paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0107043 | REF implies that MATH and MATH . Using REF it follows that MATH . Let MATH to get MATH . Since MATH we have that MATH . The last equality follows from NAME 's formula. Thus for MATH, MATH REF now follows. Similarly, MATH . We consider the cases MATH and MATH separately. In the first case it can be seen from REF that MATH and MATH. In this case MATH REF follows. For the second case it can be seen from REF that MATH and again MATH. In this case MATH and REF again follows. REF follows from REF. |
math/0107043 | Here we have used REF and the bounds on MATH and MATH. Since MATH and MATH, it follows that MATH . The last inequality follows from REF. Similarily, MATH . Here we have used REF and the fact that the bound on the angle between MATH and MATH implies that MATH. Using REF and the bound on MATH it follows that MATH . Since MATH or MATH it follows that MATH . Finally, REF follows from REF. |
math/0107043 | With the notation of REF , let MATH and set MATH. Let MATH be one of the infinitely many convergents satisfying REF and set MATH. Then MATH and MATH . For the last inequality we have used the condition on the MATH's in REF in the same way that the condition on the MATH's in REF was used in REF and the fact that chord length is shorter than arc length. Let MATH or MATH. By REF it follows that MATH and MATH . By REF, with MATH as above, MATH and MATH, it follows that MATH . Thus MATH . Similarily, MATH . It is not difficult to show that MATH is an uncountable set and from the remark following REF , it follows that it has measure zero. Thus MATH is an uncountable set of measure zero. |
math/0107043 | Let MATH be such that MATH. Recall that MATH, where MATH is the integer consisting of a tower of MATH twos with a MATH an top. Modulo REF, the convergents in the continued fraction expansion of MATH are MATH where once again the bar indicates that the convergents repeat modulo REF in this order. In particular, there are two fractions, MATH and MATH say, such that REF holds. Thus it is sufficient to show that MATH for all MATH, where MATH is the MATH-th convergent in the continued fraction expansion of MATH. In particular REF will hold and likewise a similar inequality when MATH is replaced by MATH, where MATH and MATH, are the two sequences of convergents corresponding to MATH and MATH. This in turn will ensure that MATH so that MATH will not converge generally by REF . We will show that, for MATH, MATH . This will be sufficient to prove the result. Indeed, let MATH denote the MATH-th tail of the continued fraction expansion for MATH. Then MATH . Thus all that remains is to prove REF. The proof of this inequality is similar to that of REF. MATH where the notation indicates that the last integer consists of a tower of MATH twos with a MATH on top. It can be easily checked that the second inequality holds for MATH. Suppose it holds for for MATH. Then MATH . Thus the first inequality in REF holds for all positive integers MATH and the result follows. |
math/0107044 | Note that MATH where of course MATH. This enables us to give a bijection MATH between MATH and the set of partitions of MATH, by induction. Let MATH be the empty partition. Let the first block of MATH be the set of letters of MATH, and let the rest of the blocks of MATH be as in MATH. |
math/0107044 | This case is analogous to the previous one. We have MATH . We give a bijection MATH between MATH and the set of partitions of MATH, by induction. Let MATH be the empty partition. Let the first block of MATH be the set of letters of MATH, and let the rest of the blocks of MATH be as in MATH. |
math/0107044 | It is plain that a permutation avoids MATH if and only if it avoids MATH (see CITE). Note that MATH where MATH means that any letter of MATH is greater than any letter of MATH. Hence we get a unique labelling of the binary tree corresponding to MATH, that is, if MATH and MATH then MATH. It is well known that there are exactly MATH (unlabelled) binary trees with MATH (internal) nodes. The validity of the last statement can be easily deduced from the following simple bijection between NAME words and binary trees. Fixing notation, we let the set of NAME words be the smallest set of words over MATH that contains the empty word and is closed under MATH. Now the promised bijection is given by MATH and MATH . |
math/0107046 | Consider MATH. For all MATH and for MATH, we have MATH, which implies that MATH is non-micro-characteristic of type MATH. |
math/0107046 | Fix an order MATH such that MATH. Then, MATH is the reduced NAME basis of MATH with respect to MATH. For each MATH, by applying the same method with the proof of CITE, we can prove that MATH is generated by MATH and MATH, MATH. Define the MATH-filtration MATH of MATH by MATH . We compute the restriction of MATH to MATH by NAME 's algorithm (see, for example, CITE). Since MATH, the MATH-function of MATH along MATH is MATH. Therefore, by CITE, we have the following isomorphism as left MATH-modules: MATH . |
math/0107046 | We fix some notation: CASE: MATH and MATH. CASE: MATH, MATH, MATH, MATH CASE: Let MATH be the linear form with slope MATH (that is, MATH). CASE: Let MATH be linear forms. We say that MATH if MATH. CASE: We will write MATH, MATH, MATH. The operators MATH are in MATH. Let MATH be MATH. Then, we have the following claims. CASE: For all linear form MATH we have MATH and so MATH for all MATH. CASE: For all linear form MATH we have MATH CASE: For all linear form MATH we have MATH and so MATH for all MATH. CASE: So, for all MATH we have MATH and then MATH is bi-homogeneous and MATH is not a geometric slope of MATH. CASE: On the other hand we have, for MATH, MATH. Then MATH and MATH. So MATH is not a geometric slope of MATH. CASE: So, the only possible geometric slope of MATH is MATH. Now, suppose that MATH is not a slope. Then, there is no slope, which implies that the MATH-characteristic variety MATH is the same for all MATH, MATH by CITE and CITE. It follows from MATH and the NAME algorithm that MATH is a NAME basis for the order defined by the weight vector MATH, MATH and a tie-breaking term order such that MATH. Therefore, the initial ideal MATH is generated by MATH, MATH, MATH and hence the MATH-characteristic variety is equal to MATH. This fact contradicts that the MATH-characteristic variety is the same for all MATH. |
math/0107046 | Since the singular locus of MATH, (MATH, MATH) is MATH, any rational solution MATH is a NAME polynomial with poles on MATH. Take a weight vector MATH. Then, we have MATH. The initial term MATH is annihilated by MATH . Therefore, MATH. This implies MATH, because MATH has a pole only on MATH. Take a MATH-basis of MATH as MATH to construct series solutions. Since CITE holds for non-homogeneous MATH as well, formal series solution MATH of MATH satisfying MATH can be uniquely expressed as MATH . When MATH, it is a polynomial. The rest part of the theorem is easy to show. |
math/0107046 | Any curve is NAME. By applying REF, the holonomic rank of MATH is MATH for all MATH. Put MATH. Consider the left MATH-morphism MATH . It has the inverse when MATH. Therefore, we have MATH and MATH. When MATH, the system admits polynomial solution, then it is reducible. It is also easy to see that when MATH, the equation is reducible. In fact, consider the left MATH-morphism MATH . It induces a morphism to the solutions by MATH . The solution MATH of MATH is sent to zero, so the image of MATH gives a proper subspace of solutions in the solution space of MATH. To find differential equations for the subspace, take all MATH such that MATH. Then, MATH. By the isomorphism REF , we conclude that when MATH, the system is reducible. Let us prove that the system is irreducible when MATH by applying the result of CITE. For this purpose, we firstly construct convergent series solution of MATH. Take MATH. Then the degree of MATH is equal to MATH. Since MATH contains the elements of the form MATH, the radical of MATH is MATH. Therefore, the top dimensional standard pairs have the form MATH. Let MATH be the zero of the indicial ideal associated to MATH: MATH . Assume MATH or MATH. Taking the lattice basis MATH, , MATH, we have the following MATH linearly independent convergent series solutions MATH . We consider the change of variables: MATH . The inverse of this change of variables is also rational