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math/0105087 | For MATH, let MATH be the number of elements of MATH whose characteristic polynomial is MATH. One knows CITE that, since MATH, MATH . Adding up over all elements of MATH we see that MATH, and thus MATH . |
math/0105087 | These computations use the following chain of standard observations CITE. Any characteristic polynomial of an element in MATH is the characteristic polynomial of some semisimple element MATH. Moreover, the number of elements with such a characteristic polynomial is MATH where MATH is the group of elements of MATH which commute (inside MATH) with MATH. From this, the computation of MATH immediately follows. Indeed, MATH is the unique semisimple element with characteristic polynomial MATH. If MATH, then MATH, and MATH. If MATH, then the centralizer MATH is MATH, a group of dimension MATH. In either case, the lemma now follows. |
math/0105087 | The first claim is a tautology. For dimension MATH, we enumerate elements of MATH which have one as an eigenvalue. First, we index elements MATH of MATH by MATH, the order of vanishing of MATH at MATH if MATH, and half that multiplicity if MATH. To such a MATH corresponds a decomposition of MATH as MATH, where MATH and MATH are symplectic subspaces of dimensions MATH and MATH, respectively; MATH; and MATH. The factor MATH counts the number of ways of decomposing MATH. The penultimate factor MATH counts the possibilities for MATH acting on MATH, and the last factor enumerates all choices for MATH. |
math/0105087 | We treat the case MATH, and leave the remaining case for the industrious reader. REF shows that MATH, and that MATH. For MATH, MATH . Now, for any MATH, MATH. Thus, MATH . Higher order terms - those coming from MATH - contribute less than MATH to MATH; the lemma is proved. |
math/0105087 | Fix a MATH. In fact, assume MATH; for otherwise, MATH. Consider the space MATH of characteristic polynomials of elements of MATH. By considering successively the requirements for a point in MATH to satisfy MATH, we will show that MATH is a NAME open condition. The first condition is that MATH and MATH have no common root. This is clearly an open condition, as it is equivalent to the disjointness of MATH and MATH inside MATH. The second condition says that at least one of MATH and MATH is nonzero; and the final condition says that at least one of MATH, MATH, and MATH is nonzero. It is clear that MATH is nonempty if and only if MATH. So there is a constant MATH, depending on MATH but not on MATH, such that MATH . Invoking REF now proves the lemma. |
math/0105087 | If the function field satisfies REF , then we know that MATH does not divide the class number of the function field. We thus have the conditions of REF, which shows that the function field satisfies REF . On the other hand if the function field satisfies MATH, then we are in the situation of REF, and again the function field satisfies REF . |
math/0105087 | By REF, the proportion of curves with either property MATH or MATH converges to MATH, the appropriate proportion of elements of MATH. REF estimate these values for MATH and MATH, respectively. |
math/0105087 | The proof of REF uses only statements about abelian varieties, and thus applies in this setting, too. Let MATH be the sheaf of MATH-torsion of the universal abelian variety over MATH. Fix a geometric point MATH which is the Jacobian of a curve. Since the NAME locus already has full monodromy REF , the image of MATH is MATH. Thus, all the machinery exposed in this paper applies, and the result follows. By REF, as MATH the proportion of abelian varieties with a rational MATH-torsion point approaches the proportion of symplectic matrices with one as an eigenvalue. The latter ratio, or at least its leading term, is computed in REF. |
math/0105089 | For MATH we simply set MATH and observe that MATH obviously vanishes when MATH is below the support of MATH, and is zero again when MATH is large as a consequence of MATH. Thus MATH is compactly supported and MATH. Moreover, if MATH depends smoothly on parameters, so will MATH and if MATH is compactly supported in the parameters, so is MATH. Now proceed by induction on MATH. MATH clearly is compactly supported in MATH and has vanishing integral, so by the inductive assumption MATH for some compactly supported functions MATH on MATH. Take a compactly supported bump function MATH on MATH with MATH and consider MATH which is compactly supported in all its variables. Integrating in MATH we see that the integral over MATH vanishes, and so MATH. Thus MATH which completes the inductive step. |
math/0105089 | Fix MATH and cover MATH by NAME charts MATH such that only finitely many MATH intersect the support of MATH. Take a partition of unity MATH subordinate to MATH then only a finite number of MATH are non-zero. Thus MATH. MATH has vanishing integral on MATH which can be viewed as an open set in some MATH. Thus by the previous Lemma there are functions MATH, MATH with compact support such that MATH for the NAME coordinates MATH and MATH. But MATH, and if we choose a function MATH of compact support which is identically REF on the support of MATH then MATH so we see that MATH and similarly MATH. Thus MATH. Hence MATH . Since MATH, MATH and hence MATH for all MATH. Thus MATH for all MATH and hence MATH, a constant. |
math/0105089 | Neither trace can be zero, and are proportional so MATH. If MATH then there is a first MATH where MATH. Then we substitute in the normalisation condition to give MATH which implies that MATH. This contradiction shows that MATH. |
math/0105089 | A smooth trace is a multiple (in MATH) of a normalised trace. The transform of a MATH-Euler derivation by a MATH-linear automorphism is again a MATH-Euler derivation, thus the transform of a normalised trace by a MATH-linear automorphism is again a normalised trace, and so is equal to the original normalised trace. |
math/0105091 | If MATH, then MATH for some MATH. Let MATH . Using REF we see that MATH . Thus, MATH. |
math/0105091 | Let MATH. Since MATH if, and only if, MATH the first equality in REF follows easily. REF has already shown that MATH. Now choose MATH. For sufficiently large MATH, MATH. Hence, MATH. By REF , MATH. Hence, MATH. Since MATH was chosen arbitrarily, MATH and so MATH. |
math/0105091 | Let MATH. By REF , MATH. For all MATH let MATH. Note that MATH. We claim that MATH. Suppose not, so that MATH for some MATH, which we may assume satisfies MATH. It then follows that MATH . Now choose MATH such that MATH, which we may clearly do. Define MATH so that MATH . Using REF , we see that MATH . Since MATH, the right hand side is dominated by MATH. Hence, MATH . By choice of MATH, MATH. It follows that MATH . But now, REF implies that MATH, which is a contradiction. Hence, MATH for all MATH, as claimed. Since MATH is sequence of nonpositive vectors, each of which has at least one component MATH, there must be at least one coordinate, MATH, such that MATH for infinitely many MATH. But then, since MATH is decreasing, MATH for all MATH. Hence, MATH, from which REF follows. |
math/0105091 | Since MATH, this follows from REF by letting MATH and using REF . |
math/0105091 | If MATH, then by REF , MATH for all MATH, which shows that the conclusion is necessary. Now suppose that the conclusion is satisfied and let MATH, so that MATH is bounded in the supremum norm as MATH. Let MATH where the finiteness of MATH follows from the boundedness of MATH. By continuity and monotonicity of MATH, MATH . It follows from REF that MATH is nonincreasing as MATH. Since MATH is bounded as MATH, it follows from REF , that MATH is bounded too, so that MATH must converge to a limit, MATH. But then, by continuity of MATH, MATH, so that MATH, as required. |
math/0105091 | Suppose that the orbit MATH is bounded in the NAME semi-norm, so that MATH for all MATH, for some MATH. We show that MATH has an eigenvector in MATH. As discussed in REF, the rest of the argument is clear. Let MATH, so that MATH. It follows from the definition of the NAME semi-norm that MATH for all MATH. By REF , MATH and so MATH. Applying REF to MATH and considering MATH in REF , we see that MATH. By symmetry, MATH, so that MATH. Note that if MATH, then MATH. It follows that MATH . Hence MATH satisfies the conditions for REF with MATH and so MATH for some MATH. It follows that MATH, as required. |
math/0105091 | Choose MATH and MATH satisfying the conditions of REF . Since MATH, it follows from REF that MATH, for all MATH. Hence, MATH and so MATH. Putting these together, we deduce REF . |
math/0105091 | Since MATH is strongly connected, we may choose, for each pair of vertices MATH, a directed path, MATH, from MATH to MATH in MATH. Now choose MATH so that MATH is nonempty and choose MATH. Let MATH. Note that MATH, MATH and MATH. It follows from REF that MATH. Since MATH, we may choose MATH such that MATH. Let MATH be any other coordinate. By REF , using the chosen path from MATH to MATH, MATH. It follows that MATH is bounded for all MATH. Hence, MATH is also bounded, for all MATH, from which the result follows. |
math/0105091 | MATH and MATH have the same vertices, so it suffices to show that they have the same edges. It is clear that an edge of MATH is also an edge of MATH. Conversely, suppose there is an edge from MATH to MATH in MATH. Then we can find MATH such that MATH for all MATH, MATH, and MATH. Without loss of generality, assume that MATH, so that MATH. Let MATH and let MATH. We may find MATH, satisfying the convexity conditions, such that MATH. Rewriting REF , we see that MATH so that MATH. Now MATH, so using REF we see that MATH. It follows that MATH so that there is an edge from MATH to MATH in MATH. |
