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math-ph/0107030 | We will show the theorem for MATH for notational simplicity. Generalization to arbitrary MATH is immediate. Let MATH. Divide the domain into squares MATH of side MATH. This gives rise to MATH columns MATH of MATH-grid in MATH, MATH. CASE: The number of MATH-cubes having a common point with the graph of MATH in column MATH is not greater than MATH. But MATH . Therefore MATH . Thus MATH . CASE: Set MATH. From REF it follows that the number MATH of MATH-cubes in columns MATH covering the graph of MATH and the variation of MATH satisfy MATH . Thus MATH . Therefore the number MATH of boxes covering the whole graph of MATH satisfies MATH . The same can be repeated for any direction thus MATH . CASE: An immediate corollary. Generalization to arbitrary MATH is achieved by observing that MATH. MATH . |
math/0107001 | That MATH is a homomorphism is clear in view of REF , so let us now show that MATH is an isometry. Since MATH acts transitively and by isometries on MATH, it will be enough to show that MATH for any MATH. Noticing that MATH we deduce that for any MATH, MATH. Now, for MATH, there exists MATH such that MATH is diagonal with positive coefficients. Since MATH is a ring homomorphism, it is the identity over the real numbers and so: MATH . Let us now prove that the embedding is totally geodesic. Take MATH in the image of MATH, so that MATH and MATH for MATH. There is a MATH so that MATH and MATH for MATH. Then MATH and MATH. Since MATH is obviously in the image of MATH, so will MATH be (that is, the whole flat containing MATH and MATH is in the image of the embedding). Since a geodesic MATH between MATH and MATH lies in any flat containing them, we conclude that MATH. |
math/0107001 | Let MATH. Up to multiplication by an element of MATH (which is an isometry), we can assume that MATH and MATH is a diagonal matrix (see the previous Remark). We write MATH with MATH and MATH and MATH in MATH. Consider the element MATH then MATH only has real elements except for MATH, and MATH for any diagonal element MATH. Applying REF to MATH, we get an embedding of MATH in MATH which contains MATH, and thus a totally geodesical embedding MATH whose image contains MATH and MATH. |
math/0107001 | Since MATH, the stabilizer of MATH is MATH, and the action is transitive because given any MATH, there exists a MATH such that MATH is diagonal. Since MATH, the element MATH defines an element in MATH (coming from MATH), and MATH. |
math/0107001 | An embedding of MATH in MATH defines an embedding of MATH in MATH by taking the images in MATH of the coefficients of an element in MATH. It will now be enough to see that the map MATH is well defined and a group isomorphism. Remember that for MATH a unit in MATH we denoted by MATH a diagonal matrix with MATH in the first place and MATH's in the second and third place. Now, notice (from the definition of MATH) that MATH for any MATH a unit in MATH and that trivially, for MATH, MATH. This shows that the image of MATH is contained in MATH. The map MATH being a restriction, it is a group homomorphism. Our map MATH is clearly injective, and because of REF we know that MATH maps the generators of MATH onto those of MATH. |
math/0107001 | Let MATH. Up to multiplication by an element of MATH (which is an isometry), we can assume that MATH and MATH is a diagonal matrix (use REF ). We write MATH with MATH and MATH and MATH in MATH. Consider the element MATH where MATH is a unit in MATH then MATH only has real elements except for MATH, and MATH for any diagonal element in MATH. Applying REF to MATH and MATH, we get an embedding of MATH in MATH which contains MATH and MATH, and thus an embedding MATH whose image contains MATH and MATH, and which is isometric by definition of the distance on MATH. To see that the embedding is totally geodesic is basically the same argument as in the proof of REF : Take MATH in the image of MATH, so that MATH and MATH for MATH. There is a MATH so that MATH and MATH for MATH. Now, the embedding MATH comes from an embedding MATH described in REF , so that MATH and MATH. Since MATH is obviously in the image of MATH, so will MATH be (that is, the whole flat containing MATH and MATH is in the image of the embedding). Since a geodesic MATH between MATH and MATH lies in any flat containing them, we conclude that MATH. We are now reduced to find an embedding of MATH in MATH containing MATH and MATH, but we get this one because of REF , and combining it with MATH we get the sought embedding. |
math/0107001 | CASE: Because of REF , there is an isometric copy of MATH in MATH containing MATH. Thus, because of REF there exists MATH and MATH an equilateral triangle in MATH such that the paths MATH, MATH and MATH are MATH-paths. Since the embedding is totally geodesic, the triangle MATH will be equilateral in MATH as well. CASE: Without loss of generality, we can assume that MATH and that MATH is diagonal, so that MATH for a MATH such that MATH. Because of REF there is a totally geodesic embedding MATH containing MATH, MATH and MATH. But since MATH comes from an embedding MATH we have that: MATH (for a MATH and MATH) so that the whole MATH is contained in the image of MATH in MATH and we thus can see MATH and MATH in MATH. Using REF, the triangle MATH is MATH-retractable in MATH, and since the embedding is totally geodesic, it will be MATH-retractable in MATH as well. |
math/0107001 | Take MATH and set, for any MATH, MATH . Because of REF we can cover MATH with MATH balls of radius MATH centered at each point of MATH, where MATH denotes the polynomial associated to the constant MATH in the definition of REF. Now, take MATH so that MATH is a MATH-path. Since in particular MATH, we can find MATH at a distance less than MATH, and thus MATH is in the ball of radius MATH centered at MATH. It is now obvious that MATH is a MATH-path. But in each ball of radius MATH there is a uniformly bounded number MATH of elements of MATH, so that MATH and thus MATH satisfies property MATH, choosing MATH. |
math/0107001 | Because of REF , it is enough to show that one particular uniform net has property MATH. To do that, let MATH be uniform nets and look at MATH, which is a uniform net in MATH. Take MATH, MATH and MATH. We write MATH in coordinates, that is to say MATH and MATH, where MATH for any MATH. On each factor MATH, MATH will be a MATH-path as well, and since we are considering the MATH combination of norms, MATH. On each MATH there is by assumption at most MATH of those points MATH and thus on MATH we have at most MATH's at distance to MATH less than MATH and such that MATH is a MATH-path. Obviously MATH is again a polynomial (of degree the sum of the degrees of the MATH's) and thus MATH has property MATH. |
math/0107001 | CASE: For any MATH it exists an equilateral triangle MATH in MATH such that the paths MATH, MATH and MATH are MATH-paths. Indeed, this is because of REF in case MATH is MATH for MATH or MATH, because of REF in case MATH is a MATH-type building and trivially in case MATH is a hyperbolic space. Now, setting MATH and MATH we see that by construction the triangle MATH is equilateral and that the path MATH is a MATH-paths for any MATH since MATH and similarly for the paths MATH and MATH. CASE: Here we have to find a MATH and MATH, MATH in MATH such that the paths MATH and MATH as well as the paths MATH and MATH are MATH-paths. But for any MATH, the triangles MATH and MATH are equilateral triangles and thus because of REF in case MATH is MATH for MATH or MATH, trivially in case MATH is a MATH-type building or a hyperbolic space, the triangles MATH and MATH are MATH-retractable, that is there exists MATH and points MATH and MATH so that the paths MATH and MATH as well as the paths MATH and MATH are MATH-paths. Now the points MATH and MATH are the sought points on which the triangles MATH and MATH retract, for MATH. Indeed, let us check that for the path MATH: MATH and similarly for the paths MATH and MATH as well as the paths MATH and MATH. |
math/0107001 | Take MATH (for MATH as in REF ). We first have to define the MATH-invariant subsets of MATH. Let us consider MATH as explained in REF , that is, MATH is a set of indices running along the possible orientations of equilateral triangles in MATH. For any MATH, we define MATH . In other words, MATH is the set of triples of MATH which are not too far from an equilateral triangle, and since MATH acts on MATH by isometries, the sets MATH are MATH-invariant. We then set, for any MATH: MATH . Since MATH is MATH-equivariant, MATH is MATH-invariant for any MATH. Let us explain why then MATH satisfy property MATH. Because of the distance defined on MATH, it is enough to prove MATH and MATH for MATH and the sets MATH. MATH: Take MATH, we have to show that there exists MATH in some MATH so that the triple MATH retracts on MATH. But because of REF of REF , we know that there exists MATH, forming an equilateral triangle and so that the triple MATH retracts on MATH. Now MATH being a uniform net in MATH, there exists MATH three points of MATH with MATH, MATH and MATH, so that the triple MATH belongs to MATH for some MATH. We compute: MATH and similarly for the paths MATH and MATH. MATH: Take MATH and four points in MATH defining two triples in MATH, say MATH and MATH. We have to prove that both triangles MATH and MATH are MATH-retractable. By definition of MATH we can find two equilateral triangles MATH and MATH such that MATH and MATH so that obviously MATH and MATH and thus applying REF of REF we have the existence of a MATH, and two points MATH and MATH in MATH so that the paths MATH and MATH as well as the paths MATH and MATH are MATH-paths. Again, MATH being a uniform net in MATH, we can find MATH and MATH in MATH at respective distances less than MATH to MATH and MATH. We claim that the paths MATH, MATH and MATH as well as the paths MATH, MATH and MATH are MATH-paths: MATH and similarly for the other paths. |
math/0107002 | Assertion MATH is known as the NAME - NAME Theorem (see CITE and CITE). A proof of this and the next REF assertions can be found in CITE. The assertion in REF was proved independently by CITE and CITE. Finally REF are due to CITE. |
math/0107002 | Let us begin by presenting a proof of the assertion in REF . Rotating and translating if necessary, we may assume that MATH, MATH lies in the upper half plane and that the positive imaginary axis intersects the interior of MATH. In this case we may find a convex function MATH defined on an open interval containing REF whose graph gives a portion of the boundary of MATH which contains REF. Since REF is a singular point of the boundary, MATH is not differentiable at REF and since MATH is convex, it follows that MATH . Thus, the lines though MATH with slopes MATH are tangent to MATH. NAME established REF. The proof offered below seems to be new. It is convenient to rotate once more so that MATH and MATH lies in the right half plane so that the corner has the form shown below where both MATH are tangent to MATH at REF, MATH has positive slope and MATH has negative slope as shown below. Since MATH is contained in the right half plane, we get MATH and since there is a unit vector MATH such that MATH, we get MATH by the remark following REF . We may now select MATH so small that MATH is also contained in the right half plane. Arguing as above, we get that MATH and so MATH and therefore MATH. Hence, MATH must be a reducing eigenvalue for MATH. REF is due to CITE. |
math/0107002 | The assertion in MATH is due to NAME who introduced the MATH - numerical range in his thesis CITE. A proof may be found in CITE. The assertion in MATH follows from REF and a simple calculation. REF follows from the fact that MATH. The proof of REF is straight forward. For assertion MATH observe that MATH so that MATH for some real MATH. Further, since MATH is the sum of the MATH largest eigenvalues of MATH, if MATH, then MATH. |