and the change of variables induces that of MATH. We denote by MATH the operation of these change of variables of MATH and MATH. Since the irreducibility is invariant under any birational change of variables, we will prove the irreduciblity of the ideal MATH where MATH. Let MATH be the solution space of MATH spanned by the series MATH near MATH. If MATH is reducible, then there exists a proper subspace MATH of MATH such that MATH. We consider the vector space MATH. It is easy to see that MATH where MATH. Let us prove that MATH when MATH, which implies the irreduciblity of MATH by a contradiction. We restrict the series MATH to MATH, , MATH and replace MATH by MATH. Without a loss of generality, we may assume MATH. Then the restricted series has the form MATH . It is annihilated by the ordinary differential operator MATH . By replacing MATH by MATH, we obtain the generalized hypergeometric ordinary differential equation MATH . By CITE, this ordinary differerential equation of rank MATH is reducible if and only if MATH for all MATH. If one of them is an integer, MATH becomes an integer. Therefore, the ideal MATH contains the principal ideal generated by REF , which is maximal when MATH. We conclude that MATH is generated by REF and hence MATH. |
math/0107047 | From MATH and since MATH is a critical point of MATH, one has (with MATH) MATH . From the assumption that MATH one infers MATH . This implies MATH . A direct calculation yields MATH . From this (and a similar calculation for the last integral in the above formula) one infers MATH . Of course, similar estimates hold for the terms involving MATH. It then follows that MATH and the lemma is proved. |
math/0107047 | To get a contradiction, we assume on the contrary the existence of a sequence MATH in MATH and a sequence MATH in MATH such that MATH for all MATH and MATH . In particular, MATH strongly in MATH. Moreover, since MATH is bounded, we can assume also MATH as MATH. From MATH we get MATH . Therefore, MATH . From this we conclude that MATH which is clearly absurd. This completes the proof of the lemma. |
math/0107047 | We follow the arguments in CITE, with some minor modifications due to the presence of MATH. Recall that MATH . Define MATH . As in CITE, it suffices to prove REF for all MATH, where MATH. More precisely, we shall prove that for some constants MATH, MATH, for all MATH small enough and all MATH the following hold: MATH . For the reader's convenience, we reproduce here the expression for the second derivative of MATH: MATH . Moreover, since MATH is a solution of REF, we immediately get MATH . From this it follows readily that we can find some MATH such that for all MATH small, all MATH and all MATH it results MATH . Recalling REF, we find MATH . It follows that MATH . Hence REF follows. The proof of REF is more involved. We first prove the following claim. Claim. There results MATH . Recall that the complex ground state MATH introduced in REF is a critical point of mountain-pass type for the corresponding energy functional MATH defined by MATH . Let MATH be the NAME manifold of MATH, which has codimension one. Let MATH be the NAME manifold of MATH. One checks readily that MATH. Recall CITE that MATH is, up to multiplication by a constant phase, the unique minimum of MATH restricted to MATH. Now, for every MATH, the function MATH lies in MATH, and viceversa. Moreover MATH . This immediately implies that MATH is achieved at a point which differs from MATH at most for a constant phase. In other words, MATH is a critical point for MATH of mountain-pass type, and the claim follows by standard results (see CITE). Let MATH and consider a radial smooth function MATH such that MATH . We also set MATH. Given MATH let us consider the functions MATH . A straightforward computation yields: MATH and hence MATH . Letting MATH denote the last integral, one immediately finds: MATH . Due to the definition of MATH, the two integrals MATH and MATH reduce to integrals from MATH and MATH, and thus they are MATH, where MATH is a function which tends to MATH, as MATH. As a consequence we have that MATH . After these preliminaries, let us evaluate the three terms in the equation below: MATH . One has: MATH . In order to use REF, we introduce the function MATH, where MATH is the projection of MATH onto MATH: MATH . Then we have: MATH . Since MATH is orthogonal to MATH, MATH, then one readily checks that MATH and hence REF implies MATH . On the other side, since MATH it follows: MATH . Since MATH for all MATH, and since MATH as MATH, we infer MATH. Similarly one shows that MATH and it follows that MATH . We are now in position to estimate the last two terms in REF . Actually, using REF we get MATH . The same arguments readily imply MATH . Putting together REF we infer MATH . Using arguments already carried out before, one has MATH and similarly for the terms containing MATH. This and REF yield MATH . Let us now estimate MATH. One finds MATH . In a quite similar way one shows that MATH . Finally, REF and the fact that MATH, yield MATH . Recalling REF we infer that MATH . Taking MATH, and choosing MATH small, REF follows. This completes the proof. |
math/0107047 | Let MATH denote the projection onto MATH. We want to find a solution MATH of the equation MATH. One has that MATH with MATH, uniformly with respect to MATH, for MATH. Using the notation introduced in the previous section, we are led to the equation: MATH . According to REF , this is equivalent to MATH . From REF it follows that MATH . Then one readily checks that MATH is a contraction on some ball in MATH provided that MATH is small enough and MATH. Then there exists a unique MATH such that MATH. Let us point out that we cannot use the Implicit Function Theorem to find MATH, because the map MATH fails to be MATH. However, fixed MATH small, we can apply the Implicit Function Theorem to the map MATH. Then, in particular, the function MATH turns out to be of class MATH with respect to MATH and MATH. Finally, it is a standard argument, see CITE, to check that the critical points of MATH give rise to critical points of MATH. |
math/0107047 | As in the previous theorem, one has MATH. Now choose MATH in such a way that MATH. Define again MATH as in the proof of REF . Let MATH . From REF we get some MATH such that MATH for all MATH. To apply standard category theory, we need to prove that MATH is compact. To this end, as can be readily checked, it suffices to prove that MATH cannot touch MATH. But if MATH, one has MATH by the very definition of MATH, and so MATH . On the other hand, for all MATH one has also MATH. We can conclude from REF and elementary properties of the NAME - NAME category that MATH has at least MATH critical points in MATH, which correspond to at least MATH orbits of solutions to MATH. Now, let MATH a critical point of MATH. Hence this point MATH localizes a solution MATH of MATH. Recalling the change of variable which allowed us to pass from (NLS) to MATH, we find that MATH solves (NLS). The concentration statement follows from standard arguments CITE. The Proof of the second part of REF follows with analogous arguments. |
math/0107047 | By REF , we have to find critical points of MATH. Since MATH is compact, we can choose MATH so that MATH for all points MATH. From this moment, MATH is kept fixed. form MATH is obviously a non-degenerate critical manifold We set now MATH, MATH, MATH, and MATH. Select MATH so that MATH, and no critical points of MATH are in MATH, except fot those of MATH. Set MATH. From REF it follows that MATH is close to MATH in MATH when MATH is very small. We can apply REF to find at least MATH critical points MATH for MATH, provided MATH is small enough. Hence the orbits MATH consist of critical points for MATH which produce solutions of MATH. The concentration statement follows as in CITE. |
math/0107049 | Under the assumptions we make, MATH belongs to MATH. |
math/0107049 | Observe that MATH. Consequently, MATH . Since MATH is fixed, REF follows by induction. Next, MATH is contained in the union of MATH and MATH. For the latter observe that MATH with MATH, MATH and MATH means that MATH, and since MATH is symmetric MATH, that is, MATH. Consequently, MATH, such that REF follows from REF follows, since MATH for sufficiently large MATH. |
