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math/0005025 | Apply the previous lemma and REF to MATH at any MATH, which, by the NAME Fixed Point Theorem, is non-empty since MATH. |
math/0005025 | Let MATH be a rationally smooth MATH-fixed point of MATH. Using the relative order, we may without loss of generality assume that if MATH and MATH, then MATH is nonsingular at MATH. By the relative version of NAME 's Inequality and the fact that the singular locus of MATH has codimension at least two (as MATH is normal... |
math/0005027 | REF and the continuity of the concave functions MATH and MATH at MATH imply that MATH for MATH . By the definition of the norm of a NAME space, we have MATH . The NAME space with a given fundamental function is minimal among the symmetric spaces with the same fundamental function [REF, p. REF]; therefore, MATH . Suppos... |
math/0005027 | The implications MATH and MATH follow from REF and the definition of a DSS operator, respectively. Suppose that REF does not hold, that is, that MATH . Then there exist a sequence MATH and a constant MATH such that we have MATH and MATH for MATH . The functions MATH where MATH are disjoint and MATH satisfy REF . Theref... |
math/0005027 | Since any NAME space is maximal among all symmetric spaces with the same fundamental function [REF, p. REF], we have MATH . By REF , MATH for MATH therefore, by the definition of a NAME space, MATH . Hence, MATH . Suppose that MATH is not a DSS operator. Then, in particular, there exists a sequence of disjoint function... |
math/0005027 | First, MATH and an arbitrary SS MATH is embedded in MATH [REF, p. REF]. If MATH then the function MATH increases, because MATH is concave. Therefore, REF is violated if and only if MATH (that is, if and only if MATH for some MATH and MATH), and MATH . It remains to apply REF . |
math/0005027 | Let the SS MATH be the NAME space MATH with MATH where MATH . It is readily verified that MATH is an increasing concave function on MATH and MATH . Therefore, by [REF, p. REF], MATH . Let us define the space MATH. We put MATH and MATH and define a sequence of numbers MATH by setting MATH and a sequence of functions MAT... |
math/0005027 | Since MATH [REF, p. REF] implies that MATH . Therefore the relation MATH is equivalent to MATH . Since the function MATH is concave, we have MATH hence REF is equivalent to the condition MATH . Put MATH . Then MATH and [REF ] and REF imply that MATH . By the definition of upper dilation index, there exist MATH and MATH... |
math/0005027 | By REF , there exists a function MATH such that MATH . According to REF , the operator MATH has the DSS property; all the more, it has this property when regarded as an operator from MATH into MATH. This proves REF . |
math/0005027 | Let us verify that the assumptions of REF hold, that is, that MATH and that MATH . First, REF implies the existence of MATH and MATH such that MATH whenever MATH and MATH . This and the concavity of MATH give MATH therefore, MATH . Next, since the function MATH decreases, we have MATH . Therefore, to prove REF , it suf... |
math/0005033 | The proof is a simple computation which we leave to the interested reader, compare REF. |
math/0005033 | The proof is identical to the proof of REF. |
math/0005033 | The key to the proof rests in the fact that the pair MATH solves the anisotropic averaged NAME equations if and only if MATH is a solution of MATH where MATH . This expression is obtained using REF . Now it is clear that MATH maps MATH vector fields into MATH vector fields since MATH forms a multiplicative algebra, and... |
math/0005034 | First observe that the material density of an ideal homogeneous (compressible or incompressible) fluid is constant. Notice also that NAME REF differ only in the potential energy terms. Both these terms are functions of the Jacobian, which is equivariant with respect to the action of volume preserving diffeomorphisms gi... |
math/0005036 | The first condition implies that MATH for MATH, and so implies the second condition. The fact that the third condition implies the first is a well-known consequence of the NAME - NAME lemma. So we need only show that the second condition implies the third. The proof is by induction on MATH. First consider the case MATH... |
math/0005036 | Again, we need only prove that the second condition implies the third. In analogy to REF, we have MATH where we have made the change of variable MATH in the last step. The proof proceeds by induction on MATH, the case MATH being trivial. For MATH, apply REF with MATH an arbitrary homogeneous polynomial of degree MATH. ... |
