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math/9909042 | We have MATH where MATH is determined by REF , which in this case becomes MATH . The NAME expansion of MATH is determined inductively by differentiating this equation at MATH. Clearly MATH at MATH. Consider the determination of MATH resulting from differentiating REF an even number MATH times and setting MATH. The term MATH gets differentiated MATH times, so one of the two factors ends up differentiated an odd number of times, so by induction vanishes at MATH. Now MATH, so the MATH differentiations must be split between the three factors, so one of the factors must receive an odd number of differentiations. When an odd number of derivatives hits a MATH, the result again vanishes by induction. But by REF , so long as MATH, the odd derivatives of MATH vanish at MATH. |
math/9909042 | The special defining functions MATH and MATH associated to representative metrics MATH and MATH are related as in REF . We can solve REF for MATH to give MATH, where the expansion of MATH also has only even powers of MATH up through the MATH term. It is important to note that in this relation, the MATH still refers to the identification associated with MATH. Set MATH. Then MATH is equivalent to MATH, so MATH where we have used REF . For MATH odd this is MATH . Since MATH is even through terms of order MATH in MATH, it follows that this expression has no constant term as MATH. Similarly, when MATH is even, the MATH term in REF contributes MATH, so there is no MATH term as MATH. |
math/9909044 | Note that both sides of REF are zero unless MATH and MATH. Furthermore, denoting the identity REF by MATH, it enjoys the symmetry MATH. Hence we may assume MATH and MATH in the proof below. Throughout the proof we use modified MATH-binomials defined as MATH and zero otherwise. Note that MATH is zero if MATH unless MATH. Let us now show that on both sides of REF the MATH-binomials REF can be replaced by the modified MATH-binomials. Since MATH we find from REF that MATH and MATH if MATH so that MATH and MATH in REF can be replaced by the modified MATH-binomials MATH and MATH, respectively. The other MATH-binomials can be turned into modified MATH-binomials since the top entries are nonnegative by the conditions on the parameters. The proof of REF makes frequent use of the following identity which is a corollary of Sears' transformation formula for a balanced MATH series CITE MATH where MATH and the condition MATH applies. Since we need the Sears transform REF with negative entries in the MATH-binomials it is essential that REF is used here. (The above formula is not correct for all MATH with the use of REF). We start by shifting MATH, followed by MATH. This transforms the left-hand side of REF into MATH where the sum over MATH is restricted by MATH and the MATH-system is given by MATH . Since the MATH-system has become MATH-independent, only the first four MATH-binomials depend on the summation variable MATH. Hence we may apply REF with MATH, MATH, MATH, MATH, MATH, MATH and MATH to obtain MATH . Shifting MATH and MATH, which leaves the MATH-system REF and the restriction REF on the summation over MATH invariant, yields MATH where we have used the MATH-system to simplify the exponent of MATH. Shifting MATH one can apply REF with MATH, MATH, MATH, MATH, MATH, MATH and MATH, observing that MATH thanks to REF. This yields MATH . Shifting MATH and MATH, which again leaves the MATH-system REF and the restriction REF on the sum over MATH unchanged, leads to MATH . We now need the following lemma. For MATH, let MATH with MATH-system REF and MATH. Then MATH for MATH. Change MATH and apply REF with MATH, MATH, MATH, MATH, MATH, MATH and MATH, observing that MATH by summing up the first MATH components of the MATH-system REF. This leads to MATH . We now carry out the transformations MATH and MATH, which leave the MATH-system unchanged. (Here MATH.) Using MATH and MATH, as well as the MATH-system, yields MATH transforming REF into MATH as desired. REF corresponds to MATH and we can thus use the above lemma to replace it with MATH. Since MATH, the last MATH-binomial in MATH is REF and we can perform the sum over MATH using the MATH-Saalschütz sum, which is the special case MATH of the Sears transformation REF. (When MATH, the only nonvanishing term on the right-hand side of REF corresponds to MATH.) Specifically, we take MATH, replace MATH by MATH and apply REF with the same choice of parameters as in the proof of REF but with MATH, MATH and MATH. Then we get MATH . All that remains to be done is to clean up the above expression. Introduce a new variable MATH through its components as follows MATH for MATH even/odd. Also define MATH through the MATH-system REF Eliminating MATH and MATH from REF in favour of MATH and MATH, we finally get the right-hand side of REF. We also note that MATH yields MATH so that the restriction REF on the sum over MATH translates into the restriction MATH for the sum over MATH as it should. |
math/9909046 | Let MATH and MATH be as in the statement of the theorem. By the form of the automorphisms of MATH it is obvious that MATH is a rank-one projection for every rank-one projection MATH. We show that MATH is homogeneous on the set of all rank-one operators. First, let MATH be a rank-one projection and MATH. Since MATH is also rank-one, we compute MATH with some MATH. On the other hand, by the local property of MATH, it follows that there are unitary operators MATH such that MATH that is, MATH is equal to MATH times a rank-one projection. Comparing this to REF we obtain MATH and hence MATH. Now, if MATH is an arbitrary rank-one operator, then choosing a rank-one projection MATH for which MATH we have MATH . Let MATH be any vector. Since, by the local property of MATH, MATH is a self-adjoint rank-one operator, it follows that there exists a vector MATH such that MATH. For arbitrary MATH we have MATH where MATH is a complex number (depending on MATH). Since MATH preserves the operator norm (this follows from the local property of MATH), we obtain that MATH that is, MATH. Suppose that MATH. We infer MATH from which it follows that MATH. Therefore, MATH. Similarly, in the case when MATH we compute MATH which gives us that MATH. Consequently, we have MATH . We now apply NAME 's unitary-antiunitary theorem. This celebrated theorem says that any function on MATH which preserves the absolute value of the inner product is of the form MATH, where MATH is a so-called phase function (that is, MATH is a complex valued function MATH of modulus REF) and MATH is an either linear or conjugate-linear isometry (see, for example, CITE, CITE, CITE as well as CITE for a new, algebraic approach to the result). We claim that in our case MATH is linear. To see this, suppose on the contrary that MATH is conjugate-linear. Taking into account that MATH (recall that MATH), in this case we find that MATH . Now, let MATH be such that MATH. We compute MATH . On the other hand, we have MATH . This implies that MATH . Since this obviously does not hold true for every possible choice of the vectors MATH, it follows that the operator MATH cannot be conjugate-linear. Therefore, we have a linear isometry MATH on MATH such that MATH. Similarly as in REF we get that MATH if MATH. In the opposite case choose a unit vector MATH such that MATH. We have MATH . Now, let MATH be a projection of rank MATH. Choose pairwise orthogonal rank-one projections MATH such that MATH. By the multiplicativity and the local property of MATH it follows that MATH are pairwise orthogonal rank-one projections. Since MATH we infer that MATH . By the equality of the ranks of the operators appearing on the two sides of REF, it follows that we have in fact equality in the above inequality, that is, MATH . Now, let MATH be an arbitrary finite rank operator. Let MATH be a finite rank projection such that MATH. If MATH are as above and MATH, then we see that MATH . If MATH is finite dimensional, then MATH is necessarily unitary and hence we are done in that case. So, in what follows let us suppose that MATH is infinite dimensional. From REF we see that on MATH, the map MATH can be represented as MATH . Clearly, on the whole ideal MATH, MATH can be written as MATH . Let MATH and consider an arbitrary finite rank projection MATH. From the equality MATH we obtain MATH, MATH. Since MATH was arbitrary, it follows that MATH and MATH. Similarly, from the equality MATH we infer that MATH. Consequently, our map MATH is of the form MATH where MATH is obviously multiplicative and it vanishes on the finite rank operators. Up till this point MATH has been an arbitrary ideal in MATH. Suppose now that MATH is proper, that is, MATH. By the separability of MATH, the elements of MATH are all compact operators. Let MATH be an arbitrary compact operator. Denote MATH the MATH-th MATH-number of MATH which is the MATH-th term in the decreasing sequence of the eigenvalues of the positive compact operator MATH, where each eigenvalue is counted according to its multiplicity. For a fixed MATH, denote MATH. It is well-known that MATH is a norm (somtimes called NAME norm) on MATH (see CITE). Therefore, we have MATH for any compact operators MATH. By the local property of MATH it now follows that MATH for every MATH. This gives us that the MATH-numbers of MATH are all zero and hence MATH. Consequently, we have MATH which yields MATH for every MATH. It remains to consider the case when MATH. The function MATH appearing in REF is a multiplicative map which vanishes on the set of all finite rank operators. Suppose that MATH. It is easily seen that MATH for any infinite rank projection MATH. Indeed, this follows from the fact that for any infinite rank projection MATH there is a coisometry MATH such that MATH. Choosing uncountably many infinite rank projections with the property that the product of any two of them has finite rank (see, for example, CITE) and taking their values under MATH we would get uncountably many pairwise orthogonal nonzero projections in MATH. Since this is a contradiction, we obtain MATH. So, just as in the case when MATH, we have MATH. But due to the local property of MATH we have MATH. Therefore, the isometry MATH is in fact unitary. This completes the proof of the theorem. |
math/9909046 | It follows from our theorem that there is an isometry MATH such that MATH. Since MATH, there exists an operator MATH with dense range. By the local property of MATH, MATH must also have dense range. This gives us that MATH is surjective. |
math/9909047 | First suppose that MATH is unital, that is, MATH. According to a well-known theorem of CITE every unital order automorphism of a MATH-algebra is a NAME *-automorphism. Thus MATH is a NAME *-automorphism of MATH. Since the closed NAME ideals of a MATH-algebra coincide with its closed associative ideals, following the arguments used in the proofs of CITE, we obtain the form REF with MATH. Otherwise, define MATH. We show that the unital map MATH defined by MATH is an order automorphism. To see this, we only have to prove that MATH is continuous. But this follows from the continuity of the inverse and the square-root operations. It is now trivial to complete the proof. |
math/9909047 | Let MATH be a linear map which is a local order automorphism of MATH. Clearly, we may suppose that MATH. We claim that in this case MATH is an algebra endomorphism of MATH. To see this, let MATH be fixed and consider the linear functional MATH. Since this is a positive linear functional, by the NAME representation theorem we have a regular probability measure MATH on the NAME sets of MATH such that MATH . Suppose that there are two disjoint closed sets MATH in MATH which are of positive measure. Then, by NAME 's lemma, we can choose continuous functions MATH with disjoint support such that MATH, MATH and MATH, MATH. Consider the function MATH. Clearly, its integral is a complex number with nonzero real and imaginary parts. On the other hand, from the local property of MATH it follows that MATH should be a positive scalar multiple of one of the values of MATH. Since the range of MATH lies in MATH, we arrive at a contradiction. This shows that either MATH or MATH. Since this holds for every pair of disjoint closed sets of MATH, by regularity we obtain that every NAME set in MATH has measure either REF or REF. This implies that MATH is a NAME measure. In fact, it follows that the integral (as a number) of every simple function is contained in the range of that function. Then, approximating continuous functions by simple ones, we find that the integral of every continuous function MATH belongs to its range. Therefore, MATH is a linear functional on MATH which sends REF to REF and has the property that MATH belongs to the spectrum of MATH. By the famous NAME theorem it follows that MATH is multiplicative and, therefore, a point-evaluation which gives us that its representing measure is a NAME measure. Hence we obtain that MATH is a unital endomorphism of MATH. But the form such morphisms of MATH is well-known. Namely, there is a continuous function MATH such that MATH . It remains to prove that MATH is bijective. The surjectivity of MATH follows from the injectivity of MATH. To the injectivity of MATH observe that using NAME 's lemma, by the first countability of MATH, we have a nonnegative function MATH on MATH which vanishes at exactly one point. If MATH was noninjective, we would obtain that MATH vanishes at more than one point. But this is a contradiction due to the local property of MATH. This completes the proof. |
math/9909047 | Let MATH be a linear map which is a local order automorphism. As before, we can suppose that MATH. By the proof of REF we obtain that there is a surjective function MATH such that MATH. We assert that MATH. Let MATH denote the space of all complex sequences converging to REF. Observe that under the identification of MATH and MATH, MATH can be considered as a subalgebra of MATH. Since the set of all isolated points of MATH is exactly MATH, every homeomorphism of MATH maps MATH onto MATH. Let MATH be defined by MATH. By the local property of MATH it follows that the set of all nonzeros of MATH is MATH. On the other hand, since MATH, we obtain that the set of all nonzeros of MATH is the preimage of the set of nonzeros of MATH under MATH. Consequently, we have MATH. Clearly, MATH can be considered as a map from MATH into itself. We obtain MATH . Here MATH is treated as a function from MATH into itself. Since MATH is injective, we obtain that MATH maps onto MATH. We are done if we prove that MATH is injective as well. To see this, observe that for any sequence MATH with entries all zero but one, MATH must have the same property. The proof is now complete. |
