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math-ph/0011050 | This follows immediately from the strict convexity of MATH and MATH. |
math-ph/0011050 | The proof is similar to that of REF . Let MATH be a minimizing sequence. By REF we have MATH . With NAME theorem we therefore extract a subsequence, still denoted by MATH, such that MATH . Furthermore MATH is bounded in MATH, which implies that there exists a further subsequence, again denoted as MATH, with MATH . (Thi... |
math-ph/0011050 | The uniqueness of the minimum, the spherical symmetry of the functional REF and the fact that MATH implies the continuity of MATH away from the origin. With MATH REF has a meaning in the sense of distributions on the domain MATH. Consider the set MATH . If MATH, then MATH and MATH . For all MATH we find MATH. Let MATH,... |
math-ph/0011050 | Denote MATH . On this domain we have MATH with MATH, since MATH. So we conclude from standard elliptic arguments (for example, REF) that MATH is bounded, hence continuous everywhere. From MATH we get the two times differentiability of MATH away from the origin and as long as MATH. By means of REF and a standard bootstr... |
math-ph/0011050 | Suppose by contradiction MATH. We do not assume MATH to be finite. Then, as in REF , we get that there is some MATH and a MATH, such that MATH for MATH. Since MATH, we conclude MATH for MATH large enough. Hence, from REF we get MATH which is a contradiction to MATH. |
math-ph/0011050 | As we have already argued above, the symmetry of MATH follows from the symmetry of the functional and the uniqueness of MATH. Denote MATH the non increasing rearrangement of MATH. (For definition see for example, REF.) From the fact that MATH and REF we get MATH. REF implies MATH and again from CITE we get MATH, which ... |
math-ph/0011050 | Inserting MATH into REF yields the following equation for the potential, away from the origin: MATH where we replaced the constants by one. Since MATH is spherical symmetric we can use the ansatz MATH and obtain by REF the following fourth order equation: MATH . We can see that in the surrounding of each point MATH the... |
math-ph/0011050 | Let MATH denote the angular momentum operator parallel to the magnetic field. Since MATH, we have MATH which implies that the eigenvectors of the operator MATH are of the form MATH. Hence we can write the sum of the negative eigenvalues as MATH showing that the MATH's are the eigenvectors of the one-dimensional operato... |
math-ph/0011050 | First of all we note that for any comparison wave function MATH and fixed integer MATH we get MATH with MATH . If we set MATH, add and subtract MATH in the scalar product and use MATH, where MATH is the eigenvector corresponding to the MATH-th lowest eigenvalue MATH of the one-particle operator MATH as comparison wave ... |
math-ph/0011050 | Since MATH, we get MATH . Minimizing over MATH, we get the first part of the lemma. The proof of second part works analogously to REF . |
math-ph/0011050 | Using the scaling relation REF and defining MATH we get MATH . |
math-ph/0011050 | Let us rewrite MATH and define the unitary operator MATH . With MATH and the fact that unitary transformations do not change traces we derive MATH where MATH . So we have the self-adjoint NAME operator MATH and its symbol MATH . Using the relation MATH and by means of some cut off function MATH, REF yields MATH which t... |
math-ph/0011051 | We take the bracket of both sides of REF with MATH to obtain MATH . It follows that MATH is divisible by MATH. Since MATH is of degree less than MATH in MATH and since MATH is monic of degree MATH we must have MATH and MATH . Similarly, we find MATH and also that: MATH . These expressions also allow one to compute the ... |
math-ph/0011051 | In view of REF the entries of MATH can be written in the form MATH . Note also that, by using the MATH action, we can assume that MATH, MATH, and that MATH is a disjoint union of MATH, with MATH. Let MATH. Then MATH has the following form MATH . On the upper right corner the matrix has entry MATH, and the blocks MATH,M... |
math/0011001 | The proof is straightforward; see CITE and CITE. |
math/0011001 | It is sufficient to prove the statement for two reduced decompositions that can be identified by applying a braid relation once. Therefore, it is sufficient to check the statement for rank REF NAME algebras MATH. In this case the only element that has two different reduced decompositions is the maximal element MATH. So... |
