By the NAME reflection principle MATH vanishes on the whole real axis. Therefore MATH for all MATH. Hence, MATH is invariant under the action of MATH and the lemma follows. |
math-ph/9911023 | By the inclusion REF MATH is invariant under the action of MATH. Since the set MATH generates the algebra MATH we have MATH. |
math-ph/9911023 | MATH is open and nonvoid. Let MATH be another nonvoid open set, such that MATH for some MATH. The function MATH is entire analytic in MATH and vanishes on MATH if MATH with MATH. Therefore we get MATH whenever MATH, and as a consequence MATH. This already proves the lemma, since in our chart MATH is just the shift MATH. |
math-ph/9911023 | The distribution MATH is a distributional solution to the NAME equation, and by NAME formula a solution to the equation MATH where MATH is the spinor NAME and NAME the scalar curvature. The map MATH shifts in MATH direction, and it is easy to see that therefore MATH solves this equation as well. We note that all coefficients of the differential equation are independent of MATH, and the principal part of the operator MATH is diagonal with real coefficients. If we replace in MATH all derivatives MATH by MATH we obtain an elliptic second order differential operator MATH and clearly MATH solves the equation MATH . Hence, MATH is smooth (see for example . CITE). If MATH are the components of MATH we can therefore write the equation in the form MATH such that A is an elliptic second order differential operator, and MATH is a Matrix of first order differential operators. It follows that the MATH satisfy the inequalities MATH on MATH for some constant MATH. Furthermore, by REF , the MATH vanish on the open nonvoid set MATH. Hence, by the result of CITE (see especially REF ) we get MATH and MATH. |
math-ph/9911025 | It suffices to take MATH, for the general case follows by scaling MATH. Write the difference on the left side of REF as MATH with MATH . The missing term MATH is zero because of REF . Since MATH for MATH, we have MATH . For MATH we have in any case MATH . Moreover, MATH . Hence MATH . Combining the estimates for MATH and MATH gives REF . |
math-ph/9911025 | Put MATH, MATH. By REF we have MATH . Integration over MATH, using the NAME inequality to estimate MATH, gives REF . |
math-ph/9911025 | The non-positivity of MATH is straightforward from the definition by an appropriate choice of MATH. Note that this also holds in the case where some of the MATH variables coincide. The lower bound on MATH follows from REF together with the operator inequality MATH which is obtained by ignoring the positive two-body interactions. |
math-ph/9911025 | We shall prove this by showing that MATH as a distribution. We shall use that MATH is a lower semicontinuous function satisfying the mean value inequality MATH for all MATH if MATH is small enough, where MATH is the volume of the unit ball in MATH. For MATH it follows from the lower semicontinuity of MATH that we have either MATH or there exists MATH such that MATH. In the first case we obviously have MATH since MATH for all MATH. If MATH we also conclude the above inequality since MATH . Note now that for any MATH we have for any MATH that MATH for some constant MATH and in fact this limit holds in the topology of MATH. Thus if MATH we have MATH by REF . Hence MATH. |
math-ph/9911025 | We follow closely the proof of REF, which stated the superharmonicity of the ground state energy of a one-body operator which can be considered as a mean field approximation of MATH. It is clearly enough to prove that MATH is superharmonic in MATH (on the region MATH) for MATH fixed. We shall prove this by showing that MATH satisfies the mean value inequality around any given point MATH. Let MATH. Choose a sequence of MATH normalized functions MATH such that MATH as MATH. For MATH denote by MATH the function MATH . We clearly have MATH . If MATH is close to MATH we shall use MATH as a trial function for MATH. We then obtain MATH . Hence MATH . The potential appearing in MATH, that is, MATH is a superharmonic function of MATH. Writing MATH we see that MATH is superharmonic in MATH away from the line MATH. Since MATH is independent of MATH and MATH for all MATH we have that MATH is superharmonic in MATH away from the line MATH. Moreover, we also have that the two limits MATH are independent of MATH. This is true simply because the contribution from the terms in the Hamiltonian depending on MATH tend to zero as MATH. We may therefore apply the above lemma to the function MATH. We conclude that the function MATH is superharmonic for MATH. Moreover by REF this function is bounded below if MATH is bounded away from MATH. Now using NAME 's Lemma we see from REF that the average of MATH over the set MATH is non-positive for all MATH small enough. |
math-ph/9911025 | To avoid problems at points MATH with MATH for some MATH, we replace the repulsive potential MATH by the smaller potential MATH. We denote the corresponding Hamiltonian by MATH and its ground state energy by MATH. It is obvious that MATH, so a lower bound on MATH gives a lower bound on MATH. Let MATH be a normalized, symmetric wavefunction in MATH. Since MATH we have to estimate the matrix elements of the difference MATH. Using REF , together with the NAME inequality for the integration over MATH and MATH respectively, we obtain MATH where MATH and MATH are respectively the minimum and the maximum value of MATH, MATH, with MATH, and MATH is the kinetic energy of MATH. Now if MATH, MATH is a minimizing sequence of normalized wave functions for MATH, then we may assume that the corresponding kinetic energy is uniformly bounded in MATH, MATH and MATH. In fact, we may assume that MATH is a bounded sequence. If we use the bound from REF in CITE, we obtain MATH where we have saved half of the the kinetic energy MATH of MATH. For large MATH, MATH is bounded and hence we see that MATH is bounded. The error term REF with MATH thus tends to zero as MATH, uniformly in MATH, and the lemma is established. |
math-ph/9911025 | Define the sets MATH . Since MATH is compact and MATH is lower semicontinuous (being superharmonic, in fact, superharmonic in each variable) we may find MATH such that MATH . Clearly, MATH . By the superharmonicity of MATH in each variable MATH we know that each coordinate MATH of the point MATH satisfies either MATH or MATH. Moreover, since MATH is invariant under permutations of the coordinates of MATH we may assume that MATH for all MATH. By possibly going to a subsequence we may assume that there exists an integer MATH such that for MATH large enough MATH . Moreover, we may assume that MATH converges as MATH for MATH. Since we may ignore the variables MATH, MATH, which tend to infinity we have MATH . Since MATH is lower semicontinuous we conclude that there exists a point MATH with MATH for all MATH such that MATH . By REF we have that MATH . Since MATH and hence MATH we have proved the lemma. |
math-ph/9911025 | The projector MATH on the lowest NAME band is the MATH-th tensorial power of the projector MATH that operates on MATH and is given by the integral kernel MATH where MATH and MATH is the the projector on vectors in MATH with spin component MATH. The kernel MATH is a continuous function with MATH and MATH for all MATH. A wave function in the lowest NAME band has the representation MATH. After writing MATH as an integral operator REF follows from the NAME inequality, using REF . |
math-ph/9911025 | For fixed MATH let MATH be a normalized wave function in the lowest NAME band. By REF we have MATH . We split the integral over MATH into an integral over MATH (defined in REF ) and its complement in MATH. By REF we have only to consider the latter. Using the estimate REF the task is to bound terms of the form MATH from above. If MATH we carry out the integration over all MATH with MATH and use REF for the remaining variable MATH. For small MATH, MATH and the term can be estimated by MATH . For MATH we split the integration over MATH into two parts, namely MATH and MATH. For the first part we obtain the following bound, after transforming variables and using REF , this time for MATH, MATH . For the integral over MATH we estimate MATH by its maximum value, MATH and obtain for this part of the integral the upper bound MATH where we have used REF again. We see that REF is bounded above by MATH, for MATH large enough. Since MATH is arbitrary this completes the proof. |
math-ph/9911025 | The Hamiltonian MATH is the symmetrization of MATH with respect to the group MATH with MATH elements, generated by the reflections MATH, MATH. For MATH let MATH denote the corresponding unitary operators on MATH . Then MATH for any MATH, so MATH . If MATH we may take the square integrable ground state wave function of MATH, given by REF , as a test state for MATH. It satisfies MATH for all MATH, so MATH . But MATH is not an eigenfunction of MATH if MATH, so MATH is strictly below MATH. |
math-ph/9911025 | With the simple reformulation to MATH we are on well known territory: The Hamiltonian on the left side shall have no negative eigenvalue. By the scale transformation MATH, this inequality is transformed to MATH . This inequality will hold for MATH if it is true for the larger potential MATH because the Hamiltonian in REF is bounded from below by MATH . This Hamiltonian has MATH as a positive symmetric solution to the NAME equation - as a differential equation - with zero energy. Now, if REF would have a square integrable ground state wave function MATH, this wave function would also be symmetric under reflection MATH, and the delta-function would dictate the same value for MATH as it does for MATH at MATH. So the question of the existence of MATH can be dealt with by the methods which are used for proving NAME 's comparison theorem: We assume that MATH exists, with negative energy MATH. The Wronskian MATH is zero at MATH. Its derivative is determined as MATH. If MATH is chosen positive, then MATH is negative, which implies that MATH is negative for MATH, and MATH. This inequality can be integrated to give MATH, a contradiction to the assumption of the square-integrability of g REF . Therefore we know that the Hamiltonian REF has no negative eigenvalue. And so the operator inequality holds. |