math/9909047 | Let MATH be a nontrivial subspace with the above property. Clearly, we can assume that MATH. Let MATH be arbitrary. Since MATH is a scalar multiple of a positive operator, it follows that for every MATH, the spectrum MATH of MATH lies on a straight line of the plane going through REF. This trivially implies that the spectrum of MATH consists of one element, which, by the normality of MATH, gives us that MATH is a scalar. |
math/9909047 | Let MATH be a linear map which is a local order automorphism of MATH. Clearly, we may assume that MATH. Fix MATH and consider the map MATH on MATH. By REF , this is a linear map whose values are lying in MATH. Since MATH, from REF it follows that the map MATH can be considered as a linear transformation on MATH into itself. We assert that it is an automorphism of MATH. By the local form of MATH, for every MATH there exists a function MATH whose values are invertible and positive, and a homeomorphism MATH such that MATH . Since the values of MATH are scalars, we easily obtain that MATH can be chosen to be scalar valued as well. So, the map MATH can be considered as a local order automorphism of MATH. By our assumption on MATH it follows that this map is an order automorphism of MATH. Since we have supposed that MATH, we infer that the above map is in fact an automorphism of MATH, that is, there exists a homeomorphism MATH for which MATH . Let us fix a point MATH and consider the map MATH on MATH. By the local form of MATH, the above map is a local order automorphism of MATH. By CITE, every local order automorphism of MATH is an order automorphism. Therefore, it follows that there is a positive invertible operator MATH and a NAME *-automorphism MATH of MATH such that MATH . Since MATH is unital, we infer that MATH which gives us that MATH. Hence, there is a function MATH such that MATH . Clearly, MATH is strongly continuous. We claim that MATH . To see this, pick MATH. Let MATH be positive and invertible. Consider the linear map MATH . The image of this map is a linear subspace in MATH containing the invertible operator MATH. So, by REF it follows that every member of the above linear subspace is a scalar multiple of MATH. Fixing MATH for a moment and considering any other positive invertible operator MATH, we have constants MATH such that MATH . By the additivity of MATH, in case MATH are linearly independent, we find that MATH. Clearly, this is the case also when MATH are linearly dependent. This yields that for every MATH and MATH, there is a constant MATH such that MATH holds true for every positive invertible operator MATH. Putting MATH we obtain MATH . This means that MATH holds for every MATH, MATH, and positive invertible MATH. Since every operator in MATH is a linear combination of positive invertible operators, it follows that REF holds true for every operator MATH. This gives us that MATH for every MATH and MATH. Since the set of all elements of the form MATH is dense in MATH, by the continuity of MATH we obtain MATH for every MATH. Up to this point, everything was done in order to be able to prove that MATH is surjective. This is now easy. In fact, we first verify that MATH is strongly continuous. Since every NAME *-automorphism of a MATH-algebra is an isometry, this follows from the equality MATH and from the strong continuity of MATH. To show the surjectivity of MATH, it is now enough to check that for every MATH, the function MATH belongs to MATH. But this follows from the inequality MATH using the continuity of MATH and the strong continuity of MATH. The proof is now complete. |
math/9909048 | The idea of the proof is very simple. First extend MATH to a linear map on MATH (this will be denoted by MATH) which preserves the rank and then apply a result on the form of linear rank preservers. So, the idea is obvious but we have to work hard to reach the desired conclusion. Let MATH be pairwise orthogonal rank-one projections. Let MATH. By the properties of MATH, MATH are pairwise orthogonal rank-one idempotents and MATH is a rank-MATH idempotent. Since we have MATH and MATH it follows that MATH. But the idempotents on both sides of the latter inequality have rank MATH which implies that MATH . Let MATH denote an arbitrary MATH-dimesional subspace of MATH. Consider the natural embedding MATH and for any MATH let MATH be defined by MATH. Taking REF into account we easily obtain that MATH is a measure on MATH. We assert that MATH is bounded on MATH. Indeed, suppose on the contrary that there is a sequence MATH of unit vectors in MATH such that MATH. Since MATH is finite dimensional, MATH has a convergent subsequence. We can assume without any loss of generality that this subsequence is the original sequence MATH. Let MATH. Then MATH is a unit vector and by the continuity of MATH we have MATH which is an obvious contradiction. So, for any MATH, MATH is a so-called MATH-bounded measure on MATH. By NAME 's theorem CITE this implies that, in case MATH, there exists a linear operator MATH on MATH such that MATH . Our aim now is to extend MATH to a linear transformation of MATH. Let MATH be unit vectors (the pairwise orthogonality of the MATH's is not assumed) and let MATH be real numbers. Define MATH . We have to check that MATH is well-defined. To see this, let MATH be unit vectors and MATH such that MATH . Let MATH be a finite dimensional subspace of MATH of dimension MATH which contains MATH. Let MATH be any vector. Let MATH denote the linear operator on MATH corresponding to MATH (see REF). We compute MATH . Since this holds true for every MATH, we obtain that MATH is well-defined. REF now clearly implies that MATH is a real-linear operator on the set of all self-adjoint finite-rank operators. Clearly, MATH sends projections to idempotents. It is now a standard argument to verify that the extension MATH of MATH defined by MATH for any self-adjoint operators MATH is a NAME homomorphism of MATH. See, for example, the proof of CITE. Since MATH is a locally matrix ring, it follows from a celebrated result of CITE that MATH can be written as MATH, where MATH is a homomorphism and MATH is an antihomomorphism. Let MATH be a rank-one projection. Since MATH is also rank-one, we obtain that one of the idempotents MATH is zero. Since MATH is a simple ring, it is now easy to see that this implies that either MATH or MATH is identically zero, that is, MATH is either a homomorphism or an antihomomorphism of MATH. In what follows we can assume without loss of generality that MATH is a homomorphism. We show that MATH preserves the rank. Let MATH be a rank-MATH operator. Then there is a rank-MATH projection MATH such that MATH. Thus, MATH which proves that MATH is of rank at most MATH. If MATH is any rank-MATH projection, then there are finite rank operators MATH such that MATH. Since MATH and the rank of MATH is MATH, it follows that the rank of MATH is at least MATH. Therefore, MATH is rank preserving. We now refer to NAME 's work CITE. It follows from the argument leading to CITE (which is in fact a standard 'preserver-argument' already) that there are linear operators MATH on MATH such that MATH is of the form MATH (recall that we have assumed that MATH is a homomorphism). We claim that MATH are bounded. This will follow from the following lemma. Let MATH be linear operators on MATH with the property that the map MATH is continuous on the unit ball of MATH. Then MATH are bounded. If MATH and MATH, then we have MATH . This implies that MATH which yields MATH . Consequently, the map MATH is continuous at any point different from REF. Now, let MATH and pick a nonzero vector MATH for which MATH (observe that if MATH, then there is nothing to prove). Using the polarization identity MATH we see that MATH which gives us that MATH is continuous at REF, that is, MATH is bounded. The boundedness of MATH is now obvious. To continue the proof of REF , we infer from REF that MATH for every unit vector MATH (MATH sends rank-one projections to idempotents). Clearly, this implies that MATH. Consequently MATH. We have MATH for every rank-one projection MATH. By the additivity property of MATH appearing in REF, it follows that MATH holds true for every finite-rank projection MATH as well. Denote MATH. Clearly, MATH. Let MATH be an arbitrary projection. Choose a monotone increasing sequence MATH of finite-rank projections which weakly converges to MATH. We compute MATH and MATH . So, MATH is an idempotent commuting with the range of MATH. Therefore, MATH can be written as MATH where the maps MATH and MATH are multiplicative. We see that MATH and thus MATH vanishes on the set of all finite-rank projections. This completes the proof of the theorem. |
math/9909048 | If we consider MATH only on MATH, then by REF it follows that MATH for every finite rank projection MATH, where MATH are either both bounded linear operators or both bounded conjugate-linear operators with MATH. In what follows we can suppose without loss of generality that MATH are linear. Let MATH be a rank-one operator. Then there is another rank-one operator MATH such that MATH. Since MATH preserves the rank, it follows from the equality MATH that MATH with some scalar function MATH. If MATH is a rank-one operator with MATH and MATH is the scalar function corresponding to MATH, then we have MATH which implies that MATH. If MATH is a rank-one operator and MATH, then we can choose a rank-one operator MATH such that MATH and MATH. This gives us that MATH. Therefore, the scalar function MATH does not depend on the rank-one operator MATH. In what follows this function will be denoted by MATH. It follows from the equality MATH that MATH is a continuous multiplicative function. We show that it is additive as well. Let MATH be orthogonal unit vectors. Since MATH is additive on the set of finite rank projections, we compute MATH . Multiplying by MATH from the left we can compute MATH . It follows that MATH, that is, MATH is additive. Therefore, MATH is a continuous ring endomorphism of MATH with MATH. It is well-known that this implies that MATH is either the identity or the conjugation on MATH. We show that in our case MATH is the identity. Suppose on the contrary that MATH. Let MATH be non-orthogonal unit vectors. Since MATH, we compute MATH . So, we have MATH . Now, let MATH be unit vectors for which MATH. We then have MATH . On the other hand, MATH . Comparing these two equalities we arrive at MATH . Since this equality obviously does not hold true for every possible choice of MATH, we obtain that MATH is really the identity. Now, the same argument that has led to REF shows that MATH if MATH are non-orthogonal unit vectors. If MATH are orthogonal, then choosing a unit vector MATH such that MATH and MATH we have MATH . Since MATH is the identity, we thus obtain MATH for every rank-one operator MATH. If MATH and MATH is a finite-rank projection such that MATH and MATH are pairwise orthogonal rank-one projections such that MATH, then it follows that MATH . Similarly as in REF in the proof of REF we see that the operator MATH is an idempotent commuting with the range of MATH and MATH. Therefore, MATH can be written as MATH where the maps MATH and MATH are multiplicative and MATH vanishes on the set of all finite rank operators. We claim that MATH is identically REF. Indeed, if MATH is not zero, then MATH. If MATH is a projection of infinite rank, then due to the fact that in that case there is a coisometry MATH such that MATH, it follows that MATH. Choosing an uncountable set of infinite rank projections in MATH with the property that the product of any two of them has finite rank (see the first part of the proof of CITE) and taking the values of those projections under MATH, we would obtain uncountably many pairwise orthogonal nonzero idempotents in MATH which contradicts the separability of MATH. This shows that MATH. So, MATH for every MATH. This completes the proof. |
math/9909048 | We prove that MATH preserves the rank of projections. First suppose that MATH for every finite rank projection MATH. Since MATH, just as in the proof of REF we see that MATH for every infinite rank projection MATH and then we arrive at a contradiction in the same way as there. So, let MATH be the smallest positive integer with the property that MATH whenever MATH is of rank MATH (observe that by the multiplicativity of MATH, we have MATH if MATH). We claim that the rank of MATH is REF for every such MATH. Indeed, let MATH be a rank-one projection and MATH be a rank-MATH projection such that MATH is of rank MATH. Then MATH and MATH are orthogonal and we have MATH. Since the corank of MATH is REF, this gives us that the rank of MATH is REF. We show that MATH. Suppose on the contrary that MATH. Let MATH be a projection of rank MATH. Similarly as just before, we can verify that the rank of MATH is at most REF. On the other hand there are rank-MATH projections MATH such that the product of any two of them is a rank-MATH projection. Consequently, MATH are orthogonal and MATH. Therefore, we have MATH. This gives us that MATH and hence MATH sends rank-one projections to rank-one idempotents. Let now MATH be a rank-MATH projection. Since MATH, MATH are orthogonal idempotents and MATH has corank MATH, we obtain that MATH has rank at most MATH. Now, if MATH are pairwise orthogonal rank-REF projections, then like in the proof of REF we see that MATH . Since the idempotent appearing on the left-hand side of this inequality has rank MATH and the one on the right-hand side has rank at most MATH, we infer that MATH . Therefore, MATH preserves the rank of projections. Similarly to the argument in the proof of REF before REF we get that MATH is rank-preserving. By REF we have the form REF of MATH. Since MATH, we also obtain MATH. |