math/0107002 | Since MATH is an extreme point of MATH, it has the form MATH where MATH is a projection in MATH by CITE. If MATH has rank MATH, then MATH and so MATH, where MATH. Hence, MATH. |
math/0107002 | If MATH is a projection of rank MATH, then it an extreme point of MATH and so it is also an extreme point of MATH. For the converse suppose MATH is in MATH, but MATH is not a projection and write MATH for the eigenvalues of MATH. Since MATH we get that MATH and since MATH is not a projection we have MATH for at least one index MATH. Since MATH, there must also be an index MATH such that MATH. Relabeling if necessary, we may assume that MATH. Since MATH and MATH we may select MATH and MATH such that MATH . Now write MATH . Observe that MATH because MATH. Since the diagonal entries of MATH and MATH lie in MATH and sum to MATH, these matrices are elements of MATH. As MATH there is a real number MATH with MATH, such that MATH. Next, note that MATH and therefore MATH. Thus MATH is not an extreme point. |
math/0107002 | If MATH, then by part-MATH of REF there is a projection MATH with rank MATH such that MATH and the point MATH is in MATH because MATH. Thus, MATH . So, MATH maps MATH into MATH, and MATH is clearly a one-to-one map. Now suppose that MATH so that MATH for some MATH. By REF , the NAME - NAME Theorem CITE and CITE we have that MATH is a convex combination of projections of rank MATH. Since MATH is convex by REF and points of the form MATH are in MATH by REF, we get that MATH. |
math/0107002 | Fix an extreme point MATH of MATH. If MATH is a projection, then the proof is complete. So, assume that this is not the case, let MATH denote the distinct eigenvalues of MATH and let MATH denote the corresponding eigenprojections. Since MATH is not a projection, there is at least one index MATH such that MATH. If there were another index MATH with MATH, then by adding (respectively, subtracting) and small amount to MATH and subtracting (respectively, adding) a small amount to MATH we would get two new elements of MATH whose average is MATH so that MATH would not not be an extreme point of this set. Hence, MATH and all other indices are REF or REF. If MATH did not have dimension one, then it would contain two nonzero orthogonal projections and, arguing as in the previous paragraph, we would again get that MATH is not an extreme point. Hence, MATH has the indicated form. |
math/0107002 | As MATH has dimension two, there are two distinct sectorial tangent lines of support of MATH in the plane of MATH at the corner MATH. Denote these tangent lines by MATH and MATH. Each MATH meets B only in the boundary of the isotrace slice MATH and so they are each disjoint from the interior of B. By CITE, there are distinct planes MATH and MATH such that MATH and each MATH is disjoint from the interior of MATH. Since each plane contains MATH, they are planes of support for MATH. Thus, if we write MATH, then MATH is a sharp face of MATH. Since we are now regarding MATH as a subset of MATH we may use the results in CITE and CITE to get that each hyperplane of support is determined by a spectral pair of the form MATH, where MATH is a real number and MATH is a nonzero vector in MATH . Specifically, by CITE if MATH is a hyperplane of support for MATH, then there is a spectral pair MATH such that points MATH satisfy an equation of the form MATH . The constant MATH is determined as follows. We write MATH and let MATH and MATH denote the spectral projections of MATH corresponding to the intervals MATH and MATH. With this, we have MATH . Observe that MATH is a normal vector for this plane With this, since MATH and MATH are distinct faces of MATH that have nonempty intersection, their normal vectors are linearly independent and so there exist linearly independent spectral pairs MATH and MATH such that each MATH has the equation MATH where MATH. Since the point MATH lies in each plane, we get MATH . Now fix a point MATH on the tangent line MATH and observe that since MATH lies in the plane MATH, we get MATH. Since this point also lies in the plane MATH, we get MATH and so, MATH . If the vectors MATH and MATH were linearly dependent, then it would follow that the tangent lines MATH and MATH are parallel. Since these lines intersect at a corner of the isotrace slice MATH, this is impossible and therefore the MATH's must be linearly independent. Applying CITE, we get that MATH has the form MATH, where MATH are central projections. Next, since MATH is a face in MATH, the end points MATH are extreme points of MATH. Since the projections MATH are central, the points MATH are isolated extreme points of MATH by CITE and so assertions MATH and MATH hold. |
math/0107002 | Since MATH is a face, its intersection with MATH is a face of MATH, which consists of the single extreme point MATH , by REF and the fact that faces of MATH are transverse to isotrace slices by CITE. If MATH were not an isolated extreme point of MATH, then there would be a sequence MATH of extreme points in MATH that converges to MATH. Since the inverse images of extreme points contain extreme points, each of these points would have the form MATH or MATH, where MATH and MATH are as described in REF . We shall assume that, after passing to a subsequence, MATH and further that MATH is one dimensional. The other cases are handled similarly, but somewhat more easily. Since the dimension of MATH is finite, the range of the trace on the projections in MATH is finite and since MATH for each MATH, the MATH's also form a finite set. So, by passing to another subsequence, we may assume that all MATH and MATH. We can now find another subsequence (which we continue to index with MATH's) such that MATH converges to MATH and MATH converges to MATH . It follows readily that MATH and MATH are orthogonal projections, and MATH has dimension one. Since MATH we have MATH and since MATH is one dimensional, we get that MATH. Hence, MATH is a scalar multiple of MATH for MATH by CITE. Since MATH is central, MATH is a multiple of MATH. Thus, MATH and so this element is central. But, since MATH this means that both MATH and MATH are central. This is impossible since the central projections are isolated in the set of all projections. Hence, MATH must be an isolated extreme point of MATH. |
math/0107002 | As in the proof of MATH in REF and the remark following this theorem, we have that MATH because the boundary of MATH is convex. Now suppose that MATH holds. In this case MATH lies at a corner of the isotrace slice MATH and so we may apply REF to get that REF and MATH of REF are true. Hence, if MATH and MATH hold, then for each MATH, there are faces MATH and MATH in MATH of dimension two such that MATH by CITE and REF . If MATH has dimension zero so that it is a single point, then we must have MATH and so MATH is true in this case. Now suppose that MATH has dimension one, fix MATH in the relative interior of MATH and consider the face MATH. If MATH were a single point, then it would be an extreme point of MATH, which is impossible because it lies in the relative interior of MATH. Hence MATH has dimension one. Since MATH is contained in MATH we must have MATH. Since it is obvious (and straightforward to prove) that three distinct faces of dimension two in MATH cannot intersect in a face of dimension one, we must have that MATH is either MATH or MATH. Hence REF holds in this case. Thus, in all cases we get that the faces MATH have dimension two. Since such faces are transverse to the isotrace slices of MATH by CITE, each line MATH intersects MATH in its relative interior. Thus, MATH. Next, if MATH is true, then it is clear that MATH is an isolated extreme point of MATH and so MATH. If MATH holds, then it is also clear that MATH holds and so these four conditions are equivalent. |
math/0107002 | Since each MATH is a multiple of the isotrace sliced MATH of MATH, the theorem is equivalent to showing that MATH has a finite number of extreme points if and only if MATH is normal and this is precisely what is asserted in CITE . |
math/0107002 | This is well known. See CITE, for example. |
math/0107002 | Since MATH has no repeated factors, its discriminant is not identically zero. Since the discriminant is a polynomial in MATH, it has a finite number of roots. Let MATH denote the roots of MATH. If MATH, then MATH and so MATH has MATH distinct roots. |
math/0107002 | This is basically due to CITE, although it is not explicitly stated there. He observed that if MATH, then MATH (The formula presented in CITE appears to contain a typographical error since it differs from the one offered above by a minus sign in the second coordinate). Hence, if MATH, then MATH is differentiable, and if MATH is differentiable, then it follows from MATH that this limit exists. Next, NAME showed in CITE that if MATH where MATH comes from moving counterclockwise around MATH, then MATH . Hence, if the boundary of MATH contains a line segment with end points MATH, then MATH does not exist and so MATH is not differentiable by the first part of the proof. Conversely, if MATH is not differentiable, then MATH does not exist and so MATH . Since MATH, the line segment joining MATH and MATH lies on the boundary of MATH. |
math/0107002 | We use the notation introduced in REF . Let MATH denote an open interval in MATH whose end points are adjacent points in MATH. We have then that the functions MATH are real analytic and completely distinct on MATH by REF . Hence, we may re-index these functions as MATH for MATH and assume that MATH for each MATH. Similarly, with additional re-indexing, we may assume that for each fixed MATH the eigenvalue MATH has multiplicity MATH. Hence, we may find positive integers MATH and MATH such that MATH and MATH . With this we get that MATH, which is the sum of the MATH largest eigenvalues of MATH by REF , is precisely MATH and so the map MATH is real analytic on MATH by REF . Since the set MATH is finite, this argument shows that MATH is differentiable on MATH, with the possible exception of a finite subset of MATH. Hence,the boundary of MATH contains a finite number of line segments by REF and the remaining boundary points are contained in a finite number of curved real analytic arcs. |
math/0107002 | If MATH is a face in MATH of dimension two, then MATH is transverse to the isotrace slices of MATH by CITE. Hence, it must intersect at least one isotrace in a line segment by CITE. Since each isotrace slice contains only a finite number of line segments by REF , MATH can have only a finite number of faces of dimension two. |
math/0107003 | First recall the fusion procedure in the case of one input and MATH outputs. Suppose that MATH and MATH are two sets of vectors such that MATH for any MATH, MATH. Then we have the natural projection MATH. Therefore the dual spaces form an inductive system and we can consider the limit MATH . The action of the algebra MATH on MATH can be extended to the action of the algebra MATH on the limit MATH, where summands correspond to points MATH. The statement that the set of representations MATH form a minimal model means that the representation MATH decomposes into the direct sum of external tensor products MATH . Emphasize that the sum goes over the set of representations whose product in the NAME algebra is MATH. Clearly, the space MATH is dual to the space of MATH-invariants in MATH. As the subspace MATH is dense with respect to the topology of direct limit in the subspace MATH, we have MATH that is equivalent to the statement of the theorem. |
math/0107003 | Note that MATH contains the subalgebra MATH therefore the space of MATH-invariants in MATH is spanned by several vectors MATH, MATH. Taking the invariants of MATH and MATH selects the vectors MATH with MATH. We need to show that these vectors are invariant. Note that if MATH then MATH. And we know that the vector MATH is annihilated by the subspace MATH. |