math/0105091 | If MATH is a topical function for which, for some MATH, MATH is unbounded in the NAME semi-norm, we can find a sequence MATH such that, MATH for all MATH and MATH. Since MATH is compact for the usual topology, we may, possibly after replacing MATH by a subsequence, assume that MATH converges in MATH. Let MATH, and MATH. By construction, MATH, and, since MATH for all MATH, MATH. Moreover, for all MATH, MATH, which shows that MATH is decomposable. Conversely, let us assume that MATH is decomposable, so that REF holds for some non-trivial partition MATH. Possibly after a reordering of indices, we may assume that MATH, MATH. Let MATH and let MATH, MATH. We may write MATH, where MATH and MATH. Let MATH. The decomposability of MATH implies that MATH. Now choose MATH. If MATH, then, for all MATH, MATH. Moreover, taking any MATH with MATH, and letting MATH, we see that MATH. If we now choose MATH, then MATH, for all MATH. Hence, MATH is unbounded in the NAME semi-norm. |
math/0105091 | For MATH this follows from REF . There are only finitely many triples MATH, where MATH is a strongly connected component of MATH and MATH. For each triple, either MATH or we can find a path from MATH to MATH in MATH. In the latter case, REF gives a function MATH; in the former case, simply take the identity function. The pointwise maximum of these functions, over all triples, gives a function MATH which satisfies REF . Now assume that MATH. As before, there are finitely many triples of the form MATH, where MATH is a strongly connected component of MATH and MATH. It is sufficient, by the same process of maximization as before, to choose a function MATH satisfying REF for each triple MATH. Let MATH and MATH be the vertices of MATH such that MATH and MATH. If MATH, which includes the case MATH, then choose the function, MATH, provided by the inductive hypothesis. Now assume that MATH. Then there is a path MATH in MATH, such that MATH and MATH. By REF we can find MATH such that, for MATH, MATH . For all MATH, MATH and MATH, define the maps MATH by MATH with the usual convention that MATH. (So that MATH.) For MATH, if MATH and MATH, then by REF, MATH for all MATH. We claim that if MATH and MATH then MATH which provides the required function MATH for this triple. To conclude the proof, it suffices to show REF. Choose the vertex MATH for MATH, so that MATH. Since MATH, we have MATH and so MATH . It follows that, for MATH, MATH . Moreover, by the inductive hypothesis, we have, for MATH, MATH while at the ends of the path, MATH . Composing the inequalities in REF we get REF, as claimed. |
math/0105091 | Assume that MATH is strongly connected and let MATH be such that MATH. If MATH and MATH then, arguing as in the proof of REF , we can use REF to show that MATH. Hence, all nonempty super-eigenspaces of MATH are bounded in the NAME semi-norm and so, by REF , MATH is indecomposable. Conversely, let us assume that MATH is not strongly connected. Recall the partial order of accessibility defined above. Choose a vertex MATH of MATH which is minimal in this partial order. Let MATH and MATH. For all MATH, there is a strongly connected component MATH of MATH such that MATH. Note that MATH. If MATH, then by REF there would be an edge from MATH to MATH in MATH, which contradicts the minimality of MATH. Therefore, for all MATH, MATH, which implies by REF that MATH is decomposable. |
math/0105091 | It suffices to show that the aggregation process stabilises by MATH. Let MATH and MATH be strongly connected components of MATH for which there is an edge MATH in MATH. By REF , MATH such that MATH. It follows that MATH such that there is an edge MATH in the syntactic graph, MATH, for otherwise, MATH would not depend on any of the components MATH for MATH. By REF , there is an edge MATH in MATH. Since MATH and MATH are strongly connected components of MATH, there is then a path in MATH from any vertex of MATH to any vertex of MATH. It follows that MATH cannot have any strongly connected components other than single vertices, so that the aggregation process must have stabilised by MATH. |
math/0105091 | If MATH is unbounded in the NAME semi-norm, then we can find a sequence MATH such that MATH for all MATH and MATH. We may assume that MATH for all MATH. Since MATH, then, possibly after replacing MATH by a subsequence, we may assume that MATH as MATH, for some MATH. Note that MATH and MATH, so that MATH is not of the form MATH. By REF , MATH . Dividing by MATH and letting MATH, so that MATH, we see that MATH . Since MATH, MATH. Dividing by MATH and letting MATH, we see that MATH. Hence MATH has a non-trivial eigenvector. The result follows. |
math/0105092 | According to REF, it is sufficient to show that if MATH is MATH-paracompact MATH-space then the twisted product MATH is also MATH-paracompact. Indeed, since MATH and MATH are compacts, MATH is MATH-paracompact. Therefore the twisted product MATH is MATH-paracompact. |
math/0105092 | Let MATH be any equivariant-movable MATH-space. With respect to REF there is a closed and equivariant embedding of MATH-space MATH to some MATH-space MATH. Let's consider all open MATH-invariant neighborhoods of type MATH of MATH-space MATH in MATH. By REF this neighborhoods will form a cofinal family in the set of all open neighborhoods of MATH in MATH, in particular, in the set of all open and MATH-invariant neighborhoods of MATH-space MATH in MATH-space MATH, which is MATH-space by REF . Hence from MATH-movability of above mentioned family follows their MATH-movability. That is, from MATH-movability of MATH-space MATH follows the MATH-movability of MATH-space MATH. |
math/0105092 | Let MATH be a MATH-space. By REF it is sufficient to prove the theorem in the case MATH. That is, we must prove that MATH is MATH-space. By REF we can consider MATH as a closed and MATH-equivariant subspace of a MATH-space MATH where MATH is a MATH-space, MATH is a space of continuous maps from MATH to MATH with compact-open topology and with an action MATH of group MATH and MATH is a closed ball of finite-dimensional Euclidean space MATH with orthogonal action of group MATH. At first, let us prove that the set MATH of all fixed points of MATH-space MATH is MATH-space. The spaces MATH and MATH are normed-spaces. Since the actions of group MATH on MATH and MATH are linear, the sets MATH and MATH will be closed and convex sets of local-convex spaces MATH and MATH, respectively. Therefore by well known theorem of CITE MATH and MATH are absolute retracts for metrizable spaces. By theorem of CITE they will be absolute retracts for MATH-paracompact spaces, too. The last conclusion is true for closed ball MATH since the set MATH is closed and convex in MATH. Since the action of group MATH on product MATH is coordinate-wise MATH . Hence, MATH is MATH-space as a product of MATH-spaces. Now let's prove, that MATH is MATH-space. Since MATH is MATH- space it will be MATH-retract of product MATH. Therefore, MATH will be a retract of MATH-space MATH and hence - MATH-space. The neighborhood case is proved similarly. |
math/0105092 | Let MATH be MATH-movable space. By REF it is sufficient to prove the theorem in the case MATH. So, we must prove the movabilities of space MATH of all fixed points. The MATH-space MATH we consider as a closed and MATH-equivariant space of some MATH-space MATH CITE. The family of all open, MATH-invariant MATH-type neighborhoods MATH of MATH-spaces MATH in MATH, will be cofinal in the set of all open neighborhoods of MATH in MATH CITE. It consists from MATH-spaces. The intersections MATH are MATH-spaces REF . They form a cofinal family of neighborhoods of space MATH in MATH. Indeed, for any neighborhood MATH of set MATH in MATH there is a neighborhood MATH of the set MATH in MATH such that MATH. Then the set MATH will be a neighborhood of set MATH in MATH, moreover MATH. There is a MATH such that MATH and therefor MATH. So the family of neighborhoods MATH is cofinal. Since MATH is MATH-movable for every MATH there is a neighborhood MATH that for any other neighborhood MATH there exist MATH-equivariant homotopy MATH such that MATH and MATH for any MATH. It is not difficult to make sure that the induced by MATH homotopy MATH satisfies the condition of movability of MATH. |
math/0105092 | The triviality of shape of space MATH is proved by method of CITE. Let's prove that the space MATH is not MATH-movable. Consider the set MATH of all fixed-points of MATH-space MATH. It is obvious that MATH. Hence, by REF it is sufficient to prove the following proposition. |
math/0105092 | Since the movability of inverse system is unchanged during the functorial transitions, it is sufficient to prove the non-movabilities of inverse sequence of groups MATH where MATH is fundamental group of space MATH and MATH is a homomorphism induced by mapping MATH. It is known that for sequences of groups the movability is equivalent to REF: The inverse system MATH of MATH category is movable if and only if the following condition, which called the NAME condition, is hold: MATH . For every MATH there exist MATH such that MATH for any MATH CITE. Thus we must prove that the sequence MATH does not satisfy the condition MATH. Let's observe that MATH is a free group with two generators MATH and MATH, and MATH is homomorphism defined by REF . It is easy to verify that MATH is monomorphism but not epimorphism. Hence for any natural MATH and MATH if only MATH. It means that the inverse sequence MATH does not satisfy condition MATH. |