math/0107049 | Let MATH, and let MATH be an injective group homomorphism. We claim that there exists a right MATH-compatible map MATH, that is, MATH for all MATH and all MATH. Since MATH is free as right MATH-set, we can construct MATH as follows. Choose a right MATH-transversal in MATH, and map each element of this transversal arbitrarily to an element of MATH, and extend by the MATH-action to all of MATH. Now let MATH, and choose MATH such that MATH. Define MATH by MATH for all MATH. Since MATH is right MATH-compatible, we see that MATH is independent of the choice of MATH. For all MATH, MATH, MATH . We now show that MATH restricted to MATH is an injective group homomorphism. Suppose that MATH, so MATH, and MATH . Since MATH is injective, we see that MATH is injective on MATH. Since MATH is an arbitrary MATH-orbit in MATH, we have proved that REF holds. Now suppose that REF holds. It is straightforward to check that we have the desired MATH-bi-set structure on MATH, and that it is MATH-free and MATH-free. Finally, for any MATH, there is a well-defined, injective map from MATH to MATH, with MATH mapping to MATH, for all MATH. It is surjective, since MATH for all MATH. |
math/0107049 | The proof of REF is done by an elementary estimate of integrals. Fix MATH. Then MATH . This implies that MATH . For REF we use REF and the fact that MATH. |
math/0107049 | First observe that, in the situation of REF, we have to discuss the relation between MATH and MATH. Since the action of MATH on MATH is free, MATH is defined not only for MATH, but even for MATH, where MATH is the group NAME algebra of MATH, that is, the weak closure of MATH in MATH, or equivalently (using the bicommutant theorem) the set of all bounded operators on MATH which commute with the right MATH-action. Let MATH be the orthogonal projection onto MATH. Then MATH (where MATH is the orthogonal projection onto MATH, as explained above). Let MATH be the orthogonal projection of MATH onto the set of functions with values in MATH and supported on MATH. We need to compare MATH and MATH. Let MATH be the support of MATH, that is, the set of all MATH such that the coefficient of MATH in at least one entry in the matrix MATH is non-zero, and fix MATH such that MATH. Also let MATH be the matrix of MATH with respect to the basis MATH. A calculation shows that MATH . It follows that the value of the difference is determined by the values MATH on the MATH-neighborhood MATH of the boundary of MATH. We have MATH by the invariance REF . Since MATH, MATH and MATH, we have MATH . To simplify the notation we identify the functions on MATH supported on MATH with functions on MATH. With this convention, MATH and MATH. Consider MATH, that is, MATH. It follows that MATH . By REF , the right-hand side is determined uniquely by the restriction of MATH to MATH. Thus, if the restriction of MATH to MATH is zero, then MATH . The bounded MATH-equivariant operator MATH restricts to a bounded operator MATH. Let MATH be the orthogonal complement of the kernel of this restriction inside MATH. Then MATH restricts to an injective map MATH with image MATH. Consequently, MATH CITE. Let MATH be the orthogonal projection, and let MATH be the restriction map to the subset MATH (this again is the orthogonal projection if we use the above convention). The bounded MATH-equivariant linear map MATH restricts to a map MATH. We claim that MATH is injective. In fact, if MATH and MATH then by REF MATH. Therefore, MATH. Since MATH this implies MATH. But the restriction of MATH to MATH is injective, hence MATH as claimed. It follows that MATH . Since MATH is a free MATH-space, MATH, therefore MATH . By REF , the first summand on the right hand side tends to zero for MATH. Consequently MATH . |
math/0107049 | This follows since MATH has the corresponding properties. |
math/0107049 | The construction of MATH involve only algebraic (ring)-operations of the coefficients of the group elements, and every MATH is a ring-homomorphism, which implies the first and second assertion. Taking the adjoint is a purely algebraic operation, as well: if MATH then MATH, where for MATH we have MATH. Since these algebraic operations in the construction of MATH commute with complex conjugation, the statement about the adjoint operators follows. |
math/0107049 | This follows immediately from the definitions. |
math/0107049 | If MATH then MATH, and correspondingly for matrices. The assertion follows from the finite support condition and from MATH-equivariance MATH for MATH. |
math/0107049 | Assume MATH. With MATH we get MATH . We estimate MATH . Here we used the NAME inequality, but took the size of the support into account. It follows that MATH . Observe that MATH where MATH is the MATH-th row of MATH. Consequently, for each fixed MATH, there are not more than MATH elements MATH with MATH, and hence MATH . Now REF give the desired inequality MATH . |
math/0107049 | This is obvious from the definition, since we only remove entries in the original matrix MATH to get MATH. |
math/0107049 | Using the fact that MATH is a free left MATH-set, and a set of representatives of the translates of MATH, we see that the matrix for MATH has copies of MATH on the ``diagonal" and zeros elsewhere (compare the proof of CITE). Therefore the two matrices have the same shape and the statement follows from REF . |
math/0107049 | The argument is exactly the same as for REF , replacing the free MATH-set MATH there by the free MATH-set MATH here. |
math/0107049 | In REF , we actually prove that MATH for MATH sufficiently large. Observe that only finitely many elements of MATH occur as coefficients in MATH. Let MATH be the corresponding subset of MATH. If MATH is the inverse limit of MATH and MATH restricted to MATH is injective (which is the case for MATH sufficiently big) the claim follows because the matrices describing MATH and MATH then have the same shape, and we can apply REF . If MATH is the direct limit of the groups MATH then, since MATH has finite support, there is MATH such that for MATH the map MATH is injective if restricted to the support of MATH. We arrive at the conclusion as before. In REF , first observe that by REF MATH. Because of REF , on the other hand, MATH. |
math/0107049 | For REF, this is proved in CITE. For REF, this is essentially the content of CITE. Actually, there it is proved that MATH, but observe that by Subsection REF, MATH. Moreover, the assumption that MATH is a generalized amenable extension of MATH is slightly more general than the assumptions made in CITE, but our assumptions are exactly what is needed in the proof given there. |
math/0107049 | This is a direct consequence of positivity of the trace, of the definition of spectral density functions and of the fact that MATH by the definition of a normalized trace. |
math/0107049 | The proof only depends on the key REF . These say (because of REF ) CASE: MATH CASE: For every polynomial MATH we have MATH. For each MATH choose polynomials MATH such that REF is fulfilled. Note that by the first key lemma we find the uniform upper bound MATH for the spectrum of all of the MATH. Then by REF MATH . We can take MATH and MATH and use the second key lemma to get MATH . Now we take the limit as MATH. We use the fact that MATH is normal and MATH converges strongly inside a norm bounded set to MATH. Therefore the convergence even is in the ultra-strong topology. Thus MATH . For MATH we can now conclude, since MATH and MATH are monotone, that MATH . Taking the limit as MATH gives (since MATH is right continuous) MATH . Therefore both inequalities are established. |
math/0107049 | We specialize REF to MATH and see that MATH . On the other hand, by REF , MATH and consequently MATH . As remarked above, the conclusion for an arbitrary MATH follows by considering MATH. |
math/0107049 | Set MATH. Then, by REF , MATH and MATH. Hence, MATH . If this is (by assumption) MATH, then since MATH by our normalization MATH . We want to establish the same estimate for MATH. If MATH then MATH . Since this holds for every MATH, we even have MATH . The second integral would be infinite if MATH. It follows that MATH. Since we can play the same game for every subnet of MATH, also MATH that is, the approximation result is true. For the estimate of the determinant note that in the above inequality MATH . Therefore MATH . |
math/0107049 | The proof of the induction principle is done by transfinite induction in a standard way, using the definition of MATH. The result corresponds to CITE. |