math/0107003 | From REF we can deduce that all relations between MATH hold for their images in MATH. So REF defines a homomorphism. As the image contains all the generators of MATH, this map is surjective. |
math/0107003 | It follows from the observation that the highest weight vector of MATH satisfy the same annihilation conditions as MATH. |
math/0107003 | Note that MATH contains linear combinations of the form MATH for MATH. Therefore MATH contains MATH for MATH hence (according to REF) it contains the image of MATH. |
math/0107003 | Indeed any set of polynomials MATH satisfying REF can be expressed in the following way: MATH where MATH and MATH are certain polynomials in MATH. In our case we need only to check that MATH and MATH. |
math/0107003 | REF follows from the fact that MATH if MATH. REF follows from the fact that MATH if MATH and MATH. |
math/0107003 | Suppose that MATH and MATH. Then adding REF for MATH twice and subtracting REF for MATH (if MATH) and MATH (if MATH; if MATH then for MATH and MATH) we obtain MATH. |
math/0107003 | Proof is similar. Here we add REF for MATH and for MATH then subtract REF for MATH. |
math/0107003 | Suppose that MATH and MATH. Then adding REF for MATH twice and subtracting REF for MATH we obtain MATH . Taking into account MATH and MATH we deduce MATH. |
math/0107003 | Identity REF follows from the identity REF applied to the factor MATH in each summand. |
math/0107003 | First let us check that the sum in the right hand side is indeed finite. Note that if MATH then the third factor is always zero, so entire sum is zero. If MATH then either MATH or MATH. In the first case we have non-zero summands only if MATH, in the second case only if MATH. Let us denote the polynomial in the right hand side by MATH. The idea of proof is to apply REF to MATH as a set of polynomials indexed by MATH and MATH. It means that we prove that MATH for any MATH,MATH,MATH,MATH and MATH for any MATH, MATH and certain MATH, MATH. To prove REF fix for a moment MATH, MATH, MATH, MATH and introduce the notation MATH . Then substitution MATH into MATH leads to the identity MATH . Using this identity together with REF we obtain MATH where sums go over MATH. Proof of the second part of REF is similar. To check REF we take MATH and any negative MATH. We have only one non-zero summand in MATH, namely for MATH, and it coincides with the MATH-binomial given in REF. If MATH and MATH then MATH and, as described above, the result is zero. So REF implies the lemma. |
math/0107003 | First let us check that the sum in the left hand side is indeed finite. As MATH we have either MATH or MATH. In the first case we have non-zero summands only if MATH, in the second case only if MATH. Let us denote the polynomial in the left hand side by MATH. Clearly, we have MATH and the same identity for the right hand side. Therefore we can restrict ourselves to the case MATH and prove that MATH . Applying REF to the second factor in MATH, we obtain MATH that is REF for the right hand side of REF. Therefore it remains to check REF for MATH. If MATH then the only non-zero summand in MATH is for MATH and it is equal to the right hand side of REF . |
math/0107003 | First, note that due to the action of automorphisms MATH we can consider only coinvariants of the representation MATH. The main idea of the proof is to check that the main REF is indeed an equality. Here we have a chain of inequalities, where we know the first element by REF and the last element. As soon as we check that these two numbers are equal we obtain that other numbers in REF are also equal. Suppose that the vector MATH satisfies the well-balance REF . Then this equality follows from REF by setting MATH and MATH. But if REF are not satisfied then this equality can be violated. So we should deduce general case from the well-balanced one. To this end consider the family of automorphisms MATH. They preserve the representation MATH and permute the subalgebras MATH. Namely, MATH preserves MATH, increases MATH by MATH and decreases MATH by MATH. Recall that MATH. First consider the case MATH. Then we should set MATH. Let MATH be the minimal number such that MATH. So we have MATH and as MATH we obtain MATH. Hence after applying the automorphism MATH the well-balance conditions will be satisfied. In the case MATH we set MATH. Let as above MATH be the minimal number such that MATH. So we have MATH and as MATH we obtain MATH. Hence after applying the automorphism MATH the well-balance conditions will be satisfied. |
math/0107003 | Note that MATH acts on the character by substituting MATH into MATH. Therefore it is enough to prove the theorem in the case MATH. First suppose that MATH satisfies the well-balance REF . Then REF follows from REF ; REF can be deduced from REF by REF . We know that general case can be reduced to the well-balanced case by the action of MATH. As MATH acts on the Right-hand side of REF by the same substitution MATH into MATH, we have REF in general case. REF can be deduced from REF by REF . |
math/0107004 | REF are well known. (Use MATH and induction on the degree for REF and MATH for REF.) Then REF follows from REF as any object in MATH is isomorphic to some MATH. As for REF, MATH gives a functor in one direction and properly interpreted MATH will be a quasi-inverse. Now, evaluation gives a natural morphism MATH in the category of sets. I claim that this map is a bijection. Indeed, we may assume that MATH for some MATH and using REF that MATH. As already MATH separates points MATH is injective. Since MATH, any ring homomorphism MATH is determined by what it does to MATH and so MATH is surjective. We now want to show that any ring homomorphism MATH induces, through MATH and MATH, a numerical map MATH. We reduce to MATH and MATH so we want to show that if MATH is a ring homomorphism then MATH is a numerical function which is obvious by the definition of MATH. Hence MATH maps the subcategory of rings isomorphic to some MATH into MATH and by what we have just proved it is a quasi-inverse to MATH. |
math/0107004 | Indeed, REF follows immediately from the map between the spectral sequences of REF for the TCP and the induced TCP in MATH. As for REF, the trivial nature of the limit in the definition of a special numerical space allows us to assume that MATH equals some MATH and then by REF that MATH, for some MATH-complex MATH. As MATH is concentrated in one degree, MATH is homotopic to a bounded complex MATH, indeed to one concentrated in REF degrees, and then using the naive truncations of MATH and REF again we may assume that MATH for some object MATH in MATH. Now there is a numerical (indeed a MATH-) PTCP MATH with MATH-contractible so by induction on MATH and NAME 's comparison theorem (compare , NAMEREF]) we may assume that MATH. We have a MATH-PTCP MATH so we may assume that MATH. Hence we are reduced to showing that MATH is an isomorphism. A numerical MATH-cocycle (for K(,REF)) is just a numerical function MATH such that MATH so MATH for some MATH and a REF-coboundary is of the form MATH. As we have exactly the same description for set theoretical REF-cochains and REF-coboundaries we get an isomorphism for MATH. As MATH is trivial it only remains to show that MATH for MATH as this is true in the set case. Now MATH and all the face operators are projections or sums of two adjacent coordinates. We can grade MATH by MATH and then MATH becomes a graded complex as MATH . Hence to show the required vanishing we can replace MATH by MATH as the cohomology can be computed degree by degree. This gives us a complex MATH, say. Now, MATH where MATH and this equality respects maps induced by projections and sums of coordinates (but it is not a ring isomorphism). Therefore, MATH is additively isomorphic to the standard cochain complex of MATH and so MATH and the latter group is clearly concentrated in degree REF. That a special numerical space is NAME follows by induction on the NAME tower and a trivial passage to the limit. Finally, to prove REF it suffices to prove that if MATH is a MATH-complex with a single non-zero homology group MATH, then MATH and MATH. This, however, is clear by the principal divisor theorem. |
math/0107004 | Indeed, it will be easier to prove a stronger statement. Let MATH be pointed objects and MATH . NAME and put MATH, where MATH denotes based maps. As above, if MATH is a TCP of NAME objects then MATH is a fibration and we get a long exact sequence of homotopy. If MATH is a PTCP then we get as usual the extra precision that the fibers of MATH are the orbits under an action of MATH and the sequence extends to MATH (compare CITE). We now consider a sequence of PTCP's MATH as in the definition of a special numerical space. We want to prove by induction on MATH that MATH is a bijection for all MATH. The case MATH certainly causes no problem and in general we have the PTCP MATH with fiber some MATH. The extra precision given to the long exact sequence is exactly what is needed to make REF-lemma work and we reduce hence to showing that MATH is a bijection. Now, MATH is homotopic to a bounded complex, so we may assume that MATH is bounded. By the same dévissage as before we reduce to MATH and then this bijection is true by the definition of numerical space. Putting MATH we then get the lemma for MATH replaced by MATH. To pass from MATH to MATH we use the NAME exact sequence MATH the similar sequence for MATH etc and the MATH-lemma. |
math/0107004 | Let MATH be a minimal principal NAME system, that is, the MATH are principal fibrations with fiber some MATH using the notations of REF in order of increasing MATH and MATH. We will step by step replace, up to homotopy, MATH by some MATH and MATH by MATH of a PTCP MATH. Assume that we have done this up to MATH. We then have a cartesian diagram, where MATH is the fiber of MATH, MATH . Here the right hand fibration is the standard one with MATH. Choose a resolution MATH by free f.g. abelian groups such that MATH, MATH and MATH for MATH. There is then a numerical PTCP MATH such that MATH applied to it is homotopic to MATH. Hence by REF there is a morphism MATH such that MATH is homotopic to MATH. Let MATH be the PTCP induced by MATH from MATH. Then MATH is homotopic to MATH and we put MATH. Then there is a morphism MATH which by construction is a homology equivalence and so a homotopy equivalence as MATH and MATH are nilpotent. If MATH is a homology equivalence, where MATH is a numerical space, then by obstruction theory applied to the MATH there is a map MATH s. CASE: MATH commutes up to homotopy. By REF there is a morphism MATH such that MATH is homotopic to the given MATH. In case MATH also is special another application of REF shows that MATH is a numerical homotopy equivalence. |
math/0107004 | The proof is easily reduced to the following statement. If MATH is a PTCP for a simplicial group MATH with MATH if MATH and MATH is a PTCP for a simplicial group MATH with MATH for MATH and for which MATH is injective then for any commutative diagram MATH there is a factorisation MATH of the diagram. This again amounts to saying that MATH is constant on a fibre of MATH. In proving this we immediately reduce to the case when MATH, and then need only show that the map is constant on the fibre over MATH, the unique non-degenerate MATH-simplex of MATH. As PTCP's over a simplex are trivial we may assume that MATH and MATH are trivial PTCP's. Hence we reduce to showing that any map MATH is constant on MATH. In general a map MATH is the same thing as an additive map MATH and what we want to show is that MATH is mapped to zero for MATH. If MATH then this is obvious. When MATH want to show that the kernel MATH of the map MATH induced by projection maps to zero in MATH. To prove this is equivalent to showing that MATH is zero. Now, MATH is a complex of free abelian groups, MATH is zero when MATH and MATH is zero when MATH so that any map MATH is null-homotopic. However, MATH is zero when MATH so any homotopy is zero. |