math/0105092 | Without losing generality one may suppose that MATH is closed and MATH-invariant subset of some MATH-space MATH CITE. It is obvious that MATH is closed and MATH-invariant subset in MATH which is MATH-space. Let MATH be family of all MATH-invariant neighborhoods of MATH in MATH. Let us consider the family MATH, where each MATH and is MATH-invariant neighborhood of MATH in MATH. Let's prove that the family MATH is cofinal in the family of all neighborhoods of MATH in MATH. Let MATH be arbitrary neighborhood of MATH in MATH. By one theorem of CITE there exist MATH-invariant neighborhood MATH laying in MATH. Let's denote MATH where MATH is MATH-orbit projection. It is evidently that MATH is MATH-invariant neighborhood of space MATH in MATH and MATH. So in any neighborhood of space MATH in MATH there is the neighborhood of type MATH where MATH - MATH-invariant neighborhood of MATH in MATH. Now let's prove the MATH-movability of space MATH. Let MATH be MATH-movable. It means that the inverse system MATH is MATH-movable. We must prove that the induced inverse system MATH is MATH-movable. Let MATH be any index. By MATH-movability of inverse system MATH there is MATH such that for any other index MATH there exist MATH-mapping MATH which makes the following diagram MATH-homotopic commutative REF . It turn out that for given MATH the obtained index MATH satisfies a condition of MATH-movability of inverse system MATH, too. It is obvious since from MATH-homotopic commutativity of REF follows the MATH-homotopic commutativity of the next diagram REF . where MATH is induced by mapping MATH. So, the MATH-movability of space MATH is proved. |
math/0105092 | In the case MATH from the last theorem we obtain that the orbit space MATH with the trivial action of group MATH is MATH-movable. Therefore it will be movable by REF . |
math/0105092 | The orbit space MATH is closed in MATH which is MATH-space. Let MATH be arbitrary invariant neighborhood of MATH in MATH. By assumption of theorem it follows that there exist cofinal family of neighborhoods of MATH in MATH which orbits have the same type. Therefore one may suppose that all orbits of neighborhood MATH have the same type. The orbit set MATH will be a neighborhood of MATH in MATH. From movabilities of MATH follows that for neighborhood MATH there is a neighborhood MATH of space MATH in MATH which lies in neighborhood MATH and where it contracts to any predesigned neighborhood of space MATH. Let's denote MATH where MATH is orbit projection. It is evident that MATH is invariant neighborhood of space MATH lying in MATH. Let's prove that MATH in MATH contracts to any predesigned invariant neighborhood of MATH. Let MATH be any invariant neighborhood of MATH in MATH. We must prove the existing of equivariant homotopy MATH which satisfies a condition MATH for any MATH. Since MATH is neighborhood of space MATH in MATH there is homotopy MATH such that MATH for any MATH. The homotopy MATH keeps the MATH-orbit structure, because MATH and all orbits of MATH have the same types. Let's consider REF . MATH is homotopy of including MATH, keeping the orbit structure. By covering homotopy theorem of CITE there is an equivariant homotopy MATH which covers homotopy MATH and satisfies MATH. That is one have a commutative diagram REF . The equivariant homotopy MATH is unknown. For this it remains to verify that MATH. But it immediately follows from REF and commutativity of REF . |
math/0105092 | The necessity in more general case was proved in REF . Let's prove the sufficiency. Let the orbit space MATH be movable. The MATH-space MATH one can consider as a closed and invariant subset of some MATH-space MATH. Let MATH be any orbit. From the existence of slice follows that around MATH there is such invariant neighborhood MATH in MATH that MATH for any orbit MATH from MATH CITE. Since the action of group MATH on MATH is free MATH for any orbit MATH lying in MATH. Let's denote MATH. It is evident that MATH is invariant neighborhood of space MATH in MATH and that all it's orbits have the same type. Hence, by REF MATH is equivariant-movable. |
math/0105092 | It is necessary to prove the following two equalities: MATH for any MATH and MATH from MATH. Let's prove the first. Consider some cases: CASE: MATH. Then MATH. CASE: MATH. Then MATH. CASE: MATH. Then MATH. CASE: MATH. Then MATH. In this manner the second equality of REF is proved. |
math/0105092 | MATH is MATH-space as a inverse limit of MATH-spaces MATH and equivariant mappings MATH. The uniqueness of fixed-point is evident. The connectedness, compactness and metrizability follows from the properties of inverse systems CITE. MATH-nonmovability follows from REF . Let's prove that MATH from which, in particular, follows the movability of orbit space MATH. Let MATH. MATH is equimorphic to the orbit space MATH. Indeed, let's define a mapping MATH in the following way: MATH where MATH, and MATH are selected from the classes MATH in such way that MATH or what is the same MATH for any MATH. Let's prove that the mapping MATH is defined correctly. Let MATH be some other representatives of classes MATH, respectively, with the conditions MATH for any MATH. Since each class MATH has two representatives: MATH and MATH where MATH hence either MATH or MATH. But it is obvious that if for some MATH then for any MATH by equivariantness of MATH. Thus in case of another choice of representatives of classes MATH we have MATH . MATH is bijection and continues and so - homeomorphism CITE. Thus MATH where, as it is not difficult to see, MATH and the mapping MATH defined by formulas: MATH for any MATH. Thus, we have obtained that the orbit space MATH is a limit of inverse sequence MATH . By REF the mapping MATH induces a homomorphism MATH, which acts in the next way: MATH where MATH is generator of group MATH. From the above formula it follows that MATH is a null-homomorphism and hence MATH. For any MATH is also null-homomorphism and hence MATH. Therefore by NAME classic theorem CITE all MATH are null-homotopic and so MATH. |
math/0105093 | First observe that the terms MATH for MATH in the expansion sum REF are well defined in view of REF. Observe furthermore that the second expansion sum in REF converges absolutely and uniformly for MATH in compacta of MATH due to the Gaussian type factor MATH (use for example, CITE to control the convergence of the MATH's in the expansion sum). In particular, the second expansion sum in REF is analytic in MATH. We can now verify the second identity in REF term-wise using the explicit expression for the NAME polynomial as a balanced MATH series (see REF), and using the identities MATH and MATH for MATH. So it remains to prove the first identity in REF. Denote MATH for the right hand side of REF, which we consider for arbitrary, fixed MATH as an analytic, inversion-invariant function in MATH. Recall that the defining expansion sum for MATH converges absolutely and uniformly for MATH in compacta of MATH. In particular, when applying the polynomial NAME transform MATH to MATH, it is allowed to interchange summation and integration. Combined with the orthogonality relations REF for the NAME polynomials, we obtain for MATH, MATH where the last equality follows from REF . Since any analytic, inversion-invariant function MATH is uniquely determined by its image under the polynomial NAME transform MATH, we conclude that MATH for all MATH, as desired. |
math/0105093 | Let MATH and MATH. It follows from the polynomial reduction REF of the NAME function, REF, that MATH . Now MATH and MATH, so we conclude from REF and from the duality REF of the NAME polynomials that MATH . Furthermore, MATH, hence MATH as desired. |
math/0105093 | In view of the previous proposition, it suffices to prove the explicit formula for MATH. Let MATH, then MATH is given by a series expansion in NAME polynomials MATH REF which converges absolutely and uniformly on compacta of MATH, compare with the proof of REF . In particular, MATH is an inversion-invariant, analytic function. Furthermore, when applying MATH to MATH we may interchange summation and integration. The orthogonality relations REF for the NAME polynomials then show that MATH . On the other hand, REF shows that MATH is the image under MATH of the analytic, inversion-invariant function MATH . Since any inversion-invariant, analytic function MATH is uniquely determined by its image under the polynomial NAME transform MATH, we conclude that MATH as desired. |
math/0105093 | Observe that the factor MATH appearing in the formulas is invariant under the parameter involution MATH. In view of REF it thus suffices to prove the explicit evaluation formula for MATH with MATH. We show that it is in fact a reformulation of REF. Observe that MATH, since it is a compactly supported function (use here that MATH vanishes on the discrete mass points MATH of the measure MATH). For the moment we assume that the parameters MATH satisfy REF and that they are generic. Recall that REF on the parameters MATH imply that MATH and MATH. By NAME 's Theorem we thus conclude that MATH for analytic, inversion-invariant functions MATH, where MATH is given by REF. Since MATH vanishes on the discrete mass points MATH of the measure MATH, and since MATH we obtain for MATH, MATH where the last equality is due to REF. The desired identity now follows from REF in view of the explicit REF of the constant MATH. The generic conditions on the parameters MATH can be removed by continuity. |
math/0105093 | For generic parameters MATH satisfying REF and satisfying MATH and MATH, we compute MATH in two different ways. The first way is by substituting the definition of the NAME function transform MATH and using the polynomial reduction REF for the NAME function. This gives the left hand side of REF with MATH, multiplied by the constant MATH . The second way to compute MATH is by using REF (take MATH and MATH in the first formula of REF). This gives the explicit infinite product evaluation MATH . Combining both expressions for MATH gives REF with MATH. The conditions on the parameters can be removed by analytic continuation. Due to its independent interest, two alternative, direct proofs of REF are given in REF. |