math/0011001 | The proof of existence and uniqueness of MATH is straightforward (see for example, CITE). To prove the invertibility, it is sufficient to observe that for an irreducible module MATH and large MATH, the map MATH is nonzero. Indeed, the tensor product of a NAME module with a finite dimensional module does not contain fin... |
math/0011001 | The lemma is an easy consequence of the definitions: it expresses the fact that the operation of fusion of intertwiners commutes with the operation of restriction of intertwiners to submodules. |
math/0011001 | Consider the intertwiner MATH. It satisfies the relation MATH . Therefore, MATH . This implies after a straightforward calculation that MATH . Now let us consider the intertwiner MATH. It satisfies the relation MATH . Therefore, MATH . This implies that MATH as desired. |
math/0011001 | Let MATH. Consider REF in the weight subspace of weight MATH in the tensor product MATH. Let us identify this weight subspace with the opposite one in any way, and take the determinant of both sides of REF . Since the fusion matrix is triangular with the diagonal elements equal to MATH, its determinant is MATH. Therefo... |
math/0011001 | The existence follows from the above explicit computation of MATH. The uniqueness is obvious, since the function is defined at infinitely many points. The invertibility follows from REF . |
math/0011001 | REF from CITE implies that as MATH, one has MATH and MATH. Thus, going to the limit MATH in REF , and using REF from CITE, we obtain the proposition. |
math/0011001 | Straightforward from REF . |
math/0011001 | We prove the relations for MATH; the relations for MATH follow automatically since these two operators are proportional. If MATH is REF-dimensional, then MATH is known, and it is straightforward to establish the result. From this and REF it follows that the result is true if MATH is the tensor product of any number of ... |
math/0011001 | First of all observe that in MATH, one has MATH (where MATH is the iterated coproduct). On the other hand, using the expression for MATH for REF-dimensional MATH, and the expression for MATH given in REF , we get MATH . But MATH is the submodule of MATH generated by MATH. Thus, for MATH . Now the result follows from RE... |
math/0011001 | The proof of this proposition is straightforward from the previous results. |
math/0011001 | This is proved by a straightforward calculation with intertwiners, which generalizes to the q-case the calculation of CITE. Another proof is given as a remark in REF. |
math/0011001 | The statement follows easily from the definitions by induction on the length of MATH. |
math/0011001 | If MATH is large dominant, the statement is clear from REF , since MATH is independent of the reduced decomposition MATH of MATH by the quantum NAME identities REF . For an arbitrary MATH, the statement follows from the fact that a rational function is completely determined by its values at large dominant weights. |
math/0011001 | As for MATH, the lemma is an easy consequence of the definitions: it expresses the fact that the operation of fusion of intertwiners commutes with the operation of restriction of intertwiners to submodules. |
math/0011001 | It is enough to establish the first formula. The proof of this formula is obtained by specializing REF to the case MATH, dualization of the second component, and multiplication of the components. |
math/0011001 | Let MATH be a reduced decomposition. Using the Main Definition, we have MATH (in the last equality we used that MATH depends only of MATH). This implies that MATH . But for MATH we obviously have MATH on MATH. This, together with the Main Definition implies the result. |
math/0011001 | The proof follows easily from the results of the previous section and the Main Definition. |
math/0011001 | Clear. |
math/0011001 | To prove the statement for any NAME algebra, it is enough to do so for MATH, in which case the result is an easy consequence of the results of REF. |
math/0011001 | This follows at once by comparing the actions of both sides on basis vectors of MATH. |
math/0011001 | The result follows immediately from the equality MATH and REF . |
math/0011001 | This is immediate from REF and the definition. |
math/0011001 | It is sufficient to prove the result for MATH. In this case, we have a unique simple reflection MATH, and MATH . But MATH on MATH. Therefore, by the explicit formula for MATH we have MATH, as claimed. |
math/0011001 | Follows immediately from REF . |
math/0011001 | The proof of this proposition is straightforward. |
math/0011001 | Straightforward from the definition. |