math-ph/9911025 | We use the operator REF with MATH, (MATH and MATH will finally be chosen as appropriate powers of MATH) to bound MATH from below. For each MATH we use it twice; one time with MATH, and a second time with MATH. Then we add these inequalities and divide by two: MATH . At this point the positive definiteness of MATH becomes essential. It implies, that for any real valued integrable function MATH: MATH . Expanding this expression and integrating the delta-functions we get MATH . Combining this with REF gives MATH with the one-particle operators MATH . The parameters are now chosen as MATH . The fraction of kinetic energy per particle that we borrowed in REF then decreases as MATH, and the functions MATH become MATH . In the mean field limit MATH with MATH fixed the sequence MATH is a MATH-sequence, and MATH. If MATH is smooth with MATH, then MATH. The one particle Hamiltonians MATH, with smooth MATH, converge as quadratic forms pointwise (that is, for each test function) to MATH . Moreover MATH . Since the ground state energies of operators of the type MATH are concave functions of MATH and hence continuous in MATH, the ground state energies of the right side of REF converge in the limit MATH. The ground state energy of MATH is a concave functional MATH, and the lower bound REF , when divided by the number of electrons MATH, gives MATH . Inserting the mean field density MATH for MATH (that is, the minimizer of REF which satisfies REF) gives the mean field energy, divided by MATH, as a lower bound to the limit of the energy per electron. |
math-ph/9911030 | The proof is based on the following fact CITE. Let MATH and MATH then MATH . |
math-ph/9911030 | Using the relations REF , one obtains in an explicit form that MATH . This is a MATH-valued first order differential operator. At the same time, MATH . |
math-ph/9911030 | The isomorphism REF is given by the assignment MATH . |
math-ph/9911031 | If MATH, then MATH where the NAME equality was used, MATH . Thus MATH is a positive definite selfadjoint operator in the NAME space MATH if REF holds. Note that, since MATH, one has MATH as MATH, so REF implies MATH A positive definite selfadjoint operator in a NAME space is boundedly invertible. REF is proved. |
math-ph/9911031 | Let MATH. One has MATH where we have used the assumption MATH. Similarly, MATH . Finally, using NAME 's equality, one gets: MATH where MATH . Since MATH one gets from REF the estimate: MATH . To prove REF , one notes that MATH . Estimate REF is obtained. REF is proved. |
math-ph/9911031 | Define MATH for MATH or MATH greater than MATH. Let MATH. Then REF imply MATH . If REF holds, then REF have solutions in MATH, and, since MATH, it is clear that this solution belongs to MATH and consequently to MATH, because MATH. The proof of REF shows that such a solution is unique and does exist. From REF one gets MATH . For any fixed MATH one sees that MATH as MATH, where the norm here stands for any of the three norms MATH. Therefore REF imply MATH since MATH for any fixed MATH and MATH. Also MATH where MATH are some constants. Finally, by REF , one has; MATH . From REF relation REF follows. REF is proved. |
math-ph/9911031 | Differentiate REF and get MATH . We will check below that MATH and MATH . Thus, by REF , MATH . Therefore REF can be written as MATH . By REF one gets MATH . Thus MATH . From REF one gets REF . REF can be obtained from REF by changing MATH to MATH. REF is proved if REF are checked. To check REF , use MATH and compare the equation for MATH, MATH with REF . Let MATH. Then REF can be written as MATH which is REF for MATH. Since REF has at most one solution, as we have proved above, REF is proved. To prove REF , differentiate REF with respect to MATH and get: MATH . Set MATH in REF , multiply REF by MATH, compare with REF and use again the uniqueness of the solution to REF . This yields REF . REF is proved. |
math-ph/9911031 | From REF one gets MATH . Using REF again one gets MATH REF is proved. |
math-ph/9911031 | The function MATH defined in REF solves REF and satisfies the conditions MATH . The first condition is obvious (in CITE there is a misprint: it is written that MATH), and the second condition follows from REF : MATH . Let MATH be the NAME solution to REF which is uniquely defined by the asymptotics MATH . Since MATH and MATH are lineraly independent, one has MATH where MATH, MATH are some constants independent of MATH but depending on MATH. From REF one gets MATH . Indeed, the choice of MATH and MATH guarantees that the first REF is obviously satisfied, while the second follows from the Wronskian formula: MATH . From REF one gets: MATH . Comparing REF with REF yields the conclusion of REF . |
math-ph/9911031 | If REF hold, then, as has been proved in CITE (and earlier in a different form in CITE), there is a unique MATH which generates the given MATH-matrix MATH. It is not proved in CITE that MATH defined in REF (and obtained as a final result of REF ) - REF generates the scattering matrix MATH with which we started the inversion. Let us now prove this. We have already discussed the following diagram: MATH . To close this diagram and therefore establish the basic one-to-one correspondence MATH one needs to prove MATH . This is done by the scheme REF . Note that the step MATH requires solving NAME REF with the boundary condition MATH. Existence of the solution to this problem on all of MATH is guaranteed by REF. The fact that these assumptions imply MATH is proved in CITE and CITE. REF is proved. |
math-ph/9911032 | Proof can be found in CITE. We sketch only the idea of the proof of REF. Using the known formula MATH where MATH is the transformation kernel corresponding to the potential MATH, MATH see also REF below, and substituting REF into REF , one gets after a change of order of integration a homogeneous NAME integral equation for MATH. Thus MATH. |
math-ph/9911032 | REF can be proved in several ways. One way CITE is to recover the spectral function MATH from MATH, MATH. This is possible since MATH, MATH, and MATH where MATH are the bound states of the NAME operator MATH in MATH, MATH is the delta-function, and MATH . Note that MATH and the number MATH in REF can be found as the simple poles of MATH in MATH and the number of these poles, and MATH where MATH, so MATH. It is well known that MATH determines MATH uniquely CITE, CITE. An algorithm for recovery of MATH from MATH is known REF . In CITE a characterization of the class of MATH-functions corresponding to potentials in MATH, MATH is given. Here we give a very simple new proof of REF (compare CITE): Assume that MATH and MATH generate the same MATH, that is, MATH. Subtract from REF for MATH this equation for MATH and get: MATH . Multiply REF by MATH and integrate by parts: MATH where we have used REF to conclude that at infinity the boundary term vanishes. From REF and property MATH REF it follows that MATH. REF is proved. |
math-ph/9911032 | This result is due to CITE. We give a new short proof based on property C CITE. We prove that data REF determine MATH uniquely, and then REF follows from REF . To determine MATH we determine MATH and MATH from data REF . First, let us prove that data REF determine uniquely MATH. Suppose there are two different functions MATH and MATH with the same data REF . Then MATH . The left-hand side in REF is analytic in MATH since MATH and MATH are, and the zeros of MATH in MATH are the same as these of MATH, namely MATH, and they are simple. The right-hand side of REF has similar properties in MATH. Thus MATH is an entire function which tends to REF as MATH, so, MATH and MATH. The relation MATH follows from the representation MATH . Various estimates for the kernel MATH in the formula MATH are given in CITE. We mention the following: MATH where MATH here and below stands for various estimation constants. From REF follows. Thus, we have proved MATH . Let us prove MATH . We use the Wronskian: MATH . The function MATH and therefore MATH, where the overbar stands for complex conjugate, we have already uniquely determined from data REF . Assume there are two functions MATH and MATH corresponding to the same data REF . Let MATH . Subtract REF with MATH in place of MATH from REF and get MATH or MATH is analytic in MATH and vanishes at infinity and MATH is analytic in MATH and vanishes at infinity. If this claim holds, then MATH, and therefore MATH, so MATH. To complete the proof, let us prove the claim. From REF one gets: MATH . Taking MATH in REF , integrating by parts and using REF , one gets: MATH . Thus MATH . Since MATH is uniquely determined by data REF , so is the constant A MATH (by REF ). Therefore REF imply: MATH . It remains to be checked that REF implies MATH . This follows from REF : if MATH and MATH are the same, so are MATH, and MATH as the difference of equal numbers MATH . REF is proved. |
math-ph/9911032 | Our proof is new and short. We prove that, if MATH is compactly supported or decays faster than any exponential, for example, MATH, MATH, then MATH determines uniquely MATH and MATH, and, by REF , MATH is uniquely determined. We give the proof for compactly supported potentials. The proof for the potentials decaying faster than any exponentials is exactly the same. The crucial point is: under both assumptions the NAME function is an entire function of MATH. If MATH is compactly supported, MATH for MATH, then MATH is an entire function of exponential type MATH, that is MATH CITE. Therefore MATH is meromorphic in MATH (see REF ). Therefore the numbers MATH, MATH, can be uniquely determined as the only poles of MATH in MATH. One should check that MATH . This follows from REF : if one takes MATH and uses MATH, then REF yields MATH . Thus MATH. Therefore MATH determines uniquely the numbers MATH and MATH. To determine MATH, note that MATH as follows from REF . Thus the data REF are uniquely determined from MATH if MATH is compactly supported, and REF implies REF . |
math-ph/9911032 | We claim that MATH determines uniquely MATH if REF holds. Thus, REF follows from REF . To check the claim, note that MATH, so MATH and use REF to get MATH for MATH, so MATH . From REF the claim follows. REF is proved. |
math-ph/9911032 | If MATH for MATH, then MATH for MATH, MATH, so data REF determine MATH and, by REF , MATH is uniquely determined. REF is proved. Of course, this theorem is a particular case of REF . |
