paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0003192 | Let MATH be the quadratic form determined by the Hessian at the origin. Denote the eigenvalues of MATH by MATH and note that each MATH has nonnegative real part. REF : change coordinates to make MATH exactly equal to the quadratic form MATH. Indeed since MATH, and the Hessian is nondegenerate, there is a locally smooth... |
math/0003192 | Passing to MATH if necessary, we may assume MATH. Setting MATH for all but one index MATH, we cannot obtain the zero function (by minimality), and so some term in the expansion around MATH is a pure power of MATH, and we denote the minimal degree such term by MATH. If MATH is not a homogeneous point, then there is some... |
math/0003192 | Let MATH be any point on the boundary of MATH. For MATH and MATH the power series for MATH is convergent at MATH. As MATH and MATH therefore, MATH is finite and increasing. On the other hand, the power series for MATH is not absolutely convergent on MATH, since we know MATH to have some singularity on this torus. Hence... |
math/0003193 | We show that REF implies REF . To do this, we first prove that if MATH is another splitting of REF as above, with associated linear operator MATH on MATH, then for any MATH, MATH for all MATH. Indeed, from REF we have MATH for all MATH. Taking MATH and using the fact that MATH is primitive, REF gives MATH . But MATH is... |
math/0003193 | Let MATH be a nonzero element of the kernel of MATH; the notation MATH refers to the direct sum decomposition REF , with respect to some splitting. We claim that MATH must be in MATH. Indeed, MATH for some MATH and MATH implies MATH since MATH vanishes. Also, by the classical hard NAME theorem, MATH. Now, if MATH denot... |
math/0003193 | We shall see that REF implies REF except when MATH is exponentially large compared to MATH. For large MATH, the NAME polynomials are close approximations of classical orthogonal polynomials, in this case the NAME polynomials, and we know how to bound these. By NAME 's inequality, REF gives MATH . So, REF holds whenever... |
math/0003195 | Expanding MATH as a geometric series and using unique factorization in MATH, one sees that the coefficient of MATH in the reciprocal of the left hand side is MATH times the number of monic polynomials of degree MATH, hence REF. Comparing with the reciprocal of the right hand side completes the proof. |
math/0003195 | Write the NAME expansion MATH. Then observe that MATH. |
math/0003195 | Recall that MATH is cyclic precisely when its characteristic polynomial and minimal polynomials are equal. From Subsection REF, these polynomials are equal when all MATH have at most one part. In the cycle index for MATH set MATH if MATH has at most REF part and MATH otherwise. It follows that MATH . By REF this equati... |
math/0003195 | The proof of REF shows that MATH . A matrix is separable if and only if all MATH have size MATH or MATH. Hence MATH . The result follows. |
math/0003195 | Taking coefficients of MATH on both sides of REF gives the relation MATH . Since MATH for all MATH, it follows that MATH as desired. |
math/0003195 | MATH is clearly non-negative when MATH and MATH. CITE established an equation which is equivalent to the sought identity MATH . As some effort is required to see this equivalence, we derive the identity directly using NAME 's line of reasoning. First observe that unipotent elements of MATH corresponding to nilpotent MA... |
math/0003195 | Recall the cycle index factorization MATH . Setting all MATH equal to MATH and using REF shows that MATH . Taking reciprocals and multiplying by the cycle index factorization shows that MATH . This proves the first assertion of the theorem. For the second assertion, use REF from Subsection REF. |
math/0003195 | From the above formula for NAME polynomials, it is clear that the only surviving term in the specialization MATH is the term when MATH is the identity. The rest is a simple combinatorial verification. (Alternatively, one could use ``principal specialization" formulas for NAME polynomials on page REF). |
