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hep-th/9908012 | We have MATH as (densely-defined) symmetric forms. Thus MATH as symmetric forms. Thus MATH . We have MATH . However, since the first form is positive and symmetric we have MATH as forms, and so MATH . |
hep-th/9908012 | We make use of the microlocal properties of the two-point functions, discovered by NAME REF and Junker REF . A summary adequate for understanding this proof is in the paper of NAME REF . For the general theory of wave-front sets, see NAME REF . The antilinear part of MATH is got by projecting its positive-to-negative and negative-to-positive ``frequency" parts using the two point function; it corresponds to the two-point kernel MATH . Here MATH is the two - point function. The wave - front set of MATH is MATH . Here and in what follows, it is understood that MATH; and MATH iff there is a null geodesic from MATH to MATH with covector MATH at MATH and MATH at MATH. Thus MATH . When this is integrated against MATH to form MATH, the result has MATH which is empty. Since the two-point function has a smooth kernel, so does its restriction to act on initial data. |
hep-th/9908012 | Consider MATH as a perturbation of its MATH-linear part. The term MATH is MATH-real anti-self-adjoint and so has purely imaginary spectrum. On the other hand, the compactness of MATH means that any element of MATH must be an eigenvalue. (To see this, suppose one has MATH and a sequence MATH of unit vectors in the domain of MATH with MATH. Multiplying by MATH we have MATH. Since MATH is compact, this implies MATH has an eigenvalue unity. However, this implies MATH is an eigenvalue of MATH.) However, then the positivity of MATH implies, as in the proof of REF , that any such eigenvalue is imaginary. |
hep-th/9908012 | We consider MATH as a perturbation of MATH; the latter is MATH - unitary and so has spectrum on the unit circle. It follows from standard perturbation theory (NAME and NAME REF ) that MATH is compact. Applying the argument of the previous proof, we see that any element of the spectrum of MATH not already in the spectrum of MATH must be an eigenvalue; but then that MATH be a symplectomorphism implies that eigenvalue lies on the unit circle. |
hep-th/9908012 | This proof is somewhat technical, and makes use of the theory of NAME hyperfunctions. This is the class of generalized functions MATH dual to MATH are both smooth of exponential decay MATH. (Here MATH denotes the NAME transform of MATH.) The results we use are contained in the papers of NAME and NAME REF and NAME REF . The topology on MATH is given by the family of seminorms MATH. A sequence MATH in MATH if for some MATH one has MATH as MATH. Put MATH, where MATH is a real parameter (which may acquire a small imaginary part). Let MATH denote the jump in MATH from above the real MATH-axis to below; this jump is by definition a hyperfunction. (One defines its pairing with a test function by integrating slightly above and slightly below the axis; compare NAME REF.) We shall interpret this hyperfunction in the strong sense, that is, as applied to any vector in MATH. All integrals of operators in what follows are also to be interpreted in the strong sense. We now show that MATH is not just a hyperfunction, but in fact a NAME hyperfunction. This means that for any MATH are both smooth of exponential decay MATH, the integral MATH is defined and varies continuously with MATH. We have MATH the limit MATH being understood. This is well-defined, since the class MATH is invariant under NAME transform. To see that it depends continuously on MATH, we note that if MATH is a sequence tending to zero in MATH, then there exists MATH such that MATH tends to zero as MATH. However, since MATH (say - using the NAME - NAME - NAME estimate and the fact that the spectrum of MATH is purely imaginary), the integrals tend to zero as MATH. Some comments are in order at this point: CASE: We have just shown that the jump in the resolvent is the NAME transform of MATH. CASE: The integral displayed above must lie in the domain of MATH (since the class MATH is invariant under differentiation). CASE: These results are independent of positivity properties of MATH. In general, then, groups of type zero (in the semigroup sense) have generators which can be analyzed in terms of NAME hyperfunctions. These generalized functions admit a useful microlocalization. A direct argument to establish positivity is probably possible, but partly for technical reasons and partly for its future utility, I shall give another. Every NAME hyperfunction can be viewed as an initial datum for the heat equation. If the corresponding solution for MATH is everywhere non-negative, then the hyperfunction is a measure. This result is due to NAME REF , and can be thought of as an extension of the NAME - NAME theorem. (In this context, the initial datum is called an NAME trace.) The solution to the heat equation with initial value MATH is MATH . For MATH, this maps to the domain of MATH. We also note the identities MATH and MATH where the overline indicates the complex conjugate. Using these, we have MATH and this is positive. We know at this point that for any MATH, the quantity MATH is an exponentially tempered measure. It is easily verified that the form MATH is Hermitian, so by polarization MATH is an operator-valued exponentially tempered measure (in the strong sense). If we could divide this by MATH, we could conclude that MATH is a measure. However, this is not obviously possible. For any test function MATH, write MATH. Then MATH is MATH times a smooth function. Using this decomposition, it is easy to see that MATH extends to a linear form on those continuous functions of exponential decay which are MATH at the origin. In particular, the NAME hyperfunction MATH is a distribution. Since MATH is a measure, we must have MATH for some MATH and some measure MATH. (We are following the convention where distributions are represented by ``generalized functions" under the integral sign, and so the measure is represented as MATH or MATH. This is only a symbolic representation, and is not meant to assert any regularity of the measure with respect to NAME measure. The prime is present so as to have the formal correspondence MATH, but we we are not interested in trying to construct MATH or giving MATH meaning oas a derivative. We should more carefully absorb the MATH's into MATH and MATH and write simply MATH.) We caution that MATH is known to be locally finite only on MATH. The coefficient MATH must vanish. To see this, note that we have (strongly) MATH and hence MATH . Passage from the first to the second line is justified by use of the NAME transform for MATH, or by the identity MATH; passage from the second to the last by the fact that MATH is a measure. Now let MATH and let MATH. It is easy to see that as a distribution in space - time, the quantity MATH vanishes on MATH. On the other hand, since MATH, the local positivity of the form MATH implies that MATH vanishes on the interior of MATH. (To see this, note that we may replace MATH in the integrand with MATH, which will annihilate anything supported outside of the support of MATH.) Thus if MATH were known to be smooth, we would have MATH. However, it is easy to check that for any MATH we have MATH. By the arguments just given, this vanishes for smooth MATH; since the smooth MATH's are dense in MATH, it vanishes always and MATH. Hence MATH. We shall now write MATH. It is a projection-valued distribution which is locally a projection-valued measure. To see that it is projection-valued, note that for any test function MATH we have MATH . To establish the compatibility of the decompositions with the symplectic structure, notice that MATH . For a subspace MATH to be real, the set MATH must be symmetric (up to terms of MATH-measure zero). For symmetric sets, the equation above shows that MATH if MATH. Thus the spectral decomposition by MATH respects the symplectic structure. That MATH must be strongly non-degenerate on each MATH follows. To establish MATH strongly on MATH, we consider MATH . Take, for instance, the function MATH and MATH. Then MATH . Substituting this into the previously displayed equation and integrating by parts, we get MATH . Taking MATH gives the required result. |
hep-th/9908012 | In order to make use of the results of previous sections on classically positive Hamiltonians, we shall introduce a modified family of operators MATH which are classically positive and tend to MATH as MATH. Let MATH be a continuous bump function, supported on MATH, identically unity in a neighborhood of the origin, and symmetric. Let MATH. We may regard MATH either as an operator on MATH, or, when convenient, on the subspace MATH. The discussion of the compatibility between MATH and MATH at the end of the last proof shows that MATH restricts to be non-degenerate on MATH, and then MATH is classically positive on MATH. We have MATH as MATH for MATH, since MATH is a measure (and the mass of MATH, that is, the coefficient MATH in the previous proof, is zero). Now let MATH be any NAME state of norm unity. This means that in the holomorphic representation MATH is a polynomial whose coefficients are represented by smooth fields on space - time; in particular, these coefficients lie in (tensor products of) MATH. Thus we may compute MATH where MATH is the Hamiltonian defined from MATH by normal ordering. (NAME REF showed that MATH may be defined by normal ordering.) But we know a lower bound of MATH is MATH. Now in fact the lower bounds are monotonically decreasing with MATH. This follows from the fact that for any fixed MATH, a fixed MATH can be found which simultaneously provides a similarity of all MATH with MATH to generators of orthogonal groups. It follows that MATH where MATH denotes the infimum of the spectrum. Similarly, it follows from the formula for MATH in REF that MATH is a family of symmetric positive forms, which are (as forms) increasing as MATH. But as we know that MATH strongly, we have MATH . |