math/0107049 | Remember that MATH and MATH in this case is the product of all non-zero eigenvalues of MATH, which is positive since MATH is a non-negative operator. If MATH is the characteristic polynomial of MATH and MATH with MATH then MATH. In particular it is contained in MATH. Moreover MATH for MATH. The product of all these numbers is fixed under MATH, therefore a rational number and is an algebraic integer, hence a (non-zero) integer. Taking absolute values, we get a positive integer, and consequently MATH . Using REF we get the desired estimate of the determinant. Since MATH is flat over MATH, MATH . Choose MATH. Since MATH is a field automorphism, MATH which by the same reasoning as above coincides with MATH. |
math/0107049 | The result is well known if MATH has no kernel and MATH. By scaling, we may assume that MATH. If MATH has a non-trivial kernel, choose a basis whose first elements span the kernel of MATH. Then MATH and MATH has a smaller dimension than MATH. It is obvious from the definitions that MATH, where MATH is the characteristic polynomial of MATH. Therefore MATH (since this is the first non-trivial coefficient in the characteristic polynomial). By induction MATH. Since MATH, the result follows. Observe that the statement is trivial if MATH. |
math/0107049 | In case of REF or REF, since MATH in this situation, by the induction hypothesis we have MATH . However, by REF , MATH if MATH is sufficiently large, moreover, MATH by REF , again for MATH sufficiently large. REF follows. It remains to treat REF . Here, MATH . On the other hand, MATH is to be considered not as an element of MATH, but as an element of MATH. Consequently, by the induction hypothesis, MATH . Dividing this inequality by MATH, and using the fact that by REF MATH, again we conclude that REF is true. |
math/0107049 | We shall repeatedly use the fact that MATH for a general matrix MATH. Set MATH. Then we have (compare REF ) MATH (for MATH). If MATH is transcendental, then by the very definition of the algebraic eigenvalue property, we see that MATH is not an eigenvalue of MATH. Therefore MATH and consequently MATH. We now deduce from CITE that MATH and we conclude that MATH. This proves the first statement and the second statement in the case MATH is transcendental. Finally when MATH is algebraic, the second statement follows from REF . |
math/0107049 | Suppose MATH,MATH are transcendental real numbers, with MATH . By the definition of the algebraic eigenvalue property, the spectral density function MATH is continuous at transcendental MATH, and so by REF MATH . As MATH is in a gap of the spectrum of every MATH, MATH . Therefore MATH . Consider a point MATH in MATH, and sequences MATH and MATH of real transcendental numbers that converge to MATH from above and below respectively. Then by REF , MATH . These constitute a strictly decreasing sequence of closed, non-empty subsets of the real line, and so have non-empty intersection. MATH . As this holds for any MATH in the spectrum of MATH, REF follows. |
math/0107049 | This follows immediately from CITE and from CITE. |
math/0107049 | For an arbitrary group MATH, let MATH indicate the ring of operators affiliated to the group NAME algebra MATH CITE and CITE; this is a ring containing MATH in which every element is either a zero divisor or invertible. Also let MATH denote the division closure of MATH in MATH, that is the smallest subring of MATH which contains MATH and is closed under taking inverses. Of course if MATH, then MATH is naturally a subring of MATH and hence MATH is also a subring of MATH. Furthermore if MATH are in distinct right cosets of MATH in MATH, then the sum MATH is direct and in particular the sum MATH is also direct. We describe the concept of a free division ring of fractions for MATH where MATH is a field, as defined on CITE. This is a division ring MATH containing MATH which is generated by MATH. Also if MATH denotes the division closure of MATH in MATH for the subgroup MATH of MATH, then it has the following property. If MATH, MATH is infinite cyclic and generated by MATH where MATH, then the sum MATH is direct. Now let MATH be an ordered group which satisfies the NAME conjecture over MATH. Then MATH is a division ring by CITE, and by the first paragraph is a free division ring of fractions for MATH. Also we can form the NAME division ring MATH as described in CITE. The elements of MATH are power series of the form MATH with MATH whose support MATH is well ordered. If MATH indicates the division closure of MATH in MATH, then MATH is also a free division ring of fractions for MATH, so by CITE there is an isomorphism from MATH onto MATH which extends the identity map on MATH. Therefore we may consider MATH as a subring of MATH. Let MATH be a positive integer and let MATH. We shall repeatedly use the following fact without comment: if MATH is a division ring and MATH, then MATH is a zero divisor if and only if it is not invertible. Also MATH is a right or left zero divisor if and only if it is a two-sided zero divisor, and is left or right invertible if and only if it is two-sided invertible. By CITE, the division closure of MATH in MATH is MATH. Suppose MATH has a transcendental eigenvalue MATH. Then MATH is not invertible in MATH, so MATH is not invertible in MATH. Therefore MATH is a zero divisor in MATH and we deduce that MATH is not invertible in MATH. Thus MATH is not invertible in MATH and we conclude that MATH is a zero divisor in MATH. Let us write MATH for the infinite cyclic group generated by MATH. Then MATH embeds in MATH and we deduce that MATH is a zero divisor in MATH. Let MATH denote the division closure of MATH in MATH. Then MATH is not invertible in MATH and hence is a zero divisor in MATH. Now MATH is just the NAME power series ring with respect to the ordered group MATH, where we have given MATH the lexicographic ordering; specifically MATH means MATH or MATH and MATH. Note that MATH is a free division ring of fractions for MATH. We may also form the NAME power series with respect to the ordering MATH means MATH or MATH and MATH. The power series ring we obtain in this case is MATH. Since the division closure of MATH in MATH is also a free division ring of fractions for MATH, we may by CITE view MATH as a subring of MATH. We deduce that MATH is a zero divisor in MATH, which we see by a leading term argument is not the case. |
math/0107049 | Let MATH. By REF , as MATH, MATH . The MATH are matrices over MATH, and so if MATH has the algebraic eigenvalue property, MATH for all transcendental MATH. So MATH that is, MATH has only algebraic eigenvalues. MATH therefore has the algebraic eigenvalue property. The same argument applies for the case where MATH has the algebraic eigenvalue property for rational matrices. |
math/0107049 | Since the trivial group has the algebraic eigenvalue property, REF implies the statement for amenable groups. The proof of the algebraic eigenvalue property for groups in MATH is done by transfinite induction, following the pattern of CITE. For ordinals MATH define the class of groups MATH as follows: CASE: MATH is the class of free groups. CASE: MATH is the class of groups MATH such that MATH is the directed union of groups MATH, or MATH is the extension of a group MATH with elementary amenable quotient. CASE: MATH when MATH is a limit ordinal. Then a group is in MATH if it belongs to MATH for some ordinal MATH. The algebraic eigenvalue property holds for groups in MATH by REF . Proceeding by transfinite induction, we need to establish that groups in MATH have the algebraic eigenvalue property given that all groups in MATH have the property for all MATH. For limit ordinals MATH, this follows trivially. When MATH for some MATH, a group in MATH either is an extension of a group in MATH with elementary amenable quotient, and thus has the algebraic eigenvalue property by REF , or is a directed union of groups MATH in MATH. Note that MATH can be regarded as a matrix MATH in MATH where MATH is a finitely generated subgroup of MATH, generated by the finite support of the MATH in MATH. By REF, the spectral density functions of MATH and MATH coincide. As subgroups of a group with the algebraic eigenvalue property also have the property, it follows that a group has the algebraic eigenvalue property if and only if it holds for all of its finitely generated subgroups. If MATH is the directed union of groups MATH, it follows that every finitely generated subgroup of MATH is in some MATH and so has the algebraic eigenvalue property. MATH therefore has the algebraic eigenvalue property. |