math/0107004 | Let us first prove that a polynomial with rational coefficients mapping MATH to MATH will map MATH to MATH. If MATH is a subring of MATH then it is an intersection of the localisations MATH that contains it so in that case one is reduced to MATH and then to MATH as MATH. Our polynomial defines a continuous function MATH, MATH is a closed subset and MATH is a dense subset. As MATH is mapped into MATH it follows that MATH is mapped into MATH. Conversely, if we have a polynomial with MATH-coefficients mapping MATH into MATH it maps in particular MATH into MATH and a slight modification of the argument in the proof of REF show that it is a MATH-linear combination of products of binomial polynomials. The proof of the second part is entirely analogous to the same statement for MATH given in the proof of REF . |
math/0107004 | The first part uses the fact that locality is stable under fibrations, that MATH is local if MATH is a finitely generated free MATH-complex which is clear as its homotopy groups are and that by speciality one may reduce to such MATH's. For the second part we again reduce to MATH's for MATH a finitely generated free MATH-complex. |
math/0107004 | Let MATH be the characteristic of MATH modulo its maximal proper ideal. For evident reasons we will have to carefully distinguish between equality versus homotopy of maps and we will start off with some observations. They will apply equally well to simplicial sets as to numerical spaces but for simplicity we will speak only of simplicial sets;N in any case the numerical case may be deduced from the set-theoretic one using REF . To begin with, if MATH is a simplicial abelian group with a single homotopy group MATH in degree MATH then isomorphism classes of PTCP's with structure group MATH over a simplicial set MATH correspond to elements of MATH (compare , CITE). It follows from that proof together with the use of the mapping cone construction that if MATH is a map of simplicial sets then the relative cohomology MATH correspond to equivalence classes of PTCP's over MATH together with a trivialisation of its pullback to MATH where two of them are equivalent if they are isomorphic over MATH by an isomorphism whose pullback to MATH is homotopic to one that preserves the given trivialisations. Let us also note that as we are dealing with principal fibrations, giving a trivialisation is the same thing as giving a section. The way the NAME tower MATH fits into this description is that MATH is universal for PTCP's in degree MATH over MATH that are provided with a trivialisation over MATH. From this we can construct the maps by induction over MATH. We therefore may assume we have the following diagram that is assumed to commute up to homotopy: MATH and MATH is an isomorphism. Let MATH respectively, MATH be the complexes for which MATH respectively, MATH are MATH- respectively,MATH-PTCP's. As MATH respectively, MATH are MATH respectively, MATH, MATH induces an isomorphism between them. We lift this isomorphism to a map of complexes MATH . Now, as MATH is minimal, the image of MATH in MATH is contained in MATH and hence by NAME 's lemma the map MATH is a surjection. As MATH also is minimal, the rank of MATH is the same as that of MATH and so the map MATH is an isomorphism. This implies that MATH is an isomorphism. Now, the pullback of MATH along the composite MATH has a section and hence a trivialisation. As the diagram is homotopy commutative we get a trivialisation of the pullback of MATH along the composite MATH. This in turn, by the universality of MATH, gives a mapping from MATH to the pullback of MATH along MATH covering the isomorphism MATH. This again is nothing but a map MATH of PTCP's covering MATH. As the latter as well as the base map, MATH, are isomorphisms so is MATH. By construction it gives rise to a homotopy commutative diagram MATH and thus finishes the induction step. |
math/0107004 | It is clear that with the binomial functions is a numerical ring, in fact the axioms were set up precisely to ensure this. Hence MATH certainly is a numerical ring with pointwise operations. Furthermore, MATH is clearly stable under all operations and is hence a sub-numerical ring. Therefore, if MATH there is a map of numerical rings MATH taking MATH to MATH. By REF this map is surjective and if we prove that MATH where the sum is not necessarily direct, then we are finished. As MATH it is sufficient to show that MATH is a numerical subring of MATH. That MATH is a subring follows from REF and stability under-MATH follows from the rest of the axioms. |
math/0107004 | If MATH and MATH are to be morphisms of numerical rings then the definition of MATH are forced by the axioms so we begin by showing that they are well-defined. As was remarked above any numerical ring is also a special MATH-ring. This means that if MATH is a numerical ring and if we put MATH then MATH is a ring homomorphism where MATH is given the ring structure of CITE with multiplication denoted by *. This gives us ring homomorphisms MATH and MATH coinciding on MATH and hence we get a ring homomorphism MATH showing that the operations MATH are well-defined on MATH. To show that we get a numerical ring we reduce to MATH and MATH and MATH free numerical rings on finite sets and conclude by REF as these show that MATH then is again a (free) numerical ring. Clearly MATH has the required universal property. |
math/0107004 | REF is proved by successive liftings (and is essentially a numerical cosimplicial version of the proof of the similar property for cdga's). The lifting at one accomplished as follows. By the definition of REF MATH will consist of normalised cochains in MATH and the subcomplex MATH of MATH generated by MATH and MATH is the mapping cone of a map of complexes MATH. We then extend the lifting of MATH to MATH. This in turn gives a lift of MATH which then gives a lift of MATH. Then REF follows similarly. Let us turn to REF. We will build a NAME tower. Note to begin with that when building this tower we must kill homology and not cohomology. Hence we assume that we have MATH such that if we consider MATH and MATH as complexes then MATH if MATH and MATH torsion free. Let MATH be a free complex such that MATH and MATH. Then there is a morphism of complexes MATH inducing the identity on MATH and the natural inclusion on MATH. Hence there is a morphism of cosimplicial groups MATH and by composition a morphism MATH. (Note that MATH fits in to a distinguished triangle MATH.) The composite MATH is, by construction, nullhomotopic. By adjunction we get MATH whose composite with MATH is again nullhomotopic as adjunctions preserve homotopies. Therefore there is a NAME MATH and a lifting MATH. I claim that MATH for MATH and that MATH is torsion free. To do this one has to say something about the cohomology of MATH. When MATH is finitely generated this has already been done and the general case is done by approximating MATH by finitely generated subcomplexes. After that one has to look at the NAME spectral sequence for the NAME constructed above. A small conceptual problem arises as we want to kill homology but are working with cohomology. This can certainly be overcome by brute force, but we will instead choose a hopefully more conceptual approach. This entails, however, the introduction of pro-(finitely generated abelian groups) and the reader who is unfamiliar with the concept of pro-objects will have no problem in translating the proof to follow into one using only cohomology. If MATH is a complex of torsion free abelian groups then we define its homology by MATH (compare CITE), where MATH runs over all finitely generated subcomplexes of MATH. We have the usual universal coefficient sequences expressing cohomology and homology in terms of each other if we put, for an abelian group MATH, MATH respectively, MATH equal to MATH respectively, MATH, where MATH runs over all f.g. subgroups of MATH and, for a pro-object MATH, MATH respectively, MATH equal to MATH respectively, MATH. Hence our assumptions imply that MATH if MATH and we want to prove that MATH if MATH. Furthermore, we may present MATH as a direct limit of complexes MATH which are finitely generated free, concentrated in degrees MATH and MATH with MATH free and MATH torsion. Then MATH and by REF MATH . By the well-known computation of the cohomology of NAME spaces we get that MATH if MATH or MATH (as MATH) and MATH. Finally, as above we get a NAME s.s. MATH . This and the information we have on MATH gives an exact sequence MATH and isomorphisms MATH for MATH. By construction we have an exact sequence MATH . Combining these two sequences we get that MATH is an epimorphism and that MATH is an isomorphism for MATH that is, that MATH for MATH. In case MATH is only connected we only get that MATH is zero so we may have to continue an infinite number of times just to kill homology in one degree and the end result will not necessarily be a special CNR. It will still have the lifting property of REF though. |
math/0107004 | The results clearly follow from REF . |
math/0107004 | The first part clearly amounts to showing that the reduction mod MATH map CITE MATH is a bijection. This can no doubt be done directly but the ``real" reason why it is true is the following. As we have seen, MATH is the NAME dual of MATH, the product of MATH copies of the formal multiplicative group, and so CITE for any ring MATH, MATH and it is well known CITE that MATH for MATH as well as MATH. As for the second part we note that the ring of continuous functions from MATH equals the MATH-adic completion of MATH, this is NAME 's theorem CITE. Hence the complex of continuous cochains is the completion of the complex of numerical cochains. This gives rise to short exact sequences MATH but as the MATH are finitely generated MATH the left hand side is zero and the right hand side is MATH. |
math/0107004 | It is clear that a MATH-module MATH is unipotent if and only if it is annihilated by some power of the augmentation ideal of the group ring MATH. By definition a map MATH is MATH-numerical if and only if it is annihilated by a power of the augmentation ideal and as the augmentation ideal is two-sided this is true if and only if it generates a submodule that is. Finally, as MATH modulo any power of the augmentation ideal is a finitely generated MATH-module, any unipotent submodule of MATH generated by one element is finitely generated. |
math/0107004 | If we extend the scalars of MATH to MATH we get a polynomial ring over MATH which is the affine algebra of an algebraic group MATH over MATH. Hence CITE applies and we conclude that MATH is unipotent and hence that MATH is a uniquely divisible nilpotent group. Clearly, MATH is a subgroup of MATH and hence is nilpotent and torsion-free. To prove finite generation we note that there as a finite dimensional faithful subrepresentation MATH of the representation of MATH on its affine algebra. We now want show that there is finitely generated subgroup MATH of MATH stable under the action of MATH. As MATH is finite dimensional it contains a finite number of vectors spanning it as MATH-vector space. After possibly multiplying them by a non-zero integer we may assume that they are contained in MATH. It is therefore sufficient to show that each MATH lies in a finitely generated subgroup of MATH invariant under MATH. For this we consider the product map MATH and write the pullback MATH as MATH. This means that for MATH and by REF acts on MATH by MATH. Hence we get that MATH so that the translates of MATH by the elements of MATH lies in the finitely generated group spanned by the MATH and hence is finitely generated. Thus, MATH is a subgroup of the subgroup MATH of elements of MATH stabilising MATH. As a subgroup of a finitely generated nilpotent group is finitely generated it is enough to show that MATH is finitely generated. This will be done by induction over the dimension of MATH (with MATH changing during the induction). As MATH is unipotent MATH contains a MATH-dimensional subspace MATH on which MATH acts trivially. We may use the induction hypothesis on the image of MATH in MATH and the image MATH of MATH in MATH to conclude that the image of MATH in MATH and it is then enough to show that the kernel of this map is finitely generated. However, that kernel is a subgroup of the abelian group of additive maps MATH which is finitely generated. |
