paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9907045 | REF is clear. For REF , the inclusion MATH follows from REF . Equality will come by showing that the NAME functions are the same. Consider the sequences MATH . Note that MATH. Thanks to REF we then have the following calculation. (We use the notation MATH for the MATH-th difference of the NAME function of MATH. This is standard notation, but in any case see the discussion at the end of REF.) MATH . Hence the NAME functions agree. |
math/9907045 | REF follows from REF also uses REF and its proof. |
math/9907045 | We first prove that MATH is a union of linear varieties, and hence that MATH is radical. We know from REF that we can write MATH in the form MATH where each MATH is a complete intersection of the form MATH. Using REF , it is clear that each lifted ideal MATH defines a reduced complete intersection in MATH. Finally, from REF we get that MATH is reduced and a union of linear varieties. We have seen in REF that we cannot hope that MATH will always be a generalized stick figure. Indeed, by REF , if MATH is not empty then all components of MATH will pass through MATH, so if there are too many of these components then it will fail to be a generalized stick figure. For REF , consider the lifting matrix MATH . Consider a primary decomposition of MATH and write it as MATH where, for MATH, MATH is the intersection of the primary components of codimension MATH. By REF , the components of MATH of codimension MATH are defined by the lifting MATH of MATH. Let MATH be the scheme defined by MATH (possibly empty if MATH is NAME) and let MATH be the scheme defined by MATH. Any component of MATH is a linear variety defined by MATH entries of the lifting matrix, no two from any given row, and it vanishes (set theoretically) on a component of MATH. Because the linear forms in MATH were chosen generally, subject only to the condition that the entries of the MATH-th row are linear forms in the variables MATH, it follows that the intersection of any MATH components has codimension MATH in MATH, at least away from the linear space MATH defined by the vanishing of the variables MATH. (See also REF .) But the intersection of MATH with this linear space is exactly MATH. This proves REF . For REF , the fact that MATH is arithmetically NAME follows from REF , while the fact that it is a generalized stick figure follows from REF , since MATH is empty. For REF , let MATH be the lifting of MATH. Note that MATH, by REF , MATH is unmixed by REF , and hence MATH has exactly MATH components passing through MATH. Furthermore, MATH has codimension MATH in MATH. Then using REF , we see that the only way that the intersection of three components of MATH could fail to have codimension two in MATH is if MATH and MATH for some MATH. |
math/9907045 | Let MATH be a reduced MATH-lifting of MATH. Then we clearly have MATH. On the other hand we have for the NAME functions: MATH. Thus, the equality MATH follows. |
math/9907045 | Since MATH, we have by REF that MATH . But MATH, so since MATH is a lex-segment ideal we get that MATH as well. Hence MATH as claimed. |
math/9907045 | For both parts of the theorem we have only to prove the sufficiency of the condition. We continue to use the same notation MATH for the linear form and for the corresponding hyperplane. Without loss of generality we can assume that MATH neither empty nor all of MATH. Let MATH be the ideal generated by the set of all polynomials which are products of the form MATH and which vanish on MATH. Remove generators from this set to obtain a minimal generating set. Any generator of this form corresponds, via REF , to a monomial. Let MATH be the ideal generated by the set of all such monomials. MATH is NAME since it contains a power of each of the variables, and clearly MATH is the pseudo-lifting of MATH. Hence MATH is the saturated ideal (thanks to REF ) of a scheme MATH with MATH. We have only to show that MATH, and for this we use the condition of REF . Let MATH be a component of MATH which is not in MATH (OK since MATH is not all of MATH). We will produce a polynomial MATH in MATH which does not vanish on MATH. Being a component of MATH, MATH is of the form MATH as above. Thanks to the condition of REF , we see that for each MATH, replacing the coordinate MATH by MATH for any MATH with MATH gives a component of MATH which is also not in MATH. Let MATH be the set of indices MATH for which MATH and let MATH be the set of indices MATH for which MATH. Since MATH is not empty, MATH is not the empty set. Let MATH . Clearly MATH does not vanish on MATH. However, any component MATH of MATH which is which is not in the vanishing locus of MATH has entries which satisfy MATH for MATH and MATH for MATH. Since MATH, we thus have MATH as well. For the second part of the theorem, let MATH be monomials of the same degree such that MATH. Then there is a smallest integer MATH such that MATH for MATH and MATH. Suppose MATH. We have to show that MATH. Without loss of generality we may assume that MATH. Then we have MATH . Since MATH, REF implies that MATH. Hence the condition of REF gives that MATH as well, so that again by REF we have MATH as claimed. |
math/9907045 | The proof of REF shows that MATH is a lifting of MATH, hence it is a complete intersection, and that MATH is reduced. Then REF applies. |
math/9907045 | The ``generic" NAME function is the MATH-times differentiable NAME function of dimension MATH whose corresponding MATH-vector is MATH . Any NAME monomial ideal with this NAME function is forced to be lex-segment. In any other case one can find a monomial ideal which is not lex-segment. |
math/9907046 | By enumerating sets in MATH, it suffices to consider only sets of ordinals. So, suppose that MATH has a closure point at MATH, so that MATH is MATH-generic, MATH, MATH is MATH-strategically closed", and all the MATH-approximations to a set MATH over MATH are in MATH. We want to show that MATH itself is in MATH. If MATH is not in MATH, then by chopping MATH off at an earlier ordinal if necessary we may assume that all the initial segments of MATH are in MATH. It follows that MATH, for otherwise MATH would be a fresh sequence added by forcing with a closure point contrary to the NAME Lemma of CITE, mentioned just earlier. So in MATH we may write MATH, where MATH and MATH is a continuous, increasing sequence of ordinals in MATH. Since MATH for all MATH, it follows by the closure of the forcing MATH that MATH and so MATH for some MATH-name MATH. Let MATH and MATH. These sets are uniformly definable in MATH, and their union MATH forms a tree under end-extension. Since the elements of MATH give rise to incompatible values for MATH and therefore to an antichain in MATH, it must be that MATH, and therefore also MATH. Define now MATH . In other words, MATH is the set of all possible branching points for branches through the tree MATH. Since we already observed that MATH has at most MATH many members, MATH also has size at most MATH. Thus, MATH is a MATH-approximation to MATH over MATH and hence by REF. Now we are nearly done. The set MATH determines a branch through MATH, and the information about which direction that branch turns at every possible branching point in MATH is precisely contained in the set MATH. Using MATH as a guide in MATH, therefore, we can direct our way through MATH in exactly the same way that MATH winds through MATH. So MATH, as desired. |
math/9907046 | Suppose that the forcing admits a closure point at MATH. By the Approximation Lemma, it suffices to show that every MATH-approximation to MATH is in MATH. So suppose MATH has size at most MATH, and consider MATH. We may assume that every member of MATH is a subset of MATH, since these are the only possible members of MATH. Let MATH be obtained by closing MATH under complements in MATH. Since MATH is a collection of at most MATH many sets in the filter, it follows by the MATH-completeness of MATH that MATH is in MATH. In particular, MATH is nonempty, and so we may choose an element MATH. Observe now that if MATH then MATH and consequently MATH. Conversely, if MATH and MATH then because MATH it follows that MATH and so MATH; by the assumption that MATH measures every set in MATH, therefore, we conclude that MATH. Thus, we have proved for MATH that MATH. So MATH is precisely the set of all MATH with MATH, and this is certainly in MATH. |
math/9907046 | Suppose that MATH is weakly compact in MATH, a forcing extension with a closure point at MATH, and that MATH is collapsed to MATH there. We may exhibit the gap explicitly by writing MATH, where MATH, MATH, and MATH is MATH-strategically closed". Let MATH, and choose MATH of size MATH, closed under MATH-sequences, containing MATH, MATH, MATH, MATH as well as every element of MATH and every element of MATH, both of which by assumption have size MATH in MATH. The NAME collapse of MATH has the form MATH, where MATH is MATH-generic for forcing with a closure point at MATH. Furthermore, since MATH was closed under MATH-sequences in MATH, the same holds for MATH. And since MATH has size MATH and MATH is weakly compact, there is an embedding MATH with critical point MATH. Consider now the set MATH. Since MATH was collapsed to MATH in MATH, it follows that there is a relation MATH in MATH on MATH such that MATH. And since MATH is fixed by the NAME collapse of MATH, it follows that it is also in MATH. By the elementarity of MATH, if MATH is the MATH element with respect to the relation MATH on MATH, then MATH is the MATH element with respect to the relation MATH on MATH. Thus, MATH is precisely the set MATH in MATH. So MATH. Let us now show that MATH. Since the forcing MATH admits a closure point at MATH, it suffices by the Approximation Lemma to show that every MATH approximation to MATH over MATH is actually in MATH. So suppose that MATH has size MATH, and consider MATH. This set has size MATH, so it must have the form MATH for some MATH. By the closure of MATH we know that MATH; further, since the MATH forcing is MATH-strategically closed, it must really be that MATH. Since the MATH forcing has size MATH, we conclude that MATH for some MATH of size MATH. Thus, MATH . Consequently, MATH. But MATH has size MATH and so we have MATH. Furthermore, since MATH we know MATH. Thus, MATH is also in MATH, as we had desired. So by the Approximation Lemma, MATH. Let MATH. This is a pre-filter on MATH which measures every set in MATH. In particular, since by design MATH includes every element of MATH, we know that MATH measures every subset of MATH in MATH. And since MATH is closed under MATH-sequences in MATH, it follows that the filter generated by MATH is MATH-complete in MATH. Thus, by the Filter Extension Lemma, the set MATH must be in MATH. It remains only to check that MATH is a normal fine measure on MATH in MATH. Note that because MATH, we know that MATH. Certainly MATH is a MATH-complete measure on MATH in MATH, because MATH is MATH-complete in MATH and measures every set in MATH. It is a fine measure because MATH contains the element MATH for every MATH, and so for any given such MATH, the filter MATH concentrates on the set of MATH containing MATH. To see that MATH is normal, suppose that MATH is regressive in MATH. Thus, MATH, and so MATH for some MATH. So MATH for MATH-almost every MATH. So MATH is a normal fine measure on MATH in MATH. We conclude that MATH is MATH-supercompact in MATH, as the theorem asserts. |
math/9907052 | For MATH, there exists a unique geodesic ray MATH such that MATH is perpendicular to MATH and passes through MATH. Parametrize MATH by MATH, where MATH measures the distance from MATH. Recall that for the nearest point retraction map MATH which retracts MATH along MATH for MATH, we have MATH. Because MATH is strictly distance decreasing on MATH (see REF), given a homotopically non-trivial loop MATH based at MATH, MATH will be a homotopically non-trivial loop based at MATH of strictly shorter length. So MATH is strictly increasing in MATH. |
math/9907052 | For MATH, there exists a homotopically non-trivial loop MATH based at MATH of length MATH. Because MATH is a local isometry, the projection MATH of the loop to MATH has length MATH. If MATH is homotopically trivial in MATH, then MATH bounds a disk MATH. By lifting this disk to MATH, we see that MATH bounds a disk MATH in MATH, which contradicts the assumption that MATH was homotopically non-trivial in MATH. Thus, the loop MATH must be homotopically non-trivial in MATH so that MATH. |
math/9907052 | Because MATH is the component of MATH associated to MATH, it has a product structure MATH with nearest point coordinates. By the Lifting Theorem, the inclusion map MATH lifts to a map MATH. Let MATH. Then MATH is a homeomorphism. Recall that MATH is the image of some pleated surface MATH. Then there exists a lift MATH of MATH such that MATH. Let MATH be the pleating locus of MATH. Recall that a pleated surface MATH is totally geodesic in the complement of MATH, and MATH maps the leaves of MATH geodesically. NAME REF have shown that the lifts to the universal cover of the leaves of MATH have endpoints that lie in MATH. Therefore, MATH. If MATH is not a boundary component of MATH, then there exists a geodesic arc MATH in MATH with endpoints in MATH, that is, MATH is a geodesic arc with endpoints in MATH whose interior lies outside of MATH. This contradicts the fact that MATH is a boundary component of a convex set MATH. Therefore, MATH is a boundary component of MATH. So MATH is a component of MATH. Let MATH be a component of MATH. Then for each point MATH on MATH, there exists a geodesic arc MATH perpendicular to MATH and passing through MATH. Lift MATH to a point MATH. Because MATH is an isometry on MATH, this geodesic arc lifts to a geodesic arc MATH that is perpendicular to MATH and passes through MATH. Then the set of points in MATH a distance MATH from MATH lifts precisely to the set of points in MATH a distance MATH from MATH. Hence MATH is a component of MATH. |
math/9907052 | Recall the deformation retract MATH. This map lifts to the deformation retract MATH. Then given a component MATH of MATH, we can consider a component MATH of the preimage of MATH in MATH. Here, MATH will be a boundary component of MATH. This component MATH corresponds to a component MATH of MATH, where MATH. Let MATH be the portion of MATH that is bounded by MATH and MATH. Let MATH be the portion of the preimage of MATH in MATH that is contained in MATH. Then MATH covers MATH, and MATH covers MATH. Furthermore because MATH is a neighborhood of the geometrically finite end MATH, MATH must contain a neighborhood of MATH in MATH. Because MATH, each component of MATH is contained in a component of MATH. Suppose MATH is not a component of MATH. Then there exists a point MATH in MATH such that MATH is also contained in MATH. We know that pairs of hyperbolic fixed points are dense in MATH (REF , CITE), that is, for a pair of points MATH, there exists a sequence of hyperbolic isometries whose fixed points approach MATH. So there exists a hyperbolic isometry MATH whose fixed points lie in a neighborhood MATH of MATH such that MATH. Furthermore MATH must contain a neighborhood of MATH. Thus, MATH. Then because MATH covers MATH, we can conclude that MATH does not embed in MATH, but this is a contradiction. So MATH is a component of MATH. Then MATH is also a boundary component of MATH, and for MATH, MATH. So we can conclude that MATH embeds in MATH, MATH is a component of MATH, and MATH is a boundary component of MATH. |
math/9907052 | Because MATH is topologically tame, there exists a relative compact core MATH of MATH with associated parabolic locus MATH such that each component of MATH is incompressible and each component MATH of MATH is homeomorphic to a component of MATH. (see REF) Choose a level surface MATH in each MATH so that MATH lies to one side. Then each MATH is incompressible and properly isotopic to a component of MATH. Thus, the MATH bound a relative compact core MATH of MATH such that MATH. Furthermore, the components of MATH are topologically a product. |
math/9907052 | By the argument given in REF , there exists a relative compact core MATH such that each component of MATH is topologically a product and MATH. Select a surface MATH in a component of MATH. Let MATH be the subset of MATH bounded by MATH and MATH. If MATH is not homeomorphic to MATH, then MATH does not surject onto MATH. (REF , CITE) Then there exists a homotopically non-trivial loop MATH in MATH that is not homotopic in MATH to a curve in MATH. Because MATH is incompressible and separating in MATH, MATH is not homotopic in MATH to a curve in MATH. But MATH is a compact core of MATH, so this is a contradiction. Then for each MATH, MATH is homeomorphic to MATH. |
math/9907052 | MATH: We know that MATH is a subgroup of MATH so that MATH. Thus, MATH, and hence MATH. Therefore, MATH. MATH: We will show that for a point MATH, we have MATH. If MATH, then this is automatically true. If MATH, then let MATH be a component of MATH that separates MATH from MATH. Let MATH be the component of MATH that does not contain MATH. So in particular, MATH contains MATH. Then MATH is a locally convex submanifold of MATH with boundary MATH, and hence MATH is convex. (REF , CITE) Using the MATH coordinates on each component of MATH adjacent to MATH, we can extend the homeomorphism MATH to a homeomorphism MATH where MATH is the parabolic extension of MATH. Without loss of generality, we can assume that MATH is small enough so that MATH intersects each component of MATH totally geodesically. Thus, we know that MATH. Let MATH be a component of MATH which intersects MATH. Then because MATH is a homeomorphism, for large enough MATH, MATH embeds in MATH. Then by REF , MATH is a component of MATH with boundary MATH. Therefore, MATH is a convex submanifold of MATH. We can construct a deformation retract MATH by letting MATH and MATH where MATH is the nearest point retraction map. Then MATH is a convex submanifold of MATH which carries the homotopy of MATH. Therefore, MATH. By REF , we know that for the component MATH of MATH corresponding to MATH, MATH is a component of MATH. Then because MATH is an isometry, MATH projects homeomorphically to MATH. Thus we can conclude that MATH. So if MATH, then MATH. Then MATH and hence MATH. |
math/9907052 | For each point MATH, there exists an embedded disk MATH of radius MATH in MATH. Because the curvature of MATH is MATH, in MATH, MATH, the latter being the area of a disk with the same radius in hyperbolic space. Since MATH, for any MATH in the image of a simplicial hyperbolic surface, MATH where the constant MATH depends on the NAME characteristic of MATH. |
math/9907052 | Let MATH and MATH be parabolic elements in MATH with distinct fixed points. Suppose for contradiction that MATH maps the fixed point of MATH and the fixed point of MATH to the same point in MATH. Then MATH and MATH commute in MATH. But MATH and MATH represent parabolic elements which do not commute in MATH. Therefore, MATH is not MATH-injective, but this is a contradiction. So indeed, MATH maps no two parabolic fixed points to the same point. |
math/9907052 | Let MATH be the universal cover of MATH, and let MATH be a lift of MATH. Let MATH be the component of the pre-image of MATH to MATH which contains MATH. Let MATH be a lift of MATH. By hypothesis, there exists a geodesic axis MATH which is properly homotopic in MATH to MATH. We can properly homotop MATH to a map MATH which maps MATH to a point on the geodesic covered by MATH. Using NAME 's construction, MATH is a MATH-injective simplicial pre-hyperbolic surface. Because the lift MATH of MATH maps MATH to a point on MATH, the internal vertex MATH satisfies MATH with respect to MATH so that MATH is a MATH-injective useful simplicial hyperbolic surface with distinguished edge MATH. By construction, MATH is properly homotopic to MATH. |
math/9907052 | Let MATH be the universal cover of MATH, and let MATH be a lift of MATH. Let MATH and MATH be the components of the pre-images of MATH and MATH to MATH which intersect at MATH. Using the notation of the MATH extension map, because MATH is MATH-injective, we can construct a well-defined extension map MATH such that if MATH is the fixed point of a parabolic element MATH, then MATH is the fixed point of MATH. By an argument similar to that in REF , we can show that whereas MATH and MATH share MATH as an endpoint, the other endpoints of MATH and MATH do not coincide. Hence, there exists a geodesic axis MATH which is properly homotopic in MATH to MATH. We can properly homotop MATH to a map MATH which maps MATH to a point on the geodesic covered by MATH. Using NAME 's construction, MATH is a MATH-injective simplicial pre-hyperbolic surface. Because the lift MATH of MATH maps MATH to a point on MATH, the internal vertex MATH satisfies MATH with respect to MATH so that MATH is a MATH-injective practical simplicial hyperbolic surface with marked edges MATH and MATH. By construction, MATH is properly homotopic to MATH. |
math/9907052 | Consider the MATH extension map of MATH, MATH. By definition, MATH maps the parabolic fixed points of MATH into MATH. By REF maps all the internal vertices of MATH into MATH. Thus, MATH maps all lifts of internal vertices into MATH. Recall that MATH maps a face with endpoints MATH to a geodesic triangle in MATH spanned by the points MATH. Because MATH is convex, we can conclude that MATH, and hence MATH. Suppose MATH is a useful simplicial hyperbolic surface with distinguished edge MATH. Then MATH maps MATH to a closed geodesic. Because all closed geodesics lie in the convex core, and the sole internal vertex MATH lay on MATH, we know that MATH. Then by the above argument, MATH. Suppose MATH is a practical simplicial hyperbolic surface with marked edges MATH and MATH. By definition, MATH maps each component of the pre-image of REF to MATH to a geodesic that has endpoints at parabolic fixed points associated to the ideal vertices contained in the marked edges. Thus, MATH maps each component of the pre-image of REF to MATH to a geodesic axis in MATH. Since the lift MATH of the sole internal vertex lay on this axis, MATH, and hence MATH. Then by the above argument, MATH. |
