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math/9910135 | Let MATH. Then we have MATH where the summation is taken over alternate MATH-tensor products of MATH and MATH. If we apply the isomorphism in REF from outside in each summand (note that MATH commutes with MATH and does not touch on the MATH-action), we end up with a realization of a summand in MATH with the tensor component MATH at a prescribed position: two realizations with MATH at adjacent positions correspond to the choices of isomorphisms MATH or MATH at the last stage of reductions. There are two ways of ambiguity other than the ones related by shift isomorphisms as indicated by the following diagram MATH . This is however nothing but two ways of reductions in the right hand side of MATH and the commutativity of the diagram is eventually reduced to the associativity of multiplication in the algebra MATH. It is now straightforward to check that the MATH action on MATH corresponds to the left and right multiplications of MATH on MATH. |
math/9910135 | Let MATH be such that MATH (MATH for example). Then MATH is realized as a MATH invariant closed subspace of MATH. Since the MATH action is continuously extended to the MATH action on MATH by previous lemmas, the same holds on MATH. |
math/9910135 | Since MATH is a C*-functor, it suffices to consider the case MATH. From the above definition of MATH, given MATH and MATH, the subspace MATH of MATH is invariant under the action of MATH and is equivalent to the obvious irreducible representation of MATH on the vector space MATH. Since the family MATH is a complete set of irreducible representations of MATH, we conclude that MATH is a faithful representation of MATH. |
math/9910135 | Let MATH be the algebraic free product subalgebra. For MATH of the form MATH with MATH, MATH, MATH and MATH (here MATH and MATH), we have MATH (MATH is with respect to the canonical trace MATH on MATH) and hence MATH shows that MATH. |
math/9910135 | Set MATH. Then, from the operator-valued inner product formula, MATH . Thus we can define an isometry of MATH to MATH by the same formula as MATH, which is obviously MATH linear and has the range MATH. Now, for MATH, MATH, writing MATH and MATH with MATH, MATH elements in alternate MATH-tensor products of MATH and MATH, we have MATH . |
math/9910135 | We first show the associativity on the subspace MATH, which is checked by MATH . Now the associativity is extended to the whole space as follows: By the density of MATH and MATH in MATH, we see that MATH . Since MATH's are MATH linear, the above density ensures the overall validity of the desired associativity. |
math/9910135 | Recall that the MATH linear unitary map MATH is defined by MATH where MATH denotes the NAME transform of MATH. Let MATH, MATH and write MATH as before. Then we have MATH . Thus MATH defines a unitary map MATH. |
math/9910135 | By the naturality of NAME duality, we have MATH, which is utilized in the following way: MATH . |
math/9910135 | By MATH, we shall verify the equivalent relation MATH, which is checked on the MATH-cyclic subspace in the following way: MATH which coincides with MATH if we use MATH which is an easy consequence of NAME transforms. |
math/9910135 | We compute as MATH which implies MATH by the density argument and we are done as MATH. |
math/9910135 | By the ergodicity property of free products REF , we have MATH and hence MATH . Taking into account the commutativity with elements in MATH of the form MATH with MATH and MATH, we first reduce an element in MATH to the form MATH . Now the commutativity with the NAME projection MATH reveals that MATH is further restricted to MATH . To see the independance of MATH on MATH, let MATH, MATH and MATH. Then the commutativity of MATH with MATH means MATH for MATH. Since MATH and MATH, this implies MATH for any MATH, MATH. Thus, letting MATH, we see that MATH belongs to MATH. |
math/9910135 | NAME is the equality MATH. By the additivity of MATH and MATH, we may restrict ourselves to the case of a simple MATH, which is checked by showing the positivity of MATH: MATH which obviously belongs to the positive cone of MATH. |
math/9910137 | For MATH we have MATH, respectively, MATH. Also MATH. Further with REF MATH . Hence the symbol of MATH vanishes. But this implies MATH. The claim follows by trivial induction from REF . |
math/9910137 | Take any MATH. For the scalar product we calculate (MATH is the projector defined in REF ) MATH . Hence the claim. |
math/9910137 | CASE: Recall that the identification of the sections of MATH with equivariant functions on the circle bundle MATH is an isomorphy. Hence the definition of adjoint operators agree. For the global NAME operator we obtain MATH. The star product MATH is given via the asymptotic expansion of MATH . For the asymptotic expansion of the last expression we have MATH . But this is the complex conjugate of the asymptotic expansion which defines MATH. This shows REF |
math/9910137 | Let us start with a real valued MATH. Then the operator MATH and the components MATH are self adjoint (see REF ). Let MATH and let MATH be the eigenvalues of the restriction of MATH on MATH. In particular, these are also the eigenvalues of MATH on MATH. Following REF let MATH be the discrete spectral measure. By REF it converges weakly to the limit measure MATH with a universal constant MATH only depending on the manifold MATH. An important intermediate result there is the asymptotic expansion REF MATH . For MATH we obtain MATH . To calculate MATH we evaluate REF for MATH (that is, MATH) and obtain MATH . Note that (see p. REF) MATH . Hence MATH . In particular the coefficient depends only on the dimension of MATH. This shows the claim for real valued MATH. For complex valued MATH it follows from linearity by considering real and imaginary part separately. In CITE for MATH the restriction MATH of the NAME form was used to define the volume. Here we have to work with the form MATH. Because the deRham classes of both forms coincide and because the NAME forms are closed the volume will be the same. |
math/9910137 | By MATH-linearity it is enough to show this for MATH. The element MATH is given by the asymptotic expansion of MATH. Hence MATH is given by the expansion of MATH . But for every MATH this vanishes. Hence REF follows. |
math/9910139 | It is enough to consider two segments MATH and MATH that intersect transversally at the middle point. We choose MATH and remove the crossing point as follows: MATH where MATH and MATH are small positive numbers. We have then a MATH-cycle of imbedded pairs of segments with fixed end-points. Let us now take MATH. If we replace MATH with MATH in the first line of REF, then we have a homotopy between the cycle REF and the cycle obtained by modifying only MATH. Analogously if we replace MATH with MATH in the second line of REF, we have an homotopy between REF and the cycle obtained by modifying only MATH. |
math/9910139 | If MATH, then any finite collection of (piecewise) imbedded loops can be isotopically deformed to a trivial link. Hence the connected components of MATH are determined uniquely by the position of the double points. Their pre-images are points on a circle that are identified in pairs, that is, chords. In the case of MATH, one has just to take care of the additional information given by the initial point. |
math/9910139 | First we consider graphs of odd type and review the proof given in CITE. Let us consider the contraction of MATH, that is, of the edge or portion of circle between MATH and MATH. If we exchange MATH and MATH or reverse an arrow, we get a minus sign; in both cases the roles of MATH and MATH are interchanged and we have MATH. Therefore MATH is compatible with such an exchange. Let us choose another vertex MATH, and exchange MATH and MATH. We can assume MATH and that MATH is oriented from MATH to MATH. First we suppose MATH and MATH. If we contract MATH we get a factor MATH; if we exchange MATH and MATH and then contract MATH we get a factor MATH. Obviously the underlying graph is the same in the two cases. We want to prove that the relabelling of one the two decorated graphs yields the other one, that is, the two decorated graphs define the same element in MATH. The indices lowered by one are, in the first case, all those greater than MATH and in the second case all those greater than MATH. The set of vertices that, in the first case, are labelled as MATH, are labelled, in the second case, as MATH and the sign of the relevant permutation is MATH. In summary: CASE: if we contract MATH we get a sign MATH; CASE: if we swap MATH and MATH, contract MATH and then relabel to get the same graph as in the previous case, we get the sign MATH. Similarly, we can treat the case MATH. All other cases (contraction of MATH and swapping of MATH with MATH) are trivial. Now let us prove that MATH by showing that contracting two different pairs MATH and MATH in opposite order yields the opposite sign. If we have MATH and MATH, then we can always assume that MATH, MATH and MATH. Contracting MATH gives a sign MATH and lowers MATH by one, so contracting MATH gives MATH. If we contract MATH first, then MATH is not lowered by one and the global sign is MATH. If MATH and MATH, we can pass from a (double)-contraction to the other one by exchanging MATH and MATH, with a change in sign in MATH. The same holds for MATH and MATH. Next let us consider MATH. Again we have to show that a permutation of external vertices or of edges does not affect MATH. The case when we swap external vertices is identical to the case of MATH, so we just have to verify what happens when we swap two edges labelled by MATH and MATH, with MATH. We claim that if we contract the edge MATH we obtain the same result as if we swap the edge MATH with the edge MATH and subsequently contract the edge MATH. In fact in the first case the result is MATH times a graph whose edges are: MATH, while in the second case the result is MATH times a graph whose edges are MATH. We now permute the labels of the edges of this last graph and obtain MATH; this permutation has order MATH. The total sign is therefore: MATH, that is, the same result obtained by contracting the edge MATH. Finally, we have to show that MATH. As in the odd case, the proof consists in showing that if we make two contractions in different order, then we have opposite signs and hence the relevant graphs cancel. This is obviously true if we contract two pairs of external vertices or two edges. Now we contract the arc between two consecutive external vertices (say MATH and MATH, with MATH) and an edge MATH. Remember that the number MATH of external vertices appears in REF . If we contract MATH first, we do not change the labels of the external vertices and we get the global sign MATH. If we contract MATH first, then the number of edges is lowered by one and so the global sign is MATH. |