math/0005036 | The components of MATH are linear functions of MATH and MATH, so if MATH is a polynomial of degree at most MATH, then MATH is of degree at most MATH in MATH and MATH, that is, MATH. Therefore MATH. |
math/0005036 | Assume to the contrary that MATH and MATH. We prove that MATH by induction on MATH. The case MATH being true by assumption, we consider MATH and show that the monomials MATH and MATH belong to MATH for MATH. From the identity MATH we see that for MATH, the monomial MATH is the sum of a polynomial which clearly belongs ... |
math/0005038 | Let MATH be the arc from MATH to MATH along MATH, then back to MATH along MATH. Let MATH be the sum of the winding numbers of MATH around each of the puncture points. If MATH is closer to MATH than MATH along MATH then MATH. If not, then MATH. But MATH for all MATH, so in either case MATH. |
math/0005038 | By definition, MATH is the sign of the volume form MATH at the point MATH in MATH. This is determined by the following three things: CASE: the sign of the intersection of MATH with MATH at MATH, CASE: the sign of the intersection of MATH with MATH at MATH, CASE: which of MATH and MATH is closer to MATH along MATH. Thes... |
math/0005038 | Suppose MATH is maximal. Then MATH is maximal among all the integers MATH. By REF it follows that MATH and MATH are maximal among all the integers MATH. Thus MATH. We now show that MATH. Suppose not. Then MATH. Let MATH be an embedded arc from MATH to MATH along MATH. Let MATH be an embedded arc from MATH to MATH along... |
math/0005038 | There exists a disc MATH such that MATH contains MATH, MATH contains MATH for all MATH with MATH, and MATH. Let MATH be the set of unordered pairs of distinct points in MATH. Let MATH be the pre-image of MATH in MATH. Then MATH is a free MATH-module with basis consisting of all MATH with MATH and MATH. But MATH can be ... |
math/0005043 | It is easy to make MATH transversal to MATH. Then MATH is a union of circles. It remains to homotope MATH so that we get only one component. We do this using a standard technique in MATH - manifold topology related to NAME `binding ties' CITE, as in CITE. Suppose MATH has more than one component. Let MATH be an arc joi... |
math/0005043 | The image MATH of the NAME polynomial of MATH in MATH is the so called NAME - NAME invariant, which by results of CITE and CITE is known to depend only on the homology class of MATH. Further, results of CITE show that this depends only on the NAME torsion (given an identification of homology groups). But MATH is a simp... |
math/0005043 | Let MATH denote the NAME polynomial of MATH. If the NAME module is cyclic, MATH. Since MATH, there is a NAME polynomial MATH so that MATH. Thus, MATH generates MATH. In the general case, the NAME polynomial is the determinant of the presentation matrix (with respect to some system of generators), say MATH, for the NAME... |
math/0005043 | It follows readily from the previous lemma that MATH can be expressed as MATH, for a lift of some curve MATH in MATH. Now take the band connected sum of MATH with itself (pushed off using, for instance, the MATH - framing) along a curve that goes once around MATH (see REF - here the two dotted arcs are parallel, but ma... |
math/0005043 | We use a standard construction CITE to construct a knot whose NAME polynomial is given by surgery on an unknot. Namely, we drag a piece of a curve MATH around a knot MATH and then across itself (see REF ). If MATH was a sphere, or a homology sphere, this leads to a change in the homology class of a lift of MATH in the ... |
math/0005043 | Note that one can readily change a crossing between MATH and MATH by dragging an arc of MATH along a closed curve representing MATH and then crossing MATH. This changes the intersection number with MATH by the coefficient MATH of MATH in the boundary MATH of the NAME surface. This also changes the linking of MATH at MA... |
math/0005043 | Pick a NAME surface for MATH. We shall perform MATH surgeries along curves disjoint from the NAME surface, which must thus be homologically trivial. Namely, it is well known that there are such curves that form a dual basis to a basis of curves on the NAME surface with respect to the linking pairing. Hence, one can fin... |