math/0003195 | The first assertion is clear. The second assertion is well known in the theory of partitions, but we argue probabilistically. Multiply both sides by MATH. Then note from the first assertion that the left hand side is the MATH chance of having a partition of size MATH. Now use the second equation in the proof of REF in ... |
math/0003195 | From the formula for MATH, the conditioned measure for MATH is the same as for MATH. Now let MATH be the number of times that coin MATH comes up heads in the NAME Tableau Algorithm. Letting MATH denote conditioning, it suffices to show that MATH . In fact (for reasons to be explained later) we compute a bit more, namel... |
math/0003195 | Suppose we know that MATH is a probability measure and that MATH . Then the MATH probability of choosing a partition with MATH for all MATH is MATH . Thus it is enough to prove the (surprising) assertion that MATH for all MATH. One calculates that MATH . Similarly, observe that MATH . Thus the ratio of these two expres... |
math/0003195 | This is immediate from REF . |
math/0003198 | For all MATH, we have MATH, hence MATH is a dual basis for MATH as a right MATH-module. |
math/0003198 | Define MATH as follows: for MATH, let MATH be given by MATH . Given MATH, put MATH . We invite the reader to verify that MATH and MATH are well-defined and that they are inverses of each other. |
math/0003198 | Let MATH be a finite dual basis of MATH as a right MATH-module. Then for all MATH and MATH, MATH . Define MATH by MATH, with MATH for all MATH. To show that MATH is a left MATH-linear and right MATH-linear map, take any MATH, MATH and compute MATH . Conversely, for MATH define MATH . Then for all MATH that is, MATH. Fi... |
math/0003198 | The result is a translation of REF in terms of MATH and MATH, using REF (for REF) and REF (for REF). We prove one implication of REF. Assume that MATH is a NAME pair. From REF , we know that MATH is finitely generated and projective. Let MATH and MATH be as in REF of REF, and take MATH, MATH. For all MATH and MATH, we ... |
math/0003198 | Consider MATH and MATH. Due to the naturality of MATH and REF there is a commutative diagram MATH . Write MATH and MATH. Then it follows that MATH . We have seen before that MATH. It is easy to prove that MATH - the left MATH-action is induced by the multiplication in MATH - and MATH is a morphism in MATH. Thus MATH an... |
math/0003198 | We leave the details to the reader; the proof relies on the fact that MATH is left and right MATH-linear. |
math/0003198 | This follows immediately from REF . |
math/0003198 | Suppose that MATH is a NAME pair. Then there exist MATH and MATH such that REF hold. Let MATH, and MATH. Then REF can be rewritten as MATH for all MATH. Taking MATH, MATH, one obtains REF. For all MATH and MATH, one has MATH and REF can be written as MATH for all MATH and MATH. Taking MATH and MATH, one obtains MATH . ... |
math/0003198 | Let MATH and MATH be as in REF. Then for all MATH, MATH . Write MATH and for all MATH consider the map MATH . Then for all MATH and MATH, MATH so MATH is a finite dual basis for MATH as a left MATH-module. |
math/0003198 | CASE: With notation as in REF , let MATH be the MATH-module generated by the MATH. Then for all MATH, MATH . Since MATH is faithfully flat, it follows that MATH, hence MATH is finitely generated. CASE: From descent theory: if a MATH-module becomes finitely generated and projective after a faithfully flat commutative ba... |
math/0003198 | We first show that MATH. For all MATH and MATH, we have MATH proving that MATH is left MATH-linear. It is also right MATH-linear because MATH . Notice that the dual basis for MATH satisfies the following equality (the proof is left to the reader): MATH . Using this equality one computes MATH . This proves that MATH is ... |
math/0003198 | We have to show that MATH is left and right MATH-linear and right MATH-colinear. For all MATH and MATH, MATH proving that MATH is right MATH-linear. The proof of left MATH-linearity goes as follows: MATH . Next one needs to show that MATH is right MATH-colinear. Using REF, one finds MATH . Conversely, let MATH and put ... |