math/9910139 | The coboundary operator MATH has been defined in subsection REF in such a way that it corresponds to the coboundary operator MATH via NAME 's Theorem if one considers only principal faces (see REF and in particular REF ). The fact that MATH is actually a chain map then follows from REF according to which we can neglect hidden faces. As for the degree of these maps, this is easily computed: if the graph MATH has MATH edges, MATH internal vertices and MATH external vertices, then, by REF, MATH. On the other hand the degree of the corresponding differential form is MATH where MATH is the order of the graph as defined in REF. |
math/9910139 | Let us write MATH, where MATH. We can assume that MATH. Then MATH, MATH, interpolates between MATH and MATH. We now define MATH . This form is still closed and symmetric: MATH. Denoting by MATH the inclusion at MATH, we also have MATH . Using this extended symmetric form, we can define extended tautological forms by MATH . If we now replace the edges of a graph by these extended tautological forms, after integrating over the configuration space we will get a form on MATH. Denote by MATH this map. Observe that, denoting by MATH the inclusion at MATH, we have MATH . If MATH is a cocycle, then the results of REF (in particular REF ) also show that MATH is a closed form in MATH. As a consequence, MATH where MATH is the projection MATH, and we have used again the generalized NAME formula. |
math/9910139 | Let MATH be the minimum number of internal vertices among the graphs MATH. The first statement is equivalent to saying that MATH. In fact, assume on the contrary that MATH, and let MATH be a graph (which does not vanish in MATH) with exactly MATH internal vertices. Since we consider only connected graphs, there will be at least one internal vertex connected by one edge (call it MATH) to an external vertex. In MATH there will then be a graph MATH obtained by contracting the edge MATH. First of all, notice that MATH does not vanish in MATH. In fact, if it did, then there would be an automorphism MATH of the graph underlying MATH that would yield MATH. Observe that the only REF-valent vertex in MATH has to be mapped to itself by MATH. We can now extend this to an automorphism of MATH by decollapsing MATH after the application of MATH and deciding that MATH is mapped into itself and each of its end-points are mapped into themselves (notice that we cannot interchange the end-points of MATH since one is internal and the other is external). So by the extended automorphism we would prove that also MATH vanishes in MATH. Since MATH is a cocycle, there must be other graphs such that their images under the application of MATH contain MATH. But there are only two possible graphs with this property, namely, those obtained by splitting the unique four-valent external vertex in MATH in the two possible ways. But both these graphs have MATH internal vertices, so MATH cannot be the minimum. To prove the second statement, observe first that MATH (in the odd case since the chord diagram is not closed, in the even case since the chord diagram vanishes by symmetry). So assume MATH. This means that the number of vertices is greater than REF. Since we consider only connected graphs, this means that, if a graph MATH contains a short chord, then there is only one arc (call it MATH) that has the same end-points as the short chord. Let MATH be the graph in MATH obtained by contracting MATH. Notice that MATH contains an external small loop. Thus, since we do not allow internal small loops, there is no other graph whose image under the application of MATH may contain MATH. But as above we can prove that MATH does not vanish (unless MATH is zero itself.) So MATH cannot be a cocycle. |
math/9910139 | Let MATH be the given cocycle of trivalent graphs. Let us denote by MATH REF the image of MATH REF under REF or REF. So MATH is a closed MATH-form on MATH. The forms MATH are given by integrals of products of tautological forms over the corresponding configuration spaces MATH with MATH, and MATH if and only if MATH is one of the chord diagrams which are necessarily contained in MATH by REF . We integrate first over the internal vertices and denote the result by MATH; then we have MATH. We consider MATH, and to each double point MATH, of MATH we associate a ball MATH of radius MATH. The intersection MATH is given by MATH, where MATH, MATH, are two closed segments as in REF . We now define MATH as the open subset of MATH such that the image through MATH of all its projections over MATH does not intersect MATH. We also define MATH . In other words, this complementary set MATH is equal to the subset of MATH for which all the projections over MATH yield, through MATH, one and only one element in each MATH. Thus, MATH unless MATH. Next we define accordingly MATH . In the limit MATH, we recover the whole configuration space MATH, so MATH becomes MATH. We now associate, once and for all, an index MATH to the MATH-th crossing and construct according to REF a MATH-cycle in MATH which we call MATH, where MATH and MATH are small but positive. Different choices of the parameters MATH and MATH will yield homologous cycles in MATH, as far as they do not become too large. Any other choice of the indices MATH-s will also produce a homologous cycle in force of REF . We denote this new cycle by MATH, for a suitable MATH-chain MATH. We now want to compute the integral MATH and show that MATH . In fact, if in the explicit choice of MATH we have the same MATH as in MATH, then the above integral is zero for any MATH. Otherwise, it is equal to MATH by NAME 's Theorem. The form MATH is not closed. But the main point here is that it is defined also on the space MATH given by the elements MATH for which MATH is the only double point, and MATH. This means that the above integral is well-defined also for MATH where it vanishes by dimensional reasons. But this value is also equal to the limit MATH of the integral. Next we consider all the sets MATH simultaneously. They are not disjoint, but we can redefine them (in an obvious way) so as to make them disjoint. With this we have MATH . Observe now that this expression is independent of MATH for MATH small enough; in fact, different values of MATH correspond to homologous cycles. In particular, to compute the right-hand side we can take the limit MATH. But in this limit the first term on the l.h.s. vanishes as proved above, and the integral of MATH over MATH is like the integral of a NAME current concentrated on the points of MATH whose projections on MATH yield exactly the set of those distinct points that are in pairs identified by MATH. Saying that only the forms on MATH survive is tantamount as saying that only the chord diagrams contained in MATH survive. Our final task is to prove that, when each chord connects two points on MATH that are directly identified by MATH (that is, that are in the pre-image of the same double point), then the corresponding integral is MATH, otherwise it is zero. A chord connecting two (small) intervals that contain no other vertices can be seen as the MATH-form obtained by integrating MATH over MATH with some identifications. To each point in MATH we assign a way of lifting one of these two intervals and this generates a sphere MATH. The total integral associated to this chord will then be given by the linking number of this sphere MATH with the other interval (seen as an indefinite line). This linking number is zero if the image of the two intervals does not contain the double point and, in force of REF , is MATH otherwise. |
math/9910139 | If MATH then the restriction map MATH is surjective and so, by NAME - NAME theorem, MATH is a free module over MATH, the action of the only non trivial generator of the cohomology of MATH on MATH being the product by MATH. Hence MATH cannot be exact. If MATH, then the class of the normalized top form MATH of MATH is half the NAME class and so MATH is exact. |
math/9910139 | The coboundary operator MATH has been defined in such a way that MATH is equal to MATH if one neglects hidden faces. Thus, we have to prove that hidden faces do not contribute. But, since forms corresponding to crosses are basic in the fibrations MATH, we may rely again on the results of REF. |
math/9910139 | In REF we showed that no element of MATH contains a graph with short chords. Since MATH is equal to MATH on elements of MATH not containing short chords, we immediately deduce that MATH . |
math/9910139 | First, let us write MATH where MATH . We have MATH where the map MATH denotes the restriction to the boundary MATH and MATH is the antipodal map on MATH. |
math/9910139 | MATH is the product of MATH and MATH. |
math/9910139 | Since MATH is principal, it contains at most two vertices. Thus, by our definition of graphs, there is at most one edge in MATH if it of type I or III, and no edge if MATH is of type II. If there are no edges, then there is nothing to be integrated. However, the fiber corresponding to MATH has dimension strictly positive (equal to MATH) if it is of type I or II, or if it is of type III with MATH. If however the face is of type III with MATH the fiber dimension is zero, and we get MATH. If there is exactly one edge and MATH is of type I or it is of type III with MATH, we have to integrate the MATH-dimensional form MATH over the fiber which is MATH. This yields MATH. If MATH is of type III with MATH and contains exactly one edge (to which corresponds MATH), then the fiber is zero-dimensional, and MATH. |
math/9910139 | Let MATH be the MATH-valent vertex and MATH the corresponding point. We perform the MATH-integration first. Let us compute the dimension of this fiber. Since MATH is hidden, we can use the other points to fix translations and scalings (or just scalings if MATH is of type II). So the fiber is MATH-dimensional if MATH is internal and MATH-dimensional if MATH is external. The form to be integrated over this fiber is however of degree zero since no edge ends at MATH. |
math/9910139 | As in the previous proof, we perform the MATH-integration first, where MATH is an internal univalent vertex now. As before, the fiber is MATH-dimensional. The form to be integrated over this fiber is the tautological form corresponding to the edge ending at MATH. Thus, it has degree MATH. |
math/9910139 | Now we perform the MATH-dimensional integration over MATH, where MATH is an internal bivalent vertex. Let MATH and MATH be the end-points of the two edges starting at MATH. If MATH, then the integrand form is MATH. So assume MATH. The MATH-integration can be the extended to MATH with only MATH and MATH blown up. Let us denote by MATH this space. We must then compute MATH . Following CITE, we consider the involution MATH of MATH that maps MATH to MATH. Observe that MATH is orientation preserving (reversing) if MATH is even (odd). Moreover, MATH . Thus, MATH . Therefore, MATH. |
math/9910139 | We proceed as in the proof of REF till the paragraph just before REF. Now we observe that, by dimensional reasons, MATH vanishes if MATH for MATH of type I or II, or MATH for MATH of type III. But these inequalities are again satisfied by REF. |