math/9907052 | We begin by outlining the construction of the continuous family given by NAME, then we will show that its image lies in the convex core. Let MATH be a homotopy between MATH and its geodesic representative MATH. Let MATH be the universal cover of MATH, and let MATH be a lift of MATH. Let MATH and MATH be the components of the pre-images of MATH and MATH to MATH which intersect at MATH. Lift both MATH and MATH to maps MATH and MATH such that MATH. Let MATH, and let MATH. Let MATH be the unique common perpendicular joining MATH to MATH. Let MATH, and let MATH be a geodesic arc in MATH joining MATH to MATH. Let MATH and MATH be the projections to MATH of MATH and MATH respectively. By dragging MATH along MATH, we obtain a continuous family of simplicial hyperbolic surfaces MATH. In fact, this is a continuous family of useful simplicial hyperbolic surfaces because MATH is a geodesic arc contained in a lift of MATH and hence MATH maps MATH to a closed geodesic for MATH. By dragging MATH along MATH, we obtain a continuous family of simplicial hyperbolic surfaces MATH such that MATH lies on the closed geodesic in the homotopy class of MATH. Then MATH is a useful simplicial hyperbolic surface with associated triangulation MATH and distinguished edge MATH. Finally, MATH is obtained by concatenating MATH and MATH. This completes the sketch of the argument given by NAME. Because MATH are useful simplicial hyperbolic surfaces for MATH, their image automatically lies in MATH. Since the sole internal vertex of each simplicial hyperbolic surface in the family MATH for MATH is dragged along the arc MATH, by REF , it suffices to show that this arc lies in the convex core. Let MATH be the distinguished edge of MATH. Here, MATH is defined to be the unique common perpendicular between two components of lifts of MATH and MATH, each of which is a collection of axes of hyperbolic elements. Then MATH is a geodesic that joins two axes of hyperbolic elements. Because MATH is convex, we know that MATH lies in MATH, and hence MATH. Thus, the image of MATH lies in the convex core of MATH. |
math/9907052 | We begin by outlining the construction of the continuous family given by NAME, then we will show that its image lies in the convex core. Let MATH be the universal cover of MATH, and let MATH, MATH, MATH, MATH, MATH, and MATH be lifts of MATH, MATH, MATH, MATH, MATH, and MATH to MATH such that MATH, MATH, MATH, and MATH form a quadrilateral with diagonals MATH and MATH. Let MATH be a lift of MATH. Let MATH denote the geodesic arc joining the endpoints of MATH. Notice that MATH, MATH, MATH, MATH, MATH, and MATH form a tetrahedron MATH in MATH. Let MATH denote the intersection of MATH and MATH, and let MATH. Let MATH be the geodesic in MATH joining MATH and MATH. We obtain a new triangulation MATH of MATH by adding the edge MATH to MATH so that MATH now has an additional vertex at MATH. Let MATH be the projection to MATH of MATH. By dragging MATH along MATH we obtain a continuous family of simplicial hyperbolic surfaces MATH with associated triangulation MATH. After removing the vertex MATH and the edge MATH, MATH is a useful simplicial hyperbolic surface with associated triangulation MATH. This completes the sketch of the argument given by NAME. Note that MATH and MATH are useful simplicial hyperbolic surfaces, and hence their image lies in the convex core. The triangulation MATH has two internal vertices, MATH and MATH. To show that MATH, by REF , it suffices to show that for MATH, the two vertices MATH and MATH are mapped into the convex core. Note that for all MATH, MATH is a fixed closed geodesic in MATH. Because MATH, the vertex MATH is mapped into MATH for all MATH. In the construction, the vertex MATH is dragged along a geodesic arc MATH, so it suffices to show that MATH. Since the endpoints of MATH are in MATH, it is clear that MATH. |
math/9907052 | First let us give NAME 's proof of the existence of a continuous family of simplicial hyperbolic surfaces. Let MATH be a proper homotopy of MATH into a cusp of MATH. Let MATH be the universal cover of MATH. Let MATH be a lift of MATH, and let MATH be the component of the pre-image of MATH to MATH that contains MATH. Lift both MATH and MATH to maps MATH and MATH such that MATH. Let MATH such that MATH. Let MATH be a geodesic ray properly homotopic to MATH, relative to MATH. Note that MATH has an endpoint in the sphere at infinity that corresponds to the fixed point of a parabolic element. Drag MATH along the projection MATH of MATH to obtain a continuous family of simplicial pre-hyperbolic surfaces MATH. For the remainder of the proof, we will consider the upper half plane model of MATH, normalized so that MATH is a geodesic ray with endpoint at infinity. To check that we have a continuous family of simplicial hyperbolic surfaces, we will utilize REF to verify that the internal vertex MATH satisfies NLSC with respect to MATH for MATH. Recall that the distinguished edge MATH is mapped to a closed geodesic. Let MATH be the component of the pre-image of MATH that intersects MATH. Let MATH. Let MATH be the hyperbolic isometry with axis MATH, and let MATH be the parabolic isometry whose fixed point is the endpoint of MATH. There are four edges of MATH which have one endpoint at MATH have other endpoints at MATH, and MATH. By construction MATH lies on MATH. Note that MATH lies on the same horocycle as MATH and MATH. Therefore, the geodesic spanned by MATH and MATH intersects MATH above MATH. Now because MATH, MATH, and MATH are all exactly a radial distance MATH from the axis of MATH, we know that the geodesic joining MATH and MATH lies inside the MATH-neighborhood of the axis of MATH. Then this geodesic will intersect MATH below MATH. Hence, MATH lies in the tetrahedron spanned by the four other endpoints. By REF , the vertex MATH satisfies NLSC with respect to MATH for MATH. Thus, MATH is a continuous family of simplicial hyperbolic surfaces. Now let us show that the length of MATH converges to REF. As MATH, MATH is dragged along MATH, and for each MATH, MATH maps MATH to a geodesic arc joined at its endpoints. In the universal cover MATH, for each MATH, the component of the pre-image of MATH which intersects MATH is a piecewise-geodesic line. Within the fundamental domain in MATH containing MATH, the pre-image of MATH is a geodesic arc of bounded Euclidean length whose height approaches infinity as MATH approaches REF. So the hyperbolic length of MATH converges to REF. This completes the sketch of the argument given by NAME. Now we will show that the image of the continuous family MATH lies in the convex core. Since MATH is a useful simplicial hyperbolic surface, by REF , its image lies in MATH. Because the vertex is dragged along MATH, by REF , it suffices to show that MATH. One endpoint of MATH is a lift of MATH. Since the distinguished edge MATH passes through the vertex MATH, MATH lies in the convex core of MATH. The other endpoint of MATH is a parabolic fixed point. Because MATH is a geodesic ray, we know that MATH, and hence MATH. Thus, MATH. Now consider the image MATH. Each triangle in the finite triangulation MATH on MATH is mapped to a geodesic triangle in MATH with sole internal vertex MATH. Let MATH be the image in MATH of a triangle MATH in MATH. Let MATH be a lift of MATH. Then MATH consists of triangles whose vertices lie on three geodesic rays which are conjugates of MATH. As MATH, we drag the vertices of MATH along the conjugates of MATH towards conjugates of MATH. In doing so, each component of MATH converges to an ideal triangle with vertices at three conjugates of MATH. Thus, each component of MATH lies in a totally geodesic triangular prism with three endpoints at lifts of MATH and the three remaining (ideal) endpoints at conjugates of MATH. Hence MATH lies in a finite collection of such geodesic prisms. Here, the ideal endpoints of the prisms are conjugates of MATH which are the fixed points of the parabolic elements associated to rank one cusps MATH of MATH. So we can conclude that MATH has compact closure in MATH. Now let MATH be a component of MATH. Here, we consider MATH to be a surface with an (additional) ideal vertex associated to MATH. Note that the restriction to MATH of the triangulation of MATH forms a system of curves on MATH that does not constitute a triangulation, because there are homotopic edges. However, for MATH, MATH still maps each edge of the system of curves to a geodesic arc and maps each face geodesically. Let us investigate the behavior of the map MATH restricted to the subsurface MATH. Let MATH be the MATH extension map of MATH. We will consider the restriction of MATH to the component MATH of the preimage of MATH in MATH. We can lift the triangulation MATH on MATH to a triangulation MATH on MATH and consider the restriction MATH of the triangulation MATH to MATH. Let MATH. Because MATH maps each edge to a geodesic arc and maps each face geodesically, by an argument similar to that in REF , to show that MATH lifts to MATH, it suffices to show that MATH maps all internal vertices of MATH into MATH. Furthermore, since MATH respects the group action of MATH, it suffices to show that MATH maps MATH into MATH. Because MATH is nonempty, the convex hull of MATH must contain a geodesic ray MATH with endpoint at infinity. Let MATH be a point of height MATH on MATH. Recall that MATH is also a geodesic ray with endpoint at infinity. Then for a fixed MATH, there exists MATH such that MATH so that MATH. Therefore, MATH lifts to MATH. Because MATH is a fixed Euclidean distance from MATH, as MATH, MATH, and hence MATH. Therefore, as MATH, MATH. Furthermore, because we have eliminated the accidental parabolic associated to MATH, for MATH, the map MATH has fewer accidental parabolics than MATH. |
math/9907052 | The proof of this lemma mimics that of REF ; we give a brief outline here. Let MATH be the universal cover of MATH, and let MATH be a lift of MATH. Let MATH and MATH be the components of the pre-images of MATH and MATH to MATH which intersect at a point MATH. Lift MATH to a map MATH. Let MATH. Because MATH is a MATH-injective simplicial hyperbolic surface, REF guarantees that the MATH extension MATH maps no two parabolic fixed points to the same point. Thus there exists a geodesic axis MATH in the proper homotopy class of MATH. Take MATH to be the unique common perpendicular joining MATH to MATH. Let MATH, and let MATH be a geodesic arc in MATH joining MATH to MATH. Let MATH and MATH be the projections of MATH and MATH respectively. We construct MATH by dragging MATH along MATH, and then MATH. The remainder of the proof is similar to that of REF . |
math/9907052 | The proof of this lemma mimics that of REF ; we give a brief outline here. Let MATH, MATH, MATH, MATH, MATH, and MATH be lifts of MATH, MATH, MATH, MATH, MATH, and MATH to MATH such that MATH, MATH, MATH, and MATH form a quadrilateral with diagonals MATH and MATH. Let MATH be a lift of MATH. Let MATH denote the geodesic arc joining the endpoints of MATH. Notice that MATH, MATH, MATH, MATH, MATH, and MATH form a tetrahedron MATH in MATH. Let MATH denote the intersection of MATH and MATH, and let MATH. Let MATH be the geodesic in MATH perpendicular to MATH and passing through MATH. We obtain a new triangulation MATH of MATH by adding MATH to MATH which now has an additional vertex at MATH. Let MATH be the projection to MATH of MATH. By dragging MATH along MATH we obtain a continuous family of simplicial pre-hyperbolic surfaces MATH with associated triangulation MATH. The remainder of the proof is similar to that of REF . |
math/9907052 | The proof of this lemma is virtually identical to that of REF . |
math/9907052 | The proof of this lemma also mimics that of REF . We outline the argument here. Let MATH be the universal cover of MATH, and let MATH and MATH be components of pre-images of MATH and MATH to MATH which intersect at a point MATH which is a lift of MATH. Lift MATH to a map MATH. Let MATH. Because MATH is a MATH-injective simplicial hyperbolic surface, REF guarantees that the MATH extension MATH maps no two parabolic fixed points to the same point. Then there exists a geodesic axis MATH in the proper homotopy class of MATH. Let MATH be the unique common perpendicular joining MATH to MATH. Let MATH, and let MATH be a geodesic arc in MATH joining MATH to MATH. Let MATH and MATH be the projections to MATH of MATH and MATH respectively. We construct MATH by dragging MATH along MATH, then MATH. The remainder of the proof is similar to that of REF . |
math/9907052 | The proof of this lemma mimics that of REF . We give a brief outline here. Let MATH be a proper homotopy of MATH into a cusp of MATH, and let MATH be a proper homotopy of MATH into a cusp of MATH. Let MATH and MATH be components of the pre-images of MATH and MATH to MATH, respectively, that intersect at MATH, a lift of MATH. Lift the maps MATH, MATH, and MATH to maps MATH, MATH, and MATH such that MATH and MATH, and such that MATH for some MATH. Let MATH be the geodesic line properly homotopic to MATH. Then the endpoints of MATH at the sphere at infinity are distinct parabolic fixed points. Note that MATH may not lie on MATH. Let MATH be a proper homotopy between MATH and MATH. Let MATH be such that MATH. We can properly homotop MATH to a map MATH which maps MATH to MATH. Using NAME 's construction, MATH is a MATH-injective simplicial pre-hyperbolic surface. We then drag MATH along the projection MATH of MATH to obtain a continuous family of simplicial pre-hyperbolic surfaces MATH. Now we will show that MATH is in fact a continuous family of simplicial hyperbolic surfaces. We will use REF to verify that the internal vertex MATH satisfies MATH with respect to MATH for MATH. Without loss of generality, we can renormalize so that MATH has endpoints at MATH and MATH, and that the parabolic element with fixed point at MATH is MATH and the other parabolic element with fixed point at MATH is MATH. There are four edges of MATH which have one endpoint at MATH have other endpoints at MATH, and MATH. By construction MATH lies on MATH. Note that MATH lies on the same horocycle based at MATH as MATH and MATH. Therefore, the geodesic spanned by MATH and MATH intersects MATH above MATH. Note that MATH lies on the same horocycle based at MATH as MATH and MATH. Therefore the geodesic spanned by MATH and MATH intersects MATH below MATH. Hence, MATH lies in the tetrahedron spanned by the four other endpoints. By REF , the vertex MATH satisfies NLSC with respect to MATH for MATH. Thus, MATH is a continuous family of simplicial hyperbolic surfaces. The remainder of the proof is similar to that of REF . |
math/9907052 | We first give an outline of the proof. We will use induction on the number of accidental parabolics in the weakly type-preserving isomorphism. This will divide the proof into three cases, each of which will be proved by dividing the convex core into a compact core and its complement. Using the induction hypothesis and continuous families of simplicial hyperbolic surfaces, points in both the compact core of the convex core and the complement will have bounded injectivity radius. This completes the outline of the proof. Recall that CITE has shown that for a finitely generated Kleinian group, there can be at most a finite number of conjugacy classes of maximal abelian subgroups associated to accidental parabolics. We will use induction on the number of accidental parabolics. The base step in the induction is the case where there are no accidental parabolics. In this case, MATH is isomorphic to MATH via a type-preserving isomorphism. Then by REF , we know that there exists a constant MATH such that the injectivity radius for points in the convex core is bounded above by MATH. Thus, we are done. Now let us deal with the induction step. Suppose that the induction hypothesis is true for a weakly type-preserving isomorphism with fewer than MATH accidental parabolics. Let MATH be a Kleinian group that is weakly type-preserving isomorphic to a Fuchsian group MATH with fewer than MATH accidental parabolics. Let MATH and MATH. Then by the induction hypothesis, for MATH, MATH. Formally, we are assuming the existence of an upper bound MATH on injectivity radius for points in MATH where the bound depends on the number MATH of accidental parabolics in the isomorphism between MATH and MATH, and we will use this bound to establish an upper bound MATH on injectivity radius for points in MATH where MATH and MATH are weakly type-preserving isomorphic with MATH accidental parabolics. Here, MATH, the bound obtained from REF . For ease of exposition, we will let MATH denote the upper bound on injectivity radius for points in MATH for all weakly type-preserving isomorphisms between MATH and MATH with less than MATH accidental parabolics, and we will use this bound to establish an upper bound MATH on injectivity radius for points in MATH, where MATH and MATH are weakly type-preserving isomorphic with MATH accidental parabolics. Let MATH. Let MATH be a relative compact core of MATH with associated parabolic locus MATH, and let MATH be a compact core of MATH. Then because MATH is isomorphic to MATH, MATH is homeomorphic to MATH. (REF , CITE) Let MATH be a homeomorphism where MATH is in the homotopy class induced by the weakly type-preserving isomorphism and MATH is a collection of components of MATH. We can extend the homeomorphism MATH to a homeomorphism MATH on MATH as follows: Let MATH be a proper embedding such that MATH for MATH, and let MATH where MATH is the parabolic extension of MATH. Recall that every parabolic element of MATH can be associated to a closed curve in the parabolic locus MATH of MATH. In particular, every accidental parabolic can be associated to a closed curve in a component of MATH. Let the accidental parabolics that are associated to a curve on MATH be called accidental parabolics associated to MATH; let the others be called accidental parabolics associated to MATH. Let us first assume that there are accidental parabolics associated to both MATH and MATH; we will discuss the proof in the case where accidental parabolics are associated to only one of MATH and MATH later. CASE: Accidental NAME Associated to MATH and MATH. Let us give an outline of the proof in this case. We will divide the convex core into a compact core and its complement. Then we will cover the convex core with continuous families of simplicial hyperbolic surfaces and projections of convex cores of covers with fewer accidental parabolics (which, by the induction hypothesis, have bounded injectivity radius), so that points in the compact core will have bounded injectivity radius. The remaining portions of the convex core will either lie in a MATH-thin part, or will lie in the projection of the convex core of a cover with fewer accidental parabolics. This completes the outline of the proof in this case. The first step of the proof in REF will be to find a continuous family of simplicial hyperbolic surfaces which shrinks an accidental parabolic associated to MATH and an accidental parabolic associated to MATH. We will want points in the image of this continuous family to lie in the convex core and to have bounded injectivity radius. Consider a triangulation MATH containing an edge which represents an accidental parabolic associated to MATH. Either there exists an edge in MATH which represents an accidental parabolic associated to MATH, or not. In the latter case, we can perform the modified MATH sequence of elementary moves altering MATH to a triangulation MATH containing an edge representing an accidental parabolic associated to MATH. Let MATH be the ``last" triangulation in the sequence in which there is an edge MATH representing an accidental parabolic associated to MATH. Let MATH be the ``first" triangulation in the sequence in which there is an edge MATH representing an accidental parabolic associated to MATH. There are two possibilities: CASE: MATH and MATH are adjacent, or REF they are separated by triangulations whose edges are not accidental parabolics associated to either MATH or MATH. In either of these cases, we say that the sequence of triangulations has MATH, that is, there exist triangulations MATH and MATH in our sequence such that: CASE: the edges of MATH may only represent accidental parabolics associated to MATH, CASE: the edges of MATH may only represent accidental parabolics associated to MATH, and CASE: the edges of each intermediary triangulation do not represent accidental parabolics associated to either MATH or MATH. Note that when MATH and MATH are adjacent, there are no intermediary triangulations. Thus, we can see that either a sequence of triangulations has MATH, or there exists a triangulation with edges which represent accidental parabolics associated to both MATH and MATH. Case IA. There Exists a Sequence of Triangulations with MATH. Suppose our sequence of triangulations has MATH. Now there are two cases as well: either MATH and MATH are separated by at least one intermediary triangulation, or not. Case IA REF . MATH and MATH . Separated by at Least one Intermediary Triangulation. In this case, there exist triangulations MATH and MATH such that exactly one edge MATH of MATH is an accidental parabolic associated to MATH, and exactly one edge MATH of MATH is an accidental parabolic associated to MATH, such that the edges of each intermediary triangulation are not accidental parabolics associated to either MATH or MATH. The fact that there is exactly one edge in each triangulation associated to an accidental parabolic follows directly from the fact that an elementary move alters exactly one edge in the triangulation. If MATH is a REF-punctured sphere, then there is exactly one nonperipheral edge in its suitable, one-vertex triangulation. Thus, in MATH, this nonperipheral edge is MATH; similarly, in MATH, this nonperipheral edge is MATH. Let MATH and MATH be any two edges in MATH joining distinct ideal vertices to the sole internal vertex. We can use REF to construct a MATH-injective practical simplicial hyperbolic surface MATH with marked edges MATH and MATH such that MATH is properly homotopic to MATH. Let MATH and MATH be any two edges in MATH joining distinct ideal vertices to the sole internal vertex. Similarly, construct a MATH-injective practical simplicial hyperbolic surface MATH with marked edges MATH and MATH such that MATH is properly homotopic to MATH. By repeatedly using REF , we can construct the MATH interpolation MATH such that MATH is the map MATH and MATH is the map MATH. (Note that we have henceforth abandoned the notation MATH from REF.) Here, we are using REF to perform the elementary moves in the modified MATH sequence of triangulations which do not involve the existing marked edges, and we are using REF to change the marked edges when necessary. Now we can use REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. Similarly, we can apply REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. If MATH is not a REF-punctured sphere, then there exists a nonperipheral edge MATH in MATH which does not represent an accidental parabolic, and hence can be mapped to a closed