math/0003198 | MATH. Let MATH and MATH be as in REF. Then MATH and MATH are morphisms in MATH, and MATH . The fact that MATH and MATH are right MATH-linear and left MATH-linear implies that MATH. Similarly, for all MATH, MATH . Since MATH and MATH are left MATH-linear, MATH. MATH. Obvious, since MATH and MATH are in MATH. MATH. Let M... |
math/0003198 | Details are left to the reader. Given MATH, for MATH, the natural map MATH is MATH . |
math/0003198 | We show that MATH is well-defined, leaving other details to the reader. Consider a commutative diagram MATH . The map MATH is left and right MATH-colinear. Write MATH. Then MATH . Applying MATH to the second factor, one finds MATH . The right MATH-colinearity of MATH implies that MATH and hence proves REF. To prove REF... |
math/0003198 | We only prove REF. If MATH is a NAME pair, then there exist MATH and MATH such that REF hold. Take MATH and MATH corresponding to MATH and MATH and write down REF applied to MATH with MATH, MATH . Taking MATH, MATH, and applying MATH to the first factor, one obtains REF. Conversely, if MATH and MATH satisfy REF, then R... |
math/0003198 | Observe first that MATH is a left-left entwining structure. This means that REF hold, but with MATH and MATH replaced by MATH and MATH. In particular, MATH . This can be seen as follows: rewrite REF as commutative diagrams, reverse the arrows, and replace MATH by MATH. Then we have REF in diagram form. Now fix MATH suc... |
math/0003198 | We first prove that MATH is well-defined. CASE: MATH is right MATH-linear: for all MATH, MATH and MATH, we have MATH . CASE: MATH is right MATH-colinear: for all MATH and MATH, we have MATH . CASE: MATH is left MATH-colinear: for all MATH and MATH, we have MATH . The proof that MATH satisfies REF is left to the reader.... |
math/0003198 | We first show that MATH is well-defined. Take MATH, and let MATH. CASE: MATH is right MATH-linear since for all MATH and MATH, MATH . CASE: MATH is right MATH-colinear since for all MATH and MATH, MATH . CASE: MATH is left MATH-colinear since for all MATH and MATH, MATH . Conversely, given MATH, we have to show that MA... |
math/0003198 | We only prove MATH in REF. First we show that MATH is a left inverse of MATH. Since MATH is right MATH-linear, it suffices to show that MATH . To show that MATH is a right inverse of MATH we use that MATH is right MATH-colinear and conclude that it suffices to show that for all MATH and MATH, MATH . Both sides of the e... |
math/0003208 | The steady state REF with MATH has general solution MATH, where we have used that MATH has an extremum at MATH. Hence the period is MATH, and for all small MATH we see that the perturbed function MATH is another positive periodic steady state solution. Thus MATH is not asymptotically stable in MATH with respect to even... |
math/0003208 | The proof depends on a number of elementary differential equations and inequalities that we derived in Section REF, and the reader may wish to skim those sections before proceeding. If MATH then MATH, as one sees directly from the formula in REF , using that MATH and MATH. So we assume MATH from now on, and MATH. By de... |
math/0003210 | Let MATH be a MATH-admissible tuple and MATH. Set MATH, MATH, and MATH, MATH. Using REF , one deduces that MATH; MATH with MATH; MATH and MATH for MATH. Therefore MATH. Assume now that MATH and MATH. Choose integers MATH as follows. Set MATH. Assume that MATH is chosen. If there is no MATH such that MATH, we set MATH a... |
math/0003210 | This follows from REF . |
math/0003210 | Applying REF , and REF, we conclude that MATH is a composition factor of MATH and for each MATH-admissible tuple MATH the set MATH, so MATH. |
math/0003210 | One can assume that MATH. MATH . Let MATH. Since MATH, we have MATH. This implies that MATH. MATH . Let MATH. Then MATH. Let MATH. Assume that MATH. Since MATH, we have MATH, which is impossible. Therefore MATH, so MATH and MATH. |
math/0003210 | One can rewrite the formula in REF as follows. MATH where MATH . Recall that by convention MATH for all MATH, and MATH is the trivial one-dimensional MATH-module. So REF holds for MATH. Assume now that MATH and REF is valid for all MATH. We shall prove it for MATH. Then the theorem will follow by induction. It follows ... |