hep-th/9908012 | It only remains to note that since MATH has a smooth kernel, it is trace-class. (See for example, NAME REF.) |
hep-th/9908170 | Because of MATH the gauge transformations generated by MATH lead to the addition of an exact two-form to the dual two-form MATH of MATH. First we show that two different symmetric electric fields MATH and MATH cannot lie in the same MATH-gauge class. By assumption the difference MATH of these fields fulfills MATH (MATH, MATH and MATH are spherical coordinates). If that difference was a gauge transformation generated by some MATH, this function had to obey the equations MATH . This, in turn, would imply MATH that is, the difference of the electric fields would be a constant which had to vanish due to the boundary condition. Therefore, each gauge class contains at most one spherically symmetric electric field. We now prove that each class contains at least one spherically symmetric electric field. In order to show this we need a vector field MATH on MATH which is spherically symmetric, that is, MATH, and which is subject to the conditions MATH and MATH for all MATH-orbits MATH in MATH. Such a field exists because, for the symmetry reduced theory to be non-trivial, we have to assume that there is at least one spherically symmetric electric field MATH, by means of which we can construct MATH . The existence of such a non-trivial field MATH depends on the topology of MATH. It always exists in the manifolds of REF. The properties of MATH postulated above follow from the ones of MATH and the fact that MATH does not depend on MATH (due to MATH). Let MATH now be a field vanishing at infinity. Due to the properties of MATH the averaged field MATH is spherically symmetric and fulfills the NAME constraint MATH. Furthermore, we have MATH, which remains valid after replacing MATH by an arbitrary closed surface. According to NAME duality of homology and cohomology groups the difference of the two fields is cohomologically trivial and, therefore, exact: MATH with an appropriate MATH. An electric field and its spherically symmetric average, therefore, lie in the same gauge class. Summarizing, we have proved that each gauge class contains exactly one spherically symmetric electric field. |
hep-th/9908170 | The cylindrical consistency condition for the measure and its normalization as well as diffeomorphism invariance follow from properties of the NAME measure. If MATH and MATH are two monomials, then MATH proving orthonormality. |
hep-th/9908170 | The proof is the same as in Ref. CITE except for an obvious generalization to MATH variables. |
hep-th/9908207 | As before, we denote by MATH the orthogonal projection onto MATH restricted to the subspace MATH . Given MATH, we can represent elements MATH and MATH as the determinant elements of linear operators MATH and MATH with MATH and MATH of trace-class; consequently, the operators MATH and MATH are trace-class, too. We define MATH by MATH where MATH are the projections on MATH, and MATH is the determinant line of MATH. That this operator differs from MATH by an operator of trace-class (in order that REF be well-defined) follows immediately from that fact that the operators MATH and MATH are trace-class. The canonical isomorphism on the right-side of REF is expressed via the diagram of commutative maps with exact rows and NAME columns MATH where the horizontal maps are the obvious ones. We know from CITE that such a diagram defines an isomorphism between the determinant line of the centre map with the tensor product of the lines defined by the outer columns, mapping the determinant elements to each other. Hence since MATH canonically, the isomorphism follows, and in particular with MATH and MATH we have MATH where MATH, which will be a relevant fact later in this Section. For the general REF , suppose initially that MATH and choose MATH and MATH identified with the determinant elements of linear operators MATH and MATH. Define MATH . On the other hand, from REF mean that there is an exact sequence MATH and this fits into the commutative diagram with NAME columns MATH where we modify REF by composing the injection MATH with the involution MATH, and the following surjection to MATH, while the lower maps are again the obvious ones. The central column is MATH . Because MATH and MATH are by trace-class, the operators MATH and MATH, where MATH differ by only trace-class operators and so MATH, while from the diagram we have MATH. Thus by duality (that is, take adjoints in the above diagrams, reversing the order of the columns and rows and the direction of the arrows) we have a canonical isomorphism MATH, and so composition with MATH completes the proof of REF in the case MATH. In the general case, replace MATH in REF and the lower row of the commutative diagram by MATH and repeat the argument used in the proof of REF. Finally, we note for later reference that in the `vacuum case' MATH and MATH one has MATH and MATH . |
hep-th/9908207 | First notice that in the finite-dimensional case there is a natural isomorphism between the NAME space (the exterior algebra) and its dual defined by the pairing MATH while in the infinite-dimensional case the pairing using the CAR construction follows directly from the definition MATH . For the geometric NAME space MATH, the construction of the pairing from the determinant bundle pairing MATH on MATH is entirely analogous to the construction of the inner-product MATH on MATH from the determinant bundle pairing MATH on MATH in REF . Indeed, in the case of the vacuum elements the two pairings are canonically identified (see REF below and REF). Let us deal first with REF . We give first the invariant definition, and then the `constructive' definition along the lines of MATH in REF. Invariantly, in the case MATH, the pairing MATH defines an embedding MATH by MATH, and hence a map MATH, MATH. This gives us a pairing MATH with MATH, and by duality the asserted pairing, since MATH in the topology of uniform convergence on compact subsets of MATH (MATH compare REF). The general case follows in the same way with MATH replaced by MATH. Constructively, recall that any section in MATH can be written as a linear combination of the MATH, with MATH. Hence for MATH we can define the NAME pairing by setting MATH and then extending by linearity. In particular, from REF we have MATH and MATH and so MATH where the final equality is REF . Notice further that if we extend MATH in REF to a map MATH by MATH then the NAME space inner-product becomes a Hermitian pairing MATH and with respect to the identification MATH we have MATH and MATH . The pairing REF now follows from REF. Alternatively we can define it directly as MATH, where MATH is the pairing REF. Note that if REF do not hold then MATH. REF is now just by construction, and REF follows easily from REF. |
hep-th/9908207 | We just need to recall a couple of facts. From REF we have a pairing MATH with MATH, where MATH is the operator of REF. But from REF there is a canonical isomorphism MATH, with MATH. This proves the first statement. The second statement follows similarly upon recalling from CITE REF that there is a canonical isomorphism MATH, again preserving the determinant elements. |
math-ph/9908002 | Let us rewrite REF in terms of the elliptic amplitude MATH, defined by MATH . We thus obtain the expressions MATH where we have set, for convenience, MATH . Let us concentrate, for definiteness, on the eigenfunctions of type REF. We rewrite REF in the form MATH where we take MATH when MATH is odd. Taking into account that, by the recurrence relation REF - REF, all the coefficients MATH are real, we conclude from REF that MATH . Let us now introduce the polynomial family MATH in terms of which the coefficients MATH are simply expressed by MATH . Defining two additional families of univariate polynomials MATH and MATH by MATH from REF - REF we have MATH (where MATH denotes the integer part). From REF - REF , it immediately follows that the polynomials MATH satisfy the three-term recurrence relation MATH with the initial conditions MATH where the coefficients MATH are defined by REF . It is easy to see that REF implies that the polynomial families MATH and MATH satisfy the same three-term recurrence relation REF as the family MATH. Moreover, from REF we obtain MATH . From this, and the fact that the coefficients MATH and MATH satisfy a three-term recursion relation of the type REF, we conclude that the vanishing of MATH or of MATH automatically implies the vanishing of MATH or of MATH, respectively, for all values of MATH. Thus we can write MATH . To complete the proof of the Proposition, we simply note that REF - REF and the fact that MATH and MATH satisfy the same three-term recursion relation as MATH imply that MATH and, in particular MATH . The proof for the eigenfunctions of type REF is totally analogous, and will therefore be omitted. |