geodesic. Similarly, there exists a nonperipheral edge MATH in MATH which does not represent an accidental parabolic, and hence can be mapped to a closed geodesic. Using REF , there exists a MATH-injective useful simplicial hyperbolic surface MATH with distinguished edge MATH such that MATH is properly homotopic to MATH. Similarly, there exists a MATH-injective useful simplicial hyperbolic surface MATH with distinguished edge MATH such that MATH is properly homotopic to MATH. By repeatedly using REF , we can construct a continuous family of simplicial hyperbolic surfaces that performs the elementary moves indicated in the modified MATH sequence of triangulations that interpolates between MATH and MATH. We can also use REF to construct a continuous family that alters the distinguished edge from MATH to MATH as needed in the modified MATH sequence. By concatenating all of these continuous families, we construct the MATH interpolation MATH such that MATH is the map MATH and MATH is the map MATH. Now we can use REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. Similarly, we can apply REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. Case IA REF . MATH and MATH . Not Separated by at Least one Intermediary Triangulation. In this case, we can conclude that MATH and MATH are adjacent triangulations, that is, the elementary move alters the edge MATH to the edge MATH. If MATH is a REF-punctured sphere, then because its suitable, one-vertex triangulation contains exactly one nonperipheral edge, we know that in MATH, MATH is that nonperipheral edge, and that in MATH, MATH is that nonperipheral edge. Let MATH and MATH be any two edges in MATH joining distinct ideal vertices to the sole internal vertex. We can use REF to construct a MATH-injective practical simplicial hyperbolic surface MATH with marked edges MATH and MATH such that MATH is properly homotopic to MATH. Let MATH and MATH be any two edges in MATH joining distinct ideal vertices to the sole internal vertex. Similarly, construct a MATH-injective practical simplicial hyperbolic surface MATH with marked edges MATH and MATH such that MATH is properly homotopic to MATH. We can use REF to perform the elementary move that separates the triangulations MATH and MATH. We can also use REF to construct a continuous family that alters the marked edges from MATH and MATH to MATH and MATH. After concatenating all of these continuous families, we construct the MATH interpolation MATH such that MATH is the map MATH and MATH is the map MATH. Now we can apply REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. If MATH is not a REF-punctured sphere, then there exist edges MATH of MATH and MATH of MATH, each of which can be mapped to a closed geodesic. Therefore, we can use REF to construct a MATH-injective useful simplicial hyperbolic surface MATH with distinguished edge MATH such that MATH is properly homotopic to MATH. Similarly, construct a MATH-injective useful simplicial hyperbolic surface MATH with distinguished edge MATH such that MATH is properly homotopic to MATH. We can use REF to perform the elementary move that separates the triangulations MATH and MATH. We can also use REF to construct a continuous family that alters the distinguished edge from MATH to MATH. After concatentating all of these continuous families, we construct the MATH interpolation MATH such that MATH is the map MATH and MATH is the map MATH. Now we can use REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. Similarly, we can apply REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. Case IB. There NAME Not Exist a Sequence of Triangulations with MATH. If we cannot construct a sequence of triangulations with MATH, then as we argued before, there exists a triangulation that contains edges that are accidental parabolics associated to both MATH and MATH. Using REF we can construct a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH as MATH, and that shrinks the accidental parabolic MATH as MATH. Let MATH (or MATH in Case IB) be the union of the images of the continuous families of simplicial hyperbolic surfaces. REF , and REF guarantee that MATH. Furthermore, we also know that for any MATH lies in the image of a simplicial hyperbolic surface, and hence, by REF , MATH where the constant MATH depends on the NAME characteristic of MATH. Now we have bounded the injectivity radius in a portion of the convex core. Because MATH is isomorphic to a surface group which contains no MATH subgroups, MATH possesses only rank one cusps. There exists a retraction map, MATH, such that MATH is the identity on MATH, and MATH retracts each component of MATH onto MATH. More explicitly, MATH can be defined as follows: Let MATH denote a component of MATH. Each of the MATH can be parametrized as MATH using the coordinates MATH. Then MATH is the map MATH. Note that MATH, because MATH is convex. Now we will show that MATH is compact. NAME REF has shown that for a simplicial hyperbolic surface MATH whose vertices are MATH, MATH where MATH denotes the MATH-coordinate of MATH. Therefore, to show that MATH is compact, it suffices to show that MATH has compact closure in MATH. Because MATH has compact closure in MATH for MATH and MATH is compact, we know that the image MATH in MATH is compact. Thus, we need only consider the images MATH, MATH, and MATH. By REF , and REF, we know that MATH, MATH, and MATH have compact closure in MATH, and hence MATH is compact. The next step in the proof of REF involves finding a compact core MATH of MATH, where MATH represents the collection of rank one cusps of MATH associated to MATH, such that points in MATH have bounded injectivity radius. Let MATH denote the map MATH. Let MATH be the components of MATH, and let MATH be the components of MATH. Let MATH and MATH be covers of MATH with covering maps MATH and MATH respectively. Let MATH and MATH be the rank one cusps of MATH associated to MATH and MATH respectively. Let MATH. Given MATH. By REF , there exists a relative compact core MATH of MATH such that MATH and such that MATH. Let MATH be the components of MATH such that each MATH is adjacent to MATH, and let MATH be the components of MATH such that each MATH is adjacent to MATH. Let MATH be the components of MATH such that each MATH is adjacent to MATH, and let MATH be the components of MATH such that each MATH is adjacent to MATH. Then by construction, MATH, and MATH. In the following lemma, we will choose a compact core MATH of MATH such that MATH contains MATH, MATH lies in MATH, and points in MATH have bounded injectivity radius. REF may help to keep track of all of the different surfaces used in the proof. For all MATH, there exists a compact core MATH of MATH bounded by surfaces MATH and MATH such that the following are true: CASE: MATH, CASE: if MATH, then MATH CASE: for each MATH, MATH, and for each MATH, MATH, CASE: for each MATH, there exist MATH such that MATH is a homeomorphism, and MATH, and CASE: for each MATH, there exist MATH such that MATH is a homeomorphism, and MATH. Let us begin with an outline of the proof. We will first choose MATH and MATH to be incompressible surfaces in MATH such that subsurfaces of MATH and MATH lift to MATH and MATH, and MATH and MATH bound a submanifold of MATH. For each MATH, we will construct an embedded surface MATH that is properly homotopic to MATH via a homotopy MATH whose image lies in MATH such that the MATH bound a submanifold that contains MATH and MATH. For each MATH, let MATH. The MATH bound a compact core MATH of MATH such that MATH contains MATH. We will show that MATH is contained in the union of MATH and the images of the homotopies MATH, so that points in MATH will have bounded injectivity radius. Because MATH is not an accidental parabolic in MATH, MATH has fewer accidental parabolics than MATH. Therefore, by the induction hypothesis, for every MATH, MATH where MATH depends on the surface MATH. Similarly, for every MATH, MATH, where MATH depends on MATH. Let MATH be a subsurface of MATH obtained by cutting MATH along MATH. Then MATH is a surface with boundary containing of one or two copies of MATH. Let MATH be the analogous subsurface of MATH, obtained by cutting MATH along MATH, with boundary consisting of one or two copies of MATH. By REF (or REF if MATH is a REF-punctured sphere), there exists MATH such that for every MATH and MATH, MATH lifts to MATH and MATH lifts to MATH. We also require that MATH and MATH. Let MATH, and let MATH. Now we will construct the surfaces MATH and MATH. Let MATH be the parabolic locus associated to MATH. Because the fundamental group of MATH is isomorphic to a surface group, we know that MATH is homeomorphic to MATH. (REF , CITE) Because MATH is a disjoint collection of incompressible annuli in MATH such that MATH, we can conclude that the components of MATH are incompressible in MATH and hence also in MATH. Note that MATH is also a relative compact core for MATH where the components of MATH are incompressible in MATH. Then we can apply REF to show that each MATH possesses a product structure. By the Lifting Theorem, the inclusion map of the product structure into MATH lifts to an inclusion map of the product structure into MATH. Let MATH be the image of the lifted inclusion map in MATH. Then MATH has a product structure, and MATH is a homeomorphism. Note that MATH is also a relative compact core for MATH where the components of MATH are incompressible in MATH. Then we can apply the argument of REF to MATH to show that each MATH possesses a product structure. By the Lifting Theorem, the inclusion map of the product structure into MATH lifts to an inclusion map of the product structure into MATH. Let MATH be the image of the lifted inclusion map in MATH. Then MATH has a product structure, and MATH is a homeomorphism. Let MATH be the rank one cusp of MATH associated to MATH where MATH. Similarly, let MATH be the rank one cusp REF of MATH associated to the ideal vertices of MATH where MATH. We will let MATH. Now we will show that for each MATH, MATH. Suppose not. Then there exists MATH and a component MATH of MATH such that MATH separates MATH from MATH. Consider a geodesic ray MATH in MATH that is perpendicular to MATH and passes through MATH. Let MATH be the portion of MATH beginning at MATH. Note that by REF , the injectivity radius strictly increases out a geometrically finite end, so the ray MATH is entirely contained in MATH. Let MATH, and let MATH be the lift of MATH to MATH. Without loss of generality, let us assume that MATH intersects MATH, transversely. Then one of the three following cases occurs: CASE: MATH intersects MATH an odd number of times, REF MATH intersects a different boundary component of the closure of MATH, or REF all but a compact portion of MATH is completely contained in MATH. Suppose MATH intersects MATH an odd number of times. Because MATH is a properly embedded surface separating MATH, MATH has two sides. Let the component of MATH containing MATH be the positive side of MATH. Then the last time MATH intersects MATH, MATH passes from the positive to the negative side of MATH. Let MATH. Downstairs, we can also let the side of MATH containing MATH be the positive side of MATH. Recall that because MATH, and MATH is a component of MATH, then we know that MATH lies on the negative side of MATH. Because MATH is a homeomorphism and MATH lifts to MATH, we can conclude that MATH lies to the negative side of MATH. Note that MATH is an incompressible separating surface of MATH, and that MATH is homologous to MATH. Also note that MATH leaves every compact set of MATH and all but a compact portion of MATH lies in MATH. Therefore, if MATH intersects MATH, then MATH must also intersect MATH. But MATH and MATH. So this is a contradiction. Suppose MATH intersects a different boundary component of the closure of MATH. Then MATH intersects some component MATH of MATH such that MATH is a boundary component of the closure of MATH in MATH. By REF , MATH lifts to a component MATH of MATH. Because MATH intersects MATH, there is a portion of MATH that contains MATH and joins two components of MATH. Because MATH is convex, MATH must lie in MATH, which is a contradiction. Then all but a compact portion of MATH is contained in MATH. Using the coordinates MATH on each component of MATH adjacent to MATH, we can extend the homeomorphism MATH to a homeomorphism MATH where MATH and MATH are the parabolic extensions of MATH and MATH respectively. Let MATH be the closure of the component of MATH that contains MATH. Let MATH. Let MATH be the closure of MATH in MATH. Then because MATH is compact, and MATH for every MATH, we can guarantee that MATH for large enough MATH. Then for large enough MATH, MATH is a component of MATH which embeds in MATH. Then by REF , MATH is the component of MATH with boundary MATH. Because MATH, MATH. Then MATH. But by hypothesis, MATH so this is a contradiction. Thus, for each MATH, MATH. For each MATH, choose MATH to be a level surface in the product structure of MATH such that the compact surface MATH lies to one side of MATH. Let MATH be embedded annuli in MATH with boundary components MATH and MATH. Also, for each component of MATH, let MATH be infinite annuli in MATH. Note that because MATH and MATH is convex, MATH. Let MATH. By construction, the surface MATH is embedded in MATH and properly homotopic in MATH to MATH. For each MATH, we can define MATH to be a proper homotopy between MATH and MATH so that MATH and MATH. Let MATH be a homotopy constructed from MATH such that for MATH, MATH is a geodesic arc with the same endpoints and in the same homotopy class as MATH. Then MATH is a ruled homotopy between MATH and MATH. Because MATH is convex, MATH. By the induction hypothesis, for MATH, MATH. For each MATH, let MATH so that MATH is a level surface in the product structure of MATH, let MATH, let MATH, and let MATH. Then MATH. For each MATH, we can let MATH. This map will be a proper homotopy between MATH and MATH, where MATH. By REF , for MATH, MATH. Now let MATH glued along MATH. We can construct the analogous surface MATH by repeating this process so that each MATH is the union of embedded annuli MATH in MATH, infinite annuli MATH in MATH, and a level surface MATH in the product structure of MATH such that MATH lies to one side of MATH. For MATH, the surfaces MATH are embedded and properly homotopic to MATH by construction. Construct a proper homotopy MATH between MATH and MATH such that MATH for all MATH. By construction, the image of MATH lies in MATH. Thus for MATH, MATH. Construct the analogous proper homotopy MATH between MATH and MATH such that MATH for all MATH. Let MATH . Then by construction, MATH. Because MATH, we can conclude that MATH. In particular, MATH. Also, by construction, MATH. The analogous conclusions hold for MATH and MATH for each MATH. By construction MATH and MATH are properly homotopic in MATH to the components of MATH, where MATH denotes the components of MATH associated to MATH. So MATH, MATH and the boundary components of MATH span a product structure in MATH. (REF , CITE) Hence, MATH and MATH are boundary components of a new set MATH which is a compact core of MATH. Now we will show that MATH . Because MATH and MATH are embedded, incompressible surfaces that are properly homotopic in MATH, we can conclude that MATH and MATH are embedded, incompressible surfaces that are properly homotopic in MATH. So MATH and MATH span a product structure in MATH. (REF , CITE) Then there exists a proper isotopy MATH between MATH and MATH such that MATH is an embedding in MATH. Using the inverse map MATH, we can extend MATH to a proper isotopy MATH between MATH and MATH such that MATH is also an embedding. So MATH is homeomorphic to MATH. Using standard degree arguments, MATH is contained in the image of any proper homotopy between MATH and MATH. (see REF , CITE) In particular, MATH. So for MATH, MATH. Note that in Case IB, we must use REF to find MATH, and replace the maps MATH and MATH for MATH with MATH and MATH, keeping the domains the same. This completes the proof of REF . Now we have bounded the injectivity radius for points in a compact core MATH of MATH which contains MATH. Consider again the surface constructed in the previous lemma MATH which was a MATH-injective, separating, level surface in the product structure of MATH. Let MATH be the closure of the component of MATH such that MATH. Let MATH be the subset of MATH such that MATH is a homeomorphism. Then because MATH and MATH, by REF , we know that MATH. The analogous conclusions hold for MATH and MATH, the closure of the component of MATH such that MATH. If MATH, then MATH for some MATH, or MATH for some MATH. Because MATH, we know that if MATH, then MATH. So by the induction hypothesis and REF , MATH. Similarly, if MATH, then MATH. Finally, if MATH, then MATH. Therefore, for MATH, MATH. We can do this for all MATH, so that for MATH, MATH. This completes the proof of Case I. CASE: Accidental NAME Associated Only to MATH. Now let us assume that there are accidental parabolics associated only to MATH. Then MATH is empty so that the end MATH of MATH associated to MATH is either simply degenerate or geometrically finite. Case IIA. End MATH . Is Simply Degenerate. We begin with a sketch of the proof of Case IIA. Again we will divide the convex core into a compact core and its complement. Using a continuous family of simplicial hyperbolic surfaces and projections of convex cores of covers with fewer accidental parabolics, points in the compact core will have bounded injectivity radius. The remaining portions will lie in a MATH-thin part, lie in the projection of a cover which has fewer accidental parabolics, or lie in the image of a simplicial hyperbolic surface. This completes the sketch of the proof in this case. The first step in the proof of Case IIA will be to find a continuous family of simplicial hyperbolic surfaces which shrinks an accidental parabolic associated to MATH. We will want points in the image of this continuous family to lie in the convex core and to have bounded injectivity radius. In REF, NAME has shown that if MATH is a simply degenerate end of MATH, then there exists a sequence of useful simplicial hyperbolic surfaces which are properly homotopic to MATH and which exit every compact set in MATH. Recall that MATH, a relative compact core of MATH with associated parabolic locus MATH, is homeomorphic to MATH, where MATH is a compact core of MATH. Therefore, the sequence of useful simplicial hyperbolic surfaces have the form MATH. Let MATH be a useful simplicial hyperbolic surface in MATH with distinguished edge MATH. In the triangulation MATH, either there exists an edge which represents an accidental parabolic associated to MATH or not. In the latter case, let MATH be a triangulation of MATH such that there exists an edge in MATH which is an accidental parabolic associated to MATH. We can perform the modified MATH sequence of elementary moves altering MATH to MATH. Let MATH be the ``first" triangulation in the sequence in which there is an edge MATH representing an accidental parabolic associated to MATH. There are two cases: CASE: MATH and MATH are adjacent, or REF they are separated by triangulations whose edges are not accidental parabolics associated to MATH. In either of these cases, we say that the sequence of triangulations has MATH, that is, there exist triangulations MATH and MATH in our sequence such that: CASE: the edges of MATH do not represent accidental parabolics, CASE: the edges of MATH may only represent accidental parabolics associated to MATH, and CASE: the edges of each intermediary triangulation do not represent accidental parabolics associated to MATH. Note that when MATH and MATH are adjacent, there are no intermediary triangulations. Thus, we can see that either a sequence of triangulations has MATH, or there exists an edge in the triangulation MATH which represents an accidental parabolic associated to MATH. Case IIA REF . There Exists a Sequence of Triangulations with MATH. Suppose our sequence of triangulations has MATH. Now there are two cases: either MATH and MATH are separated by at least one intermediary triangulation, or not. Case IIA REF a. MATH and MATH . Separated by at Least one Intermediary Triangulation. In this case, there exist triangulations MATH and MATH such that no edge of MATH is an accidental parabolic, and exactly one edge MATH of MATH is an accidental parabolic associated to MATH, such that the edges of each intermediary triangulation are not accidental parabolics associated to MATH. The fact that there is exactly one edge of MATH associated to an accidental parabolic follows directly from the fact that an elementary move alters exactly one edge in the triangulation. If MATH is a REF-punctured sphere, then there is exactly one nonperipheral edge in its suitable, one-vertex triangulation. Thus, in MATH, this nonperipheral edge is MATH; in MATH, this nonperipheral edge is MATH. Let MATH and MATH be any two edges in MATH joining distinct ideal vertices to the sole internal vertex. We can use REF to construct a MATH-injective practical simplicial hyperbolic surface MATH with marked edges MATH and MATH such that MATH is properly homotopic to MATH. Let MATH and MATH be any two edges in MATH joining distinct ideal vertices to the sole internal vertex. Similarly, we can construct a MATH-injective practical simplicial hyperbolic surface MATH with marked edges MATH and MATH such that MATH is properly homotopic to MATH. First we can use REF to construct a continuous family of simplicial hyperbolic surfaces which interpolates between the useful simplicial hyperbolic surface MATH with distinguished edge MATH and the practical simplicial hyperbolic surface MATH with marked edges MATH and MATH. Then by repeatedly using REF and concatenating, we can construct the MATH interpolation MATH such that MATH is the map MATH and MATH is the map MATH. Now we can use REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. If MATH is not a REF-punctured sphere, then there exists a nonperipheral edge MATH in MATH which does not represent an accidental parabolic, and hence can be mapped to a closed geodesic. Then by REF , we can construct a useful simplicial hyperbolic surface MATH with distinguished edge MATH. By repeatedly using REF , we can construct a continuous family of simplicial hyperbolic surfaces that performs the elementary moves indicated in the modified MATH sequence of triangulations that interpolates between MATH and MATH. We can also use REF to construct a continuous family that alters the distinguished edge from MATH to MATH as needed in the modified MATH sequence. By concatenating all of these continuous families, we have