math/0011001 | Straightforward from the definition. |
math/0011001 | The first statement is immediate from REF , as for MATH, the operator valued function MATH of MATH is regular at integer values of MATH when restricted to a weight subspace of a generic NAME module. Let us now prove the second statement. Because MATH is generic, any nonzero homogeneous vector MATH of weight different f... |
math/0011001 | It suffices to assume that MATH is homogeneous with respect to the weight decomposition. Let MATH be the NAME generators of MATH corresponding to some reduced decomposition of the maximal element of MATH. By the PBW theorem (see CITE), the submodule MATH is given by MATH, where MATH is the algebra of polynomials of MAT... |
math/0011001 | We have to check that the operators MATH satisfy the braid relations. This can be checked on finite dimensional NAME algebras of rank REF. But in this case, according to REF , everything reduces to the case when MATH is finite dimensional, where the statement is known. |
math/0011001 | The statements are obtained from REF by passing sending MATH to infinity. Namely, the first equation follows from REF. The second equation follows from REF. |
math/0011001 | We will first transform the equality to a convenient form, and then show that both sides satisfy the same ABRR equation CITE, which has a unique solution. This will imply that the two sides are equal. Let us make a change of variable MATH. Then the equation to be proved takes the form MATH . Let MATH. Then the equation... |
math/0011001 | This follows directly from REF , and REF . |
math/0011001 | The proposition is immediate from REF . |
math/0011001 | The proof is analogous to the proof of the NAME character formula using the approach of CITE; it is based on the fact that in the NAME group of the category MATH, an irreducible module is an alternating sum of NAME modules. |
math/0011001 | The proof follows from REF , and REF . |
math/0011001 | We have MATH (for brevity we drop the subscripts indicating the modules in which the operators act). From REF we easily obtain MATH . Let us substitute this equation into the previous equation, and use the fact that in the second component we are restricting to the zero-weight subspace. It is easy to see that the MATH-... |
math/0011001 | The proof is similar to the proof of the symmetry of MATH, given in CITE. It suffices to assume that MATH. Let MATH denote the right hand side of the equality to be proved. By REF , MATH, like MATH, is a solution of the NAME equations and the dual NAME equations. Moreover, both MATH, MATH have the form: MATH times a fi... |
math/0011001 | This follows by applying REF several times. |
math/0011001 | REF follows from REF and the definitions of CITE. REF can be checked directly using REF . |
math/0011001 | This Proposition is well known, but we will give a proof for the reader's convenience. REF follows from the fact that any finite dimensional representation has a weight decomposition with respect to any (quantum) MATH-subalgebra corresponding to a simple root (by representation theory of quantum MATH). Let us prove REF... |
math/0011001 | Let us call the operator defined by the right hand side by MATH. Using the identity MATH, we get that MATH for MATH. Since the statement that MATH is NAME group invariant, and MATH, this is sufficient. |
math/0011001 | We need to show that MATH lands in MATH and that it is a homomorphism. Let us prove the first statement. So let MATH, MATH, and let us show that MATH. It is clear that if MATH (where MATH is the dual vertex to MATH) then MATH is mapped under MATH to MATH, with MATH. Thus, we need to show that MATH and MATH are also map... |
math/0011001 | Straightforward, as in REF ; see also REF. |
math/0011001 | The lemma is proved by arguments similar to those in CITE. Namely, similarly to CITE, one can write down an explicit formula for MATH, and show that its poles are all of first order and can occur only on hyperplanes MATH for positive roots MATH and such MATH that MATH. If a dominant weight MATH belongs to such a hyperp... |
math/0011001 | The proof is analogous to the proof of NAME REF. |
math/0011001 | The proof is the same as that of REF . |
math/0011001 | Clear. |
math/0011001 | This is, after some transformations, the content of REF. This is also the multicomponent version of the ABRR equation for affine NAME algebras, projected to the product of loop representations (see CITE). |