math/0105093 | Let MATH. Let MATH be the meromorphic function satisfying MATH. Let MATH be the intersection of the polar divisor of MATH with the set MATH of discrete mass points of MATH. It is easy to verify that MATH . The closure of MATH in MATH is exactly the sub-space of functions MATH which vanish on MATH, since MATH is compactly supported. Hence the orthocomplement of MATH in MATH is a MATH-dimensional sub-space of MATH. Equivalently, the orthocomplement of MATH in MATH is a MATH-dimensional sub-space of MATH. The identities in REF , which were proven for generic parameters MATH, are valid for all MATH since they are regular at MATH. By REF the previous results on MATH thus imply that the orthocomplement of MATH in MATH is a MATH-dimensional sub-space of MATH. |
math/0105093 | Let MATH. Observe that the condition MATH implies MATH, so MATH for MATH iff MATH, where (recall) MATH. Hence we can define a positive, discrete measure MATH supported on MATH (MATH), with weights MATH . The corresponding MATH-space MATH is isomorphic to MATH via the surjective isometric isomorphism MATH defined by MATH . Furthermore, the image of MATH under MATH is exactly the space MATH of polynomials with complex coefficients. It thus follows from the previous lemma that MATH is dense in MATH iff MATH. A direct computation using REF shows that MATH with MATH the positive, discrete measure defined by REF and with MATH a strictly positive, bounded function with bounded inverse. We conclude that MATH is dense in MATH iff MATH. Choose now MATH and MATH arbitrarily. Then there exist parameters MATH such that MATH . This proves the corollary in case MATH. The corollary for MATH follows from the result for MATH, using the surjective isometric isomorphism MATH defined by MATH. |
math/0105093 | We first prove the last part of the proposition. Suppose that the parameters MATH satisfy REF and MATH, where MATH is the unique integer such that MATH. The dual parameters MATH then satisfy REF in view of CITE. It remains to verify the inequality MATH, where MATH is the unique integer such that MATH. By the definition of dual parameters the condition MATH is equivalent to the condition MATH. In particular, MATH. But then MATH which is the desired inequality. We now focus on the first part of the statement. We fix parameters MATH satisfying REF . Via the change of variable MATH we can rewrite the measure MATH as a positive measure MATH on MATH supported on MATH, where MATH with MATH. Here MATH is the unique integer such that MATH. Under the change of variable MATH, the NAME space MATH is isomorphic to the NAME space MATH of MATH-functions with respect to the measure MATH (compare with the identification of MATH and MATH as discussed at the beginning of this section). Consider MATH and MATH as linear sub-spaces of MATH, where MATH and MATH are the meromorphic functions satisfying MATH . Then MATH is dense iff MATH is dense. Let MATH (respectively MATH) be the restriction of the measure MATH to MATH (respectively MATH), and denote MATH (respectively MATH) for the associated MATH-space. We define surjective, continuous linear mappings MATH by MATH and MATH. Observe that MATH is non-zero MATH-a.e. due to REF on the parameters MATH. Since the measure MATH is compactly supported, we conclude that the sub-space MATH is dense, compare with the proof of REF . Let MATH be the closed sub-space of functions MATH with support contained in MATH. Then MATH since MATH vanishes on MATH, and MATH is a surjective isometric isomorphism. It follows that MATH is dense. Since MATH is dense we have that MATH is dense iff MATH is dense. It thus suffices to prove that MATH is dense iff the parameters MATH satisfy the extra condition MATH. Observe first that MATH by REF on the parameters MATH and by the definition of the integer MATH. Furthermore, for any discrete mass point MATH (MATH), we have MATH with MATH the measure REF and with MATH a bounded function with bounded inverse (we have use here the explicit expression for the weights MATH, see CITE). It follows that MATH is dense iff MATH is dense, compare the proof of REF . The desired density result is now a direct consequence of REF . |
math/0105093 | Denote the left hand side of the desired identity by MATH. It is clear that MATH is a polynomial of degree MATH in MATH. In the expansion MATH the coefficients MATH are explicitly given by MATH . Changing the summation variable to MATH and simplifying the sum yields MATH . Now applying CITE to the MATH-shifted factorial MATH appearing in the right hand side of the last formula for MATH, leads to MATH . The terminating MATH is summable by the MATH-Vandermonde formula CITE. Simplification of the resulting expression then shows that MATH as desired. |
math/0105095 | It is obvious that MATH. Suppose that MATH with MATH. We will show that MATH for all MATH. By taking out the degree MATH part, we may assume that MATH for all MATH. Let MATH. Suppose MATH is not empty. Then there exists a minimal element MATH in MATH (with respect to the partial order by reverse inclusion). Let MATH and write MATH with MATH. Let MATH. Because of the minimality of MATH, one has MATH. Thus there exists MATH such that MATH. Let MATH be the prime ideal of MATH generated by the polynomial functions vanishing on MATH. Then MATH. Thus MATH where MATH and MATH. Multiply MATH to the both sides of MATH to get MATH . Since MATH and MATH is a primary ideal, one has MATH . This is a contradiction because MATH . Therefore MATH. |
math/0105095 | First assume that MATH spans MATH and MATH. Then MATH is equal to the set of MATH-bases for MATH which are contained in MATH. Thus MATH is generated over MATH by MATH . Similarly MATH is spanned over MATH by MATH . Then REF is exactly the desired result. Next let MATH and MATH. Regard the dual vector space MATH as a subspace of MATH and the symmetric algebra MATH of MATH as a subring of MATH. Then MATH is a subset of MATH and MATH spans MATH. Consider MATH and MATH which are both contained in MATH. Note that MATH can be regarded as a MATH-module because MATH annihilates MATH. Denote the zero vector of MATH by MATH. Then it is not difficult to see that MATH . Since there exists a natural graded isomorphism MATH one has MATH . |
math/0105097 | Take any MATH such that MATH on MATH and MATH. If MATH is quasiconvex then the function MATH has a minimum in MATH. Therefore MATH and MATH. Straightforward computation shows that MATH anyway and MATH reads: MATH . With the notation MATH remark that MATH, because MATH is a vectorspace and MATH for any MATH. It follows that there is MATH such that: MATH . Integration by parts shows that MATH has more symmetry, namely: MATH which turns to be equivalent to MATH. Therefore there are MATH such that MATH. All it has been left to prove is that for any MATH there is a MATH and a vector field MATH such that MATH on MATH and MATH. For this suppose that MATH is the unit ball in MATH, take MATH a MATH map, such that MATH and define: MATH . It is a matter of computation to see that MATH is well chosen to prove the thesis. |
math/0105097 | The point REF has a straightforward proof by translation and rescaling arguments. For REF let us consider MATH and an arbitrary open bounded MATH with smooth boundary. The application which maps MATH to MATH is well defined and bijective. By REF, if the function MATH is MATH-quasiconvex then we have MATH . The change of variables MATH resumes the proof of REF. With MATH like in the hypothesis of REF, the application which maps MATH to MATH is well defined and bijective. The proof resumes as for the point REF. |
math/0105097 | Let MATH be an open bounded set and MATH. The vector field MATH verifies the condition that almost everywhere MATH is invertible. Therefore, if MATH is quasi-convex in MATH, we derive from the inequality: MATH . We implicitly used the chain of equalities MATH . We have proved that quasi-convexity implies MATH-quasiconvexity. In order to prove the inverse implication let us consider MATH such that almost everywhere MATH is invertible. We have therefore MATH and MATH. We use now the hypothesis that MATH is MATH-quasiconvex in MATH and we find that MATH is also quasi-convex. |
math/0105097 | For MATH sufficiently small consider the set: MATH . From the NAME covering theorem and from the fact that MATH is bi-Lipschitz we deduce that there is a sequence MATH such that: CASE: MATH CASE: for any MATH is approximatively differentiable in MATH and MATH CASE: we have MATH . Choose MATH such that MATH . We have therefore: MATH where the quantities MATH are given below, with their estimates. MATH where MATH. We have the estimate: MATH . Indeed, by changes of variables we can write: MATH . The difference MATH is majorised like this MATH . The function MATH is rank one convex and satisfies the growth condition MATH for any MATH. Therefore this function satisfies also the inequality: MATH . Use now this inequality, the properties of the chosen NAME covering and the uniform bound on NAME norm of MATH, MATH, to get the claimed estimate. MATH . By the change of variable MATH and the hypothesis we have MATH . Put all the estimates together and pass to the limit with MATH and then MATH. |
math/0105097 | Indeed, remark that in the proof of the previous step it is used only the fact that MATH is a group of invertible maps. |