math/0107004 | REF follows from REF and induction over the length of the ascending central series,. REF is obvious. As for REF consider first a function MATH that is numerical with respect to the given numerical structure. Let MATH be the algebraic group over MATH whose ring of regular function is MATH. Then MATH generates a finite dimensional subrepresentation of MATH which is unipotent as MATH is by CITE. Hence MATH-numerical by REF . Assume conversely that MATH is MATH-numerical. Again by REF it generates a unipotent module MATH. We may choose a MATH-invariant filtration of MATH whose successive quotients are free of rank MATH with trivial MATH-action. Having done this, the MATH-action on MATH corresponds to a group homomorphism from MATH to MATH, the group of unipotent upper triangular integer MATH-matrices, where MATH is the rank of MATH. Furthermore, MATH is a numerical group (in fact an algebraic one) and MATH is the composite of a numerical map MATH and the group homomorphism MATH. It will therefore suffice to show that the group homomorphism MATH is numerical. The ascending central series of MATH is given by MATH, where MATH is defined by MATH and MATH is clearly also a numerical group. We now prove by descending induction on MATH that the composite MATH is numerical. The homomorphism MATH is a central extension. The obstruction for lifting the homomorphism MATH to MATH is an element of MATH that is zero as the morphism is known to lift and the obstruction for lifting it to a numeric homomorphism is an element of MATH. The latter group maps bijectively to the former by REF and hence the numerical homomorphism MATH lifts to a numerical homomorphism MATH. The set of group homomorphism liftings are classified by MATH and the set of numerical group homomorphism liftings are classified by MATH. Again by REF the map between these groups is a bijection and hence every lifting is numerical. In particular the given one is which finishes the induction step. Finally, REF follow from REF. |
math/0107004 | This follows from the previous results of this section together with REF . |
math/0107004 | This follows directly from the spectral sequence MATH and the corollary. |
math/0107004 | The first part is implicit in CITE but can be found explicitly in CITE. The second part is clear when MATH as MATH is the constant simplicial object with constant value MATH. For larger MATH it follows from the first part, induction and the exact sequences MATH coming from the acyclicity of MATH. |
math/0107004 | If we begin with the first part and we consider a fibre of MATH over a MATH-simplex MATH, an extension to MATH is completely determined by the restriction of MATH to MATH. Furthermore, MATH applied to any element MATH of MATH is determined as it has to be the already prescribed image of MATH. That means that if MATH and MATH are two extensions of MATH then MATH maps MATH into MATH. Conversely given an extension MATH and a linear map MATH there is an extension MATH of MATH such that the restriction of MATH to MATH is the given map. Hence, the map MATH is a principal homogeneous space over the simplicial group MATH. To show that it is a PTCP we need to find a section the restriction MATH to the subcategory MATH of MATH consisting of those increasing maps MATH that take MATH to MATH (compare , CITE). This is obtained by the following observations CASE: Given a map MATH to extend it to MATH one needs to find a map MATH such that MATH for all MATH. This is possible as MATH and MATH is acyclic. It can be done explicitly given a contraction of MATH. CASE: Given a MATH-point MATH we may use it as origin and use the algebraic contraction MATH and the usual integration formulas to construct a contraction of MATH. This contraction is natural for affine maps preserving the chosen basepoints. CASE: The restriction of the cosimplical scheme MATH to MATH has a base point and hence the restriction of MATH to MATH has a contraction. To turn to the second part it is clear that the only of the equivalence relations MATH on MATH that does not factor through MATH is MATH. As for MATH it is clear that two MATH-simplices MATH and MATH are equivalent if their restrictions to MATH are and if the restrictions to MATH of MATH and MATH are equal. This combined with the first part now gives the second. |
math/0107004 | We leave to the reader to prove, in a fashion analogous to CITE, that if MATH is a minimal model then MATH is a homotopy equivalence and that the formal scheme structure on MATH induces a special numerical space structure on MATH and will assume that MATH is minimal. That means that there is a filtration MATH of MATH by sub-cdga's such that MATH, where MATH is concentrated in a single degree and MATH maps MATH into MATH. Furthermore, the degree of MATH tends monotonically to infinity with MATH. One now proves by induction that MATH is a homotopy equivalence and that MATH is a minimal MATH-numerical space using REF . One then concludes by noting that MATH is the inverse limit of MATH and that this system is eventually constant in each degree. |
math/0107004 | REF is clear as a special morphism is flat so MATH and MATH are both exact. As for REF one verifies it by induction over a NAME tower of MATH. Finally, REF follows as in the simplicial case CITE. |
math/0107005 | MATH has a presentation: MATH (see for example CITE) where MATH and MATH correspond to matrices MATH, MATH respectively. (It is a classical fact that MATH. Hence MATH has a presentation: MATH. One can use a map MATH: MATH to show that these two presentations define isomorphic groups.) By definition of MATH the following sequence is exact. MATH . Consider a homomorphism MATH defined by the formulas: MATH. If we denote elements MATH; MATH; MATH by MATH and MATH respectively we see that these elements MATH generate MATH and the relations MATH are satisfied. To find all defining relations for MATH we need to find how MATH acts on the generators MATH and MATH of MATH by conjugation. First note that MATH and MATH. Hence MATH . |
math/0107005 | We will give a presentation of the factor group MATH that coincides with the above presentation of MATH. By MATH and MATH we denote the generators of MATH as above. Consider isotopy classes MATH and MATH of the following diffeomorphisms of MATH: MATH . Define map MATH by the identities: MATH, MATH and extend it linearly to the whole group MATH. We will show that MATH is a well defined homomorphism from MATH to MATH. First we check that MATH: MATH . Thus MATH and hence MATH. From now on we will denote a diffeomorphism and the isotopy class of it by the same capital letter, omitting the brackets. To prove the equality MATH (MATH stands for the identity diffeomorphism of MATH) we will need some auxiliary results. Consider the following diffeomorphisms MATH: MATH . If we choose spheres MATH and MATH as generators of the group MATH, it is obvious that diffeomorphisms MATH and MATH preserve these spheres and act trivially on homology of MATH. Isotopy classes of diffeomorphisms MATH and MATH generate the group MATH . Let us compute MATH. Since MATH is a parallelizable manifold and the normal bundle of MATH in MATH is trivial we need to find an element of the group MATH that corresponds to the differential of MATH. Take a point MATH. We can identify the fiber of the normal bundle MATH at MATH with the fiber of the tangent bundle MATH at this point. Furthermore, via the projection MATH we can identify this tangent fiber at MATH with the tangent fiber at MATH (NAME algebra MATH of the group MATH). Since MATH the map MATH will correspond to the adjoint representation MATH, and MATH. Thus MATH. If we choose an element MATH then it is well known that MATH. This map is a generator of the group MATH (see CITE, ch.I REF) and therefore the isotopy class of diffeomorphism MATH is a generator of the group MATH. In a similar way one can show that the isotopy class of diffeomorphism MATH is the other generator (corresponding to the map MATH) of the group MATH. Now we show that MATH, MATH in the factor group MATH. The first equality follows from the results of NAME, since as we just saw, MATH and MATH generate the abelian subgroup of the group MATH. To prove the other equalities we use the group structure on MATH: MATH as required. MATH. MATH and hence MATH that is, MATH. Identities MATH and MATH imply MATH. Then from the above equalities we get MATH. Similarly we see that MATH, hence MATH which proves the claim. Using these identities we can show that MATH. MATH. It implies that exact sequence REF splits; the factor group MATH has four generators MATH and the following set of defining relations: MATH. In particular, we see that groups MATH and MATH have the same presentations and therefore isomorphic. |
math/0107005 | Take a sphere MATH. We will denote by MATH the diffeomorphism of MATH which is the identity outside an embedded disk MATH and corresponds to the element MATH of the mapping class group. To show that MATH it is enough to show that MATH is diffeomorphic to MATH. We construct an h-cobordism between these two manifolds. Take the handlebody MATH and remove from it an interior disk MATH. The resulting manifold is denoted by MATH. Boundary components of MATH are MATH and MATH. Take disks MATH in these two components and connect them by a tube MATH embedded into MATH. Next extend MATH in an obvious way (by the identity outside the tube) to a diffeomorphism MATH of MATH. Consider now two manifolds: MATH and MATH. Present the boundary of MATH as the union: MATH and the boundary of MATH as the union: MATH . Using diffeomorphism MATH we can glue MATH and MATH together along the common submanifold MATH to obtain a cobordism (after smoothing the corners) MATH between MATH and MATH. It is clear that MATH is simply connected. Using NAME exact sequence of the union MATH we see that MATH . In a similar way we can get homology groups of MATH: MATH . Thus, by the h-cobordism theorem CITE two homotopy spheres MATH and MATH are diffeomorphic. |