math/0011001 | We have seen that the operator MATH conjugates the operators MATH to the diagonal operators MATH, and the dynamical action of the braid group to the multicomponent dynamical action. It is easy to see that the multicomponent dynamical action commutes with the operators MATH. This implies the desired statement. |
math/0011001 | This is the main result of CITE; see also REF (where the simply laced case is treated). This is also the multicomponent version of the ABRR equation for quantum affine algebras, projected to the product of loop representations (see CITE). |
math/0011001 | Analogous to REF . |
math/0011001 | The proof is easy. |
math/0011001 | Recall that any automorphism of an algebra acts on the set of equivalence classes of representations of this algebra. All we need to show is that the representation MATH is stable under the automorphism MATH for any MATH. It follows from NAME 's highest weight theory of finite dimensional representations of MATH that t... |
math/0011001 | We have (dropping MATH from the subscripts and MATH from the superscripts for brevity): MATH . Now recall that in the braid group MATH we have MATH, and hence MATH (with the length of both being MATH in the affine NAME group). This implies that the product MATH is symmetric under interchanging MATH and MATH (REF-cocycl... |
math/0011001 | REF is exactly REF follows from the fact that the operator MATH commutes with the KZ (qKZ) equations. |
math/0011001 | This is immediate from the previous results. |
math/0011001 | The proposition follows immediately from REF and the definitions. |
math/0011001 | The proof is by a straightforward comparison of the two systems. |
math/0011008 | Suppose that MATH does not hold, then MATH is clean. Suppose also that MATH does not hold: then by REF , for MATH, there is a point MATH with MATH clean in MATH. This MATH also satisfies the slope condition MATH and is hence, by REF , reachable from MATH. |
math/0011008 | By REF , for MATH, there is a point MATH with MATH clean in MATH. This MATH also satisfies the slope condition MATH and is hence, by REF , reachable from MATH. |
math/0011008 | We will see that all the properties in the condition follow essentially from the form of our definitions. Note that when an existential or universal quantifier is applied to a family of monotonically increasing (decreasing) events, the result is monotonically increasing (decreasing) event (as a function of the sequence... |
math/0011008 | Let MATH be the sequence of maximal external intervals of MATH, of size MATH. (We consider MATH, so that MATH is automatically of size MATH.) Let MATH be the intervals betwen them. By REF of MATH, each MATH can be covered by a sequence of neighbors MATH in MATH. Every wall of MATH intersects an element of this sequence... |
math/0011008 | If MATH contains no walls of MATH then it is a hop of MATH and we are done. Let MATH be the union MATH of all walls of MATH in MATH. The inner cleanness of MATH in MATH implies that MATH is farther than MATH from its ends. REF applied to MATH implies that MATH is spanned by a sequence of neighbor walls MATH of MATH. Si... |
math/0011008 | We will use MATH which follows from REF. Consider an interval MATH of size MATH containing no walls of MATH. Let MATH be the middle third of MATH. By REF , it is contained in a hop of MATH. REF implies that MATH is covered by a sequence MATH of light neighbor walls of MATH separated from each other by hops of MATH of s... |
math/0011008 | Suppose that this is not the case. Then the operation of creating emerging barriers would turn every interval of the form MATH with MATH into an emerging barrier. Both MATH and MATH can be chosen to be clean. This choice would define an emerging wall. Since we assumed that we are at the point of the construction when n... |
math/0011008 | For MATH, let MATH be the event that MATH is realized by a hole ending at MATH but is not realized by any hole ending at any MATH. Let MATH be the event that MATH is strongly right-clean. Then MATH, MATH, the events MATH and MATH are independent for each MATH, and the events MATH are mutually disjoint. Hence MATH . |
math/0011008 | If MATH then we can apply the hole lower bound REF; suppose therefore that this does not hold. Let MATH. Then REF is applicable to MATH, and we get MATH . Consider the event MATH that MATH is strongly right-clean and the interval MATH contains no barriers. Then MATH. Event MATH is decreasing with MATH. By the FKG inequ... |