math-ph/9911032 | First, assume MATH. If there are MATH and MATH which produce the same data, then as above, one gets MATH where MATH, MATH, MATH. Thus MATH . The function MATH is an entire function of MATH of order MATH (see REF with MATH), and is an entire even function of MATH of exponential type MATH. One has MATH . The indicator of MATH is defined by the formula MATH where MATH. Since MATH, one gets from REF the following estimate MATH . It is known CITE that for any entire function MATH of exponential type one has: MATH where MATH is the number of zeros of MATH in the disk MATH. From REF one gets MATH . From REF and the known asymptotics of the NAME eigenvalues: MATH one gets for the number of zeros the estimate MATH . From REF it follows that MATH . Therefore, if MATH, then MATH. If MATH then, by REF , MATH. REF is proved in the case MATH. Assume now that MATH and MATH . We claim that if an entire function MATH in REF of order MATH vanishes at the points MATH and REF holds, then MATH. If this is proved, then REF is proved as above. Let us prove the claim. Define MATH and recall that MATH . Since MATH, the function MATH is entire, of order MATH. Let us use a NAME lemma. CITE If an entire function MATH of order MATH has the property MATH, then MATH. If, in addition MATH as MATH, then MATH . We use this lemma to prove that MATH. If this is proved then MATH and REF proved. The function MATH is entire of order MATH. Let us check that MATH and that MATH . One has, using REF and taking into account that MATH: MATH . Here we have used elementary inequalities: MATH with MATH, MATH, and REF . We also used the relation: MATH . Estimate REF implies REF . An estimate similar to REF has been used in the literature (see for example . CITE). REF is proved. |
math-ph/9911032 | NAME REF - REF to get MATH . Assume that there are MATH and MATH which generate the same data MATH, MATH. Let MATH. Subtract from REF with MATH, similar equations with MATH, and get MATH . Multiply REF by MATH, where MATH, MATH, MATH, integrate over MATH and then by parts on the left-hand side, using REF . The result is: MATH . Note that MATH is an entire function of MATH. Since MATH and is compactly supported, the function MATH is an entire function of MATH, so it has a discrete set of zeros. Therefore MATH where MATH for almost all MATH. REF imply MATH. REF is proved. |
math-ph/9911032 | NAME REF to get MATH where MATH . It follows from REF that MATH where MATH is the NAME solution to REF . From REF one gets MATH where MATH. From REF one obtains MATH . From REF one concludes MATH . From REF one gets MATH . Thus MATH is known for all MATH. Since MATH is compactly supported, the data MATH determine MATH uniquely by REF is proved. |
math-ph/9911032 | Let MATH be arbitrary, MATH, MATH if MATH. Suppose MATH . Denote by MATH and MATH the transformation operators corresponding to potentials MATH and MATH which generate spetral functions MATH and MATH, MATH. Then MATH where MATH and MATH are NAME operators. REF implies: MATH where MATH is the adjoint operator and the norm in REF is MATH-norm. Note that MATH and MATH . From REF it follows that MATH where MATH is a unitary operator in the NAME space MATH. If MATH is unitary and MATH are NAME operators then REF implies MATH. This is proved in REF below. If MATH then MATH, therefore MATH and MATH. Here we have used the assumption about MATH being in the limit-point at infinity case: this assumption implies that the spectral function is uniquely determined by the potential (in the limit-circle case at infinity there are many spectral functions corresponding to the given potential). Thus if MATH, then MATH. REF is proved. |
math-ph/9911032 | From REF one gets MATH and, using MATH, one gets MATH . Denote MATH where MATH are NAME operators. From REF one gets: MATH or MATH . Since the left-hand side in REF is a NAME operator of the type REF while the right-hand side is a NAME operator of the type REF , they can be equal only if each equals zero: MATH and MATH . From REF one gets MATH or MATH. Thus MATH and MATH as claimed. REF is proved. |
math-ph/9911032 | If MATH and MATH have the same spectral function MATH then MATH for any MATH, where MATH the function MATH solves REF with MATH, and MATH, satisfies first two REF , and MATH where MATH is the transformation operator: MATH . Note that MATH . From REF it follows that MATH . Since Range MATH, REF implies that MATH is unitary (an isometry whose range is the whole space MATH). Thus MATH where MATH is a NAME operator of the type REF . Therefore MATH and this implies MATH. Therefore MATH and MATH. REF is proved. |
math-ph/9911032 | CASE: Step MATH is done by REF . Let us prove MATH. Assume there are MATH and MATH corresponding to the same MATH. Then MATH . Therefore MATH . By REF relation REF implies MATH, so MATH. CASE: Step MATH is done by solving REF for MATH. The unique solvability of this equation for MATH has been proved below REF . Let up prove MATH. From REF one gets MATH . Let MATH in REF and write REF as MATH . Note that MATH. Thus REF can be written as: MATH . This is a NAME integral equation for MATH. Since it is uniquely solvable, MATH is uniquely recovered from MATH and the step MATH is done. CASE: Step MATH is done by REF . The converse step MATH is done by solving the NAME problem: MATH . One can prove that any twice differentiable solution to REF solves REF with MATH given by REF . The NAME REF -REF is known to have a unique solution. REF -REF is equivalent to a NAME equation (CITE, CITE). Namely if MATH, MATH, MATH, then REF take the form MATH . Therefore MATH . This NAME equation is uniquely solvable for MATH. REF is proved. |
math-ph/9911032 | CASE: The step MATH is done by REF as we have already mentioned. The step MATH is done by finding MATH, MATH and MATH from the asymptotics of the function REF as MATH. As a result, one finds the function MATH . If MATH and MATH are known, then the function MATH is known. Now the function MATH can be found by the formula MATH . So the step MATH is done. CASE: The step MATH is done by solving REF for MATH. This step is discussed in the literature in detail, (see CITE, CITE). If MATH (actually a weaker condition MATH is used in the half-line scattering theory), then one proves that conditions REF - REF are satisified, that the operator MATH is compact in MATH and in MATH for any fixed MATH, and the homogeneous version of REF : MATH has only the trivial solution MATH for every MATH. Thus, by the NAME alternative, REF is uniquely solvable in MATH and in MATH. The step MATH is done. Consider the step MATH. Define MATH . The function MATH determines uniquely MATH by the formula: MATH and consequently it determines the numbers MATH as the only zeros of MATH in MATH, the number MATH of these zeros, and MATH. To find MATH, one has to find MATH. REF allows one to calculate MATH if MATH and MATH are known. To find MATH, use REF and put MATH in REF . Since MATH is known for MATH, REF allows one to calculate MATH. Thus MATH, are found and MATH can be calculated by REF . Step MATH is done. The above argument proves that the knowledge of two functions MATH and MATH for all MATH determines MATH uniquely. Note that: REF we have used the following scheme MATH in order to get the implication MATH, and REF since MATH and MATH, we have proved also the following non-trivial implication MATH. CASE: REF is done by REF . The converse step MATH is done by solving the NAME problem: MATH REF -REF is equivalent to a NAME integral equation for MATH (see CITE). MATH . One can prove that any twice differentiable solution to REF solves REF with MATH given by REF . A proof can be found in CITE, CITE and CITE. REF is proved. |
math-ph/9911032 | Clearly, every MATH solution to REF solves REF . Let us prove the converse. Let MATH solve REF . Define MATH . We wish to prove that MATH solves REF . Take the NAME transform of REF in the sense of distributions. From REF one gets MATH and from REF one obtains: MATH . Add MATH to both sides of REF and use REF to get MATH . From REF one gets: MATH REF is equivalent to REF since all the transformations which led from REF to REF are invertible. Thus, REF hold (or fail to hold) simultaneously. REF clearly holds because MATH since MATH are zeros of MATH. REF is proved. |
math-ph/9911032 | The solution to REF is MATH where MATH, MATH, MATH, MATH, MATH, MATH is defined in REF , MATH is defined in REF and MATH is defined in REF . The functions MATH are the data REF . Since MATH when MATH, REF implies MATH, so one knows MATH . From REF one derives MATH and MATH . From REF one gets MATH and MATH . Eliminate MATH from REF to get MATH where MATH REF is a NAME problem for the pair MATH, the function MATH is analytic in MATH, MATH and MATH is analytic in MATH. The functions MATH and MATH tend to one as MATH tends to infinity in MATH and, respectively, in MATH, see REF . The function MATH has finitely many simple zeros at the points MATH, MATH, where MATH are the negative eigenvalues of the operator MATH defined by the differential expression MATH in MATH. The zeros MATH are the only zeros of MATH in the upper half-plane MATH. Define MATH . One has MATH where MATH is the number of negative eigenvalues of the operator MATH, and, using REF , one gets MATH . Since MATH has no negative eigenvalues by REF , it follows that MATH. In this case MATH (see REF below), so MATH, and MATH is uniquely recovered from the data as the solution of REF which tends to one at infinity. If MATH is found, then MATH is uniquely determined by REF and so the reflection coefficient MATH is found. The reflection coefficient determines a compactly supported MATH uniquely by REF . If MATH is compactly supported, then the reflection coefficient MATH is meromorphic. Therefore, its values for all MATH determine uniquely MATH in the whole complex MATH-plane as a meromorphic function. The poles of this function in the upper half-plane are the numbers MATH. They determine uniquely the numbers MATH, MATH, which are a part of the standard scattering data MATH, MATH, MATH, MATH, where MATH are the norming constants. Note that if MATH then MATH, otherwise REF would imply MATH in contradiction to REF . If MATH is meromorphic, then the norming constants can be calculated by the formula MATH, where the dot denotes differentiation with respect to MATH, and MATH denotes the residue. So, for compactly supported potential the values of MATH for all MATH determine uniquely the standard scatering data, that is, the reflection coefficient, the bound states MATH, and the norming constants MATH. These data determine the potential uniquely. REF is proved. MATH . |