math/9909048 | Let MATH. By the properties of MATH, MATH is a normal operator whose spectrum does not contain any scalar different from MATH. Therefore, MATH. This gives us that MATH is homogenous. We prove that for any orthogonal projections MATH we have MATH. Let MATH be of infinite rank such that MATH. Pick a scalar MATH. We have MATH. We distinguish three cases. First suppose that MATH. Since MATH preserves normality, this yields MATH. Taking powers, we obtain MATH . Using the continuity of MATH we have MATH. Since MATH, we get MATH. On the other hand, MATH are equivalent projections and it follows that MATH which is a contradiction. Next suppose that MATH. Then we have MATH. Taking powers again, we have MATH . Thus, we infer MATH which gives us that MATH, a contradiction. Consequently, we have MATH. This implies that MATH where MATH are nonzero projections such that MATH. We show that in this case MATH holds for every MATH. By what we have just proved, for an arbitrary MATH we can write MATH as MATH where MATH are orthogonal nonzero projections with MATH. Since MATH clearly preserves the commutativity, we get that MATH and MATH commute. Referring to the spectral theorem we obtain that MATH are pairwise commuting. Furthermore, since MATH it follows that MATH is of the form MATH. Because of the equality MATH and the fact that the spectrum of MATH is MATH we obtain that MATH. Therefore, MATH and MATH. Since MATH, it follows that MATH and MATH. So, we have MATH for every MATH. Sending MATH to REF, we get MATH. Since MATH, MATH preserves the inverse operation. This yields that MATH . By the homogenity of MATH we infer that MATH . If MATH, we arrive at MATH. Consequently, we have MATH . If MATH are projection such that MATH and MATH, then multiplying REF by MATH we arrive at MATH . Therefore, we have REF whenever MATH are orthogonal, either both infinite or both finite rank projections. If MATH is of finite rank and MATH is of infinite rank, then we can write MATH where MATH are orthogonal and they are of infinite rank. The argument leading to REF gives us that MATH and MATH. We then have MATH . Hence, MATH is additive on the set of all projections. Since MATH sends projections to projections, MATH is bounded on MATH. By a deep result due to CITE it follows that MATH can be extended to a bounded linear transformation MATH on MATH. Since MATH sends projections to projections and MATH is continuous, it is a standard argument to verify that MATH is a NAME *-homomorphism (once again, see the proof of CITE). We next refer to the proof of CITE. Similarly to the argument followed there, we obtain that there is a central projection MATH in the MATH-algebra generated by the range of MATH such that MATH is a *-homomorphism and MATH is a *-antihomomorphism. This gives us that MATH is the direct sum of the maps MATH and MATH where MATH are isometries with pairwise orthogonal ranges and MATH denotes the transpose with respect to a fixed orthonormal basis in MATH. Consequently, MATH can be represented as MATH . We show that the *-antihomomorphic part of MATH is in fact missing, and hence MATH is a *-homomorphism. Let MATH be pairwise orthogonal projections and let MATH be scalars. We compute MATH . By the continuity of MATH and the spectral theorem we get that MATH holds for every normal operator MATH. Suppose that MATH do appear in REF. If MATH are self-adjoint operators such that MATH is normal, then by the multiplicativity of MATH we have MATH which yields that MATH. Let MATH be orthogonal unit vectors and MATH, MATH. It is trivial to check that MATH is normal but not self-adjoint. Therefore, we obtain that MATH, that is, MATH is a *-homomorphism. It is easy to see that every rank-one operator is the scalar multiple of the product of at most three rank-one projections. This gives us that MATH and MATH coincide on the rank-one operators. To complete the proof, let MATH be arbitrary. Choose a maximal set MATH of pairwise orthogonal rank-one projections in MATH. We compute MATH where we have used the weak continuity of MATH which clearly holds by REF. Finally, since MATH, we have MATH. This completes the proof. |
math/9909049 | Let first MATH be a projection and let MATH denote a modular orthonormal basis in the closed submodule MATH. By CITE we have MATH . Since MATH and MATH for MATH, we obtain MATH. Now, let MATH be a modular orthonormal set and denote MATH the closed submodule generated by this set. We show that MATH is a modular basis in MATH. Since this collection is a modular orthonormal family, if this was not maximal, then we could find a nonzero element MATH which is modular orthogonal to MATH, that is, MATH for every MATH. But this is a contradiction, since every element of MATH can be approximated by finite sums of the form MATH and hence we would obtain that MATH is modular orthogonal to itself. By the first part of the proof we obtain that the orthogonal projection onto MATH is equal to MATH, so this operator is a MATH-linear projection. |
math/9909049 | See CITE and its proof. |
math/9909049 | It is clear that MATH is a NAME algebra since it is the commutant of the set MATH in the full operator algebra over MATH as a NAME space. To show that MATH is a factor, it is sufficient to verify that the central projections in MATH are all trivial. Let MATH be a nonzero central projection. Let MATH be a modular unit vector in MATH. For any MATH we have MATH . This implies that MATH . The element MATH is a rank-one projection. Hence, every element of MATH is the sum of MATH-type elements and hence we obtain that MATH for every MATH. Thus MATH. So, MATH is a factor. We next prove that MATH is type I. Let MATH be a modular unit vector. Since MATH (see CITE), for any MATH we compute MATH where MATH is scalar such that MATH (the existence of such a scalar follows from the fact that MATH is a rank-one matrix). This shows that the projection MATH is abelian. So, every nonzero central projection in MATH contains a nonzero abelian projection which means that MATH is type I. Suppose that the modular dimension of MATH is greater than REF. To see that MATH is not isomorphic to MATH it is now enough to show that the linear dimension of MATH is greater then REF. Let MATH be a modular orthonormal set in MATH. Denote MATH. If MATH, then there are elements MATH such that MATH is independent for every MATH. It is easy to check that MATH is linearly independent. Therefore, the algebraic dimension of MATH is at least REF. If MATH, then the statement is trivial. |
math/9909049 | Let MATH be such that MATH. Define MATH . It is easy to see that MATH is an isometry and MATH. Let MATH be an arbitrary unit vector. Then MATH is a rank-one projection, so it is of the form MATH with some unit vector MATH. Since MATH we obtain that MATH is equal to MATH multiplied by a scalar of modulus REF. Therefore, MATH. Since this holds true for every unit vector MATH, by linearity we have the first assertion of the lemma. As for the second statement, we can apply a similar argument. Choosing MATH such that MATH, define MATH . One can verify that MATH is an antiisometry (that is, a conjugate-linear isometry), and then prove that MATH. Considering an antiunitary operator MATH for which MATH and defining MATH, we conclude the proof. |
math/9909049 | Clearly, we may assume that MATH. Taking traces on both sides of the equality MATH, we obtain MATH. On the other hand, we also have MATH . By NAME inequality MATH . So, there are equalities in the NAME inequalites MATH. This implies the assertion. |
math/9909049 | We define an orthoadditive projection-valued measure MATH on the lattice MATH of all MATH-linear projections as follows. If MATH is a modular orthonormal set, then let MATH . Observe that by REF, MATH is also modular orthonormal and, hence, by REF MATH belongs to MATH. We show that MATH is well-defined. Let MATH and MATH generate the same closed submodule MATH. We claim that the same holds true for MATH and MATH. Indeed, if MATH, then due to the fact that MATH is a modular basis in MATH we see that MATH. This implies that MATH which, by REF , gives us that MATH belongs to the closed submodule generated by MATH. It is now obvious that MATH is an orthoadditive MATH-valued measure on MATH. Let us suppose that the modular dimension of MATH is greater than REF. By REF we can apply a deep result of CITE. It states that every bounded finitely orthoadditive, NAME space valued measure on the set of all projections in a NAME algebra without a summand isomorphic to MATH can be uniquely extended to a bounded linear transformation defined on the whole algebra. Let MATH denote the transformation corresponding to MATH. Since it sends projections to projections, it is a standard argument to verify that MATH is a NAME *-endomorphism of MATH, that is, we have MATH, MATH (see, for example, the proof of CITE). We prove that MATH for every MATH. Let MATH, where MATH's are nonnegative real numbers and MATH's are pairwise orthogonal rank-one projections. Define MATH. We have MATH and MATH if MATH, that is, MATH is modular orthonormal. Then MATH since MATH implies that MATH (see REF ). So, we have MATH . But MATH if MATH. Indeed, we compute MATH for every MATH. Hence, MATH . So, the question is that whether the equality MATH holds true. Clearly, MATH is modular orthonormal. We compute MATH which, by REF , implies that MATH. We know that MATH. Similarly, MATH. Since MATH is a rank-one projection, we obtain that MATH is also rank-one. Furthermore, as MATH is a scalar multiple of MATH we can infer that MATH, where MATH is a scalar of modulus REF. Therefore, we have MATH . But similarly as above, for MATH we have MATH . Therefore MATH . But MATH. Indeed, since MATH is a modular unit vector, we have MATH (see CITE). Consequently, we obtain MATH and this was to be proved. So, we get MATH for every MATH. We assert that MATH is either a *-homomorphism or a *-antihomomorphism. By REF the minimal projections in MATH are exactly the operators of the form MATH, where MATH is a modular unit vector. Clearly, MATH sends minimal projections to minimal projections. By CITE the linear space generated by the minimal projections in MATH is MATH. Since MATH is a type I factor, it is isomorphic to the full operator algebra MATH on a NAME space MATH. Since *-isomorphisms preserve the minimal projections, MATH corresponds to the ideal MATH of all finite rank operators in MATH. Under this identification, we obtain a NAME *-homomorphism MATH on MATH corresponding to MATH which sends rank-one projections to rank-one projections. Since MATH is a local matrix algebra, by CITE we obtain that MATH is the sum of a *-homomorphism and a *-antihomomorphism. As MATH preserves the rank-one projections, from the simplicity of the ring MATH it follows that MATH is either a *-homomorphism or a *-antihomomorphism. Obviously, the same holds for MATH. Let us suppose that the modular dimension of MATH is greater than MATH. By CITE, there are vectors MATH such that MATH. The map MATH is either a *-homomorphism or a *-antihomomorphism. First consider this latter case. Referring to REF we have an operator MATH with MATH and a *-antiautomorphism MATH of MATH such that MATH. We define MATH where MATH are fixed and such that MATH. Clearly, MATH is a conjugate-linear operator. We have MATH that is, MATH. We compute MATH . Since MATH is conjugate-linear, by polarization we obtain MATH . We show that MATH which will imply MATH (MATH is the projection onto the range of MATH). Let MATH. In the previous part of the proof we have learnt that MATH is a linear combination of operators of the form MATH, where MATH's are modular unit vectors. We have MATH and, MATH being a *-antiautomorphism, MATH is a minimal projection. Therefore, MATH with some modular unit vector MATH and hence MATH. Now let MATH, where MATH and MATH. We have MATH . This gives us that MATH which shows that MATH. We next prove that MATH is surjective. Let MATH be arbitrary. Since MATH is a *-antiautomorphism of MATH, we can find an operator MATH such that MATH. We compute MATH . Since MATH was arbitrary, we have the surjectivity of MATH. We compute MATH . This gives us that MATH . Replacing MATH by MATH, we obtain MATH . Since every element of MATH is a linear combination of elements of the form MATH, it follows that MATH holds for every MATH. This implies that for every MATH, the matrices MATH and MATH are linearly dependent. It requires only elementary linear algebra to verify the following assertion. If MATH are vector spaces and MATH are linear operators such that for every MATH, the set MATH is linearly dependent, then either MATH and MATH have rank at most one or MATH is linearly dependent. Since the rank of the linear operator MATH is clearly greater than REF if MATH, we have a scalar MATH (depending only on MATH) such that MATH. This gives us that there is a function MATH such that MATH which results in MATH. It follows from the properties of MATH that MATH is of modulus REF. Finally, we have MATH . Since this must hold true for every MATH, it follows that for every rank-one matrix MATH we have MATH. But this is an obvious contradiction. Since we have started with assuming that MATH is a *-antihomomorphism, we thus obtain that it is in fact a *-homomorphism. Pushing the problem from MATH to the full operator algebra MATH, we see that there is a MATH-isometry MATH such that MATH. This gives us that MATH for every MATH. Similarly as before, this implies that MATH which yields MATH. We next compute MATH which gives us that MATH . Just as above, it follows that MATH is a scalar multiple of MATH. Therefore, there exists a MATH-isometry MATH and a phase-function MATH such that MATH . This completes the proof in the case when the modular dimension MATH of MATH is greater than MATH. We now treat the low dimensional cases, that is, when MATH. Let MATH denote the MATH-dimensional complex Euclidean space. Then MATH can be considered as a NAME MATH-module. Here, the module operation is MATH and the generalized inner product is defined by MATH. Clearly, the modular dimension of this module is REF. It now follows from the structure of our NAME MATH-modules (see, for example, CITE) that MATH is isomorphic to the MATH-fold direct sum of MATH with itself. So, we may assume that MATH. The definition of the module operation and that of the inner product on this direct sum is defined as follows MATH . Let us describe the elements of MATH. Since every element of MATH is a linear operator on the direct sum of vector spaces, it can represented by a matrix MATH where MATH's are linear operators acting on MATH. Now, MATH-linearity means that MATH holds for every MATH and MATH. It is easy to see that this is equivalent to MATH which means that MATH's are scalars. Consequently, MATH is isomorphic to MATH. Suppose that MATH. If MATH is any vector in MATH, then let MATH denote the element of MATH whose coordinates are all REF except for the MATH-th one which is MATH. Fix a unit vector MATH. We have MATH . From REF we infer that for every MATH, there is a scalar MATH such that MATH. Clearly, the columns of the matrix MATH are unit vectors. Since MATH for MATH, it follows that the columns of our matrix are pairwise orthogonal as well. So MATH is a unitary matrix and hence it defines a MATH-unitary operator MATH on MATH. Considering MATH instead of MATH, we can assume that MATH is equal to MATH for every MATH. If MATH is any vector in MATH, then considering the equality MATH we obtain MATH with some scalars MATH of modulus REF. We claim that all the MATH's are equal. Fix a MATH whose coordinates are pairwise orthogonal unit vectors in MATH (recall that MATH). It is apparent that if we multiply MATH from the left by a MATH-unitary operator whose matrix is diagonal, then the so obtained transformation still has REF . So we may assume that MATH. Let MATH be arbitrary. We have MATH . This implies that MATH which gives that MATH . So, if MATH, then we have MATH. Suppose now that MATH but MATH. Let MATH be any nonzero vector and consider MATH. By what we have just proved, it follows that MATH is a scalar multiple of MATH. We compute MATH which clearly gives us that MATH. Therefore, we obtain that for any vector MATH, MATH is equal to MATH multiplied by a complex number of modulus REF. The assertion of the theorem now follows for the case MATH. Finally, suppose that MATH, which means that MATH. Our problem is to describe those maps MATH for which MATH. But this equality clearly implies that MATH is equal to MATH multiplied by a scalar of modulus REF. The proof of the theorem is now complete. |