the MATH interpolation MATH such that MATH is the map MATH and MATH is the map MATH. Now we can use REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. Case IIA REF b. MATH and MATH . Not Separated by at Least one Intermediary Triangulation. In this case, we can conclude that MATH and MATH are adjacent triangulations, that is, the elementary move alters an edge of MATH to the edge MATH. If MATH is a REF-punctured sphere, then because its suitable, one-vertex triangulation contains exactly one nonperipheral edge, we know that in MATH, MATH is that nonperipheral edge, and in MATH, MATH is that nonperipheral edge. Let MATH and MATH be any two edges in MATH joining distinct ideal vertices to the sole internal vertex. We can use REF to construct a MATH-injective practical simplicial hyperbolic surface MATH with marked edges MATH and MATH such that MATH is properly homotopic to MATH. Let MATH and MATH be any two edges in MATH joining distinct ideal vertices to the sole internal vertex. Similarly, construct a MATH-injective practical simplicial hyperbolic surface MATH with marked edges MATH and MATH such that MATH is properly homotopic to MATH. We can use REF to construct a continuous family of simplicial hyperbolic surfaces which interpolates between the useful simplicial hyperbolic surface MATH with distinguished edge MATH and the practical simplicial hyperbolic surface MATH with marked edges MATH and MATH. By using REF , we can construct a continuous family of simplicial hyperbolic surfaces that performs the elementary move that separates the triangulations MATH and MATH. We can also use REF to construct a continuous family that alters the marked edges from MATH and MATH to MATH and MATH. After concatenating all of these continuous families, we construct the MATH interpolation MATH such that MATH is the map MATH and MATH is the map MATH. Now we can use REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. If MATH is not a REF-punctured sphere, then there exists a nonperipheral edge MATH in MATH which is unaltered by the sole elementary move and hence can be mapped to a closed geodesic. Therefore, we can use REF to construct a MATH-injective useful simplicial hyperbolic surface MATH with distinguished edge MATH such that MATH is properly homotopic to MATH. We can use REF to perform the elementary move that separates the triangulations MATH and MATH. We can also use REF to alter the distinguished edge from MATH to MATH. After concatenating all of these continuous families, we can construct the MATH interpolation MATH such that MATH is the map MATH and MATH is the map MATH. Now we can use REF , setting MATH, to obtain a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. Case IIA REF . There NAME Not Exist A Sequence of Triangulations with MATH. If we cannot construct a sequence of triangulations with MATH, then as we argued before, there exists an edge MATH in the triangulation MATH that represents an accidental parabolic associated to MATH. Using REF we can construct a continuous family of simplicial hyperbolic surfaces MATH that shrinks the accidental parabolic MATH. Let MATH (or MATH in Case IIA REF ) be the union of the images of the continuous families of simplicial hyperbolic surfaces. REF , and REF guarantee that MATH. Furthermore, we also know that for any MATH lies in the image of a simplicial hyperbolic surface, and hence, by REF , MATH where the constant MATH depends on the NAME characteristic of MATH. Now we have bounded the injectivity radius in a portion of the convex core. Note that MATH because MATH is convex, and that by a result of NAME (REF , CITE) and REF , MATH is compact. The next step in the proof of Case IIA involves finding a compact core MATH of MATH, where MATH represents the collection of rank one cusps of MATH associated to MATH, such that points in MATH have bounded injectivity radius. Let MATH denote the map MATH. Let MATH be the components of MATH. Let MATH be covers of MATH with covering maps MATH. Let MATH be the rank one cusp of MATH associated to MATH. Let MATH. Given MATH. By REF , there exists a relative compact core MATH of MATH such that MATH and such that MATH. Let MATH be the components of MATH such that each MATH is adjacent to MATH, and let MATH be the component of MATH which is a neighborhood of the simply degenerate end MATH. Let MATH be the components of MATH such that each MATH is adjacent to MATH. Then MATH. Let MATH be the component of MATH such that MATH. In the following lemma, we will choose a compact core MATH of MATH such that MATH contains MATH, MATH lies in MATH, and points in MATH have bounded injectivity radius. For all MATH, there exists a compact core MATH of MATH bounded by surfaces MATH and MATH such that the following are true: CASE: MATH, CASE: if MATH, then MATH CASE: for each MATH, MATH, and CASE: for each MATH, there exist MATH such that MATH is a homeomorphism, and MATH. The proof of this lemma mimics that of REF . We give an outline of the steps. Let MATH, where MATH lifts to MATH, and let MATH. By REF , we know that MATH, MATH, and each MATH and MATH possesses a product structure. Construct a surface MATH with the following properties: REF each MATH is the union of embedded annuli MATH in MATH, embedded infinite annuli MATH in MATH, and a level surface MATH in the product structure on MATH such that MATH lies to one side of MATH; and REF the surface MATH is properly homotopic to MATH via a proper homotopy MATH whose image lies in MATH. Then for a point MATH in the image of MATH, MATH. Let MATH be the union of a level surface MATH in the product structure of MATH such that MATH lies to one side, and infinite annuli MATH in MATH. Let MATH be a proper homotopy between MATH and MATH. Consider a point MATH that is in the image of MATH, but not in the image of MATH nor MATH. Note that MATH has a product structure, that MATH is disjoint from MATH, and that MATH is properly homotopic to MATH. Then by a result of NAME (REF , CITE) and standard degree arguments (see REF , NAME, CITE), there exists a useful simplicial hyperbolic surface MATH with image in MATH such that MATH lies in the image of any proper homotopy between MATH and MATH. Then, in particular, MATH lies in the image of the MATH interpolation MATH joining MATH and MATH. REF , and REF guarantee that MATH. Furthermore, by REF , if MATH, MATH. Therefore, for a point MATH that is contained in the image of MATH but that is not contained in the image of MATH nor MATH, MATH. Let MATH and MATH. We can show that MATH and that MATH. Furthermore, we can show that there exists MATH such that MATH is a homeomorphism, and MATH. Then we can show that MATH and MATH bound a compact core MATH of MATH such that MATH. By covering MATH with MATH and the images of MATH and MATH, we can show that for MATH, MATH. Note that in Case IB, we must replace the maps MATH and MATH with MATH and MATH, keeping the domains the same. This completes the sketch of the proof of REF . Now we have bounded the injectivity radius for points in a compact core MATH of MATH which contains MATH. Consider again the surface constructed in the previous lemma MATH which was a MATH-injective, separating, level surface in the product structure of MATH. Let MATH be the closure of the component of MATH such that MATH. Let MATH be the closure of the component of MATH such that MATH. Let MATH be the subset of MATH such that MATH is a homeomorphism. Then because MATH and MATH, by REF , we know that MATH. If MATH, then MATH for some MATH, or MATH. Because MATH, we know that if MATH, then MATH. So by the induction hypothesis and REF , MATH. Suppose MATH. Recall that MATH has a product structure, that MATH is disjoint from MATH, and that MATH is properly homotopic to MATH. Then by a result of NAME (REF , CITE) and standard degree arguments (see REF , NAME, CITE), there exists a useful simplicial hyperbolic surface MATH such that MATH lies in the image of any proper homotopy between MATH and MATH. Then, in particular, MATH lies in the image of the MATH interpolation MATH joining MATH and MATH. REF , and REF guarantee that MATH. Furthermore, by REF , if MATH, MATH. Finally, for MATH, MATH. Therefore for MATH, MATH. We can do this for all MATH, so that for MATH, MATH. This completes the proof of Subcase A. Case IIB. The End MATH is Geometrically Finite. The basic outline of the argument for Case IIB is still the same. We will divide the convex core into a compact core and its complement. However, to fill up the convex core this time, we will use an interpolation between a boundary component of the convex core and a simplicial hyperbolic surface, a continuous family of simplicial hyperbolic surfaces, and projections of convex cores of covers with fewer accidental parabolics. The remaining portions will, again, lie in a MATH-thin part, or lie in the projection of a cover which has fewer accidental parabolics. The first step in the proof of Case IIB will be to find a homotopy between the boundary of the convex core and a useful or practical simplicial hyperbolic surface, and a continuous family of simplicial hyperbolic surfaces which shrinks an accidental parabolic associated to MATH. We will want points in the image of the homotopy and continuous family to lie in the convex core and to have bounded injectivity radius. Let MATH be the pleated surface whose image is the boundary component of MATH associated to the geometrically finite end MATH. The following result of NAME guarantees the existence of a ``short" homotopy between a pleated surface and a simplicial hyperbolic surface. (REF , CITE) Given a pleated surface MATH and MATH, there exists a homotopy MATH such that MATH, the map MATH is a simplicial hyperbolic surface, and the trajectories MATH have lengths at most MATH. Furthermore, the triangulation MATH of MATH corresponding to MATH contains a closed curve that is mapped to a closed geodesic, and there is an upper bound MATH on the number of vertices in MATH, depending only on the NAME characteristic of MATH. Without loss of generality, by pre-composing with the nearest point retraction map, we can assume that MATH. As the image of a simplicial hyperbolic surface, by REF , for MATH, MATH. Then because the trajectories MATH have lengths at most MATH, for MATH, MATH. We can use REF to find a bounded length homotopy between the simplicial hyperbolic surface MATH obtained above and a useful simplicial hyperbolic surface MATH. (REF , CITE) Let MATH be a simplicial hyperbolic surface whose associated triangulation MATH has a closed curve MATH which is mapped to a closed geodesic and has at most MATH vertices. Then there is a homotopy MATH such that the following are true: CASE: MATH, CASE: MATH, CASE: MATH is a useful simplicial hyperbolic surface with distinguished edge MATH, and CASE: MATH is contained in the MATH-neighborhood of the image of MATH, for an independent constant MATH. Because MATH is a homotopy whose flow lines are contained in the MATH-neighborhood of the image of MATH, for MATH, MATH, where the constant MATH depends only on MATH. Now we have bounded the injectivity radius in a portion of MATH. The triangulation MATH on MATH from REF either contains an edge which represents an accidental parabolic associated to MATH or not. In the latter case, let MATH be a triangulation of MATH such that there exists an edge in MATH which is an accidental parabolic associated to MATH. We can perform the modified MATH sequence of elementary moves altering MATH to MATH. Let MATH be the ``first" triangulation in the sequence in which there is an edge MATH representing an accidental parabolic associated to MATH. There are two cases: the sequence of triangulations has MATH or not. If the sequence has MATH, then using the argument given in Case IIA REF , we can construct a continuous family of simplicial hyperbolic surfaces MATH and MATH such that MATH interpolates between the useful simplicial hyperbolic surface MATH and a useful or practical simplicial hyperbolic surface MATH, and such that MATH shrinks the accidental parabolic MATH. If the sequence does not have MATH, then using the argument given in Case IIA REF , we can construct a continuous family of simplicial hyperbolic surfaces MATH such that MATH shrinks the accidental parabolic MATH which is represented by an edge in MATH. Let MATH (or MATH if the sequence does not have MATH) be the union of the images of homotopies and continuous families of simplicial hyperbolic surfaces. REF , and REF guarantee that MATH. Furthermore, we also know that for any MATH that also lies in the image of a simplicial hyperbolic surface, by REF , MATH where the constant MATH depends on the NAME characteristic of MATH. To summarize, for MATH, MATH. Now we have bounded the injectivity radius in a portion of the convex core. Note that MATH because MATH is convex, and that by a result of NAME (REF , CITE) and REF , MATH is compact. The next step in the proof of Subcase B involves finding a compact core MATH of MATH, where MATH represents the collection of rank one cusps of MATH associated to MATH, such that points in MATH have bounded injectivity radius. Let MATH denote the map MATH. Let MATH be the components of MATH. Let MATH be covers of MATH with covering maps MATH. Let MATH be the rank one cusp of MATH associated to MATH. Let MATH. Given MATH. By REF , there exists a relative compact core MATH of MATH such that MATH and such that MATH. Let MATH be the components of MATH such that each MATH is adjacent to MATH, and let MATH be the component of MATH which is a neighborhood of the geometrically finite end MATH. Let MATH be the components of MATH such that each MATH is adjacent to MATH. Then MATH. Let MATH be the component of MATH such that MATH. In the following lemma, we will choose a compact core MATH of MATH such that MATH contains MATH, MATH lies in MATH, and points in MATH have bounded injectivity radius. For all MATH, there exists a compact core MATH of MATH bounded by surfaces MATH and MATH such that the following are true: CASE: MATH, CASE: if MATH, then MATH, CASE: for each MATH, MATH, and CASE: for each MATH, there exist MATH such that MATH is a homeomorphism, and MATH. The proof of this lemma mimics that of REF . We give an outline of the steps. Let MATH, where MATH lifts to MATH, and let MATH. By REF , we know that MATH, MATH, and each MATH and MATH possesses a product structure. Construct a surface MATH with the following properties: REF each MATH is the union of embedded annuli MATH in MATH, embedded infinite annuli MATH in MATH, and a level surface MATH in the product structure on MATH such that MATH lies to one side of MATH; and REF the surface MATH is properly homotopic to MATH via a proper homotopy MATH whose image lies in MATH. Then for a point MATH in the image of MATH, MATH. Let MATH and MATH. We can show that MATH and that MATH. Furthermore, we can show that there exists MATH such that MATH is a homeomorphism, and MATH, Then we can show that MATH and MATH bound a compact core MATH of MATH such that MATH. By covering MATH with MATH and the image of MATH, we can show that for MATH, MATH. Note that if the sequence does not have MATH, then we must replace the maps MATH and MATH with MATH and MATH, keeping the domains the same. This completes the sketch of the proof of REF . Now we have bounded the injectivity radius for points in a compact core MATH of MATH which contains MATH. Consider again the surface constructed in the previous lemma MATH which was a MATH-injective, separating, level surface in the product structure of MATH. Let MATH be the closure of the component of MATH such that MATH. Let MATH be the closure of the component of MATH such that MATH. Let MATH be the subset of MATH such that MATH is a homeomorphism. Then because MATH and MATH, by REF , we know that MATH. Recall that MATH is the component of MATH associated to the geometrically finite end MATH. Note that we chose MATH, and that MATH is a component of MATH not adjacent to MATH. So MATH. Suppose MATH. Then MATH for some MATH. Because MATH, we know that if MATH, then MATH. So by the induction hypothesis and REF , MATH. If MATH, then MATH. Therefore for MATH, MATH. We can do this for all MATH, so that for MATH, MATH. This completes the proof of Subcase B. Thus, the proof of REF is complete. |
math/9907052 | If MATH is orientable, then REF proves the existence of MATH. If MATH is non-orientable, then let MATH be an orientable double cover of MATH, and let MATH be a double cover of MATH associated to MATH. Using REF , there exists a constant MATH such that for MATH, MATH. Let MATH, and let MATH. Then since MATH is finite, we know that the two limit sets MATH and MATH are equal. (II.K. REF) Thus, MATH is a double cover of MATH. Then, by REF , for MATH, MATH. |
math/9907054 | Pinch in CITE has proved that any MATH may be written in the form MATH for some non negative integer MATH. Therefore MATH. If MATH is equal to REF, then MATH is divisible by MATH, otherwise MATH divides MATH. |
math/9907054 | Note that the roles of MATH and MATH are symmetric. Thus it suffices to show that MATH divides MATH. If MATH is a divisor of unity then the assertion is true. Assume the opposite, that is, MATH is not a divisor of unity and MATH does not divide MATH. Choose MATH. Since MATH is a divisor of unity, the scaling factor MATH does not divide neither MATH, nor MATH. According to REF this implies MATH and MATH. Therefore MATH divides MATH, thus a contradiction. |
math/9907054 | From the relation REF it is clear that all elements of MATH are polynomials in MATH. Since MATH, it satisfies a quadratic equation with integer coefficients, namely MATH, that is, MATH. Using this quadratic equation any polynomial MATH from REF may be reduced to the form MATH. Therefore clearly MATH. The condition MATH implies that we need MATH. This can be satisfied only if MATH for some integer MATH. The restriction MATH gives only two admissible values for the parameter MATH. CASE: Let us consider MATH, that is, MATH. Here MATH or MATH. REF states that both MATH and MATH divide REF in MATH. Since MATH, MATH is a divisor of unity and hence divides REF automatically. Let us assume that MATH divides REF. It means that there exists an element MATH, such that MATH . It follows that MATH and MATH. Since MATH is a positive integer, this may happen only for MATH and MATH. These are the two cases given in the statement. CASE: Let us now consider MATH, that is, MATH. Here MATH or MATH. Again from REF , MATH divides REF. It means that there exists an element MATH, such that MATH . This implies MATH and MATH. For the equation MATH we consider MATH, so that the only solution is MATH. |
math/9907054 | Consider the operations MATH with MATH. For simplicity, denote MATH . The statement of the lemma is equivalent to the fact that any point MATH in the interval MATH can be generated from MATH, and MATH using operations MATH, MATH. We first show that MATH can be generated using all operations MATH, with MATH. In the second step we find expression for MATH, MATH, as a composition of operations MATH, MATH. CASE: Consider MATH to be the solution of the equation MATH, MATH. According to REF, any MATH has a finite MATH-expansion MATH . By induction on MATH we show that MATH belongs to MATH. Clearly, for the first step of induction we have MATH for any MATH. Suppose that the MATH-expansion of a point MATH is of the form MATH . Then MATH can be rewritten as the combination MATH . By induction hypothesis, MATH could be generated from MATH and MATH using the given operations; that is, MATH. Since MATH one has the following relation, MATH . Therefore MATH . The later inclusion is valid due to the first step of induction (see REF). Assume that the coefficient MATH of MATH in its MATH-expansion is equal to MATH. Necessarily, from the properties of MATH-expansions, MATH is equal to REF. In this case we use MATH . Thus we have shown that all points in MATH can be generated starting from REF, and REF, using the operations MATH, MATH. The following relation can be used to reduce the number of necessary operations from MATH to MATH. The relation is valid for any MATH, MATH, independently of MATH. MATH . CASE: Let now MATH be the root of MATH, MATH. In this case, if we want to use the MATH-expansions of numbers, we have to encounter a complication: There exist points in the ring MATH which do not have a finite MATH-expansion. Recall that a MATH has a finite MATH-expansion if and only if MATH, (see REF). For us, only MATH are of interest. The following statement is valid. Let MATH. Then either MATH or MATH has a finite MATH-expansion. Indeed, since MATH is an integer and MATH, necessarily MATH. Therefore MATH if and only if MATH. Now we can proceed, similarly as for REF by induction on the length of the MATH-expansion in order to show that any element of MATH can be generated by corresponding operations MATH, MATH. For elements MATH, we have MATH. Therefore MATH can be generated from REF using the given operations. In the corresponding combination we replace all REF by REF and vice versa, which gives the desired combination for element MATH. The coefficients in a MATH-expansion take one of the values MATH. For a MATH, whose MATH, MATH, the procedure is the same as in REF . Consider a MATH . Since the segment MATH of a MATH-expansion is strictly lexicographically smaller than the sequence MATH, the string MATH is strictly smaller than MATH. Therefore MATH is the MATH-expansion of MATH. By induction hypothesis MATH, therefore MATH. By that we have shown that any MATH can be generated by given operations MATH, MATH. The number of operations can be reduced from MATH to MATH by the following relation, which is valid for any MATH, MATH, for all MATH. MATH . |
math/9907054 | In order to prove the statement of the proposition, we have to find MATH, such that MATH. We show that this property is satisfied by the convex hull of MATH, so we put MATH. We have to justify that any element of MATH can be generated using the given operations starting from points of MATH. Consider MATH. Since MATH, then also MATH, and hence MATH, as a consequence of REF . Therefore, if MATH, then MATH belongs to MATH. Without loss of generality let MATH. Observe that the points of MATH cover densely the interval MATH. Therefore one can find a finite sequence of intervals MATH, such that the elements MATH belong to MATH, and it holds that MATH, MATH and MATH. According to REF, one has MATH . Therefore MATH can be generated from points of MATH by the operations MATH, hence it is contained in MATH. Thus we have proven that MATH. It suffices now to justify that MATH is bounded. This is true, because otherwise MATH would not be uniformly discrete. |
