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math/0105046 | Let MATH be a uniformly locally finite covering of MATH by NAME charts as in REF . Let MATH be the bundle over hyperbolic space associated to the same MATH or MATH representation as MATH, and for each MATH, let MATH be the operator defined by MATH. Since the metric MATH is uniformly MATH equivalent to MATH in NAME coordinates by REF , it follows that the coefficients of the MATH-th derivatives of MATH that appear in MATH are uniformly bounded in MATH. Therefore, using REF, we have MATH with an analogous estimate in the NAME case. |
math/0105046 | This is true for MATH by the fact that MATH is formally self-adjoint. The general result follows by density. |
math/0105046 | Under the hypotheses of REF is locally in MATH by interior elliptic regularity, so only the estimate REF needs to be proved. Let MATH be a uniformly locally finite covering of MATH by NAME charts as in REF , and let MATH as in the proof of REF . Since the coefficients of the highest-order terms in MATH are constant-coefficient polynomials in MATH and MATH, and MATH is uniformly equivalent to the hyperbolic metric by REF , MATH is uniformly elliptic on MATH. Moreover, since the coefficients of MATH are uniformly bounded in MATH by the same lemma, we have the following standard local elliptic estimate CITE: MATH where the constant MATH depends on MATH, MATH, MATH, and MATH, but is independent of MATH and MATH. Thus, using REF again, we have MATH . The argument for REF is similar, using interior NAME estimates CITE. |
math/0105046 | Because of the density of MATH in MATH, MATH is densely defined, and REF shows that its domain is exactly MATH. Clearly the domain of its NAME space adjoint contains MATH. On the other hand, if MATH is in the domain of the adjoint, then there exists MATH such that MATH for all MATH. This means in particular that MATH as distributions, which by REF implies that MATH. Thus the domain of the adjoint is equal to the domain of MATH. |
math/0105046 | The argument in the forward direction is fairly standard; see, for example, CITE and CITE. Let MATH and MATH be precompact open subsets of MATH such that MATH, and let MATH be a smooth bump function that is equal to MATH on MATH and supported in MATH. It is easy to check that multiplication by MATH is a bounded map from MATH to itself for any MATH. On the compact set MATH, the MATH norm is uniformly equivalent to the standard MATH norm, and MATH is uniformly elliptic. First observe that REF extends to all MATH by continuity. For any MATH, we can write MATH, where MATH is supported in MATH and MATH is supported in MATH. We estimate as follows: MATH . For the first term, REF gives MATH where in the last line we have used the fact that MATH is an operator of order MATH supported on MATH. For the second term, standard interior elliptic estimates yield MATH . Finally, standard interpolation inequalities on the compact set MATH (see CITE) allow us to replace MATH by MATH with an arbitrarily small constant MATH. Absorbing the MATH term on the left-hand side, we conclude MATH . Now suppose MATH is a sequence in MATH. Normalize MATH so that MATH. By the NAME lemma on the compact set MATH, some subsequence converges in MATH. From REF we conclude that this subsequence is NAME and hence convergent in MATH, and therefore MATH is finite-dimensional. Next we need to show that MATH has closed range. Since MATH is finite-dimensional, it has a closed complementary subspace MATH. I claim there is a constant MATH such that MATH . If not, there is a sequence MATH with MATH . By the NAME lemma again, there exists a subsequence (still denoted by MATH) that converges in MATH. Then REF shows that the subsequence also converges in MATH. However, the limit MATH then satisfies MATH and MATH by continuity, and is therefore a nonzero element of MATH, which is a contradiction. Now if MATH with MATH in MATH, we can assume without loss of generality that each MATH, and then REF shows that MATH converges in MATH, which shows that MATH has closed range. Conversely, suppose MATH is semi-Fredholm. Let MATH be a compact subset chosen as follows: If MATH has the unique continuation property, MATH can be any compact subset of MATH with nonempty interior; otherwise, let MATH, where MATH is chosen small enough that no element of the finite-dimensional space MATH vanishes identically on MATH. The key fact is that there exists MATH such that MATH whenever MATH is supported in MATH and MATH. To see this, note that our choice of MATH ensures that MATH is a norm on MATH. Since all norms on a finite-dimensional vector space are equivalent, there exists a constant MATH such that MATH for all MATH. Suppose now that MATH is supported in MATH, MATH, and MATH. If MATH, then by the reverse triangle inequality MATH . If on the other hand MATH, then because MATH vanishes on MATH, MATH . REF (with MATH) then follows for general MATH by homogeneity. As above, MATH has a closed complementary subspace MATH, and both MATH and MATH are NAME spaces with the induced norms. Then MATH is bijective, and its inverse MATH is bounded by the open mapping theorem. This means there exists a constant MATH such that REF holds. Let MATH be supported in MATH, and write MATH, with MATH, MATH. It follows from REF that MATH and therefore, by REF with MATH, MATH which is REF. Finally, suppose that both REF hold. To show that MATH is actually NAME, all that remains to be shown is that the range of MATH has finite codimension in MATH. Recall that MATH is dual to MATH under the standard MATH pairing. The argument above, using REF instead of REF, shows that MATH has finite-dimensional kernel. Any MATH that annihilates the range of MATH in MATH satisfies in particular MATH for all MATH, so is a distribution solution to MATH. By REF , MATH. Thus there is at most a finite-dimensional subspace of MATH that annihilates the range of MATH. Since MATH is closed in MATH, it has finite codimension. |
math/0105046 | If MATH is an isomorphism, then clearly it is NAME. Conversely, if MATH is NAME, then by REF MATH satisfies MATH whenever MATH is supported in the complement of some compact set MATH. Suppose MATH is any smooth, compactly supported section of MATH. There is a NAME transformation MATH such that MATH, so MATH satisfies REF. Because MATH and the MATH norm are preserved by MATH, MATH itself satisfies the same estimate. Therefore, by continuity, REF holds for all MATH, so MATH is trivial. Since MATH is self-adjoint as an unbounded operator on MATH, its index is zero, which means that it is also surjective. |
math/0105046 | The isometry invariance of MATH implies that MATH has the following equivariance property for any orientation-preserving hyperbolic isometry MATH: MATH . Also, since MATH is defined purely in terms of the hyperbolic distance function, it is clearly isometry invariant: MATH . Therefore, to prove REF, it suffices to show that MATH . Note that MATH defines a smooth section of the bundle MATH over MATH, whose fiber at MATH is the vector space MATH. The group of orientation-preserving hyperbolic isometries that fix MATH is exactly MATH, acting linearly on the unit ball as isometries of both the hyperbolic and Euclidean metrics. Let MATH be the ray MATH along the MATH-axis. The subgroup of MATH that fixes MATH pointwise is MATH, realized as the group of linear isometries acting in the first MATH variables only. Observe that for each MATH, MATH acts orthogonally (or unitarily if MATH is a complex tensor bundle) on the fiber MATH by pulling back. Let MATH be an orthogonal decomposition of MATH into irreducible MATH-invariant subspaces. We extend this to a decomposition of the bundle MATH over MATH as follows. First, for each MATH, let MATH be the subspace of MATH obtained by parallel translating MATH along MATH with respect to the Euclidean metric MATH; since MATH acts as Euclidean isometries, it follows that MATH is an orthogonal irreducible MATH-decomposition of MATH. Then for an arbitrary point MATH, let MATH, where MATH is the unique point of MATH such that MATH, and MATH satisfies MATH. Since MATH is invariant under MATH, MATH does not depend on the choice of MATH. Since MATH can be chosen locally to depend smoothly on MATH (by means of a smooth local section of the submersion MATH), this results in MATH smooth subbundles MATH of MATH over MATH. For each pair of indices MATH, we choose a MATH-equivariant linear map MATH as follows: If MATH and MATH are isomorphic as representations of MATH, let MATH be a MATH-equivariant Euclidean isometry from MATH to MATH, extended to be zero on MATH for MATH; and otherwise let MATH be the zero map. By NAME 's lemma, the nonzero maps MATH form a basis for the space of MATH-equivariant endomorphisms of MATH. Let us renumber these nonzero maps as MATH. Next we extend each map MATH to a section MATH of MATH over MATH in the same way as we extended the spaces MATH: First, for each point MATH, define MATH to be MATH followed by MATH-parallel translation along MATH from MATH to MATH; and then for arbitrary MATH, define MATH, where MATH and MATH satisfies MATH. I claim that there are smooth functions MATH such that MATH for all MATH. To see this, first note that for any point MATH, the equivariance REF implies that MATH is a MATH-equivariant linear map from MATH to MATH, and therefore by NAME 's lemma it can be written as a linear combination of the maps MATH: MATH . For any other point MATH, let MATH be the point of MATH such that MATH, and let MATH