math/0105097 | Because MATH is a group, it is sufficient to make the proof for MATH. Let MATH be a sequence weakly * convergent to MATH on MATH and MATH. For MATH sufficiently small and MATH we have MATH . It is not restrictive to suppose that MATH exists and it is finite. For any MATH there is MATH such that for any MATH. Take a minimal NAME extension MATH . The NAME norm of this extension, denoted by MATH, is smaller than some constant independent on MATH. Now, for any MATH define: MATH . According to NAME REF. CITE, there is a solution MATH of the problem MATH . Let MATH . Note that MATH. The following estimate is then true: MATH is MATH-quasiconvex, therefore: MATH . We put all together and we get the inequality: MATH . The proof finishes after we pass MATH to REF. |
math/0105097 | Given such a MATH, consider the solution of the o.d.e. problem: MATH . This is an one-parameter group in the diffeomorphism class MATH. Define then: MATH . The MATH quasiconvexity of MATH implies that MATH has a minimum in MATH. That means MATH and MATH. The first condition is trivially satisfied and the second is, by straightforward computation, just the conclusion of the proposition. |
math/0105097 | We shall use the notations from the proof of the preceding proposition. We see that MATH . Therefore there is a MATH such that MATH . Using integration by parts we find that for any indices MATH we have: MATH which implies that MATH has rank one. Hence there are MATH such that MATH. Use the definition of MATH rank one convexity to prove that REF implies the MATH rank one convexity. |
math/0105097 | We have to prove that if MATH is rank one affine then MATH. It is sufficient to prove the thesis for any MATH in an open dense set in MATH. We shall use the following maps: MATH . Take arbitrary MATH and perpendicular MATH. If MATH is MATH rank one affine then the mapping MATH is linear for any MATH. We have used here the relation and the equality MATH, for any orthogonal MATH. Rank one convexity of MATH means that the second derivative of MATH with respect to MATH vanishes for any choice of MATH and MATH. We express MATH in terms of the coordinates MATH and MATH. After some elementary computation we obtain the following minimal system of equations for the function MATH: MATH . From REF we find that MATH has the form MATH . From REF we obtain the equation MATH . From here we derive that MATH and MATH. We update the form of MATH, use REF to get MATH and REF to get MATH. We collect all the information and we obtain that MATH has the expression: MATH which proves the theorem. |
math/0105097 | The map is polyconvex hence it is rank one convex. It is not quasi-convex though. To see this fix MATH, MATH and MATH. There is a NAME solution to the problem MATH . We have then, for MATH: MATH . |
math/0105097 | Take MATH like in the hypothesis. Then for any (continuous) MATH we have MATH by straightforward computation. Use now the proof of REF to deduce the first part of the conclusion. For the second part use REF . |
math/0105097 | It is clear that REF implies the hypothesis of point REF . Indeed, the conclusion of REF can be written like this: for any MATH such that MATH we have the inequality MATH . Take a sequence of mapping MATH uniformly convergent to MATH. The previous inequality and the continuity of MATH imply: MATH . Apply now REF and obtain the thesis. |
math/0105105 | By using REF and MATH, we have: MATH . The other equality can be proven the same way. |
math/0105105 | Let MATH and MATH satisfy REF . We have MATH . This shows that MATH satisfies REF as well. |
math/0105105 | Let MATH be an algebra map. It is obvious that MATH is an antialgebra map. We have MATH . On the other hand, let MATH be an antialgebra map that satisfies REF . We define MATH. It is evident that MATH is an algebra map. We verify that MATH. Indeed, by our assumption we have MATH . Applying MATH, where MATH denotes the multiplication map of MATH, to both sides of the above equation, we obtain MATH or, equivalently, MATH . |
math/0105105 | Thanks to REF , we just need to check REF for the generators of MATH. Let MATH. We just check the validity of this condition for generators MATH, the rest being straightforward to check. We know that MATH . So the right hand side of REF converts to MATH . Now by replacing MATH and MATH in the above expression by their equals MATH and MATH respectively, and using the fact that MATH is multiplicative, we find that the above expression is equal to MATH . After cancelling the identical terms with opposite signs we obtain MATH, which is MATH . |
math/0105105 | We should verify the following identities: MATH . The cosimplicial relations are not hard to prove. We just verify the cyclic relations MATH and MATH, and leave the others to the reader. This is evident for MATH because MATH. Let MATH, and define MATH . We have MATH . On the other hand we can compute MATH in terms of the diagonal action of MATH on MATH, that is, MATH . Using REF , one has MATH . We can simplify MATH as follows: MATH . By repeating the same argument as above we obtain: MATH . Applying MATH to both side, we obtain: MATH . Now since MATH, we have MATH . Next, we check the identity between MATH and MATH, that is, MATH. MATH . Finally we check the relation between MATH and MATH. Using REF one has: MATH . The converse is evident because MATH . |
math/0105105 | Let MATH be the map MATH . It is easy to check that MATH is a contracting homotopy for the NAME complex of MATH and hence MATH for MATH and MATH. The rest follows from NAME 's long exact sequence relating NAME and cyclic cohomology. |
math/0105109 | For each MATH there exists an even polynomial MATH such that MATH and MATH . Hence MATH . The right - hand side converges almost surely (respectively, in the expectation value) to MATH. By taking the limit MATH both parts of the lemma follow. |
math/0105109 | The first part of the propositions follows under additional assumption that MATH almost surely from recent results of CITE. For the general case observe that since for all unitary matrices MATH and MATH the distributions of random variables MATH and MATH coincide, hence it is enough to prove the first part under assumption that every matrix MATH is almost surely diagonal. REF can be generalized to this case CITE. The second part of the proposition was proved by CITE. |
math/0105109 | Let us consider a probability space MATH, a standard matrix Brownian motion MATH and for each MATH we find the solution of the system of stochastic differential equations MATH together with initial conditions MATH . We have that for each MATH and MATH the joint distribution of random variables MATH, MATH coincides with the joint distribution of singular values of the random matrix MATH. The theorem will follow from the following stronger statement: for almost every MATH and every MATH we have MATH . From REF it follows that for almost every MATH we have that MATH has a continuous derivative (see REF). For a fixed MATH let MATH be the smallest MATH such that REF does not hold. Trivially we have MATH. There exists an index MATH such that MATH and for every MATH we have MATH. REF gives us MATH . It is easy to see that if there exists at least one index MATH such that MATH then MATH so it follows that for small MATH we have MATH for MATH. This contradicts the minimality of MATH. We define MATH. If we replace in REF MATH by MATH then it becomes a system of non - stochastic ordinary differential equations for MATH. If for all indexes MATH we have MATH then MATH and the solution exists and is unique in some (backward) interval. This contradicts the minimality of MATH. |
math/0105109 | We can regard MATH as a MATH - dimensional real Euclidean space equipped with a scalar product MATH. As usually we define the Laplacian to be MATH, where MATH is the orthonormal basis of this space and MATH is a derivative operator in direction MATH. Notice that MATH. We can regard MATH as a holomorphic function of MATH complex variables (=entries of the matrix). On the other hand it is a known - fact that if MATH is a holomorphic function then the Laplacian of its logarithm is a positive measure. NAME and NAME formula imply the first part of the proposition. For the second part we construct a sequence MATH, where MATH, such that MATH converges in MATH - moments to MATH and such that MATH and apply REF for the sequence MATH. This shows that MATH is nondecreasing. On the other hand the inequality MATH can be proved similarly as in REF . |
math/0105109 | Let us fix MATH. For any MATH we define functions on MATH . Each function MATH is well defined on MATH and MATH converges to the function MATH pointwise as MATH tends to MATH. Each function MATH is increasing and MATH converges pointwise to MATH as MATH tends to MATH. Since the function MATH has a polynomial growth at infinity, therefore there exist even polynomials MATH and MATH such that MATH holds for every MATH and furthermore MATH, MATH hold for every MATH. We apply REF for a pair of matrices MATH and MATH and obtain a probability space MATH and Gaussian random matrices MATH such that MATH holds for every MATH, where as usually MATH. For simplicity here and in the following we skip the obvious dependence of random variables on MATH. We have that MATH . REF show that MATH holds almost surely. Hence by taking the limit MATH we obtain that the inequality MATH holds almost surely. The upper estimate MATH follows from REF , what finishes the proof. |