math/9907058 | For the sake of simplicity, let us assume that MATH is convex co-compact. Let us start with an outline of the sketch of the proof in this case. The idea is to choose a REF-complex MATH which will be a union of surfaces MATH which are isotopic to the boundary components MATH of MATH. We will then consider a map MATH such that MATH and such that MATH is in the appropriate homotopy class so that each of the components of MATH will lie in the image of the convex core of a manifold whose convex core has bounded injectivity radius, thus proving the theorem. Now we will give a more complete sketch of the proof. First we construct a REF-complex MATH in MATH. For each MATH, let MATH be an annulus whose boundary components are the core curve of MATH and the core curve of MATH. Let MATH be REF-complex formed by gluing the boundaries of MATH to those of MATH. Then by construction, the inclusion of MATH into MATH is a homotopy equivalence. Let MATH be the MATH-th boundary component of MATH, and let MATH, glued along their boundaries. Then MATH is isotopic to MATH, and the component of MATH containing MATH is homeomorphic to MATH. Let MATH be our plain book of MATH-bundles described above. Let MATH, and let MATH be a hyperbolic REF-manifold homeomorphic to MATH, via a map MATH. Let MATH be a cover of MATH with covering map MATH. Let the vertex set MATH on MATH consist of exactly one internal vertex on the core curve of MATH. Triangulate MATH. Let MATH be a map such that for each MATH, MATH is a useful simplicial hyperbolic surface so that MATH. Moreover, construct MATH so that it is homotopic to MATH and hence MATH-injective. We further require that for each MATH, the map MATH lifts to MATH such that MATH. By REF , a boundary component MATH of MATH lifts to a boundary component MATH of MATH. We can construct a homotopy MATH between MATH and MATH such that MATH. Because the image of the homotopy lies in the convex core of a surface group, by REF , for MATH, MATH where MATH depends on MATH. So by REF , for MATH, MATH. With some work, we can show that MATH . Thus, for MATH, MATH. |
math/9907058 | The outline of the proof is the similar to that for a plain book of MATH-bundles. The idea is to use the image of a REF-complex MATH to divide MATH into portions, each of which lies in the image of the convex core of a manifold whose convex core has bounded injectivity radius. In the following lemma we will choose a REF-complex MATH such that the inclusion of MATH into MATH is a homotopy equivalence, and we will construct maps MATH whose images can be glued along MATH to form a REF-manifold homeomorphic to MATH. Let MATH be the boundary components of MATH. Then there exists a REF-complex MATH such that MATH is a deformation retract of MATH. Furthermore, there exist maps MATH with the following properties: CASE: for each MATH, the image MATH, CASE: the map MATH is a homeomorphism onto its image, CASE: the image MATH, and CASE: the manifold MATH is homeomorphic to the quotient space MATH, where MATH if and only if MATH. Recall that MATH, where MATH is a MATH-bundle over MATH, and MATH is a disjoint collection MATH of surfaces with boundary. Here, MATH is a disjoint collection MATH of solid tori. The boundary of each MATH consists of a collection of MATH parallel annuli MATH, which we can order cyclically MATH. Let MATH be the bundle projection map. Given a component MATH of MATH, for each component MATH of MATH, MATH is identified with some annulus MATH on the boundary of some MATH. In fact, there is a one-to-one correspondence between the set MATH and the set MATH is a boundary component of some MATH. We define two annuli MATH and MATH to be adjacent on MATH if they are ordered consecutively on MATH. Note that if MATH and MATH are adjacent annuli on MATH, then they will be physically separated by the parallel annulus MATH to which no MATH-bundle is identified. Given a MATH-bundle MATH over MATH, either MATH, or MATH is a twisted MATH-bundle over MATH. In the latter case, MATH, where MATH is a free involution and MATH. For a component MATH of MATH, let MATH denote the middle surface of the MATH-bundle over MATH, where MATH if MATH, and MATH if MATH is a twisted MATH-bundle over MATH. Let MATH denote the set of all the middle surfaces MATH. Let MATH be a compact REF-manifold, where MATH is a MATH-bundle over MATH and MATH is a disjoint collection of solid tori as before. For each MATH, MATH is decomposed into parallel annuli MATH. Given a component MATH of MATH with boundary component MATH, if MATH is identified with MATH on MATH in MATH, then identify MATH with MATH on MATH in MATH. Consider the quotient map MATH defined such that MATH is the identity map, MATH is a homeomorphism, and MATH identifies MATH to the core curve MATH of MATH. Now we will construct a disjoint union of surfaces, MATH, such that MATH will be the desired REF-complex MATH. For a fixed annulus MATH in MATH, construct the annulus MATH. Then for fixed MATH, let MATH where the union is taken over all components of MATH. Then MATH is a surface with boundary MATH. Then we define MATH, and MATH. There is a natural product structure on each component of MATH. For adjacent annuli MATH and MATH, we can extend this product structure so that the component of MATH containing MATH also has a product structure as shown in REF . This product structure induces a product structure on all of MATH. Note that as a result, we can deformation retract MATH along the product structure of MATH onto MATH. Now let us construct the maps MATH. Consider a boundary component MATH of MATH. Then for each MATH, let MATH be a homeomorphism along the product structure of MATH where MATH. We can let MATH for MATH. Then for all MATH, MATH. By identifying the images of the maps in MATH, we see that MATH is homeomorphic to the quotient space MATH, where MATH if and only if MATH. Let MATH be a hyperbolic REF-manifold homeomorphic to MATH via a homeomorphism MATH. For ease of exposition, let MATH be defined by MATH for MATH. In the next lemma, we will construct a map MATH such that for each MATH, the image MATH lifts to the MATH-neighborhood of the convex core of a cover of MATH. Let MATH be a hyperbolic REF-manifold homeomorphic to MATH via a homeomorphism MATH. Let MATH be the inclusion map, and let MATH. Let MATH, and let MATH. Then there exists a map MATH such that for each MATH, MATH lifts to a map MATH such that MATH. Recall that MATH where MATH, a disjoint union of surfaces with boundary. We will define the map MATH by constructing a map MATH, where for each MATH, the map MATH is a simplicial hyperbolic surface that factors through the quotient map MATH. Let us construct a triangulation on MATH that will induce a ``triangulation" on MATH. We put ``triangulation" in quotes, because MATH is a REF-complex, not a surface. First choose a vertex set MATH on MATH to be exactly one internal vertex MATH on each MATH. Let MATH be the vertex set on MATH. Include the arcs MATH in the edge set of the triangulation on MATH. For a fixed MATH and MATH, if MATH is homotopic to a MATH-curve on MATH, then there are MATH vertices on MATH, and the edge set MATH in MATH is a MATH-to-REF cover of MATH in MATH. Triangulate the remainder of MATH, and call it MATH. Then MATH is a ``triangulation" on MATH. Because MATH is a deformation retract of MATH, the inclusion map MATH is a homotopy equivalence. Then MATH is MATH-injective. Now we will construct a map MATH such that MATH will be homotopic to MATH, and MATH will be a simplicial hyperbolic surface which is a ``straightening" of MATH. For each MATH, MATH represents either a hyperbolic element or a parabolic element in MATH. If MATH represents a hyperbolic element, then its geodesic representative, MATH, lies in MATH. Let MATH be a map that is homotopic to MATH such that MATH is a point on MATH. If MATH represents a parabolic element, then MATH has no geodesic representative in MATH. Normalize so that the fixed point of the parabolic element which MATH represents is at infinity in the upper half space model of MATH. Recall that MATH is the core curve of MATH. Consider the collection MATH of boundary components of MATH such that MATH. Consider the collection of covers MATH of MATH associated to the MATH, where each MATH. Because each MATH is non-empty, the convex hull of each MATH must also contain a geodesic ray MATH with endpoint at infinity. Consider a horoball MATH about infinity. Under the action of the normalized parabolic element MATH, the horoball MATH induces a rank one cusp in MATH, the boundary of which is an infinite annulus. Then for large enough MATH, there exists an arc MATH on MATH such that MATH is a closed curve in MATH on the infinite annulus MATH, and such that MATH is contained in the MATH-neighborhood of each MATH so that MATH for each MATH. Let MATH be a map that is homotopic to MATH such that MATH is a point on MATH. Using NAME 's construction, let MATH. Then for each MATH, MATH is a simplicial pre-hyperbolic surface. Because MATH is in the homotopy class of MATH, MATH is a MATH-injective map. By construction, the map MATH respects the quotient map MATH, so there exists a map MATH which is a simplicial pre-hyperbolic REF-complex, that is, MATH weakly preserves parabolicity, maps every edge in MATH to a geodesic arc, and maps each face of MATH to a non-degenerate totally geodesic triangle in MATH. Furthermore, by construction, MATH is homotopic to MATH and hence is MATH-injective. Now we will show that MATH lifts to MATH. Because MATH is homotopic to MATH, by the Lifting Theorem, MATH lifts to a map MATH. Because MATH is a subset of MATH, MATH is a simplicial pre-hyperbolic surface. Then MATH is also a simplicial pre-hyperbolic surface. Because MATH is also convex, by an argument similar to that in the proof of REF, in order to show that MATH, it suffices to show that for each internal vertex MATH in the triangulation on MATH, MATH lifts to MATH. Let MATH be an internal vertex in the triangulation on MATH. For each MATH, MATH represents either a hyperbolic element or a parabolic element in MATH. If MATH represents a hyperbolic element in MATH, then MATH. The closed geodesic MATH lifts to MATH to an axis of a hyperbolic element of MATH which lies in MATH. Then MATH lifts to MATH. If MATH represents a parabolic element in MATH, then, by construction, MATH lifts to MATH to a point on MATH which lies in MATH. Therefore, MATH lifts to MATH, and we have shown that MATH. The next step in the proof of the book of MATH-bundles case involves finding a compact core MATH of MATH, such that the points in MATH have bounded injectivity radius. Given MATH. By the results of CITE and CITE and REF , there exists a compact core MATH of MATH such that MATH, and such that MATH. In the next lemma, we will choose a compact core MATH of MATH such that MATH contains MATH, MATH lies in MATH, and such that points in MATH have bounded injectivity radius. Let MATH, and let MATH be a cover of MATH with covering map MATH. Let MATH be the components of MATH. For all MATH, there exists a compact core MATH of MATH bounded by the surfaces MATH such that the following are true: CASE: MATH, CASE: if MATH, then MATH, and CASE: for each MATH, there exists MATH such that MATH is a homeomorphism, and MATH, Let us start with a sketch of the proof. First we will construct surfaces MATH such that for each MATH, there exists a homotopy between MATH and the inclusion map MATH. We will construct each homotopy so that its image lies in the image of the convex core of a manifold whose convex core has bounded injectivity radius. Then we will construct surfaces MATH such that there exists a homotopy between the inclusion maps of MATH and MATH. Again, we will construct each homotopy so that its image lies in the image of the convex core of a manifold whose convex core has bounded injectivity radius. Finally, we will show that the MATH bound a compact core MATH and that MATH is contained in the union of the images of these homotopies. Thus, by construction, all points in MATH will have bounded injectivity radius. Note that MATH, so that MATH is also a compact core of MATH. Because MATH is homeomorphic to MATH, and MATH is a compact core of MATH, we can conclude that MATH is homeomorphic to MATH. (REF , CITE) Then because MATH has incompressible boundary, MATH also has incompressible boundary. Note that MATH is also a compact core for MATH where the components of MATH are incompressible in MATH. Let MATH be the components of MATH. Then by REF , each component MATH of MATH possesses a product structure MATH. Because MATH, by the Lifting Theorem, the inclusion map MATH lifts to a map MATH. Let MATH. Then MATH is also possesses a product structure MATH, and MATH is a homeomorphism. Now we will show that for each MATH, MATH. Suppose not. Then there exists MATH and a component MATH of MATH such that MATH separates MATH from MATH. Consider a geodesic ray MATH in MATH that is perpendicular to MATH and passes through MATH. Let MATH be the portion of MATH beginning at MATH. There exists a constant MATH such that MATH. Recall that by REF , the injectivity radius strictly increases out a geometrically finite end, therefore the ray MATH is contained in MATH. Without loss of generality, let us assume that MATH intersects MATH transversely. Then one of the following three cases occurs: MATH intersects MATH an odd number of times, MATH intersects a different boundary component of the closure of MATH, or all but a compact portion of MATH is contained in MATH. Suppose MATH intersects MATH an odd number of times. Because MATH is an embedded surface separating MATH, MATH has two sides. Let the component of MATH containing MATH be the positive side of MATH. Then the last time MATH intersects MATH, MATH passes from the positive to the negative side of MATH. We can also let the side of MATH containing MATH be the positive side of MATH. Recall that because MATH and MATH is a component of MATH, MATH lies on the negative side of MATH. Because MATH is a homeomorphism and MATH lifts to MATH, MATH lies to the negative side of MATH. Note that MATH is an incompressible separating surface of MATH, and that MATH is homotopic to the inclusion map MATH. Because MATH leaves every compact set of MATH and all but a compact portion of MATH lies in MATH, if MATH intersects MATH, then MATH must also intersect MATH. But MATH and MATH. So this is a contradiction. Suppose MATH intersects a different boundary component of the closure of MATH. Then MATH intersects some component MATH of MATH such that MATH is a boundary component of the closure of MATH in MATH. By REF , MATH lifts to a component MATH of MATH. Because MATH intersects MATH, there is a portion of MATH that contains MATH and joins two components of MATH. Because MATH is convex, MATH lies in MATH, which is a contradiction. Then all but a compact portion of MATH is contained in MATH. Let MATH be the closure of the component of MATH that contains MATH. Let MATH. Let MATH be the closure of MATH in MATH. Then because MATH is compact and MATH for every MATH, we can guarantee that MATH for large enough MATH. Then MATH is a component of MATH which embeds in MATH. Then by REF , MATH is a component of MATH with boundary MATH. Because MATH, MATH. Then MATH. But by hypothesis, MATH so this is a contradiction. Thus, MATH for each MATH. For each MATH, choose MATH to be a level surface in MATH. Recall that MATH does not intersect MATH so that MATH lies to one side of MATH. By construction, MATH is embedded, and the inclusion map of MATH into MATH is homotopic to the map MATH. Let MATH be a homotopy between the inclusion map MATH and the map MATH, where MATH and MATH. Let MATH be the ruled homotopy constructed from MATH such that for MATH, MATH is the geodesic arc with the same endpoints and in the same homotopy class as MATH. Because MATH is convex, we know that MATH. Because MATH is a surface group, by REF , for MATH, MATH. For each MATH, let MATH be a level surface in MATH. Consider the homotopy MATH. Here MATH is the inclusion map MATH which is the projection of the inclusion map MATH; MATH; and MATH. By REF , for MATH, MATH. Now for each MATH, choose MATH to be a level surface in MATH such that MATH lies to one side. Let MATH be a level surface in MATH. By construction MATH. Because each of the MATH is incompressible, disjoint, and homotopic to the boundary components of MATH, the MATH and MATH span a product structure in MATH. (see REF , CITE) Hence, the MATH are the boundary components of a new compact core MATH of MATH. In particular, since the MATH lie in MATH, we can conclude that MATH . For each MATH, let MATH be homotopy along the product structure of MATH between the inclusion maps MATH and MATH, where MATH and MATH. Because MATH, by REF , for MATH, MATH. Then MATH is a product homotopy in MATH between the inclusion maps MATH and MATH, where MATH and MATH. Furthermore, MATH. By REF , for MATH, MATH. Finally we will show that MATH . By REF , we know that MATH, where MATH if and only if MATH. Let MATH be a map such that MATH and MATH. Then by construction, if MATH, then MATH. Hence we have a well-defined map MATH. Using standard degree arguments, MATH is contained in the image of any proper homotopy between MATH and MATH. (see REF , NAME, CITE) In particular, MATH. So for MATH, MATH. This completes the proof of REF . Now we have bounded the injectivity radius for points in MATH. Now consider points in MATH. If MATH, then there exists a MATH, such that MATH. Recall that in the previous lemma we showed that MATH was a closed incompressible separating surface and that MATH. By REF we know that if MATH, then MATH. So MATH for some MATH. Then we can conclude that for MATH, MATH. We can do this for all MATH so that for MATH, MATH. This completes the proof of REF . |
math/9907058 | We begin with a sketch of the proof of the acylindrical case. The proof will be by contradiction. Suppose there exists a sequence of points in the convex cores of manifolds such that the injectivity radius based at these points goes to infinity. We will find a compact core MATH in the algebraic limit which embeds in the geometric limit as MATH. The compact core MATH will pull back to a compact core MATH in each manifold MATH in the sequence such that points in MATH will have uniformly bounded injectivity radius. The complement of MATH in MATH will either be covered by the convex cores of manifolds whose convex cores have bounded injectivity radius or will have injectivity radius uniformly bounded by the injectivity radius of a fixed compact subset of the geometric limit of the sequence of manifolds. Thus, we will have found a uniform bound on the injectivity radius for points in the convex core of each manifold in the sequence, which is a contradiction. Suppose for contradiction that there does not exist an upper bound on the injectivity radius for points in the convex cores of hyperbolic REF-manifolds homeomorphic to MATH. Then there exists a sequence of representations MATH together with its corresponding sequence of manifolds MATH, and a sequence of points MATH, such that MATH diverges to infinity. By NAME 's Compactness REF , a subsequence of MATH converges algebraically, up to conjugation, to a representation MATH. Let the algebraic limit manifold be MATH. Using a result of NAME (REF , CITE), we can take a further subsequence, again called MATH, such that MATH converges to MATH geometrically. Let the geometric limit manifold be MATH. By definition of geometric convergence, there exists a sequence of MATH-approximate isometries MATH such that MATH and MATH. Furthermore, because MATH is a subgroup of MATH, there is a natural covering map from the algebraic limit to the geometric limit MATH. Let MATH be the conformal extension of MATH. CITE have an alternate definition of an accidental parabolic which, to avoid confusion, we will call an unexpected parabolic. We say that MATH has connected limit set and no unexpected parabolics if and only if every closed curve MATH in MATH which is homotopic to a curve of arbitrarily small length in MATH is homotopic to a curve of arbitrarily small length in MATH. The following lemma shows that the fundamental group of an acylindrical, hyperbolizable manifold has connected limit set and no unexpected parabolics. These properties will be useful in showing the existence of a compact core in the algebraic limit which will embed in the geometric limit. Let MATH be an acylindrical, hyperbolizable REF-manifold. Let MATH, MATH, and MATH. Then every closed curve MATH in MATH which is homotopic to a curve of arbitrarily small length in MATH is homotopic to a curve of arbitrarily small length in MATH. Therefore, MATH has connected limit set and no unexpected parabolics. The proof will be by contradiction. For MATH, CITE and CITE guarantee the existence of a relative compact core MATH of MATH with associated parabolic locus MATH such that MATH. Note that MATH, so that MATH is also a compact core of MATH. Because MATH is homeomorphic to the interior of MATH and MATH is a compact core of MATH, we can conclude that MATH is homeomorphic to MATH. (REF , CITE) Therefore, because MATH has incompressible boundary and is acylindrical, MATH has incompressible boundary and is acylindrical. Suppose for contradiction that there exists a homotopically non-trivial, closed curve MATH in MATH which is homotopic to a curve of arbitrarily small length in MATH, but that MATH is not homotopic to a curve of arbitrarily small length in MATH. Recall that CITE has shown that there exists a homeomorphism MATH which is homotopic to the canonical nearest point retraction MATH (see REF), and which is MATH-bilipschitz on MATH where MATH is independent of MATH. Suppose MATH is homotopically trivial in MATH, then MATH is a homotopically non-trivial, closed curve in MATH which is homotopically trivial in MATH. Here, MATH is a compact core of MATH so that there exists a homotopically non-trivial curve in MATH which is homotopically trivial in MATH. But MATH is an incompressible subsurface of MATH, and MATH has incompressible boundary, so we can conclude that MATH is incompressible in MATH. But this is a contradiction. Therefore MATH is homotopically non-trivial in MATH. Because MATH is MATH-bilipschitz on MATH, we know that MATH is a closed curve in MATH which is homotopic to a curve of arbitrarily small length in MATH, but that MATH is not homotopic to a curve of arbitrarily small length in MATH. Because MATH is a compact core of MATH, we know that MATH is homotopic to a curve on MATH. Therefore, without loss of generality, we can consider MATH to be a closed curve on MATH. Because MATH is homotopic to a curve of arbitrarily small