math/0107049 | The support of the elements MATH of MATH over MATH is finite, and so we can find a finite field extension MATH of MATH and a positive integer MATH such that MATH, where MATH is the ring of integers of MATH. If MATH is a NAME transcendental, then so is MATH. We can then regard MATH as being in MATH without loss of generality. Since MATH, we can apply REF to the operator MATH . Observe that MATH since by REF MATH. If MATH is an eigenvalue of MATH, then MATH will be an eigenvalue of MATH (since MATH). Observe that MATH . If MATH is the normalized dimension of the eigenspace to MATH, then MATH . Since MATH has coefficients in MATH, we can use REF , MATH . With MATH self-adjoint, MATH where MATH is some quadratic polynomial with coefficients depending only on MATH, and MATH is a constant depending only on MATH and MATH. Let MATH. Then combining REF , MATH . The right hand side becomes arbitrarily small as MATH. Consequently, MATH, that is, MATH is not an eigenvalue of MATH. |
math/0107049 | Write MATH, MATH. Since only finitely many of the MATH and MATH are non-zero, they are contained in a finitely generated subfield MATH. We may write MATH with MATH algebraically independent over MATH, and MATH algebraic over MATH (because of the theorem about the primitive element, one such MATH will do). Upon multiplication by suitable non-zero elements of MATH, we may assume that MATH is integral over MATH, that is, that there is a polynomial MATH with MATH, irreducible in MATH, and with MATH. This can be achieved because the quotient field of the ring of integral elements of MATH is MATH itself. Moreover, by multiplying MATH and MATH by appropriate non-zero elements of MATH (the common denominator of the MATH or MATH, respectively), we may assume that MATH, with MATH. We now proceed to construct a ring homomorphism MATH with MATH and MATH. Obviously, MATH, and this proves the claim. To construct MATH, observe that substitution of algebraic numbers for MATH defines a well defined homomorphism MATH. Each such homomorphism can be extended to MATH, provided MATH has a solution in MATH. Since the highest coefficient of MATH is MATH, MATH is not a constant polynomial. Consequently, MATH being algebraically closed, the required solution of MATH and therefore the extension of the ring homomorphism MATH to MATH always exists. Choose now MATH with MATH. Observe that MATH is integral over MATH since MATH is integral. The same holds for MATH. In particular, there exist irreducible polynomials MATH with MATH. Irreducibility implies in particular that MATH . By induction on MATH, using the fact that MATH is infinite and that any non-zero polynomial over a field has only finitely many zeros, we show that there are MATH with MATH. Consequently, if MATH is the corresponding substitution homomorphism MATH is not one of the zeros of the polynomials MATH or MATH. Let MATH be an extension of MATH. Then MATH is a zero of MATH, and MATH is a zero of MATH, that is, MATH. However, with MATH and MATH, MATH is the coefficient of MATH in MATH, and MATH is the coefficient of MATH in MATH. It follows that MATH, as desired, but MATH. |
math/0107049 | If MATH is elementary amenable and torsion free then the strong NAME conjecture over MATH is true for MATH and therefore MATH if MATH, hence MATH has trivial kernel on MATH. If MATH is right orderable then MATH has trivial kernel on MATH by CITE. In particular MATH has no non-trivial zero divisors. Since MATH is amenable, by an argument of CITE given in REF MATH fulfills the NAME condition. Therefore the assumptions of REF are fulfilled. |
math/0107049 | Suppose we are given MATH with MATH. Without loss of generality, we need to find MATH with MATH such that MATH. Write MATH and MATH, where MATH for all MATH. Let MATH (where MATH denotes MATH, the support of MATH). Using the NAME condition, we obtain a finite subset MATH of MATH such that MATH. Let us write MATH and MATH where MATH are to be determined. We want to solve the equation MATH, which when written out in full becomes MATH . By equating coefficients, this yields at most MATH homogeneous equations in the MATH unknowns MATH. We deduce that there exist MATH, not both zero, such that MATH. Since MATH is a domain and MATH, we see that MATH is not a zero divisor and the result is proved. |
math/0107049 | By REF holds if MATH. We use the NAME condition to reduce the case of matrices to the case MATH. Since MATH the assumptions imply that MATH does not have zero divisors. Therefore the NAME localization MATH of MATH is a skew field. Moreover, since MATH has no kernel on MATH it becomes invertible in the ring MATH of operators affiliated to the group NAME algebra of MATH CITE and CITE. Therefore MATH embeds into MATH. (By CITE MATH fulfills the strong NAME conjecture over MATH.) Fix now MATH. By linear algebra, we find invertible matrices MATH such that MATH with MATH. The NAME condition implies that we can find MATH such that MATH with MATH, for MATH. Therefore MATH . Applying the same principle to the entries of MATH and MATH, we can find MATH such that MATH and MATH with MATH. Altogether we arrive at MATH where all objects are defined over MATH (MATH). Moreover, MATH and MATH are invertible over MATH and therefore also over MATH and hence have trivial kernel as operators on MATH. REF translates to MATH (if MATH is the direct limit of the groups MATH, then the images of the left and right hand sides of the above in MATH are equal, consequently for all sufficiently large MATH the above equality will be true). By CITE MATH and MATH. The one dimensional case immediately implies MATH . We have the exact sequences MATH . Because of additivity of the MATH-dimension CITE MATH . Since all these operators are endomorphism of the same finite NAME MATH-module MATH, MATH and similarly MATH. Everything together implies REF. |
math/0107053 | Let us write MATH, where MATH. By REF, MATH. Comparing the coefficients of MATH in REF, we obtain MATH where the dots denote terms which depend on MATH with MATH. There exists the inverse matrix MATH whose coefficients are power series in MATH alone, and therefore the vector MATH is uniquely determined via MATH with MATH. |
math/0107053 | Note that MATH and that for MATH, MATH. Therefore, for MATH we have, MATH . Therefore the limit of MATH exists as MATH tends to infinity. Then it is equal to MATH by the uniqueness part of REF . |
math/0107053 | It suffices to consider the case MATH. Written out explicitly, the MATH-component of the LHS reads MATH . We apply REF to obtain seven terms corresponding to vertices in REF . Namely MATH where MATH . The lemma follows from the identities MATH, MATH. |
math/0107053 | For the proof we use the explicit formulas for each term in the Right-hand side given in REF. By a direct computation we verify that they are all defined, and that each operator MATH are also defined on them. Substitution MATH to the Right-hand side clearly gives MATH (because of the nontrivial vector part) unless MATH, when we get MATH (coming from the term MATH). By the uniqueness part of REF , it is enough to show that multiplication by MATH does not change the Right-hand side There are two things to show. First, we have to show that multiplication by MATH reproduces all the terms, and second, that the extra terms appearing cancel out. The first statement, informally speaking, can be reformulated as follows. If a monomial MATH with MATH is good, then monomial MATH itself is also good. We verify it by case checking. The nontrivial cases are MATH, MATH, MATH and MATH with MATH. We have to show that MATH and MATH occur in the Right-hand side Let us consider the case of MATH, the other case is similar. Expanding MATH into a sum of monomials, we see that all monomials in MATH are good, except for the monomial MATH, which we replace by the sum of good monomials MATH, using the identity (see REF ) MATH . For the second statement, we claim that cancellations always happen in pairs. We list the necessary equalities which are checked using the explicit formulas given in REF. In all equalities below, MATH stand for arbitrary non-negative integers. First we have cancellations of bad monomials of type MATH, where MATH is a good monomial. MATH . Then another set of MATH equalities are obtained from the above MATH by replacing the first factor MATH by MATH. They describe cancellations of bad monomials of type MATH, where MATH is a good monomial. In the case MATH, where MATH is a good monomial, we have the following equalities. MATH . The cancellations of bad monomials of type MATH, where MATH is a good monomial are given by replacing first factor MATH in the above formulas by MATH. Note that if MATH is good then MATH is also good. |