math/0003210 | By REF , MATH . It follows from REF that the factors MATH come from MATH and the factors MATH, and MATH if nonzero come from MATH. Note that MATH so MATH for all MATH. Therefore by REF , MATH . Similarly, we get MATH for MATH (and for MATH if MATH and MATH). Hence MATH . (Here the symbol MATH is extended to the zero mo... |
math/0003210 | CASE: Assume that MATH. Choose maximal MATH such that MATH. Then MATH. Take minimal MATH such that MATH and set MATH. Then MATH and MATH. One has MATH. Moreover, if MATH, then MATH. Therefore REF implies that MATH is a composition factor of MATH for MATH, and MATH. As MATH, we get a contradiction. CASE: In view of REF ... |
math/0003210 | REF yield that MATH, MATH, MATH, and MATH, MATH, MATH, are inductive systems for MATH. Let MATH be an inductive system of fundamental representations. It is clear that either for every MATH there exist MATH and MATH such that MATH, MATH, and MATH, or MATH for some MATH and MATH. In the first case we claim that MATH. In... |
math/0003210 | Since the restrictions of the fundamental representations of a semisimple algebraic group over an algebraically closed field to relevant NAME groups over arbitrary subfields remain irreducible and can be realized over these subfields, only REF (or REF), REF, and REF require some analysis. Let MATH be the MATH-module MA... |
math/0003212 | Choose a basis MATH for MATH . Let MATH denote the bilinear form defining the NAME bracket in MATH with respect to this basis. Let MATH denote the MATH-variety of upper triangular matrices of order MATH. Hence the MATH-points of MATH are just the upper triangular matrices of order MATH with coefficients in MATH. We hav... |
math/0003212 | We just have to add the constructible condition: CASE: a matrix MATH in MATH defines a MATH-submodule of MATH if and only if for each MATH there exist MATH such that MATH . |
math/0003212 | We want to put a condition on a subspace of MATH that it is a subalgebra of codimension MATH in MATH . The trouble is now that MATH is not necessarily an ideal in MATH . However we use the fact that MATH . Let MATH be a basis for MATH with MATH a basis for MATH, and MATH be a basis for MATH with MATH a basis for MATH .... |
math/0003212 | Let MATH be a finite set of generators. Then, for any fixed MATH, there exists some MATH and MATH such that MATH and MATH . So we can carry out the same analysis essentially as the previous lemma. Let MATH be a basis for MATH with MATH a basis for MATH, and MATH be a basis for MATH with MATH a basis for MATH . Let MATH... |
math/0003212 | Again we just have to add the constructible condition: REF a matrix MATH in MATH defines a MATH-submodule if and only if, for each MATH, there exist MATH such that MATH . |
math/0003212 | Suppose MATH is a closed ideal of codimension MATH in MATH . Let us show that each ideal MATH of codimension MATH in MATH contains some fixed term of the filtration. We can assume MATH. Let us consider MATH where MATH is the lower central series of MATH which might be distinct from the closure of the lower central seri... |
math/0003212 | This follows from the fact that a MATH-submodule MATH of codimension MATH in MATH must contain MATH. In REF we apply this to MATH which must have codimension bounded by MATH. Recall that the case MATH either is empty or else MATH is abelian and MATH . |
math/0003212 | Let MATH be a closed subalgebra of MATH of codimension MATH . Suppose MATH is the decomposition of the underlying simple NAME algebra. Then MATH where MATH is central, MATH is a derivation and MATH is an automorphism of MATH which has order MATH (MATH is assumed to have primitive MATH-th roots of unity) and MATH . We c... |
math/0003212 | This depends on the fact that any derivation MATH with MATH can be realised as a linear multiple of the NAME product MATH where MATH and MATH. For the NAME algebras MATH and MATH the result follows, since we can then realise the derivation in the grading of weight MATH as a NAME product of elements of weight MATH and M... |