length in MATH, there exists a closed curve MATH such that MATH is homotopic to MATH in MATH. Then since MATH and MATH is a compact core, we can conclude that MATH is homotopic to MATH in MATH. Suppose MATH is homotopic to MATH in MATH. Then MATH is homotopic into the parabolic locus of MATH in MATH. Then MATH is a peripheral curve in MATH, and hence is homotopic to a curve of arbitrarily small length in MATH. Then MATH must have been homotopic to a curve of arbitrarily small length in MATH. But this contradicts our initial assumptions about MATH. So MATH is not homotopic to MATH in MATH. Using the Homotopy Annulus Theorem (REF , CITE), we can construct a MATH-injective, proper embedding MATH such that MATH and MATH where MATH cannot be properly homotoped into MATH. Therefore, MATH is not acylindrical. But this is also a contradiction. Thus, every closed curve MATH in MATH which is homotopic to a curve of arbitrarily small length in MATH is homotopic to a curve of arbitrarily small length in MATH. Therefore, MATH has connected limit set and no unexpected parabolics. This completes the proof of REF . Next, we will show that there exists a compact core MATH in the algebraic limit MATH which embeds in the geometric limit MATH as MATH. Let MATH be an acylindrical, hyperbolizable REF-manifold. Let MATH be the covering map between the algebraic limit MATH and the geometric limit MATH. Then there exists a compact core MATH such that MATH embeds in MATH. There are two cases - either the limit set of MATH is the entire sphere or not. Suppose the limit set of MATH is the entire sphere. Because MATH satisfies NAME 's REF , MATH is topologically tame. Then REF states that the algebraic limit and the geometric limit agree, and, in this case, any compact core of the algebraic limit is also a compact core of geometric limit by default. (REF , CITE) Let MATH be a sequence of discrete faithful representations converging algebraically to MATH. If the limit set of MATH is all of MATH and MATH is topologically tame, then MATH converges strongly to MATH. If the limit set of MATH is not the entire sphere, then the domain of discontinuity of MATH has nonempty domain of discontinuity. Recall that by REF , MATH has connected limit set and no unexpected parabolics. In this case, REF guarantees that given an algebraically convergent sequence such that its associated image groups converge geometrically, we can find a compact core in the algebraic limit which embeds in the geometric limit. (Cor B, CITE) Let MATH be a finitely generated, torsion-free, nonabelian group, and let MATH be a sequence in MATH converging algebraically to MATH. Suppose that MATH converges geometrically to MATH. Let MATH, MATH, and let MATH be the covering map. If MATH has nonempty domain of discontinuity, connected limit set, and contains no unexpected parabolics, then there exists a compact core MATH of MATH such that MATH is an embedding restricted to MATH. Thus, in either case, we can find a compact core in the algebraic limit which embeds in the geometric limit. This completes the proof of REF . By REF , we know that MATH has connected limit set and contains no unexpected parabolics. The next lemma due to NAME and NAME shows that for large enough MATH, the pull back of MATH to MATH is a compact core of MATH. (REF , CITE) Let MATH be a finitely generated, torsion-free, nonabelian group, and let MATH be a sequence in MATH converging algebraically to MATH. Suppose that MATH converges geometrically to MATH. Let MATH, MATH, and let MATH be the covering map. Suppose MATH has connected limit set and contains no unexpected parabolics. Let MATH be a compact core of MATH such that MATH is an embedding restricted to MATH. Then for large enough MATH, MATH is a compact core for MATH. Now let us show that points in MATH have uniformly bounded injectivity radius, where the bound depends on the compact set MATH in MATH. For large enough MATH, and for MATH, we have MATH. Note that because MATH is a compact set in MATH, there exists a constant MATH such that for MATH, we have MATH. Then, because MATH is compact and MATH is a MATH-approximate isometry such that MATH and MATH, it is possible to choose MATH such that for MATH, we guarantee that MATH and that the closure of the MATH-neighborhood of MATH lies in MATH. Let MATH be a homotopically non-trivial loop in MATH that is based at MATH and is of length MATH. If MATH is a homotopically trivial loop, then MATH bounds a disk MATH in MATH. Consider the immersion MATH. For MATH, let MATH be the line segment in MATH joining MATH and MATH. Let MATH be the geodesic arc in MATH that is properly homotopic to the segment MATH. Fix MATH. Then the new disk MATH has diameter MATH in MATH. For MATH, we know that MATH so that MATH. Then MATH is a disk in MATH, so MATH is homotopically trivial in MATH. But this is a contradiction. Therefore, MATH is a homotopically non-trivial loop based at MATH of length MATH. Then for MATH, we know that MATH. In particular, for MATH, we know that MATH, so for MATH, we have MATH. The following lemma shows that points in MATH also have bounded injectivity radius, because they are either covered by manifolds whose convex cores have bounded injectivity radius or have injectivity radius bounded by the injectivity radius of a fixed compact subset of the geometric limit. Let MATH, and let MATH be the boundary components of MATH. Then for large enough MATH and MATH, we have MATH. Temporarily fix MATH. Because MATH is homeomorphic to the interior of MATH where MATH has incompressible boundary, the compact core MATH of MATH is homeomorphic to MATH. (REF , CITE) Since MATH has incompressible boundary, so does MATH. Let MATH be the components of MATH. Then each MATH is homeomorphic to MATH and is an incompressible separating surface of MATH. Let MATH be the component of MATH with boundary component MATH. Because MATH is a compact core of a topologically tame REF-manifold with incompressible boundary, by REF , MATH possesses a product structure MATH for each MATH. Because MATH, by the Lifting Theorem, the inclusion map MATH lifts to a map MATH. Let MATH. Then the projection map MATH is a homeomorphism. Let MATH. If MATH, then there exists MATH such that MATH. By REF , it suffices to bound the injectivity radius based at MATH in MATH. There are two possibilities: either MATH or not. If MATH, then by REF , MATH. Suppose MATH. There are two possibilities here also: either MATH or MATH. If MATH, then MATH. Suppose MATH. Then there exists a geodesic ray MATH that is perpendicular to a component MATH of MATH and that passes through MATH. Let MATH be the portion of MATH beginning at MATH. Note that by REF , the injectivity radius strictly increases out a geometrically finite end. Because MATH, the ray MATH is entirely contained in MATH. Then either MATH is contained in MATH, or MATH intersects MATH. Suppose MATH is contained in MATH. Let MATH be the closure of the component of MATH that contains MATH. Let MATH. Because MATH is compact and MATH for all MATH, we can guarantee that MATH for large enough MATH. Then MATH is a component of MATH that embeds in MATH. By REF , MATH is the component of MATH with boundary MATH. Because MATH, MATH. Then MATH. But by hypothesis, MATH so this is a contradiction. Thus MATH must intersect MATH. By REF , we know that the injectivity radius strictly increases out a geometrically finite end. Thus, it suffices to bound the injectivity radius for points MATH. Because MATH is in the image of a compact set in MATH, there exists a constant MATH such that for MATH, MATH. Because MATH and MATH is a MATH-approximate isometry, by the argument in the proof of REF , for MATH, we know MATH. Furthermore, because MATH is an isometry and MATH, for MATH, we know MATH. Because MATH is incompressible in MATH and MATH is a lift of MATH, MATH is incompressible in MATH. So for MATH, we know that MATH. Then by REF , for MATH, MATH, and MATH intersecting MATH at MATH, we have MATH. Because MATH, for large MATH, we have MATH. Thus, for large MATH and MATH, we can conclude that MATH. This completes the proof of REF . Therefore, by REF , for MATH, we know that MATH. We can do this for all MATH, so that for MATH, we have MATH. This uniform bound contradicts the assumption that there exists a sequence of points MATH such that MATH converges to infinity. This completes the proof of REF . |
math/9907058 | First let us show that if MATH is homotopy equivalent to a book of MATH-bundles, then MATH is homeomorphic to the interior of a book of MATH-bundles. Let MATH be a compact core for MATH. Because MATH satisfies NAME 's REF , MATH is topologically tame, and hence MATH is homeomorphic to the interior of its compact core MATH. Thus, it suffices to show that MATH is a book of MATH-bundles. We will use the characteristic submanifold theory developed by CITE and CITE. First let us introduce some definitions. A map MATH of an annulus or torus, MATH, into MATH is essential if MATH is MATH-injective and MATH is not properly homotopic into MATH. A map MATH of a MATH-bundle MATH into MATH is admissible if MATH is the associated MATH-bundle. A map MATH of a MATH-bundle MATH into MATH is essential if MATH is MATH-injective, and for every component MATH of MATH, MATH is an essential map of an annulus or torus into MATH. A compact submanifold MATH of MATH is a characteristic submanifold if MATH is a minimal collection of admissibly embedded, essential MATH-bundles and NAME fiber spaces such that every essential, admissible embedding MATH of a NAME fiber space or MATH-bundle into MATH is properly homotopic to an admissible map with image in MATH. In fact, every compact, orientable, irreducible REF-manifold with incompressible boundary contains a unique (up to isotopy) characteristic submanifold. We can apply the characteristic submanifold theory to MATH and to MATH to obtain the characteristic submanifolds MATH and MATH. Recall that MATH is a book of MATH-bundles if there exists a disjoint collection MATH of incompressible annuli such that each component of the manifold obtained by cutting MATH along MATH is either a solid torus, or a MATH-bundle MATH over a surface of negative NAME characteristic such that MATH is the associated MATH-bundle. Thus, we know that MATH is a collection of MATH-bundles and solid tori, and MATH is a collection of solid tori. Because MATH is a hyperbolizable REF-manifold and MATH contains no MATH subgroup, then REF MATH is a collection of MATH-bundles and solid tori. Let MATH be a homotopy equivalence. By NAME (REF , CITE), the map MATH is homotopic to a map MATH such that MATH, MATH is a homeomorphism, and MATH is a homotopy equivalence. Because MATH is a homeomorphism, each component of MATH is a solid torus. Then as a consequence of a result of NAME (REF , CITE), MATH is also a book of MATH-bundles. Thus, MATH is homeomorphic to the interior of a book of MATH-bundles MATH. Now we will show a relationship between the NAME characteristic of the boundary components of MATH and the number of generators of MATH. Let MATH be the double of MATH. Then MATH is a closed REF-manifold which has NAME characteristic MATH. So MATH or MATH. Recall MATH where MATH is the rank of MATH. Because MATH is a connected REF-manifold with boundary, MATH. Because MATH is the abelianization of MATH, MATH where MATH is the number of generators in MATH. So MATH or MATH. By applying REF to MATH and MATH, we obtain an upper bound MATH such that for MATH, MATH. Recall from the proof of REF , MATH, where the maximum is taken over all boundary components MATH of MATH and MATH is the bound obtained in REF . Let MATH where the maximum is taken over all surfaces MATH such that MATH. Note that this is a finite set of surfaces. Then MATH, and hence for MATH, MATH, where the bound MATH depends only on the number of generators of MATH. This completes the proof of REF . |
math/9907058 | The proof is a direct corollary of the Main Theorem and REF , CITE which relates upper and lower bounds on injectivity radius in the convex core to convergence of limit sets. |
math/9907058 | Let us first state and outline the proof of the result of NAME cited in the proof of the previous theorem. (REF , CITE) For MATH, let MATH be a sequence of hyperbolic REF-manifolds with base frame MATH such that: CASE: the baseframe MATH lies in MATH, CASE: the injectivity radius at MATH is greater than MATH, and CASE: for MATH, we require that MATH be bounded above by MATH. Suppose MATH converges geometrically to a limit manifold MATH. Then MATH converges to MATH in the NAME topology. Let MATH. Because each point in MATH is a bounded distance (depending only on R) away from MATH or MATH, we know that a limit point of MATH in MATH must be a point in MATH, that is, MATH. By geometric convergence, MATH converges uniformly to MATH on compact subsets of MATH. Therefore MATH. Because injectivity radius for points in the convex core is bounded above by MATH, we know that MATH. Without loss of generality, suppose the origin MATH is the basepoint of MATH for all MATH. By hypothesis, we know that MATH for all MATH, where MATH denotes convex hull. Then MATH contains all limits of rays from MATH to MATH. Therefore, MATH . Since MATH for all MATH, the result follows. From the proof of REF , we can see that if for large MATH, the basepoint of MATH lies in the MATH-neighborhood of MATH, then MATH contains all limits of rays from the basepoint of MATH to MATH. Then we know MATH and hence MATH converges to MATH in the NAME topology. Without loss of generality, suppose the origin MATH is the basepoint of MATH. We will show that for large MATH, the origin MATH lies within a uniformly bounded neighborhood of MATH. Because MATH is a limit of torsion-free Kleinian groups, MATH itself is torsion-free. (REF , CITE) Then since MATH is also nonabelian, we can conclude that MATH is nonelementary. Therefore, MATH contains a hyperbolic element MATH. (REF , CITE) Let MATH denote the axis of MATH in MATH. Then there exists a constant MATH such that MATH. As an element of the geometric limit, there exists a sequence MATH such that MATH. For large MATH, MATH is a hyperbolic element, and therefore, MATH. Then for large MATH, MATH. Because MATH for all MATH, we can conclude that MATH for large MATH. This concludes the proof of REF . |
math/9907065 | We first study the family version of the critical points of MATH on MATH, where MATH is from a set of imaginary valued co-closed REF-forms on MATH with compact support in MATH. Denote this set of perturbations as MATH. Let MATH be a critical point of MATH. We need to show that the derivative of the gradient of MATH is surjective. Namely, consider MATH which sends MATH to MATH . Suppose that MATH is orthogonal to the image of the above map, then MATH satisfies the following equations: MATH . The elliptic regularity implies that MATH is smooth. From REF , we know that MATH on MATH. The following Lemma due to CITE will ensure that MATH on MATH. Hence, there is a real-valued smooth function f on MATH, such that MATH. Using REF , we obtain MATH on MATH, which leads to MATH on MATH. By REF , we get MATH on Y. Note that MATH does not disconnect any domain in MATH (the unique continuation principle for NAME operator (see page REF)). Therefore, MATH which implies that MATH. From the NAME theorem, there is a NAME set of MATH such that all critical points of MATH in MATH are non - degenerate for a generic MATH. Now we need to prove that the reducible critical point of MATH is also non-degenerate. By the assumption, MATH admits reducible critical point if and only if MATH is a rational homology REF-sphere. From the analysis in CITE, we know that, in order to achieve the non-degenerate condition at reducible critical point, MATH is required to be away from the codimension one subset MATH where the corresponding NAME operator has non-trivial kernel. This completes the proof of the Proposition. Now we give the proof of NAME 's Lemma. CASE: Let MATH and MATH as above, where MATH is a solution to the NAME REF and MATH satisfies REF of REF . Then MATH obeys an equation of the form MATH at all points where MATH. Here MATH is the Laplacian on differential REF-forms and MATH and MATH are linear maps that depend implicitly on MATH. The set of points where MATH is path connected open dense set in MATH. The unique continuation principle applies to MATH so that MATH cannot vanish on MATH without vanishing everywhere on MATH. Proof of NAME 's Lemma: Apply the Laplacian to MATH, we have the following expression of MATH: MATH here MATH acting on spinors is MATH, and MATH is the pairing using the metric on MATH. Now invoke the NAME formula for the NAME operator, MATH where MATH is the scalar curvature on MATH. Thus, from the NAME equations for MATH and MATH, we obtain MATH here MATH. Plug these two equations into REF , and note that MATH and MATH. We get MATH . Write MATH where MATH is a real-valued function on MATH and MATH, then MATH . Hence REF can be written as MATH . To complete the proof, we only need to show that MATH and MATH can be written as combinations of linear maps on MATH and MATH. On the set of points where MATH, MATH, we write MATH where MATH is a unit-length spinor. Choose a local basis MATH for the MATH bundle, so that NAME multiplication in the local orthonormal coframe MATH for MATH is given by MATH where MATH can be expressed as MATH . Write MATH for a real-valued function MATH and a complex-valued function MATH, then MATH . On MATH, MATH defines a bundle isomorphism between MATH . Thus, we obtain that MATH, and MATH . Let MATH, then from REF of REF , we have MATH as MATH can be written in terms of MATH and MATH, so is MATH. This completes the proof of NAME 's Lemma. |
math/9907065 | We may choose a trivialization of the cotangent bundle to MATH so that, using a full-stop to denote NAME multiplication by a one form, we can make the identifications: MATH . Letting the NAME * on forms on MATH be denoted by MATH, then under the preceding identifications we have MATH hence we get MATH . The form MATH is uniquely determined, as a MATH-valued REF-form, by its MATH-part MATH. Similarly, the NAME operator on MATH can be expressed as MATH . Thus gives the NAME equation as in the Lemma. |
math/9907065 | Direct calculation shows that we have MATH . |
math/9907065 | Since MATH satisfies the NAME equation on MATH: MATH . The NAME formula for the NAME operator on MATH with flat metric then gives MATH . Take the inner product of both sides with MATH, use the NAME equation, and note that MATH. We obtain MATH . Integrating the above identity over MATH, we can write the result as MATH . Here we write MATH as a spinor on MATH and use REF for MATH. Note that MATH solves the gradient flow equation of MATH, hence MATH . |
math/9907065 | From the MATH-bound on MATH, we immediately obtain a MATH-bound on MATH from the NAME equation, MATH . By the NAME inequality, we get MATH . Here we use that MATH. In the proof the previous lemma, we found that MATH satisfies MATH . Multiplying both sides of the above equation with a cut-off function MATH which equals REF on MATH and vanishes near the boundary of MATH, and then integrating by parts, we obtain MATH where MATH is a constant depending only on the cut-off function MATH. Putting the above inequalities together we get the estimates as claimed with MATH. The finite variation of MATH along MATH for a solution on MATH is the direct consequence of adding up over a sequence of middle tubes of length MATH, namely, MATH, hence MATH . |
math/9907065 | Write the curvature of MATH on MATH as MATH, where MATH is the curvature on MATH. Then we have the following calculation: MATH which implies that as MATH, MATH . By NAME 's weak compactness result and the compactness of MATH, we know that MATH weakly converges to a flat connection. By the monopole equation on MATH, we also obtain MATH . This implies that MATH converges weakly to a point in MATH. |
math/9907065 | Write MATH, then MATH satisfies the following equation: MATH . Here MATH is second order in MATH with MATH and MATH. Note that the flowline of MATH on MATH is static, hence we can establish the analogous result as REF as follows. Let MATH denote the projection of MATH onto the eigenspaces of MATH with positive and negative eigenvalues. Let MATH be the functions on MATH given by the MATH-norm on the MATH-slice of MATH. Then we have MATH . When MATH, from the above inequalities together with the comparison principle (Compare REF), we obtain that the MATH-norm of MATH on the MATH-slice is decaying exponentially with decay rate MATH. Then the claim of the lemma follows from the standard bootstrapping argument. |
math/9907065 | From REF , we know that the variation of MATH is finite, that is, MATH is finite. Then we have the following estimate, whose proof is analogous to the proof of REF. We sketch the proof here. Claim: There exist constants MATH and MATH such that for any MATH, and for MATH any solution to the NAME equation in temporal gauge on MATH satisfies MATH then we have the estimate MATH . Proof of the Claim Let MATH be a solution to the NAME equation on MATH in temporal gauge, then from REF , we have MATH . Denote by MATH . Then we have the following estimates MATH . We proceed as in REF and differentiate the NAME equations to get MATH . The gauge fixing condition implies that MATH . Introduce a cutoff function MATH identically equal to REF on the middle third piece of MATH and vanishes near the boundary such that MATH is at most MATH where MATH is a universal constant. Set MATH. Then we can estimate the quantity MATH by MATH . Here MATH is a universal constant depending only on MATH. On the other hand, we can estimate MATH . Assume that MATH, then the NAME multiplication theorem and NAME embedding theorem imply that there are constants MATH and MATH such that MATH . Similarly, by choosing MATH appropriately, we have MATH . These inequalities imply that MATH . Standard estimate for the elliptic operator MATH can be employed to show that there is a constant MATH such that MATH . The NAME inequality and the NAME embedding theorem imply that there exists a constant MATH such that MATH . Put all these inequalities together, we have MATH . Then there is a constant MATH and a constant MATH satisfying MATH such that if MATH and MATH, there is a constant MATH with the following estimate MATH . Since on the middle third piece MATH, MATH and MATH, this implies that for any tube MATH of length MATH and any solution MATH on MATH of energy at most MATH, we have MATH . Then the estimate in the claim follows by adding up a sequence of middle third pieces of tubes (length MATH) with the constant MATH and MATH as above. With this claim and REF , we can prove REF using the method of the proof of REF and the fact that MATH is a NAME function on MATH and satisfies the NAME condition on paths coming from monopoles on MATH. |