math/0107005 | For the proof we construct a spin manifold MATH bounded by the sphere MATH and compute the MATH-invariant MATH defined by CITE. First we extend diffeomorphisms MATH and MATH to diffeomorphisms of the handlebodies MATH and MATH respectively. It can be done since in the definition (recall REF ) we can assume that MATH for diffeomorphism MATH and MATH for diffeomorphism MATH. These extensions we also denote by MATH and MATH respectively. Next we present MATH as the union of five manifolds: MATH where MATH and MATH belong to MATH. Consider manifolds MATH, MATH and MATH with boundaries presented as the unions: MATH . Using extension diffeomorphisms MATH and MATH defined above, we now construct a manifold MATH which will be used to compute MATH. Define MATH to be the manifold obtained from the union MATH by smoothing the corners. MATH . Evidently, the union MATH is a simply connected manifold with simply connected boundary MATH. Using exact sequence of NAME it is easy to see that homology groups of MATH are trivial in all dimensions MATH. Hence by the characterizations of the smooth n-disk MATH (see CITE) this union is diffeomorphic to the disk MATH. Same proof works in the second case. Thus we can write MATH. Now note that MATH where (using notations of CITE) by MATH we denote the boundary of the following union of three REF-disks: MATH . The gluing maps are (compare REF): MATH where MATH and MATH, defined by the formula: MATH for MATH and MATH. In particular, we see that our homotopy sphere MATH is diffeomorphic to the manifold MATH. Using NAME exact sequence for the manifold MATH we see that MATH, and we can apply results of NAME and NAME REF to the manifold MATH to compute the MATH-invariant of MATH. It is shown (see CITE, page REF) that MATH where MATH is the first NAME number and MATH, MATH are NAME numbers of the stable vector bundles over MATH determined by the compositions (compare REF): MATH . We show that MATH. It is well known how NAME numbers depend on the characteristic maps of the stable vector bundles over the spheres : MATH. Here MATH is the integer, corresponding to the characteristic map: MATH and MATH if MATH is even or odd respectively. In our case MATH and to find the integer MATH we need to find the homotopy class of the composition MATH . It is a standard fact (which can be deduced from REF) that inclusion MATH induces map MATH which is multiplication by REF. Hence MATH and similarly MATH. Therefore MATH. For the generator MATH of MATH we have (see REF) MATH or MATH. From REF and NAME (CITE, p. REF) we see (changing orientation of MATH if necessary) that MATH. Since MATH we can apply REF . to get MATH. It shows that MATH and finishes the proof. |
math/0107005 | We know that MATH (REF .). Using identities: MATH we find that MATH. Similarly MATH. |
math/0107005 | Equality MATH shows that MATH is a REF-cocycle. Verification of this equality is straightforward and left as an exercise. Let MATH and MATH be presentations of groups MATH and MATH respectively. Then a central extension of MATH by MATH will have a presentation: MATH with MATH and some MATH. It follows from REF that MATH for the function MATH, and that the cocycle MATH defines the extension with MATH. We denote this extension by MATH. If MATH then cocycles MATH and MATH define non equivalent extensions. Indeed, suppose that MATH and MATH are equivalent. Then there exists an isomorphism MATH that makes the following diagram commute: MATH . Using the above presentation of groups MATH and MATH one can see from this diagram that MATH for some MATH. Hence MATH is a generator of the group MATH. |
math/0107005 | Since MATH it follows from the previous lemma that MATH is a REF-cocycle. Let MATH and MATH be presentations of groups MATH and MATH. Then a central extension of MATH has a presentation: MATH with MATH and some MATH. Using the group law: MATH one can easily find that function MATH defines the extension with MATH and MATH. NAME MATH defines the extension with MATH and MATH. As in the proof of REF . one can show that if MATH then cocycles MATH and MATH define non equivalent extensions. Therefore function MATH is indeed a generator of MATH. |
math/0107005 | First we give presentations of groups MATH with generators and defining relations. Denote the generators of MATH, MATH by MATH, MATH and elements MATH, MATH of the group MATH by MATH and MATH respectively. Note: To avoid cumbersome notations we use the same letters MATH and MATH to denote elements of different groups. We hope it will not cause any confusion. Using group law MATH one can easily show (compare proof of REF . above) that MATH and MATH. Hence we get presentations: MATH, MATH, MATH. Next define maps MATH and MATH by the formulas: MATH, MATH, MATH and MATH. Map MATH induces commutative diagram MATH from which it follows that MATH is a monomorphism. Similarly one proves that MATH is a monomorphism. Since MATH is identified with MATH and MATH we obtain the following presentation of MATH with generators and defining relations: MATH . Consider now two elements: MATH and MATH. Obviously, MATH and MATH. One can easily show that the above presentation of the group MATH is equivalent to one of the NAME group MATH obtained in REF . |
math/0107005 | We prove this lemma only for MATH. In the other cases the proof is similar. Consider the first quadrant of the LHS spectral sequence MATH, for MATH. MATH . From this MATH-term we see that MATH. Hence for any MATH, MATH. Since MATH, by the Universal Coefficient Theorem MATH, and therefore MATH. |
math/0107005 | First note that it follows from the presentation of the group MATH that MATH. To compute the cohomology groups, we use a NAME exact sequence of the amalgamated product MATH. By the previous lemma we get the following fragment of this exact sequence: MATH where MATH and MATH is induced by the inclusion MATH (compare proof of REF .). Commutative REF together with the functorial dependence (compare REF) induces the functorial map MATH. Hence MATH is the map induced by multiplication: MATH. If we denote by MATH generators of groups MATH, MATH respectively then we see that MATH where MATH are generators of the groups MATH. Similarly one can show that MATH where by MATH and MATH we denoted the generators of groups MATH and MATH. Hence MATH and MATH. It is clear that MATH. We can assume that MATH and MATH where by MATH we denote the elements of finite order. Then by the Universal Coefficient Theorem MATH, that is MATH. Similarly we have MATH. Therefore MATH and MATH as required. |
math/0107005 | Choose the presentation of MATH found in REF . Then any central extension MATH: MATH has the following presentation: MATH with some MATH and MATH, MATH and MATH. Consider element MATH. Then from equality MATH we get MATH. Instead of MATH we get MATH and so on. Thus, by changing MATH to MATH we can make MATH. Similarly, changing MATH to MATH and MATH to MATH we can eliminate MATH and MATH and assume that MATH in the above presentation of MATH. Now one can easily obtain the following equalities: MATH, MATH. Therefore MATH. Analogously, if we compare MATH with MATH we find that MATH. Hence MATH. It means that we can choose set-theoretic cross-sections of MATH (that is, functions MATH so that MATH) in such a way that any central extension of MATH by MATH has a presentation: MATH and any element of MATH is cohomologous to a cocycle, defined by one of these MATH. Choose a REF-cocycle MATH that defines the extension MATH. Evidently, MATH defines the extension REF of MATH by MATH and therefore generates MATH. Hence MATH is onto and MATH generates the direct summand MATH of MATH. As for the other generator MATH of MATH, we can choose MATH (recall REF above). It can be verified that MATH defines the extension MATH and the subgroup generated by MATH is the kernel of MATH. This proves the exactness. Proof of the asserted splitting is left to the reader (compare REF). |
math/0107008 | If MATH is not hyperbolic then it fixes some vertex MATH. To rule out REF , suppose that MATH is any vertex not fixed by MATH. The subtree MATH spanned by MATH, MATH, and MATH is equal to MATH, and MATH meets MATH in a single vertex, MATH. This vertex is fixed by MATH and so MATH. As MATH, this path has even length and contains a pair of edges which are related by MATH. These two edges have the same image in MATH. This shows that neither REF nor REF can hold. To rule out REF let MATH be the edge, and look at the subtree spanned by MATH, MATH, and MATH. Then as MATH, the edges MATH and MATH are both oriented toward MATH, or both oriented away from MATH. Since any oriented path traversing MATH travels first toward MATH and then away from MATH, the two edges are incoherently oriented, violating REF . |
math/0107008 | The element MATH takes MATH to MATH and one easily checks that these two edges are coherently oriented. Then MATH is hyperbolic by REF . The subtree MATH is a linear MATH - invariant subtree, and must therefore be the axis. Now note that MATH. |
math/0107008 | First we show that the fold must be of type I or type II. Suppose that edges MATH and MATH generate the fold (so MATH) and that it is of type III. This means that MATH but MATH. Then MATH for some MATH. If MATH then MATH is hyperbolic by REF . Otherwise, if MATH for some MATH, then MATH and MATH. Again by REF , MATH is hyperbolic. We now re-define MATH to be MATH, which is hyperbolic and takes MATH to MATH. In both cases, after the fold, the image vertex of MATH is fixed by MATH. Hence MATH becomes elliptic, a contradiction. Thus we consider type I folds and type II multi-folds. We can assume that the folds are of type A, as remarked in REF. Consider first a type I fold. In this situation, MATH and MATH. Since the stabilizer of the image of MATH is MATH, this subgroup contains no hyperbolic elements. Thus it fixes an end or vertex MATH by REF . Consider the paths MATH and MATH. If MATH then MATH as MATH fixes MATH. If in addition MATH then MATH, and so MATH. The same result as the fold can now be achieved by sliding MATH over MATH and then collapsing MATH. Similarly, if MATH and MATH then the fold is equivalent to sliding MATH over MATH and then collapsing MATH. The last possibility is that MATH and MATH (the three cases correspond to whether the path from MATH to MATH joins at MATH, at MATH, or at MATH). Then MATH and MATH as before. In this case we collapse both MATH and MATH end expand a new edge with stabilizer MATH, bringing across all edges which were previously incident to MATH or MATH. The result is the same as the fold. Finally, consider a type II multi-fold, performed at the set of edges MATH. In this situation MATH is contained in a single edge orbit MATH, with MATH. Consider MATH and MATH. Assuming that MATH (otherwise the multi-fold is trivial) there is an edge MATH with MATH. The image vertex of MATH and MATH has stabilizer containing MATH and MATH, and therefore MATH contains only elliptic elements. From the Hyperbolic Segment REF it follows that MATH (and MATH). The multi-fold can be replaced by the following: collapse MATH (the rest of MATH collapses with it, by equivariance) and then expand a new edge with the appropriate stabilizer. |
math/0107008 | The implications REF are trivial, and MATH follows from REF . For REF , consider a stabilizer MATH of some vertex MATH. As a group acting on MATH it consists of elliptic elements, and by REF it has a fixed point. Thus every MATH - elliptic subgroup is MATH - elliptic, and conversely by symmetry. Now REF yields conclusion REF . |
math/0107008 | Write MATH and MATH. Choose basepoints MATH and MATH, and let MATH be an isomorphism. Setting MATH, the trees MATH and MATH are MATH - trees (via MATH in the case of MATH). The vertex stabilizers of both trees have REF , so each stabilizer has fixed points in both trees. Hence MATH and MATH have the same elliptic subgroups. REF now provides a sequence of elementary moves from MATH to MATH. There is a corresponding sequence of elementary moves of graphs of groups taking MATH to a quotient graph of groups MATH of MATH. Now, since MATH and MATH have MATH - isomorphic covering trees, there is an isomorphism MATH by CITE. |