math/0005044 | Because MATH is self-dual and contains MATH, it is the whole MATH. One has to show (REF , page REF) that the restriction of MATH to MATH is split (the uniqueness is obvious). Since MATH is reduced, it is enough to define the splitting at the level of MATH-points. Let MATH. Since MATH and MATH is divisible, there exists... |
math/0005044 | Because MATH is the pull-back of MATH by MATH, the restriction of MATH to both MATH splits (see REF , page REF again). Observe however that the splitting is determined not only by MATH but by MATH. For the splitting over MATH, proceed as in REF . |
math/0005044 | Let us construct geometrically the basis MATH. We define MATH by MATH . By construction, one has MATH . Because MATH comes from MATH, it is MATH invariant. The relations REF follow. Because MATH is irreducible, MATH is a basis. If we have another such basis MATH, the endomorphism given by MATH is MATH-equivariant and t... |
math/0005044 | REF follows immediately from REF is a special case of NAME 's singularity theorem (see for example, CITE). The differential at the origin of the separable isogeny MATH is an isomorphism MATH, which identifies with the NAME map MATH. Given MATH, it will be enough to compute the tangent space to the divisor MATH at the o... |
math/0005044 | In order to show commutativity of REF it suffices to show that the image of the injection MATH is contained in the image of the multiplication map MATH . Let MATH and MATH be the canonical Theta bases of MATH and MATH and consider the pull-back of equality REF, written for MATH, by the morphism MATH, MATH, MATH . If MA... |
math/0005044 | Let MATH be the inclusion in the first factor MATH. A standard computation modelled on REF , which involves REF, gives the equality in MATH . We observe that the composite MATH is the diagonal map, hence the LHS can be rewritten as MATH. Applying the isogeny REF , we get MATH . Comparing with REF we can conclude. Final... |
math/0005044 | First we observe that the image MATH is invariant under the Theta group MATH. Hence the equation of MATH has to be MATH-invariant. Since MATH and MATH (MATH is separable), we have MATH. The unique MATH-invariant quadric is MATH. Since MATH is non-degenerate, we see that it is not contained in a quadric. A straightforwa... |
math/0005044 | We proceed in two steps. First we consider a flat family of curves MATH, where MATH is a discrete valuation ring such that its residue field at the closed point MATH is MATH, its field of fraction MATH is of characteristic zero, and MATH. We consider the moduli scheme MATH of semi-stable rank MATH vector bundles of tri... |
math/0005044 | CASE: This follows immediately from CITE and REF . The base locus of MATH is given by the intersection MATH, which turns out (after some elementary computations) to be a unique point with projective coordinates MATH. In terms of vector bundles this point, denoted MATH, corresponds to the direct image MATH. Indeed, MATH... |
math/0005044 | It will be enough to show that any point MATH is in the image, since, twisting by a degree zero line bundle, we can always assume the determinant to be trivial. For any non-zero MATH, we choose a MATH such that MATH and we denote by MATH the moduli of rank MATH vector bundles with fixed determinant equal to MATH. Then ... |
math/0005044 | We write MATH as an extension of line bundles MATH for some line bundle MATH with MATH and MATH. The extension class MATH of the exact sequence, gotten by pull-back under MATH, that is, MATH is obtained from MATH via the linear map MATH . The last map coincides with the induced map on cohomology of the canonical exact ... |
math/0005044 | Since the proof is in the same spirit as the proof of REF , we just give a sketch. Let MATH be the vector field on MATH associated to MATH. We also denote by MATH the endomorphism of MATH obtained via the canonical MATH-integrable connection MATH REF . We observe that we have a commutative diagram MATH . The first hori... |
math/0005044 | Since MATH, the action of MATH on MATH factorizes as MATH . Since MATH, we see that in terms of the canonical Theta coordinates MATH . Let MATH be the subfield of MATH generated by the constants MATH and MATH, for MATH, be the finite field extension of MATH generated by the coordinates MATH such that MATH with MATH. We... |
math/0005046 | REF is proved by CITE for simply laced groups, and by CITE in the general case. REF follows from REF since MATH for MATH, MATH. It remains to prove REF . Since the conjugation action of MATH on MATH is trivial, its action on any connected component of MATH is scalar multiplication on the fibers by some character for MA... |