math/9907065 | The transversality argument is the same as in the closed case, see the proof of REF , namely, we look at the deformation complex REF for the parametrized moduli space MATH to get the transversality for the parametrized moduli space MATH. We then apply the infinite dimensional version of NAME theory to the projection MATH, we obtain that, for MATH in a NAME space MATH, the moduli space MATH is a smooth manifold of dimension given by virtual dimension calculated as above. We first show that for a generic perturbation MATH (a co-closed imaginary valued REF-form with compact support in MATH), the map MATH as given by REF is surjective. At an irreducible monopole MATH in MATH for MATH, we will show that MATH is surjective. Suppose that MATH is MATH-orthogonal to the image of the above map, then MATH is MATH-orthogonal to the image of the above map, hence, MATH is in MATH and satisfies REF as in the proof of REF . Hence, there is a real valued function MATH on MATH (with infinite cylindrical end) such that MATH, MATH and MATH . MATH implies that on MATH, MATH is in MATH, then by NAME inequality MATH this implies that for MATH, MATH for some constants MATH. MATH, whose asymptotic behaviour has been studied in the previous subsection, we see that there exist gauge representatives MATH and MATH of MATH and MATH, so that MATH decays to MATH exponentially at the rate at least MATH, where MATH is the smallest absolute value of the non-zero eigenvalue of MATH (Compare REF ). On MATH, write MATH with MATH, using the analysis in Appendix of CITE, we get MATH from some constant MATH. As MATH, we obtain MATH for some constant MATH. From two REF , and note that MATH, we have MATH . Now multiply MATH by MATH and integrate it by parts, we get MATH, hence MATH. The proof of that-MATH is an immersion and transversal to any given immersed curves follows from the NAME theorem. An orientation of MATH is obtained from a trivialization of the determinant line bundle of the assembled operator of the deformation complex REF . The trivialization of the determinant line bundle of the complex of REF is obtained from the orientation of MATH, the cohomology groups of MATH-decaying forms. In fact, we can deform the operator MATH with a homotopy MATH, MATH. The asymptotic operator MATH is preserved in the deformation. Thus, if the weight MATH is chosen in such a way that MATH is not in the spectrum of MATH, then REF we can ensure that the operator MATH is NAME, for all MATH. Since the NAME operator is complex linear and it preserves the orientation induced by the complex structure on the spinor bundle, a trivialization of the determinant line bundle at MATH is obtained by the orientation of MATH. This in turn determines a trivialization of the determinant line for MATH, hence an orientation of MATH. |
math/9907065 | Using the natural complex structure on MATH, we can identify MATH as the space of constant sections of MATH . For MATH, we have MATH which is a constant section. Take MATH as the coordinates on MATH, we have MATH . |
math/9907065 | This is a direct calculation using the gradient flow REF . |
math/9907065 | It follows from REF that MATH converges to some MATH as MATH. REF define a torus over MATH and, as MATH, points defined by REF approach points in MATH. |
math/9907065 | By the center manifold theorem in CITE, the restriction of any finite energy monopole MATH to the tube MATH (for a fixed large MATH) is exponentially close to a flow line in the center manifold starting from the point MATH given by the refinement boundary map REF . The exponential weight is at least a half of the smallest absolute value of the non-zero eigenvalues of MATH. REF shows that, for a generic choice of the perturbation, the moduli space MATH is a smooth manifold of dimension MATH, away from MATH. From the analysis of the center manifold theorem, since MATH is a REF-manifold with boundary MATH, we know that generically MATH, if non-empty, is a smooth manifold of dimension given by the virtual dimension: MATH, so MATH must be empty and MATH is a smooth oriented REF-dimensional manifold with boundary MATH. |
math/9907065 | In order to prove the first claim it is sufficient to check that elements of MATH give rise to approximate eigenfunctions on MATH with slowly decaying eigenvalues, that is, with eigenvalues MATH satisfying MATH at most like MATH. The first statement is then an analogue, in our case, of REF. Suppose we are given an element MATH. If MATH is a cutoff function supported in MATH satisfying MATH on MATH, we have an estimate MATH . This implies a similar estimate for the operator MATH on MATH, for MATH large enough, since we are assuming that this operator differs from MATH by terms that are exponentially small in MATH. This is the setting used in CITE. The second part of the statement can be derived from the asymptotically exact sequence MATH as in the Main Theorem of CITE. Here MATH denotes the span of the eigenvectors of MATH with small eigenvalues that decay at a rate of at least MATH. We use the notation MATH for the extended MATH-solutions, and MATH for the asymptotic values of the extended MATH-solutions. |
math/9907065 | We have MATH . Moreover, for a generic choice of the perturbation of the monopole equations on MATH, the Lagrangian subspaces MATH and MATH intersect transversely. Thus, we have MATH and MATH. The previous Proposition shows that the span of eigenvectors with slowly decaying eigenvalues is non - trivial. In fact, it shows the existence of (at least) a two dimensional family parameterized by the elements of MATH. |
math/9907065 | Let MATH be a constant such that the quadratic term satisfies the estimate MATH independent of MATH. This follows from the NAME multiplication theorem in dimension REF. On the image of MATH, the operator MATH is bounded with norm bounded by MATH. We have an estimate for MATH which implies that MATH maps the ball MATH to itself. Let MATH, we have MATH . Thus, from MATH as chosen, we obtain that MATH is a contraction on MATH. |
math/9907065 | For all approximate solutions MATH in MATH, we have an estimate on the error term MATH for MATH, which follows from the exponential decay estimate proved in REF . Thus, we can apply REF , with MATH and MATH. By REF we know that, for MATH, the projection MATH, hence the solution MATH of REF, provided by REF also satisfies trivially REF . |
math/9907066 | CASE: It is enough to show that MATH is compact, since it is then (possibly after a perturbation as in REF) a one-manifold with boundary MATH. Let MATH be a small sphere in the descending manifold around MATH, and let MATH be a small sphere in the ascending manifold around MATH. A flow line corresponds to a triple MATH such that downward flow from MATH for time MATH hits MATH. Compactness will follow from an upper bound on MATH. If MATH is unbounded, then one can show as in REF that there is a broken or degenerate flow line from MATH to MATH at some time MATH. For REF , we get a similar compactness for MATH. Since the orbits remain nondegenerate, none are created or destroyed, and the NAME signs cannot change. |
math/9907066 | Consider the limit as MATH. There exists MATH such that all critical points of MATH are nondegenerate for MATH, so that MATH for MATH. For convergence of MATH, we must show that for MATH and MATH, there exists MATH such that MATH where MATH ranges over any sequence of non-bifurcation values converging to MATH from below. We use REF . For any path MATH from MATH to MATH, we have MATH where MATH is a constant which is independent of MATH and varies continuously with MATH. Thus if MATH is a downward gradient flow line and MATH is bounded from above then MATH is also bounded from above, so if we are sufficiently close to MATH then there are no degenerate or broken flow lines from MATH to MATH by definition of semi-isolated, so MATH cannot change by REF . Similary, REF implies that the zeta functions converge. The last sentence of the lemma follows from REF . |
math/9907066 | Consider the limit as MATH. By definition we have MATH. So we have to prove that MATH where MATH is a fixed NAME structure. For MATH sufficiently small we can identify the critical points for different MATH. Fix a basis for MATH consisting of a lift of each critical point to MATH, in the equivalence class determined by MATH. For a non-bifurcation MATH, recall that MATH is the sum of the torsions of MATH. The torsion of MATH is zero if MATH is not acyclic; this criterion is independent of MATH, by the NAME isomorphism REF. Moreover, even if MATH is a bifurcation, the limiting complex MATH is acyclic if and only if MATH is acyclic for all non-bifurcations MATH, because the NAME isomorphism REF, as defined in CITE, can be extended by a limiting argument to MATH. When MATH is acyclic, we compute its torsion using REF . Choose subbases MATH and MATH as in REF for MATH. We can use these same subbases in the interval MATH for some MATH, because if the determinants in REF are nonzero in the limiting complex, then they are nonzero near MATH. The reason is that each determinant for the limiting complex has a nonzero ``leading term" involving flow lines of length MATH for some MATH, which will be unchanged near MATH by REF . For MATH we have MATH when the leading order of MATH exceeds the leading order of MATH and MATH by at least MATH. This means that a high order change in the denominator of MATH, as computed above, will change MATH by high order terms. We are now done by REF . |
math/9907066 | Since every bifurcation satisfies REF , it follows that the non-bifurcations are dense in MATH. Moreover every MATH is good. (If MATH is a non-bifurcation and fails to satisfy REF , then a compactness argument shows that MATH is a bifurcation after all, giving a contradiction.) By the assumptions and REF , we have MATH for each MATH. It follows from REF that if we fix an NAME structure MATH and MATH, then for all MATH, there exists MATH such that MATH for all non-bifurcations MATH. Since MATH is compact and the non-bifurcations are dense, it follows that MATH. Taking MATH, while keeping MATH fixed, completes the proof. |
math/9907066 | We will work with MATH families. After arranging REF , we will show that in the space of MATH families, there is a countable intersection of open dense sets, whose elements are families with the desired REF . As in CITE, we then obtain a dense set in MATH. We begin by making the graph of MATH transverse to REF-section of MATH. Then MATH is a smooth REF-dimensional submanifold of MATH. We can further arrange that MATH is a NAME function on MATH such that all critical points have distinct values. A critical point of MATH on MATH is a pair MATH where MATH has a degenerate zero at MATH. By REF we can choose (possibly time-dependent) local coordinates MATH near such a point so that MATH . We now fix the metric MATH on MATH to be Euclidean near the origin in these coordinates. This gives REF . By a standard transversality argument, we can obtain REF in a countable intersection of open dense sets. Fixing the metric near the degenerate critical points does not interfere with the transversality argument because no flow line or closed orbit is completely supported near a degenerate critical point. To obtain REF , we first arrange for the space of closed orbits to be cut out transversely. We then use a compactness argument to show that REF for each MATH, only finitely many bifurcations of length MATH occur. We can arrange that these bifurcations occur at distinct times, by intersecting with an open dense set of deformations. So in a countable intersection of open dense sets, we can arrange that REF all bifurcations occur at distinct times. Now REF imply REF . |
math/9907066 | By the definition of semi-isolated, we may choose MATH such that for all MATH with MATH, there are no degenerate or broken flow lines from MATH to MATH. As in the proof of REF , the moduli space of flow lines from MATH to MATH for MATH is compact, so MATH, since the signed number of boundary points of a compact REF-manifold is zero. For every MATH, for every pair of critical points MATH with index difference REF and MATH, the coefficient MATH likewise does not change for MATH sufficiently close to MATH. For every MATH, the coefficients in the zeta function of MATH with MATH do not change for MATH sufficently close to MATH, by REF . |
math/9907066 | The NAME complex is unchanged as in the proof of REF . To show that the zeta function does not change, the idea is that locally the zeta function looks like REF, and this is invariant because the signed number of fixed points of a map is invariant, assuming suitable compactness. More precisely, at time MATH there is an isolated irreducible closed orbit MATH, with MATH, such that MATH or some multiple cover of it is degenerate. Choose MATH, and let MATH be a disc of radius MATH transverse to MATH and centered at MATH. Let MATH be the (partially defined) first return map for the flow MATH. We restrict the domain of MATH to a maximal connected neighborhood of MATH on which it is continuous. Define MATH for non-bifurcations MATH. We claim that MATH . To prove this, given MATH, we must find MATH such that MATH. By the definition of semi-isolated, there exists MATH so that for MATH, all closed orbits MATH with MATH are nondegenerate, except for covers of MATH at time MATH. By compactness as in REF , we can choose MATH sufficiently small that no such closed orbit (other than covers of MATH) intersects MATH at time MATH. Then for MATH sufficiently small, the contribution to MATH from closed orbits MATH avoiding MATH with MATH does not change, and when moreover MATH is not a bifurcation, the contribution to MATH from all other closed orbits MATH with MATH is counted by the order MATH terms of MATH, as in REF. This proves REF. Given any positive integer MATH, as above we can choose MATH such that at time MATH, no closed orbit MATH with MATH (other than covers of MATH) intersects MATH. In particular, for MATH, the boundary of the graph of MATH does not intersect the diagonal in MATH. (Here we are compactifying the graph as in the proof of REF , see also CITE.) It follows that MATH is independent of MATH for non-bifurcations MATH close to MATH. This implies that MATH . Together with REF, this proves the lemma. |
math/9907066 | For each flow line from MATH to MATH at the bifurcation time, a flow line from MATH to MATH is created or destroyed. This follows from a standard gluing argument CITE and can also be seen using finite dimensional methods as in CITE. Similarly, for each flow line from MATH to MATH at the bifurcation time, a flow line from MATH to MATH is created or destroyed. By REF , no other flow lines or closed orbits are created or destroyed or change sign. This implies REF , after a check that the orientations are consistent. REF follows by REF , since MATH. |
math/9907066 | REF is easy. To prove REF , let MATH be a primitive MATH root of MATH. For MATH, define a ring homomorphism MATH by MATH for MATH. By CITE, we have MATH . The first of these identities implies that for MATH, MATH (which can also be seen more directly). This proves REF . To prove REF , we compute MATH . Here the middle equality holds because MATH is defined using a power series (see REF ) and MATH is a ring homomorphism. To prove REF , observe that REF implies that MATH respects the field decomposition of MATH. REF now follows from REF and the injectivity of MATH. |
math/9907066 | CASE: Every closed orbit MATH in MATH is a lift of a unique closed orbit MATH in MATH, with MATH. Conversely, if MATH and MATH, let MATH denote the period one orbit underlying MATH, and let MATH be the order of MATH in the group MATH. Then MATH lifts to MATH distinct closed orbits MATH, each of which has period MATH and NAME sign MATH. Therefore MATH . By REF , MATH . Combining the above equations and applying MATH proves REF . REF A finite free complex MATH over MATH can be regarded as a complex MATH over MATH with MATH times as many generators. Moreover, a basis MATH for MATH over MATH determines a map MATH, and if MATH then the map MATH is independent of the choice of MATH. Now we observe that if MATH, then the NAME complex MATH, with the basis determined by MATH, is obtained from MATH and MATH by this construction. So we need to show that MATH. REF implies that MATH and MATH are compatible with the decompositions of MATH and MATH into sums of fields. So we can restrict attention to a complex MATH where MATH is a field; let MATH denote the corresponding field in MATH. If MATH is not acyclic, then MATH is not acyclic either, so both torsions are zero. If MATH is acyclic, we can decompose it into a direct sum of REF acyclic complexes. Our claim then reduces to the fact that if MATH is a square matrix over MATH and MATH is the corresponding matrix over MATH, then MATH. This follows from the definition of MATH, after putting MATH into NAME canonical form over an algebraic closure of MATH. |
math/9907066 | CASE: Let MATH denote the divisibility of MATH in MATH. (Note that MATH is not a torsion class.) Let MATH be a positive integer relatively prime to MATH, and let MATH be a homomorphism sending MATH. Let MATH be the MATH-fold cyclic cover with monodromy MATH. Then the critical points MATH project to distinct points in MATH. Let MATH. By semi-isolatedness, we can find MATH such that no bifurcation of length MATH occurs between time MATH and MATH, other than the slide of MATH over MATH. Choose a smaller MATH if necessary so that the pairs MATH are admissible. Perturb the pulled back family MATH, fixing the endpoints, to satisfy the genericity conditions of REF . By a compactness argument, we can choose the perturbation small enough that no bifurcations of length MATH occur other than slides of MATH over MATH. Then iterating REF and using REF , we find a power series MATH such that MATH . (Here MATH indicates a term involving flow lines MATH with MATH.) Without loss of generality, MATH or MATH for some MATH (since otherwise REF is vacuously true for any MATH). Then REF implies that for MATH fixed, MATH is constant for large MATH. If we define MATH to be this stable value of MATH, then REF follows. REF follows from REF . Now recall that the slide of MATH over MATH comes from a single transverse crossing of ascending and descending manifolds. Under a sufficiently small perturbation of the deformation in MATH, this crossing will persist, and no other such crossing will appear, by a compactness argument. So for a sufficiently small perturbation, MATH, and hence MATH. This proves REF . |
math/9907066 | CASE: By REF , a closed orbit can be created or destroyed in the bifurcation only if it is homologous to MATH for some MATH. So MATH is a power series in MATH. Thus MATH is a power series in MATH. (A priori the coefficients MATH are rational; it's not important here, but we actually know that MATH, due to the product REF for the zeta function.) REF Let MATH be a compact ``tubular" neighborhood of the flow line MATH from MATH to itself at MATH. There is a function MATH such that MATH for some MATH. Let MATH be a level set for MATH away from MATH. The flow MATH induces a partially defined return map MATH. Closed orbits homologous to MATH near MATH are in one to one correspondence with fixed points of MATH. A fixed point of MATH is an intersection of the diagonal MATH with the graph MATH, and the NAME sign of the closed orbit equals the sign of the intersection. The graph MATH has a natural compactification (see CITE) to a manifold with corners MATH whose codimension one stratum is MATH . Here MATH and MATH are the ``first" intersections of the descending and ascending manifolds of MATH with MATH, and MATH is a component arising from trajectories that escape the neighborhood MATH. The number of closed orbits near MATH changes whenever MATH crosses MATH. This is happening at time MATH at a single point, transversely, and an orientation check shows that the sign is MATH. No other closed orbits homologous to MATH can be created or destroyed, as in REF . |
math/9907066 | By REF , we can write MATH for some MATH. We need to show that each coefficient MATH vanishes. Let MATH denote the divisibility of MATH in MATH. Let MATH be a homomorphism which sends MATH and annihilates the torsion subgroup of MATH. Let MATH be the corresponding finite cyclic cover. By REF , MATH . As in REF , we can choose MATH sufficiently large that a bifurcation of length MATH in MATH near MATH will not affect terms of order MATH in MATH or MATH. Now perturb the deformation in MATH as in the proof of REF , so that modulo bifurcations of length MATH, there are only slides of MATH over MATH. When MATH does not divide MATH, we know by REF that the torsion and zeta function in MATH do not change in such a slide. When MATH divides MATH, we apply the analogue of REF in the covering MATH, to conclude that MATH gets multiplied by MATH. It follows that MATH, as long as we know that MATH. If MATH is acyclic for at least one of the subfields MATH of MATH, then MATH, and it follows from REF that MATH, completing the proof. If MATH is not acyclic for any MATH, then MATH and we have nothing to prove. |