math/9910141 | This is a simple computation. One has MATH . Here we are abusing the notation MATH, which now also denotes the arc in MATH that is the image of an arc in MATH. We use the fact that MATH and MATH lie in the same coset of MATH (and similarly for MATH) to rewrite the right hand side of the above expression as MATH . Using the relations MATH, MATH, MATH and MATH, we have MATH . Using the relations MATH and MATH and reindexing the last set of terms of the sum, we obtain MATH . Using again the relations MATH, MATH, and MATH, we arrive at the first equality of the statement. To get the second equality, one changes the index to MATH and uses the relation MATH. |
math/9910141 | Because MATH comes from the orientation-reversing automorphism of MATH, it anticommutes with MATH. The surjectivity of the map MATH then implies MATH. The second part of the statement follows from the second equality in REF and the definitions of the symmetrization maps. The coefficient is changed to MATH because the sum is now over MATH instead of MATH. |
math/9910141 | This follows from switching MATH with MATH and MATH with MATH in REF. |
math/9910141 | It is easy to compute the logarithmic derivative of MATH using the NAME triple product formula CITE. Details are left to the reader. |
math/9910141 | This is a consequence of a more general formula of CITE. |
math/9910141 | As in REF , we get MATH . Differentiating it again with respect to MATH and plugging in MATH, we get MATH . Notice now that MATH which is an NAME series. Indeed, in the notation of CITE, it is MATH . To obtain a nice formula for MATH, recall that MATH satisfies the heat equation MATH which implies MATH . We can integrate it with respect to MATH while keeping in mind that MATH to obtain MATH which is again an NAME series. In the notation of CITE it is equal to MATH . |
math/9910141 | Notice that the functions MATH defined by MATH span the space of all functions from MATH to MATH, so the functions MATH span the space of all even functions from MATH to MATH. Then use explicit formulas for MATH and MATH from the proof of the above proposition. |
math/9910141 | Let us first explain what happens in the case where all MATH are non-integral. We are integrating over MATH the product of cohomology elements MATH where MATH is easily seen to be an NAME series due to REF . The intersection number MATH for MATH equals MATH if MATH and MATH come from adjacent cones and equals MATH otherwise. This finishes the argument. When some of the MATH are zero the argument is a bit more complicated and involves the expansion of the right hand side of REF in powers of MATH. Then up to NAME series one ends up with the integral of the product of MATH over MATH for which MATH. Details are left to the reader. |
math/9910141 | We will identify MATH with the lattice of integer row vectors with two components. Consider the fan MATH that has MATH, MATH, MATH, and MATH as generators of its one-dimensional faces, and a degree function on MATH defined by MATH and MATH. REF shows that MATH . Let us calculate the action of the NAME operator MATH on this form modulo MATH. By REF we have MATH . Let us investigate the contribution of each MATH. To get into the setup of REF , we need a subdivision MATH of the fan MATH so that the consecutive MATH form a basis. There is a standard way of doing so. For each quadrant (that is, for each two-dimensional cone of MATH), consider the set MATH of all non-zero points of MATH in that quadrant. The boundary of the convex hull of MATH consists of two half-lines and some segments REF . We ignore the half lines and add to the list of MATH all points in MATH that lie on the rest of the boundary. It is easy to show that this choice of MATH guarantees that the new toric variety is smooth. Indeed, if MATH and MATH did not form a basis of MATH then there would exist a point in MATH lying in the convex hull of MATH, MATH and MATH by Pick's Theorem CITE. Because of the symmetry, it is enough to consider the first quadrant. Notice that if MATH and MATH, then MATH, with MATH and MATH. Conversely, every pair of MATH with MATH as above generates some superlattice MATH of coindex MATH. Moreover, this pair will form a segment of the boundary of the convex hull of all non-zero points of MATH in the first quadrant. Indeed, if any other non-zero point MATH with MATH did lie below the line through these two points, then the area of the triangle with vertices MATH, MATH and MATH would be positive but less than MATH. The values of MATH on the points MATH and MATH are MATH and MATH respectively, hence by REF we have MATH . Therefore, from the definition of MATH, REF we conclude that MATH . |
math/9910141 | This follows from REF and the fact that MATH is isomorphic as a NAME to MATH CITE. |
math/9910141 | All we need to do is calculate the NAME involute of MATH. This was accomplished in CITE, up to the constant term. To compute the constant term, we remark that these forms have already appeared in the literature as the NAME forms CITE. |
math/9910141 | This follows from CITE, since MATH is associated to the linear map on the NAME algebra that maps MATH to MATH. |
math/9910141 | It is clear that if MATH is a lift of an eigenform MATH of analytic rank one or more, then MATH. Indeed, for every MATH the form MATH is a linear combination of various lifts of MATH, which implies MATH. It remains to show that if MATH is a lift of MATH then MATH is a linear combination of lifts of MATH. This follows from the commutation relation MATH for all prime MATH coprime to MATH. Indeed, from the definition of MATH CITE, MATH . |
math/9910141 | Let MATH be a cusp form, and let MATH be the corresponding element of MATH. Then by REF , MATH . Notice that we are using our convention that MATH for MATH. On the other hand, by REF the composition of all the maps in REF except for the last one yields an element of MATH given by MATH where we again apply our convention. Also, we can formally use the same expression for MATH as for the rest of MATH because the coefficient at MATH is zero. In what follows, it will be convenient for us to ignore the constant terms in all our expressions. Indeed, all our functions are modular forms, so any constant term can always be restored. We will denote all these constant terms by MATH. Using MATH . MATH and symmetry properties of MATH, we get MATH . MATH . Let us now simplify the second part of this expression to make it look more like MATH. We split it into four sums MATH and deal with each sum separately. We will give a detailed calculation for one of the sums and will indicate how to manipulate the other three. MATH . We first of all rewrite MATH as MATH. Then for the second term we make the change of variables MATH. We perform similar but easier manipulations for the remaining three sums in REF . For the first sum we make the change of variables MATH. For the second sum we make the change of variables MATH, so that it cancels the first sum in REF . For the fourth sum in REF we make the change of variables MATH. After some straightforward calculations we get MATH . As a result, MATH . MATH . It is possible to further simplify this equation to obtain MATH which is an NAME series. Indeed, it can be easily written (up to a constant) as a linear combination of MATH, MATH and the (non-modular) MATH-Eisenstein series MATH by REF . It remains to observe that the coefficient of MATH must be zero, because of the transformation properties of MATH under MATH. |
math/9910141 | We will prove this theorem by induction on MATH. For small levels there is nothing to prove, because there are simply no cusp forms. Fix MATH and assume that the statement of the theorem is true for all smaller levels. In particular, this implies that lifts of all forms of smaller levels are contained in the span of toric forms, because the space of toric forms is stable under liftings, see REF . Every new NAME eigenform MATH of analytic rank zero is contained in the image of MATH, see REF . By REF , MATH is contained in the image of MATH, and so is a toric form up to NAME series. Because the space of toric forms is NAME stable, this implies that MATH is toric. This proves that the space of toric forms contains the span of all NAME eigenforms of analytic rank zero. To prove the opposite inclusion, notice that by the induction assumption it is enough to consider MATH with MATH. By REF , there is an element MATH such that MATH. We use here that MATH, because otherwise the symbol MATH is not defined. By the definition of MATH in REF there is an element MATH such that MATH. Moreover, we can find a cusp form MATH which induces the linear map MATH on MATH. Then REF shows that MATH is proportional to MATH up to an NAME series. By REF MATH lies in the span of NAME eigenforms of rank zero and NAME series, which finishes the proof. |
math/9910144 | Let MATH be a second-countable group. Then the NAME space MATH conains a MATH-invariant separable NAME subspace MATH whose elements separate points and closed subsets of MATH. The unit ball MATH of the dual space to MATH in the weak-MATH topology, being a convex compact subset of the separable NAME (= completely metrizable locally convex) space MATH, is homeomorphic to the NAME cube MATH by force of NAME 's theorem (compare CITE). Consequently, MATH . |
math/9910144 | The proof is based on the following remarkable property of the functor MATH, first noticed and put to use by NAME: every action of a group MATH on the space MATH by isometries extends in a canonical way to an action of MATH on MATH (by left translations), and if the original action of MATH on MATH was continuous, so will be the extended action of MATH on MATH. (To better appreciate the usefulness of NAME functions, notice that in general the same action of MATH on the space of all REF-Lipschitz functions on MATH need not be continuous!) Now it is rather evident that every continuous action of a topological group MATH on a metric space MATH by isometries extends to a continuous action of MATH on MATH be isometries in a canonical sort of way. If the original action on MATH determined an embedding of topological groups MATH, then clearly MATH is a topological subgroup of MATH as well. If MATH is a separable topological group, then one can start with any separable metric space MATH whose group of isometries MATH contains MATH as a topological subgroup to obtain the desired conclusion. (For example, MATH as in the proof of Teleman's theorem, or simply MATH itself equipped with a right-invariant metric generating the topology.) The space MATH is then separable and isometric to MATH as in the statement of the theorem. |
math/9910144 | Let MATH be an arbitrary topological group. Without loss in generality, assume that MATH. Using NAME 's REF , embed MATH into a minimal, topologically simple group MATH. Let now MATH be embedded, as a topological subgroup, into MATH, where MATH. For at least one MATH, the image of MATH under the MATH-th coordinate projection MATH is non-trivial. Because of topological simplicity of MATH, the kernel of MATH is MATH, and thus the restriction MATH is a (continuous) group monomorphism. Because of minimality of MATH, the latter monomorphism is in fact a topological isomorphism. Consequently, MATH is a topological isomorphism of MATH with a topological subgroup of MATH. |