math-ph/9908003 | By CITE, the series MATH and MATH act on MATH by MATH. It follows that MATH. Indeed, we have MATH and by REF we obtain MATH. Applying MATH to this equation MATH times and using REF again we obtain MATH. Similarly, from MATH, it follows that MATH. Note that MATH, MATH and MATH, so MATH , because a scalar multiple of MATH is obtained from MATH by successive commutators with MATH. |
math-ph/9908003 | Since MATH in MATH, all MATH - graded elements of the form MATH act trivially in MATH. When acting on the highest weight vector MATH, MATH . In addition, since MATH, MATH . |
math-ph/9908003 | Consider the set of elements in MATH, MATH and MATH where MATH is an ordered monomial. Due to REF , when applied to the highest weight vector MATH, these are trivial in MATH. Let us show that an ordered monomial is the dominant monomial of an element in this set if and only if it violates one or more of REF - REF . The ordered monomials MATH and MATH are the dominant monomials of MATH and MATH, respectively, and an ordered monomial violates REF if and only if it can be represented in one of these forms. Consider MATH with MATH and MATH. Their dominant monomials violate one of REF - REF , as shown in the following table: Given any ordered monomial MATH, the dominant term in MATH also violates the conditions as listed above, hence it is not a MATH-monomial. On the other hand, suppose that an ordered monomial MATH violates one of REF - REF . It can decomposed in MATH as MATH, in such a way that MATH, and MATH violates the same condition as MATH. Then, using this table, there exist MATH and MATH such that MATH is the dominant monomial of MATH. Since MATH is the dominant monomial of MATH, any monomial which violates any of REF - REF is of the form MATH. We have shown that the set of MATH-monomials, which satisfy REF - REF span the space MATH. |
math-ph/9908003 | The proof is similar to REF . Recall that MATH is a quotient of MATH, and therefore of MATH. For any MATH, MATH, define MATH. By definition it is trivial in MATH. If MATH, MATH or MATH, then the dominant monomial of MATH is MATH. Consider the set of elements MATH together with the elements REF . Clearly, their image in MATH is zero. The dominant monomials of MATH are precisely the monomials of length larger than MATH. Therefore, our assertion follows. |
math-ph/9908003 | REF follows from REF and the fact that MATH is the sum of the dimensions of the representations appearing in the right hand side of this formula. |
math-ph/9908003 | Let MATH be the number of MATH - monomials of length MATH with MATH, MATH, MATH. REF - REF imply that MATH . Consider the matrix MATH given by MATH . Note that this matrix does not depend on MATH. Clearly, MATH. From REF we have MATH with MATH . Let us simplify this formula using REF . Note that MATH depends only on MATH, MATH, MATH, MATH and MATH. So we can apply REF to all the multi - indices MATH with the same MATH and MATH in REF . We obtain MATH . For example, if MATH, we have MATH . See also REF (with MATH) and the matrix MATH in the proof. Note that MATH and MATH do not depend on MATH, so we can transpose REF and then apply REF again. The reduced matrix is in fact equal to the matrix MATH of REF , and the reduced column is nothing but REF : MATH . |
math-ph/9908003 | Let us begin with REF . If MATH consists of distinct complex numbers then REF follows from REF . Arguments from deformation theory show that for any (not necessarily distinct) MATH we have MATH. On the other hand, REF for MATH shows that MATH. Therefore, by using REF , we have the equality of dimensions for any MATH. REF follows. REF follows from REF ; we need only to prove that MATH - monomials are linearly independent. Suppose that some linear combination of MATH is trivial in MATH. Let MATH be the maximal length of monomials MATH. Then, this linear combination is trivial in MATH, which contradicts REF . |
math-ph/9908003 | Each monomial REF contributes MATH to the character where MATH, MATH; there is also a contribution from MATH, which is equal to MATH. Let MATH be the partial character of the linear span of MATH - monomials of length MATH with MATH, MATH, MATH. Then we have MATH and therefore MATH where MATH . The coefficients MATH depend only on MATH, MATH, MATH, MATH and MATH. Therefore, we can apply REF again. We obtain MATH with MATH . Finally, noting that MATH for all MATH and MATH, we can replace the summation with respect to MATH and MATH with the multiplication of the row vector MATH. |
math-ph/9908003 | Consider the case MATH. We have MATH . Then use induction on MATH. Consider MATH . In the second line, we assumed the lemma is proved for MATH. The last term is of the desired form, and the other terms can be reduced to this form by using the following procedure: MATH . In the second step, the last two terms are again of the desired form, and the degree of MATH in the first term is less than MATH, the degree on the left hand side of the equation. Therefore if we repeat this REF finite number of times, the process will terminate when the degree is low enough so that MATH. We are left with a sum of terms with MATH acting on the left, with degree MATH. Therefore the lemma follows. |
math-ph/9908003 | The fact that MATH follows from the fact that the right hand side of REF contains only positive powers of the rightmost variable in the product of currents acting on MATH. The fact that MATH is a symmetric function in MATH and MATH implies that MATH . The same argument holds for MATH. Since MATH if MATH or MATH, we have MATH . From REF it is clear that MATH. Therefore, the lemma follows. |
math-ph/9908003 | We define an ordering among all separately symmetric NAME polynomials of the form MATH in a similar manner as for pairs of partitions MATH. Then, we have MATH . The coupling REF is equivalent to MATH . Therefore the coupling of the basis MATH to the vectors MATH is triangular. |
math-ph/9908003 | This follows from REF . |
math-ph/9908003 | We only note that we expand MATH when we compute the coupling of MATH with an element of the form MATH. |
math-ph/9908003 | Since we have REF , it is enough to show, first that a vector of the form REF belongs to the orthogonal complement to the space MATH if MATH or MATH, and second that the characters of the space of vectors of the form REF satisfying the conditions in the statement of the theorem, is equal to that of the space MATH. We only note that MATH is replaced by MATH because MATH is placed to the right of MATH. The rest of the proof is straightforward. |
math-ph/9908003 | Choose a set of vectors MATH such that their images form a basis in the space MATH. We prove by induction on MATH that the images of the elements MATH span MATH. Indeed, let MATH. Then there exists a linear combination MATH, MATH, such that the image of MATH in MATH is in the image MATH. It means that MATH, where MATH and MATH is in MATH. By induction hypothesis there exists a linear combination MATH, MATH, MATH, such that MATH. Then, MATH. Therefore, if the images of MATH form a basis in MATH then the images of the same vectors span MATH. |
math-ph/9908003 | It follows from REF applied to the vector space MATH with the filtration defined in REF, and the set of operators MATH; MATH. |
math-ph/9908003 | It follows from REF applied to the vector space MATH with the filtration induced by the MATH-grading, and the set of operators MATH where MATH. |
math-ph/9908003 | By expanding the identity MATH with MATH, we have MATH . We have, in particular, that MATH . This is equal to the dimensions of the coinvariants MATH given by NAME 's formula. So by REF we have that MATH. Note that we have the same inequalities as in REF for any graded component of these spaces. Therefore we have the equality of characters, that is the statement of the theorem. |
math-ph/9908003 | By symmetry reason it is enough to show one of these, say REF . It is clear that MATH is well-defined and injective. If MATH, the map MATH sends MATH of REF to MATH . If MATH, it is obvious that this is well-defined and surjective. If MATH, it is less obvious because in the space MATH, the first condition of REF is void and also because MATH. The first condition of REF for MATH becomes MATH . This is equivalent to MATH. Therefore, if MATH or MATH, then MATH is allowed and the map MATH is surjective. If MATH and MATH, we have MATH, and therefore, we must use the definition MATH. The rest of the proof is straightforward. |
math-ph/9908003 | Consider, for example, the map MATH. Clearly, MATH preserves the relations REF and the highest weight condition, MATH, where MATH and MATH. For example, the images of the zero vectors MATH, MATH, MATH are also zero vectors. Finally, MATH maps the subalgebra generated by MATH to the the subalgebra generated by MATH. |
math-ph/9908003 | The proof is by induction on MATH. First, by inspection we check that MATH, and MATH, is consistent with the statement of the Theorem. Next, assume that MATH . Let us prove that MATH is a basis of MATH. We use the sequences REF with MATH, MATH. We show the following: REF there exists an injection MATH satisfying REF there exists a bijection MATH satisfying MATH . In fact, the map REF is given by MATH . Then we have, MATH . The map REF is given by MATH . Similarly, we prove that MATH is a basis of MATH using the sequences REF with MATH. Finally, we prove that MATH is a basis of MATH using the sequences REF with MATH. |