math/9907066 | In the calculations below, we will omit all orientations. We first note that if MATH are disjoint from MATH, then no flow lines from MATH to MATH are destroyed, that is, there is a natural inclusion MATH. To see this, suppose to the contrary that a flow line is destroyed. Then by compactness there is a sequence of flow lines from MATH to MATH before the bifurcation converging to a broken or degenerate flow line from MATH to MATH at time MATH. There are no degenerate flow lines at MATH (by the definition of semi-isolated), so the limit flow line is broken, and the only place it can be broken is at MATH. In the neighborhood MATH, the broken flow line approaches MATH in the half space MATH and leaves MATH in the half space MATH. But such a broken flow cannot be the limit as MATH of unbroken flow lines at time MATH, because there is a ``barrier": At time MATH, a downward flow line cannot cross from MATH to MATH within the neighborhood MATH, since the downward gradient flow is in the positive MATH direction for MATH. Likewise, there is a natural inclusion MATH. To analyze what gets created, choose a small MATH and let MATH. Let MATH and MATH. For MATH small, let MATH denote the partially defined map given by downward gradient flow at time MATH. Consider a broken closed orbit obtained by concatenating flow lines MATH (in downward order) from MATH to itself. Choose MATH small enough so that each MATH crosses MATH immediately after leaving MATH and crosses MATH immediately before returning. Let MATH and MATH denote the corresponding intersections of MATH with MATH. The downward flow defines a return map MATH from a neighborhood of MATH in MATH to a neighborhood of MATH in MATH. A new closed orbit approximating the broken one gets created for each fixed point of the partially defined map MATH near MATH. We will prove below that MATH . It follows that for MATH small, the graph of REF intersects the diagonal once near MATH transversely, because MATH intersects MATH once transversely at MATH. This proves REF . (Note that no additional closed orbits can be created, because by compactness a closed orbit can be created only out of a broken closed orbit as above.) To prove REF , suppose we have a broken flow line from MATH to MATH at time MATH consisting of a flow line MATH from MATH to MATH, followed by the concatenation of MATH and a flow line MATH from MATH to MATH. Let MATH and MATH denote the corresponding intersections with MATH and MATH of the descending manifold of MATH and the ascending manifold of MATH. Let MATH and MATH. A new flow line is created for each intersection of the graph of the partially defined map MATH with MATH near MATH. We will prove below that MATH . It follows that for MATH small, the graph of REF intersects MATH once transversely near MATH, because MATH intersects MATH transversely at MATH, and MATH intersects MATH transversely at MATH. This proves REF . We now prove REF . We first note that by the local model REF, we have MATH . In general, if MATH are manifolds and MATH and MATH are any smooth maps, then MATH intersects MATH transversely in MATH and MATH where MATH denotes the projection. Using REF one proves REF together by induction on MATH. |
math/9907066 | By REF , we have MATH for MATH, and MATH . We can rewrite this as MATH . Now let MATH be a subfield of MATH, as in REF . Choose decompositions MATH as in REF . We can then get subbases for MATH satisfying the conditions of REF by taking MATH and MATH, and keeping the other subbases fixed. Let MATH denote the corresponding restrictions and/or projections of the MATH components of MATH. Using REF, we compute MATH . Putting this into REF and summing over subfields MATH, we obtain REF . To prove REF , let us write MATH where there is one MATH for each flow line from MATH to MATH at time MATH. Then MATH . The first equality is a consequence of REF ; the denominator MATH arises because summing over MATH-cycles and dividing by the period is equivalent to summing over MATH-tuples and dividing by MATH. The second equality can be verified by taking the logarithm of both sides. This proves REF . |
math/9907066 | Since the NAME complex is invariant under scaling, it makes no difference if we take MATH where MATH is small. Suppose MATH is a closed orbit of MATH. The homology class of MATH must be nonzero, since the cohomology class MATH pairs nontrivially with it. We can then put a lower bound on the length of MATH away from the critical points. Since there is a positive lower bound on MATH away from the critical points, we deduce a lower bound on MATH. If MATH is sufficiently small, then the closed orbit MATH cannot exist, or else we would get a positive lower bound on MATH, contradicting the fact that MATH. Transversality and intersection number are invariant under small perturbations, so if MATH is sufficiently small, then the critical points of MATH will be small perturbations of the critical points of MATH and remain nondegenerate, and the ascending and descending manifolds will still intersect transversely with the same intersection numbers. This implies admissibility and REF. |
math/9907066 | (compare CITE) Choose a splitting MATH where MATH is finite and MATH is free. Then MATH and MATH. The total quotient ring of MATH is a finite sum of (cyclotomic) fields, MATH . We then have MATH . A ``leading coefficients" argument shows that MATH and MATH are integral domains, so MATH and MATH are fields. Thus REF proves REF . |
math/9907067 | The case MATH follows from MATH . Now suppose by induction that MATH is odd, that MATH, and that MATH for all MATH such that MATH. We show that MATH, and this establishes the lemma by induction on (odd) MATH. First we have MATH . If we let MATH, we also have MATH . Now our hypotheses imply that MATH if MATH. So this equation implies that MATH also. Since MATH this gives MATH . But our inductive hypothesis implies that MATH for MATH. So we obtain MATH . Since REF imply that MATH, which also implies that MATH. This completes the proof of the lemma. |
math/9907068 | Let MATH denote the homothety class of MATH in MATH. Then the triple MATH depends on MATH parameters (compare REF). Let MATH, MATH, MATH and MATH be the torsion subsheaf of MATH. Further, let MATH and MATH be the inverse image of MATH in MATH. Thus we have three short exact sequences MATH in which all terms except MATH are vector bundles. The triple MATH is determined by the first and last sequences (up to homothety) and a map MATH whose cokernel is MATH. The triple MATH depends on MATH parameters and so the whole configuration depends on at most MATH parameters. Now, MATH and MATH are stable, so MATH and hence MATH and MATH. Furthermore MATH and MATH . Hence, using the bilinearity of MATH in short exact sequences, we get MATH . But now MATH and so MATH and hence MATH A simple dimension count (compare REF ) shows that, for general MATH and MATH to appear in sequences REF , it is necessary that MATH and MATH. In other words, MATH and hence MATH . Thus, subsituting this into REF , we finally deduce that MATH . This is only possible if REF MATH, REF MATH or MATH and REF unless MATH, we also have MATH. But REF means that MATH has maximal rank and then REF means that MATH is torsion-free, unless MATH. |
math/9907068 | After twisting by a suitable line bundle of positive degree, we may assume that MATH, for all MATH, and that every MATH is generated by global sections. Suppose that each MATH is of type MATH. Extending the usual dimension counting argument in the NAME MATH, we may choose MATH sections of MATH so that the induced map MATH is an isomorphism of the fibres at the general point of MATH and drops rank by only REF at other points. Thus the cokernel of MATH is the structure sheaf MATH of a subscheme MATH of degree MATH in MATH, that is, MATH is an extension of MATH on top of MATH. The parameter space of such subschemes MATH is the MATH-fold symmetric product MATH, which is an irreducible algebraic variety and which carries a universal family MATH. Since MATH is torsion, MATH for all MATH and all MATH. Hence there is a vector bundle MATH whose fibre above the point MATH is MATH and this carries a tautological family of extensions MATH . Letting MATH and MATH, we may replace MATH by the non-empty open set on which MATH and MATH are vector bundles and obtain the required irreducible familty of extensions of vector bundles. To see that every possible extension of MATH on top of MATH occurs note that every such extension has a REF step filtration with MATH on top of MATH on top of MATH. But, since MATH, the extension at the bottom of this filtration splits and so it is simply an extension of MATH on top of MATH. |
math/9907068 | To solve REF we simply need to solve MATH and set MATH. Given one solution MATH, the complete set of solutions is MATH which contains precisely one solution in the range REF . REF is provided by REF . For the main part of the proof, the first step is to construct a short exact sequence MATH with MATH of type MATH, MATH of type MATH and MATH. First suppose that MATH. Then REF implies that for generic MATH and MATH we have MATH and, since MATH, the generic map is surjective. Let MATH be a particular choice of such generic bundles and map and let MATH be the kernel. At this stage, we have MATH, but may not have MATH. On the other hand, MATH since MATH. Hence, REF also implies that, for generic MATH and MATH of types MATH and MATH respectively, MATH and, since MATH, the generic map is an injection of vector bundles. Let MATH be a particular choice of such generic bundles and map and let MATH be its cokernel. This time, we have MATH, but may not have MATH. But now we may include MATH and MATH in an irreducible family MATH by REF . Then, by REF , there is an irreducible family of extensions MATH which includes both MATH and MATH. Hence we may choose for REF a general extension in this family and both MATH groups will vanish as required. For an arbitrary value of MATH, we may obtain a sequence of the form REF by taking the direct sum of MATH copies of one for MATH. For the second step, suppose that we have a sequence of the form REF . By REF , we may include MATH in an irreducible family MATH, whose generic member is general and for which every member satisfies MATH. There is then a vector bundle MATH whose fibre at MATH is MATH and over MATH there is a tautological map MATH. Replacing MATH by the non-empty open set on which MATH is injective, we have MATH of type MATH. We may further replace MATH by the non-empty open set on which MATH. Now observe that the homomorphism MATH must be isomorphic to MATH, because a linear dependence between the MATH components of the homomorphism from MATH to MATH would imply that MATH is a summand of the kernel, which would contradict MATH. If we consider just the family MATH, then, as before, we may include this in an irreducible family MATH, whose generic member is general and such that MATH and MATH for every MATH. Hence the kernel MATH of the homomorphism MATH is general and satisfies MATH, because this was already true over MATH. Thus we have all the properties we require. |
math/9907068 | After restricting to suitable open subvarieties and taking the pullback along MATH we may assume that MATH is the identity map. We have two distinct MATH bundles. These have the same NAME class if and only if over a suitable open subvariety of MATH the associated bundles of central simple algebras are isomorphic or equivalently the two MATH bundles are isomorphic over this open subvariety. |
math/9907068 | Let MATH and MATH be vector bundles of weight MATH. Then MATH is a vector bundle of weight MATH and MATH is a vector bundle over MATH. It has a structure of a bimodule with MATH acting on the left and MATH acting on the right. Over the generic point of MATH it defines a NAME equivalence between (the generic fibres of) MATH and MATH. Hence the NAME classes defined by MATH and MATH are equal; in other words the NAME class depends only on the weight. Now, if MATH, then MATH with the diagonal action of MATH is a vector bundle of weight MATH and MATH is the MATH-th tensor power of MATH. Since the class defined by MATH is MATH, the class defined by MATH is MATH. On the other hand, MATH with the diagonal action of MATH is a vector bundle of weight MATH. In particular, MATH is the sheaf of algebras opposite to MATH and therefore the class defined by MATH is MATH and, as above, the class defined by MATH is MATH. Finally, MATH is a vector bundle of weight MATH and defines the class REF. |
math/9907068 | Let MATH be a vector bundle of weight MATH and rank MATH. It is enough to deal with the case where MATH is a division algebra since the remaining cases are all matrices over this and hence the values for MATH that arise are all multiples of this. In particular, therefore, we may assume that MATH divides MATH. If MATH, there is nothing to prove so we may assume that MATH. Thus at the generic point of MATH, there is an idempotent of rank MATH. This idempotent is defined over some open subset of MATH which is MATH-equivariant since the idempotent is MATH-invariant and gives a decomposition MATH as a direct sum of vector bundles which are MATH-equivariant one of which has rank MATH. These bundles have weight MATH since they are subbundles of MATH which has weight MATH. |
math/9907068 | The first equality follows immediately from the fact that the lower diamond in REF is an equivariant pullback. The action of MATH on MATH over MATH lifts naturally to an action of MATH on the universal subbundle MATH over MATH. Hence MATH has weight MATH and so the NAME class on MATH represented by MATH is MATH by REF . As already observed, the pullback MATH of MATH to MATH is trivial with fibre MATH and so the quotient by MATH also gives a trivial bundle MATH with fibre MATH on MATH. But then MATH is equal to the NAME class represented by MATH, which is equal to the NAME class represented by MATH, completing the proof in the covariant case. The proof in the contravariant case is identical except that now MATH and MATH are trivial with fibre MATH so that MATH is the negative of the class represented by MATH. |
math/9907068 | The idea of the proof is that since MATH is general, the set of quotients MATH such that MATH, MATH and MATH is not empty. It is bijective to the set of MATH-dimensional subspaces MATH such that MATH is injective, MATH, MATH and MATH. We fill in the details below. Consider the open set in MATH parametrizing those MATH for which MATH. Over this open set there is a vector bundle MATH whose fibre at MATH is MATH. Since MATH acts with weight REF on MATH, it acts with weight MATH on MATH. We claim that MATH is birational to MATH. To see this, consider the contravariant partial frame bundle MATH whose fibre over MATH is naturally identified with the set of maps MATH for which the induced map MATH is injective. On an open subset in MATH, the map MATH is injective as a map of bundles and its cokernel MATH gives a point in MATH. Since MATH, but not MATH, is determined by MATH we see that MATH is birational to MATH and so MATH is birational to MATH. Since MATH is MATH, we deduce that MATH is birational to MATH. We can arrange all the rational maps we have considered above into the following diagram of the form of REF . MATH . Since MATH has weight MATH, the first and last formulae in REF give MATH which completes the proof. |
math/9907068 | If MATH divides MATH, then MATH and MATH is isomorphic to MATH by tensoring with a line bundle of degree MATH. This isomorphism may be taken to be MATH and it preserves the NAME class. Otherwise, we saw in REF how to construct a map MATH with MATH and we proved in REF that MATH . We construct the map MATH, by induction on the rank MATH, as the composite of the top row of the following commutative diagram of dominant rational maps which combines MATH and the NAME correspondence described in REF. The other elements of the diagram we will explain next. MATH . The maps MATH and MATH are of the same sort as MATH and may be assumed to exist by induction, since both MATH and MATH are less than MATH. Thus they are birationally linear and satisfy MATH because MATH by REF . The central square in the diagram is a pull back. In particular, MATH is the pull back of MATH along MATH and hence, by REF , it is a NAME bundle over MATH whose twisting is measured by MATH. Thus MATH and MATH are twisted NAME bundles associated to vector bundles of weight MATH over MATH and of ranks MATH and MATH respectively (see REF ). We will prove in REF below that this implies that there is a birationally linear map MATH such that MATH and hence MATH . The pullback MATH of MATH along MATH is birationally linear and satisfies MATH . Thus MATH is birationally linear and pulls back MATH to MATH. But by REF , this means that MATH pulls back MATH to MATH as required and to complete the proof we need to show that MATH is birationally linear. This follows from REF below, because, as we saw in REF, MATH is a twisted NAME bundle of MATH-dimensional subspaces of a vector bundle MATH of weight REF over MATH and the NAME class MATH is represented by a central simple algebra of dimension MATH. Thus although MATH is not locally trivial in the NAME topology when MATH, we can show that it is birationally linear since its generic fibre is birational to a NAME over a division algebra; this is not the way it is expressed in REF though the translation to this is fairly simple. |
math/9907068 | Let MATH be the central simple algebra given by the bundle of central simple algebras MATH over the field MATH. Then by REF has a left ideal of dimension MATH which is of necessity a direct summand of MATH. Therefore, over some dense open subvariety of MATH, MATH where MATH and MATH are bundles of left ideals for MATH and MATH. We may as well assume that this happens over MATH. We obtain a corresponding direct sum decomposition of MATH, MATH where MATH and MATH are MATH stable subbundles of MATH and hence both of weight MATH. Also MATH. Now consider the vector bundle MATH. Let MATH be the universal homomorphism of vector bundles defined on MATH and consider the map of vector bundles over MATH, MATH. This representation of MATH as a subbundle of MATH defines a map from MATH to MATH which is MATH-equivariant, injective and onto an open subvariety of MATH. Hence MATH which is a vector bundle over MATH is an open subvariety of MATH. |
math/9907068 | MATH is a bundle of left modules for MATH of rank equal to MATH and since MATH there is an open subvariety of MATH on which MATH for some vector bundle of left ideals MATH since this is true at the generic point of MATH. Hence we may assume that on MATH, MATH for MATH stable subbundles MATH and MATH. Let MATH be the universal subbundle on MATH. Then MATH and MATH are both vector bundles of weight MATH on MATH. We consider the vector bundle MATH over MATH. Let MATH be the universal homomorphism of vector bundles defined on MATH and let MATH be the universal inclusion of MATH in MATH pulled back to MATH; now consider the map of vector bundles MATH defined on MATH. This gives a subbundle of MATH of rank MATH and hence defines a map from MATH to MATH. This map is injective and onto an open subvariety of MATH and it is also MATH-equivariant. Hence MATH is an open subvariety of MATH. However, MATH is a vector bundle over MATH which gives our lemma. |
math/9907073 | The proof is modelled on REF. The coequalizer above is reflexive because the input arrows admit the common section MATH. Now MATH admits the structure of left MATH-module, because by REF MATH preserves reflexive coequalizers. Moreover MATH is the coequalizer of the pair above in the category of left MATH-modules. |
math/9907073 | We observe that MATH is obtained by gluing together for each face MATH of MATH of codimension MATH a copy of MATH. This is true because the codimension of a face of the manifold with corners MATH is equal to the number of internal edges of the associated tree. Then MATH admits the structure of manifold with corners diffeomorphic to MATH, and the composition maps of both operads are described by grafting of trees. |
math/9907073 | We build an extension MATH of MATH that is a map of operads and a weak equivalence. An element MATH in MATH is represented by a labelled tree MATH on MATH. If the MATH-tuple MATH labels a vertex MATH of valence MATH, then for each MATH we associate the embedding MATH to the MATH-th incoming edge MATH of MATH. The equivalence relation defining MATH preserves this association. We observe incidentally that the multicomposition of the labels of MATH is the MATH-tuple of embeddings MATH such that for each MATH is the composition of the embeddings associated to the edges along the unique path from the MATH-th twig to the root. Suppose that the internal edges of MATH are labelled by numbers in MATH. Let MATH denote the length of an edge MATH and if MATH let MATH be the dilatation of the MATH-disc by MATH. Consider the labelled tree MATH obtained from MATH by replacing for each vertex MATH and for each MATH the embedding MATH by the rescaling MATH. Let MATH be the multicomposition of the labels of MATH, and set MATH. We have defined MATH on a dense subspace of MATH. The map MATH extends to MATH and MATH is an operad map, because the boundary and the composition of MATH are described by a limit procedure. Let MATH be the inclusion such that MATH is represented by the tree on MATH with a single vertex labelled by MATH. The map MATH is a MATH-equivariant homotopy equivalence for each MATH because MATH is such CITE and MATH. In particular MATH is a weak equivalence of topological operads, and MATH. |
math/9907073 | The set of right homotopy classes MATH in the sense of CITE is the set of MATH-homomorphisms from a cofibrant model of MATH to a fibrant model of MATH modulo right homotopy. Now MATH is a cofibrant resolution of MATH, and MATH is fibrant because every object is such. It is easy to see that the the right homotopy classes of MATH-homomorphisms from MATH to MATH are ordinary homotopy classes, because MATH is a path object. |
math/9907073 | It is sufficient to observe that MATH is homeomorphic to MATH, and apply the homotopy invariance REF. |
math/9907073 | We choose a trivialization of the tangent bundle MATH. Then the composition MATH is described by grafting of trees representing elements as in REF. |
math/9907073 | We apply the same proof of REF to show that MATH, and conclude by the homotopy equivalence MATH. |
math/9907073 | If MATH is a proper MATH-monoid, then we have relative homeomorphisms MATH for MATH, and we conclude by REF If MATH is a partial MATH-monoid, then we denote by MATH the space of reducible elements that are equivalent to an element of some MATH with MATH. For example, MATH . We have relative homeomorphisms MATH and we argue similarly. |
math/9907073 | It is sufficient to observe that there is a deformation retraction of right MATH-modules MATH. If an element MATH is represented by a finite number of labelled trees MATH based at distinct points MATH, then MATH is represented by the single tree obtained by connecting MATH to a root vertex labelled by the class MATH. |
math/9907073 | The proof makes use of the fact that a copy of the manifold with corners MATH lies inside its interior MATH, so the retraction MATH is a MATH-equivariant homeomorphism onto its image. We compare via this retraction the pushout diagram for MATH and the pushout diagram for MATH . Here we denote by MATH the subspace of MATH of those labelled configurations such that several points are concentrated in the same macroscopic location if and only if their labels are summable. The inclusion of the space on the left hand top corner of the first diagram into that of the second diagram is a homotopy equivalence, because MATH induces a common retraction onto a copy of the second space. The same holds for the spaces on the left hand bottom corner. We conclude by induction and the gluing lemma CITE. |
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