math/9910144 | Indeed, the latter group admits a continuous action without fixed points on the compact space MATH. |
math/9910145 | To explain the idea, recall first how to find integers MATH for which MATH has small order modulo MATH: The trick is to take MATH, since then MATH, and so MATH. To modify this idea to our context, assume for simplicity that the matrix MATH is ``principal", that is the action of MATH on MATH is equivalent to the action of the unit MATH on the maximal order MATH (in general we need an ideal in the order MATH, see REF). Then MATH is equivalent to MATH (in general, only the implication MATH is valid). Factor MATH as a product of prime powers: MATH where MATH means the product over primes MATH which split in MATH, MATH the product over inert primes and MATH the product over the ramified primes MATH. On the other hand, we have MATH . Write the ideal factorization of MATH as MATH . Since MATH, we get on comparing the prime exponents that MATH . Since MATH is even, we can set MATH . Then MATH where MATH is the product of all ramified primes of MATH. We have MATH and so MATH, equivalently MATH. Thus we find MATH and so MATH as required. |
math/9910145 | Let MATH and let MATH be the matrix coefficients of MATH expressed in terms of the basis MATH. From REF we have that MATH and by REF for MATH a root of unity. Thus, if MATH are the matrix coefficients of MATH in terms of the basis MATH, then MATH . If we denote by MATH the column vectors of MATH, then the MATH-entry of MATH is MATH and since MATH we get MATH . Substituting the definition of MATH and using REF, we see that MATH is given by MATH times a sum, ranging over MATH, of terms MATH where MATH has absolute value one. Now take the trace; as follows from REF (see REF), the absolute value of the trace of MATH equals MATH if MATH, zero otherwise. The result now follows by taking absolute values and summing over all MATH. (For more details, see REF.) |
math/9910145 | Let MATH be the real quadratic field containing the eigenvalues of MATH, and let MATH be the residue class field at the prime MATH, that is, MATH where MATH is a prime of MATH lying above MATH. MATH has cardinality MATH if MATH splits in MATH, or MATH if MATH is inert. We may diagonalize the reduction of MATH modulo MATH over the field MATH. In the eigenvector basis we have MATH and MATH, where the assumption of linear independence modulo MATH implies that both MATH (in MATH.) Thus REF is equivalent to the following two equations over MATH: MATH which in turn (see REF) is equivalent to MATH . Hence MATH is determined by the triple MATH. Dividing by MATH and letting MATH and MATH we rewrite the second equation as MATH . If the first (or second) factor equals zero then MATH (or MATH) since the order of MATH in MATH equals MATH. If the third factor is zero then MATH where MATH. In each case this leaves MATH possibilities for the pair MATH, and since MATH is unconstrained the total number of solutions is at most MATH. |
math/9910145 | Let MATH be a solution to REF . If MATH then REF holds with MATH replaced by MATH. Arguing as in REF one of the three factors in REF must be zero, and the vanishing factor determines which one of the three equations in REF that MATH must satisfy modulo MATH. For example, if the first factor in REF is zero, then MATH. Now, the group generated by MATH modulo MATH is cyclic and isomorphic to MATH where the MATH's are distinct primes. We will denote the MATH component of MATH by MATH and similarly for MATH. Since MATH is equal to the least common multiple of MATH there exists for each MATH at least one prime MATH such that MATH. Claim: if MATH is a solution to REF then MATH satisfies one of the equations in REF . The reason is as follows: there is a prime MATH such that MATH, thus one of the equations in REF is satisfied modulo MATH. Since MATH this implies that MATH satisfies one of the equations in REF . (Note in particular that this leaves MATH possibilities for MATH if we specify one of the equations in REF to be satisfied). Now, to each MATH there are REF different types of trivial solutions, and since MATH must satisfy one of the possibilities in REF for all MATH, we obtain that there are at most MATH solutions to REF . |
math/9910145 | By the remark after REF , linear dependence modulo MATH holds if and only if MATH, where MATH. Let MATH . Then the hypothesis in REF is satisfied for MATH, leaving MATH possible values for MATH modulo MATH. Now, an element in MATH has exactly MATH preimages in MATH. Hence there are at most MATH solutions to REF . Since MATH we get that MATH . Finally noting that since MATH is square-free, MATH we find (by REF) that MATH for all MATH, and we are done. |
math/9910145 | Let MATH be the set of integers of the form MATH, where MATH is square free, MATH, and MATH. By REF , MATH has density one. For MATH, we wish to bound the number of solutions to MATH . Since MATH is not square free we cannot apply REF directly. For MATH, MATH square-free, we further decompose MATH, so that MATH and MATH are coprime. Given MATH there are exactly MATH solutions to MATH if MATH. Thus, a solution of MATH lifts to at most MATH solutions for which MATH. This, together with REF applied to REF gives there are at most MATH solutions to REF . Clearly MATH, MATH, and since MATH, MATH are coprime, with MATH, we have MATH for all MATH (by REF). Hence the number MATH of solutions of REF is bounded by MATH . Thus we find that for MATH the number of solutions of REF is bounded by MATH and consequently we see from REF that MATH as required. |
math/9910145 | Factor MATH, MATH with MATH. Then MATH, MATH where MATH . Thus the Lemma reduces to the following easily verified inequality: For any non-negative reals MATH, MATH, we have MATH . |
math/9910145 | Suppose that MATH is divisible by a prime MATH. From the definition of MATH, this implies that there are two distinct prime divisors MATH, MATH of MATH so that MATH divides MATH, MATH. In particular, MATH. Thus we find two distinct primes such that MATH . For fixed MATH, MATH the number of MATH divisible by MATH is MATH. Thus for fixed MATH, the number of MATH satisfying REF is at most MATH . By NAME (REF - recall MATH), this is bounded (up to constant factor) by MATH. Summing over all primes MATH, we find that the number of integers MATH such that MATH is divisible by some prime MATH is at most MATH . |
math/9910145 | By REF we may assume that MATH is MATH-smooth, with MATH. For MATH, write the MATH-smooth part of MATH as MATH, with MATH square-free. Set MATH . Note that since MATH is square-free and MATH-smooth, it divides the product of all primes MATH. Thus for MATH we have: MATH . Since MATH is MATH-smooth and divides MATH, it also divides the product MATH. Thus MATH or MATH . Now for almost all MATH we have REF MATH and so by REF if MATH is large, so is MATH. Specifically, if MATH then by REF, we find MATH . We will show that this fails for almost all MATH and thus prove the Proposition. To estimate the number of MATH for which MATH, recall that by the definition of MATH there is some prime MATH dividing MATH (and hence dividing MATH) so that the MATH-smooth part of MATH is MATH and MATH (in particular if MATH then MATH). Thus there is a prime MATH for which MATH. Given MATH there are at most MATH integers MATH divisible by MATH, and hence the total number of MATH with MATH is at most MATH . By REF we have for fixed MATH and summing over MATH gives at most MATH . Thus the number of MATH for which MATH is at most MATH and we are done. |
math/9910145 | If MATH then MATH and so MATH. Thus the number of such primes is bounded by the total number of prime divisors of the integers MATH, MATH, that is by MATH where MATH is the number of prime factors of MATH. Now trivially MATH, and MATH where MATH is the largest eigenvalue of MATH. Thus we get a bound for the number of primes as above of MATH as required. |
math/9910145 | Let MATH if MATH and let MATH if MATH, and for composite integers MATH put MATH. Then MATH. Since MATH the sieve of NAME gives that MATH. Indeed, MATH . Putting MATH and noting that MATH since MATH we obtain MATH. Now following CITE, we consider the smoothed sum MATH. By partial summation we have MATH . Using the identity MATH we obtain: MATH . Now, MATH by NAME 's bound on MATH. Moreover, MATH implies that MATH. Hence MATH . However, MATH and thus MATH . |
math/9910145 | We may write MATH as MATH and by REF we may bound this sum by MATH . Putting MATH we obtain the desired conclusion. |
math/9910145 | CASE: We have MATH . CASE: By REF, MATH and hence MATH . |
math/9910145 | If MATH then MATH, and the number of MATH such that MATH for some MATH is bounded by MATH . Hence the number of MATH for which MATH for some MATH is MATH. |
math/9910148 | We find that MATH and hence to prove REF it is enough to prove that MATH, and this follows in a similar way to CITE. If MATH then MATH is a smoothing operator, and so the last statement is immediate. |
math/9910148 | We have that MATH while a direct computation yields MATH from which the equality follows. |
math/9910152 | Taking MATH instead of MATH, we can suppose that the rotation set MATH is equal to zero. It follows that there is a fixed point MATH of the extended homeomorphism in the prime end compactification of MATH. From prime end theory we can find a sequence of open arcs MATH in MATH of diameter MATH each of which extends to a closed arc in the prime end-compactification, with end points which belong to MATH and which are close to MATH, one on each side. Moreover, these arcs have the property that the closures MATH converge to the point MATH in the prime end compactification as MATH goes to infinity. Each of these arcs divides MATH into two simply connected domains. We denote by MATH the small one (that is, MATH belongs to the closure of MATH in the prime end compactification). We wish to show that MATH . If the arc MATH doesn't meet its image under MATH, the fact that MATH is fixed implies that one of the inclusions MATH or MATH is true. In the first case we can find an open set MATH contained in MATH which is wandering and has a positive orbit contained in MATH. In the second case we can construct a wandering domain whose negative orbit is contained in MATH. Since these conditions contradict our hypothesis we conclude that MATH . This fact together with the fact that the diameters of the arcs MATH tend to zero implies that any limit point in MATH is a fixed point of MATH. |
math/9910152 | Let's suppose for example that we can find a real number MATH, a sequence of integers MATH converging to MATH and a sequence MATH in MATH such that MATH . We have the inequality MATH where MATH . Any closure value MATH of the sequence MATH belongs to MATH and satisfies the inequality MATH . We have got a contradiction. |