math/9909052 | Let us fix a prime MATH and consider the MATH-adic NAME module MATH of MATH. Let MATH be the corresponding MATH-vector space of dimension MATH. There is a canonical embedding MATH and dimension arguments imply that this embedding is an isomorphism. In particular, MATH is isomorphic to the matrix algebra of size MATH over MATH. Since the center of the matrix algebra over MATH has dimension MATH over MATH, the center of MATH has dimension MATH over MATH and therefore coincides with MATH. This implies that MATH is a central simple MATH-algebra of dimension MATH. Hence, there exists a simple abelian variety MATH over MATH and a positive integer MATH such that MATH is isogenous to MATH over MATH. This implies that MATH is a division algebra over MATH and MATH is isomorphic to the matrix algebra of size MATH over MATH. In particular, MATH . Since the center of MATH coincides with MATH, the center of MATH also coincides with MATH. It follows from NAME 's classification REF that either MATH or MATH is a definite quaternion algebra over MATH. If MATH then MATH has dimension MATH. This implies that MATH is a quaternion MATH-algebra and therefore MATH . On the other hand, MATH. This implies that MATH, that is, MATH and MATH is an elliptic curve. Since MATH is the quaternion algebra, MATH is a supersingular elliptic curve and MATH. Since MATH is isogenous to MATH, it is a supersingular abelian variety. |
math/9909052 | It suffices to check that MATH. The MATH-invariant splitting MATH implies that MATH . This implies that MATH. Since MATH acts doubly transitively on MATH, we have MATH (REF on p. REF). This implies that MATH. |
math/9909052 | Let us prove first that MATH is irreducible. Let MATH be a non-zero MATH-stable subspace in MATH. Let MATH be a non-empty subset of MATH with smallest possible cardinality. Since MATH is odd, MATH. If MATH consists of MATH elements then we are done, because MATH acts doubly transitively on MATH and each subset in MATH of even cardinality could be presented as a symmetric difference (disjoint union) of MATH-element sets. So, assume that MATH consists of at least MATH elements. Pick elements MATH and MATH. Then there is an even permutation MATH such that MATH. Clearly, the symmetric difference MATH has two elements less than MATH which contradicts the choice of MATH. This proves the irreducibility of MATH. It follows from REF that MATH is absolute irreducible. |
math/9909052 | Clearly, MATH is a faithful MATH-module and MATH . CASE: MATH is a semisimple MATH-module. Indeed, let MATH be a simple MATH-submodule. Then MATH is a non-zero MATH-stable subspace in MATH and therefore must coincide with MATH. On the other hand, each MATH is also a MATH-submodule in MATH, because MATH. In addition, if MATH is a MATH-submodule then MATH is a MATH-submodule in MATH, because MATH . Since MATH is simple, MATH or MATH. This implies that MATH is also simple. Hence MATH is a sum of simple MATH-modules and therefore is a semisimple MATH-module. CASE: The MATH-module MATH is isotypic. Indeed, let us split the semisimple MATH-module MATH into the direct sum MATH of its isotypic components. Dimension arguments imply that MATH. It follows easily from the arguments of the previous step that for each isotypic component MATH its image MATH is an isotypic MATH-submodule for each MATH and therefore is contained in some MATH. Similarly, MATH is an isotypic submodule obviously containing MATH. Since MATH is the isotypic component, MATH and therefore MATH. This means that MATH permutes the MATH; since MATH is MATH-simple, MATH permutes them transitively. This gives rise to the homomorphism MATH which must be trivial, since MATH is the simple group, whose order REF is greater than MATH order of MATH. This means that MATH for all MATH and MATH is isotypic. CASE: Since MATH is isotypic, there exist a simple MATH-module MATH and a positive integer MATH such that MATH. Clearly, MATH . It is also clear that MATH is isomorphic to the matrix algebra MATH of size MATH over MATH. Let us put MATH . Since MATH is simple, MATH is a finite division algebra of characteristic MATH. Therefore MATH is a finite field of characteristic MATH and MATH divides MATH. We have MATH. CLearly, MATH is stable under the adjoint action of MATH. This induces a homomorphism MATH . Since MATH is the center of MATH, it is stable under the action of MATH, that is, we get a homorphism MATH, which must be trivial, since MATH is the simple group and MATH is abelian. This implies that the center MATH of MATH commutes with MATH. Since MATH, we have MATH. This implies that MATH and MATH is trivial if and only if MATH. Since MATH, MATH is trivial if and only if MATH, that is, MATH is an absolutely simple MATH-module. It follows from NAME density theorem that MATH with MATH. This implies that MATH is trivial if and only if MATH, that is, MATH. The adjoint action of MATH on MATH gives rise to a homomorphism MATH . Clearly, MATH is trivial if and only if MATH commutes with MATH, that is, MATH. CASE: It follows from the previous step that we are done if either MATH or MATH is trivial. Since MATH , the desired triviality follows from the following Observation. Let MATH be a positive integer. If MATH then every homomorphism from MATH to MATH is trivial. Indeed, the cardinality of MATH is strictly less than MATH and MATH order of the simple group MATH (MATH). |
math/9909056 | From the condition MATH, we have MATH . Solving this by using REF , we get MATH . In order to have MATH, MATH, , MATH, the condition in the proposition must hold. |
math/9909056 | Let us compute the ratio REF explicitly. MATH . In order to calculate MATH, it is convenient first to evaluate MATH . The factors MATH and MATH are all nonzero for a generic string solution MATH. They are canceled in the ratio REF and we find MATH . |
math/9909056 | Owing to REF , we have MATH. Since MATH, it suffices to show MATH for MATH. From REF both MATH and MATH for MATH are zero at MATH. Thus among MATH's the non-vanishing ones are only MATH, MATH, and MATH. Let MATH be a row vector of the matrix MATH. In view of the above result, the linear dependence MATH can possibly holds only when MATH is independent of MATH. Consequently we consider the equation MATH. The MATH-th component of the vector MATH is given by MATH where we have taken REF and MATH into account. Due to REF the last expression is equal to MATH. Therefore the equation MATH is equivalent to MATH for any MATH. This admits only the trivial solution for MATH if MATH. |
math/9909056 | Because of linearity, it is enough to show the statement for MATH. Consider a pair of lattices MATH, where MATH consists of the solutions MATH for the homogeneous equation MATH mod MATH, and MATH. The column vectors MATH of MATH is a basis of MATH. It is well-known CITE that MATH equals to MATH, which is MATH. |
math/9909056 | Denote MATH simply by MATH. By REF , it suffices to verify MATH. We do this by a double induction on MATH and MATH regarding MATH as non-negative variables independent of MATH. First let MATH be arbitrary and MATH. Thus MATH for any MATH, hence MATH. Next let MATH. Then MATH because of the assumption MATH. Finally let MATH and MATH be arbitrary. Then there exists MATH such that MATH. Setting MATH and MATH, one can expand the determinant as MATH. By induction the two terms in the right hand side are both positive. |
math/9909056 | Under the assumption the number of off-diagonal solutions has already been obtained in REF . By virtue of REF it is equal to MATH where we have substituted REF . By means of REF the MATH - sum can be taken, leading to MATH . |
math/9909056 | We show MATH for any MATH. Note that MATH implies MATH. Therefore the assertion follows from REF . |
math/9909056 | We prove by induction on MATH. The case MATH is due to the formula MATH . Assume MATH has the above expansion. Then from REF MATH is MATH where the sum MATH is over MATH, , MATH. Upon substituting REF , the right hand side becomes MATH . Applying REF again, we obtain the desired expansion. |
math/9909056 | Put MATH, MATH. Then it is easy to check MATH . From REF we are to show MATH. By expanding a binomial coefficient in REF , the MATH is expressed as REF MATH where MATH, MATH, MATH and MATH. By REF they can also be written as MATH, MATH, MATH and MATH. Thus the contributions containing MATH and MATH (respectively, MATH and MATH) amount to MATH (respectively, MATH). |
math/9909056 | REF . REF Put MATH in REF and apply REF . CASE: It is enough to show that the limit MATH exists in MATH. Note that MATH from REF . In the series MATH in REF , those MATH containing MATH with MATH make contributions in the order higher than MATH. It follows that MATH mod MATH. Then, we have MATH which means MATH exists. CASE: Suppose MATH satisfies REF . Setting MATH, we find MATH by REF . (MATH here is the truncated one REF .) Therefore REF specializes to MATH where MATH is over MATH, , MATH. By taking the limit MATH using REF for MATH, this leads to MATH. Since MATH's are arbitrary, we obtain MATH. Comparing this with REF we conclude MATH. |
math/9909056 | CASE: In view of REF , the two equalities are equivalent. The first one is due to REF and MATH by REF . CASE: In view of REF , the two equalities are again equivalent. To be self-contained, let us include a quick proof of the first one although this has been done in CITE. Let MATH be as in REF and MATH. In the expansion of MATH by means of REF , specialize the variables as MATH (hence MATH) and MATH. The result reads MATH where MATH is taken over MATH, MATH. Picking up the coefficient of MATH, we get MATH where MATH is used. In the limit MATH this is equivalent to MATH due to REF . |
math/9909056 | It suffices to show the former assuming that MATH is any positive integer. Notice that MATH is the number of maps MATH such that MATH if MATH and MATH belong to the same block of MATH. Similarly, MATH is the number of maps MATH such that MATH if and only if MATH and MATH belong to the same block of MATH. Since ``if" case consists of the disjoint union of ``if and only if" cases labeled by MATH, the former relation holds. |
math/9909066 | We may assume that MATH since the claim follows from REF otherwise. Let MATH be the first integer such that MATH; from the hypotheses we have MATH. Define MATH inductively by MATH. One can verify inductively that MATH for all MATH, and in particular that MATH. From REF we have MATH for any MATH. Observe that MATH . If we thus set MATH for a suitable constant MATH, we thus obtain MATH . Iterating this, one obtains MATH . The claim then follows from REF . |
math/9909066 | Let MATH. From REF for MATH we have MATH . Square summing this in MATH we obtain the first estimate, and the second estimate follows from REF for MATH and NAME 's inequality. |
math/9909066 | By the pigeonhole principle it suffices to show that MATH . From the symmetry and translation covariance of MATH we have the identity MATH . On the other hand, from REF we have MATH so from REF we have MATH and thus that MATH . The claim then follows. |
math/9909066 | By scaling it suffices to verify this when MATH. The claims follow directly from the easily verified estimates MATH . |
math/9909066 | By NAME it suffices to verify the claims when MATH. To prove REF we first use REF, and REF to obtain MATH and REF follows from REF. A similar argument gives MATH and REF follows from REF. We now prove REF. From REF with MATH replaced by MATH, we have MATH . By applying REF with MATH, MATH replaced by MATH, MATH, and then applying MATH, we obtain MATH so by REF we have MATH . Combining this with the previous estimate we obtain REF. |
math/9909066 | By translation invariance we may take MATH, and by scaling we may take MATH. The blue cone is a null surface, and so standard energy estimates will not prove this estimate. However this can be salvaged since MATH is red, so that the characteristics of MATH will be transverse to the blue cone. We turn to the details. Let MATH be the wave evolution operator defined in REF. By REF, it suffices to show that MATH for all MATH, where MATH is the slice MATH. We shall apply the MATH method. By duality the above estimate is equivalent to MATH . We square this as MATH where MATH is the usual complex inner product. By NAME and NAME 's inequality it suffices to show that MATH for all MATH. When MATH this follows from NAME and the MATH boundedness of the operator MATH, so we may assume that MATH. The convolution operator MATH behaves essentially like MATH, and its kernel MATH satisfies a similar kernel estimate to REF, namely MATH . The claim then follows from the transversality of MATH and MATH and crude estimates. |
math/9909066 | The contribution of MATH is acceptable by using REF to control MATH and REF to control MATH. Similarly for the contribution of MATH. |
math/9909066 | Let MATH have frequency MATH. From hypothesis, the frequency support of MATH is contained in a a sector MATH of width MATH. By NAME 's theorem it thus suffices to show that MATH for all MATH and MATH, where MATH and MATH denote surface measure on MATH and MATH respectively. By NAME 's inequality we have MATH so it suffices by interpolation to show that MATH which by positivity reduces to showing that MATH . But this follows from the observation that the intersection of MATH and MATH is always transverse and has MATH-dimensional measure at most MATH for any MATH. |
math/9909066 | The idea is to combine REF with the localization machinery developed in REF. Let MATH be the disk MATH. If MATH is sufficiently large, then from REF we have MATH and MATH . From the triangle inequality we thus have MATH . From REF we see that MATH has dispersion MATH. By REF we thus have MATH . The claim then follows by letting MATH range over the lifespan of MATH, averaging, and applying NAME. |