math-ph/9908003 | It follows from the remark after REF . |
math-ph/9908003 | The MATH-basis of MATH is described by REF . By the argument similar to the proof of REF , the same vectors form a basis in MATH. REF is proved by application of the map MATH to this basis. |
math-ph/9908003 | This is a special case of REF with MATH non - distinct points. |
math-ph/9908003 | For an element MATH, there exists an integer MATH such that MATH for any MATH. Therefore, the element MATH defines a map MATH and a dual map MATH. Note that the action of MATH on MATH is of level MATH. Therefore, we have an action of the central extension of MATH at level MATH on the limit MATH. Introduce the topology in the central extension of MATH where open sets are MATH. The completion in this topology is exactly MATH. Note that for any vector MATH some MATH acts trivially on MATH. Therefore, we have the action of this completion on MATH. Since as we noted the cocycle of MATH in MATH is trivial, the cocycle of MATH in MATH is equal to the negative of the cocycle in MATH. Therefore, the level of the representation of MATH obtained by the completion is MATH. |
math-ph/9908003 | Note that the subalgebra MATH belongs to MATH. The action on MATH of the MATH subalgebra of MATH generated by MATH and MATH is integrable because MATH is the inductive limit of the finite dimensional spaces MATH on which MATH acts. In order to see the integrability with respect to the MATH subalgebra of MATH that is generated by MATH and MATH, we define the subalgebras MATH . The subalgebra REF belongs to the completion of MATH. Since MATH, the module MATH is also the inductive limit of the finite dimensional spaces MATH on which the completion of MATH acts. Therefore, the integrability follows. |
math-ph/9908003 | We prove the statement by induction on MATH. If MATH, the statement is obvious. If MATH, because of REF , we have a subspace MATH invariant with respect to MATH. Define MATH by MATH. We will show that MATH. By REF , MATH is MATH - invariant, and therefore, the action of MATH on MATH is trivial. Since MATH, we have MATH. In other words, the subspace MATH is nonzero and MATH - invariant. By the induction hypothesis, the action of MATH on the MATH module MATH is nilpotent, and therefore the action on MATH itself, too. In the proof, we have also constructed a filtration which satisfies the requirement of the lemma. |
math-ph/9908003 | Since MATH if MATH, it is enough to prove the lemma for a single MATH. We use REF by setting MATH, MATH and MATH. The statement follows immediately. |
math-ph/9908003 | Note that the completion of MATH in MATH is equal to MATH . The space MATH is the space of MATH - invariants in the MATH - module MATH. We know that MATH is a direct sum of products MATH of integrable modules and that the space of MATH - invariants in integrable module at level MATH is an irreducible MATH - module of dimension at most MATH. Therefore, we have the statement of the lemma. |
math-ph/9908003 | Clearly, MATH . There is an action of MATH on the tensor product MATH through REF . Clearly, we have MATH . Therefore, we have MATH . As MATH acts on MATH by zero, we get MATH which implies the lemma. |
math-ph/9908003 | Let us set MATH. By REF it is enough to prove that the natural surjection MATH is an isomorphism. Consider the exact sequence MATH A standard computation with the NAME group shows that MATH. We have the exact sequence of coinvariants MATH . Since MATH, by REF , the first term is zero. Therefore, the second map is an isomorphism. Repeating this procedure for MATH we obtain that REF is an isomorphism. (Compare CITE for an alternative proof.) |
math-ph/9908003 | REF follows from REF . REF follows from REF because the quotient of an irreducible MATH - module by the subalgebra MATH is one - dimensional space. |
math-ph/9908010 | See CITE. |
math-ph/9908010 | In NAME coordinates MATH . |
math-ph/9908010 | This follows directly from REF : MATH . |
math-ph/9908010 | We start to build the diffeomorphism. We compute the pull-back of the twisted symplectic REF-form by the flow of MATH. The choice of this vector field is dictated by its role in the pertubation problem as shown on the previous section. The pull-back of MATH by MATH, the MATH-time flow of MATH is given by MATH where MATH is the fiber coordinate function and MATH is a fiber-independent REF-form. This is a straightforward computation using NAME 's Formula and the structure equations. The identity MATH follows from the fact that MATH is the NAME connection restricted to MATH. The Flat Case : Suppose MATH is flat. Then MATH, the bundle is trivial and we can take MATH that is, MATH. The pull-back computations reduce to MATH . MATH . MATH . We obtain MATH which simplifies to MATH giving MATH where MATH . So, for the planar case, the first diffeomorphism is already sufficient to bring the twisted symplectic REF-form to the desired form. One should compared this expression with the one found by CITE. A second diffeomorphism will be necessary if MATH. Denote MATH. First we will restrict MATH to the level set MATH. In this case we have that MATH and REF becomes MATH . For a constant MATH and a sufficiently small MATH REF there exist neighborhoods MATH and MATH (depending on MATH) of MATH, and a diffeomorphism MATH such that MATH. Moreover MATH with MATH uniformly bounded in the MATH norm. MATH has dimension REF, consequently the forms MATH and MATH degenerate. Since MATH is compact it follows from theorem of NAME (a version of NAME 's homotopy argument) that there exists a diffeomorphism MATH such that MATH. The order of the diffeomorphism is given by tracing back NAME 's homotopy argument. This is done in REF Composing the two diffeomorphims MATH and MATH we obtain on the section MATH of the principal bundle that MATH . For a fixed MATH, MATH is a family of diffeomorphisms parametrized by MATH, each defined only on its own constant section of the principal bundle. We will paste all those diffeomorphisms together. Let MATH. With respect to our local trivialization we write MATH for an arbitrary point of MATH in the given neighborhood. Define MATH by MATH . Assume MATH is a simple closed curve. For MATH sufficiently small (MATH), there exists a neighborhood MATH of MATH (depending on MATH) and a diffeomorphism MATH such that MATH for some REF-form R. Since MATH restricted to MATH is equal to MATH if follows that MATH differs from MATH by a factor of the form MATH, for some REF-form MATH, that is, we can write MATH . The one-form MATH is defined mod MATH and can be taken to be MATH . For any diffeomorphism MATH we have that MATH (see for example, CITE). Therefore MATH . So we must calculate MATH. We do this on the following lemma. MATH where MATH is a horizontal vector field relative to our trivialization. The proof of this lemma is given in REF Using this lemma we compute REF , MATH . We recall the expression for MATH REF MATH . It follows that MATH for some REF-form MATH. We have that REF can be writen as MATH . Using REF we have MATH giving MATH for some REF-form MATH. Finally this implies MATH which proves REF . REF-form REF is almost in the final form. REF-form MATH may not be necessarily exact. A final change of coordinates is needed. So far we have been working on a neighborhood of a constant level curve MATH of the magnetic field. By hypothesis this has the topology of an annulus. We introduce NAME coordinates MATH on this annulus (see for example, CITE). The NAME coordinates parametrize the MATH neighborhood MATH on the following way: Given a point MATH, consider the segment of geodesic MATH that connectes MATH to MATH and has minimal length. Let MATH denote its oriented length. Denote by MATH the oriented length from some arbitraty point MATH in MATH to the intersection of MATH with MATH. Using these coordinates we define MATH where MATH denotes the length of the level set MATH. With MATH as above there exist functions MATH and MATH such that MATH . Since MATH is MATH independent and MATH is closed REF implies that MATH . Let MATH and denote by MATH the restriction of MATH to a constant section of MATH, that is, MATH . Thus REF implies that MATH from which we see that MATH is closed. Now define a REF-form MATH on the section MATH by MATH . Thus MATH giving that MATH for any MATH homotopic to MATH (since MATH is closed). Since the section MATH has the homotopy type of an annulus we have that REF together imply that MATH for some function F. Thus we can write REF as MATH . But MATH difers from MATH by a factor of the form MATH for some function MATH and we can write that MATH proving the lemma. Using REF we can rewrite REF obtaining that MATH . Define a new variable MATH and the diffeomorphism MATH . Substituting back in REF, and denoting by MATH the pull-back of MATH by MATH we obtain that MATH that is, MATH that we write as MATH where MATH which proves REF . |
math-ph/9908010 | Choose a vector potential MATH for REF-form MATH. Thus we have by NAME theorem that MATH . Now consider the vector field MATH. Let MATH be its time MATH flow. Thus MATH implying that MATH . We can write that MATH . Differentiating with respect to MATH and using NAME 's formula we obtain that MATH . The first term on the integration vanishes since MATH integrates to zero along the closed curve MATH. To get the other term, observe that for any smooth simple closed curve with smooth normal MATH we have MATH the arc length. Since MATH and MATH we find that MATH as desired. |