math/0107053 | The identity MATH is obvious. The identity MATH is equivalent to REF with MATH and MATH. |
math/0107054 | Let MATH. Then using REF, we obtain: MATH . The assertion follows from linearity of operators MATH and REF . |
math/0107054 | We only need to show the uniqueness of the solution of REF with a given initial condition. The difference REF has the form MATH where MATH is a vector whose components are labeled by MATH satisfying REF, and MATH is a matrix whose non-zero entries are MATH with some MATH. If we expand MATH as MATH then REF reduces to the relations MATH where MATH signifies terms which contain MATH only with MATH. For MATH, MATH has an inverse matrix as a formal power series in MATH. Hence MATH is uniquely determined by MATH. This proves the proposition. |
math/0107054 | The lemma will follow if we show that, for any sequence MATH of operators MATH, MATH holds unless MATH for all sufficiently large MATH. Let us verify this statement. In view of REF, it suffices to show that for any MATH we have MATH. The vector part of MATH is MATH. Hence, for any monomial MATH of degree MATH, we have MATH, where MATH is a power series and each MATH has degree MATH in MATH. Applying further a monomial MATH of degree MATH, each component of the vector part of MATH acquires a power MATH due to the shift MATH. Therefore this expression vanishes when MATH. |
math/0107054 | The proof is only case checking. |
math/0107054 | Consider the mapping MATH given by MATH . We have MATH unless MATH, and MATH. Suppose that MATH. The trajectory of MATH by the successive mappings MATH, where MATH changes its values as MATH, etc., stays at MATH exactly at the following values of MATH. The trajectory starts at MATH for MATH. Its stays at MATH while MATH. It changes to MATH when MATH for the first time after MATH. Suppose that this is when MATH. Then, the trajectory comes back to MATH when MATH. Since MATH is good, we have MATH. It stays at MATH while MATH, and changes to MATH when MATH for the first time after MATH. This pattern repeats. In particular, the trajectory comes to MATH only if the next MATH is equal to MATH. On the trajectory of MATH at MATH, MATH appears only if MATH applies to MATH. However, since the trajectory of MATH at MATH comes to MATH only if the next mapping is MATH, this never occurs. |
math/0107054 | From REF , if MATH is good its equivalence class is uniquely determined from the corresponding configuration. Suppose MATH is bad. There exists MATH such that MATH, MATH, and MATH for MATH. Since MATH and the configurations corresponding to MATH are equivalent, we have MATH. However, this is impossible because MATH is good, and by the same argument as in REF , we can show that if MATH then MATH is necessarily MATH. |
math/0107054 | We retain the notation of REF. If MATH, then there is a MATH in the MATH-th component of MATH. This MATH will then move to the left by one step with each MATH or MATH arrow, until it reaches the second component in the array. Hence the first component of MATH is MATH if and only if MATH. |
math/0107054 | Let MATH where MATH and MATH. Suppose that MATH is the cancellation MATH-arrow. Then, MATH is a good monomial, and the path after MATH has the form MATH . Replacing the cancellation MATH-arrow by MATH, the path associated with MATH after MATH becomes MATH . This shows that the MATH in MATH is the cancellation MATH-arrow. Hence we obtain a map from MATH into MATH. The inverse map can be constructed similarly. |
math/0107054 | Suppose that MATH, and MATH. We are to show that MATH and MATH corresponding to MATH and MATH, respectively, are in fact equal. There are four REF MATH, REF MATH, REF MATH and MATH, and REF MATH and MATH. The statement is clear if MATH because the part MATH is common to MATH and MATH. For MATH since MATH, we must show that MATH. Consider the marked path MATH corresponding to the monomial MATH. We have MATH. Let MATH be such that the position of MATH in the marked vertex MATH is MATH, and that of the first MATH we find when we move from MATH to the right cyclically is MATH. It is easy to see that MATH and MATH. We have MATH and MATH. Therefore, MATH. Now consider MATH. If MATH then MATH. Take MATH such that MATH. We will show that MATH, where MATH in the LHS and MATH in the Right-hand side are at the place MATH. Since MATH and MATH are good, MATH never occurs on the trajectories of MATH in these formulas. Therefore, it is enough to show MATH for both MATH. Note that MATH unless MATH. On the other hand, MATH as shown above. Therefore, MATH. This proves the statement. |
math/0107054 | We have MATH where in the second equality we used REF . Therefore MATH . |
math/0107054 | First note the following. Suppose a marked vertex has MATH at the last column. If we apply only MATH arrows on this vertex, we will never reach a cancellation MATH-arrow. Define MATH to be the integer such that MATH has the form MATH with some monomial MATH. The existence of such MATH in MATH is assured by the above remark. We write MATH with MATH or MATH, so that MATH, MATH and MATH. Starting with the marked vertex MATH let us keep track of its changes under successive applications of MATH. For the first MATH letters MATH we obtain MATH . From REF of a good monomial follows that MATH since otherwise MATH cannot appear in MATH. Continuing with the move after MATH and noting MATH, we have MATH where MATH signifies MATH if MATH, and MATH otherwise. The next letter MATH must be MATH, since otherwise the MATH will disappear by MATH and it would mean that MATH is a good monomial. After that, only MATH (MATH times) is allowed, and we arrive at MATH . If the next letter is MATH, this is a cancellation arrow. In this case we have shown that MATH with MATH. Otherwise, MATH with MATH. We can now repeat the same process for MATH, starting with the vertex MATH . After a finite number of steps we exhaust MATH and arrive at the cancellation MATH-arrow. |
math/0107054 | To simplify the notation let us write MATH, MATH, MATH and MATH. We have seen that MATH causes a change of vertices MATH . We now continue with MATH. The leftmost MATH in MATH changes the right hand side to MATH . We distinguish the three cases. CASE: MATH, CASE: MATH, CASE: MATH. REF is impossible. Indeed, if REF takes place, then MATH changes the vertex to MATH and the next letter MATH leads to a bad vertex. In REF followed by the next MATH letters MATH change the vertex to MATH . By the same reason as above, the next MATH letters MATH must all be MATH, the case MATH inclusive. The proof is over. |
math/0107054 | Recall REF . The proof is straightforward because MATH is a good monomial, and MATH contains MATH at the right end. |
math/0107054 | For each MATH let MATH. If MATH, then MATH. If MATH, then MATH. Noting that MATH (MATH), we obtain the assertion. |
math/0107054 | REF follow from REF applied to MATH . Since MATH, we have MATH and MATH. On the other hand, MATH implies MATH, whence follows REF . |
math/0107054 | Let us calculate MATH's by applying MATH step by step. REF is easily obtained. After MATH steps, MATH is given by MATH . Hence during the next MATH steps the computation of MATH involves only MATH. Since MATH, MATH takes the form MATH . The MATH for MATH can be calculated from this. |
math/0107054 | As in REF , define the factors MATH corresponding to MATH, respectively. By REF MATH we have MATH and likewise for MATH. From REF MATH we obtain MATH . Similarly we find MATH . It is clear from REF MATH and REF that MATH . The lemma follows from these relations. |