math/0005046 | We begin by showing that MATH is symplectic. Since MATH is a regular element, MATH. Let MATH, MATH, and consider the splitting of the tangent space MATH where MATH and the second summand is embedded by the generating vector fields. By CITE, the splitting is MATH-orthogonal and the restriction of MATH to MATH is symplec... |
math/0005046 | Note first of all that the right hand side of REF is well-defined. Indeed, if MATH is replaced by MATH with MATH, the factor MATH does not change because MATH is fixed under the level MATH action of any element of MATH fixing MATH, and MATH changes by an even number. Given faces MATH of the alcove and any MATH, the sym... |
math/0005046 | The projection map MATH restricts to an equivariant diffeomorphism MATH. Let MATH be MATH-line bundle corresponding to MATH. Then MATH . The normal bundle of MATH in MATH splits MATH-equivariantly into the normal bundle MATH in MATH and the constant bundle MATH. Using MATH we obtain, MATH . Let MATH be the square root ... |
math/0005046 | By the NAME character formula, MATH . Given MATH let MATH. We claim that the sum over MATH is just MATH . Indeed, by REF and since MATH, we have MATH . But MATH since MATH. Hence MATH proving the claim. |
math/0005046 | For all MATH, with MATH we have MATH. Hence, ``quantization commutes with reduction" REF together with REF shows that MATH . Let MATH be the characteristic function of MATH. Using the NAME formula MATH the alternating sum over MATH equals MATH. |
math/0005046 | The first part follows by observing that the integral can be re-written as an integral over the cut space MATH of MATH, that is over the image of MATH in MATH: MATH . Here MATH is the ``cut" of MATH as explained in REF, and similarly for MATH and MATH. REF is expressing the integral over MATH as a sum of integrals over... |
math/0005046 | The proof is an extension of the argument given in CITE. We may assume MATH. Observe that all of the characteristic forms in REF admit MATH-equivariant extensions. Hence we can write MATH where MATH . Let us apply the NAME localization formula for orbifolds (compare CITE) to this expression. Let MATH be a fixed point m... |
math/0005047 | Since MATH is obtained from a direct product of MATH copies of MATH by passing to diagonal actions for some of the MATH-factors, it suffices to prove REF for MATH. By REF, an element MATH is fixed by MATH if and only if MATH . Both MATH and MATH belong to the exponential of the alcove MATH. Since each conjugacy class m... |
math/0005047 | The line bundle MATH is MATH-equivariant at levels MATH. Since MATH carries up to isomorphism a unique line bundle at every level CITE, it follows that MATH is the pre-quantum line bundle for the symplectic structure defined by MATH. Hence MATH is a pre-quantum line bundle for the corresponding symplectic structure on ... |
math/0005047 | The point MATH lies in identity level set of MATH, and its stabilizer in MATH is the image of the diagonal embedding of MATH. REF-form MATH restricts to a symplectic form on the tangent space MATH. By REF, MATH can be computed in terms of the symplectomorphism MATH of MATH defined by MATH: Choose a MATH-invariant compa... |
math/0005047 | We first show that the map MATH is surjective. Present MATH as a quotient of a MATH-gon MATH, with sides identified according to the word MATH . Then MATH is obtained as a similar quotient of MATH minus a disk in MATH. The sides MATH map to generators of MATH, which we also denote MATH. Given MATH, choose continuous ma... |
math/0005047 | The image MATH of an element MATH is fixed under MATH if and only if MATH. In this case, MATH fixes the maximal torus MATH pointwise, hence MATH since MATH is a regular element. Therefore MATH and MATH. |
math/0005047 | We describe the inverse map. Let MATH. Given MATH, the element MATH lies in MATH. The equation MATH means, by definition of the action of MATH on MATH, that MATH. |
math/0005047 | We have MATH, and the fixed point set is just a fusion product of the fixed point sets of the factors. Because of REF it is enough to consider the case MATH. Thus MATH is the MATH-valued Hamiltonian MATH-space MATH with MATH acting by conjugation, moment map MATH, and REF-form MATH . Using left-trivialization of the ta... |
math/0005047 | It suffices to consider the case of REF-punctured torus MATH. Let MATH be given. By REF from REF, it is possible to choose commuting lifts MATH of MATH. Then MATH and MATH. |