math-ph/9911032 | We prove MATH. The proof of the equation MATH is similar. Since MATH equals to the number of zeros of MATH in MATH, we have to prove that MATH does not vanish in MATH. If MATH, then MATH, MATH, and MATH is an eigenvalue of the operator MATH in MATH with the boundary condition MATH. From the variational principle one can find the negative eigenvalues of the operator MATH in MATH with the NAME condition at MATH as consequitive minima of the quadratic functional. The minimal eigenvalue is: MATH where MATH is the NAME space of MATH-functions satisfying the condition MATH. On the other hand, if MATH, then MATH . Since any element MATH of MATH can be considered as an element of MATH if one extends MATH to the whole axis by setting MATH for MATH, it follows from the variational REF that MATH. Therefore, if MATH, then MATH and therefore MATH. This means that the operator MATH on MATH with the NAME condition at MATH has no negative eigenvalues. Therefore MATH does not have zeros in MATH, if MATH. Thus MATH implies MATH. REF is proved. |
math-ph/9911032 | This is an immediate consequence of the following: Theorem CITE: If MATH is holomorphic in MATH, MATH is of NAME in MATH, that is: MATH and MATH where MATH then MATH. The function MATH maps conformally MATH onto MATH, MATH and if MATH, then MATH is holomorphic in MATH, MATH for MATH and MATH, and MATH . From REF and the above Theorem REF follows. |
math-ph/9911032 | Subtract from REF with MATH this equation with MATH and get: MATH where MATH . Multiply REF by MATH, integrate over MATH, and then by parts on the left, and get MATH . By the assumption MATH if MATH, so MATH and MATH vanish at infinity. At MATH the left-hand side of REF vanishes since MATH . Thus REF implies REF . |
math-ph/9911032 | One can prove CITE that the kernel MATH of the transformation operator must solve the NAME problem MATH and conversely: the solution to this NAME problem is the kernel of the transformation operator REF . The difficulty in a study of the problem comes from the fact that the coefficients in front of the second derivatives degenerate at MATH, MATH. To overcome this difficulty let us introduce new variables: MATH . Put MATH . Then REF becomes MATH where MATH is defined in REF and MATH . Note that MATH for any MATH and any MATH, where MATH is some constant. Let MATH . Write REF as MATH . Integrate REF with respect to MATH and use REF , and then integrate with respect to MATH to get: MATH . Consider REF in the NAME space MATH of continous function MATH defined for MATH, with the norm MATH where MATH is chosen so that the operator MATH is a contraction mapping in MATH. Let us estimate MATH: MATH where MATH is a constant which depends on MATH, and on MATH. If MATH then MATH is a contraction mapping in MATH and REF has a unique solution in MATH for any MATH and MATH. Let us now prove that estimate REF holds for the constructed function MATH. One has MATH . The last inequality follows from the estimate: MATH where MATH and MATH are arbitrarily small numbers, MATH . The proof of REF is complete when REF is proved. Estimate REF holds. From REF one gets MATH where MATH and MATH . Without loss of generality we can take MATH in REF : If REF is derived from REF with MATH, it will hold for any MATH (with a different MATH in REF ). Thus, consider REF with MATH and solve this inequality by iterations. One has MATH . One can prove by induction that MATH . Therefore REF with MATH implies MATH . Consider MATH . This is an entire function of order MATH and type REF. Thus MATH . From REF estimate REF follows. REF is proved. REF is proved. |
math-ph/9911032 | The idea of the proof is to consider MATH in REF as a parameter and to reduce REF to a NAME equation with constant integration limits and kernel depending on the parameter MATH. Let MATH, MATH, MATH . Then REF can be written as: MATH REF is equivalent to REF , it is a NAME equation with kernel MATH which is an entire function of MATH and of the parameter MATH. The free term MATH is an entire function of MATH and MATH. This equation is uniquely solvable for all MATH by the assumption. Therefore its solution MATH is an analytic function of MATH in a neighborhood of any point MATH, and it is an entire function of MATH CITE. Thus MATH is an analytic function of MATH in a a neighborhood of the positive semiaxis MATH. |
math-ph/9911032 | First, we prove convergence of the process REF in MATH. The proof makes it clear that this process will converge in MATH and that in final number of steps one recovers MATH uniquely on CITE. Let MATH, MATH, MATH. Let us start with The map MATH maps MATH into itself and is a contraction on MATH if MATH, MATH. Let MATH, MATH . One has: MATH . Here we have used the estimate MATH and the assumption MATH, MATH. If MATH then MATH is a contraction on MATH. Let us check that the map MATH maps MATH into itself if MATH. Using the inequality MATH if MATH, MATH, one gets: MATH . Thus if MATH then the map MATH is a contraction on MATH in the space MATH. REF is proved. From REF it follows that process REF converges at the rate of geometrical progression with common ratio REF . The solution to REF is therefore unique in MATH. Since for the data MATH which comes from a potential MATH the vector MATH solves REF in MATH, it follows that this vector satisfies REF . Thus, process REF allows one to reconstruct MATH on the interval from data REF , MATH, where MATH is defined in REF . If MATH and MATH are found on the interval MATH, then MATH and MATH can be calculated for MATH. Now one can repeat the argument for the interval MATH, and in finite number of the steps recover MATH on the whole interval CITE. Note that one can use a fixed MATH if one chooses MATH so that REF holds for MATH defined by REF with any MATH. Such MATH does exist if MATH. REF is proved. |
math-ph/9911032 | It is sufficient to prove that, for any MATH, the function MATH . Since MATH, and since MATH provided that MATH (see REF ), it is sufficient to check that MATH . One has MATH, thus MATH where MATH . From REF one obtains REF since MATH. REF is proved. |
math-ph/9911032 | The proof goes as above with one difference : if MATH then MATH is present in REF and in REF with MATH one has MATH . Thus, using REF , one gets MATH where MATH is a constant. Similarly one checks that MATH if MATH. REF is proved. |
math-ph/9911032 | Write MATH . Clearly MATH . By the NAME theorem CITE, one has MATH . Actually, the NAME theorem yields MATH. However, since MATH, one can prove that MATH. Indeed, MATH and MATH are related by the equation: MATH which implies MATH or MATH where MATH is the convolution operation. Since MATH and MATH the convolution MATH. So, differentiating REF one sees that MATH, as claimed. From REF one gets: MATH where MATH is a constant defined in REF below, the constants MATH are defined in REF and the function MATH is defined in REF . We will prove that MATH (see REF ). To derive REF , we have used the formula: MATH and made the following transformations: MATH where MATH . Comparing REF one concludes that MATH . To complete the proof of REF one has to prove that MATH, where MATH is defined in REF . This is easily seen from the asymptotics of MATH as MATH. Namely, one has, as in REF : MATH . From REF it follows that MATH . From REF it follows that MATH. REF is proved. |
math-ph/9911032 | One has MATH . From REF one gets: MATH . If MATH, then MATH . Here by MATH we mean the right-hand side of REF since MATH is, in general, not analytic in a disc centered at MATH, it is analytic in MATH and, in general, cannot be continued analytically into MATH. Let us assume MATH. In this case MATH is continuously differentiable in MATH. From the Wronskian formula MATH taking MATH, one gets MATH . Therefore if MATH and MATH, then MATH and MATH. One can prove CITE, that if MATH, then MATH is bounded as MATH, MATH. From REF it follows that MATH . From REF one gets: MATH . Since MATH is a real-valued function if MATH is real-valued this follows from the integral REF shows that MATH and REF implies MATH REF is proved. |
math/9911002 | Uniqueness is clear. In the notation used above, we may without loss of generality take the representation MATH to be so that the automorphism MATH is spatially implemented, that is, so that there is a unitary MATH with MATH for every MATH. Then equating MATH with MATH we have the unitary MATH and MATH for every MATH and MATH. Let MATH. |
math/9911002 | Because MATH, using NAME 's result CITE that MATH is monotone, we have MATH. We will use the notation from the beginning of this section and we will denote by MATH the inclusion arising from the construction. In order to show that MATH, it will suffice to show that MATH for every MATH and for every finite subset MATH of MATH. Let MATH and MATH be finite subsets of MATH and respectively MATH, so that MATH. Let MATH and let MATH be a finite subset of MATH so that MATH for every MATH; (MATH exists by amenability of MATH). Let MATH, let MATH be a positive integer and let MATH. Let MATH and MATH be completely positive contractions such that for every MATH and every MATH, MATH. Let MATH be the normalized characteristic function of MATH. Let MATH and MATH be the completely positive contractions defined in CITE. By CITE we have, for every MATH and MATH, MATH . If MATH and MATH for some MATH and MATH then for every MATH we have MATH because MATH. Using that MATH we obtain MATH . We could have chosen MATH so small that MATH for every MATH, which would have given the estimate MATH . Therefore, MATH. |
math/9911002 | The MATH - algebra MATH, being an inductive limit of exact MATH - algebras, is exact. Now CITE implies that MATH is exact, hence that MATH is exact. |
math/9911002 | Uniqueness is clear. Let MATH; then MATH as above. The commuting diagram MATH gives rise to an automorphism MATH of MATH that commutes with MATH. Let MATH be the automorphism of MATH arising from MATH via REF . Then MATH and the restriction of MATH to MATH is the desired automorphism MATH. |
math/9911002 | Let us use the notation of the proof of REF . Then we have MATH where the inequalities follow from monotonicity of MATH and the equality follows from REF . However, NAME 's result CITE on inductive limit automorphisms gives MATH. |