math/0107054 | Set MATH, MATH. We check MATH for MATH and MATH, by using REF . From REF MATH, REF , we have MATH . The same relations hold by replacing MATH with MATH. Combining them with MATH we obtain MATH, which proves REF MATH. Let us show REF MATH. Note that MATH for MATH. If MATH with MATH, then MATH by REF , and likewise for MATH. Hence REF MATH implies MATH. If MATH, then we can find some MATH with MATH such that MATH and MATH. Therefore, by REF , MATH . Finally, if MATH, one can show REF MATH directly since MATH, and otherwise by using REF and the assumption MATH. |
math/0107054 | For any MATH and MATH, MATH has the form MATH with MATH. REF implies that MATH has the same form with MATH. From this follow REF MATH and REF MATH. The relation REF MATH is clear from the definition of MATH. |
math/0107054 | We choose the representatives of good monomials of the form MATH where MATH. First we construct a mapping MATH. Suppose that MATH. We define for MATH where MATH . We set MATH . When MATH we understand this monomial in the following sense. Set MATH . Because of REF we have MATH and MATH if and only if MATH is equal to one of these MATH integers. We have, in particular, that the first (that is, left) MATH elements in MATH are MATH. Therefore, by REF we can find MATH such that MATH. In order to show that MATH given by REF is good it is enough to show that if MATH then MATH. This is clear because the only place where MATH occurs is at MATH, that is, we have MATH and MATH. Next, we construct the inverse mapping MATH. Suppose MATH. We modify MATH to MATH, which is still good. We construct MATH for MATH by the procedure given below. Then, the one corresponding to MATH is obtained from MATH by changing MATH to MATH. We write MATH where MATH and MATH and all MATH contains at least one MATH. This is, in general, possible because of the modification of MATH to MATH in the first step. Because of REF , we have MATH if MATH. For convenience we define MATH. Namely, MATH for all MATH. We say MATH is new if MATH. Besides, all MATH in MATH are new by definition. We define MATH for MATH so that new MATH's in MATH, when they are read from right to left, are MATH for some MATH such that MATH. We set MATH. Note that MATH because MATH. Thus we obtained an element MATH of MATH. Now, MATH and MATH for MATH are uniquely determined by the condition that MATH is of the form MATH where REF is satisfied. We have MATH. Therefore, we have MATH, and define MATH. By the construction REF are satisfied. Note that the first MATH elements of MATH are MATH. Since MATH, it implies MATH. Therefore, REF is satisfied. We have constructed mappings MATH and MATH. By comparing two constructions we can prove that they are inverse to each other. |
math/0107054 | We define MATH for arbitrary MATH by REF where MATH and MATH. The consistency of REF can be checked by induction on MATH for all MATH, reducing the case MATH to MATH, and MATH to MATH. We then define MATH for all MATH by REF . The validity of REF can be checked by induction on MATH, reducing MATH to MATH, and MATH to MATH. |
math/0107054 | For MATH we have MATH . Set MATH . The general case reads as MATH . The result REF is obtained by using the commutation relations MATH. The number of times that MATH occurs for MATH is given by MATH. For such MATH the counting of the powers of MATH and MATH differs from MATH. Taking this difference into account we obtain REF . |
math/0107054 | This proposition follows from REF . |
math/0107054 | We prove the statement by induction on MATH starting from MATH where the statement is valid because the ``if" statement is false. Suppose MATH where MATH and MATH. Let MATH be the smallest integer such that MATH and MATH. Since the color MATH is attained at MATH before MATH, by the way of coloring we have MATH and MATH. Note that MATH, and therefore we have MATH and MATH, which contradicts to the induction hypothesis. |
math/0107054 | We prove the statement by induction on MATH starting from MATH. Our induction hypotheses are the following. CASE: The vector part of MATH is of the form MATH if MATH, otherwise it is MATH; In the former case, all components are of the form MATH, and in the latter there exists MATH such that MATH if MATH and MATH if MATH. CASE: MATH . In the course of induction, we will also prove REF . If MATH, then MATH and MATH. We have MATH and MATH by REF . Note also that the vector part of MATH is MATH. In the below we use the projective notation MATH for the vector part MATH. Namely, MATH where MATH is independent of MATH. We use the symbols MATH, MATH and MATH to indicate the submonomial sitting in the right of these symbols. For example, if we write MATH, the vector part at MATH means that of MATH. Consider MATH such that MATH. The monomial MATH is of the form MATH . Here MATH are all MATH, but MATH can be a mixture. We have the restriction MATH because MATH is good. Note that MATH belongs to MATH. By the induction hypothesis the vector part is of the form MATH at MATH. Then, by the following MATH, the vector part changes to MATH, and after the next MATH of MATH, it further changes to MATH . At the point MATH it is of the form MATH . REF immediately follow from this. Note that the difference MATH have increased by MATH from MATH to MATH. By REF there is only one MATH between MATH and MATH. Therefore, the total length of blocks of color not MATH between MATH and MATH is equal to MATH. Therefore, REF is also proved. Next, we consider the case MATH where MATH. There are two cases: CASE: MATH for all MATH, and REF there exists MATH for some MATH. In the latter case, we have MATH, MATH and MATH for MATH. REF : the monomial MATH is of the form MATH . Here MATH are all MATH, but other dots can be MATH or MATH. We have MATH because MATH is good. MATH belongs to the block MATH. At the point MATH the vector part is of the form MATH. It changes to MATH at MATH, and then to MATH at MATH. REF follow from these observations. REF : the monomial MATH is of the form MATH where MATH and there are no MATH between the two MATH in this diagram. By the induction hypothesis, the vector part at the point MATH is of the form MATH. It changes to MATH at MATH, and then to MATH at MATH. REF follow from these observations. |
math/0107056 | Both sides of REF are linear in MATH and MATH, therefore it suffices to verify REF for some linear basis in the space of possible MATH's and MATH's. A convenient linear basis is formed by the series REF as the parameter MATH varies. Using the canonical anticommutation relation satisfied by MATH and MATH one then verifies REF directly. |
math/0107056 | First consider the expectation of MATH . Similarly, the variance of MATH behaves like MATH whence MATH, which concludes the proof. |
math/0107058 | We choose an equivalent representation of the space MATH as an inductive limit of countably normed spaces: MATH . Here we use the special system MATH of neighborhoods of MATH. Now let MATH for a MATH. Since MATH is an asymptotic function on the whole domain MATH we can use NAME 's Theorem and dominated convergence to calculate its NAME transform MATH by shifting the integration plane as follows: MATH with arbitrary MATH. Choosing MATH for MATH, we estimate MATH with certain MATH. This shows that MATH is an exponentially decreasing MATH-function, that is, an element of MATH . Thus, for the norm MATH on MATH with MATH we have MATH and similarly for every MATH, which shows continuity of MATH with respect to the inductive limit topologies of MATH and MATH . Then, by the classical NAME inversion formula, MATH is a continuous linear bijection and the continuity of MATH follows similarly as above. |
math/0107058 | Since MATH extends to linearly to sums of boundary values, it suffices to consider an asymptotic hyperfunction MATH represented by a single boundary value MATH from an orthant MATH. So, let MATH be an infinitesimal wedge of type MATH and MATH be a defining function for MATH. Then, for every compact set MATH such that MATH is compact in MATH, holds the estimate: MATH with arbitrary MATH. This shows MATH since MATH can be made arbitrarily small, leaving the integral unchanged by NAME 's Theorem. On the other hand, let MATH be such that all derivatives of MATH decrease exponentially outside the closed cone MATH, which can be achieved by eventually decomposing the original function utilizing exponentially decreasing multipliers. Then, the inverse NAME transform MATH is a boundary value MATH and thus a NAME hyperfunction, compare CITE. It is an easy calculation to show MATH locally uniformly in MATH. This shows MATH. The inversion formula for NAME hyperfunctions, see CITE, thus implies that MATH is a linear bijection from MATH onto MATH with inverse MATH. It remains to show continuity. If we choose a special exhausting sequence of compacta MATH for MATH, such that some points of the cylindrical surface MATH are contained in MATH, then we can make MATH in estimate REF small enough to conclude MATH . This yields MATH for all MATH, showing continuity of MATH in the topologies of MATH and MATH . |