math/0003212 | Let MATH and MATH be the free generators. If MATH is any NAME word in MATH and MATH we define the length MATH to be the number of terms in MATH . We take a NAME set MATH as a basis for MATH (see CITE II. REF) which is defined as follows: REF if MATH and MATH and MATH then REF MATH and REF an element MATH of length MATH... |
math/0003212 | We may assume MATH is contained in MATH. By REF applied to MATH and MATH, MATH since MATH. We deduce that MATH . The result follows since MATH and MATH. |
math/0003212 | By taking a cover of the NAME MATH by open NAME cells corresponding to different choices of bases of the lattice MATH, one deduces the result from the following elementary REF . |
math/0003212 | Let MATH be the subalgebras of codimension MATH. Then MATH and MATH . |
math/0003212 | The first line REF follows from REF and the remark that MATH if MATH is a subalgebra of MATH . Let MATH . Let MATH be in MATH and write MATH for MATH. We have MATH if and only if, for every MATH, there exist MATH such that MATH, for some finite field extension MATH of MATH. Here MATH is the bilinear mapping MATH corres... |
math/0003212 | This follows from the fact that MATH. Hence the codimension of MATH is given as usual by MATH . The codimension of MATH in MATH on the other hand is given by MATH. |
math/0003212 | Everything has already been proven, except for REF , which follows from REF . |
math/0003212 | We need to prove that the image of a MATH-closed subgroup MATH under the map MATH is a MATH-subalgebra. The proof of REF can be applied in this setting to prove that MATH . But since MATH is MATH-closed and MATH we have MATH . Hence MATH that is, MATH is additively closed. Once we have this then REF implies that MATH i... |
math/0003212 | The proof works just the same as the one in CITE for the explicit formula for MATH-adic cone integrals, using REF for performing the change of variable, compare the proof of REF. |
math/0003212 | This follows from REF and the fact that if MATH for almost all primes MATH then MATH . |
math/0003213 | Let MATH be a general line of MATH (in particular, MATH is the only component of MATH containing MATH), and let MATH be a plane containing MATH. Then MATH splits as a union MATH where MATH is a plane curve of degree MATH. So MATH has length MATH and it is formed by points that are singular for MATH, hence either tangen... |
math/0003213 | of REF The lines in MATH which are tangent to MATH are parametrized by a ruled threefold MATH: any line on MATH corresponds to the pencil of lines in MATH which are tangent to MATH at a fixed smooth point. Then MATH . MATH is a surface: otherwise, a general point MATH would be contained in infinitely many lines of MATH... |
math/0003213 | Let MATH be a line such that MATH and MATH . Let MATH be the NAME variety parametrizing the lines in MATH which intersect MATH . The only singular point of MATH is MATH . In fact, it is easily seen that MATH is the intersection of MATH with the (projectivized) tangent space to MATH at MATH . In particular, the points o... |
math/0003213 | Assume by contradiction that there exists a plane MATH such that MATH for every MATH . We perform some local computations and we use the same notations as in REF . So, let MATH be an affine chart in MATH with coordinates MATH . Assume that the origin is a general point MATH of MATH and that MATH is defined by MATH . Le... |
math/0003213 | First of all, the dimension of MATH must be at least MATH otherwise MATH would contain a MATH-dimensional family of planes, hence a MATH-dimensional family of lines, a contradiction. So assume by contradiction that MATH . But then along each fibre of the NAME map there is even a fixed tangent hyperplane, contradicting ... |
math/0003213 | Assume the contrary. Then, when MATH moves on MATH to MATH the plane MATH moves to a limit plane MATH . The intersection MATH is a curve which has the line MATH as a "double component" ; in particular, this curve is singular along MATH . Then MATH for every MATH . In fact, if MATH then MATH would be smooth at MATH cont... |