math/0107008 | To simplify the argument we use non-equivariant folds. We will prove that a factorization MATH exists, where MATH is a finite composition of non-equivariant multi-folds and MATH is an embedding. If one then replaces the multi-folds of MATH by their equivariant counterparts, the factorization is still valid. We begin by defining an equivalence relation MATH on the set MATH. Choose a basepoint MATH and let MATH denote the subtree spanned by MATH, MATH, and MATH (for MATH). Set MATH if there exists a graph isomorphism MATH fixing MATH and sending MATH to MATH and MATH to MATH, such that the diagram MATH commutes. We will call the equivalence classes MATH - types. Since the tripods MATH have bounded diameter and MATH is finite, there are only finitely many MATH - types. A subtree MATH supports a MATH - type if the MATH - type contains a pair MATH such that MATH. By taking a union of such tripods one can find a finite subtree MATH which supports all MATH - types. For such a MATH we claim that if MATH is an embedding, then MATH is an embedding. For otherwise, if MATH and MATH for some pair of edges MATH, then the MATH - type containing MATH is not supported by MATH (as MATH is not injective). A full multi-fold along MATH is the multi-fold defined by identifying all of the edges MATH to one edge. It is the ``maximal" multi-fold along MATH through which MATH factors. Now let MATH be a finite tree which supports all MATH - types, and suppose that MATH is not an embedding. Then there are two vertices MATH with the same image in MATH. Since MATH contains no circuits, the image of the path from MATH to MATH must contain a reversal. Thus there exists a pair of edges MATH such that MATH and MATH. Among all such pairs choose MATH and MATH to minimize the distance MATH from MATH to MATH. Factor MATH as MATH where MATH is the full multi-fold along MATH. We claim that MATH preserves the equivalence relation, that is, that MATH implies MATH. To show this, let MATH be the subtree spanned by the vertices having distance at most MATH from MATH, and define MATH similarly (note, MATH). By the choice of MATH, MATH is an embedding. In fact, since MATH supports all MATH - types, MATH is an embedding, and then we must have MATH. This last statement holds because if there were an edge MATH then by injectivity of MATH, the MATH - type containing MATH would not be supported by MATH. Now suppose that MATH and consider the effect of MATH on the tripods MATH and MATH. Let MATH be the graph isomorphism shown in REF . Note that since MATH is injective, the restriction MATH is the identity (and MATH). If MATH then MATH and MATH is injective on both trees (here we use non-equivariance of MATH). Therefore MATH via the isomorphism MATH (compare REF below). If MATH then MATH and MATH maps MATH bijectively to MATH. Note that MATH identifies MATH to a single edge and is bijective on other geometric edges. Hence two edges of MATH are identified by MATH if and only if their images under MATH are identified by MATH. This implies that there is a graph isomorphism MATH making the following diagram commute: MATH . Now restrict MATH to the subtree MATH to obtain a graph isomorphism MATH which realizes the equivalence MATH. Thus, MATH preserves the equivalence relation. It follows that MATH supports all MATH - types. Note also that MATH has fewer edges than MATH. By repeating this factorization process at most MATH times, we obtain a finite composition of multi-folds MATH and a morphism MATH satisfying MATH, such that MATH is an embedding and MATH supports all MATH - types. Then MATH is an embedding. |
math/0107008 | Suppose MATH contains edges MATH and MATH with MATH. First we reduce to the case where MATH. Apply the NAME REF to the subtree MATH to obtain a factorization of MATH: MATH where each MATH is a multi-fold and MATH. Each map in this factorization induces an isomorphism of quotient graphs, because their composition does. Similarly, each map preserves hyperbolicity of elements of MATH, because equivariant maps cannot make elliptic elements hyperbolic. Take MATH smallest so that MATH. Then MATH or MATH. We can now proceed with the argument using this pair of edges and the morphism MATH. We have already remarked that the hypotheses of the proposition hold for the this map. Thus, without loss of generality, MATH. Since MATH and MATH induces an isomorphism on quotient graphs, MATH and MATH are in the same orbit, so there exists MATH with MATH. Note that MATH, and also MATH. Now choose an element MATH. Suppose that MATH is in MATH as well. Then using the Hyperbolic Segment REF one finds that MATH is hyperbolic in MATH. However, MATH and MATH are both in MATH so their product is elliptic in MATH, giving the desired contradiction. Thus our strategy will be to pin down the behavior of MATH acting on MATH. Choose an element MATH. The product MATH is hyperbolic in MATH by the Hyperbolic Segment REF and hence is hyperbolic in MATH by equivariance. Consider its axis MATH. The image MATH is a MATH - invariant subtree of MATH and therefore it contains the axis MATH. Notice that MATH is the support of the following oriented path: MATH . Since MATH, the axis MATH contains an edge which is mapped by MATH to MATH, and this edge must be in the same orbit as MATH; call it MATH where MATH. Now, using equivariance, one sees that the set MATH contains the following oriented edges, arranged coherently and in order along MATH (possibly with gaps in between): MATH . This shows that the edges MATH and MATH are incoherently oriented, and so MATH fixes a unique vertex MATH. Configuration REF also shows that MATH. Hence MATH, and MATH fixes no edge of this segment. Having established the behavior of MATH we return our attention to MATH, or rather to its conjugate MATH. This element fixes MATH and takes MATH to MATH (MATH). Now MATH and MATH satisfy the Hyperbolic Segment REF using the segment MATH. This implies that MATH is hyperbolic in MATH. On the other hand, as MATH, we must have MATH by equivariance. Then MATH, MATH, and MATH are in the subgroup MATH and so MATH is elliptic in MATH, a contradiction. |
math/0107008 | Suppose first that MATH is a single loop. If the edges of MATH satisfy MATH then by REF the map MATH is an isomorphism. Otherwise suppose that MATH for some MATH. Then MATH is parabolic and hence the elliptic elements form a subgroup of MATH (see REF ). As MATH is also a single loop, this implies that MATH is also parabolic (using the Hyperbolic Segment REF ). Let MATH be an edge with image MATH or MATH, such that MATH and MATH for some MATH (for every edge MATH, either MATH or MATH has these last two properties, because MATH is parabolic). We wish to know that MATH. If MATH then this is already true. So suppose that MATH. By equivariance, we have that MATH, that is, MATH. Hence MATH, which implies that MATH for all MATH. Similarly, MATH for all MATH, because MATH. These inclusions fit into a diagram as follows. MATH . The central vertical inclusions follow from equivariance of MATH, since MATH and MATH. The rightmost vertical inclusion is a consequence of all of the other inclusions in the diagram. Now observe that MATH and MATH are both equal to MATH, the set of elliptic elements. In each case this follows from the facts that MATH is hyperbolic (by REF ) and the tree is parabolic. Therefore all of the inclusions in the bottom row of the diagram are equalities. This implies that MATH is a linear tree with all stabilizers equal to MATH, as in REF . In particular, MATH as desired. Renaming MATH as MATH if necessary, we have: MATH, MATH, MATH, MATH, and MATH. Now consider the subgroup MATH, and let MATH be the relation on MATH defined by REF using MATH, MATH, and MATH. We claim that MATH . This equivalence implies that MATH is a parabolic fold from MATH to its image. Since MATH is minimal (because it is reduced), this image is all of MATH. Note that both relations in REF are equivariant, so in its proof we can take one of the edges to be MATH. So suppose that MATH ( MATH). Then MATH for some MATH, and MATH by equivariance. Therefore MATH. Next suppose that MATH, that is, that MATH with MATH and MATH. Then MATH. The properties MATH and MATH imply that MATH whenever MATH. Thus MATH, and so MATH. This implies that MATH, and so we have shown that MATH implies MATH. Since the latter relation is transitive and MATH is the transitive closure of MATH, it now follows that REF holds. Therefore REF is proved in the case where the quotient graph is a single loop. Next consider the general case. An edge MATH is called MATH - isolated if no two edges of MATH have a common endpoint. As MATH is reduced, REF implies that every edge that is not MATH - isolated projects to a loop in MATH. We are interested in MATH - isolated edges because of the following observation: if MATH is any morphism of trees such that every edge of MATH is MATH - isolated, then MATH is an embedding. This is true because otherwise, MATH would factor through a nontrivial fold, and then the edge where the fold is performed would not be MATH - isolated. Suppose that MATH is an edge which is not MATH - isolated. Let MATH, and let MATH be oriented so that MATH. Let MATH be the connected component of MATH containing MATH. Then MATH is a parabolic MATH - tree with quotient graph a single loop (MATH is the stabilizer of MATH). Let MATH be the connected component of MATH containing MATH. Then MATH is also a MATH - tree, and MATH is a MATH - equivariant map inducing an isomorphism of quotient graphs. By the previous analysis, MATH is a parabolic fold of MATH - trees. In fact it is the parabolic fold performed at MATH using the subgroup MATH. Let MATH denote the relation on MATH given by this parabolic fold, and let MATH be its equivariant extension to MATH. Then MATH is simply the parabolic fold in MATH performed at MATH using MATH. We have that MATH if and only if MATH, for edges MATH. Since MATH is equivariant, we also have that MATH implies MATH for all MATH. It follows that MATH factors as MATH, where MATH is the parabolic fold defined by MATH. Note that MATH is an isomorphism. In particular the parabolic fold MATH is nontrivial, since MATH is not an isomorphism (as MATH is not MATH - isolated). Recall that in a parabolic fold at MATH, all identifications of vertices occur inside connected components of MATH. Therefore edges and vertices in different components remain in different components. Now let MATH. We do not know whether MATH, but no two edges of MATH are in the same connected component of MATH, and therefore MATH is MATH - isolated. Thus, by factoring through a parabolic fold, any edge may be made MATH - isolated. During this parabolic fold, the other edges are affected as follows. All translates of MATH and MATH are MATH - isolated. If MATH is a MATH - isolated edge of MATH such that MATH, then MATH is MATH - isolated, as MATH does not affect MATH away from MATH. If MATH is a MATH - isolated edge of MATH with MATH and MATH, then MATH. This is true because MATH is strictly larger than MATH, as MATH is nontrivial, and MATH (because MATH). Thus, MATH is MATH - isolated by REF . Lastly, if only one of MATH, MATH is in the orbit MATH, then MATH does not project to a loop in MATH, and hence is MATH - isolated by REF . These remarks imply that MATH factors through a composition of finitely many nontrivial parabolic folds, at most one for each vertex orbit, after which all edges are MATH - isolated. Here the factorization is MATH, where MATH now represents the sequence of parabolic folds. Then MATH is an embedding, and it is an isomorphism because MATH is minimal. |
math/0107008 | We are given a parabolic fold MATH, performed at MATH. Suppose that MATH fixes a vertex MATH. Let MATH be the subtree spanned by the edges MATH. We claim that MATH has finite diameter. If MATH then MATH for some MATH (because parabolic folds induce isomorphisms on quotient graphs). Then MATH, so MATH. We now have MATH and hence MATH (as MATH is a path joining MATH to MATH). This shows that MATH has diameter at most MATH, and the same must hold for MATH. Now, applying the NAME REF using the subtree MATH, we obtain a factorization of MATH as MATH. Here MATH is a finite composition of multi-folds and MATH is an embedding. Thus, MATH is a single edge. The same is true of translates of MATH by equivariance, and also of other geometric edges, since the original parabolic fold MATH is bijective away from the orbit of MATH. This shows that MATH is an isomorphism, and hence MATH has been factored into a finite composition of multi-folds. To see that the multi-folds are of type II, note that their composition induces an isomorphism on quotient graphs, and hence each multi-fold does as well. Finally consider the converse: suppose that the parabolic fold MATH is a finite composition of multi-folds (which must be of type II). Note that these multi-folds occur along edges MATH, so that MATH. It follows that these multi-folds preserve hyperbolicity of elements of MATH, so by REF , MATH is an elementary deformation. Now, by REF , every stabilizer in MATH fixes a vertex of MATH. In particular, MATH fixes some MATH. |