math/9910152 | We can find a complex structure on MATH which defines the same differential structure on MATH. Let's consider a conformal map MATH between the simply connected domain MATH and MATH and the maps MATH and MATH . We consider also a lift MATH of MATH to the universal covering space and the map MATH given by MATH . For every MATH we consider the essential simple closed curve MATH of MATH. Let's fix MATH such that the curve MATH is above the image of MATH. We can find a homeomorphism between the closed annulus MATH delimited by MATH and MATH and the closed annulus MATH delimited by MATH and MATH, which is equal to MATH on MATH and to the identity on MATH. We denote by MATH its extension to the infinite annulus which is equal to MATH below MATH and to the identity above MATH. If MATH is area preserving the fact that MATH is invariant tells us that the two annuli MATH and MATH have the same area and we can choose the homeomorphism MATH to be area-preserving (so is the map MATH). We write MATH for the lift of MATH which coincides with MATH on MATH. Using the condition MATH, we can find MATH such that MATH . Using an argument of compactness, we can find MATH such that MATH . Consider a MATH decreasing map MATH equal to MATH on MATH and let's define the map MATH equal outside of MATH to the identity and on MATH to the time one map of the Hamiltonian flow of MATH for the usual symplectic structure on MATH. That means the flow defined by the equations: MATH . The map MATH satisfies REF. It also fixes the curve MATH. We can choose the function MATH such that its satisfies REF and the second part of REF . Indeed, if MATH is the lift of MATH equal to identity on MATH, we have MATH for every MATH, so we have the relation MATH . If we choose the function MATH sufficiently large on MATH we will get MATH . |
math/9910152 | We first prove REF using the results of the paragraphs REF. Suppose that MATH has a bounded component MATH. It is periodic and we can suppose it fixed by taking an iterate of MATH instead of MATH. Consider the extension MATH of MATH to the end compactification MATH. If MATH is any invariant bounded component of MATH then it is topologically an open disk and must contain a fixed point since MATH is area preserving. The fact that MATH doesn't contain any fixed points implies that the set of invariant bounded components of MATH is finite, since otherwise there would be a sequence of fixed points with limit in MATH. Moreover we can find a finite family MATH of closed disks, one in each fixed component of MATH, such that every fixed point of MATH is contained in the interior of one these disks and such that MATH for every MATH. There are at least three components: MATH, MATH and MATH . This contradicts the NAME formula. We next prove REF. Using REF , we know that the rotation numbers MATH and MATH are irrational. Let's consider a rational number MATH. Applying the extension lemma to MATH and to the lift MATH, one can construct an area preserving homeomorphism MATH of MATH which coincides with MATH on a neighborhood of MATH and which fixes two essential simple curves MATH and MATH such that: MATH-there is no fixed point of rotation number MATH in MATH; MATH-there is no fixed point of rotation number MATH in MATH; MATH-the rotation number of MATH and MATH have the same sign; MATH-the rotation number of MATH and MATH have the same sign. Using the fact that MATH is filled and doesn't contain any periodic orbit we deduce that the map MATH has no fixed point of rotation number MATH in the closed annulus defined by the curves MATH and MATH. From REF we deduce that: MATH the rotation set MATH doesn't contain MATH; MATH the numbers MATH and MATH are not separated by MATH. The conclusion follows immediately. |
math/9910152 | We give a proof by contradicting the assumption that MATH has no periodic points of type MATH in MATH. If this is assumed then MATH has no periodic points of type MATH in some neighborhood MATH of MATH since the periodic points of type MATH form a closed set. It then follows from REF that MATH and MATH. Using REF we construct an area preserving map MATH, which coincides with MATH on some neighborhood MATH of MATH, which leaves invariant two essential simple closed curves MATH and MATH, and which has no fixed points of rotation MATH in MATH for the lift equal to MATH on MATH. By REF , there exists a fixed point of MATH of rotation number MATH in the annulus MATH delimited by MATH and MATH. Since it is not in MATH and not in MATH this fixed point must lie in a bounded invariant component of MATH and hence a bounded invariant component of MATH . Let MATH be a finite covering map of degree greater than the rotation number of any periodic point of MATH. Then there is a lift MATH of MATH such that MATH is a fixed point of MATH if and only if MATH is a fixed point of MATH of rotation number zero. Let MATH so that MATH has no fixed points in MATH . Since any component of MATH containing a fixed point of MATH of rotation number zero is an invariant open topological disk it follows that any component of MATH containing a fixed point of MATH is also an invariant open topological disk. As in the proof of REF , the fact that MATH and the boundary of MATH don't contain any fixed points of MATH implies that the set of bounded components of MATH containing fixed points is finite. Moreover we can find a finite family MATH of closed disks, one in each component of MATH containing such a fixed point, such that every fixed point of MATH is contained in the interior of one these disks and such that MATH for every MATH. There is at least one such disk since there was at least one fixed point of MATH of rotation number zero. This again contradicts the NAME formula for MATH . |
math/9910152 | Suppose MATH has non-empty intersection with both MATH and MATH . Let MATH be a component of MATH which contains a point of MATH. Then MATH is a periodic domain since every component of the complement of MATH is. There is an an open arc MATH whose endpoints are in MATH. We note that if MATH, then MATH and MATH are disjoint since MATH cannot be a subset of MATH (as MATH is in a branch) and MATH cannot contain an endpoint of MATH. Let MATH be the component of MATH which contains MATH . Since MATH is simply connected MATH separates it into two non-empty simply connected domains MATH and MATH each of which has non-empty intersection with MATH. It is not difficult to show that this implies MATH separates MATH into two connected open sets MATH and MATH. Choose MATH so that MATH. The map MATH is area preserving, so MATH is a non wandering open set for MATH. We deduce that there exists an integer MATH such that MATH and MATH. Moreover the closure in MATH of MATH is not contained in MATH, it meets the boundary MATH. Similarly MATH meets the closure of MATH in MATH. We deduce that MATH meets its image by MATH, which is impossible. We conclude it is not possible for MATH to have non-empty intersection with both MATH and its complement. If MATH lies in the complement of MATH it must be in a single component because it is connected. |
math/9910152 | We first will prove REF by contradiction. If REF is false then by replacing MATH with an iterate we can assume that there is an invariant simply connected domain MATH with rotation number zero. This implies that the extension of MATH to the prime end compactification of MATH has a fixed point MATH on the boundary. From the theory of prime ends and the proof of REF we know that there is a sequence of open arcs MATH in MATH, each of which is the interior of a closed arc with endpoints in MATH, which define the prime end MATH. Moreover, we can assume the sequence of diameters of MATH tends to zero as MATH tends to infinity, so by choosing a subsequence we can assume the closures of the arcs converge to a point MATH in MATH . We observed (see REF ) that MATH for each MATH and hence MATH is a fixed point of MATH. We note first that this fixed point MATH cannot be a NAME stable point. This is because a NAME stable point is in the interior of an arbitrarily small MATH invariant disk on the boundary of which MATH is minimal. Thus the boundary of such a disk must intersect the complement of MATH and hence, by minimality, the boundary is disjoint from MATH. This would imply that the entire disk is disjoint from MATH which is impossible. It follows that the point MATH must be a hyperbolic fixed point. By the NAME theorem there is a neighborhood MATH of MATH on which there are topological co-ordinates MATH with MATH at the origin and in which MATH. We will use these local coordinates on MATH and refer to the four local branches at MATH by which we mean the intersection of MATH with the positive and negative MATH and MATH axes. The contradiction we wish to derive will follow from the following lemma Replacing the sequence MATH with a subsequence, we may conclude that there is a neighborhood MATH of MATH such that one component of MATH contains the arcs MATH for MATH sufficiently large, and this component consists of one of the following: CASE: the intersection of MATH with a single open quadrant in the MATH co-ordinates, or CASE: the intersection of MATH with two adjacent open quadrants and the local branch between them, or CASE: the intersection of MATH with three open quadrants and the two local branches between them, or CASE: the complement in MATH of MATH and a single local branch. We first observe that if MATH is a local branch of MATH in MATH then by REF MATH must be entirely contained in MATH or disjoint from MATH. If it is contained in MATH there is a MATH such that a neighborhood of the interval MATH is in MATH. The collection of all iterates of this neighborhood contain the intersection of a smaller neighborhood MATH with the two open quadrants in MATH co-ordinates of which MATH is part of the boundary. Thus if one point of a local branch is in MATH then, perhaps in a smaller neighborhood, the entire branch and the open quadrants on either side of it are in MATH. We shall make repeated use of this property. The first use of this fact is to observe that it is not possible for all four local branches at MATH to be in MATH as this would imply that there is a neighborhood MATH of MATH such that MATH . This would mean that MATH which is not the case. We again use the fact that if a local branch intersects MATH then (perhaps in a smaller neighborhood) the entire branch and the open quadrants on either side of it are in MATH. This implies that either one of properties REF - REF above holds or there is a single quadrant such that the interiors of infinitely many of the arcs MATH are in that quadrant, but the two branches which bound this quadrant are not in MATH. We complete the proof of the lemma by showing that this last alternative implies that REF holds. We can suppose, without loss of generality, that it is the open first quadrant MATH which contains the interiors of infinitely many of the MATH and the positive MATH and MATH axes are in the complement of MATH. By choosing a subsequence we may assume every MATH is in the first quadrant. We need only show that there is a neighborhood MATH of MATH such that MATH is entirely in MATH. To do this, for each small MATH we consider the arc MATH made up of the two line segments MATH and MATH. These two segments form two sides