math/9909066 | The quantities MATH have the units of length, while the frequencies MATH, MATH have the units of inverse length. Finally, MATH and the margins are dimensionless. One can then verify that the entire Proposition is dimensionally correct and thus scale invariant. By scaling we may thus set MATH. By translation invariance we may assume MATH is centered at the origin. Set MATH, where MATH is chosen so that MATH. We may assume that MATH (for instance), since otherwise the claim is trivial by setting MATH and using some trivial bound such as REF to show REF. Let MATH be a maximal MATH-separated subset of MATH, and let MATH denote the lattice MATH. We define a red tube to be any set MATH of the form MATH where MATH and MATH. We let MATH denote the set of all red tubes. If MATH is a red tube, we define the cutoff function MATH on MATH by MATH . We shall need the following careful decomposition of MATH into wave packets which are concentrated on tubes in MATH. With the above notation, one can find for each MATH a red wave MATH with frequency REF and angular dispersion at most MATH, such that CASE: We have the margin estimate MATH . CASE: The map MATH is linear for each MATH, and MATH . CASE: We have MATH for all MATH and MATH in the lifespan of MATH. CASE: If MATH then MATH . CASE: We have MATH . CASE: We have the NAME inequality MATH whenever MATH runs over a finite index set and the MATH are non-negative numbers such that MATH for all MATH. Roughly speaking, MATH is the portion of MATH which has frequency support in the sector of width MATH and direction MATH, and is spatially concentrated in MATH. A naive microlocalization to this region of space and frequency, taking some care to ensure that the MATH are still waves, would obtain most of the above properties, but would probably need to replace the MATH factor in REF by a larger constant, which would then cause a similar unacceptable loss in REF and then destroy the induction. This necessitates a delicate construction of the MATH based on averaging. Because the details of the proof are technical and not particularly relevant to the rest of the argument, we defer the proof of this Lemma to REF, and continue with the proof of REF . Using REF , we can now define MATH by MATH for all MATH, where MATH and MATH . The MATH factor is only present in REF to ensure that MATH does not completely degenerate to zero. One can think of MATH as consisting of those wave packets MATH such that MATH concentrates in MATH. It is clear that MATH is a red wave with MATH and that MATH . The estimate REF follows from REF. We now show REF. Let MATH be any disk of radius MATH with MATH in the lifespan of MATH. By REF it then suffices to show that MATH . Let MATH denote the slightly larger disk MATH, and let MATH denote the even larger disk MATH. We may divide MATH . The map MATH is linear, so we may write MATH accordingly. From REF we have MATH . Next, we claim that MATH . To see this, we first consider the tubes MATH which do not intersect MATH. By REF, and NAME, their contribution is acceptable. Thus we need only consider those tubes which intersect MATH. By the triangle inequality it suffices to show MATH for each tube MATH. First suppose that MATH. Then by REF, and the assumption MATH we see that MATH . The claim then follows from REF. Now suppose that MATH. By REF, the hypothesis MATH and NAME 's inequality (or NAME embedding) we see that MATH whenever MATH. The claim then follows. This completes the proof of REF. Combining REF, we see that MATH and REF easily follows (using the trivial inequality MATH). This proves REF. We now turn to the proof of REF. By REF we have MATH so by the triangle inequality it suffices to show that MATH for each MATH in MATH. Fix MATH. We shall use a (heavily disguised) version of the arguments in CITE. If MATH intersects MATH, then MATH. By squaring REF, we thus reduce to showing that MATH . Consider a single summand from REF. From REF and the triangle inequality we have MATH . Consider the tubes MATH which do not intersect MATH. By REF, their total contribution to REF is MATH. As for the tubes MATH which do intersect MATH, we may apply REF (since MATH) to obtain MATH . The total contribution of the error term to REF is MATH by REF and the fact that there are only MATH tubes being summed here. Combining all these estimates we obtain MATH . Inserting this back into REF, we see that it suffices to show that MATH . Using the trivial estimate MATH followed by NAME, we have MATH . By REF, we have MATH where MATH . Since MATH, we see from elementary geometry that MATH . From REF and the triangle inequality we thus have MATH . Inserting all these estimates back into REF, we see that it will suffice to show that MATH . We re-arrange the left-hand side as MATH and estimate this by MATH . Since the inner sum is MATH by REF, the desired estimate REF then follows from REF. This proves REF. |
math/9909066 | Consider the NAME linear transformation MATH given by MATH and define the associated operator MATH by MATH where MATH is the adjoint of MATH. A routine computation shows that MATH, so that MATH . Also, we observe from NAME that MATH and MATH. Finally, we have MATH. Combining all these facts we see that to prove REF it suffices to do so when MATH. Set MATH. We can write MATH where MATH and MATH, and MATH is the symbol MATH . Since MATH is smooth and compactly supported, we may decompose MATH as a NAME series MATH for MATH, where MATH is some discrete lattice and MATH are rapidly decreasing co-efficients (uniformly in MATH). This implies that MATH where MATH . The first claim of REF then follows from this decomposition, the triangle inequality, REF, and the observation that MATH, MATH have the same energy as MATH, MATH respectively. The second claim is proven similarly but also uses the observation that MATH has the same magnitude as MATH. |
math/9909066 | In order to apply the complex interpolation method we shall allow the indices MATH, MATH, MATH to acquire an imaginary part MATH. For notational simplicity we keep ourselves to the case MATH, but the reader may easily verify that the following argument also works for arbitrary MATH with constants which are at most exponential in MATH. When MATH these claims were proven in CITE (see also CITE). Since we have replaced REF with the linear REF , we see by complex interpolation that it suffices to verify the claims when MATH. In this REF becomes MATH; since the operator MATH is bounded in MATH by standard multiplier theory we may assume that MATH. Let MATH, MATH be solutions to the free wave equation. By a finite decomposition and time reversal symmetry we may assume that MATH is supported in the upper light cone MATH and that MATH is supported on either the upper or the lower light cone MATH. Write MATH, MATH, where MATH, MATH have frequency supports on the region MATH, MATH respectively. We now rewrite REF as MATH . It suffices to show that MATH for all MATH and some MATH, since the claim then follows from the triangle inequality, NAME, and NAME 's inequality for sums. By symmetry we may take MATH; by scaling and REF we may take MATH. Having used REF, our arguments will no longer require this condition and we shall discard it. Since we are assuming REF to hold with strict inequality we may absorb the MATH into the MATH factors, and reduce ourselves to showing that MATH . Fix MATH. For any MATH, decompose the double light cone into finitely overlapping projective sectors MATH of angular width MATH, and let MATH, MATH be a NAME decomposition of MATH, MATH subordinate to these sectors. It suffices to show that MATH for all MATH, since the claim follows by summing in MATH using the bilinear partition of unity based on angular separation (see for example, CITE, CITE). By the triangle inequality and NAME as before, it suffices to show that MATH for each MATH, MATH with angular separation MATH. By a spatial rotation we may assume that MATH has frequency support in REF, and that either MATH or MATH has frequency support in REF. Now suppose MATH. In this case MATH is equal to MATH times a harmless multiplier on the frequency support of MATH, so by REF the left-hand side of REF is majorized by MATH . REF then follows after some algebra from REF, and the assumption that REF holds with strict inequality. It remains to consider the case MATH; by a mild NAME transformation we may make MATH. (This affects MATH slightly, but this change is irrelevant). Since MATH and MATH differ in angle by MATH, some geometry shows that MATH is at least MATH on the frequency support of MATH. We may thus majorize MATH in MATH by MATH, and by REF again the left-hand side of REF is bounded by MATH . REF then follows after some algebra from REF, and the assumption that REF holds with strict inequality. |
math/9909066 | We use the method of descent, exploiting the fact that the paraboloid in MATH is a conic section of the light cone in MATH. To prove Conjecture REF for MATH, it suffices to verify it for MATH, since the other endpoint MATH is trivial. From NAME 's theorem it suffices to show that MATH for all MATH functions MATH on MATH. Fix MATH, MATH. Let MATH be a large number, and consider the functions MATH, MATH defined on MATH by MATH and MATH . One can easily verify that MATH and MATH solve the wave equation, and that MATH . By the assumption that REF holds at REF, we have MATH for some MATH, MATH, MATH. However, a computation of the frequency support of MATH shows that MATH, MATH, MATH are all given by smooth functions comparable to REF on this support, and so we have MATH . Making the change of variables MATH, MATH, we have MATH and so we have MATH the MATH factor having cancelled against the Jacobian term. The phase factor MATH can be discarded. Letting MATH and taking limits, and then evaluating the MATH, MATH integrations, we thus see that MATH and REF follows since the MATH behaviour is trivial. |
math/9909068 | They are the unique elements in MATH and MATH respectively that do not vanish when MATH is applied. |
math/9909068 | Let MATH and MATH. Since MATH is a MATH highest element, MATH for MATH. It means that MATH and MATH for MATH. Thus we have MATH, and then MATH, that is, MATH. Repeating the same process we come to MATH, that is, MATH. Thus MATH and MATH. Thus we have MATH, and then MATH, that is, MATH. Repeating the same process we come to MATH, that is, MATH. Thus MATH and MATH. Therefore we have a condition MATH. If MATH has no MATH then MATH and this condition certainly holds. If MATH has no MATH then MATH, thus we impose the condition MATH. |
math/9909068 | For a set of operators MATH on the MATH crystals, we define MATH by MATH for MATH, MATH for MATH, MATH for MATH and so on. Let MATH or MATH. The lemma can be proved by applying the following sequence of operators MATH to the both sides of REF . In the sequel we will show MATH . In the case MATH and MATH, the value of the energy function was lowered by MATH when the first to the MATH-th MATH's were applied, and raised by MATH when the MATH-th to the last MATH's were applied. In the case MATH and MATH, in addition to the same change as in the previous case, the value of the energy function was raised by MATH when the MATH-th to the last MATH's were applied, and lowered by the same amount when the first to the MATH-th MATH's were applied. In the case MATH and MATH, the value of the energy function was lowered by MATH when the first to the MATH-th MATH's were applied, and raised by MATH when the MATH-th to the last MATH's were applied. In the case MATH and MATH, in addition to the same change as the previous case, the value of the energy function was raised by MATH when the MATH-th to the last MATH's were applied, and lowered by the same amount when the first to the MATH-th MATH's were applied. Recall that we have normalized the energy function as MATH. Thus we have MATH. |
math/9909068 | If MATH, let MATH or MATH. The lemma can be proved by applying the following sequence of operators MATH to the both sides of REF . In the case MATH and MATH, the value of the energy function was lowered by MATH when the first to the MATH-th MATH's were applied. In the case MATH, MATH and MATH (respectively, MATH), in addition to the same change in the previous case, the value of the energy function was raised by MATH when the MATH-th to the last MATH's were applied, and lowered by MATH (respectively, MATH) when the first to the MATH-th (respectively, the last) MATH's were applied. In the case MATH and MATH, the value of the energy function was lowered by MATH when the first to the MATH-th MATH's were applied. In the case MATH, MATH and MATH (respectively, MATH), in addition to the same change in the previous case, the value of the energy function was raised by MATH when the MATH-th to the last MATH's were applied, and lowered by MATH (respectively, MATH) when the first to the MATH-th (respectively, the last) MATH's were applied. If MATH, one can check that MATH. REF and the previous case of the present lemma enable us to obtain its image under the map MATH. They also tell us that now the value of the energy function is equal to MATH. Then apply MATH. Since it again turns out to hit the right component of the tensor product, the value of the energy function is raised by MATH. If MATH, one can check that MATH. REF enable us to obtain its image under the map MATH. They also tell us that now the value of the energy function is equal to MATH. Then apply MATH. Since it again hits the right component of the tensor product, the value of the energy function is raised by MATH. |
math/9909068 | Since MATH and MATH are injective, the only if part of the statement follows immediately after when the if part is proved. Without loss of generality we assume MATH. Set MATH . First consider the case MATH and MATH. Then MATH, MATH and MATH by REF . On the other hand we have MATH where MATH . By REF one has MATH under the isomorphism MATH. The energy was lowered by MATH when the first to the MATH-th MATH's were applied, and raised by MATH when the MATH-th to the MATH-th MATH's were applied. Then it was lowered by MATH when the leftmost MATH was applied. Thus we have MATH . The proof is finished in this special case from REF . Now we consider the general elements MATH and MATH that are mapped to each other under the isomorphism. Take any finite sequence MATH made of MATH's and MATH's MATH such that MATH which is equivalent to MATH . For any operator in MATH, REF determine whether it should hit the left or the right component of the tensor product. For any MATH we have MATH and MATH from REF . Thus the alternatives in REF are not changed by MATH. From REF it follows that MATH. Applying MATH (respectively, MATH ) to REF (respectively, REF ) we thus get MATH . From REF it follows that MATH under the isomorphism MATH. When comparing REF change of the value of the energy function caused by MATH is not affected by MATH. Therefore from REF we have MATH. |