math-ph/9908010 | Since the second derivative is expressed as the integral of a REF-form, the same reasoning applied to compute the first derivative can be applied to compute the second. |
math-ph/9908010 | We look to the magnetic problem given by REF-form MATH for MATH where MATH. According to REF the Hamiltonian vector field for MATH is given by a scaling of the Hamiltonian vector field for MATH. Thus their characteristics differ only by a time reparametrization. By REF there is a neighborhood MATH of MATH and a diffeomorphism MATH such that MATH . Thus the characteristic line bundle of MATH is spanned by the Hamiltonian vector field given by the Hamiltonian system with MATH-and symplectic REF-form MATH (where by abuse of notation we denote MATH by MATH). In REF we introduced action-angle coordinates on the neighborhood of MATH and reduced the dynamics of the Hamiltonian system MATH on this neighborhood to the dynamics of a twist map. REF 's twist theorem is calculated to be REF according to REF . So NAME 's twist theorem applies and invariant circles exist. By dimensionality, they trap the charge for all times MATH . |
math-ph/9908010 | Observe that each term in REF depends on diferent powers of MATH. A careful analysis of this fact (done in REF) implies that REF can not be satisfied near a critical point of MATH. So for a sufficiently large MATH we can find a small neighborhood of MATH where REF is satisfied everywhere. |
math-ph/9908010 | The proof is by contradiction. Suppose that given a neighborhood MATH of MATH we have that all the level sets contained on MATH are degenerate. This implies that exists a constant MATH such that MATH on MATH. This is equivalent to MATH . What implies that MATH . Since the right hand side is positive, the constant MATH must be negative and writing MATH it follows that MATH where MATH denotes the length of the level MATH. Thus we have that MATH where MATH is the minimum of MATH on MATH, the closure of MATH. REF implies that MATH is dominated by MATH. Since MATH was arbitrary this implies that MATH the length of the critical level set must be zero, but this contradicts the hyphothesis. |
math-ph/9908010 | We follow CITE. The first part of REF is an imediate consequence of NAME 's homotopy argument. The diffeomorphism MATH is built by realizing it as the flow of a vector field MATH on MATH. First we consider the family of REF MATH for MATH. We want a vector field MATH such that its flow MATH is such that MATH . (in what follows we will omit the MATH and MATH dependency from the notation). Differentiating this relation we obtain that MATH that is, that, MATH . Since MATH is a diffeomorphism and MATH is closed it follows that this can be satisfied if MATH . We want to solve REF for MATH. Since MATH is non-degenerate, it suffices to find a family of REF MATH such that MATH . In fact, REF imply MATH . To construct MATH we consider the restriction of the exponential map to the normal bundle MATH of the submanifold MATH with respect to the Riemannian metric on MATH. We denote this restriction by MATH . Consider the neighborhood of the zero section MATH . Then the restriction of the exponential map to MATH is a diffeomorphism onto MATH for MATH sufficiently small. Let MATH. Define MATH as MATH . Now define MATH for MATH by MATH . Then MATH is a diffeomorphism for MATH and we have MATH, MATH, and MATH . Calling MATH we define MATH . It's a direct computation to check that the family of REF MATH has the desired properties. Thus the family REF satisfy REF and it follows that MATH the flow of MATH, the vector field that satisfies REF is such that MATH . It follows that MATH where MATH stands for the tail of the expansion of MATH. Denoting MATH we have MATH that we write as, MATH where MATH and MATH. To solve REF we write MATH where MATH is the solution of MATH for MATH . We write MATH where MATH. Since MATH is a continuous vector field it follows that MATH is bounded on a neighborhood of MATH by a constant K. The equation for the flow MATH is MATH . Chosing MATH and integrating it follows MATH . Writing MATH the flow of MATH as MATH we obtain MATH giving MATH and the result follows. |
math-ph/9908010 | Using the NAME coordinates let MATH be a point of MATH. NAME the MATH dependence of MATH for notational convenience we can write that MATH . The derivative of MATH can be calculated as MATH which give us that MATH where MATH. Recalling the definitions and REF we can write that MATH . Since MATH is the flow of MATH we have MATH . Integrating this equation we have that MATH that simplifies to, MATH . Expanding MATH in MATH, inserting the resulting expression in REF and collecting terms we have MATH for some MATH. Since MATH is MATH independent, we obtain MATH . This is to say that MATH is of order at least one in MATH and we write that MATH where MATH . REF gives MATH and we write MATH where MATH proving the lemma. |
math-ph/9908010 | We will study each term of REF individually. Let MATH. Since MATH is a nondegenerate minimum (or maximum) point we have by NAME 's lemma that there is a neighborhood MATH and a system of coordinates MATH on MATH such that MATH where the plus sign is used if the point is a minimum and the minus sign is used if the point is a maximum. Let MATH and let MATH. Then we have that MATH . The proof of this lemma is straightforward. Note that REF implies that MATH on MATH. For a level set MATH not equal to MATH and such that MATH we have that MATH is constant (since B is) and we can write that MATH that is, MATH . Now we can see that the first term of REF is bounded, in fact we have that MATH but on MATH we have that MATH and MATH are bounded above and MATH is bounded below. So we can find constants MATH and MATH such that MATH on MATH, allowing us to write that MATH . This give us that the first term of REF is bounded for any level set close enough to MATH as claimed. Now we look to the second term MATH . For a constant metric one computes that MATH . REF becomes MATH considering REF , and proceeding as before we have that MATH . The case of a nonconstant metric can be reduced to the case of a constant metric by choosing normal coordinates on a neighborhood of MATH. The last term of REF can be minorated by noticing that MATH for some constant MATH (since MATH and MATH are bounded on MATH). Now observing that MATH, REF implies MATH . And we can write MATH for some constant MATH such that MATH on MATH. It follows that MATH . Now considering REF we see that REF implies MATH which can not be satisfied if MATH is small enough. Thus for a sufficiently small neighborhood of MATH all the level sets are non-degenerate. |
math-ph/9908018 | This can be obtained as follows. Let consider the grand canonical probability MATH with MATH where MATH is the i-th one dimensional stick that we are decomposing our volume in, and where MATH is the grand-canonical partition function. Clearly, we have MATH . Define MATH and we have MATH . The idea now is to make use of the local central limit theorem for the probability distribution of the occupation number in the i-th stick (see REF .). Let MATH. For any integer MATH, consider, the probability MATH . Due to the factorization property of MATH, the MATH's are independent identically distributed random variables. For centered i.i.d. random variables MATH with variance MATH, the local central limit theorem guarantees that the random variable MATH is close to a Gaussian in the sense that the quantity MATH fulfills the bounds MATH where MATH is the constant MATH . By applying REF to the centered quantity MATH, we obtain the following bounds on the ratio of probabilities: MATH where MATH . In terms of the non-centered variables MATH we have MATH where MATH is the average number of particles of a REFD stick MATH, in the grand canonical ensemble with chemical potential MATH. From this and REF , we obtain MATH . Note that in case MATH is chosen so that MATH or MATH then we can replace MATH by MATH, with the result that MATH may be replaced by MATH as well. Also, from REF , we have MATH . Using REF (and observing that MATH), we have MATH and MATH . Changing to base MATH then leads to REF. By the derivation of REF, we have the bounds on the variance for the number of particles in a REFD stick: MATH . In conjunction with the remark about replacing MATH by MATH, this gives REF . |
math-ph/9908018 | (Proof of Corollary) It follows from the monotonicity of MATH proved in REF, that the equation MATH always has a unique solution for MATH. Then, REF follows immediately from REF , once we observe that MATH as MATH in the sense prescribed in the corollary. For REF , take MATH, with base MATH, and MATH such that MATH where MATH denotes the largest integer MATH. Then, MATH solving REF , is easily seen to converge to MATH, and REF follows from REF . |