satisfy MATH. Then REF yields MATH . Since MATH is a diffeomorphism, there are functions MATH such that MATH whenever MATH, and then REF is equivalent to REF. The smoothness of MATH implies that the functions MATH are smooth. Now the equation MATH reduces to an analytic system of ordinary differential equations for the functions MATH, and the fact that MATH is uniformly degenerate implies that this system has a regular singular point at MATH. Because the sections MATH of MATH extend smoothly to MATH, our definition of the indicial map of MATH guarantees that the characteristic exponents of this system of ODEs are precisely the characteristic exponents of the operator MATH. Therefore, by the standard theory of ODEs with regular singular points, each coefficient function MATH satisfies MATH for some characteristic exponent MATH and some nonnegative integer MATH. If MATH is any tensor in one of the summands MATH, then the images MATH lie in different summands of MATH and are therefore orthogonal, so MATH where MATH is the Euclidean metric on MATH. Since MATH is a Euclidean isometry onto its image, MATH is independent of MATH. Therefore, if MATH, then on MATH we have MATH for some positive constant MATH. Because MATH is in MATH away from MATH, by REF we must have MATH for each such MATH. Since for each MATH there is some MATH such that MATH, the same inequality holds for every MATH. By definition of the indicial radius MATH, this implies that in fact MATH and MATH. Using REF, we conclude that for any MATH there is a constant MATH such that MATH . This in turn implies MATH whenever MATH, which was to be proved. |
math/0105046 | We use the following standard integral representation for hypergeometric functions CITE: MATH which is valid if MATH and MATH. The hypergeometric function MATH is analytic for MATH and satisfies a second-order ODE that has a regular singular point at MATH with characteristic exponents MATH and MATH CITE. As long as MATH, therefore, it satisfies MATH . Applying this with MATH, MATH, and MATH proves the lemma. |
math/0105046 | By an isometry, we can arrange that MATH and MATH is on the positive MATH-axis. Substituting MATH into the integral, we must estimate MATH . Parametrize the ball by the map MATH given by MATH so that MATH is the Euclidean distance from MATH and MATH is the angle from the positive MATH-axis. In these coordinates, the hyperbolic metric is MATH where MATH represents the standard metric on MATH. The hyperbolic volume element is therefore MATH where MATH is the volume element on MATH. In these coordinates, we have MATH, MATH and MATH. The integrand in REF is constant on each sphere MATH, so we can immediately integrate over MATH and write MATH as a constant multiple of MATH . Since we are only interested in estimates up to a constant multiple, we will write MATH to mean that MATH is bounded above and below by positive constants depending only on MATH, MATH, and MATH. Thus, for example, MATH, MATH, MATH, and MATH . The MATH integral can be simplified by the substitution MATH to obtain MATH where MATH . Because our hypothesis guarantees that MATH and therefore MATH, REF shows that the right-hand integral in REF is bounded by a constant multiple of MATH. Substituting this into REF yields MATH . A computation shows that MATH . Inserting this into REF, we conclude that MATH . REF then shows that this is bounded by a multiple of MATH. |
math/0105046 | Since MATH by self-adjointness, the two inequalities are equivalent, so it suffices to prove the second one. We will write MATH and estimate each term separately. For the first term, we observe that MATH is uniformly locally integrable near MATH: Since MATH satisfies MATH, and the NAME delta function is in the NAME space MATH for MATH (defined as the dual space to MATH, MATH), local elliptic regularity implies that MATH. Therefore, if MATH is arbitrary and MATH is any NAME transformation sending MATH to MATH, the change of variables MATH yields MATH . Using the triangle inequality together with the elementary fact that MATH for MATH, we estimate MATH . It follows that MATH on the set where MATH. By symmetry, the same inequality holds with MATH and MATH reversed. Thus MATH . For the second term, choose MATH small enough that MATH . Then with MATH, we have MATH and MATH, so we can use REF to conclude MATH . |
math/0105046 | Using REF , it suffices to prove that MATH for all MATH. Because MATH is dense in MATH, it suffices to prove this inequality for MATH. Since MATH in that case, it suffices to prove the estimate MATH . Put MATH so that MATH . By NAME 's inequality and REF , we estimate MATH . Therefore, MATH . By REF again, we can evaluate the MATH integral first to obtain MATH . |
math/0105046 | Injectivity is an immediate consequence of REF. To prove surjectivity, let MATH be arbitrary, and let MATH be a sequence such that MATH in MATH. Set MATH, so that MATH. Then each MATH is in MATH by REF, and in MATH by REF , and REF shows that MATH is NAME in MATH. It follows that MATH satisfies MATH as desired, so MATH is surjective. The continuity of the inverse map then follows from REF. |
math/0105046 | By REF , MATH which implies MATH . |
math/0105046 | To prove surjectivity, let MATH be arbitrary and set MATH, so that MATH by REF . An easy computation shows that MATH in the distribution sense, so MATH by REF . To prove injectivity, choose MATH close to MATH and MATH large such that MATH and MATH. Then MATH by by REF . Since MATH is injective on MATH by REF , it is injective on the smaller space MATH. |
math/0105046 | Because MATH is not precompact in MATH, we have to interpret the MATH norm on the left-hand side of REF as an intrinsic NAME norm, defined by REF. For each point MATH, we have a NAME chart MATH defined by MATH . Then we need to get an upper bound for MATH . Since MATH is a hyperbolic isometry, the norm above is the same as MATH . In MATH coordinates, we have MATH . Let us abbreviate this composite map as MATH, so that MATH . Since the MATH norm in REF is just the norm of the components in MATH-coordinates, and MATH is uniformly bounded on MATH together with all its derivatives, it suffices to show that MATH for any function MATH that vanishes at MATH. Moreover, the MATH coordinates are uniformly MATH-equivalent to the original background coordinates MATH, because the coefficients MATH, the matrix MATH, and its inverse are uniformly bounded. Thus the MATH norm of MATH in MATH coordinates is uniformly bounded by the global norm MATH. To bound the sup norm of MATH, we use the mean value theorem and the fact that MATH to estimate MATH where MATH is some point on the line between MATH and MATH. For any coordinate MATH, we have MATH . The NAME norm of the first derivatives is estimated as follows: MATH . The general case now follows by induction on MATH, using the fact that MATH. |
math/0105046 | The fact that MATH in MATH is just a computation: MATH . To check the mapping properties of MATH, we begin by observing that the fact that the functions MATH are uniformly bounded in MATH implies by REF that multiplication by MATH is a bounded map from MATH to itself for each MATH and all MATH, with norm bounded independently of MATH and MATH. The fact that MATH maps MATH to itself then follows from REF , because the factors of MATH and MATH introduced by MATH and its inverse cancel each other. Moreover, REF implies REF whenever MATH is supported in MATH, for some constant MATH independent of MATH and MATH. The mapping properties of MATH will follow from a similar argument once we show that the commutator MATH maps MATH to MATH. Observe that each term in the coordinate expression for MATH is a product of four factors: a constant, a MATH-th covariant derivative of MATH, a MATH-th covariant derivative of MATH, and a polynomial in the components of MATH, MATH, and their derivatives up through order MATH, with MATH and MATH. Since MATH is uniformly bounded in MATH, the result follows. The argument for the NAME case is identical. |
math/0105046 | Just choose MATH small enough that REF holds with MATH. It then follows that MATH has a bounded inverse. We just set MATH and then REF follows immediately from REF. Once again, the argument for the NAME case is identical. |
math/0105046 | If MATH, the result is a trivial consequence of REF, so assume MATH. Consider REF . By REF , it suffices to show that MATH. For any small MATH, by means of a bump function we can write MATH, where MATH is compact and MATH. Local elliptic regularity gives MATH. Since MATH agrees with MATH off of a compact set, MATH, so by REF , if MATH is small enough, MATH where MATH. Iterating this argument finitely many times, we conclude that MATH. By REF , this implies that MATH as claimed. The argument for the NAME case is the same. |