math/9909068 | Thanks to REF it suffices to verify the above claims only when MATH or MATH has no MATH. Such a case can be reduced to REF by the argument as follows. Let MATH be an element of MATH, which is not necessarily a MATH highest element and either MATH or MATH is free of MATH. Let MATH under the isomorphism MATH. There exists a sequence MATH such that MATH is a MATH highest element. Then we have MATH. Let MATH. Since MATH does not change MATH, MATH or MATH also has no MATH. Therefore from REF we have MATH where MATH or MATH has no MATH. Since MATH and MATH, we conclude that MATH or MATH has no MATH. Thus REF is indeed valid as MATH. By REF is equivalent to MATH . Here MATH is defined in REF . Regarding MATH as an element of a MATH crystal, we apply REF , to get MATH which is equivalent to MATH showing REF . Since MATH does not change the shape of the tableaux CITE and MATH, REF follows from REF . |
math/9909068 | They are the unique elements in MATH and MATH respectively that do not vanish when MATH is applied and do not vanish when MATH is applied. |
math/9909068 | REF is due to REF and the fact that the irreducible decomposition of the MATH module MATH is multiplicity-free (for generic MATH). We call an element MATH of a MATH crystal a MATH highest element if it satisfies MATH for MATH. To show REF , it suffices to check it for MATH highest elements. Then the general case follows from REF because MATH for MATH if MATH. We assume MATH with no loss of generality. Suppose that MATH is a MATH highest element. In general it has the form: MATH where MATH and MATH are arbitrary as long as MATH. Applying MATH to the both sides of REF , we find MATH . Here MATH. In the course of the application of MATH's, the value of the energy function has changed as MATH . Thus according to our normalization REF we have MATH. On the other hand for this highest element the column insertions REF lead to a tableau whose first row has the length MATH. This completes the proof of REF . |
math/9909070 | REF is established by taking MATH as the powers of the augmentation ideal are invariant under all automorphisms of MATH. To simplify the notation, write MATH for the generators of the free group MATH. By commuting generators and double points one at a time, it is sufficient to verify REF for the generators MATH, their inverses, and the set of elementary singular braids MATH. It is clear that if MATH the generator MATH commutes with MATH. Hence, it remains to verify the lemma for MATH and MATH. We will do the calculations only for MATH as the case of MATH and MATH can be treated in the same manner. The following relations are easily checked: MATH . Thus, MATH where MATH and MATH . Finally, MATH where MATH and MATH are as above, so MATH and MATH. |
math/9909070 | In view of REF it is clear that REF implies an analogous statement for the NAME expansion of braids. Namely, for any braid MATH with at least MATH double points MATH has no terms of degree less than MATH and the MATH-th degree term is the image of MATH under the natural projection MATH. The proof proceeds by induction on the type of an invariant MATH. Invariants of type zero are just constants and the existence of a homomorphism MATH such that MATH for them is trivial. Suppose that homomorphisms of MATH to MATH which correspond to invariants of type less than MATH have been found. Let MATH be of type MATH. As MATH vanishes on MATH, it defines a homomorphism MATH and can be trivially extended to a homomorphism MATH. Then MATH is an invariant of type MATH and, hence by the inductive hypothesis there exists a MATH such that MATH . Now, setting MATH we obtain a homomorphism MATH which represents MATH. The uniqueness is straightforward. |
math/9909070 | By duality, the product on MATH gives rise to a coproduct on MATH. This is also the coproduct induced by the natural coproduct on MATH, namely MATH for each pure braid MATH. From a result of CITE, the space of indecomposable elements of a commutative algebra is dual to the space of primitive elements in the dual coalgebra, so MATH is equal to the dimension of the subspace of primitive elements in MATH. Now, denote by MATH the abelian group MATH. The graded group MATH is a NAME algebra with the NAME bracket induced by the group commutator. REF says that the algebra MATH is the universal enveloping algebra of of the NAME algebra MATH. However, the space of primitive elements of a universal enveloping algebra is naturally isomorphic to the original NAME algebra. This means that MATH is equal to the rank of the abelian group MATH. The rank of MATH can be found from NAME 's work CITE or expressed via the ranks of MATH using a result of CITE. Alternatively, notice that if MATH is considered as a coalgebra with the coproduct induced by the natural coproduct on MATH, the combing identification MATH is an isomorphism of coalgebras. Hence, the primitive elements in MATH and MATH are in one-to-one correspondence. Now, according to NAME 's theorem, MATH is equal to the rank of the abelian group MATH. Observe that for any groups MATH and MATH the abelian group MATH is isomorphic to MATH. Hence MATH . The theorem then follows from NAME 's formula for the ranks of the quotient groups MATH, see CITE: MATH . |
math/9909070 | The left hand side can be seen as the number of maps from a MATH element set to the following set, such that the image is `vertical'. MATH . The proof proceeds by counting these maps in a different way. So, MATH . The third equality follows from a simple identity and the final equality comes from relabelling and the convention that MATH. |
math/9909070 | MATH . The theorem then follows as the matrix MATH is invertible - it has inverse MATH. |
math/9909076 | By the hypotheses on MATH and MATH we may write MATH with MATH finite index sets. Next one decomposes MATH and MATH with respect to their poles located in the interior and exterior of the bounded domain encircled by MATH, MATH . Then straightforward residue computations yield MATH . Here the last integral in REF vanishes since the integrand is analytic inside MATH and the first integral in REF vanishes by symmetry. Moreover, we used the symbol MATH to indicate summation only over those MATH and MATH with MATH, since only first-order poles contribute in this calculation. |
math/9909076 | Given MATH, introduce the partition MATH of the closed interval MATH, MATH and denote by MATH the piecewise continuous function MATH where MATH is the characteristic function of the set MATH. In the NAME space MATH introduce the (possibly unbounded) operators MATH . We note, that by REF , the spectrum of MATH, MATH, in the interval MATH consists of finitely many eigenvalues (possibly of infinite multiplicity). By REF one infers that the sequence of operators MATH converges in norm resolvent sense to MATH in the NAME space MATH, MATH. This convergence, in turn, combined with the hypothesis MATH, MATH, implies the convergence MATH where in obvious notation MATH . Thus, in order to prove REF it suffices to check that every term on the left-hand side of REF is nonnegative. In the following it is useful to decompose the NAME functions MATH in the form MATH where MATH are NAME functions associated with the spectral subspaces MATH and MATH of MATH, MATH. According to the decomposition REF, the NAME functions MATH are rational operator-valued functions of the form MATH where MATH . Given MATH, one decomposes the integrals on the left-hand side of REF as a sum of four terms MATH . According to REF, the first integral MATH in REF can be represented as follows MATH . Applying the residue theorem to the integrals on the right-hand side of REF one infers MATH since MATH is a nonincreasing differentiable function and since the inequalities MATH hold (due to the fact that the operators MATH, MATH, MATH, are nonnegative by REF). The integral MATH vanishes, MATH since MATH is holomorphic in MATH. The remaining integrals MATH and MATH can also be evaluated by the residue theorem and one obtains MATH . We recall that MATH, MATH, by REF. At the same time MATH by REF, while MATH by REF. Hence, MATH and thus, MATH and MATH. Together with REF (combined with REF) this proves REF. |
math/9909076 | By REF , the operators MATH, MATH, are well-defined (compare REF). In particular, MATH . Since MATH one infers MATH uniformly with respect to MATH. Thus, combining REF, one concludes that MATH is differentiable with respect to MATH in MATH-topology. Along with the existence of a continuous MATH on MATH, this yields the differentiability of the function in REF then follows by a straightforward computation using REF. |
math/9909076 | Since MATH is nonnegative, MATH, MATH and hence MATH by REF. Applying REF to the operator-valued NAME functions MATH and MATH, one obtains the inequality MATH . Combining REF one infers MATH by REF , proving the assertion. |
math/9909076 | Introducing the sequence MATH of spectral projections of MATH, MATH one concludes that for fixed MATH the bounded operators given by MATH converge to MATH in the strong resolvent sense and therefore, MATH by REF. Since by REF is a bounded function, the family of operators MATH is uniformly bounded with respect to MATH and hence by REF, MATH . Given MATH and MATH, one can always find a closed oriented NAME contour MATH in MATH, and a bounded interval MATH such that the collection MATH satisfies REF . Since by REF is a nonnegative nonincreasing function on MATH and REF holds, one concludes by REF that the function MATH is nonincreasing in some neighborhood of MATH and hence on the whole interval MATH since MATH was arbitrary. Therefore, the function MATH is also nonincreasing on MATH as a pointwise limit REF of the nonincreasing functions in REF. |
math/9909076 | First, one observes that by REF the integrated spectral shift function MATH given by REF is uniformly bounded, that is, MATH . Moreover, since MATH is semibounded, one concludes that MATH by REF of the set MATH and hence for all MATH, MATH . Thus, using REF, one infers MATH . Next, combining REF, an integration by parts in the trace REF yields MATH . Given MATH, the integrated spectral shift function MATH is concave with respect to MATH by REF and hence the left-hand side of REF is also a concave function of MATH by REF (as a weighted mean of concave functions with a positive weight). |
math/9909077 | For MATH consider the variety MATH. It follows from REF that the set of its irreducible components of dimension MATH is in a bijection with MATH. Thus, for MATH, we set MATH to be the closure in MATH of the corresponding irreducible component of MATH. The fact that this map is a bijection satisfying all the required properties is straightforward. |
math/9909077 | The filtration MATH is compatible with the direct sum decomposition MATH. Let MATH (respectively, MATH) denote the corresponding subspace (respectively, sub-quotient) of MATH. By REF, we can identify MATH with the cohomology MATH. In addition, the filtration MATH on MATH coincides with the filtration on the compactly supported cohomology induced by the decreasing sequence of open subsets in MATH: MATH . Therefore, MATH. The assertion of the proposition follows now from REF - REF of the bijection MATH. |
math/9909077 | Consider the variety MATH . According to REF, its set of irreducible components of dimension MATH can be identified with the LHS of REF. Now, for MATH, let us denote by MATH the following scheme: MATH where MATH is as in REF. It is easy to see that the natural map MATH is a locally closed embedding. Similarly, for MATH and MATH we define the sub-scheme MATH of MATH as MATH. We have a commutative diagram MATH . Therefore, to each element of the set MATH we can attach an irreducible component of dimension MATH in MATH. By taking its closure in MATH we obtain an irreducible component of MATH and it is easy to see that the map we have just described is a bijection. This proves our assertion. |
math/9909082 | Let MATH be the MATH-stable decomposition. Then MATH and therefore MATH is orthogonal to MATH. That is, MATH is a characteristic of MATH. Consider the algebraic NAME algebra MATH. Then MATH as MATH-vector space. Using algebraic NAME decompositions for MATH and MATH, one easily shows that MATH contains MATH, a reductive NAME subalgebra of MATH (see CITE). Hence MATH and we are done. |
math/9909082 | CASE: Identify the nodes of MATH with the elements of a basis of a MATH-dimensional MATH-vector space MATH. If MATH, then MATH stands for the corresponding basis vector in MATH. It is convenient to assume that MATH, if MATH. Define MATH (respectively, MATH) to be the operator corresponding to the horizontal (respectively, vertical) arrows in MATH: If MATH then MATH and MATH. Obviously, MATH are nilpotent endomorphisms of MATH. From REF it follows that MATH. Define the semisimple endomorphisms of MATH by the formulae: MATH, MATH. The condition on the barycentre of MATH is equivalent to that MATH, MATH. It is easily seen that MATH satisfy relations REF . Since MATH is connected, a straightforward computation shows that MATH is the set of all traceless endomorphisms of MATH that are diagonal in the basis MATH. It is then easily seen that MATH. Therefore MATH is distinguished by REF . To prove uniqueness, assume that MATH is another basis of MATH and MATH is the corresponding distinguished pair. Clearly, there exist MATH such that MATH for all MATH and some MATH. Then MATH (MATH). CASE: Let MATH be a sw-Young graph and MATH the source of it. Then MATH is a cyclic vector in MATH relative to MATH. Therefore MATH is generated by all non-zero powers MATH, MATH. It is easily seen that MATH if and only if MATH. Whence MATH. If MATH is an ne-Young graph, we may consider the dual endomorphisms MATH of MATH. Obviously, the pair MATH corresponds to the sw-Young diagram MATH and MATH. |