math/9911002 | Since MATH - subalgebras of exact MATH - algebras are exact and since MATH, if MATH is exact then MATH is exact. Assume now that MATH is exact, and let us show that MATH is exact. There is a net, ordered by inclusions, of pairs MATH where each MATH is a separable MATH - subalgebra of MATH and where each MATH is a seperable closed linear subspace of MATH such that the restiction of the usual operations makes MATH a NAME bimodule over MATH and such that MATH and MATH. From the inclusions MATH mentioned early in this section, we see that MATH is the direct limit of the MATH. Hence we may and do assume without loss of generality that MATH and MATH are separable. Let MATH. Since MATH, it will be sufficient to prove that MATH is exact. Denote by MATH the vector MATH, and let MATH. Then MATH satisfies the following relations: MATH . The MATH - algebra MATH is the closed linear span of the set of all elements of the form MATH where MATH, MATH, MATH, MATH and where MATH if MATH. Consider the unitary MATH given by MATH, MATH). Denote by MATH the resulting automorphism MATH of MATH. Thus MATH, MATH, REF and MATH, (MATH. Note that MATH is an action of the group MATH on MATH. Let MATH be the fixed point subalgebra of MATH, that is . MATH. MATH is the closed linear span of the set of operators of the form REF for which MATH. If MATH is of the form REF then MATH. Hence if MATH then MATH. The map MATH is a faithful conditional expectation from MATH onto MATH; letting MATH be as above, if MATH then MATH while otherwise MATH. If MATH, then MATH can be approximated by a linear combination of operators of the form REF. Since MATH by assumption, it then follows that MATH can be approximated by a linear span of operators of the form REF for which MATH. This proves REF . MATH. It is sufficient to show that any operator MATH of the form REF can be written as MATH or MATH for some MATH and MATH. Let MATH. If MATH, then MATH and we are done. If MATH then since MATH we have MATH and MATH. If MATH, then MATH, and MATH. This proves REF . MATH defines an injective endomorphism of MATH. MATH is isomorphic to the (universal) crossed product of MATH by this endomorphism, namely MATH. MATH is an injective endomorphism because MATH and MATH. Let MATH and let MATH denote the isometry arising from the crossed product contruction and implementing MATH; thus we have MATH, MATH. As is well known, there exists a continuous family MATH of automorphisms MATH of MATH, such that MATH, and MATH for every MATH, (where we identify the circle MATH with MATH). Let MATH be the conditional expectation MATH. Then MATH is faithful. By universality of MATH, there is a surjective map MATH, such that MATH for MATH, and MATH. Let MATH. Then MATH. Since MATH, we have MATH, so that MATH. But MATH is injective, so MATH and hence MATH and MATH. It follows that MATH. This proves REF . The MATH - algebra MATH is exact. Denote by MATH the subspace of MATH that is the closed linear span of the set of words of the form MATH with MATH, and for which MATH. Note that, in light of REF , we may without loss of generality assume that in MATH, MATH and MATH. Now it is easily seen that MATH is a MATH - subalgebra of MATH and that MATH is an increasing sequence of subalgebras with MATH dense in MATH. Hence it will suffice to prove that each MATH is exact. Denote by MATH the restriction and compression of the representation of MATH on MATH to MATH. Since in the decomposition MATH acts on MATH as MATH, it follows that MATH is a faithful representation. Let MATH be the closed linear span of the set of all words of the form MATH. Using the relations REF one easily sees that MATH is a closed two sided ideal of MATH. Moreover, observing the action of MATH on MATH one easily sees that, with respect to the decomposition MATH, we have MATH, where MATH denotes the algebra of compact operators on the NAME MATH - bimodule MATH. Now the quotient MATH is canonically identified with the closed linear span of all words MATH in MATH for which MATH. Thus MATH is isomorphic to MATH, and the canonical inclusion MATH provides a splitting for the short exact sequence MATH. We now show exactness of MATH by induction on MATH. For MATH, MATH is exact by assumption. Having restricted to the separable case, we have that MATH is separable and hence MATH is an exact MATH - algebra by the NAME stabilization lemma CITE. Using the induction hypothesis that MATH is exact, we find that in the split exact sequence MATH the algebras MATH and MATH are exact. By CITE, (see also CITE), MATH is exact. This completes the proof of REF . Now we can finish the proof of the theorem. By REF , MATH is isomorphic to the universal crossed product MATH of an exact MATH - algebra MATH by an injective endomorphism MATH. By REF , MATH is exact. |
math/9911002 | The NAME algebra MATH is defined as a certain quotient of MATH (see CITE for details). If MATH is exact then MATH is exact, and it was proved by CITE that quotients of exact MATH - algebras are exact. Since MATH contains a copy of MATH as a MATH - subalgebra, exactness of MATH implies exactness of MATH. |
math/9911002 | Let us abuse notation by writing MATH for all of the corresponding unital subalgebras MATH, MATH and MATH arising from the free product and tensor product constructions, and similarly for MATH. The unitary MATH generates a copy of MATH, the continuous functions on the circle, on which MATH is given by integration with respect to NAME measure. Thus MATH, where MATH is NAME measure. By the main result of CITE, without loss of generality we may and do assume that MATH. It is easily checked that in MATH the family MATH is free; letting MATH be the MATH - subalgebra of MATH generated by MATH, conjugation by MATH acts as the free shift on MATH. As MATH generates MATH and as MATH for every nonzero integer MATH, we see that MATH and the GNS representation of the restriction of MATH to MATH is faithful on MATH. Therefore, we may use the uniqueness of the free product construction to see that MATH. Regarding MATH, we have the embeddings MATH and these satisfy MATH. Hence by the main result of CITE there is an injective homomorphism MATH extending each MATH and satisfying MATH. |
math/9911002 | Let MATH be the NAME MATH - bimodule associated to MATH. This means that MATH is obtained by separation and completion of the algebraic tensor product MATH with respect to the norm induced by the MATH - valued inner product MATH or, in other notation, MATH. Denote by MATH the vector MATH. Let MATH be the canonical vacuum expectation given by compression with the projection MATH. Consider MATH. By CITE, the restriction of the conditional expectation MATH to MATH is scalar-valued; we denote this restriction by MATH. In fact, as is easily seen, MATH is isomorphic to the algebra of NAME operators generated by the nonunitary isometry MATH and MATH is the state whose support is MATH. We have by CITE that MATH is a reduced free product, MATH, because MATH satisfies MATH for all MATH. The algebra MATH contains a unitary MATH with the property that MATH for all MATH; for example, MATH can be obtained by contiunuous functional calculus from the semicircular element MATH. Then by REF there is a injective homomorphism MATH satisfying that MATH. |
math/9911002 | Since each MATH is canonically embedded as a MATH - subalgebra of MATH, exactness of MATH implies exactness of every MATH. Suppose that every MATH is exact and let MATH be the minimal tensor product of MATH - algebras. Then MATH is exact, as is easily seen from the definition of exactness and by taking inductive limits if necessary. By REF , MATH is isomorphic to a MATH - subalgebra of MATH, for some NAME MATH - bimodule MATH. By REF MATH is exact, and it thus follows that MATH is exact. |
math/9911002 | Let MATH and consider the conditional expectation MATH and the completely positive map MATH given by MATH. Note that MATH takes values in MATH and that MATH. We now consider the MATH - creation operator MATH; this construction was introduced by CITE and CITE, and proceeds as follows. We take the NAME MATH - bimodule MATH which is obtained by separation and completion of the algebraic tensor product MATH equipped with the natural left and right actions of MATH and with the inner product MATH . We let MATH be the element corresponding to MATH; taking the NAME space MATH we let MATH be the corresponding creation operator. As is well known and is easily seen using REF , MATH for every MATH. Let MATH be the conditional expectation defined by compression with the orthogonal projection MATH and let MATH. Then MATH and MATH are free with respect to MATH. This claim is proved by applying CITE. In fact, from its proof, we see that REF of that theorem imply REF of that theorem. Hence in order to apply CITE to show freeness of MATH and MATH, we need only show CASE: MATH whenever MATH, MATH and MATH; CASE: MATH for every MATH. Indeed, REF together show that MATH is distributed with respect to MATH as a MATH - creation operator, showing REF holds, while REF is REF is REF. But REF follow from the facts that MATH is distributed as a MATH - creation operator and MATH for MATH. This finishes the proof of REF . Now let MATH and MATH. MATH is a projection, MATH is a partial isometry, MATH and MATH. The creation operator MATH is easily seen to be an isometry; thus MATH is a projection. Clearly MATH. Straightforward computations reveal that MATH and then that MATH. So REF is proved. Let MATH and let MATH be the GNS representation of MATH arising from the conditional expectation MATH. Then MATH; hence MATH is unitary We must show that MATH for all integers MATH and MATH and all MATH. We will show REF by induction on MATH. For MATH, clearly MATH is in the kernel of MATH, hence also of MATH. Suppose now MATH. Writing MATH and similarly for MATH and then distributing, we may assume that each MATH and each MATH lies either in MATH or in MATH. For every MATH we have MATH, where MATH is the automorphism of MATH given by MATH. Indeed, we may take MATH and then MATH becase MATH for MATH. From MATH it follows that MATH . Note also that MATH. If MATH and MATH then MATH; similarly if MATH and MATH then MATH. If MATH for some MATH then MATH and we may use the induction hypothesis to conclude that REF holds; similarly if MATH for some MATH then REF holds. Hence we may assume that MATH for every MATH and MATH for every MATH. Writing MATH and similarly for MATH, distributing and letting MATH and MATH, we find that MATH is equal to a sum of MATH terms which are obtained by replacing each MATH variously with MATH and with MATH, and each MATH variously with MATH and with MATH. If MATH is one of these terms where where at least one MATH or MATH has been replaced by its expectation under MATH then we can see that MATH by using the induction