of a rectangle MATH the other sides of which lie in the MATH and MATH axes. This rectangle contains infinitely many of the arcs MATH and in particular points of MATH. Also if MATH is sufficiently small we may assume this rectangle is disjoint from MATH and in particular cannot contain all of MATH. It follows that the arc MATH has non-empty intersection with MATH. We want next to show there is an open sub-arc MATH of MATH with the property that it is in MATH but its endpoints are in the complement of MATH, and that MATH. To see this we note that MATH separates MATH and MATH for a large MATH. Hence some component MATH of MATH must separate MATH and MATH in MATH. Clearly MATH is an open sub-arc of MATH with endpoints in the complement of MATH. Let MATH and MATH be the components of MATH which contain MATH and MATH respectively. The fact that MATH and MATH implies MATH for MATH . Since MATH is area preserving MATH cannot be a proper subset of MATH. We conclude that MATH. Since MATH and MATH consists of the single point MATH we conclude that both MATH and its inverse image MATH are in MATH . Thus the arc MATH made of the two line segments MATH and MATH is entirely in MATH. Similarly for any MATH we have MATH and indeed for any MATH . This easily implies that MATH, which means that REF holds. We next observe that this lemma and the assumption that REF is false do lead to a contradiction. This is because from the lemma we see that there is a neighborhood of the prime end MATH corresponding to MATH in the prime ends of MATH which consists of accessible prime ends corresponding to points in either stable or unstable branches of MATH. It also follows that there are finitely many fixed prime ends and they all have this property. We can conclude that any non-fixed prime end MATH, which is between the fixed prime ends MATH and MATH must correspond to a point on a branch of MATH and a branch of MATH where MATH is the fixed point in MATH corresponding to the prime end MATH. Since this is true for the entire interval of prime ends between MATH and MATH there is a saddle connection between MATH and MATH which contradicts our hypothesis. We conclude that REF must hold. REF of the theorem is a consequence of REF . As before, by taking an iterate we may assume the point in question, MATH, is fixed. The closure of one branch of this point is a compact connected invariant set MATH. From REF we know that another branch of MATH is either a subset of MATH or disjoint from it. It cannot be disjoint since if MATH denotes the component of the complement of MATH which contains MATH then MATH would be an invariant simply connected domain and the fixed point MATH would be an accessible point of MATH. This would mean that the rotation number of MATH is rational and hence REF would be contradicted. We conclude MATH. Since we have shown any branch of MATH is a subset of the closure of any other, REF follows. |
math/9910152 | Let MATH be a hyperbolic fixed point of MATH and let MATH be the union of all closed topological disks in MATH whose boundaries consist of finitely many segments of the branches of MATH. Clearly MATH . REF implies MATH is in the interior of MATH. Since any closed segment on a branch of MATH can be mapped by a power of MATH into a small neighborhood of MATH we see that these segments are in the interior of one of the closed disks whose union forms MATH . It follows that any of these closed disks is contained in the union of the interiors of finitely many other such disks, so MATH is also equal to the the union of the interiors of these disks and is open. This proves REF. We next show REF. This is equivalent to showing that MATH inessential implies the complement of MATH in MATH is connected. If this is not the case then there is a simple closed curve MATH in MATH which separates two components, call them MATH and MATH, of the complement of MATH . Since MATH is the union of the interiors of closed disks whose boundaries are branch segments, the curve MATH is covered by the interiors of finitely many such disks. Let MATH denote the union of these open disks and let MATH be the frontier of the component of MATH containing MATH . Then MATH is a simple closed curve which separates MATH and MATH and consists of finitely many segments of branches of MATH. If MATH were essential in MATH then MATH would be essential. Hence we know MATH bounds a disk in MATH which must contain one of MATH or MATH. This would imply that one of MATH or MATH is in MATH which is a contradiction. Hence we have shown that MATH inessential implies that MATH is simply connected. If MATH is essential then a similar argument shows that the complement of MATH has exactly two components so REF follows. More precisely, the fact that MATH is essential implies MATH is not simply connected so MATH has at least two components and in particular there are distinct components MATH and MATH each of which contains one of the components of MATH. If REF does not hold then the complement of MATH has at least three connected components: MATH and MATH and one more MATH which is a subset of MATH . There is a simple closed curve MATH in MATH which separates MATH from MATH and hence from MATH. As above we can find MATH a simple closed curve which separates MATH and MATH and which consists of finitely many segments of branches of MATH. Since MATH does not separate the two components of MATH it bounds a closed disk in MATH. The interior of this disk contains MATH but is also in MATH, a contradiction. We conclude that MATH has only two components. Finally we show REF. If MATH does not intersect both MATH and the complement of MATH then either MATH or MATH . But if MATH does intersect both MATH and its complement then either MATH is entirely contained in the interior of a disk whose boundary consists of branch segments (in which case MATH ) or MATH contains part of a branch of MATH . If MATH contains a point of a branch of MATH then it contains the entire branch by REF and hence all branches of MATH. Thus if MATH is any closed disk in MATH whose boundary consists of branch segments then the boundary of MATH is in MATH and inessential in MATH. This boundary cannot be essential in MATH as MATH is either simply connected or annular and essential in MATH . Hence the boundary of MATH is contractible in MATH. Since MATH we conclude that MATH. Since this is true for any MATH we have shown that MATH . |
math/9910152 | From the definition of essential hyperbolic fixed point we know that there is an essential curve MATH in MATH consisting of segments of branches of MATH which intersect transversely. The fact that any stable branch of the fixed point has non-empty transverse intersection with any unstable branch allows us to deduce that this curve can be chosen to lie in the union of any pair of branches, one stable and one unstable. Let MATH be the frontier of one of the components of the complement of MATH which intersects one component of MATH and let MATH be the frontier of the component of the complement of MATH which contains the other component of MATH . Then MATH consists of segments of branches of MATH which intersect transversely and is simple. Since it separates the two components of MATH it is essential. Since MATH is in the interior of MATH the branches of MATH contain points from each component of the complement of MATH . |
math/9910152 | Taking MATH instead of MATH and MATH instead of MATH we can suppose that MATH. Replacing MATH by a lift to a finite cover we may assume that there are no fixed points except those which project to fixed points with rotation number MATH. Note that a continuum is essential if and only if a lift to a finite cover is essential. Hence it suffices to prove the result for this lift since it has an essential continuum which projects to MATH. By abuse of notation we refer to the lift as MATH. Consider MATH the extension of MATH to MATH the end compactification of MATH, that is, the compactification obtained by collapsing each of the two components of MATH to a point, obtaining a space MATH homeomorphic to MATH. We will denote the two points added in this compactification by MATH and MATH. If MATH and MATH are the components of MATH containing MATH and MATH respectively then they are simply connected MATH-invariant domains. We can apply REF to show these domains have irrational rotation number despite the fact that at the points MATH and MATH the homeomorphism MATH is not smooth and neither is the invariant measure. The proof of REF of REF is still valid. The theory of prime ends allows us to conclude that there is a closed disk MATH which contains all the fixed points of MATH which are in MATH and such that MATH. We can find MATH with the analogous property. By REF to every inessential hyperbolic fixed point MATH, we can associate an invariant simply connected domain MATH. The same proposition implies there are a finite set of such domains which are disjoint and which contain all inessential fixed points of MATH (all of which, by the remarks above, can be assumed to have rotation number zero). Moreover they are either disjoint from the annular regions MATH and MATH or contained in them. In each such domain MATH which is disjoint from MATH we may choose a closed disk MATH which contains all the fixed points in MATH and such that MATH. We write MATH, MATH, for the family of disks obtained in this way. We next consider the finite family MATH, MATH, of fixed points of MATH which are outside all the disks MATH . By the NAME formula, we have MATH and we can write MATH . REF tells us that one of the integers MATH or MATH is MATH. We know that MATH for every MATH and that MATH if MATH is elliptic. We deduce that MATH and that there exists a fixed point MATH such that MATH. This point has to be hyperbolic. It is essential because it is not contained in the disks MATH. The analogous result for MATH follows from similar but easier reasoning. |
math/9910152 | If MATH is not irrational (see REF ) it contains a periodic point MATH. By REF we can suppose that it is an essential hyperbolic periodic point. The set MATH contains an essential simple closed curve MATH (in MATH), and a point in each component of the complement of MATH by REF . We deduce that the open set MATH meets the two components of MATH and doesn't meet MATH. The contradiction comes from the connectedness of MATH. |
math/9910152 | We first prove REF . Suppose that the sequence MATH of irrational invariant essential continua converges to MATH. This set is an essential invariant continuum. Let's suppose that it is not irrational. It contains an essential hyperbolic periodic point MATH, an essential simple closed curve MATH, a point above that curve and a point below. For MATH big enough, MATH also contains a point in each component of the complement of MATH. We deduce that MATH meets the curve and hence must contain the point MATH. We have a contradiction to the assumption that MATH is irrational. Let's suppose that the sequence MATH converges to MATH, where MATH is the rotation number of MATH. We can choose for any MATH an invariant measure MATH. Any limit point MATH of the sequence MATH belongs to MATH and its rotation number is MATH. It is also equal to MATH. We deduce easily the continuity of the map MATH from this argument. REF is straightforward. If MATH is a sequence in MATH, we can choose for every MATH an irrational invariant essential continuum MATH which contains MATH. Any limit point of the sequence MATH is an irrational invariant essential continuum which contains MATH. REF is also very simple. Any connected component of MATH is closed, invariant, essential and does not contain any periodic orbit. So it is irrational. Of course, it is maximal among the irrational invariant essential continua. |