math/0005047 | Recall that MATH carries a unique equivariant line bundle MATH at level MATH CITE. Moreover, this line bundle carries a connection MATH whose curvature is the two-form defined by MATH. The pull-back of MATH to MATH is isomorphic to MATH. The average of MATH over MATH descends to a connection on MATH. For REF , we may a... |
math/0005047 | Let MATH be the image of MATH in MATH, and MATH the connected component containing MATH. By REF, the square root MATH coincides with the square root of the eigenvalue of the action of MATH on the symplectic vector space MATH. As in the simply connected case, it is enough to consider the case of REF-punctured torus MATH... |
math/0005047 | Let MATH be the surface obtained from MATH by capping off the boundary component. Let MATH denote the kernel of the restriction map MATH. As explained in CITE, it suffices to show that the cocycle is trivial over the subgroup MATH of MATH. We would like to define a map MATH with coboundary condition MATH . Given MATH, ... |
math/0005047 | If MATH is contained in the identity component of the stabilizer MATH, then the formula follows from the pre-quantum condition. Our strategy is to reduce to this case, using finite covers. Since MATH fixes MATH, there exists a flat connection MATH mapping to MATH and a gauge transformation MATH restricting to MATH on t... |
math/0005047 | It is well-known that evaluation at MATH, MATH restricts to an injective map MATH. The image MATH of this map is the centralizer of the image MATH of the homomorphism MATH defined by parallel transport using MATH. Since MATH and MATH is a torus, MATH. It follows that MATH is contained in the identity component of MATH. |
math/0005047 | Let MATH, and MATH the unique element projecting to MATH. By REF , the weight for the action of MATH on MATH is MATH. On the other hand, by REF , MATH, which completes the proof. |
math/0005047 | Recall MATH. Since MATH is fixed by MATH, the stabilizer of MATH is equal to the stabilizer of MATH, using the embedding MATH induced by MATH. By REF, the fixed point components for the action of MATH on MATH are the sub-manifolds MATH with MATH. Since MATH is diffeomorphic to MATH, we have MATH. The normal bundle MATH... |
math/0005047 | This is a special case of a result for non-abelian groups proved in CITE. In the abelian case, the following much simpler argument is available. Notice that MATH with diagonal MATH-action has moment map MATH not only for the fusion form MATH but also for the original symplectic REF-form MATH. Suppose MATH is a (weakly)... |
math/0005047 | We may assume that MATH is the exponential of a Hamiltonian MATH-space MATH. Rescaling by MATH, we obtain a family of Hamiltonian spaces MATH, together with their exponentials. Let MATH be the corresponding fusion forms. We claim that MATH give the required isotopy of symplectic forms. Indeed, each MATH is symplectic, ... |
math/0005047 | Decomposing into irreducible factors we may assume MATH is simple. For any simple group other than MATH with MATH even, the center MATH is a cyclic group and the claim follows by choosing a lift of the generator. It remains to consider the case MATH even which has center MATH. We use the usual presentation CITE of the ... |
math/0005052 | Suppose MATH is REF-avoiding. Choose a reduced expression for MATH for which a pair of MATH's is as close together as possible for some MATH. These two copies of MATH must be separated by at least one of MATH, otherwise our expression would not be reduced. But then our reduced expression looks like MATH where MATH. If ... |
math/0005052 | Let MATH be the smallest index for which MATH. Such a MATH must exist by our stipulation that MATH. Consider the sequence MATH. Since MATH is reduced, MATH. Hence, MATH. We now investigate the differences MATH for MATH. There are four possibilities (note that in each case, MATH): CASE: MATH, MATH. Then MATH, MATH. So M... |
math/0005052 | A picture is given in REF The claim follows immediately from Lateral Convexity by applying REF to the pairs MATH and MATH. Proof of REF . First consider the case where MATH for MATH. This is illustrated in REF By REF, MATH is in MATH. Since MATH, MATH contains the indicated hexagon by REF. Alternatively, we can have MA... |
math/0005052 | Assume that MATH is not a forest - that is, MATH contains a cycle. We will assume that MATH is REF-avoiding and show that if MATH contains a cycle then MATH contains a hexagon. Note that since MATH is REF-avoiding, REF (Lateral Convexity) holds. Let MATH be a minimal subset such that the subgraph MATH of MATH spanned b... |