hypothesis; indeed, we write MATH for each MATH appearing in MATH and similarly for each MATH and then we distribute; this expresses MATH as a sum of terms to each of which the induction hypothesis applies to show has expectation zero under MATH. We are left to show only that MATH . But this holds by the freeness proved in REF . Thus the proof of REF is finished. The restriction of MATH to MATH is faithful. This follows from the fact that MATH and MATH have faithful GNS representations. The inner product arising from the GNS construction for MATH gives a map MATH which, upon identifying MATH with MATH becomes a conditional expectation MATH; moreover, we have MATH. Let MATH and let MATH. Then CASE: MATH is a unitary satisfying MATH and MATH. Consider the subalgebras MATH, MATH and MATH. Then CASE: MATH, (MATH); CASE: MATH is isomorphic to MATH via an isomorphism that sends MATH to MATH and conjugates MATH to MATH, (MATH); CASE: MATH and MATH are free with respect to MATH. REF follows from REF and the fact that MATH for all MATH; note that REF holds for the same reason. For REF , the isomorphisms are MATH which we denote MATH and MATH, respectively. That MATH sends MATH to MATH and conjugates MATH to MATH is straightforward to see. We have MATH because MATH for every MATH, and this also shows that MATH is mapped by MATH onto MATH. We must show that MATH for every MATH. But MATH . Write MATH. It is easily seen using REF that MATH. This and the freeness result proved in REF show that MATH, while since MATH we have MATH . Hence REF holds and REF is proved. To prove REF it will suffice to show that MATH whenever MATH where MATH, MATH and MATH. Writing MATH we find that MATH where MATH and MATH . Rewriting MATH as a product of MATH's alternating with MATH's and MATH's, using that each MATH and MATH lies in MATH, that MATH and the freeness proved in REF , we find that MATH. This finishes the proof of REF . We now finish the proof of REF . By properties of the free product construction, MATH is isomorphic to the image of MATH under the GNS representation of MATH, while MATH is itself a subalgebra of a quotient of a subalgebra of MATH. |
math/9911002 | Since MATH and MATH are MATH - subalgebras of MATH, exactness of MATH implies that of MATH and MATH. For the converse, suppose MATH and MATH are exact and let MATH. Then MATH is an exact MATH - algebra. By REF , MATH is an exact MATH - algebra whenever MATH is a NAME MATH - bimodule. Using REF , the well known fact that MATH - subalgebras of exact MATH - algebras are exact and NAME 's result CITE that exactness passes to quotients, we have that MATH is exact. |
math/9911002 | Let MATH be minimal projections in MATH such that MATH. We may without loss of generality assume that for every MATH, MATH for some MATH. Let MATH be an enumeration of MATH and let MATH be such that MATH. Note that MATH and let MATH . Then either MATH or MATH. We now recursively define elements MATH so that for all MATH we have MATH, MATH and if MATH then MATH. For the recursive step, if MATH, if MATH have been defined and if MATH has at least MATH elements then let MATH. Then for every MATH we have MATH. Note that MATH; let MATH . Letting MATH does the job. |
math/9911002 | Let MATH be a finite or countable set such that the submodule of MATH generated by MATH is dense in MATH. Let MATH be the set obtained form MATH using the NAME - NAME procedure of NAME REF. We shall define MATH by MATH where when MATH is infinite we shall show that the sum converges in MATH. Suppose first that MATH is finite. Then easy calculations show that MATH, MATH and consequently MATH. Now suppose that MATH is infinite and enumerate it by MATH. For every positive integer MATH and MATH let MATH. Then by the result for the case of MATH finite we have MATH . For every state MATH on MATH, the sequence MATH is a bounded and increasing sequence of positive numbers, hence converges. Therefore the sequence MATH converges in MATH. Then for MATH we have that MATH and the right - hand - side tends to zero as MATH. Therefore the sequence MATH is NAME in MATH, hence converges in MATH. We may thus define MATH. From the corresponding facts for the finite dimensional case we obtain that MATH, MATH and MATH. It remains to show that MATH. If MATH is finite then this is clear, so suppose MATH is infinite. Given MATH we have MATH and MATH, so MATH. Given MATH then for all MATH there are MATH and MATH such that MATH. But MATH and hence MATH. Therefore MATH. |
math/9911002 | The NAME automorphism MATH arises from a MATH - linear map MATH and an affiliated automorphism MATH as in REF . Let MATH and MATH. Let MATH be defined by MATH, (MATH); note that MATH and MATH together satisfy the conditions in REF , so that we have the NAME automorphism MATH of MATH. Now MATH is canonically embedded in MATH and the restriction of MATH to MATH is MATH; by the monotonicity of MATH, which was proved in CITE, it will therefore suffice to show that MATH. As in the proof of REF , MATH is isomorphic to the crossed product MATH - algebra MATH, where MATH with MATH and where MATH is the endomorphism of MATH given by MATH with MATH. Since MATH, we have MATH and hence the restiction of MATH to MATH is an automorphism of MATH that commutes with the endomorphism MATH. It thus follows from REF that MATH. Let MATH denote the automorphism MATH of MATH. Let MATH be a faithful unital representation of MATH on a NAME space MATH and let MATH denote the representation of MATH which is the inclusion MATH followed by the representation MATH of MATH on the NAME space MATH. We must show MATH and to do so it will suffice to show that MATH for every MATH and every finite subset MATH of MATH. Given a finite subset MATH there are MATH and a MATH - subbimodule MATH of MATH that is finite dimensional as a MATH - vector space and such that MATH where MATH . Recall the definition of MATH from the proof of REF and let MATH denote the projection onto MATH. Consider the completely positive contractions MATH where MATH is the unitary operator canonically defined by the decomposition in REF in the proof of REF . Note that MATH for every MATH. For every integer MATH let MATH; then MATH . Moreover, MATH is a MATH - subbimodule of MATH whose MATH - linear dimension satisfies MATH. Let MATH . Clearly MATH is a finite dimensional MATH - subbimodule of MATH. By REF , there is MATH which is the projection onto MATH; note that MATH commutes with the left action of MATH on MATH. Consider the completely positive contractions MATH where MATH maps MATH to the submodule MATH of MATH via MATH. Since MATH whenever MATH, we see that MATH for every MATH. As MATH is a finite dimensional MATH - algebra, we have that MATH . As the MATH - algebra MATH can be faithfully represented on the NAME space MATH, making a crude estimate we get MATH . Because this upper bound grows subexponentially as MATH, we conclude that MATH. |
math/9911006 | We have the projectively normal embedding of MATH given by MATH. Therefore, the projective curve MATH is normal and thus smooth. Consider the exact sequence MATH of NAME divisors arising from viewing MATH as a cone over MATH CITE. Since MATH is finitely generated, so is MATH. In particular the Jacobian MATH is trivial (MATH denotes degree zero divisor classes). Therefore the genus of MATH is MATH, or equivalently MATH. Using the normality of MATH once again we get MATH for some very ample line bundle MATH on MATH. But due to the equality MATH such a line bundle is a positive multiple of MATH, and hence MATH is the NAME subalgebra of the polynomial algebra MATH of some level MATH. |
math/9911006 | In case MATH it is easy to see that MATH. Consider the case MATH. We write MATH and denote the retraction MATH by MATH. Consider the set MATH of monomials. There is a unique face MATH such that MATH and MATH is naturally isomorphic to MATH. Then MATH is a composite of the two retractions MATH where MATH is the homomorphism induced by MATH. Observe that MATH is in fact a retraction as it is split by MATH. Therefore we can from the beginning assume that MATH. In this situation MATH extends (uniquely) to the normalizations MATH . This extension is given by MATH . It is known that the semigroup MATH is normal for all natural numbers MATH CITE. Therefore, by restricting the retraction REF to the MATH-th NAME subalgebra for such a number MATH, we get the retraction MATH . Let us show that MATH is finitely generated for MATH. We choose a lattice point MATH of MATH that is in the interior of the cone MATH. By localization REF gives rise to the retraction MATH . Since MATH is a localization of the NAME polynomial ring MATH, it is a factorial ring. Then its retract MATH is factorial as well (for example, see CITE). By NAME 's theorem CITE MATH is generated by the classes of the height REF prime ideals of MATH containing MATH - a finite set. It is also clear from REF that MATH is generated in degree REF. Consequently, by REF for each MATH there is a natural number MATH and an isomorphism MATH . We now fix such a number MATH. Restricting MATH and MATH to the iterated NAME subalgebra MATH we obtain two isomorphisms of MATH with MATH. It follows that there exists MATH with MATH and MATH, and furthermore the restrictions of MATH and MATH differ by an automorphism of MATH. However, each automorphism of MATH can be lifted to an automorphism of MATH, and then restricted to all NAME subrings of MATH. (This follows from the main result of CITE.) Therefore we can assume that the restrictions of MATH and MATH coincide. Then they define an isomorphism of the subalgebra MATH of MATH generated by the elements in degree MATH and MATH to the corresponding subalgebra of MATH; see REF below. Taking normalizations yields an isomorphism MATH. But then MATH as well, because MATH and MATH coincide in degree REF (being retracts of algebras with this property). |
math/9911006 | One checks easily that one only needs to verify the following: if MATH for homogeneous elements MATH, MATH, then MATH. As MATH is reduced, it is enough that MATH. Since MATH this follows immediately from the hypothesis that MATH and MATH coincide on MATH. If MATH is reduced, then every non-zero homogeneous ideal in MATH intersects MATH non-trivially, and this implies the second assertion. |