math/0011008 | We will use the following inequality, which can be checked by direct calculation. Let MATH, then for MATH we have MATH . In view of REF , for the first statement of the lemma, we only need to consider the case MATH. Let MATH, then we have MATH . Let MATH . REF is applicable to MATH and also MATH by REF. We have MATH, h... |
math/0011008 | This follows directly from the reachability REF , if we observe that MATH, so that the the minslope is MATH. |
math/0011008 | Recall the definition of an emerging barrier. Suppose that there are MATH with MATH, MATH, MATH and a light barrier type MATH such that no hole of type MATH is cleanly contained in MATH. Then MATH is a barrier of MATH. This definition implies that if MATH is an emerging barrier then there is a light barrier type MATH s... |
math/0011008 | Assume REF. As MATH is passed by MATH, there is a path from MATH to MATH . Due to REF there is a path from MATH to MATH. As MATH is passed by MATH, there is a path from MATH to MATH. The total slope of the combination of these three paths is clearly at most REF. (For reachability, the total slope condition is not impor... |
math/0011008 | Assume MATH. then MATH, and MATH. REF turn into the true inequalities MATH. Assume MATH, then MATH, and MATH, MATH. REF will turn into the true inequalities MATH. Assume MATH, then MATH, MATH, MATH, MATH. REF will turn into the true inequalities MATH. Assume MATH, then MATH, MATH. Given MATH and MATH, what we need to c... |
math/0011008 | For fixed MATH, let MATH be the event that a compound barrier of any type MATH with MATH, distance MATH between the component barriers, and size MATH appears at MATH. For any MATH, let MATH be the event that a barrier of rank MATH and size MATH starts at MATH. We can write MATH where events MATH, MATH are independent. ... |
math/0011008 | Let MATH, MATH. For each MATH, let MATH be the event that there is a MATH such that MATH is a jump of MATH, and a hole of type MATH starts at MATH and ends at MATH, and that MATH is the smallest possible number with this property. Let MATH be the event that there is a MATH such that MATH is a jump of MATH, and a hole o... |
math/0011008 | We have MATH which is satisfied by the choice MATH in REF. |
math/0011008 | We can choose MATH last, to satisfy REF, so consider just the other inequalities. Choose MATH to satisfy REF; then REF will be satisfied if MATH. This is achieved by MATH . |
math/0011008 | For the moment, let us denote the largest existing rank by MATH. Emerging types got a rank equal to MATH, and the largest rank produced by the compound operation is at most MATH (since the compound operation is applied twice), hence MATH. Since also MATH (since there is only one rank in MATH), we have for MATH: MATH . ... |
math/0011008 | Let us prove REF, which says MATH. We have MATH which is smaller than REF if MATH is sufficiently large. For REF , note that MATH which because of REF, is clearly less than REF if MATH is large. |
math/0011008 | The probability that a point MATH is strongly clean in MATH but not in MATH is clearly upperbounded by MATH, which upperbounds the probability that a barrier of MATH appears in MATH: MATH . For sufficiently large MATH, we will always have MATH. Indeed, this says MATH, which is satisfied if MATH. This implies that if RE... |
math/0011008 | Recall REF . Let MATH. For any point MATH, the expression MATH is an upper bound on the sum, over all MATH, of the probabilities that an emerging barrier of type MATH (with rank MATH) starts at MATH. We have MATH . Due to REF, this expression grows exponentially in MATH, and MATH decreases double exponentially in MATH.... |
math/0011008 | Let MATH be two types with ranks MATH. Assume without loss of generality that MATH and that MATH is light: MATH. With these, according to REF of the scale-up algorithm, we can form compound barrier types MATH, as long as MATH. This gives a type of rank MATH, for all MATH. The bound REF and the definition of MATH in REF... |
math/0011008 | By REF , each rank MATH occurs for at most a constant number MATH values of MATH. For every such value but possibly the last one, the probability sum can only be increased as a result of the two operations of forming compound types. According to REF , the increase is upperbounded by MATH. After these increases, the pro... |
math/0011008 | We will show that compound hole types in MATH satisfy REF if their component types do (they are either in MATH or are formed in the process of going from MATH to MATH). Consider the compound hole type MATH where MATH . Let MATH, then MATH. Let MATH and MATH. Following the notation of REF , let MATH be the event that th... |
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