math/0003213 | By REF there are two tangency points of MATH to MATH on MATH and two on MATH. The point MATH is singular for MATH, but it cannot be singular for MATH because, otherwise, letting MATH and MATH vary, every point of MATH would be singular. So MATH is a tangency point of MATH to MATH. Hence, MATH is tangent to MATH at exac... |
math/0003213 | Assume by contradiction that MATH contains the lines MATH. Then the residual curve of MATH in MATH splits as MATH. Hence, by REF , there is a new tangency point on MATH, against REF . |
math/0003213 | We assume by contradiction that MATH is a curve. Then every point of MATH belongs to infinitely many lines of MATH. The curve MATH is not a line because every line of MATH meets MATH in MATH points, and MATH . If MATH is general, from MATH it follows that through MATH there are two secant lines of MATH say MATH and MAT... |
math/0003213 | The first assertion of MATH is clear. To prove the second, it is enough to observe that exactly MATH lines of MATH, different from MATH, pass through a general point of MATH, and that these lines are separated by the blow-up of MATH along MATH. The assumption of MATH means that, for every MATH, the lines of MATH inters... |
math/0003213 | To evaluate MATH we choose the lines MATH and MATH so that they intersects at a point MATH smooth for MATH . Since MATH by REF we can also assume that MATH and MATH are the only lines of MATH contained in the plane MATH, so that the lines intersecting both MATH and MATH are those passing through MATH. The conclusion fo... |
math/0003213 | Note first that MATH is equal to the degree of the curve, intersection of MATH with a hyperplane MATH. We can assume MATH then MATH splits in the union of MATH with MATH other lines meeting MATH. Indeed, if MATH and MATH, there exists a line passing through MATH and meeting MATH, which is necessarily contained in MATH.... |
math/0003213 | It was already remarked in the Introduction that for the degree MATH of MATH we have MATH. Let MATH be a general point of MATH and fix a system of affine coordinates MATH such that MATH. Let MATH be the equation of MATH . As usual, we assume that MATH is defined by MATH and, moreover, we write MATH for MATH . The polyn... |
math/0003213 | The first part of the statement was already shown in the proof of REF . Moreover, in the same proof we saw that, if MATH is not tangent to MATH at some point of MATH then MATH and MATH are transversal inside MATH . In fact, the GCD of the polynomials MATH in MATH has degree exactly MATH . Hence the double structure on ... |
math/0003213 | We analyze in detail the case MATH. A similar proof can be given if MATH. For a different proof of this last case, see CITE. Let us recall that MATH, so given MATH general and skew, there are three lines MATH meeting both MATH and MATH. The lines MATH are pairwise skew, otherwise MATH would fail to be skew. Since MATH ... |
math/0003213 | Since we assume that MATH is not a quadric bundle we have that the dimension of MATH is MATH by REF . Then, through a general point of MATH there are infinitely many curves MATH and, by REF we conclude MATH . |
math/0003213 | The algebraic system of dimension two MATH on the surface MATH is linear because there is exactly one curve of the system passing through two general points (MATH). Also the self - intersection is equal to MATH, therefore MATH is a homaloidal net of rational curves, which defines a birational map MATH from MATH to the ... |
math/0003213 | Let MATH be general and set MATH for simplicity. Let MATH denote a normalization of MATH . The proof of the proposition is based on the following two lemmas. The curve MATH is irreducible, hyperelliptic of genus MATH . Hence MATH can be embedded into MATH as a smooth quintic. Let MATH be the surface parametrizing the s... |
math/0003213 | The proof is divided into several steps. CASE: There is a birational map MATH where MATH denotes the symmetric product of the curve MATH by itself. On MATH there is the algebraic system of curves MATH of dimension MATH . Since MATH, there are exactly MATH curves of the system containing two fixed general points on MATH... |