math/9911006 | Since polytopal algebras are precisely the standard graded affine semigroup rings coinciding with their normalizations in degree REF, we only need to show that MATH is a semigroup ring. REF reduces the proposition to the assertion on binomial ideals MATH discussed in REF where the ideal MATH to be considered is the kernel of a surjection MATH. We set MATH. By hypothesis there exists a linear transformation MATH, MATH such that MATH is binomial. For all MATH we consider the surjection MATH . We define the linear endomorphisms MATH and MATH, MATH of MATH by MATH . There exists MATH for which MATH is an automorphism. In fact, let MATH denote the MATH matrix MATH. If MATH is not an automorphism, then the MATH submatrix MATH of MATH consisting of the intersection of the first MATH rows with the first MATH columns of MATH has rank MATH and the larger submatrix MATH consisting of the first MATH rows has rank MATH (otherwise MATH.) There exists MATH such that the MATH-th column of MATH lies in the vector space MATH spanned by the remaining MATH columns of MATH, but the MATH-th column of MATH does not belong to MATH. If we replace the MATH-th column of MATH by its sum with MATH-st column of MATH, we obtain a MATH matrix of rank MATH, and this matrix defines the automorphism MATH. There exists a unique epimorphism MATH making the diagram MATH commutative. Since MATH for MATH, we have MATH, and MATH maps binomial ideals to binomial ideals. |
math/9911006 | If MATH is not homogeneous, then among the relations between the irreducible elements of MATH there occurs an equation MATH for some MATH. But then MATH in MATH, where MATH denotes the residue class of MATH in MATH. Since these are nonzero classes we get a contradiction. The other implication is clear. |
math/9911006 | As MATH and MATH are retracts of MATH they have to be normal domains. We denote by MATH and MATH the affine normal toric varieties corresponding to MATH and MATH. The two irrelevant maximal ideals will be denoted by MATH and MATH respectively. Let MATH be any smooth point. Then for the local ring of the point MATH we have MATH as graded algebras (the MATH are indeterminates over MATH). On the other hand, identifying MATH with the variety MATH we see that there is a maximal ideal MATH such that MATH . By an iterated use of REF we get MATH for some positive normal affine semigroup MATH, an integer MATH, and a maximal ideal MATH, containing MATH. Since MATH is algebraically closed there are elements MATH such that MATH is generated by the maximal monomial ideal MATH and MATH. Therefore one has a natural graded isomorphism MATH for MATH. Consequently, MATH as graded algebras. By REF MATH is a homogeneous affine semigroup. Then, clearly, MATH for some normal polytope MATH and we get the graded isomorphism MATH (MATH denotes the MATH-unit simplex). By an iterated use of REF we can split off the polynomial extension on the left. That is, MATH is polytopal. By symmetry MATH is also polytopal. |
math/9911006 | That each of the column vectors of the polytopes serves as a column vector for MATH is clear. Now let MATH. If MATH is parallel to either MATH or MATH then either MATH or MATH since MATH. So without loss of generality we can assume that MATH is parallel neither to MATH nor to MATH. Since MATH and MATH span MATH they cannot be contained simultaneously in the base facet of MATH. But then either MATH or MATH for suitable vertices MATH and MATH. We get a contradiction because one of the points MATH or MATH is outside MATH for interior lattice points MATH, MATH. |
math/9911006 | We prove REF by induction on MATH. REF will follow automatically from the description of MATH derived below. For MATH there is nothing to show. Assume the theorem is proved for semigroups of MATH and choose a segmentonomial MATH. Then MATH contains a minimal prime MATH over the principal ideal MATH. Assume that MATH is a binomial ideal. By REF , MATH for some affine semigroup MATH and such that monomials in MATH go to monomials in MATH. But then segmentonomials in MATH are likewise mapped to segmentonomials in MATH. This holds true because affinely independent terms lift to affinely independent terms. By induction hypothesis the image of MATH in MATH is binomial. Since binomials can be lifted to binomials in MATH, we conclude that MATH is binomial. The general situation thus reduces to the case in which MATH for some segmentonomial MATH and MATH. If MATH contains a monomial, then MATH is a height REF monomial prime ideal, and we are done. Otherwise MATH. Consider the localization MATH. It is a height REF prime ideal in the NAME polynomial ring MATH. Therefore, MATH for some prime element MATH. Also MATH is segmentonomial. In fact, we have MATH for some MATH implying the equality MATH for the corresponding NAME polytopes. Since MATH is excluded, MATH. Multiplying MATH by a suitable term from MATH we can achieve that the origin MATH is one of the end-points of MATH. Let MATH denote a rational line containing MATH. In a suitable basis of the free abelian group MATH the line MATH becomes a coordinate direction. Therefore, we can assume that MATH and that MATH is a monic polynomial in MATH. Since MATH is algebraically closed, if follows that MATH for some MATH. Since MATH does not contain a monomial, one has MATH . Thus MATH is the kernel of the composite homomorphism MATH . This is a homomorphism mapping the elements of MATH to NAME monomials in MATH, and therefore MATH is generated by binomials. |
math/9911006 | We only need to derive REF from REF . Let MATH be an affine hull of MATH in MATH. We have to show that MATH is a lattice polytope with MATH. Consider the subsemigroup MATH . Then MATH as well. On the other hand MATH . Thus the restriction MATH is also an isomorphism. It follows that MATH, and every element in MATH is a product of elements of MATH. Furthermore MATH since any rational point of the complement MATH gives rise to elements in MATH. |
math/9911006 | It is not hard to check that there is no restriction in assuming that MATH. Note that MATH. In fact, if a monomial is mapped to MATH by MATH, then MATH contains a monomial prime ideal MATH of height MATH. Since MATH in turn contains all monomials in the interior of MATH, it must also contain monomials from MATH, which is impossible. Thus MATH can be extended to the normalization MATH; on MATH the extension is the identity. Set MATH, and let MATH be the intersection of the MATH-vector subspace of MATH generated by MATH with MATH. Choose a basis MATH of a complement of MATH in MATH. Since MATH contains elements of degree MATH (given by the last coordinate), we can assume that MATH for MATH. In sufficiently high degree we can find a lattice point MATH in MATH such that MATH. We have the relation MATH. It follows that MATH, equivalently MATH, for some MATH and MATH. After a toric `correction' leaving MATH fixed we can assume MATH for all MATH. After the inversion of the elements of MATH, we can further extend the homomorphism MATH to a map defined on the NAME polynomial ring MATH. Then we have MATH . The vectors MATH are also a basis of a complement of MATH, and thus part of a basis of MATH. Therefore the elements MATH generate a prime ideal of height MATH in MATH. It is now clear that MATH (after the toric correction) is just the retraction MATH where MATH is the affine hull of MATH in MATH and MATH is the sublattice of MATH generated by the vectors MATH upon the identification of MATH with the degree MATH sublattice of MATH. |
math/9911006 | The case MATH is an easy exercise. Now we use induction on MATH. Assume MATH and assume the claim has been shown for pyramids of dimension MATH. Consider any MATH-dimensional subspace MATH perpendicular to the base MATH. For a polytope MATH let MATH denote the image of MATH in MATH under the orthogonal projection MATH. Then MATH is a MATH-dimensional pyramid and we have the NAME sum representation MATH . By induction hypothesis there are homothetic transformations of MATH transforming MATH into MATH and MATH respectively. Considering all the possible subspaces MATH we conclude that CASE: both MATH and MATH are MATH-pyramids (provided none of them is just a point - in this situation the lemma is obvious) such that the cones they span at corresponding vertices are parallel shifts of the cone spanned by MATH at its apex MATH, CASE: the corresponding bases of MATH and MATH are parallel to MATH. That is exactly what we wanted to show. |
math/9911006 | As in the proof of REF we can assume MATH, and, furthermore, MATH, for otherwise MATH itself passes through a facet retraction. Thus MATH can be extended to the NAME polynomial ring MATH, MATH, and in particular to a retraction of MATH with image MATH. The latter restricts to retractions MATH for all MATH. The kernel of the extension MATH is a height MATH prime ideal and thus principal; MATH and MATH for some element MATH. Since MATH is a base of MATH, MATH contains the elements MATH, MATH, MATH, and MATH is a linear form on the points of MATH. Then MATH is a NAME summand of the pyramid MATH with vertex at MATH. One can shift MATH by an integer vector into MATH such that the image MATH satisfies MATH . Evidently MATH is the NAME polytope of MATH for some MATH. Replacing MATH by MATH, we can assume that MATH satisfies MATH. By REF MATH is homothetic to MATH. Clearly, MATH is a base of MATH. The corresponding apex of MATH is some MATH. Consider the valuation MATH determined by the conditions: MATH . We claim: MATH and MATH for any MATH and MATH. In fact, for MATH big enough there is an element MATH such that MATH. Since MATH is a base of the induced retraction MATH there exists a linear form MATH on MATH such that MATH. Thus MATH is a NAME summand of MATH. Because of the condition MATH we conclude MATH. Hence MATH. Now choose MATH. Since MATH is a base of MATH, we can write MATH for some linear form MATH on the points of MATH. Therefore, the pyramid MATH is a NAME summand of the pyramid MATH which has its apex at MATH. By REF the cones spanned by these pyramids at their vertices are the same modulo a parallel shift. This observation in conjunction with the already established equality MATH makes the claim clear. We have shown that the vectors MATH, MATH, are column vectors for MATH. Now, by REF there exists MATH such that MATH for some MATH. Therefore, MATH is the monomial prime ideal MATH, and this finishes the proof of REF . |