math/9909082 | CASE: The argument here is analogous to that in REF . Consider a basis MATH of an even-dimensional MATH-vector space MATH indexed by the nodes of MATH. The symmetric bilinear form on MATH is defined by MATH unless MATH and MATH, and MATH. For any MATH, set MATH . The verification of all required relations is easy. CASE: Let MATH be an odd-dimensional MATH-vector space with a basis indexed by the nodes of MATH, MATH. According to REF , each MATH represents a distinguished pair MATH, with characteristic MATH, in MATH. Then straightforward computations show that the direct sum MATH is a distinguished pair in MATH in both cases: CASE: Here MATH and MATH. Hence MATH and MATH. Since MATH as MATH- and MATH-module and MATH, we obtain MATH, that is, MATH is a NAME subalgebra. It is also easily seen that MATH. CASE: Here MATH. Hence MATH. The assumption MATH implies that MATH is REF-dimensional. Therefore MATH, that is, MATH is a NAME subalgebra. For MATH, we have the similar formula MATH . It follows from the connectedness of MATH that MATH (MATH). Finally, as both skew-graphs have more than REF node, the space MATH is not MATH-stable. Whence MATH. We conclude by REF . CASE: Let MATH REF be a MATH-vector space with a basis indexed by the nodes of MATH and MATH. The spaces MATH are supposed to be pairwise orthogonal. Construct distinguished pairs in MATH and MATH according to the recipes in REF respectively. Define MATH to be the direct sum of these two pairs. Again, we use the decomposition MATH to show that MATH and hence MATH is a NAME subalgebra of MATH. The crucial equality here is MATH (compare proof of REF). Then the problem of proving that MATH becomes ``local", that is, it reduces to MATH and MATH, that is, to preceding REF . We conclude by REF . CASE: As in REF , one proves that the distinguished pairs constructed in REF , and REF form a single MATH-orbit. If one of the connected components of MATH has an odd number nodes, then this orbit is actually a MATH-orbit (compare REF). But if MATH is connected and hence contains an even number if nodes, then this orbit is a union of two MATH-orbits (the conjugating transformation MATH with MATH can not be corrected). |
math/9909082 | CASE: For a non-integral rectangle, the proof is similar to that given in REF and is easy. Therefore we omit it. Suppose MATH is a near-rectangular diagram. Let MATH, the symmetric form, and MATH be the same as in REF . The only thing that has to be verified is that MATH. Here MATH has no cyclic vectors and the previous proofs do not apply. However, for all possible shapes, there is an argument reducing the problem to a smaller rectangle. CASE: Consider a near-rectangular diagram of the first shape: Four letters inside of squares stands for the corresponding basis vectors in MATH. By assumption, the length of the last and first row is odd, say MATH. Without loss of generality, we may assume that MATH, MATH, and MATH. Then MATH. Set MATH. Obviously, MATH is again a quadratic space and MATH. Put MATH, MATH. Consider MATH and the natural map MATH. Then MATH. We are interested in MATH. Since MATH is a cyclic vector in MATH relative to MATH and MATH, we see that any MATH takes MATH to MATH and all other basis vectors of MATH to zero. If MATH, then MATH and, as MATH is skew-symmetric, MATH. Consequently, MATH (actually, MATH). Claim: MATH. Indeed, let MATH. Since MATH for MATH, we have MATH. Now MATH. Whence MATH and therefore MATH. It follows from the claim that MATH. On the other hand, MATH induces the nilpotent pair MATH in MATH, which corresponds to the rectangular diagram MATH. Hence MATH is principal in MATH. Since MATH, we obtain MATH. Thus, MATH is principal, too. CASE: Up to the permutation of MATH and MATH, a near-rectangular diagram of the second shape is the same a one of the first shape. CASE: If MATH is a near-rectangular diagram of the third shape, the same idea can be used. In place of MATH, one has to consider the linear span of all basis vectors corresponding to the rightmost column of MATH. The details are left to the reader. CASE: We use the notation from the proof of REF : If MATH is rectangular, then MATH (the space MATH corresponds to the sinks of MATH). Next, MATH is principal in MATH by REF . Whence MATH. CASE: Since MATH has no vertical arrows and MATH has no horizontal arrows, we have MATH and MATH. That is, MATH and MATH operate only in MATH. Moreover, these are regular elements in MATH. Now MATH and MATH is REF-dimensional. Therefore MATH. Thus MATH is a principal pair. |
math/9909083 | CASE: If MATH belongs to MATH, then MATH belongs to MATH, so that MATH is locally absolutely continuous, and MATH is a function of class MATH on MATH, with values in MATH. By an obvious induction argument, MATH is infinitely differentiable. REF implies that MATH tends to MATH at infinity. Therefore, MATH tends to zero at MATH: by NAME 's formula, MATH the right hand side of this relation tends to MATH at infinity; therefore, the left hand side must also tend to MATH. Assume MATH. System REF is equivalent to a system of four ordinary differential equations of the first order in MATH, and MATH is a critical point of this system. We apply the theory of invariant manifolds in a neighborhood of MATH; the matrix of the linearized system at that point has two double eigenvalues MATH, where MATH. We make the convention that MATH. Therefore, MATH, MATH and MATH decrease exponentially fast at infinity. Multiply REF by MATH, and integrate over MATH; we obtain after an integration by REF . This shows that MATH vanishes. CASE: Define MATH . We let MATH; then MATH and MATH solve the linear differential equation MATH and they cannot both vanish identically. If MATH is a solution of REF which does not vanish identically, then MATH and MATH cannot vanish simultaneously; if MATH and MATH are two solutions of REF which vanish at infinity, their Wronskian MATH is constant and vanishes at infinity; therefore, any two solutions of REF are proportional. Thus MATH and MATH are linearly dependent, which proves the desired assertion. CASE: Thanks to REF , if MATH is a solution of REF, with MATH, we may assume that it is real without loss of generality. Thus, it solves REF; moreover, MATH and MATH cannot vanish simultaneously by uniqueness of solution of the NAME problem for REF. Therefore, we can find a continuous determination of the angle MATH such that MATH . It is immediate that this determination is also of class MATH. Let us multiply REF by MATH, and integrate, remembering that MATH and MATH is real; we obtain MATH . By taking the limit of the left hand side of REF, we see that the constant on the right hand side of REF vanishes. We use REF, and we can see after substituting into REF and dividing by MATH that MATH . Assume that MATH is strictly negative. As MATH tends to infinity, the upper limit of the left hand side of the above equation is at most equal to MATH, while the right hand side of this equation tends to MATH; thus, we have a contradiction. CASE: The proof of REF shows that MATH can be taken real; define a function MATH by MATH then MATH is a real solution of MATH which decays at infinity. Moreover, if we define MATH REF implies MATH . In order to have a non trivial solution of REF which tends to MATH at MATH, we must choose MATH so that MATH can vanish for non zero values of MATH; letting MATH it is immediate that MATH can vanish for positive values of MATH if and only if MATH . Henceforth, we will write MATH . The level curves of MATH are represented at REF ; this figure shows that there exist exactly two homoclinic orbits of REF through REF for MATH; on one of them, MATH takes only positive values, and on the other one MATH takes only negative values. Therefore, we choose the homoclinic orbit which is situated in the half plane MATH. The solution MATH is still defined only up to translation. When MATH reaches a maximum at some point MATH, MATH, and MATH vanishes. Therefore, MATH . Since REF is invariant by the reflexion MATH, we can see that MATH; if MATH has more than one maximum, it is periodic, which is possible only if MATH vanishes identically. Therefore, MATH has exactly one maximum. If we choose the translation parameter so that MATH attains its maximum at MATH, then MATH is even and positive over MATH, which concludes the proof of the lemma. |
math/9909083 | We infer from REF the relation MATH it is also immediate that MATH satisfies the ordinary differential equation MATH which can also be found by passing to the limit as MATH tends to MATH in the differential equation satisfied by MATH: MATH . REF of MATH and a direct computation give MATH from which we infer the identity MATH . Thanks to REF, we deduce immediately REF from REF. If we differentiate REF with respect to MATH, we find MATH . Then, with the help of REF, relation REF is clear. By a passage to the limit as MATH tends to MATH, or by a direct computation, we see that REF hold. As MATH tends to MATH, MATH tends to REF and MATH tends to MATH. |
math/9909083 | Clearly, MATH is self-adjoint. We have MATH . The essential spectrum of MATH is contained in MATH (see NAME, CITE,REF ), and the spectrum of MATH is bounded from below by MATH; for all MATH, the intersection of the spectrum of MATH with MATH contains only eigenvalues of finite multiplicity; since MATH is a NAME operator in one-dimensional space, these eigenvalues are all simple; moreover, since MATH is even, MATH is even, and the eigenfunctions of MATH are even or odd; in particular, an eigenfunction corresponding to the lowest eigenvalue MATH of MATH does not change sign, thanks to the maximum principle; therefore, it is even, since such an eigenfunction cannot vanish together with its derivative. Denote by MATH an eigenfunction of MATH corresponding to MATH; this MATH can be chosen so that MATH . Differentiating REF with respect to MATH, we obtain MATH . It is clear that MATH belongs to MATH; therefore, it is an eigenvector of MATH, corresponding to the eigenvalue MATH. On the other hand, MATH is odd; therefore, we have proved that the infimum of the spectrum of MATH is a strictly negative number, MATH. |
math/9909083 | CASE: Denote by MATH (respectively, MATH) the subspace of even functions belonging to MATH (respectively, MATH). Define a mapping MATH from MATH to MATH by MATH . This mapping is of class MATH, we have MATH, and MATH, which is an isomorphism from MATH to MATH. Therefore, the implicit function theorem applies, and MATH is a MATH function of MATH with values in MATH. CASE: If we divide MATH by MATH we obtain a NAME operator whose differential part is constant, and whose potential part depends in a MATH fashion on MATH, according to REF of this proof; therefore, we can apply the results of NAME, CITE IV. REF, which enable us to conclude. |
math/9909083 | The differentiation of REF with respect to MATH gives MATH which implies immediately REF. An analogous computation implies MATH and MATH . Differentiating REF, we find MATH and MATH . Observe that MATH so that: MATH moreover we have also MATH . We use the information given by REF, to infer that MATH the conclusion of the lemma follows immediately. |
math/9909083 | Let us compute MATH by differentiating the equation MATH with respect to MATH; we obtain MATH where MATH . To prove our assertion, we will have to calculate MATH and MATH. From REF , we can see that MATH and from estimate REF, we have MATH but relations REF imply that MATH we infer from relation REF that MATH so that MATH . On the other hand, MATH . Thus, we obtain from REF the relation MATH . The conclusion of the lemma is a direct consequence of REF. |
math/9909083 | Define MATH . It is clear that MATH satisfies the NAME equation MATH . As MATH is strictly negative for all MATH, this means that MATH is the lower bound of the spectrum of the operator MATH defined by MATH . As MATH and MATH, the essential spectrum of MATH is included in MATH, and thus the second eigenvalue of MATH is some number MATH. We use now the twisting trick of CITE: define the operators MATH let MATH be an infinitely differentiable function from MATH to MATH which vanishes for MATH and is equal to MATH for MATH and let MATH. Define now a unitary transformation in MATH by its matrix MATH . We compute MATH and we find that MATH . An elementary calculation shows now that MATH where MATH and MATH are matrix valued functions whose norm tends to MATH as MATH tends to infinity. It is plain that the resolvent sets of MATH and MATH are identical. If MATH and MATH are NAME spaces, MATH is the space of linear bounded operators from MATH to MATH and MATH the corresponding norm; if MATH, we abbreviate MATH to MATH. If MATH is the space MATH we abbreviate MATH to MATH. Let MATH belong to the resolvent set of MATH; then for every MATH in MATH, MATH belongs to MATH, and the function MATH is continuous on the resolvent set of MATH. It is equivalent to solve MATH and MATH the previous considerations show that for all MATH in the resolvent set of MATH, we can find MATH such that for all MATH, MATH is also in the resolvent set of MATH. In particular, if we take MATH, we can find MATH such that for all MATH, the segment MATH is included in the resolvent set of MATH. In particular, we see also that for all MATH, the generalized eigenspace of MATH relative to the interval MATH contains exactly two eigenvalues; the minimum of MATH is equal to MATH; the above considerations show that these two eigenvalues tend to MATH as MATH tends to MATH. |
math/9909083 | We observe that for MATH, MATH is an isomorphism from MATH to MATH which transforms even functions into even functions, and odd functions into odd functions. Therefore, since MATH is even by construction, MATH is even too. We rewrite the identity MATH as MATH . Hence, by integration along MATH, MATH . Since MATH, we obtain MATH . If MATH is small enough, REF implies that MATH does not vanish identically. Thus, by construction it is an eigenfunction of MATH relative to the eigenvalue MATH or to the eigenvalue MATH. But we know that MATH is even; thus, it has to be an eigenfunction of MATH relative to MATH, and MATH . But MATH and, on the other hand MATH . This proves REF. In order to obtain REF which holds in MATH norm, while REF holds in MATH norm, we subtract the relation MATH from the relation MATH and we obtain MATH . For the last assertion of the theorem, we remark thanks to NAME 's theorem, the last term in REF can be rewritten as MATH and MATH is the unique solution of MATH . Subtracting REF from REF, we find that MATH and this implies immediately estimate REF, which concludes the proof. |
math/9909083 | For the first estimate, we differentiate the relation MATH with respect to MATH; we multiply scalarly the result MATH by MATH and we obtain MATH . Since MATH we may write MATH . Thanks to REF, MATH here we have used REF. Therefore, MATH . At this point it is clear that there exists a number MATH such that MATH but the explicit calculation of MATH is tedious. It can be avoided by integrating REF with respect to MATH: we see that MATH and we infer from REF that MATH. The other asymptotics are obtained as follows: we differentiate the relation MATH with respect to MATH: MATH . We deduce from REF that for MATH therefore, it suffices to find an asymptotics for MATH . But MATH and it is clear now that REF holds. |