math-ph/9908018 | Let MATH be determined by REF , and define MATH as follows: MATH where MATH. We will obtain the equivalence of ensembles by combining two facts. The first is that MATH is approximately equal to MATH, and the second is an estimate showing that MATH . But first, let us recall the definitions of the expectation of an observable MATH: MATH . Since MATH is an observable localized in MATH, we note that MATH. Moreover, we may decompose the grand canonical state into a superposition of canonical states: MATH . Since MATH commutes with MATH, it does not have off-diagonal matrix elements between these canonical states with all different values of the total spin. Therefore, MATH . Note also, that since we have a decomposition MATH and using the previously described properties, we have MATH . This differs from the definition of MATH only by the final factor, which is a ratio of partition functions hence amenable to our activity bounds. In fact, we have MATH where MATH, which equals MATH for our choice of MATH. Thus we obtain MATH where MATH . Now we use the activity bounds REF , but replacing MATH by its actual value, MATH. We arrive at the bounds MATH where MATH . Therefore, MATH. We now use the triangle inequality and the fact that the exponent is negative to obtain: MATH so that MATH . Similarly, MATH . We will use the NAME inequality to control the expectation term in REF . Specifically, for any MATH, MATH . In REF we show that MATH. One choice for MATH is MATH. This gives the bound MATH . The leading order term in the bound is MATH for fixed MATH, strictly between REF. Also, let MATH which is greater than both MATH and MATH. Then MATH. In particular, MATH, which is to say that MATH. Then, using the triangle inequality, we have MATH . So, defining MATH, the theorem is proved. |
math-ph/9908018 | We begin by calculating how a two-site hamiltonian MATH acts on the perturbed state. We consider the decomposition of our lattice into the relevant bond MATH and everything else MATH. Thus MATH where MATH is the unit vector from REF on the pair MATH, and MATH . Here MATH is as would be defined by REF , but with MATH replaced by MATH and MATH replaced by MATH. For example MATH. But MATH is orthogonal to MATH and MATH, since MATH lies in the sector of total spin REF. And MATH . Now it is straightforward to see MATH where we have defined MATH . Then we may write MATH . Actually, MATH depends on MATH only through MATH. So from here on, we'll write it as MATH, and observe the following: MATH where MATH. Let us estimate the term MATH. We have an inequality MATH (which is actually an equality in the limit MATH for our ansatz). Also, MATH . In fact, using the inequality MATH one may conclude that the error in REF is bounded by MATH. Incorporating this estimate into the inequality of REF , we have MATH . Finally, as MATH, the sum over each MATH becomes increasingly well-approximated by the integral over MATH, we is proved in REF immediately following this proof. The lemma gives us a bound MATH where MATH is the Laplacian and MATH is the maximum radius for the NAME domain. (Note that by its definition, as the trace of the second-derivative tensor, the Laplacian enjoys the bounds MATH which may be combined with error term in REF .) Combining REF gives us the theorem, modulo the term MATH, for which we derive the necessary in REF . |
math-ph/9908018 | For each MATH, MATH . This clearly leads to the bound MATH . From this, the lemma follows easily. |
math-ph/9908018 | Recall MATH . The ratio of partition functions in the equation above is clear: It is the probability of finding one particle shared by the sites of MATH, and MATH particles shared by the sites of MATH, conditioned on finding MATH total particles on MATH. We consider the operator MATH . Then MATH and MATH where MATH, where MATH, and MATH is a bond in the stick containing the origin, which we denote by MATH. The restriction of the state in MATH with MATH spins down is of the form MATH where MATH is any observable commuting with MATH, as is, for example, MATH, and the MATH are non-negative numbers summing up to one. We will now derive an upper bound for MATH, that is independent of the coefficients MATH. We start from MATH where MATH denotes the probability in the ground state with MATH spins down for a one-dimensional system on MATH, the sites of which we label by MATH. Each term in the Right-hand side of REF can be estimate as follows. MATH REF gives the following bounds MATH . Combining these inequalities and summing over MATH yields MATH for all MATH. Together with REF this concludes the proof. |
math-ph/9908018 | The proof of REF is a direct consequence of the equivalence of ensembles because, since MATH is a unitary operator, MATH. Let us now consider the proof of REF . We wish to bound the denominator from below; that is, to demonstrate that MATH is not too small. This is tantamount to showing that MATH is not too close to REF. Furthermore, we know this quantity lies between REF. We estimate the actual canonical average with the grand canonical average, and take the logarithm in order to exploit the factorization properties of the grand canonical ensemble. First, we note MATH . Recall the definition MATH. This allows us a more convenient form in place of REF MATH . We partition the product over planes and estimate the logarithm, thus: MATH . We may approximate MATH by MATH, with an error no larger than MATH which is the same as MATH. In this case MATH . We may approximate the sum over MATH with an integral such that the error is bounded by MATH. We may bound the sum MATH from below by its largest term (since all the terms are positive). The largest term occurs for that integer MATH which is closest to MATH. Thus, defining MATH, we see MATH . Using these bounds, we may continue the estimate of MATH. We arrive at MATH . Since MATH and since MATH, we have REF . |
math-ph/9908018 | The periodicity of MATH follows directly from its definition. To prove REF , define MATH for MATH as MATH . Then, MATH . And clearly the remainder term MATH satisfies REF . For the bounds, we first restrict ourselves to MATH. For MATH, we note that REF implies MATH . Then we use REF in combination with this bound to also get the upper bound for MATH. MATH . Due to the peridicity REF , the upper and lower bound are automatically extended to all real MATH. The special values stated in REF are straightforward from REF . |
math-ph/9908027 | Let MATH be the symmetric-decreasing rearrangement of MATH (see CITE). Then MATH. |
math-ph/9908027 | Using the NAME product formula it suffices to show that the solutions MATH of the equations MATH are log concave, if MATH is log concave. The first follows from the fact, that the convolution of two log concave functions is log concave, the second follows easily from convexity of MATH, and the third is shown in CITE. |
math-ph/9908027 | Since MATH it is clear that MATH. For the converse we write MATH in the form MATH and obtain MATH . In particular, MATH, and with MATH we obtain MATH . (Regularizing MATH, if necessary, we may assume that MATH.) In the limit MATH the gradient term vanishes, and thus the limit of the energies is proved. Now MATH . Since MATH is the minimum of MATH, the second term vanishes for MATH, and it follows that MATH is a minimizing sequence for MATH. Since both terms in the functional are nonnegative, they must converge individually, in particular MATH converges to MATH. On the other hand MATH converges weakly to MATH by uniqueness of the minimizer. Together with the convergence of the norms this implies strong convergence. The solution of the variational equation for MATH is simply MATH with MATH given by MATH. |
math-ph/9908027 | Define MATH by MATH . Because MATH is the minimizer of MATH, it must be true that MATH . This leads to the virial REF . |
math-ph/9908027 | Using MATH as a test function for MATH, where MATH denotes the characteristic function of MATH, we immediately get MATH . Let MATH be a MATH function on MATH, with MATH outside MATH, and MATH inside MATH. We use MATH as a test function for MATH. Since MATH is bounded and MATH (because MATH tends to infinity and MATH is bounded by REF), we have MATH . This completes the proof of REF. |
math-ph/9908027 | Let MATH be the ground state wave function of MATH. As test wave function for MATH we take MATH where MATH and MATH are defined as in REF, that is, MATH is essentially the zero energy scattering solution and MATH is the distance of MATH from its nearest neighbor. We have MATH . For MATH one uses the estimates MATH and for the derivatives one has MATH . After division by the norm of MATH REF becomes MATH . MATH does not depend on MATH. One integrates first over MATH and then over the remaining variables. In analogy with the estimates REF one gets MATH . This implies MATH . By scaling, MATH and REF follows. |