math/0105046 | If MATH with MATH, then MATH whenever MATH by REF . Since such MATH and MATH can be chosen that also satisfy MATH, it suffices to prove the proposition under the hypothesis that MATH with MATH. For the rest of the proof, we assume this. First we treat REF . Assume that MATH with MATH, and let MATH be the following set: MATH . Clearly MATH by hypothesis. We will show that MATH. It will then follow from REF that MATH, which will prove REF : If MATH, then MATH. To prove this, assume MATH and MATH. The fact that MATH means that there is some MATH with MATH such that MATH. By REF , MATH for any MATH such that MATH. Choosing MATH so that MATH is sufficiently close to MATH, we can ensure that MATH. This implies that MATH as desired. CASE: If MATH and MATH satisfies MATH and MATH where MATH then MATH. The assumption that MATH means that MATH for some MATH with MATH. Choose MATH satisfying MATH . By virtue of the first inequality above, REF implies that MATH . Since the two inequalities of REF guarantee that MATH, REF shows that MATH. Our restriction on MATH guarantees that MATH so REF implies that MATH. Now REF implies that MATH . Therefore, using REF, we obtain MATH which proves that MATH as claimed. CASE: MATH. If MATH, this follows immediately from REF together with the obvious fact that MATH. Otherwise, just iterate REF , starting with MATH. After finitely many iterations, we can conclude that MATH. Finally we turn to REF . Suppose MATH with MATH, and choose MATH large and MATH close to MATH satisfying MATH . Then by REF , MATH. If we choose MATH sufficiently close to MATH, we have MATH, and thus MATH by REF above. If MATH is also chosen large enough that MATH, the NAME embedding theorem REF implies MATH, and if MATH is sufficiently close to MATH we will have MATH. Then REF implies MATH. |
math/0105046 | We begin with the NAME case, REF . First consider the special case MATH, MATH, and MATH (in which case this is basically the standard construction of a partial inverse for a NAME operator on MATH). As noted above, the assumption of a MATH estimate REF near the boundary implies that MATH is NAME by REF . By definition, MATH is precisely the orthogonal complement of MATH in MATH, so we have an orthogonal direct sum decomposition MATH. Since MATH is formally self-adjoint, any MATH satisfies MATH for all MATH, so MATH. On the other hand, if MATH is orthogonal to MATH, then for any smooth, compactly supported section MATH of MATH we have MATH, so MATH is a distributional solution to MATH, which means MATH. This shows that MATH, and since MATH is closed in MATH we have MATH. Now MATH is bijective and bounded, so by the open mapping theorem it has a bounded inverse MATH. Define MATH by MATH and define MATH to be the orthogonal projection onto MATH. Then REF are immediate from the definition of MATH, and REF (for MATH) and REF (for all MATH) follow by considering MATH and MATH separately. Next consider the case of arbitrary MATH satisfying MATH, still with MATH and MATH. If MATH and MATH, then MATH, so MATH by REF . Thus the restriction of MATH to MATH takes its values in MATH, as does MATH by REF. In this case, REF - REF are satisfied because they are already satisfied on the bigger space MATH (or MATH in case of REF), and MATH. On the other hand, if MATH, we can use the fact that MATH to extend the definition of MATH and MATH to MATH by duality: For any MATH, let MATH and MATH be the elements of MATH defined uniquely by MATH for all MATH. In other words, MATH are defined to be the dual maps of MATH. Since MATH and MATH are self-adjoint on MATH (MATH because it is an orthogonal projection, and MATH because MATH is self-adjoint as an unbounded operator), these are indeed extensions of the original maps MATH and MATH. To see that these extended operators satisfy REF - REF, we observe that REF implies that MATH for MATH exactly when MATH for all MATH in the image of MATH; since this image is exactly MATH, it follows that MATH, which is REF. Since the restriction of MATH to MATH is the identity, it follows that MATH. On the other hand, for any MATH, we have MATH for all MATH, which means that MATH is a weak solution to MATH. Thus MATH, which proves REF then follow easily from our definitions by duality. Next we generalize to MATH and MATH, still with MATH. If MATH, we can choose MATH such that MATH and MATH, so that MATH. Arguing as above, we see that the restrictions of MATH and MATH map MATH to MATH. On the other hand, for MATH, we can extend MATH and MATH to maps from MATH to itself by duality as above. In both REF - REF are satisfied, by restriction or duality as appropriate. Now consider the general case of MATH with MATH, MATH, and MATH. Since MATH, it is clear that MATH restricts to a map of MATH to itself. If MATH for MATH, observe as above that MATH, so MATH by REF . For MATH, we have MATH. Thus in each case MATH and MATH restrict to maps from MATH to itself, and REF - REF are satisfied by restriction. Finally, consider REF , and assume that MATH, MATH, and MATH. We can choose MATH satisfying MATH and MATH, so that MATH and the results of REF apply to MATH. Then the restrictions of MATH and MATH map MATH to itself by the same argument as above, and REF - REF are automatically satisfied by restriction. |
math/0105046 | Let MATH be arbitrary, and let MATH be any neighborhood of MATH in MATH. Since the characteristic exponents are constant on MATH by REF , there is a tensor MATH such that MATH. We can extend MATH to a MATH tensor field MATH on a neighborhood of MATH in MATH, still satisfying MATH, as follows. Shrinking MATH if necessary, we can choose background coordinates on MATH, and for each MATH, let MATH be the matrix of MATH. By REF , the matrix entries of MATH are MATH functions of MATH. If MATH is any smooth, positively-oriented closed curve in MATH whose interior contains MATH but no other eigenvalues of MATH, then the projection onto the kernel of MATH can be written as MATH. (Here we use the fact that the eigenvalues of MATH and their multiplicities are independent of MATH.) We define MATH to be the tensor field on MATH whose coordinate expression is MATH which is a MATH tensor field along MATH satisfying MATH. If we extend MATH arbitrarily to a MATH tensor field on MATH, and let MATH where MATH is any smooth cutoff function that is positive at MATH and supported in MATH, then MATH, and MATH which implies MATH so MATH. On the other hand, MATH for MATH by REF . Now choose countably many points MATH and disjoint neighborhoods MATH of MATH. For each MATH we can construct MATH as above with support in MATH, so the space spanned by MATH is clearly infinite-dimensional. |
math/0105046 | It follows immediately from REF that MATH is NAME as a map from MATH to MATH if and only if MATH satisfies an estimate of the form REF. By REF , the kernel and range of MATH as an unbounded operator on MATH are the same as those of MATH. The proposition follows. |
math/0105046 | We will prove REF together. The proof of sufficiency is identical for the NAME and NAME cases, so we do only the NAME case. Suppose MATH, MATH, and MATH, and let MATH be as in REF . We have already remarked that the kernel of MATH is equal to MATH, which is finite-dimensional, and in fact is the same as the MATH kernel. We will show that the range of MATH is closed by showing that it is equal to MATH. If MATH, then clearly MATH for all MATH, so MATH. On the other hand, if MATH, then MATH by REF, so MATH. Since every MATH can be written MATH, where MATH and MATH, it follows that MATH. Therefore, MATH which is finite-dimensional. This also shows that the cokernel and kernel of MATH have the same dimension, so MATH has index zero. Next we will prove the necessity of the stated conditions on MATH. In fact, we will show that MATH has infinite-dimensional kernel when MATH is strictly below the NAME range, and infinite-dimensional cokernel when MATH is strictly above; in the borderline case MATH, we will also show that MATH has infinite-dimensional kernel, and in all other borderline cases, we will show that it fails to have closed range. First we address the NAME case below the NAME range. Assume MATH, MATH, and MATH, and consider MATH. We will prove that MATH is not NAME in this case by showing that it has an infinite-dimensional kernel. The definition of the indicial radius MATH and the symmetry of the characteristic exponents about MATH imply that MATH has a characteristic exponent MATH with MATH. Let MATH be the subspace of MATH given by REF for this characteristic exponent. (The compact set MATH is irrelevant in this case.) For any MATH, MATH by REF . If we choose MATH large enough, then MATH will satisfy MATH, so there exist operators MATH satisfying REF - REF. Let MATH be the linear subspace defined by MATH . Because MATH takes its values in the finite-dimensional space MATH, the space MATH is also infinite-dimensional. Note that for MATH, MATH implies MATH by REF. Define MATH by MATH. It follows from REF that MATH . Therefore, MATH by REF . Moreover, MATH is injective because MATH implies MATH, which implies that MATH by REF . Thus we have shown that MATH is an infinite-dimensional subspace of MATH. Since MATH whenever MATH and MATH, it follows that MATH has infinite-dimensional kernel on MATH in all such cases. Next consider the NAME case below the NAME range. When MATH, MATH, and MATH, we have MATH by REF , so MATH has infinite-dimensional kernel in MATH as well. Now we consider the exponents strictly above the NAME range, beginning with the NAME case. Suppose MATH, MATH, and MATH. Recall that the dual space to MATH is MATH (where MATH is the conjugate exponent, MATH), acting by way of the standard MATH pairing. Since MATH, the argument above shows that MATH has infinite-dimensional kernel. Each element MATH of the infinite-dimensional space MATH thus defines a continuous linear functional on MATH by MATH, and each such linear functional annihilates MATH by REF . It follows that the range of MATH has infinite codimension in MATH. For the NAME case, suppose MATH, MATH, and MATH. Choose MATH close to MATH and MATH sufficiently large that MATH. It follows that MATH, which implies as above that MATH has an infinite-dimensional kernel in MATH. The fact that MATH implies that MATH. As