math/9909083 | The second equation at MATH can be written MATH . Therefore, if we let MATH, we have to find MATH and MATH such that REF holds, that is, MATH . Define an unbounded operator MATH in MATH by MATH . Clearly, MATH is self adjoint. By construction MATH . The lower bound of the essential spectrum of MATH is MATH; since MATH is positive, the lower bound of the spectrum of MATH is zero, which is a simple isolated eigenvalue. Therefore, solving REF will be possible if and only if its right hand side MATH is orthogonal to MATH. This defines MATH thanks to MATH . Moreover, we define completely MATH by requiring MATH . It is clear that the MATH we obtained belongs to MATH. Since the equation is invariant by the transformation MATH, MATH has to be even. |
math/9909083 | Let us compute the difference between MATH and MATH: MATH . But an explicit computation and REF give MATH that is MATH . In the same fashion, MATH . Therefore, MATH which implies immediately REF. Since MATH is a MATH function of MATH with values in MATH, and it is integrable as well as MATH, MATH is a MATH function of MATH. Let us compute MATH; we have MATH . But MATH . Similarly, MATH . Therefore, the numerator of REF is equal to MATH . With the help of REF, we obtain REF. |
math/9909083 | From REF, we obtain MATH . For MATH, we have thanks to REF MATH . But relation REF implies also that for MATH . Therefore, MATH . On the other hand, MATH . Therefore, we have also MATH . Thus we have obtained the asymptotic for MATH . Relation REF is an immediate consequence of REF. In order to have an asymptotic for MATH, we will give another asymptotic for MATH, which will be much less precise, but more usable at this level of the discussion; nevertheless, the precise asymptotic REF will be useful later. For MATH, MATH . Therefore, for MATH, MATH . In the same fashion, for MATH, MATH . By integration in MATH, we get now MATH . The value of MATH is determined by the condition that MATH should be orthogonal to MATH (see REF ). Therefore, MATH . At this point we observe that MATH . This yields the first relation in REF. The second relation in REF is a consequence of the first together with REF. Estimates REF are a direct consequence of REF of MATH, of the equivalents REF and of REF. |
math/9909083 | If MATH is an eigenvalue of MATH, it is odd, thanks to the general NAME theory. Define an operator MATH in MATH by MATH then MATH is the first eigenvalue of MATH. In particular, for all MATH, the following inequality holds: MATH . Define a test function MATH by MATH . For all large enough MATH, MATH on MATH; therefore an integration by parts on MATH shows that MATH . On the other hand, MATH therefore, we see now that MATH . Conversely, define MATH by MATH then, thanks to REF MATH the function MATH changes sign only at MATH; therefore, there exists MATH such that MATH increases on MATH and decreases on MATH. CASE: MATH. Write MATH; since MATH and MATH cannot vanish simultaneously and MATH vanishes, there exists a continuous determination of the angle MATH such that MATH . A classical calculation gives MATH . There exists a constant MATH such that for all MATH large enough, MATH if we let MATH, MATH satisfies the differential inequality by integration of the inequality MATH which we integrate as: MATH . Thanks to REF, MATH; therefore thanks to REF, MATH but the definition of MATH and the condition on MATH imply immediately that MATH which implies immediately the desired estimate REF MATH. We have to give a different argument: multiply the equation MATH by MATH and perform two integration by parts on the term involving a second derivative; we eventually obtain MATH . Normalize MATH so that MATH; then, by concavity on MATH, MATH therefore, there exists MATH such that MATH on the other hand, MATH . Therefore MATH which implies immediately REF. This proves REF . |
math/9909083 | We have MATH . This gives REF. The second object is more difficult. If we define MATH estimate REF implies that MATH. But, thanks to REF, MATH can be rewritten as MATH . Thus, we have MATH . We transform this expression using REF: MATH and by integration by REF . But MATH . From estimate REF over MATH, and the following facts MATH we can see that MATH and from asymptotic REF, we obtain MATH . This implies REF. The last of the three estimates concerns MATH . Thanks to REF, the second term will contribute a MATH. Let us estimate MATH, which will be equivalent to MATH. More precisely, MATH . From the asymptotics REF on MATH and REF on MATH, we can see that MATH . We obtain now MATH which is precisely the last of the three estimate of our list. |
math/9909083 | We have already obtained an expression for MATH: it is given by MATH . We can see that MATH . With the help of REF, we obtain MATH . An expression for MATH is given by MATH . Arguing as above, MATH . Relations REF imply the assertion of the lemma. |
math/9909083 | We first study MATH as an operator from MATH to MATH. Given MATH, we first solve MATH where MATH is to be found in MATH. REF is equivalent to the system MATH . Since MATH is perpendicular to MATH, the first equation gives MATH . But MATH is precisely equal to MATH defined at REF. Therefore, there exists a unique MATH satisfying REF and MATH . Restricted to even functions perpendicular to MATH, MATH is invertible, and its inverse is bounded thanks to REF . Therefore MATH and by classical elliptic estimates MATH . For the second equation of REF to have a solution, the following orthogonality condition must be satisfied: MATH . Hence MATH which implies immediately that MATH . The operator MATH restricted to MATH is invertible, and the norm of its inverse is estimated by MATH, thanks to REF . Therefore, the unique MATH orthogonal to MATH which solves the second equation of REF satisfies MATH . By classical elliptic estimates MATH and therefore MATH . The full system MATH can be rewritten MATH . Clearly, MATH . Therefore, for MATH small enough and MATH, we can see that MATH and our assertion is proved. |
math/9909083 | The vector MATH satisfies the equation MATH . The first equation of REF yields MATH . Since MATH we can see that REF holds. Therefore, MATH which together with the orthogonality MATH implies REF. In a similar fashion, MATH so that MATH . Hence REF holds, and REF is inferred immediately from REF. |
math/9909083 | We just have to translate the condition on MATH from the condition on MATH. We know from REF that there exist MATH, MATH and MATH such that MATH . Since the function MATH is increasing on MATH for MATH small enough, we can see that if MATH the existence theory works; but it is clear that for MATH small enough, the right hand side of REF is at least equal to MATH, and the theorem is proved. |
math/9909083 | From REF , we know that the spectrum of MATH inside MATH contains only the semisimple double eigenvalue MATH and the simple eigenvalue MATH. We know also that MATH is the closest element of the remainder of the spectrum of MATH to MATH. Define MATH . We define a twice punctured disk MATH by MATH . Then we have the following estimate on the resolvent MATH of MATH: MATH . It will be convenient to embed MATH into a holomorphic family of operators MATH defined by MATH and to do the same type of calculations as in REF; however, the results quoted therein cannot be directly applied, since our eigenvalues are not uniformly isolated with respect to the remainder of the spectrum. The NAME perturbation series for MATH is the expression MATH this series converges provided that MATH . We infer from REF that there exist MATH and MATH such that MATH without loss of generality, we may assume that for all the MATH that we consider, MATH is larger than MATH; thus the function MATH is holomorphic with values in MATH on the set MATH and the following estimate holds over MATH: MATH . If MATH is the circle defined in the statement of the lemma, any element of MATH satisfies for MATH large enough MATH. Therefore, the circle MATH is included in the resolvent set of MATH provided that MATH . This will hold for MATH and MATH large enough, and the first assertion of the lemma is proved. Since MATH is holomorphic in a neighborhood of MATH, it admits a NAME series of the form MATH with MATH where MATH is a circle about MATH of radius strictly less than MATH. The first term of the expansion is equal to MATH hence our notations are coherent. If we take MATH of radius MATH, we have the estimates MATH and also, for all MATH such that MATH, and for all MATH, MATH . The projection on the total eigenspace of MATH relative to the eigenvalues inside MATH is given by MATH . We infer from REF that the operator MATH defined by MATH satisfies the estimate MATH for all MATH such that MATH and for all MATH, we have MATH . For MATH, MATH is a holomorphic function of MATH. For MATH, relation REF implies that MATH is the projection on the space spanned by MATH, MATH and MATH; therefore, it is of dimension MATH. A classical argument shows that a continuous family of projections which is of finite rank for some value of the parameter is of finite rank for all its values. This proves the second assertion of the lemma. |
math/9909083 | It is clear that MATH, MATH and MATH are holomorphic functions of MATH if MATH. By construction, they belong to MATH. For MATH, MATH, MATH and MATH are orthogonal vectors spanning MATH; therefore, there is a neighborhood of MATH for which MATH, MATH and MATH constitute a basis of MATH. Let MATH denote the MATH-bilinear product on MATH defined as follows: if MATH belong to MATH, we let MATH . In order to estimate the size of the neighborhood of MATH where MATH, MATH and MATH constitute a basis of MATH, we introduce a quasi NAME matrix MATH defined by MATH . The terms MATH and MATH vanish because MATH and MATH are even functions, while MATH is odd. We also remark that MATH is symmetric, not hermitian, and that MATH . We infer from the hypothesis MATH and from REF that there exist MATH and MATH such that MATH . Therefore, if MATH, MATH is invertible, its inverse is analytical in MATH and it satisfies the estimate MATH . Denote by MATH the function from MATH to MATH complex matrices whose columns vectors are respectively equal to MATH, MATH and MATH. We define three vectors MATH, MATH and MATH by MATH . Then, for MATH, MATH, MATH and MATH are holomorphic in MATH and they constitute a basis of MATH which is dual to the basis MATH, MATH, MATH of MATH, that is, MATH . Thus, MATH is given by MATH . It is immediate that MATH, MATH and MATH are analytical in MATH with values in MATH: in fact, they are polynomial in MATH. The study of REF and in particular estimates REF, completed by estimate REF for MATH and estimate REF show that there exist numbers MATH and MATH such that the following estimates hold: MATH . Standard methods of analytic functions give REF and analogous estimates for MATH, MATH and MATH. Finally, differentiating with respect to MATH the relations MATH we find that MATH and MATH are eigenvectors of MATH relatively to the eigenvalue MATH. Thus MATH and MATH belong to MATH. Therefore the form of the matrix MATH is clear, and this concludes the proof of the lemma. |
math/9909083 | The first statement of the lemma is an immediate consequence of the definition of the MATH's and of the special form of MATH, MATH and MATH. Let us calculate MATH: MATH is equal to MATH and thus MATH . Hence, thanks to the theorem of residues, MATH . Similarly, according to the definition of MATH, MATH and MATH . |
math/9909083 | From the analyticity properties we infer that MATH . From REF and the value of MATH we infer immediately REF. Relation REF can be rewritten with the help of REF as MATH . It is immediate that MATH vanishes. Moreover, we perform the scalar product of the second component of the above identity with MATH; since MATH, we obtain the second relation of REF. Finally, REF implies that MATH . If we perform the scalar product of this relation with MATH and if we observe that MATH, we obtain REF. |
math/9909083 | According to REF , we have MATH and in virtue of REF, we can see that MATH . Therefore MATH and MATH . Similarly, according to REF , MATH and thanks to REF we have MATH . Therefore MATH and in particular, MATH . There remains to calculate MATH . We infer from REF that MATH and we have the following relation MATH . Therefore MATH . On the other hand, MATH . But according to REF, a NAME expansion gives MATH and thanks to REF , this expression is equal to MATH. Hence, from REF MATH . Now we obtain the expansion MATH . We observe that MATH therefore, with the help of REF, we are left with MATH . This concludes the proof of the theorem. |
math/9909083 | We recall REF; we expect that the dominant term in MATH will be MATH the remaining terms being small with respect to MATH, as we will check later. The number MATH is also equal to MATH . Thanks to estimate REF, MATH and by integration by parts: MATH . Now, thanks to REF, we obtain MATH . We differentiate REF with respect to MATH, and we recall REF of MATH; then MATH . In the above expression, the principal term is the first term on the right hand side; the other terms are estimated as follows: it is immediate that MATH thanks to REF, MATH thanks to REF, we also have MATH therefore, MATH . Thanks to REF, MATH it is straightforward that MATH . Thus, we have proved that MATH . In order to give an asymptotic of the integral in the above expression, we cut it into three pieces: one piece from MATH to MATH, which is MATH the second piece is MATH which we integrate thanks to the change of variable MATH: it is thus equal to MATH . The last piece is MATH since for MATH, MATH, this last piece is a MATH. Finally, we have obtained MATH . There remains to estimate the other terms in MATH. It is easy to see that all of them are of order MATH, and therefore negligible before the error given in the above formula. This completes the proof of the Theorem. |
math/9909083 | The asymptotic REF is a consequence of all the previous results. In order to find the critical value of MATH, we have to solve the equation MATH . After replacing MATH and MATH by their respective values, and defining the new unknown MATH we find the following equation in MATH: MATH . It is clear that there is a solution MATH of this equation which satisfies MATH . Using relations REF, we can see now that at the critical value of MATH, MATH . Relation REF is an immediate consequence of the above, together with MATH. |
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