math/0005052 | Partition MATH where MATH consists of all masks in MATH ending in MATH for MATH. There are natural bijections MATH and MATH given by MATH. So, to prove the lemma, we need only compare MATH to MATH. If MATH, then MATH. In this case, if MATH (MATH), then MATH, so MATH. Alternatively, if MATH (MATH), then MATH and MATH. T... |
math/0005052 | CASE: Assume MATH is REF-hexagon-avoiding. We need to show that the MATH are the NAME polynomials. Now, every MATH has three critical zeros. Furthermore, by REF, no point is a critical zero for REF distinct defects. So the number of edges in MATH equals the number of shared critical zeros. Hence, MATH . Now, by REF, MA... |
math/0005052 | As a permutation, MATH . This is easily seen to be REF-hexagon-avoiding. So by REF, MATH. The claim is true for MATH. The proof is by induction. The situation of the general case is illustrated in REF for some MATH. Let MATH. In REF, no new defect is introduced by MATH, so MATH. In REF, we have MATH. The claim follows ... |
math/0005052 | We only sketch the proof. We see that MATH is clearly REF-hexagon-avoiding, so by REF, MATH. The idea is to use recursion on MATH. From REF, it is easy to see that MATH. Similar recurrences can be found for MATH where MATH. Solving these recurrences for MATH yields REF. |
math/0005052 | It has been proved by CITE that for MATH and MATH, MATH is smooth on the NAME cell MATH if and only if MATH. By REF, MATH for every MATH. So to show that MATH is singular, we need only show that MATH contains a mask of positive defect. Let MATH correspond to MATH. Since every defect must have two critical zeros (in add... |
math/0005054 | By translating the sets MATH if necessary, we may assume that each MATH contains the origin. By hypothesis, there exists a packing of MATH into each MATH, so we may choose MATH for each MATH. The set MATH is closed and bounded, hence compact, and so by REF the space MATH is also compact. Since the sets MATH all contain... |
math/0005054 | For any MATH and any pair MATH, MATH of elements of MATH, we have MATH . We certainly have MATH by REF of the metric MATH. On the other hand, all entries of the matrix MATH are also at most MATH in absolute value, while the entries of the vector MATH are at most MATH in absolute value. Therefore each entry of MATH is b... |
math/0005054 | Since MATH and MATH are open sets that are not disjoint, we can choose a point MATH and a positive number MATH such that MATH. Using REF we may set MATH and MATH, so that MATH and MATH; we also set MATH . Then for MATH or REF, for any MATH such that MATH the upper bound REF tells us that MATH so that MATH by REF . In p... |
math/0005056 | For such a choice of MATH, the operator MATH descends to give a homogeneous differential operator on MATH. The symbol of the NAME operator is NAME multiplication, which is invertible, so the operator is elliptic. We may therefore apply REF to compute its index, and by REF we have MATH using the homogeneous NAME REF in ... |
math/0005056 | As in the proof of REF , we use the homogeneous form REF of the NAME theorem to decompose the kernel of MATH as the direct sum MATH over the finite dimensional NAME operators MATH acting on the spaces MATH where all but finitely many of these kernels vanish since the operator MATH is NAME. Ignoring the signs of the hal... |
math/0005057 | To compute the central charge of the spin representation MATH, we extend it to obtain the spin representation associated to the entire NAME algebra MATH. Since the construction of spin representations is multiplicative, we have MATH . These two spin representations have the same central charge since they differ only by... |
math/0005057 | We first note that the construction of the spin representation is multiplicative, provided that the underlying vector spaces are even dimensional. In our case, the positive and negative energy subspaces pair off, while for the zero modes, the maximal rank condition implies that the complement of MATH in MATH and the co... |
math/0005057 | First, we show that the operators MATH and MATH are super-derivations with respect to the NAME multiplication REF . For MATH and MATH, we have MATH . Now, to prove the identities REF , we need only verify them for the generators MATH, but it follows from the definition of the NAME algebra that MATH and by applying REF ... |