above, each linear functional on MATH defined by an element of MATH annihilates the range of MATH, so once again we conclude that MATH has infinite-dimensional cokernel. Next we consider the borderline cases. The lower borderline NAME case MATH was already treated above, when we showed that MATH has infinite-dimensional kernel in MATH whenever MATH. In all remaining cases, we will show that MATH does not have closed range. We begin with the upper borderline NAME case, MATH with MATH. Let us assume that MATH has closed range and derive a contradiction. If MATH is chosen sufficiently close to MATH that MATH, the argument at the beginning of this proof showed that MATH is finite-dimensional. Since MATH, we see that MATH is finite-dimensional as well, so MATH is semi-Fredholm. It follows from REF that there is a compact set MATH and a constant MATH such that MATH when MATH is supported in MATH. By definition of MATH, MATH has a characteristic exponent MATH whose real part is equal to MATH. Let MATH be any element of the space MATH defined in REF corresponding to this characteristic exponent, with MATH, MATH, but MATH. Let MATH be a family of cutoff functions as in REF , and define MATH. Note that for any MATH, MATH is in MATH and compactly supported, so it is in MATH for all MATH. Because MATH and MATH uniformly on compact sets as MATH, we have MATH. On the other hand, MATH . If we can show that MATH remains bounded as MATH, we will have a contradiction to REF. The fact that MATH implies that MATH in the MATH norm, so the first term in REF is clearly bounded in MATH. For the second term, observe that the commutator MATH is an operator of order zero with coefficients that are uniformly bounded in MATH and supported on the set where MATH. It follows by induction that MATH is an operator of order MATH with bounded coefficients supported in the same set, and therefore MATH . Since MATH is integrable for MATH, each integral above goes to zero as MATH by the dominated convergence theorem. This contradicts REF and completes the proof that MATH does not have closed range in this case. Next consider the upper borderline NAME case, MATH. The argument is almost the same as in the NAME case, except in this case we have to set MATH and show that MATH while MATH remains bounded below by a positive constant. The details are left to the reader. The only case left is the lower borderline NAME case, MATH with MATH. Since MATH, we showed above that MATH does not have closed range, and thus neither does MATH considered as an unbounded operator (since its domain is exactly MATH). Since a closed, densely defined operator has closed range if and only if its adjoint does (compare CITE), this implies that MATH does not have closed range, and then it follows from the regularity results of REF that MATH does not either. Finally, to prove REF , just observe that MATH being NAME on MATH implies that it is NAME from MATH to MATH. By REF , this in turn implies that MATH. |
math/0105046 | Since MATH is independent of MATH, we need only consider the case MATH. As in the proof of REF, we compute (in background coordinates) MATH . Therefore, MATH . To compute the second term above, assume MATH is a tensor of type MATH with MATH, and let MATH be the difference tensor MATH. The components of MATH are MATH . Let us introduce the shorthand notations MATH and MATH. Using REF for the components of MATH, together with the fact that MATH, we obtain MATH . Therefore, MATH . Inserting this back into REF, we obtain MATH which implies the result. |
math/0105046 | By REF , to determine MATH it suffices to compute MATH. The cases MATH and MATH follow from REF. For the general case, suppose MATH. Assuming MATH is trace-free, symmetric, and smooth up to the boundary, and using the notation of the preceding proof, we compute MATH . As in the preceding proof, we have MATH and therefore, MATH . Using REF again, we compute MATH . Inserting these above and using the fact that MATH is symmetric and trace-free, we obtain MATH . Suppose MATH is any indicial root of MATH and MATH is a corresponding unit eigentensor. Using the formula above for MATH together with REF , and observing that MATH is trace-free, we compute MATH which implies that each indicial root satisfies MATH . It follows that the indicial radius of MATH is at least MATH. On the other hand, if MATH is chosen so that MATH along MATH (that is, MATH is purely tangential), we find that MATH. Solving MATH for MATH, we find that two of the indicial roots of MATH are MATH . This proves that the indicial radius is exactly MATH as claimed. |
math/0105046 | Comparing the formulas given by REF for MATH and MATH, we find that MATH . Observe that MATH is self-adjoint by REF , so it has real eigenvalues. Now MATH is a characteristic root of MATH precisely when MATH is a solution to the quadratic equation MATH for some eigenvalue MATH of MATH. If some eigenvalue were nonpositive, this equation would have a root with real part equal to MATH, which would imply MATH. Therefore, the assumption MATH means that all of the eigenvalues MATH of MATH are strictly positive, and therefore the characteristic roots of MATH are MATH, with MATH. Since MATH, the characteristic roots of MATH are MATH, and the one with smallest real part greater than MATH is MATH. Thus the indicial radius of MATH is MATH as claimed. |
math/0105046 | Observe first that MATH preserves the splitting of symmetric MATH-tensors into trace and trace-free parts: MATH where MATH is the bundle of symmetric covariant MATH-tensors, MATH is the subbundle of tensors that are trace-free with respect to MATH, and MATH is the real line bundle of multiples of MATH. On MATH, MATH, so MATH acts as the ordinary Laplacian: MATH . It follows from REF (or REF ) that the indicial radius of MATH on functions is MATH, so the indicial radius of MATH is MATH, which is greater than REF. Therefore, it suffices to show that MATH acting on trace-free symmetric MATH-tensors has indicial radius given by REF. The asymptotic REF for the NAME curvature tensor implies that the action of MATH and MATH on MATH near the boundary is given by MATH . Thus the indicial radius of MATH on MATH is the same as that of MATH, which is MATH by REF . |
math/0105046 | Let MATH. The divergence theorem gives MATH . Thus MATH which is equivalent to REF . |
math/0105046 | REF implies MATH which extends continuously to compactly supported functions in MATH. If MATH is a smooth, compactly supported tensor field, the function MATH is NAME, hence in MATH. NAME 's inequality says that MATH almost everywhere (see CITE, where this is proved for scalar functions; the proof extends easily to tensors). Therefore MATH . |
math/0105046 | REF through REF are standard, and can be found, for example, in CITE. For REF, choose a point MATH and a frame MATH such that MATH at MATH. Then, computing at MATH and using REF, and REF, we have MATH . The computation for REF is similar. |
math/0105046 | Let MATH. Using REF , we compute MATH . |
math/0105046 | We will use REF for forms MATH supported near the boundary. Let MATH be a smooth defining function. Using REF, we compute MATH . The tensor MATH acts as the identity on MATH-forms, and since we extend it to act as a derivation, it acts on MATH-valued MATH-forms as MATH times the identity. Using REF, therefore, the action of MATH on a MATH-valued MATH-form MATH can be written MATH . Thus with MATH, where MATH is a constant to be determined later, the integrand on the right-hand side of REF can be estimated as follows: MATH . Given MATH, we can choose MATH small enough that the absolute value of the MATH factor above is bounded by MATH on MATH. If MATH, we then have for MATH compactly supported in MATH . This estimate is optimal when MATH, so as long as MATH, which is to say MATH, we obtain the conclusion of the lemma with MATH. If on the other hand MATH, we use the fact that MATH approaches MATH uniformly at MATH. We choose MATH small enough that the MATH factor in REF is bounded by MATH and MATH on MATH, and conclude that MATH . This in turn is optimal when MATH. Thus as long as MATH, so that MATH, we obtain the conclusion of the lemma with MATH. |
math/0105046 | This is easiest to see in components, noting that the last index of MATH is considered to be the MATH-form index. MATH . Applying the NAME identity to the commutator MATH yields the result. |
math/0105046 | Using REF, we compute that the curvature operators MATH and MATH have the following asymptotic behavior on trace-free symmetric MATH-tensors near MATH: MATH . To prove the lemma, just use REF with MATH together with REF to obtain MATH . |
math/0105046 | This follows immediately from the preceding lemma together with the asymptotic REF for MATH and MATH. |
math/0105046 | The operators MATH and MATH are NAME on MATH for the claimed values of MATH by virtue of REF . On the other hand, the indicial radius computations at the beginning of this chapter show that these are precisely the values of MATH for which these operators have positive indicial radius, so these are the only values of MATH for which the operators are NAME. The result for the NAME Laplacian follows similarly from REF in the case of scalar forms. The claims about the essential spectrum follow immediately from the NAME results: Since each of these operators is self-adjoint on MATH, its spectrum is contained in MATH, and a real number MATH is in the essential spectrum if and only if MATH is not NAME. For the vector Laplacian MATH, NAME proved in CITE that the following asymptotic estimate holds: MATH . It follows from REF that MATH is NAME on MATH, and then the rest of the claims follow from REF . |