math/0105109 | For the proof of the first part of the theorem let let MATH be a compact set. In the following MATH will denote either MATH or the matrix MATH. Let MATH be a smooth enough function with a compact support MATH. From the definition of the NAME measure we have MATH . Since twice differentiable functions MATH are dense in the set of all continuous functions MATH therefore the almost certain convergence of measures MATH in the weak topology to the measure MATH would follow if the sequence of functions MATH converges to the function MATH in the local MATH norm almost surely. Therefore it would be sufficient to show that for almost every MATH we have (for simplicity here and in the following we skip the obvious dependence of random variables on MATH) MATH . From REF and the NAME theorem follows that for almost every MATH we have MATH for almost all MATH. Now it is sufficient to show that MATH holds almost surely in order to apply the majorized convergence theorem. Note that MATH is subharmonic CITE and hence it is a local MATH function; therefore we only need to find estimates for the first summand in REF . REF gives us that for almost every MATH hence REF implies that MATH is uniformly bounded from below over MATH. From NAME theorem follows that for almost every MATH we have MATH . From the simple inequality MATH which holds for every MATH we have MATH . By REF we have that MATH converges almost surely, hence the family of functions MATH is almost surely uniformly bounded from above, what finishes the proof of the first part of the theorem. From the first part of theorem follows that there exists a decreasing sequence MATH of positive numbers which converges to MATH and such that for any compact MATH holds almost surely. REF implies that the majorized convergence theorem can be applied (we recall that MATH is always a local MATH function) hence MATH . The above two equations combine to give MATH almost surely. The convergence of empirical distributions of eigenvalues follows now exactly as in the proof of the first part. |
math/0105109 | Let MATH be a sequence given by REF . Since MATH almost surely CITE, hence for almost every MATH is the wanted sequence. |
math/0105109 | The square of a circular element is a free NAME element with parameter MATH. The probability density of this element can be explicitly calculated CITE and the integral MATH can be computed directly. Let us fix MATH. Let MATH be random vectors in MATH which are defined to be columns of the matrix MATH. We define MATH where MATH is the standard hermitian form on MATH. The above matrix MATH is the complex analogue of the NAME matrix; therefore - informally speaking - we can regard MATH to be the ``complex volume" of the ``complex parallelepiped" defined by vectors MATH. Of course MATH is equal to the product of MATH and MATH, where MATH is the length of the projection of the vector MATH onto the orthogonal complement of the vectors MATH. Since MATH it follows that MATH . It is easy to see that the distribution of MATH coincides with the distribution of the length of a random Gaussian vector with an appropriate covariance in the complex MATH - dimensional space and therefore MATH and hence NAME inequality gives us MATH . Above we have used that random variables MATH are independent and simple inequality MATH for MATH. Since MATH NAME - NAME lemma implies MATH almost surely. This together with REF gives us the first part of the proposition. It is possible to find a constant MATH such that for every MATH we have MATH . If MATH is the distribution of the random variable MATH then integration by parts gives MATH and the second part follows. |
math/0105111 | We have MATH . One may assume that MATH is of the form MATH with MATH and MATH, since any MATH can be uniformly approximated by such functions, and denote accordingly MATH the MATH norm of MATH's first MATH components. We shall prove the lemma in this case by showing that both sums in REF contain finitely many nonzero terms, this number being uniformly bounded on some open neighborhood of a given MATH, and that MATH and MATH are continuous for every MATH. For the second sum these two facts are trivial: MATH and MATH coincide in their first MATH components MATH, since splitting a component doesn't affect the ordering of the larger ones, and thus MATH for MATH. Moreover, MATH's continuity follows from equicontinuity of MATH. As for the first sum, given MATH with positive components (the necessary modification when MATH has zero components is straightforward), let MATH be such that MATH and consider MATH's open neighborhood MATH. In particular, for all MATH, MATH and thus, when MATH , MATH, which means that MATH and MATH coincide in their first MATH components, or that MATH for every MATH and MATH. Finally, each MATH is continuous because the series defining it converges uniformly. Indeed, for MATH and uniformly in MATH. For a given MATH, choose MATH such that MATH whenever MATH. Then MATH which proves the uniform convergence. |
math/0105111 | Let MATH, and consider the random variable MATH on MATH which counts the intervals longer than MATH. We first prove REF . (The value MATH is used in the subsequent proof of REF .) Assume that MATH is finite which is always the case for MATH since on MATH, MATH and is also true for MATH if only finitely many MATH are non zero. Then the expected (conditioned on MATH) increment MATH of MATH after one step of the underlying NAME process is well-defined. It equals MATH . The right hand side of REF converges as MATH tends to REF to MATH . Since MATH is MATH-invariant we have MATH for all MATH. Now REF follows from REF by dominated convergence since MATH. For the proof of REF note that for all MATH, MATH has full measure on sequences MATH for which the number MATH of nonvanishing components is finite because we start with MATH-a.s. and MATH can increase at most by one in each step. Given such a MATH, the expected increment MATH of MATH equals (see REF ) MATH. Therefore for MATH, MATH . Summing over MATH yields MATH . The left hand side of REF is nonnegative due to MATH and MATH. This proves REF . |
math/0105111 | Define MATH and MATH as in REF . Since MATH is compact, MATH is compact. Consequently, there are MATH and a strictly increasing sequence MATH of positive integers such that MATH converges weakly towards MATH as MATH. This limiting measure MATH is invariant under MATH by the following standard argument. For any continuous function MATH, MATH . Hence it remains to show that MATH has full MATH-measure, that is, MATH. To prove this observe that MATH (unlike MATH) is a continuous function on MATH. Therefore by REF , weak convergence and REF , MATH by which the first inequality is an equality, and thus MATH . |
math/0105111 | Let MATH be an integer and MATH. Given MATH, consider the expected increment MATH of MATH after one step of the process driven by MATH: MATH . Note that MATH is finite because of MATH. Therefore, by invariance, MATH, which implies MATH . Hence, for any MATH, MATH . Taking MATH we conclude that MATH. This proves the second claim. In addition, we have MATH for all MATH. Obviously, REF also holds true for MATH. Extend MATH to probability measures on MATH which are supported on MATH. It is enough for the proof of MATH to show that for all continuous real valued functions MATH on MATH which vanish on MATH the integrals MATH coincide for MATH. Fix such a MATH and choose a sequence of polynomials MATH which converges uniformly on MATH to MATH as MATH. Then MATH converges uniformly on MATH to MATH. Since MATH vanishes on the support of MATH and MATH we get for MATH, MATH which is the same for MATH and MATH due to REF . |
math/0105111 | We start by proving that MATH implies MATH. Fix an arbitrary MATH and consider the random variables MATH . After one step of the process MATH may increase, decrease or stay unchanged. If we merge two intervals then MATH cannot decrease, but may increase by the mass of one or two intervals which are smaller than MATH but become part of an interval which is bigger than MATH. If we split an interval then MATH cannot increase, but it decreases if the original interval was larger than MATH and at least one of its parts is smaller than MATH. Thus given MATH, the expected increment MATH of MATH after one step of the process is MATH . We bound MATH from below by MATH where MATH and MATH from above by MATH . Since MATH is invariant by assumption, MATH and therefore MATH . Consequently , MATH which is finite by assumption. Thus by NAME, MATH is MATH-a.s. eventually (for large MATH) less than REF/REF. However, MATH converges MATH-a.s. to REF as MATH tends to MATH. Thus even MATH is MATH-a.s. eventually less than REF/REF, which means that MATH is MATH-a.s. eventually empty, that is MATH. Now we assume MATH in which case there exist some MATH and MATH such that MATH . By MATH successive merges of the positive parts and MATH's invariance we obtain MATH . Next, we assume MATH and note that MATH and thus, defining MATH, one obtains an invariant measure supported on MATH. The chain determined by MATH on MATH is MATH-irreducible, and has MATH as invariant measure, with MATH. Therefore, NAME 's recurrence theorem CITE yields MATH. Finally, we assume MATH and show MATH. If MATH, then MATH, and when MATH we write MATH, where MATH has distribution MATH. Furthermore, restricted to MATH and conditioned on MATH, MATH - a.s., where in terms of the chain's sampling and merge/split interpretation, MATH is the first time a marked part of size MATH is sampled, that is, a geometric random variable with parameter MATH. Thus MATH . |
math/0105111 | In view of REF , for the study of invariant measures it is enough to restrict attention to the state space MATH, where the NAME chain MATH is MATH-irreducible, implying, see CITE, the uniqueness of the invariant measure. |