math/0003213 | Since MATH and MATH there are MATH secant lines of MATH through a general point of MATH and MATH secant lines of MATH contained in a general plane of MATH . Therefore, the class of MATH in the NAME group MATH is MATH with traditional notations. It follows that the degree of MATH is MATH this means that there are MATH s... |
math/0003213 | Let us remark first of all that the curves MATH and MATH are birational. Indeed let MATH . If MATH and MATH then the plane MATH intersects MATH at the points MATH . We get a birational map MATH by setting MATH . We fix now a general secant line MATH of MATH . Starting from the just constructed map MATH, we can also con... |
math/0003213 | If MATH is a fixed line of MATH, then the lines of MATH meeting it generate the rational ruled surface MATH having MATH as simple unisecant. Hence MATH results to be a rational surface. Similarly for MATH. There are two possibilities regarding the algebraic system MATH, whose dimension is two (because MATH is not a qua... |
math/0003213 | For every pair of indices MATH and general MATH, the surface MATH is a smooth quadric and it is clear that the linear systems MATH and MATH coincide: we call it MATH. We want to study the intersection of two quadrics belonging to two families of the form MATH and MATH, MATH. Let us remark first that, if MATH, MATH are ... |
math/0003216 | Given MATH, we may write MATH, where MATH and MATH, for some constant MATH depending on MATH. Then MATH and MATH by the NAME Embedding Theorem, with MATH the norm of the embedding MATH, MATH by the diamagnetic inequality (see CITE). The lemma follows from this. |
math/0003216 | The proof is similar to that in CITE. The following formal argument for deriving REF is instructive, and will be helpful for obtaining the analogous result in MATH for MATH. The set of Hamiltonian quaternions MATH is the unitary MATH-algebra generated by the symbols MATH with the relations MATH . Multiplication is asso... |
math/0003216 | CASE: Let MATH. Then MATH . Thus REF follows by continuity, and this implies REF once REF is established. CASE: Let MATH and MATH. Then MATH whence MATH by the diamagnetic inequality. Thus REF is established, and so REF . |
math/0003216 | Let MATH. Then MATH . Hence, MATH if and only if MATH with MATH. Moreover, for any MATH whence MATH . The result follows since MATH. |
math/0003216 | This is quite standard, but we give the short proof for completeness. We show that MATH is compact. Let MATH be a sequence which converges weakly to zero in MATH, and hence in MATH by REF . Then, in particular MATH, say. Given MATH, set MATH where MATH with support MATH and MATH say, and MATH. Then MATH . The first ter... |
math/0003216 | For MATH and MATH where MATH . From REF and since MATH and MATH we have MATH for some constant MATH, and so MATH satisfies MATH and extends to an operator in MATH, the space of bounded linear operators on MATH. Thus, there exists a neighbourhood MATH of MATH such that for MATH in MATH. It follows that MATH and MATH not... |
math/0003216 | The kernel of MATH is finite-dimensional, and we have the orthogonal decomposition MATH . With respect to this decomposition, we can represent MATH as MATH where MATH and MATH . We are required to prove that, for any MATH, there exists a neighbourhood MATH such that MATH for all MATH. We can write MATH where MATH, MATH... |
math/0003216 | We already know from REF that MATH in MATH. It is therefore sufficient to prove the theorem for MATH, where MATH. Define MATH . The theorem will follow if we prove that MATH, in view of the compactness of MATH. We shall prove that MATH is both open and closed. Since MATH, we know that MATH. It is clear from REF that MA... |
math/0003216 | Let MATH in REF be denoted by MATH and set MATH. We shall prove that MATH is an open subset of MATH; the theorem will then follow from REF since the density of REF is a consequence of REF . For MATH, let MATH be magnetic fields which satisfy MATH. Then, if MATH are the associated vector potentials given in REF , MATH f... |
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