math/0105046 | If MATH, then MATH . By the divergence theorem and the fact that MATH, MATH . Substituting this into REF, we obtain MATH . Given MATH, choose MATH small enough that MATH whenever MATH is smooth and compactly supported in MATH, and such that MATH on MATH. Applying REF to MATH and using REF, we obtain MATH which implies REF as long as MATH. |
math/0105046 | The proof is by induction on MATH. For MATH, consider first the case MATH, in which case the claim is MATH . We will prove this claim by induction on MATH. Observe that MATH and MATH are in MATH by REF . Thus to prove REF, it suffices to show that the difference between the two terms is MATH. For MATH, the only value of MATH that satisfies the hypotheses is MATH. Since MATH is a coordinate function, we consider separately the cases MATH and MATH. When MATH the right-hand side of REF is zero. On the other hand, the left-hand side is in MATH and vanishes on MATH (since MATH vanishes on MATH and MATH is tangent to MATH), so it is MATH. When MATH, MATH by REF, and by definition of the derivative, MATH . It follows that MATH and vanishes on MATH, so it too is MATH. This proves REF in the MATH case. Now let MATH, and suppose REF holds for all smaller values of MATH. Observe that MATH by REF. Therefore, by the chain rule and the inductive hypothesis, MATH . Finally, for MATH, the product rule gives MATH . This completes the proof of the MATH step. Now suppose the proposition is true for some MATH. For any MATH-tuple MATH, beginning with the induction hypothesis in the form MATH we apply the chain rule to obtain MATH where in the last line we have used the induction hypothesis again to evaluate MATH. |
math/0105046 | Write MATH in background coordinates as MATH where MATH is vector-valued and the coefficient functions MATH are matrix-valued. If MATH, then an easy computation shows that MATH which implies that MATH . On the other hand, REF shows that MATH which proves the result. |
math/0105046 | Define a nonlinear operator MATH by MATH . It is clear that REF is equivalent to MATH. It follows from REF that MATH has the following expression in background coordinates: MATH where MATH is a universal polynomial in the components of MATH, MATH, their inverses, and their coordinate derivatives of order less than or equal to MATH. It follows that MATH takes its values in MATH and is a smooth map of NAME spaces. We will recursively construct a sequence of metrics MATH satisfying MATH. In fact, we will prove a bit more, namely that MATH . Begin with MATH. It follows from REF that MATH . Since the intersection of REF with MATH is MATH we have REF in the MATH case. Assume by induction that for some MATH, MATH, we have constructed MATH satisfying the analogue of REF. Then letting MATH, we see from REF that MATH is a MATH section of MATH. If MATH, the same argument as in the proof of CITE shows that MATH . By REF , we have MATH . Because the indicial radius of MATH is MATH, MATH is invertible provided MATH. Thus as long as MATH, there is a unique MATH tensor field MATH along MATH such that MATH. By REF , there is a tensor field MATH such that MATH. Inserting this back into REF, we conclude that MATH, and thus actually satisfies REF. Setting MATH completes the inductive step. After the MATH step, we set MATH, and REF is satisfied. The smoothness of the operator MATH is immediate from the remark following REF above. |
math/0105046 | The proof follows closely that of REF. We define an open subset MATH by MATH . Define a map MATH by MATH where MATH is defined in REF and MATH in REF . It follows just as in CITE that MATH is a smooth map of NAME spaces. Since MATH and MATH is an NAME metric, MATH. As is shown in CITE, the linearization of MATH about MATH is the linear map MATH from MATH to MATH given by MATH where MATH . Because MATH preserves the splitting MATH, to show that it satisfies the hypotheses of REF , we need to show that it has trivial MATH kernel on each of these bundles separately. Since MATH, the action of MATH on sections of MATH is just given by the scalar Laplacian: MATH . Since MATH, it follows easily from integration by parts that MATH acting on scalar functions has trivial MATH kernel. Thus the assumption that MATH has trivial MATH kernel on MATH is sufficient to allow us to apply the results of REF . In particular, MATH is an isomorphism for MATH, which is to say for MATH. This is true for MATH, so the linearization of MATH has a bounded inverse given by MATH . Therefore, by the NAME space inverse function theorem, there is a neighborhood of MATH on which MATH has a smooth inverse. In particular, for MATH sufficiently MATH close to MATH, there is a solution MATH to MATH. Putting MATH and MATH, we have MATH. Moreover, if the MATH neighborhood of MATH is sufficiently small, then MATH will be uniformly MATH close to MATH, and therefore MATH will have strictly negative NAME curvature. By REF, this implies that MATH is NAME. Note that MATH, which is contained in MATH by REF . Since MATH by construction, it follows that MATH is asymptotically hyperbolic of class MATH as claimed. It remains only to prove that MATH has trivial MATH kernel on MATH under the assumptions stated in REF . First suppose that MATH has nonpositive sectional curvature. A simple algebraic argument (see CITE) shows that if MATH is an NAME metric on a MATH-manifold with scalar curvature MATH, its NAME curvature operator MATH acting on trace-free symmetric MATH-tensors satisfies the following estimate at each point MATH: MATH where MATH is the maximum of the sectional curvatures of MATH at MATH. (In CITE, this is attributed to a hard-to-find REF paper of NAME; however, the argument was already given in REF by CITE.) Since the NAME assumption implies that MATH, we have MATH (see REF). Therefore, if MATH has sectional curvatures everywhere bounded above by MATH, for any MATH we have MATH . The same is true for MATH because MATH is dense in that space. If MATH, it follows immediately that MATH has trivial MATH kernel. On the other hand, if MATH, the sequence of inequalities above implies that if MATH is a solution to MATH, the nonnegative quantity MATH must vanish identically on MATH. Since the sectional curvatures of MATH approach MATH at infinity, there is some compact set MATH such that MATH for MATH, and then REF implies that MATH on MATH. Since MATH is a NAME operator, it satisfies the weak unique continuation property CITE, and therefore MATH is identically zero. Finally, suppose that the conformal infinity MATH has nonnegative NAME invariant. Then the result of CITE shows that the Laplacian satisfies the following MATH estimate for smooth, compactly supported scalar functions MATH: MATH . (The proof in CITE required MATH to have a MATH conformal compactification. However, using REF, it is easy to reduce that to MATH. See also CITE for a different proof.) By REF , therefore, the same is true when MATH is a smooth, compactly supported tensor field, and by continuity for all MATH. Suppose MATH has sectional curvatures bounded above by MATH. Then REF gives MATH . If MATH is a solution to MATH, therefore, MATH . It follows as before that the nonnegative function MATH must be identically zero, and since the operator MATH is positive definite outside a compact set, MATH must be identically zero by analytic continuation. |
math/0105048 | If the NAME superalgebra MATH is nilpotent the existence of MATH such that MATH and MATH is obvious. For the converse, assume that there exist MATH such that MATH and MATH, then every operator MATH with MATH is nilpotent. Let MATH, as MATH is an element of MATH, then MATH is nilpotent. This implies that MATH is nilpotent for every MATH. By NAME 's theorem for NAME superalgebras CITE, this implies that MATH is nilpotent NAME superalgebra. |
math/0105048 | Using REF , for every filiform lie superalgebra we have an adapted base MATH such that the product of MATH is given by : MATH where MATH for every vector MATH. This product satisfies the NAME identity MATH . Let MATH be two vectors of the adapted base MATH of MATH. We have MATH because MATH. The relation REF becomes : MATH . Also we have for every MATH : MATH . As the superalgebra is nilpotent, MATH shows : MATH for all MATH, where MATH is the graded NAME bracket. This implies that MATH. The filtrations : MATH associated to the graduations MATH and MATH shows that MATH. From CITE, we also have MATH . |
math/0105048 | It is clear that MATH can be decomposed into a sum of three homogeneous maps : MATH . As MATH, we have : MATH where MATH is an even element and MATH an odd element of MATH. This prove that MATH is a cocycle of the filiform NAME algebra MATH. As MATH is nilpotent, MATH has no component on MATH. This implies that MATH and MATH . This prove that both maps MATH and MATH are cocycles. |
math/0105048 | Let MATH be a cocycle such that : MATH and MATH must satisfy MATH then by induction on MATH and MATH we prove that MATH for MATH. |
math/0105048 | Let MATH be a cocycle of MATH, such that MATH for MATH. We can prove by induction on MATH that if MATH for MATH then MATH. We can assume that MATH, if not, we consider the cocycle MATH such that MATH. It is easy to see that there exists a linear combination of MATH such that MATH satisfies MATH. Using the previous paragraph, we have that MATH we deduce that MATH, and if MATH was not zero, MATH will be MATH this prove that MATH and MATH are generator. As they are linearly independent, we have a base. |