math/0005057 | All of the commutation relations follow immediately from REF and the above discussion with the exception of that for MATH. For example, we derive MATH . To compute MATH, we note that the fundamental REF-form is closed, so we have MATH . Written in terms of an orthonormal basis MATH for MATH, the fundamental REF-form is... |
math/0005057 | Since the NAME group acts by isometries, the weights MATH satisfy the identity MATH and it follows from REF that the NAME operator MATH vanishes on any MATH-invariant subspace of MATH transforming like MATH. To complete the proof, it remains to show that each of these representations occurs exactly once in the domain o... |
math/0005057 | The highest weight space of an irreducible representation of MATH is always one dimensional, so the weight MATH appears with multiplicity REF in MATH. Now consider the complex spin representation MATH associated to the orthogonal complement of a NAME subalgebra MATH in MATH. Given a positive root system for MATH, the c... |
math/0005057 | If MATH is a weight of MATH, then MATH is a weight of the tensor product MATH. Since the NAME group acts simply transitively on the NAME chambers, there exists an element MATH such that MATH is dominant, where we recall that a weight MATH is dominant if and only if MATH for all positive roots MATH. Note that every weig... |
math/0005057 | The bracket MATH is simply the definition of the NAME algebra, while the brackets MATH and MATH and MATH follow immediately from REF . By the NAME identity, for any MATH and MATH we have MATH which shows that MATH is a projective representation of MATH on MATH. In REF , we established that this spin representation has ... |
math/0005057 | In order to simply our calculations, we first note the following identites: MATH . We now show that the commutator of MATH with an element MATH is MATH . It follows that the operator MATH commutes with the action of MATH, and therefore takes a constant value on each irreducible representation. Acting on the minimum ene... |
math/0005057 | Since MATH is MATH-equivariant, it respects the decompositions of MATH and MATH into their constant energy subspaces, and it can be written in the block diagonal form MATH, with MATH. If both MATH and MATH are of finite type, then each of the subspaces MATH and MATH is a finite dimensional MATH-module, and so the MATH-... |
math/0005060 | The estimates in REF are immediate. The proof of REF is also an easy estimate, which can be found in CITE, for example. The arguments for REF are also quite standard. We leave the proof for the reader. Let us see that REF holds. If MATH and MATH are concentric, the identity MATH is a direct consequence of the definitio... |
math/0005060 | Let MATH be the biggest cube centered at MATH with side length MATH, MATH, such that MATH. Then, MATH. Otherwise, MATH and since MATH and MATH we get MATH which contradicts the choice of MATH, assuming MATH. Now we have MATH . Thus MATH . Let MATH be the smaller doubling cube of the form MATH, MATH. Then MATH. Also, MA... |
math/0005060 | Suppose, for example, MATH. Then, MATH. Let MATH be the smallest doubling cube of the form MATH, MATH. We have MATH . Thus MATH . We also have MATH . Since MATH and MATH have comparable sizes, MATH, and so MATH . Therefore, MATH . By REF , the proposition follows. |
math/0005060 | Let MATH be an atomic block supported on some cube MATH, with MATH, where MATH are functions supported on cubes MATH such that MATH. We will show that MATH. First we will estimate the integral MATH . For MATH and MATH, since MATH, we have MATH . Thus MATH . Now we will show that MATH and we will be done. If MATH and MA... |
math/0005060 | The arguments are quite standard. For any MATH there exists some cube MATH which contains MATH, with MATH and such that MATH. Then by NAME 's Covering Theorem, there are points MATH such that MATH and so that the cubes MATH, MATH, form an almost disjoint family. Observe that REF cannot be applied to the cubes MATH (the... |
math/0005060 | Both statements are a straightforward consequence of REF , since MATH and MATH . |
math/0005060 | We can assume MATH. Let MATH be some fixed constant. If MATH, then MATH. So, if MATH is a cube centered at MATH with side length MATH, we have MATH. By REF we get MATH . Since MATH if we set MATH, we obtain MATH . Then for MATH small enough we have MATH . This implies MATH, which is not possible. |
math/0005060 | By the previous lemma, MATH and MATH. This gives MATH. |
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