math/0105111 | The statement about positive recurrence is included in REF . The idea for the proof of the transience statement is to show that under REF the event that the size of the smallest positive part of the partition never increases has positive probability. By MATH we enumerate the times MATH at which the value of the NAME chain changes. Denote by MATH the (random) number of instants among the first MATH steps of the NAME chain in which some interval is split. Since MATH is the number of steps among the first MATH steps in which two parts are merged and since this number can never exceed MATH if MATH, we have that MATH-a.s., MATH . Let MATH denote the times at which some part is split. This part is split into two parts of sizes MATH and MATH with MATH. According to the model the random variables MATH are independent with common distribution MATH. Further, for any deterministic sequence MATH, let MATH denote the law of the process which evolves using the kernel MATH except that at the times MATH it uses the values MATH as the splitting variables. Note that MATH . Now denote by MATH the size of the smallest positive part at time MATH. We prove that for MATH, MATH . (Here and in the sequel, we take MATH if MATH). Indeed, we need only consider the case MATH, in which case there exists a MATH such that MATH, and MATH. But clearly MATH, and then the condition MATH and the fact that MATH imply MATH, as claimed. Next, fix some MATH where MATH. We will prove by induction over MATH that MATH . For MATH the left hand side of REF equals the probability that the unit interval is split in the first step with the smaller part being smaller than MATH which equals MATH. Assume that REF has been proved up to MATH. Then, with MATH, MATH . Now choose MATH minimal such that MATH. One possibility to achieve MATH is not to merge the part MATH in the next step in which the NAME chain moves. The probability to do this is MATH . Therefore REF is greater than or equal to MATH . By REF this can be estimated from below by MATH . This is due to REF greater than or equal to MATH . Along with the induction hypothesis this implies REF for MATH. Taking expectations with respect to MATH in REF yields MATH . By independence of MATH from MATH the right hand side of REF equals MATH . Observe that REF implies MATH. By NAME 's inequality and monotone convergence, REF can be estimated from below by MATH with some positive constant MATH. Since MATH for MATH this is greater than MATH where we used that due to independence MATH . Due to REF and therefore also the left hand side of REF are positive. This implies transience of MATH. |
math/0105111 | Note that we work with the MATH topology, and hence have to modify the proof in REF . The MATH topology makes the mapping MATH continuous (when MATH is equipped with the induced MATH topology). Fix MATH. By REF we have MATH . Note that for MATH is of the form MATH, with MATH, and MATH denoting the standard duality pairing. In stating this we have used the facts that MATH is continuous and bounded, and that all the mappings MATH and MATH are contractive. The continuity of MATH , then follows from MATH after observing that MATH and MATH remain bounded in any MATH neighborhoods of MATH. The continuity of MATH is obvious being the product of two continuous functions of MATH. It has thus been shown that MATH. |
math/0105111 | By REF it suffices to verify that MATH is reversing for the kernel MATH, which for simplicity will be denoted by MATH for the rest of this proof. We thus need to show that MATH . Because MATH and MATH are absolutely continuous with respect to MATH, it follows from REF that if MATH are uniformly bounded functions such that MATH, then MATH. Thus, by standard density arguments we may and shall assume MATH and MATH to be continuous. Define for each MATH the truncated intensity measure MATH, and the corresponding NAME measure MATH, with expectation operator MATH. Alternatively, if MATH is distributed in MATH according the MATH, then MATH is the distribution of MATH, that is, MATH. Observe that MATH, MATH implying that the measures MATH converge weakly to MATH as MATH. To prove REF we first write MATH and conclude that the first and third terms in REF converge to MATH as MATH by virtue of the weak convergence of MATH to MATH and MATH's NAME property, established in REF . It thus remains to be shown that, for all MATH and MATH, MATH . The truncated intensity MATH has finite mass MATH, and thus MATH, MATH - a.s. In particular each MATH can be naturally represented as a sequence MATH of symmetric MATH's MATH, with MATH for each MATH. As a result, and in terms of the expectation operators MATH of MATH conditioned on MATH, we may write MATH while by REF of MATH and the properties stated above of the NAME random measure conditioned on MATH, MATH where MATH. Our goal is to prove that this expression, after summing in MATH, is roughly symmetric in MATH and MATH (as stated precisely in REF ). Obviously MATH, and in addition we aim at showing that MATH (with an error appropriately small as MATH). This will be achieved by a simple change of variables, including the splitting coordinate MATH in MATH. In the integral of the MATH - th term in MATH perform the change of variables MATH given by MATH (or MATH). More precisely, MATH for which MATH and MATH, so that MATH (MATH is as the term preceding it but with the MATH and MATH integrals taken in MATH, and the notation MATH means that the variable MATH has been eliminated from the denominator) MATH (by MATH's symmetry, the sum's MATH terms are equal, hence the last equality). On the other hand, and for the same reason of symmetry, the MATH terms in MATH are all equal so that MATH . Comparing REF with REF , and observing that by REF, we conclude that there exists a MATH such that, for MATH, MATH . Applying REF via REF twice, once as written and once reversing the roles of MATH and MATH, and noting that MATH, we have MATH from which REF follows immediately since MATH. |
math/0105111 | CASE: The NAME law MATH is reversing by REF , and hence invariant. We now show that it belongs to MATH. Note first that MATH is absolutely continuous with respect to MATH: for any MATH with MATH, it holds that MATH where we used the fact that under MATH, MATH and MATH are independent, with MATH being distributed according to the Gamma law MATH of density MATH (see CITE). It thus suffices to compute the limit MATH where all MATH are pairwise distinct and nonzero, to have MATH . For such MATH, set MATH and MATH. Define MATH . By the memoryless property of the NAME process, for any NAME subset MATH, MATH where REF , as above, is due to CITE. Further, recall that MATH and MATH are independent. Recall that the density of the NAME process at MATH is MATH, where MATH. Performing the change of variables MATH, one finds that the Jacobian of this change of coordinate is MATH (in computing this Jacobian, it is useful to first make the change of coordinates MATH where MATH are considered as independent coordinates, and note the block-diagonal structure of the Jacobian). It follows that MATH which is real analytic on MATH. Thus, MATH. In passing, we note that MATH on MATH when MATH. CASE: First we show that the family of functions MATH associated with MATH, determines MATH. To this end, define for MATH functions MATH by MATH . Note that any function MATH with MATH can be written after expansion of the product as a (finite) linear combination of functions MATH with MATH. Since we have by the definition of MATH that MATH the family MATH therefore determines the expectations MATH. Consequently, MATH determines also the joint laws of the random variables MATH under MATH. We claim that these laws characterize MATH. Indeed, let MATH be the distribution of the random variable MATH under MATH. Since MATH is injective it suffices to show that the distributions of MATH under MATH determine MATH. But, since any continuous test function MATH on the compact space MATH can be uniformly approximated by the local function MATH, this is true due to MATH . CASE: For MATH, the set of numbers MATH with MATH and MATH yare enough to characterize MATH, and hence by the first part of the proof of REF, to characterize MATH. It is thus enough to prove that MATH uniquely determines these numbers. Toward this end, first note that MATH . To simplify notations, we define MATH and extend MATH to a function on MATH by setting it REF on the complement of MATH. For MATH we have MATH . Indeed, for MATH this is REF while for MATH, and arbitrary MATH, MATH which implies REF . Now we fix MATH, apply MATH to the test function MATH, with MATH having pairwise distinct coordinates, and take MATH, which yields the basic relation MATH . Here the left hand side represents mergings and splittings that produce a new part roughly at one of the MATH-s; the right hand side represents parts near one of the MATH-s that merge or split. After multiplying by MATH, rearranging and using REF to get rid of the integral with upper limit REF, we obtain the equality MATH . We now proceed to show how REF - REF yield all the required information. As starting point for a recursion, we show how to compute MATH for all MATH. Taking in REF - REF all MATH except for MATH and using the continuity of the functions MATH yields MATH . Letting MATH we get MATH. With MATH as start of the recursion this implies MATH . For the evaluation of the derivatives of MATH we proceed inductively. Recall the functions MATH defined in REF , and write MATH, MATH, for MATH. Fix MATH such that MATH, with MATH. Our analysis rests upon differentiating REF - REF MATH times with respect to MATH; to make this differentiation easy, call a term a G term of degree MATH if it is a linear combination of terms of the form MATH and MATH and MATH and MATH . Note that REF - REF contains one G term of degree MATH in REF and that differentiating a G term of degree MATH once yields a G term of degree MATH. Thus, differentiating the G term in REF MATH times and substituting MATH, we recover a constant multiple of MATH. Similarly, call a term an H term of degree MATH if it is a linear combination of terms of the form MATH . Observe, that differentiating an H term of degree MATH produces an H term of degree MATH. If we differentiate