math/0105048 | Let us prove that MATH for every MATH. For MATH we have : MATH . Suppose that the relation is true up to MATH. For MATH we have : MATH then MATH for all integer MATH and MATH. |
math/0105048 | Let MATH be a symmetric bilinear map. It is easy to see that there exists coefficient MATH such that MATH for MATH, where MATH for MATH. Using REF we deduce that this equality vanishes for every pair MATH. This proves that MATH . Using the fact that MATH and MATH if MATH, the family MATH is free. |
math/0105048 | A cocycle MATH satisfies the two relations REF . A consequence of this is that MATH. We deduce that MATH. To satisfy relation REF , MATH has to satisfy the relation MATH for MATH. Such a map MATH does not have a component on MATH in its image, hence MATH for MATH. This prove that relation REF is satisfied and that every map MATH satisfying REF is a cocycle. |
math/0105048 | Using REF every cocycle of MATH is given by : MATH with MATH. We can write MATH like : MATH where MATH has a component on MATH and MATH does not have any component on MATH. As MATH is a cocycle, we have : MATH . Let us consider the component on MATH, we have : MATH . This proves that MATH is a cocycle, as MATH. Using the cocycle MATH, we prove by induction that MATH is given by a linear combination of the cocycles MATH. |
math/0105048 | Let be MATH and MATH be a non zero cocycle. If we can decompose MATH, we have the smallest cocycle, if not we choose the smallest integer MATH, MATH such that MATH . If MATH is indecomposable, we stop. If not, we have MATH with MATH, this sequence is increasing and has an upper bound, therefore it exists MATH such that MATH is indecomposable. To proof the uniqueness, suppose that MATH and MATH are smallest. Then MATH is a cocycle. We have MATH with MATH the smallest integer MATH such that MATH is in MATH. As MATH is the smallest, we must have MATH and then MATH. This proves the uniqueness. |
math/0105048 | Let MATH be such that MATH. The cocycles MATH span MATH because every cocycle MATH can be written as a linear sum of MATH, MATH. Let be MATH such that : MATH . Note that : MATH . We have for MATH : MATH then MATH for MATH. By induction on MATH, every coefficient vanishes, and the MATH are linearly independent. |
math/0105048 | Let MATH be a cocycle from the proposition. If MATH then MATH . Let MATH be such that MATH and MATH, we will proof that MATH is not a cocycle. We have MATH . If MATH is a cocycle, we have MATH . This implies MATH . Thus MATH, and MATH. We have MATH and MATH. This proofs that MATH cannot be a cocycle. |
math/0105048 | Let MATH be such that MATH, MATH and MATH such that MATH. Let's proof that there exists MATH such that MATH is a cocycle. We have MATH, then for MATH to be a cocycle, it must satisfy MATH equations given by REF . Suppose MATH, if MATH is vanishing then MATH as MATH. This is possible only if MATH. As MATH, there exists a value of MATH and MATH such that MATH. This proofs that in the MATH linear equations given by REF , every coefficient MATH appear. This proof that this system admits a solution. |
math/0105048 | Using REF , we can compute a lower bound of the number of non empty sets MATH. For each of this sets, there exists a unique cocycle MATH (see REF ) which is a vector of the base of MATH this is established in REF . |
math/0105049 | It is due to the fact that MATH with MATH for group law. |
math/0105053 | We have MATH . Now assume that MATH. Then there exists MATH satisfying the relation MATH . (Here we regard MATH, and MATH). This implies that MATH . In particular, we have MATH, or equivalently MATH. Under the notation in REF , we have MATH . This implies that MATH unless MATH. Here note that the condition MATH is equivalent to MATH, and similarly for MATH. The second assertion follows from this. We now assume that MATH and MATH. Then by REF , we have MATH . This shows the first assertion, and the lemma is proved. |
math/0105053 | By the previous remark, MATH permutes the factors MATH if MATH is not a multiple of MATH, and so MATH. Now assume that MATH. Then we can write MATH . Since MATH as MATH-modules, we have MATH. We note that MATH . In fact, MATH is an irreducible MATH-module isomorphic to MATH, and it is known from REF that MATH. MATH has a basis consisting of weight vectors, on which MATH acts diagonally. Here for a vector MATH, the weight of MATH is given by an elemenet MATH, where MATH is the number of MATH occuring in the expression of MATH. On the other hand, MATH acts on MATH, by permuting the weight vectors; if MATH is a weight vector with weight MATH, then MATH is also a weight vector with weight MATH. Hence, in order to compute MATH, we have only to consider the weight vectors in MATH whose weights are of the type MATH such that MATH for each MATH. Note that monomials MATH obtained as weights of MATH produce the NAME function MATH. Then the corresponding weight for MATH is given by MATH for MATH as above (the product is taken for MATH such that MATH). Hence MATH is obtained by picking up the monomials of this type from MATH, which coinicdes with MATH. This proves REF . To prove the lemma, it is enough to show that MATH for MATH. We show REF . Since MATH and MATH are both extensions of MATH, there exists a linear character MATH of the cyclic group MATH such that MATH. On the other hand, since MATH on MATH, we see that the action of MATH on MATH corresponds, under the isomorphism MATH of MATH-modules, to the action of MATH on MATH multiplied by MATH. This implies REF , and the lemma follows. |
math/0105053 | It is known that the ``character table" MATH of MATH is a non-singular matrix. (Here MATH runs over all the elements in MATH, and MATH runs over all the representatives of MATH-orbits in MATH). Since NAME functions form a basis of MATH, we see that MATH is also a basis by REF . In particular, they are all distinct. This proves REF . The second statement is then immediate from REF . |
math/0105053 | First we show REF . We fix MATH and put MATH as before. Let us consider the expansion of MATH by making use of the first equality of REF . By applying the operator MATH on both sides of this expansion, we have MATH where MATH runs over MATH-partitions of any size. Take MATH. In view of REF , we may only consider MATH such that MATH in the sum REF , where MATH. By making use of the formula for MATH which is similar to REF , we have MATH . Now there exists MATH such that MATH. Then MATH coinicdes with MATH. Moreover the action of MATH on MATH is compatible with that on MATH. Then the last formula of REF can be written as MATH . It follows that MATH where MATH runs over all the elements in MATH. This shows the first equality of REF . The second equality is shown similarly. Next we show REF . We consider the expansion of MATH by REF . By applying MATH on this equality, together with REF , we have MATH where MATH runs over MATH-partitions of any size such that MATH (see the discussion in REF). MATH is the function defined similar to REF for a MATH-partition MATH, by replacing MATH by MATH. We fix MATH as above. Then there exists MATH such that MATH. Hence MATH satsifies REF , and MATH. Let MATH be the number of all the pairs MATH for a fixed MATH. Then by REF , we have MATH where the sum is taken over all the pairs MATH for a fixed MATH. Now using the explicit description of MATH in REF and subsequent parts, one can check that MATH . Substituting REF into REF , together with REF , we have MATH where MATH runs over all the elements in MATH. This implies REF , and the proposition is proved. |
math/0105053 | By substituting MATH into REF , we have MATH where MATH for MATH. On the other hand, starting from the expansion of MATH by means of NAME functions MATH as given in REF , by using a similar argument as in the proof of REF , one can show that MATH . It follows that one can define a hermitian form on MATH satisfying the properties that MATH (MATH), and that MATH (MATH). Hence the seaquiliner form on MATH is reduced to the hermitian form on MATH by substituting MATH. Now the arguments in the proof of REF can be applied to our situation, and one can show that there exist unique functions satisfying REF in the proposition, respectively. So, we have only to show that MATH satisfy the properties in the proposition. First we note that MATH unless MATH by REF . It follows from REF that we have MATH unless MATH. This shows the proerty REF . Next we shall show REF . Clearly REF is equivalent to the statement that MATH for any MATH, where MATH unless MATH and MATH. We consider REF for a fixed MATH with respect to the MATH sign. Take MATH such that MATH and MATH. Taking the scalar product with MATH on both sides of REF , we have MATH . When MATH runs over all the elements in a fixed similarity class, REF can be regarded as a system of equations with unknown variables MATH. As in the arguments in the first part of the proof (compare REF ), the matrix MATH is non-singular. Hence REF has a unique solution MATH. Put MATH using thus determined MATH for MATH and MATH. Then clearly we have MATH . On the other hand, it follows from REF that we have MATH . This implies, in view of REF , that MATH . Now REF implies that MATH, and REF is proved for MATH. The proof for MATH is similar. Finally, we show REF . By REF , we have MATH . This implies, by REF , that MATH. The formula MATH is shown similarly. |