math/9910152 | First we prove REF. The set MATH is an invariant essential continuum which is not included in MATH. So it contains a periodic point. This periodic point belongs to MATH because MATH and MATH are irrational invariant essential continua. The rotation numbers of MATH and MATH being irrational, the set MATH is a segment of length MATH. The extension of MATH to the prime-end compactification of MATH satisfies the hypothesis given at the beginning of Paragraph REF and we can apply REF to the essential continuum MATH itself: for every rational number MATH there exists in MATH an essential hyperbolic periodic point of type MATH, where MATH and MATH and MATH are relatively prime. But this point belongs to MATH and not to the boundary curves. We now prove REF . Let MATH be an essential invariant continuum contained in MATH and not a subset of MATH. The boundary of MATH is an irrational invariant essential continuum. It cannot meet MATH (since it is irrational) and must be contained in MATH. More precisely it is contained in MATH because it is connected and because MATH is not a subset of MATH. In the same way the boundary of MATH is contained in MATH. Hence we conclude that MATH meets the two boundary components of MATH. If MATH is an essential hyperbolic periodic point, we can find an essential curve contained in MATH. The set MATH meets that curve and must contain MATH. In particular, if MATH is another essential hyperbolic periodic point contained in MATH the set MATH contains MATH. The proof of REF is exactly the same. |
math/9910152 | The connected component MATH of MATH which contains MATH is an invariant essential continuum and the boundary of MATH is an irrational invariant essential continuum. This boundary is connected, doesn't meet MATH and doesn't contain MATH: it is contained in MATH. We deduce that MATH. |
math/9910152 | We write MATH for the upper end of MATH and MATH for the lower end. Let's consider the extension MATH of MATH to the end compactification MATH. The end MATH is a fixed point of MATH and the sequence MATH of NAME indices is well defined and constant equal to MATH. Indeed, the rotation number of MATH is irrational. Let's consider a closed disk MATH which contains MATH in its interior and which is contained in MATH. The connected component of MATH which contains MATH is equal to MATH by REF . So, we can find a closed disk MATH which contains MATH in its interior and such that MATH. We deduce that the set MATH, which is contained in MATH, doesn't meet the boundary of MATH. Such a set is called an isolating block. To such a set we can associate a sequence of integers MATH and this sequence is non positive for infinitely many values of MATH (see CITE or CITE). The integer MATH is equal to the sum of the NAME indices (for MATH) of all the fixed points of MATH whose MATH orbit lies entirely in MATH if there are a finite number of such points. For any value MATH such that MATH there exists a fixed point of MATH distinct from MATH, whose orbit lies entirely in MATH. To prove that there is a homoclinic point in MATH we consider an arbitrary essential fixed point MATH of MATH and let MATH be an essential simple closed curve in MATH consisting of segments of MATH and MATH. We define MATH and let MATH be the closure of the component of the complement of MATH in MATH which contains MATH . We note that MATH so MATH . From this it follows that MATH and hence that MATH is forward invariant under MATH and similarly under MATH . We conclude MATH is invariant and from REF that MATH and since the family MATH is a nested sequence of compact sets we know that MATH for MATH sufficiently large. The frontier of MATH consists of finitely many segments of MATH and MATH and hence must contain a homoclinic point of MATH . |
math/9910152 | If MATH is an essential simple closed curve in MATH and if MATH is the component of its complement (in MATH) which contains MATH, a well-known result of NAME (see REF) implies that the component of the compact set MATH containing MATH, must meet MATH. This component, MATH, is a positively invariant set containing the set MATH but not equal to it. REF tells us that the globally invariant set MATH is reduced to MATH. So for any point MATH we have MATH. REF tells us that the connected globally invariant set MATH meets MATH. If MATH is an essential hyperbolic periodic point, REF says there exists an essential closed curve contained in the union of any unstable branch of MATH and any stable branch. If MATH is large enough, the set MATH must meet this curve. We deduce that MATH meets both unstable branches of MATH. More precisely, there exists an open interval MATH of MATH which contains MATH is disjoint from MATH and whose boundary points belong to MATH. This arc is contained in a component MATH of MATH and separates MATH into two components (because MATH is connected) which are connected components of MATH. Similarly we can construct a negatively invariant compact set MATH in MATH, containing the set MATH but not equal to it. For any point MATH we have MATH. Also as before, MATH meets both branches of MATH. Consider two points MATH, one in each branch of MATH . For large MATH the points MATH will be close to MATH, one in each component of MATH. Hence the connected set MATH meets the two components of MATH. The set MATH doesn't intersect MATH, so it intersects MATH. Any point of intersection of MATH and MATH satisfies the proposition. |
math/9910152 | Let MATH be a branch of MATH for the lift MATH . We can assume without loss of generality that it is an unstable branch. Since MATH is essential there is a MATH such that MATH intersects MATH transversely. The MATH-lemma then implies the closure of MATH contains the closure of MATH. The MATH-lemma also implies that the set MATH consisting of all values of MATH for which MATH intersects MATH is closed under addition. Another straight-forward MATH-lemma argument and the fact that the rotation number of MATH is in the interior of the rotation set for MATH shows that MATH contains both positive and negative elements. From this we may conclude that MATH is a subgroup of MATH say the subgroup MATH . If we replace MATH and MATH by a MATH-fold cover of MATH and a lift of MATH to it then clearly nothing has changed in what we want to prove except that now MATH has been replaced by MATH. That is, we may assume without loss of generality that MATH or that MATH . Hence we may assume that the closure of MATH contains the closure of MATH for every MATH. Now let MATH be an element of MATH and let MATH . The fact that MATH is dense in MATH implies that for any MATH there is a point MATH of MATH which is within MATH of MATH. Equivalently, there is an integer MATH and a point MATH of MATH which is within MATH of MATH . Hence MATH is dense in MATH . |
math/9910152 | It is sufficient to prove this result in the case where MATH is finite and the MATH are connected. In fact, appealing to induction, it is sufficient to prove that the union of two invariant simply connected domains MATH and MATH is simply connected. The proof is obvious if one of the complements MATH or MATH is empty or reduced to a point. We will suppose that none of these properties occurs. The boundaries MATH and MATH are irrational invariant continua and we can write MATH where MATH and MATH are some invariant simply connected domains. If the sets MATH and MATH meet, their union is an irrational invariant continuum and the complement MATH has exactly two components, by REF. At least one of the sets MATH or MATH is not empty (because MATH has no interior and cannot contain MATH). We deduce that MATH is either empty or connected and that MATH is simply connected. If MATH is contained in MATH, then one of the open sets MATH or MATH doesn't meet the connected set MATH and is contained in MATH. In the first case MATH is simply connected, in the second case we have MATH so MATH is simply connected. If MATH is contained in MATH, then one of the open sets MATH or MATH doesn't meet the connected set MATH and is contained in MATH. In the first case MATH and MATH are disjoint and MATH is simply connected, in the second case we have MATH and MATH is simply connected. |
math/9910152 | By REF , we know that MATH is connected. It is clear that MATH and MATH are connected components of MATH. If MATH is a periodic continuum such that MATH and MATH are components of MATH, then the other components are in the family MATH, so MATH contains MATH. To see that MATH is invariant if MATH and MATH are we observe that MATH is MATH-invariant in that case. We still have to prove that MATH is periodic, but this follows by considering MATH where MATH is a common period for MATH and MATH and noting that MATH is the same for MATH and MATH, so our previous remark implies MATH . |
math/9910152 | We can suppose that MATH and MATH are invariant. If MATH has no periodic point it is an irrational continuum, so the components of MATH are exactly MATH and MATH and we deduce that MATH. Let's prove now the second part of the proposition. There is no periodic point in MATH and in MATH because these two sets are periodic irrational continua. So every elliptic periodic point MATH has a periodic simply connected neighborhood, bounded by a NAME curve, which meets neither MATH nor MATH. We deduce that MATH. |
math/9910152 | The set MATH of periodic points of type MATH in MATH is finite. If it is disjoint from MATH, there exists a smallest finite family MATH of components of MATH which covers MATH. Every MATH is MATH-invariant and every fixed point of MATH in MATH belongs to MATH. The contradiction comes from the NAME formula MATH . |