twice the term MATH in REF we get an H term of degree REF. Therefore differentiating this term MATH times results in an H term of degree MATH. Since the term MATH in REF is an H term of degree -REF, differentiating this term MATH times produces also an H term of degree MATH. Thus both terms produce after MATH-fold differentiation and evaluation at MATH a constant multiple of MATH. The H term MATH in REF is treated more carefully. It is easy to see by induction that its MATH-th derivative equals MATH. Evaluating it at MATH gives MATH. Moreover, the terms in REF for MATH vanish when differentiated twice with respect to MATH. Summarizing the above, we conclude by differentiating REF - REF MATH times with respect to MATH and subsequent evaluation at MATH that there are some constants MATH, such that MATH . For MATH only the first three lines do not vanish and give a recursion which allows us to compute starting with REF all derivatives MATH. Further differentiating with respect to MATH, one concludes that there exist constants MATH such that MATH where MATH unless MATH possesses at least one component which is zero. We now compute iteratively any of the MATH, with MATH: first, substitute in REF MATH to compute MATH, for all MATH. Then, substitute MATH REF to compute iteratively MATH from the knowledge of the family MATH, etc. More generally, having computed the terms MATH, we compute first MATH by substituting in REF MATH, and then proceed inductively as above. |
math/0105113 | If MATH is the canonical projection, then we have an isomorphism MATH for every MATH (we put MATH). Indeed, for a MATH-algebra MATH, a MATH-valued point of MATH is a ring homomorphism MATH such that MATH for all MATH. Here MATH is an equation defining MATH. Therefore we can write MATH, and MATH is a homomorphism if and only if the classes of MATH in MATH define a MATH-valued point of MATH. But MATH is uniquely determined by the classes of MATH in MATH, so this proves the isomorphism in REF . Recall that we have MATH. An easy application of REF shows that for every MATH, we have MATH. We deduce that if MATH, then MATH . A recursive application of this inequality shows that for every MATH, we have MATH, if MATH, and MATH, if MATH. This implies MATH . By REF , there is MATH such that MATH and we get MATH. |
math/0105113 | We know that MATH and MATH. By REF , there is MATH such that MATH . We first show that if MATH and REF holds for MATH, then it holds also for MATH. Since MATH, it follows from REF that MATH . The isomorphism REF in the proof of REF implies MATH . On the other hand, we have MATH . Since MATH, we deduce from REF that we have equality in REF , hence in REF . Therefore REF holds also for MATH. The above argument shows that REF holds for MATH, so that we have MATH . REF in the proof of REF gives MATH. Since MATH for every MATH, we deduce that there is a closed subset MATH with MATH such that MATH for all MATH. Fix MATH. If MATH, then REF would give MATH, a contradiction. Therefore we must have MATH. By a linear change of coordinates we may assume that MATH and we write the equation MATH of MATH as MATH, with MATH homogeneous of degree MATH for all MATH. Since MATH for MATH, we deduce that MATH for MATH, so that MATH, where MATH is the hypersurface defined by MATH in MATH. Since MATH, by applying inductively the above argument we get a homogeneous hypersurface MATH such that MATH. Note that our assumption on the singular locus of MATH implies that MATH. The converse is standard, and in fact we will prove a slightly stronger statement in the next proposition. |
math/0105113 | An equivalent statement with that of the proposition is that MATH. Since MATH, we have MATH. This implies that every irreducible component of MATH dominates MATH, so that MATH. The fact that this number is MATH is well-known. To see this using jet schemes we can use the analogue of REF in the proof of REF together with the fact that MATH where MATH is the projection corresponding to MATH. By induction we get MATH, for every MATH, and REF gives MATH. |
math/0105114 | If MATH then this follows from CITE; see REF above. Suppose MATH. Note that MATH is taken so that MATH. Applying REF, we have MATH . |
math/0105114 | Let MATH and let MATH and MATH be MATH - factors such that MATH . Then using REF we have MATH . |
math/0105114 | For REF , if MATH then this is REF . If MATH but MATH then this is REF . Suppose MATH and MATH. Let MATH be such that MATH. Then applying REF twice gives MATH . Now the proofs of REF - REF are obtained by rescaling both sides of the desired isomorphisms by the same MATH which is small enough and applying REF and perhaps REF . For example, to prove REF let MATH and use REF three times to get MATH . For REF , if MATH is large enough then by CITE, MATH is the interpolated free group factor MATH. By CITE, MATH . |
math/0105114 | Use REF followed by REF . |
math/0105114 | For MATH , if MATH then by REF , MATH while MATH can be seen be choosing MATH. For MATH , if MATH then using REF , MATH . Taking MATH shows MATH . |
math/0105115 | If MATH for some MATH then this follows directly from the proof of REF. Suppose MATH is not a reciprocal integer. From the proof of REF, MATH where MATH are some partial isometries with MATH and MATH. Moreover, the family MATH of subalgebras of MATH is free over a common two - dimensional subalgebra MATH and MATH where MATH . Let MATH . Note that MATH. We will find MATH such that MATH. First suppose MATH, that is, MATH . Then MATH, and it suffices to take MATH where MATH is the hyperfinite MATH - factor. Now suppose MATH . Then MATH so MATH and MATH where MATH. If we can find MATH with MATH and such that MATH has no central and minimal projections of trace MATH, then we will have MATH, as required. Since MATH it will suffice to let MATH where MATH and REF holds. We must show this is possible. Noting MATH setting MATH and solving yields MATH . But the lower bound REF gives that MATH. We can take MATH for suitable MATH and MATH making REF hold, and this yields MATH. Finally, suppose MATH . Then MATH and MATH, so REF holds. In the isomorphism REF, MATH. So letting MATH we find that MATH. Therefore, in every case we have MATH where MATH with MATH and MATH, each MATH is a tracially identical copy of MATH and the family MATH is free. Then MATH and the family MATH is free over MATH. But MATH is in MATH both freely complemented by MATH and unitarily equivalent to MATH. Hence MATH is freely complemented in MATH by an algebra isomorphic to MATH. Similarly, as MATH is in MATH both freely complemented by an algebra isomorphic to MATH and unitarily equivalent to MATH, we conclude that MATH is freely complemented in MATH by an algebra isomorphic to MATH. Altogether, we have that MATH is freely complemented in MATH by an algebra isomorphic to the algebra REF. |
math/0105115 | There is MATH and there are MATH such that MATH, MATH and MATH (MATH). Define MATH by MATH . |
math/0105115 | Let MATH denote the inclusion. Let MATH and let MATH denote the embedding arising from the free product construction. By REF and the hypothesis on MATH, there is an isomorphism MATH such that MATH. By REF , MATH extends to an isomorphism MATH such that MATH. |
math/0105115 | Let MATH be so large that MATH is isomorphic to the interpolated free group factor MATH for both MATH and MATH. Let MATH be a system of matrix units in MATH and let MATH . Then MATH . By CITE, MATH and MATH are free. Choosing any isomorphism MATH and taking the identity maps on MATH and MATH, we construct the desired isomorphism MATH. |
math/0105115 | Let MATH . Let MATH and MATH be the embeddings arising from the free product construction. Let MATH . Then by CITE, MATH is the MATH - factor MATH. Let MATH and let MATH be such that MATH and MATH. By CITE, MATH and MATH are free. Let MATH, MATH, let MATH and let MATH be the restriction of the free product trace on MATH to MATH. Then the pair MATH satisfies the desired properties. Moreover, if the MATH are fixed then MATH is clearly unique up to trace - preserving isomorphism. However, using partial isometries in MATH, the projections MATH may be chosen arbitrarily so long as MATH. This shows the desired uniqueness. |
math/0105115 | By the results of CITE, the free product of a MATH - factor with any NAME algebra is a factor. Hence if MATH then MATH is a factor. By induction, it follows that MATH is a factor whenever MATH is finite. For MATH infinite, factoriality of MATH can be proved by transfinite induction on the cardinality of MATH. Let MATH be a well - ordering of MATH with the order structure of the least ordinal having the same cardinality as MATH. Given MATH, let MATH and let MATH. Then MATH . By the induction hypothesis, each MATH is a MATH - factor. As MATH it follows that MATH is a factor. |
math/0105115 | In order to prove freeness over MATH of MATH and MATH, it will suffice to show MATH where for subsets MATH and MATH of an algebra, MATH is the set of all words which are products MATH, of elements MATH, satisfying MATH. Let MATH, MATH and MATH. Then MATH is the weak closure of the linear span of MATH, where MATH . The set MATH is the weak closure of the linear span of MATH and MATH, respectively MATH, (MATH), is the weak closure of the linear span of MATH, respectively MATH, where for MATH, MATH is the set of words in MATH . Note that every element of MATH has at least one letter from MATH. We have that MATH is the weak closure of the linear span of MATH . Thus, in order to prove REF, it will suffice to show MATH . However, beginning with a word MATH from the left hand side of REF, one can erase parentheses and regroup to show that MATH is equal to a word from MATH. Then MATH follows by freeness. |
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