math/0105053 | REF follows from REF by a standard argument, (for example, [M, I, REF]). We show REF . We write MATH. Since MATH, REF implies that the transition matrix between two bases MATH and MATH is a block diagonal matrix with respect to the partition by similarity classes. Hence its inverse matrix is also block diagonal, that is, MATH is written as MATH, where MATH unless MATH. Substituting this into the second formula in REF , we obtain REF . Thus the corollary is proved. |
math/0105053 | Let MATH be the right hand side of REF . We shall compute MATH. Let MATH be the element in MATH corresponding to MATH. Since MATH is the character table of MATH, we have MATH by REF , where MATH is the diagonal matrix with diagonal entries MATH. It follows that MATH . Now it is known that if MATH, we have MATH . Hence by REF , MATH. In particular, the MATH-entry of MATH is equal to MATH . Therefore MATH and the theorem follows. |
math/0105053 | By using the relation REF for NAME functions with respect to the variables MATH, we can exprese MATH in terms of a linear combination of MATH with MATH. Then by applying MATH on both sides, we have MATH where MATH, and MATH in the right hand side runs over all MATH such that MATH and that MATH. Hence if we choose a representative MATH from each MATH-orbit, the above formula turns out to be MATH . On the other hand, by using REF , we can express MATH in terms of a linear combination of MATH. In particular, we have MATH fpr any MATH such that MATH. Now by making use of REF and a similar formula for MATH, we can rewrite the above formula as MATH for any MATH such that MATH. Note that the left hand side is understood to be REF in the case where MATH is not a multiple of MATH. Since MATH are linearly independent in MATH, by comparing with REF , we have MATH for each MATH. Now, REF holds for any MATH and any MATH such that MATH. Hence it can be translated to a matrix equation MATH . This proves the proposition. |
math/0105054 | The proof is a simple variant of an unpublished trick due to NAME. Let MATH be the set of free lozenge tilings of the region MATH, where MATH consists of an equilateral triangle of side MATH (recall that a lozenge has side length MATH). Then MATH is partitioned into equivalence classes, two tilings being equivalent if they have the same set of tiles which protrude beyond the boundary. The number of equivalence classes is less than MATH for some constant MATH, so some equivalence class MATH must satisfy MATH . Reflect the region MATH along one of its edges MATH. Given any two tilings MATH there is a tiling of the union of MATH and its reflection, where the MATH is filled with MATH and its reflection is filled with the reflection of MATH. This is because the tiles which protrude across the boundary edge MATH are fixed under this reflection. Take the union of MATH reflections of MATH so as to make the parallelogram of REF . Given any MATH elements of MATH, we can fill them in the MATH triangles to make a tiling of this whole parallelogram. Furthermore, the configurations of tiles protruding beyond opposite boundaries of this parallelogram are translates of each other. As a consequence we can glue opposite sides of the parallelogram together to make a tiling of the MATH by MATH torus (tilings of which correspond to perfect matchings of MATH). We therefore have (where MATH is the set of tilings of MATH) MATH . This implies that MATH which completes the proof. |
math/0105054 | This follows from REF and the convergence argument above. |
math/0105054 | We show for MATH that MATH; by the dihedral MATH-fold symmetry of MATH the result will then follow. From REF we have MATH where we used the substitution MATH. |
math/0105054 | This follows from the definition of the height function. On the vertical path of faces from MATH to MATH, the height increases by MATH for edge of MATH not present in the matching, and decreases by MATH for every edge present in the matching. So the height difference is MATH. |
math/0105054 | The proof follows from REF , the fact that MATH (see the comments after REF ), and a convergence argument analogous to that preceding REF . |
math/0105055 | Using REF we calculate MATH . Thus, we obtain MATH . Let MATH be any point and let MATH be an orthonormal frame in a neighbourhood of MATH with MATH. Then, we have at the point MATH . Inserting this into MATH we obtain REF . |
math/0105055 | It remains to show that, for the first eigenvalue MATH of MATH, in REF equality can not occur. Let us assume the counterpart. Then any eigenspinor MATH corresponding to MATH satisfies the equation MATH with the optimal parameter MATH. By REF , MATH implies MATH . Moreover, the limiting case of REF implies that in the inequality for MATH we have an equality, that is, MATH . Hence, MATH is a space of constant sectional curvature. In particular, MATH is NAME MATH and MATH implies MATH. Consequently, the NAME tensor vanishes and MATH is flat, a contradiction. |
math/0105058 | Let MATH be an arbitrary object. It appears that for an unknown object (see REF ) we may take any object MATH and for morphism MATH it is necessary to take a zero-morphism MATH. Indeed, let MATH be an arbitrary morphism. It is clear that zero-morphism MATH is the morphism we seek, that is MATH, which follows from REF . |
math/0105058 | Let MATH be any object. Let's consider the initial object MATH of category MATH. We denote by MATH the single morphism from the object MATH to the object MATH. Now it is not difficult to note that the object MATH and the morphism MATH are unknowns. Indeed: let MATH be any object and MATH be any morphism of category MATH. It is clear that the single morphism MATH satisfies a condition MATH. |
math/0105058 | Let condition MATH be satisfied. We must prove that MATH is movable. Let's consider spectrum MATH which is associated with the topological space MATH. Let MATH be any element and MATH be the natural projection REF . For the map MATH let MATH be a MATH-space, and MATH and MATH be homotopy classes satsifying the condition MATH (their existence follows from the condition MATH). Since MATH , there exists MATH, MATH and MATH such that MATH . It is not difficult to verify that MATH . Indeed: MATH . From the equality REF and the definition of "association" we infer the existence of an index MATH, MATH for which MATH . The obtained index MATH satisfies the condition of association of the spectrum MATH with the topological space MATH. Indeed, let MATH, MATH be any element. For the maps MATH and MATH (with the condition MATH) there exist a map MATH , which satisfies to the equality MATH . (see the condition MATH ). Now it is easy to see that the map MATH is the map we seek, i. e. the next condition is satisfied: MATH . Indeed: MATH . MATH REF . MATH . Now we must prove the converse. Let MATH be movable topological space and MATH some associated with MATH inverse spectrum. Let's prove that the condition MATH is satisfied. To this end, consider any homotopy class MATH REF . From the association of the spectrum MATH with the space MATH it follows that there exist an index MATH and homotopy class MATH such that MATH . For the index MATH let's consider an index MATH, MATH, which satisfies the condition of movability of the spectrum MATH. From REF we get MATH . Now let's prove that MATH-space MATH, homotopy classes MATH and MATH satisfy the condition MATH. Indeed, let MATH be any MATH-space, MATH and MATH homotopy classes, which satisfy the condition MATH . For the homotopy class MATH there exist an index MATH, MATH and homotopy class MATH that MATH . It is clear that MATH . Therefore, according to the definition of association, we can fined an index MATH, MATH such that MATH . By the movability of spectrum MATH we can select the homotopy class MATH with the condition MATH . Let's define MATH. It is remains to note that the homotopy class MATH satisfies the condition MATH . MATH REF . Indeed: MATH . |
math/0105059 | By REF , we can pick a finite set of generators: MATH for MATH where MATH are connected. Let MATH denote the collection of all targeted morphisms from the empty set to MATH. For any ring MATH, let MATH denote the MATH-module of formal linear combinations of elements of MATH with coefficients in MATH, and with only finitely many non-zero coefficients. Define MATH on generators by MATH . MATH takes values in MATH, by the almost MATH-integrality property. We also have a surjective MATH-module map from MATH which sends MATH to MATH . MATH factors through MATH, and induces a surjection between MATH and the image of MATH. This surjection is injective by REF . Thus we have an isomorphism between MATH and the image of MATH . But a submodule of MATH is necessarily finitely generated and projective CITE. This proves REF . Let MATH be the kernel of MATH and let MATH be the kernel of the canonical map MATH which sends MATH to MATH. Then we have short exact sequences: MATH . MATH . We will show that MATH . As MATH is projective, the sequence REF splits CITE, and we may obtain an exact sequence by tensoring REF with MATH: MATH . Comparing the exact sequences REF and using REF , we obtain the natural isomorphism of REF . We now show that REF holds. By the proof of REF above, MATH is the kernel of MATH. Similarly, MATH is isomorphic to the kernel of MATH . So we have exact sequences MATH . MATH . As MATH is projective, we may, as above, obtain an exact sequence by tensoring sequence REF with MATH: MATH . Comparing the exact sequences REF, we see that MATH. REF follows as the composition of two targeted morphisms is again targeted. |