math/9910152 | The periodic continuum MATH is contained in MATH. If it doesn't separate MATH and MATH, the component of MATH which contains MATH and MATH is a periodic simply connected domain MATH and MATH since MATH has a fundamental system of neighborhoods whose boundaries are in the branches of MATH. We deduce that the component of MATH which contains MATH is a periodic domain disjoint from MATH and MATH. It is disjoint from MATH by definition of this set so we have a contradiction. Let's prove now that MATH is invariant by showing that the orbit of MATH is contained in MATH. Consider the components MATH and MATH of MATH which contain respectively MATH and MATH. The set MATH is a MATH-invariant irrational continuum and meets its image because it separates MATH and MATH and MATH is area preserving. We deduce that MATH is a MATH-invariant irrational continuum. The orbit of MATH doesn't meet the component MATH of MATH which contains MATH, it is contained in the other component which is disjoint from MATH. Similarly the orbit doesn't meet MATH. It doesn't meet any other component of MATH because it is contained in MATH. We have proven that it is contained in MATH. We deduce that any stable branch of MATH in the orbit of MATH meets any unstable branch of MATH. We deduce from this fact and from the MATH-lemma the end of the proposition in the case where MATH is a periodic point of period MATH. But if MATH is a fixed point then we choose a hyperbolic periodic point MATH of period at least two and observe that MATH . The fact that MATH intersects MATH transversely and MATH intersects MATH transversely allows us to apply the MATH-lemma once again and obtain a simple NAME curve separating MATH and MATH which is a subset of MATH . |
math/9910152 | If the complement of MATH has at least two invariant domains, this is a consequence of REF . In the other case the NAME formula tells us that there exists at least one fixed point in MATH. If MATH contains a hyperbolic fixed point, the result is obvious. If MATH contains an elliptic fixed point MATH, we can find two minimal invariant curves MATH and MATH of different irrational rotation number surrounding MATH in a neighborhood of that point. These curves meet MATH and are contained in it. By the NAME theorem, we can apply REF to the domain MATH containing MATH bounded by the closest curve and to the domain MATH which doesn't contain MATH and bounded by the other curve. There exists MATH, such that MATH is invariant and a simple curve separating MATH and MATH and contained in MATH. This curve meets the connected set MATH and we have MATH. |
math/9910152 | This is an immediate consequence of REF and of the fact that there is no strict inclusion MATH, for MATH. |
math/9910152 | Recall that the set MATH consists of invariant continua. The result is a consequence of REF and of the following facts: CASE: the set MATH is closed in the NAME topology; CASE: any continuum MATH, MATH, is isolated in MATH; CASE: for any MATH, the set of continua MATH which contain MATH is open in MATH. |
math/9910152 | Indeed, any convergent sequence MATH, MATH, of invariant continua converges either to a trivial continuum MATH, MATH, or to an irrational invariant continuum, or to an invariant continuum which contains an invariant set MATH, MATH. In the last situation the set MATH is contained in MATH, for MATH big enough. The sequence MATH is constant and equal to MATH. |
math/9910152 | The fact that it is closed is a consequence of REF . We prove that it is connected by showing that any component MATH of MATH is simply connected. We first observe that MATH is not invariant. If it were invariant, its closure would be an invariant continuum which is not irrational (because MATH). It would contain an invariant continuum MATH, MATH, which is disjoint from MATH (because MATH). A contradiction arises from the fact that there is a fundamental system of neighborhoods of MATH whose boundary is in MATH. Next we prove that MATH is simply connected. Any NAME curve MATH contained in MATH is disjoint from its image. The map MATH being volume-preserving, one of the two domains bounded by MATH is disjoint from its image. Any component of MATH being invariant, we deduce that this domain is disjoint from MATH and contained in MATH. We conclude that MATH is simply connected. We have proven that any component of MATH is a strictly periodic simply connected domain of period MATH. Conversely, a strictly periodic domain MATH of period MATH doesn't contain any fixed points. It cannot meet any fixed continuum MATH, MATH since otherwise it will meet MATH and the closure MATH which is disjoint from its image will contain MATH. Finally we must show that MATH cannot meet any irrational invariant continuum MATH. If it did then we could not have MATH so MATH . It follows that MATH is a periodic irrational continuum and by our remarks above that MATH is a periodic irrational continuum. Since MATH has no interior, the complement of MATH would necessarily have at least MATH components since it would contain a point from MATH, a point from MATH and a point in neither of them. Each of these would be in different components of MATH . This contradicts REF. |
math/9910152 | We can write MATH and MATH, where MATH is the union of the MATH-invariant irrational continua and of the elliptic fixed points of MATH and MATH the union of the sets MATH, MATH, which are MATH-invariant. Each set MATH is closed and each set MATH is a MATH, moreover MATH is closed, connected and contains MATH. The first assertion is clear as is the connectedness of MATH. The last assertion is an immediate consequence of REF . The density of MATH follows from the inclusion MATH that we will prove now. For any decreasing sequence MATH of strictly periodic domains which contains MATH such that the sequence of periods tends to infinity, the volume of MATH tend to zero. So MATH which is contained in MATH meets a given neighborhood of MATH if MATH is big enough. Since MATH we deduce that MATH . If we show MATH then it follows that MATH and we can conclude REF . If MATH is in MATH then by REF MATH where MATH is a strictly periodic simply connected domain of high period (and hence of small measure). Given MATH we may choose MATH so that any point of MATH is within MATH of MATH . We also know that MATH where MATH is a strictly periodic simply connected domain of higher period than MATH . The annular domain MATH is invariant under some iterate of MATH and contains periodic points, so we may apply REF to its prime end compactification to conclude that it has an essential periodic point MATH. Then MATH separates MATH and MATH . It follows that there must be a point of MATH within MATH of MATH and hence MATH. |
math/9910152 | It is sufficient to prove that among the generic diffeomorphisms studied in this section (that is, the NAME generic diffeomorphisms, see REF), the ones which don't have any periodic open annulus without periodic points is generic. For such a map the interior of the sets MATH, MATH, is empty as is the interior of the union MATH. We deduce the second part of the proposition from the inclusion MATH. Thus it suffices to prove that there is a generic set of diffeomorphisms in MATH which have no periodic open annulus which contains no periodic points. Consider an open connected and simply connected subset MATH . We will first show that for a dense open subset of MATH the set MATH is not contained in a periodic open annulus which contains no periodic points. Consider the set of diffeomorphisms MATH defined by the property that MATH if and only if there are two disjoint simple closed curves MATH and MATH contained in one component of MATH with the property that there are hyperbolic periodic points of MATH in each of the three components of MATH . It is clear that if MATH then MATH is not contained in a periodic open annulus which contains no periodic points since MATH and MATH would have to be contained in such an annulus and this is impossible. We will show that MATH is dense and open. The fact that MATH is open is clear since for any diffeomorphism MATH close to MATH we know MATH is contained in one component of MATH because MATH is compact. Also if MATH is close to MATH it will still have hyperbolic periodic points in the three components of MATH . So MATH is open. To see that MATH is dense consider any MATH. We can approximate it arbitrarily closely by a NAME generic diffeomorphism MATH. We will show either MATH or it can be approximated by an element of MATH . Consider MATH. If MATH contains a periodic point of MATH then it contains infinitely many hyperbolic periodic points since for NAME generic diffeomorphisms hyperbolic periodic points are not isolated and elliptic periodic points are in the closure of hyperbolic periodic points. It is thus clear that if MATH contains any periodic points we may choose simple closed curves MATH and MATH contained in one component of MATH such that each component of MATH contains a hyperbolic periodic point. Thus in this case MATH . So we may assume MATH contains no periodic points. We choose a periodic component MATH of MATH, say of period MATH, so MATH . If the complement of MATH has three or more components containing periodic points we can, as before, choose MATH and MATH showing MATH . It is not possible for the complement of MATH to have only one component containing periodic points, since the complement of that component would be a simply connected MATH invariant domain with no periodic points. All that remains is the case that the complement of MATH has exactly two components containing periodic points. We choose a simple closed curve MATH in MATH which separates these two components. The complement of these two components is an open MATH invariant annulus MATH containing MATH and in which MATH is essential. We choose a small annular neighborhood MATH of MATH in MATH. Perturbing MATH by the time MATH map of an area preserving MATH flow MATH supported in the interior of MATH we may continuously vary the mean rotation number of MATH on the annulus MATH. However a result of CITE says that if this number is rational there must exist a periodic point. Clearly if MATH is small the perturbation is MATH close to MATH. One point on the new periodic orbit must lie in the interior of MATH since that is where the perturbation is supported. By a slight further perturbation we obtain MATH close to MATH with a hyperbolic periodic point in the interior of MATH. Choosing MATH and MATH to be the boundary components of MATH we observe that MATH. Thus we have shown that MATH is open and dense. If we now choose a countable basis MATH for the topology of MATH consisting of open disks and let MATH is the desired dense MATH. |
math/9910156 | The first part is proved in CITE. We now want to prove that the following diagram MATH commutes. Remark first that it clearly commutes on MATH. Put MATH. The upper part of the diagram gives a morphism MATH which induces the identity on MATH. It thus factorizes uniquely through MATH to give a morphism MATH equal to MATH on MATH. It follows that MATH (indeed, MATH is injective because MATH has no torsion supported on MATH, and therefore is onto as MATH is holonomic). |
math/9910156 | Remark first that there is a natural isomorphism of functors MATH . Indeed, denoting by MATH the direct image of MATH-modules, recall that one has MATH: indeed, put MATH and consider the natural morphism of right MATH-modules MATH such that, for any local section MATH of MATH, the image of MATH evaluated on any function MATH is equal to MATH; this morphism is an isomorphism, as can be seen from a local computation; going from right to left, one gets the assertion. It follows that MATH . As MATH and MATH are exact functors, we obtain from REF an isomorphism MATH and thus the result, as MATH is an equivalence. |
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