math/0107058 | We need only consider the case MATH due to the tensor product decomposition property for MATH , see REF , and the corresponding property of MATH . If MATH then MATH for some MATH and a neighborhood MATH of MATH. Without loss of generality, we can assume that MATH. Under this conditions, it follows from elementary properties of holomorphic functions on wedge-shaped domains, see CITE, that MATH for every neighborhood MATH of MATH which is relatively compact in MATH. This shows MATH for every MATH and thus MATH. The inclusion is continuous, since the topology of MATH is weaker than that of MATH , which in turn is a consequence of NAME 's estimates. It is also finer by definition, and the inclusion has dense image. |
math/0107058 | We first consider the case MATH. By decomposing MATH linearly into components decreasing exponentially outside chosen cones, we can assume that the NAME transform of MATH can be represented as a single boundary value MATH and such that the holomorphic function MATH on an infinitesimal wedge MATH of type MATH is exponentially decreasing outside another closed cone MATH. Let MATH be an exhausting sequence of compact sets in MATH for MATH. Then, every one of the functions MATH is infra-exponential and we can apply the two cited Lemmata to conclude that there exists a positive, infra-exponential, entire function MATH and constants MATH such that MATH holds for MATH. By REF , MATH is a MATH-function on the real axis. Consider the function MATH which is for MATH and MATH absolutely integrable in MATH and its restriction to the real axis is MATH. Thus the inverse NAME transform of MATH are well-defined in the sense of NAME hyperfunctions as a boundary value MATH on an infinitesimal wedge of type MATH which is furthermore exponentially decreasing outside MATH. Now, MATH can be extended to a continuous function on the real axis, which can be shown to be of asymptotic decay, since it is the inverse NAME transform of a smooth function. Set MATH, then REF shows the claim. The case MATH follows by similar reasoning, except for the following modification: The NAME transform MATH becomes a distribution of finite order in this case and one can apply the well-known structural theorem for these, see, for example, CITE, to represent it as a finite order differential operator applied to a continuous function. After dividing by MATH and inverse NAME transformation one can conclude that MATH becomes a continuous function with a certain polynomial, that is, tempered, growth. |
math/0107058 | We have to show that the remainder of the MATH-th order expansion is in the remainder space of the same order: MATH for every MATH. The partial sum MATH is trivially a hyperfunction with support in the origin. Since the exponentially decaying NAME - NAME kernel MATH of the last section induces a continuous embedding of MATH into MATH , we can regard MATH as an asymptotic hyperfunction. Thus the difference MATH is again asymptotic and MATH follows immediately from the definition of the remainder. |
math/0107058 | Similar to the proof of REF , we represent MATH as an inductive limit MATH, MATH, with neighborhoods MATH. A locally convex topology on MATH can be generated by the system of seminorms MATH for MATH. This topology is apparently weaker than the original topology of the space MATH , but has the following property, which is sufficient for our purpose: If MATH is a sequence which is bounded in MATH and converges to zero in the seminorms MATH for every MATH, then it converges to zero in MATH for every MATH, compare REF . That means, the systems of seminorms MATH induces on the bounded subsets of MATH the original topology of that space. Now, for any MATH the set MATH is bounded in MATH and therefore it is also bounded in some MATH . We can thus argue as follows: Let MATH such that MATH for MATH. Then, there is a constant MATH for which MATH . If MATH then MATH, and thus MATH for all MATH. Since the topology of MATH is generated by all seminorms for which the restriction to every subspace MATH is again a continuous seminorm, the assertion follows. |
math/0107058 | If MATH and MATH are as above then MATH . We write the NAME expansion of MATH in MATH around MATH with remainder as MATH . Since MATH, it follows MATH . Now, MATH is in MATH and MATH, MATH. Consequently, the assertion follows from the above lemma, the definition of MATH and the fact that MATH is a continuous linear functional on MATH . |
math/0107058 | By REF , we can calculate the derivatives of MATH at MATH explicitly: MATH . That is, the stated asymptotic expansion is identical to the NAME expansion of the smooth function MATH around MATH and we have MATH for MATH by definition of the remainder space MATH . |
math/0107058 | By REF , the assertion is equivalent to: There exists a function MATH such that MATH, since then MATH fulfills the original condition. Choose the sector MATH in REF large enough to contain MATH and a function MATH as there. This function is analytic and therefore MATH in MATH and has a MATH-continuation to the point MATH. Furthermore, it is asymptotic along the real axis, and thus we can choose MATH. |
math/0107058 | We only have to show the bounds on the derivatives of MATH. The proof follows CITE. As in REF , we can assume MATH to be represented by a single boundary value MATH, MATH, such that MATH decreases exponentially outside a certain cone MATH. A coarse estimate for the NAME transform of MATH is MATH compare the proof of CITE. Since the derivatives MATH exist for arbitrary MATH, we can calculate that of, for example, MATH: MATH . Since this final result is finite for every MATH, we can replace the denominator by MATH for a suitable constant MATH. This proves the assertion. |
math/0107058 | We first calculate the NAME moments as in the proof of REF , to obtain MATH . Since MATH is MATH, we can write down its NAME series up to the order MATH with remainder around the point MATH as MATH with MATH and MATH for a suitable MATH. Inserting this in the formula above yields, since we have MATH for MATH, the exact form of MATH , namely: MATH proving the assertion. |
math/0107058 | By CITE, the MATH-th partial sum MATH of the NAME asympÂtoÂtic expansion has a unique preimage under the NAME transformation. By the same theorem, this preimage MATH is a hyperfunction with compact support and therefore especially in MATH . Thus, the remainder MATH is asymptotic and its image under MATH is exactly the remainder of the NAME asymptotic expansion. The NAME moments of the remainder vanish up to order MATH by definition. |
math/0107058 | If MATH has support in MATH then MATH vanishes for MATH and all MATH. The estimate is then immediate from REF . Let conversely MATH be a sequence of polynomials satisfying NAME 's condition and the estimate REF. From REF and the formula for the NAME moments found in the proof of REF it follows that there is an asymptotic hyperfunction MATH such that MATH. By the assumption and REF , MATH can be chosen such that every component MATH has compact support in MATH. Then MATH itself vanishes for MATH by REF . |
math/0107058 | Let MATH be a bounded sequence. Then, we have the estimate MATH which holds, with constants MATH, MATH, for every compact set MATH, every open neighborhood MATH of MATH in MATH, and uniformly in MATH. Under these conditions, CITE implies that there is a subsequence MATH converging to a limit MATH uniformly on compact subsets of MATH. The second inequality above shows that MATH converges to MATH in MATH . This shows the assertion. |
math/0107058 | Repeat the proof of CITE, or CITE with minor modifications. |
math/0107058 | We have a continuous inclusion MATH. On the other hand, an application of NAME 's integral formula as in the proof of CITE yields an embedding MATH. This suffices to show equivalence of the inductive limits in the tempered case. The asymptotic case follows by similar considerations. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.