math/9911006 | Let MATH be such a closed torus. Then there is an algebraic embedding MATH and a maximal intermediate torus MATH. By NAME 's theorem MATH for some MATH. Let us show that MATH is a based retraction. By the previous theorems this completes the proof. We can assume that MATH is not a monomial ideal, or, equivalently, MATH contains no monomials. Otherwise MATH factors through a facet projection, and we are done. Claim. MATH is a binomial ideal. We know that MATH is stabilized by the codimension REF subtorus MATH. Since MATH a is non-monomial height REF prime ideal of MATH there is an element MATH such that MATH . (Compare with the proof of REF .) In this situation for any MATH we have MATH. Since MATH is a subtorus of the embedded torus, the latter equation is equivalent to the existence of an element MATH such that MATH. (MATH acts naturally on coordinate ring MATH.) This means that for any MATH the terms in the canonical MATH-linear expansion of MATH are multiplied by the same scalar from MATH when MATH is applied, a condition equivalent to to the segmentonomiality of MATH. In fact, if MATH we would have three non-collinear terms MATH, MATH and MATH in the MATH-linear expansion of MATH, and the condition MATH on MATH would be equivalent to the condition that each MATH is a solution to two independent binomial equations on MATH. Hence, MATH would be a sub-torus of codimension at least MATH - a contradiction. By this argument we have shown that MATH is a segmentonomial and, hence, a binomial by REF . Therefore, by REF we can assume MATH, MATH. In particular, MATH where MATH is the natural map induced by MATH and MATH is the embedded torus of MATH corresponding to MATH. The equality holds true because monomial preserving automorphisms are mapped to monomial preserving automorphisms. Identifying MATH and MATH via MATH and MATH and MATH via MATH we get a MATH-equivariant embedding MATH . Assume MATH be the set of degree REF terms in MATH. Then for any MATH there exist MATH such that MATH . Arguments similar to those in the proof of the claim above show that the polytopes MATH, MATH, are segments parallel to MATH, maybe some of them degenerated into points. Let MATH be such that the line MATH is parallel to all the MATH. Denote by MATH, MATH the upper end-points of MATH in the direction MATH. Then standard arguments with NAME polytopes ensure that the MATH are subject to the same affine binomial dependencies as the MATH. It is now clear that the subpolytope MATH is a base for the retraction MATH. |
math/9911006 | Clearly, without loss of generality we can assume MATH. Let MATH be an embedding. We write MATH . Thus we have the equations MATH for MATH. Put MATH . In the NAME polynomial ring MATH we can write MATH with coprime MATH. The equality MATH (and the factoriality of MATH) imply that MATH divides MATH and MATH divides MATH. CASE: Both MATH and MATH are monomials in MATH. In this situation the NAME polygon MATH is the parallel shift of MATH by the MATH-th multiple of the vector representing the support term of the monomial MATH. But then the existence of the desired embedding MATH is obvious. CASE: At least one of MATH and MATH, say MATH, is not a monomial. Then MATH can be embedded in any of the edges of the polygon MATH. Since MATH for some MATH, we get MATH and the existence of an embedding MATH is evident. |
math/9911006 | Let MATH be a codimension REF retraction. By REF there is an isomorphism MATH for some MATH, and REF then yields an embedding MATH. Let MATH denote its image. If the restriction of MATH to the subalgebra MATH is injective then it is evidently bijective, and we are done by REF . Therefore we can assume MATH . Clearly, MATH is segmentonomial. Since MATH the ideal MATH is a minimal prime ideal of MATH. If MATH contains a monomial, then MATH factors through a facet projection MATH, and then MATH is a base for MATH. In the other case it is enough to show there is yet another embedding MATH, say with image MATH, such that MATH and MATH are not parallel. REF shows that we then have found a base. Let MATH denote the line spanned by MATH. By REF we have MATH . We use the notation MATH . (As in the proof of REF we write MATH.) Evidently there is no loss in generality in assuming that MATH since, roughly speaking, the retraction can be restricted to the lattice polytope on the right hand side. CASE: MATH. For every linear form MATH we consider all triples MATH such that MATH, MATH and MATH . In particular MATH and MATH are vertices of MATH and MATH respectively. We have MATH . It follows that MATH has a unique maximum on MATH, necessarily taken at MATH, MATH. Thus we get the system MATH of collinear lattice points. Subcase MATH. There is a triple MATH such that MATH is not parallel to MATH, and in particular MATH. Then the points MATH are pairwise different, and so we find MATH successive lattice points in MATH on a line not parallel to MATH. Subcase MATH. In the other case one notes first that the edges of MATH and MATH are pairwise parallel under an orientation preserving correspondence (in other words, MATH and MATH have the same normal fans; see CITE for this notion). Moreover, if MATH is a vertex of MATH and MATH is the corresponding vertex of MATH, then MATH or the segment MATH is parallel to MATH. Let us choose MATH, MATH such that MATH is parallel to MATH, and let MATH be a non-zero linear form with MATH. Furthermore we consider the upper and lower boundary of MATH in direction MATH. (A point MATH on MATH belongs to the lower boundary if the ray in direction MATH through MATH enters MATH at MATH, and the upper boundary is defined analogously; see REF .) Each of these subsets has two endpoints MATH and MATH respectively (which may coincide). We choose the indices such that MATH and MATH. It is clear that MATH takes its minimum on MATH in MATH and MATH and its maximum in the other two points. Applying the same construction to MATH we obtain the points MATH and MATH. The correspondence between the vertices of MATH and those of MATH pairs MATH with MATH etc. If MATH and MATH, then MATH by our initial observation in this subcase. But we are assuming that MATH in REF . For example, suppose that MATH. Then we can find a linear form MATH such that MATH and MATH are the unique maxima of MATH on MATH and MATH respectively. So the triple MATH defines MATH lattice points MATH in MATH on a line MATH parallel to MATH, and one sees easily that MATH intersects MATH in an edge: it is impossible that MATH for some MATH, MATH. Since MATH there is a lattice point MATH such that the triangle MATH spanned by MATH and MATH successive lattice points MATH has MATH. We apply an integral affine transformation MATH such that MATH, and MATH for some MATH. There exists MATH such that MATH. The segment MATH clearly contains exactly MATH equidistant lattice points and is not parallel to the image of MATH under the transformation MATH. Again we are done. CASE: MATH. Then it follows from the equation MATH, MATH that all the MATH coincide and are therefore identical with MATH. If MATH has an edge containing at least MATH lattice points, then either this edge is not parallel to MATH or there is a triangle MATH as in subcase MATH, and in both cases we are done. Only the case in which every edge of MATH has lattice length strictly less than MATH is left. In this situation we resort to the general result below REF which we have singled out because of its independent interest. |
math/9911006 | Assume to the contrary that such MATH exists. For a subset MATH and an element MATH we put MATH . Let MATH be an edge. It is impossible to find a graded MATH-algebra embedding of MATH into MATH (by NAME function reasons or REF ). In particular, the homomorphism MATH is not injective. (As usual, MATH is the facet projection MATH.) It follows that MATH as graded MATH-algebras. Thus there is an element MATH such that MATH . Consider the toric automorphism MATH determined by MATH . Then MATH where MATH. Applying the same arguments to the embedding MATH and an edge MATH sharing an end-point with MATH we conclude MATH . (Because of the common end-point there is no need to further `correct' MATH by a toric automorphism.) We can assume MATH. Going once around MATH we finally arrive at the equations MATH where MATH denotes the boundary of MATH. After an integral affine change of coordinates we may assume MATH and that MATH. Then MATH can be considered as a subalgebra of MATH generated by monomials MATH. Moreover, after a parallel translation of MATH we can assume that MATH contains a lattice point MATH, or equivalently, MATH contains a monomial MATH. For any element MATH we have the retraction MATH for all MATH. The image of MATH under MATH is a polytopal algebra MATH generated by the elements MATH with MATH. Due to the infinity of the field MATH there is MATH such that the elements MATH are pairwise different. In fact, it is clear that for any finite set of polynomials MATH there is an element MATH such that MATH maps MATH to a set of MATH pairwise different polynomials from MATH. Set MATH . In view of the fact that the elements MATH coincide `along MATH', we see that the coefficients MATH coincide for all MATH, and the same holds for the coefficients MATH. The MATH matrix MATH has two linear dependent rows, and since MATH, its rank is MATH. It follows that MATH is not injective. Thus we see as above that there exists MATH in MATH with MATH . Then however MATH. By a suitable choice of MATH we can furthermore achieve that MATH. Then it follows immediately that MATH. This is a contradiction since we have chosen MATH such that the MATH are pairwise different. |
math/9911013 | Suppose MATH and MATH. We prove the assertion of the lemma by induction on MATH. The case MATH is trivial. Assume now that MATH and that the assertion holds for MATH and all MATH, MATH with MATH. For an arbitrary MATH, we set MATH and MATH . We observe that MATH, for all MATH. We also have that MATH, where MATH. It follows now, by the induction hypothesis, that MATH, for all MATH. Therefore, MATH as MATH. The proof of the lemma is now complete. |
math/9911013 | By induction on MATH. The case MATH is trivial. Assume the assertion holds for MATH. Let MATH. We shall show that MATH, for every MATH and MATH. To this end choose MATH, successive members of MATH so that MATH and MATH. Let also MATH be an enumeration of the set MATH. We define MATH and observe that MATH, for all MATH, where, MATH are chosen so that MATH, for all MATH. Therefore, MATH and thus MATH . Finally, MATH and hence MATH. We conclude, since MATH was arbitrary, that MATH, as claimed. |
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