math/0105059 | By CITE, MATH where MATH is some ideal in MATH . Thus MATH . It is easy to check that the corresponding basis for MATH has the claimed property. |
math/0105059 | There is an ideal MATH in MATH such that MATH CITE. So the characteristic polynomial of MATH lies in MATH . The characteristic polynomial of MATH is MATH times the characteristic polynomial of MATH . |
math/0105059 | MATH is the closure of an endomorphism in MATH . So MATH . This is a coefficient of the characteristic polynomial, and so, by REF , lies in MATH . |
math/0105059 | We have MATH so that MATH and MATH . Thus MATH and MATH . Choose MATH as in the proof of REF . Let MATH . MATH is then an endomorphism of a free MATH-module. Moreover MATH and MATH . But MATH where MATH denotes the list the eigenvalues of MATH with multiplicity. Each MATH is itself an algebraic integer as it satisfies a polynomial in MATH namely the characteristic polynomial of MATH . Moreover MATH . The result follows as MATH is MATH times an algebraic integer. |
math/0105059 | Recall that MATH and that MATH is a prime ideal. Using CITE, the map MATH (below) is surjective and its kernel is the infinite cyclic subgroup generated by MATH . It is easy to check that MATH (below) is surjective. The rest of the commutative diagram with exact rows and columns can be completed by diagram chasing. Here MATH denotes the trivial multiplicative group. MATH . |
math/0105059 | If MATH then MATH is isomorphic to MATH which is free. By REF and the preceding discussion, then MATH is a free MATH-module. |
math/0105059 | Let MATH denote MATH . Thus MATH if and only if MATH is one. Otherwise MATH is even. Note that MATH, so we can identify MATH with MATH. Consider the NAME sequence for MATH as the union of MATH and MATH . MATH . Let MATH denote the dimension of the cokernel of the first map, which is given by MATH . We have that MATH . The result now follows. |
math/0105059 | Let MATH and MATH be even morphisms in MATH . We wish to show that MATH is even. We may reduce to the case that MATH . Let MATH denote a simply connected REF-manifold with boundary MATH and signature MATH . Let MATH . From the long exact sequences of the pairs for MATH and MATH and from REF-manifold interpretation of the weight of a composition, we have that MATH . Since each MATH is even, by REF , we have that MATH . To show that MATH is even, it suffices, by REF , to show that MATH . Because MATH is connected and each component of MATH and MATH is simply connected, we conclude using a NAME sequence that MATH . Thus by NAME duality and the universal coefficients theorem, MATH . Therefore, using the long exact sequence for the pairs MATH . We have the NAME sequence for MATH as the union of MATH and MATH: MATH . Thus by the exactness of sequence REF MATH . Recall we also have that MATH . We can derive REF , by substituting REF , and REF into REF . |
math/0105059 | The first statement follows from REF-manifold interpretation of weights. The second statement holds because the parity change between MATH and MATH is the same as the parity change between MATH and MATH . |
math/0105059 | It suffices to prove this when MATH is connected. If we equip MATH with the weight MATH then MATH is even. So it is the closure of an even morphism MATH . So MATH is the closure of the even morphism MATH . It follows that MATH . The result then follows from REF . |
math/0105059 | It is enough to prove this for MATH connected. If MATH is even, then MATH . By REF , MATH is odd. So MATH . The second statement is proved similarly. |
math/0105059 | The pairing MATH has an adjoint: MATH . Take bases for MATH and MATH consisting of linearly combinations over MATH of MATH-vacuum states. Consider the matrix for MATH with respect to these bases. By REF , the entries of the matrix all lie in MATH . Let MATH denote the determinant of this matrix. MATH is a unit from MATH and MATH . As MATH we conclude that MATH . In other words, MATH is a unit in MATH . Let MATH and MATH denote the MATH submodules spanned by the chosen bases for MATH and MATH . We have that MATH . The restriction of MATH to MATH is an injective map MATH . But its further restriction to MATH is an isomorphism as it is given by a matrix over MATH whose determinant is a unit. It follows that MATH equals MATH and so MATH is free. |
math/0105059 | If MATH is isomorphic to MATH which is free. By REF and the preceding discussion, MATH is a free MATH-module. MATH is isomorphic to MATH and so is a free MATH-module. |
math/0105059 | First we consider the case when all the colors of MATH are even. Then we can work in a situation where MATH has the trace property. Assign MATH the weight MATH so that MATH is even. Applying REF with MATH, and MATH we obtain MATH or MATH . By REF , MATH . Now suppose that MATH. If MATH is even, then MATH . Thus MATH . If MATH then MATH differs from an integer times MATH by an element in MATH . Now suppose some colors of MATH are odd. We can replace MATH by a MATH-linear combination of banded links (colored one) without changing MATH . Next replace MATH by same linear combination of the inverse image of these links. This will not change MATH . This makes use of the idempotent property of the NAME idempotents. Next we recolor each link component MATH without changing the quantum invariants upstairs or downstairs CITE. The result now follows by the earlier case. |
math/0105061 | Straightforward. See REF for a full argument. |
math/0105061 | Because MATH and MATH we have four possible situations depending on the choice of the signs in the above relations. The choice of a plus sign in both relations translates in MATH and MATH which by REF and our hypothesis implies that MATH and MATH (contradiction). The same argument shows that the choice of a minus sign in both relations is impossible. Now, we can suppose that MATH and MATH, the other case being treated similarly. Using the minimal length properties of MATH and MATH we can write MATH which shows that MATH. Our conclusion follows from the uniqueness of MATH. |
math/0105061 | We can prove the statement for MATH with the same arguments as in REF . The remaining statement was essentially proved in REF. |
math/0105061 | Let MATH be any element of MATH. Using the intertwining relation REF we get MATH . Therefore, MATH is an element of MATH. A short computation shows that MATH therefore MATH acts as a constant on MATH. It is easy to see that our hypothesis implies that this constant is nonzero, showing that MATH and consequently MATH is an isomorphism. |
math/0105061 | The proof is very similar with the proof of the corresponding result REF in CITE. The only difference is that we have to use the fact that MATH is a set of representatives for the orbits of the affine action of MATH on MATH, and the fact that MATH is in MATH for MATH. From the proof also follows that an element in MATH is uniquely determined by the coefficient of MATH in MATH. |
math/0105061 | The result follows from the characterization REF . |
math/0105061 | An easy calculation as in REF shows that MATH for any MATH, hence MATH for all MATH. This implies that MATH is MATH-invariant, and so it must be a multiple of MATH. Moreover, MATH acts as identity on MATH . |
math/0105061 | When MATH the statement follows straightforward from REF and from REF . For the remaining case, using REF , all we need is to compute the coefficient of MATH in MATH which by REF can be shown to be MATH. |
math/0105061 | The statement is obvious for MATH. Now, we know from REF that if MATH we have MATH . Therefore, if MATH is a reduced decomposition MATH is also reduced. Henceforth, using the definition of MATH and MATH and the recursion REF our conclusion follows. |
math/0105061 | By REF we know that the irreducible representation of MATH with highest weight MATH occurs in the decomposition of MATH with multiplicity one. Let us denote by MATH the copy of the irreducible representation of MATH with highest weight MATH embedded in MATH and by MATH the irreducible representation of MATH with highest weight MATH. It is easy to see that MATH is the MATH-module generated by the weight space MATH , where MATH. From the fact that the space MATH is one dimensional and from MATH we deduce that MATH is the lowest weight space of MATH, and therefore MATH both being equal with the MATH-module generated by the weight space MATH. By the same argument the MATH-module MATH is generated by the one dimensional weight space MATH which also generates MATH as a MATH-module. Our conclusion follows. |
math/0105062 | Consider the long exact sequence of the pair MATH with coefficients in MATH. Since the local system MATH has trivial monodromy about the distinguished hyperplane MATH, the restriction of MATH to MATH is itself a local system, MATH. So the long exact sequence of the pair is of the form MATH . Thus it suffices to show that MATH for each MATH. Let MATH be a tubular neighborhood of MATH in MATH, where MATH is the distinguished hyperplane. The neighborhood MATH admits the structure of a trivial bundle over MATH with fiber MATH. Identify the zero section of this bundle with MATH. Then the complement of the zero section, MATH, may be identified with MATH. By excision, we have MATH. Furthermore, the triviality of the bundle MATH yields MATH. The restriction, MATH, of the local system MATH to MATH is necessarily trivial. Hence, we may compute the cohomology MATH using the NAME formula: MATH . Since MATH and the restriction of MATH to MATH is MATH, we have MATH, as was required. |
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