math-ph/9908027 | As in CITE the lower bound will be obtained by dividing space into cubic boxes with NAME conditions at the boundary, which only lowers the energy. Moreover, interactions among particles in different boxes are dropped. Since MATH, this, too, lowers the energy. For the lower bound one has to estimate the energy for a definite particle number in each box and then to optimize over all distributions of the MATH particles among the boxes. REF (Finite box): The first step is to show that all the particles can be assumed to be in some large, but finite box. Since MATH tends monotonically to MATH with MATH, one knows that the energy of a particle outside a cube MATH of side length MATH and center at the origin is at least MATH. Hence MATH where MATH denotes the energy of MATH particles in MATH, with NAME conditions at the boundary. We now apply REF (which holds also in a cube with NAME conditions). Applying the theorem MATH times and noting that MATH is monotone in MATH we have MATH . Hence there is a constant MATH (that depends only on MATH), such that MATH implies that the infimum is obtained at MATH. This is fulfilled for all sufficiently large MATH, independently of MATH if MATH is fixed. So we can restrict ourselves to estimating the energy in MATH with NAME boundary conditions. REF (Trading MATH for MATH): We shall now use the GP equation to eliminate MATH from the problem, effectively replacing it by MATH. We write the wave function in MATH as MATH where MATH denotes the the minimizer of the GP functional in MATH; since it is strictly positive, every wave function can be written in this form. Note also that MATH and MATH obey NAME conditions. We have MATH where the integrals are over MATH. Using the GP REF this becomes MATH . Inserting the value REF for MATH gives MATH . REF (Division into boxes): MATH is a normalized quadratic form on the weighted MATH-space MATH, and can be minimized by dividing the cube MATH into smaller cubes with side length L, labelled by MATH, distributing the MATH particles among the boxes and optimizing over all distributions. We therefore have MATH where the infimum is taken over all distributions of the particles with MATH, and MATH is given by MATH where the integrals are over MATH in the box MATH, MATH. Note that here MATH, and REF is the same as REF with MATH replaced by MATH and MATH with the box MATH. We now want to use REF and therefore we must approximate MATH by constants in each box. Let MATH and MATH, respectively, denote the maximal and minimal values of MATH in box MATH. With MATH one has MATH . This holds for all MATH, and if we use MATH in REF, we get MATH where MATH is the energy in a box without an external potential. Remark: If we had not taken REF and used instead the division into boxes directly on the original Hamiltonian REF we would be considering the minimization of MATH . Such a procedure, however, would not lead to the GP energy. To see this, consider the special case of no interaction, that is, MATH and hence also also MATH. The minimum of REF is then simply MATH, whereas the GP energy is in this case MATH times the ground state energy of MATH. REF (Minimizing in each box): Dropping the subsidiary condition MATH can only lower the infimum. Hence it is sufficient to minimize each MATH separately. To justify the use of REF, we have to ensure that MATH is large enough. But if the minimum is taken for some MATH, we have MATH and using REF , which states that MATH we see that MATH is at least MATH. We shall later choose MATH, so the conditions needed for REF are fulfilled for MATH large enough, since MATH and hence MATH, MATH and MATH. Thus we have (for large enough MATH) MATH . We now use MATH, and drop the requirement that MATH has to be an integer. The minimum of REF is obtained for MATH . This gives for REF MATH . Now MATH is differentiable by REF , and strictly positive. Since all the boxes are in the fixed cube MATH there are constants MATH, MATH, such that MATH . Therefore we have, for MATH and MATH small, MATH with suitable constants MATH and MATH. Also, MATH and hence MATH . As last step it remains to optimize the length MATH. Recall that MATH and MATH is fixed. The exponents of MATH in both error terms in REF are the same for MATH . The final result, therefore, is MATH . |
math-ph/9908027 | The second term in REF is linear, the third quadratic in MATH. So it suffices to show that the first term is convex. Let MATH and MATH be given, with MATH and MATH in MATH. Then also MATH for all MATH. We have MATH . Hence MATH . |
math-ph/9908027 | Let MATH be a minimizing sequence in MATH, that is, MATH. It is clear that there exists a constant MATH, such that MATH, MATH and MATH for all MATH (recall that MATH is nonnegative). Hence the sequence belongs to a weakly compact set in MATH, as well as in the NAME space MATH, and in the space MATH, defined by the MATH norm MATH. Thus, there exists a MATH and a weakly convergent subsequence, again denoted by MATH, such that MATH . Because the MATH norm, the NAME norm, and the MATH norm are all weakly lower semicontinuous, we have MATH and it remains only to show that MATH. Since MATH converges to MATH in MATH it is clear that MATH. Moreover, MATH for all bounded regions MATH. If MATH with MATH, then there exists a constant MATH for all MATH, such that MATH for all MATH. Since MATH, this would imply MATH, which is impossible because MATH is a minimizing sequence for the functional MATH. Hence MATH. The uniqueness of MATH follows immediately from strict convexity, REF . |
math-ph/9908027 | Pick a function MATH. The stationarity of MATH at MATH implies MATH with a NAME parameter MATH to take account of the subsidiary condition MATH. With MATH real valued one obtains MATH and an analogous equation for MATH with MATH purely imaginary. The value of MATH is obtained by multiplying the GP equation with MATH and integrating. By the same argument, every solution MATH to the GP equation satisfies MATH and is thus a minimizer. |
math-ph/9908027 | Since MATH (by an analogous computation as in the proof of REF ), we know that MATH is a minimizer and hence a solution to the GP equation. It is thus an eigenstate of the Hamiltonian MATH with MATH (recall that MATH is unique), and since it is nonnegative, it must be a ground state. Since MATH solves the same equation it is also a ground state. Now MATH and MATH, so the ground state of MATH is unique up to a phase and without zeros (see CITE, XIII. REF). |
math-ph/9908027 | Put MATH and let MATH. The GP equation implies MATH . Using the NAME potential MATH we can rewrite this as MATH . Since MATH, and MATH for MATH with MATH large enough, we also have MATH . Now MATH, and hence MATH . |
math-ph/9908027 | The last lemma and the GP equation imply MATH. Thus MATH exists and is NAME continuous (see CITE). The MATH property follows by a bootstrap argument. The differentiability with respect to the parameter MATH may be shown by a NAME type argument like analogous statements (for example, differentiability with respect tonuclear charges) in TF theory CITE. REF follows immediately from REF and MATH. |
math/9908002 | To take the homotopy pushout we must replace MATH by its mapping cylinder MATH where the last term denotes the disc bundle of MATH. Hence when we take the ordinary pushout MATH we are simply gluing MATH to the disc bundle of MATH, which gives a space homeomorphic to MATH. |
math/9908002 | Observe that the NAME sequence for MATH which we get from REF is MATH . However, since MATH, and the map MATH is an isomorphism, the map MATH in the NAME sequence is surjective in every degree, forcing the coboundary map to vanish. Hence the long exact sequence splits into short exact sequences MATH . Given MATH, let MATH. Clearly the pair MATH maps to zero in the NAME sequence, and so defines a unique element MATH, satisfying REF . |
math/9908002 | In the definition of a residue, clearly REF is satisfied iff MATH defines a map from MATH to MATH. Since MATH REF will be satisfied iff MATH for all MATH and MATH. The former says that MATH defines a map from MATH to MATH, while the latter requires that this map split the inclusion of MATH into MATH. The set of such splittings is just MATH. |
math/9908002 | The map MATH evidently vanishes on MATH and MATH since every element of either set has no pole at MATH. On MATH, MATH so MATH . So MATH extends MATH as desired, hence is the required residue map. |
math/9908002 | The residue theorem implies that for a function MATH with poles only at MATH, MATH. Hence the two maps agree on MATH, so MATH is a residue map. However it is easy to see that they disagree on MATH. In fact a simple calculation shows that MATH . |
math/9908002 | Recall that ordinary localization on MATH gave REF : MATH . Now MATH, so applying MATH to REF gives (writing MATH in place of MATH) MATH because MATH is a MATH-module map. The last term is just MATH, so the theorem is proved. |
math/9908002 | We will apply REF to the manifold with boundary MATH, whose boundary is MATH, and apply the ordinary localization theorem to all of MATH. Clearly MATH so REF gives MATH while MATH so MATH . We will analyze the first term; the second is similar. Recall that by our conventions, the weights of MATH for MATH are positive. Hence, letting MATH be the weight MATH isotypic part of MATH, the MATH-Euler class is MATH . Each factor is a polynomial in MATH with constant term MATH, invertible leading coefficient, and degree MATH. The sum of these degrees is MATH. Let MATH be a ring. Let MATH have invertible leading coefficient and constant coefficient and degree MATH. Then MATH is invertible in the NAME ring MATH, and moreover, MATH actually lies in MATH. When MATH, this is clear from examining the poles and order of vanishing of MATH. The same holds for any ring MATH by direct calculation: MATH . Hence MATH and MATH . Therefore, for MATH, MATH . Given REF , MATH so MATH . Similarly, REF implies that for MATH, MATH . Hence MATH . |
math/9908002 | MATH splits MATH into MATH, MATH. For any fixed point component MATH, we have MATH so MATH which is negative on MATH, positive on MATH. Hence both conditions in REF are satisfied, and MATH . |
math/9908003 | The translation group MATH acts on the space of packings in MATH, and the action is continuous relative to our NAME topology. This group action may be interpreted as a dynamical system with a MATH - dimensional time variable. The notions of limits and uniform recurrence are then familiar from the dynamical systems point of view. To say that MATH is a limit of MATH is to say that MATH lies in the closure of the orbit of MATH. A limit equivalence class is an invariant set (not necessarily closed) in which each orbit is dense. If such an invariant set is closed, it is called a minimal set. A uniformly recurrent packing is a point in a minimal set in this dynamical system. Since the space of packings is compact, minimal sets exists in the closure of every orbit. The set of uniformly dense packings is unfortunately not compact, but it is exhausted by invariant compacta in the following way: If MATH is uniformly dense, the proportion of a large disk covered by MATH must approach the packing density MATH at a certain rate. That is, there is some function MATH which converges to REF as MATH goes to infinity and such that the density in any ball MATH is within MATH of MATH. The set of packings that are uniform using any fixed MATH is compact, and it is also invariant under the action. |
math/9908003 | Since MATH is self-nesting, we can loosen any packing MATH by homothetic expansion: We expand the packing and each copy of MATH by a factor of MATH, and then we replace each copy of MATH by a copy of MATH contained in its interior. For every MATH and MATH, there is a MATH such that in the loosened packing MATH, no two copies of MATH are within MATH of each other. Intuitively, if we loosen MATH, we decrease the density by expansion, but we may then increase it by re-saturation. We choose the constant MATH so that the latter would outweigh the former if MATH failed to satisfy the conclusion of the theorem. More precisely, we choose MATH so that MATH . Let MATH be the constant in the previous paragraph. Consider a maximal packing of balls MATH such that MATH is within MATH of an unsaturated packing in each ball. By maximality, the corresponding collection MATH covers MATH. In the loosened packing MATH, we can cram in at least one extra copy of MATH in each expanded ball MATH. If MATH had upper density greater than MATH, the new packing MATH would have greater density than that of MATH, contradicting the assumption that MATH is dense. |
math/9908003 | Let MATH and MATH, and choose the corresponding MATH according to REF . For each MATH, there is a non-zero measure of points MATH such that MATH is simultaneously at least MATH away from unsaturated in MATH for all MATH. Let MATH be one such point. The sequence of translated packings MATH must have a convergent subsequence. The limit MATH of this subsequence has the same property for all MATH; in particular it is completely saturated. By construction MATH is a limit of MATH. Since MATH is uniformly recurrent, the two packings are limit-equivalent. |
math/9908003 | The result follows from the fact that the unique smallest possible NAME region in a circle packing is a regular hexagon CITE. For every MATH, the union of the NAME regions that are more than MATH away from the regular circumscribed hexagon must have density REF in the plane in a dense packing. Consequently there are arbitrarily large disks in which every NAME region is within MATH of the optimal shape. Taking the limit MATH, the conclusion is that there are arbitrarily large regions that converge to the hexagonal packing. Thus the hexagonal packing is a limit of every dense packing. Conversely, any periodic packing is the only limit of itself. In particular, the hexagonal circle packing has this property. |
math/9908003 | Assume that MATH has diameter REF. Then the maximum diameter of a connected union of MATH copies of MATH is MATH. Let MATH be the measure with density function MATH . If MATH is the union of MATH interior-disjoint copies of MATH, MATH is the union of MATH such copies, and MATH is connected, then we claim that MATH . The reason is that if we express MATH and MATH as integrals, the ratio of volumes of MATH to MATH is MATH but the ratio of the integrands is at least MATH. Suppose that MATH is formed from MATH by replacing the components of MATH with the components of MATH. Then MATH as desired. |
math/9908003 | The argument of REF actually shows that its conclusion is ``true by a margin", in the following sense: Let MATH be a bounded domain and consider the measure MATH from the lemma. Then there is a constant MATH such that if MATH is obtained from MATH by a connected replacement of MATH copies of MATH by MATH, and if this replacement intersects MATH, then the diffused measure in MATH increases by at least MATH: MATH . If the connected replacement does not intersect MATH, the diffused measure in MATH at least does not decrease. At the same time, the total measure in MATH is bounded above by MATH. Thus only finitely many of the replacements in the sequence MATH may meet MATH. |
math/9908003 | Apply the process of REF to a uniformly dense packing. The usual notion of uniform density is that the limit MATH converges uniformly in MATH. If MATH is a finite measure, then it is equivalent to demand that MATH converges uniformly in MATH. Since the diffused measure MATH strictly increases under connected replacements of MATH copies, and since the density is already maximized, the limit in REF does not change and continues to converge uniformly. |
math/9908009 | We construct analytic discs with boundaries contained in MATH. For each MATH, we claim that the non-singular quadric MATH meets the square pillar MATH in an analytic disc MATH with piecewise smooth boundary in the folding screen MATH . In fact, the disc in question, with its boundary, may be parameterized by MATH for it is easy to check that any such MATH may be written uniquely both as MATH and MATH . Since MATH it follows that the disc MATH is contained in the hypersurface MATH which intersects the folding screen as MATH . This intersection is empty unless MATH and, in this case, it is easily verified that MATH . Since MATH is the intersection of MATH with the folding screen, from these inequalities it follows that MATH for all MATH. The following diagram shows what is happening in the MATH-plane when MATH. It is clear that every point in MATH, not already in MATH, lies in MATH for some MATH. A similar family of discs sweeps out MATH. The result thus follows from the maximum modulus principle. |
math/9908009 | Consider the linear change of coördinates: MATH where MATH . We claim that MATH. Were this to be shown, the result would follow immediately from REF . Since MATH, MATH . Since MATH, MATH . Thus, MATH since MATH. Altogether, MATH as required. |
math/9908009 | The result follows from REF provided we can find MATH so that MATH. Suppose MATH. Then MATH and MATH, so MATH. If we take MATH and MATH, then MATH which is one of the conditions forcing MATH to be in MATH. If we also take MATH, then MATH which is the first condition for MATH. Now MATH. Therefore, MATH. Thus, MATH which we can arrange to be less than MATH by taking MATH sufficiently small and MATH sufficiently large. This leaves one condition to be ensured. It is that MATH when MATH. Now MATH on MATH so it suffices to show that MATH when MATH. Choosing MATH, this expression is strictly bounded below by MATH which, for MATH, is bounded below by MATH . Taking MATH, we may neglect the second term, leaving MATH which is non-negative if we choose MATH. |
math/9908009 | We first translate MATH to the origin, and then by a complex linear transformation, we may assume that MATH . Then, near the origin, MATH for smooth functions MATH and MATH whose NAME series begin with terms of order at least two. Let MATH and MATH denote their third order NAME polynomials. Then the complex polynomial change of coördinates MATH is invertible near the origin and, in these new coördinates, MATH osculates MATH to third order, as required. We shall now assume that this is done and drop the hats. The equation for MATH takes the form MATH where MATH is a MATH . Hermitian matrix (the NAME form), MATH is a real symmetric matrix, MATH is a real vector, and the ellipsis MATH indicates cubic and higher order terms. The change of coördinates MATH preserves MATH and gives a new equation MATH as required. |
math/9908009 | First observe that REF imply that MATH so it follows easily upon choosing MATH small enough that MATH lies in a small neighborhood of the origin. The second line of REF implies that MATH, so it suffices to show that MATH. If MATH denotes the aperture of MATH, this follows from the inequalities MATH and MATH . Now MATH so the first inequality is satisfied if MATH and MATH. For the second inequality, we will need to estimate MATH . Observe first from REF that MATH where MATH is a constant depending only on MATH and a bound for the matrix MATH. Therefore we obtain MATH . If we choose MATH so that MATH and MATH so that MATH then the second inequality holds as well. |
math/9908009 | If MATH and MATH, then MATH, which implies that MATH . Therefore we can guarantee the first condition defining MATH by choosing MATH and MATH so small that MATH. For the second condition, we have MATH . By definition of MATH, the first term is at most MATH. The second term is equal to MATH . Now REF give MATH for some constant MATH, so the second term is at most MATH . Collecting the terms, we deduce that MATH . Since MATH and MATH can be made small by choosing MATH and MATH small, the result follows. |
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