math/0107008 | We are given that MATH fixes MATH. Thus MATH fixes the path from MATH to MATH, including the first edge MATH on this path, and so MATH (where MATH). Since MATH is reduced, there is an element MATH of MATH taking MATH to MATH. Then MATH. The element MATH is hyperbolic by REF , using either of the first two criteria. |
math/0107008 | Note that MATH and that MATH has translation length MATH for positive MATH. Thus: MATH . |
math/0107008 | We will define the morphism MATH in the following manner. First we define MATH on a finite set of vertices of MATH, one representing each vertex orbit. Having done this correctly, the map then extends equivariantly to MATH in a unique way. Once the map is defined on MATH, each edge MATH is sent to the unique path in MATH joining MATH to MATH. We need to be certain that these two vertices are not the same. Then MATH is subdivided so that preimages of vertices in MATH are vertices in MATH, and we have an equivariant morphism MATH (here MATH is MATH, after subdivision). Let MATH be a subtree whose edges map bijectively to the edges of MATH. For each vertex MATH of MATH, the stabilizer MATH may or may not be maximal among the elliptic subgroups of MATH. Label the vertices of MATH as MATH so that the vertices having maximal elliptic stabilizers occur first in the list. Also arrange that all vertices in the same orbit occur consecutively. Consider the MATH orbit arising in the list. It has a first vertex MATH and various translates MATH for MATH. Set MATH and choose elements MATH such that MATH for each MATH. Thus, whenever MATH and MATH are in the same orbit, there is a chosen element MATH taking MATH to MATH. Also, for each MATH the element MATH takes MATH to a chosen representative of its orbit. If MATH then MATH is hyperbolic, by REF . We wish to define MATH on MATH with the following properties: CASE: MATH is injective, CASE: MATH for each MATH, CASE: whenever MATH and MATH are in the same orbit, MATH. We proceed one orbit at a time. Suppose that MATH has already been defined on MATH, satisfying all three properties. Consider MATH and its translates MATH. The stabilizer MATH fixes a vertex MATH of MATH. If MATH then we can define MATH for each MATH. REF - REF are satisfied in this case. However, REF might not hold. So suppose that MATH for some MATH and MATH. We consider two cases: either MATH is maximal among elliptic subgroups, or MATH is not maximal elliptic. The third possibility, that MATH is maximal elliptic and MATH is not, was ruled out when we ordered the vertices. In the first case note that MATH by REF for MATH, and by maximality of MATH these stabilizers are equal. Also, since MATH, conjugation by MATH yields MATH. Therefore MATH. In the second case, as MATH is not maximal elliptic, there is an elliptic subgroup MATH properly containing MATH. Since MATH is elliptic it is contained in MATH for some MATH and we have MATH. In each of the two cases we have found that MATH is contained in the stabilizer of a vertex of MATH other than MATH. By REF , there must be a hyperbolic element MATH such that MATH. Now observe that MATH . We can simplify forthcoming notation by setting MATH, so that we have MATH. This inclusion will allow us to replace MATH by some MATH to arrange that REF holds. Note that MATH is hyperbolic. For any pair of indices MATH and MATH consider the vertices MATH and MATH, and the element MATH. Applying REF with this data we find that MATH for sufficiently large MATH. Fix MATH so that REF holds for every MATH and MATH. Now define MATH for each MATH. Trivially, REF holds. For REF note first that since each MATH is hyperbolic, MATH is injective on the set MATH. It is injective on the set MATH by assumption, and REF implies that the MATH - images of these two sets are disjoint. Thus MATH is injective. REF holds as follows. For each MATH we have MATH . Thus MATH satisfies REF - REF , and proceeding inductively on orbits, such a MATH can be defined on all of MATH. REF imply that MATH has a (unique) equivariant extension to MATH. Now consider any edge MATH, with endpoints MATH, MATH. REF implies that MATH and so MATH can be mapped to a nontrivial embedded path in MATH. By equivariance this holds for every edge of MATH, as MATH meets every edge orbit. Therefore the equivariant morphism MATH can be constructed as required. |
math/0107008 | We have seen in REF that elementary moves do not change the set of elliptic subgroups of MATH. Thus one implication is clear. For the other implication, we are given cocompact MATH - trees MATH and MATH with the same elliptic subgroups. In particular MATH and MATH define the same partition of MATH into elliptic and hyperbolic elements. We can assume without loss of generality that MATH and MATH are reduced (by performing collapse moves on both). By REF , there is an equivariant morphism MATH, where MATH is the result of subdividing MATH. Note that since subdivision is a special case of an expansion move, it suffices for us to show that MATH and MATH are related by a deformation. As MATH is reduced, it is minimal, and hence MATH is surjective. By REF , MATH factors as MATH, where MATH induces an isomorphism on quotient graphs and MATH is a finite composition of folds. By REF the map MATH factors as a finite composition of parabolic folds, as MATH. Now the original map MATH has the form MATH . Suppose the parabolic fold MATH is performed at MATH. Consider the stabilizer MATH in MATH. By our assumptions on MATH and MATH, this subgroup has a fixed point MATH. By equivariance of the maps, MATH fixes the vertex MATH of MATH, and hence MATH fixes a vertex of MATH. Now we can apply REF to the map MATH and factor it into a finite composition of type II multi-folds. Repeating this argument for each MATH, we conclude that MATH factors into a finite composition of folds and type II multi-folds. Recall that the map MATH preserves hyperbolicity of elements of MATH. It follows that each individual fold or multi-fold also has this property, as equivariant maps cannot make elliptic elements hyperbolic. Therefore, applying REF to each fold and multi-fold, MATH is a finite composition of elementary moves. This concludes the proof of the theorem. |
math/0107008 | Note that every MATH maps MATH to itself. Then if MATH fixes MATH, it also fixes the path from MATH to MATH, which contains MATH. Thus MATH. |
math/0107008 | First note that MATH because the right hand side is a MATH - invariant subtree of MATH (equal to the subtree spanned by the MATH's). To prove the lemma it remains to show that every MATH - telescope is contained in MATH. Note that every MATH - telescope has the property that its endpoint stabilizers are strictly larger than any of its edge stabilizers (recall that MATH meets MATH and MATH only in its endpoints). The Hyperbolic Segment REF shows that such a segment is contained in the axis of a hyperbolic element. Since MATH contains every hyperbolic axis, we are done. |
math/0107008 | Suppose that MATH. The geometric edges MATH and MATH in MATH are separated by at least one edge. Consider the smallest segment in MATH containing these two edges. Without loss of generality assume that its vertices, in order, are MATH for some MATH. Let MATH be the smallest tree containing MATH for each MATH, and let MATH be the span of MATH. Then each MATH is the union of the characteristic subtrees in REF and the MATH - telescopes joining them. More specifically, for each MATH we have that MATH as the right hand side is certainly a (connected) tree containing the required subtrees. Similarly, MATH. Now suppose that MATH. This implies that MATH . Together with the previous remarks we now have that MATH . Consider the MATH - telescope MATH. It is contained in MATH and its interior meets no MATH, MATH. Therefore REF implies that MATH . Thus every edge of MATH is contained in another MATH - telescope, contradicting REF . |
math/0107008 | Let MATH be a MATH - telescope with injective sub-telescope MATH. If MATH then it is proper because MATH and MATH are maximal. If MATH for some MATH and MATH, then MATH because MATH contains a MATH - minimal edge. As MATH, we also have MATH. Let MATH be the edge with initial vertex MATH and terminal vertex MATH. Then MATH has initial vertex MATH, and hence MATH. Note that MATH and MATH, so we have MATH. As MATH is strongly slide-free, MATH. But then MATH is inverted by an element of MATH, a contradiction. Thus MATH cannot hold, and similarly MATH, so MATH is proper. |
math/0107008 | Suppose that MATH is MATH - minimal for the MATH - telescope MATH. If MATH is contained in another MATH - telescope, then by REF the two telescopes have a common endpoint. Suppose the second MATH - telescope is MATH. Then since MATH and MATH, we must have MATH, as MATH is a telescope. Now since MATH is strongly slide-free, MATH and MATH for some MATH. Then MATH, fixing the endpoint in MATH. Since MATH, MATH is fixed by MATH. But MATH as MATH, and hence MATH, a contradiction. |
math/0107008 | Let MATH for MATH and MATH. Let MATH be the image of MATH under the collapse move. Then the image segment MATH is monotone with MATH minimal if and only if MATH for MATH. Of course, MATH and MATH may be the same vertex for various values of MATH. MATH : REF implies that MATH for each MATH. Now suppose that an element MATH as in REF exists. There are two cases. First, if MATH then MATH. Recall the elementary fact that if MATH is any segment with MATH then MATH for every MATH. Thus, as MATH, we have that MATH, contradicting REF . The second case is that MATH, meaning that MATH for some MATH, so MATH. Notice that MATH, so again we have that MATH, contradicting REF . CASE: If MATH then no vertex stabilizer changes during the collapse move. Thus MATH for all MATH and so MATH for each MATH. Otherwise, if MATH, then the assumption REF implies that for every MATH and every MATH, either MATH or MATH or MATH. If MATH then MATH, so MATH. If MATH with MATH then either MATH or MATH. If MATH then MATH and so MATH. If MATH then MATH, so MATH. Combining all cases we have that MATH for MATH, and hence the segment MATH is monotone with MATH minimal. |
math/0107008 | Let MATH for MATH and MATH. Let MATH be the image of MATH under the collapse move. Then MATH for each MATH. For each MATH, MATH and MATH are not both in MATH; this would imply that MATH, but this set is empty because MATH admits a collapse move. Similarly, MATH and MATH are not both in MATH. If MATH then there is a MATH which fixes MATH but takes MATH to MATH. But then MATH, contradicting the assumption that MATH, and so MATH and MATH are not both in MATH. Finally suppose that MATH, for some MATH. Notice that since MATH collapses to MATH, MATH. The previous cases imply that MATH and so MATH is not collapsed, and has the same stabilizer after the move. Then monotonicity of MATH implies that MATH. Hence MATH. There is a MATH fixing MATH REF and taking MATH to MATH. Then MATH, and therefore MATH fixes MATH, as this edge lies between MATH and MATH. This contradicts MATH, and hence MATH and MATH are not both in MATH. These facts imply that if MATH is collapsed (that is, MATH) then MATH is not collapsed, unless MATH and MATH. Now consider an edge MATH. If MATH is not collapsed then MATH by monotonicity, and so MATH. If MATH is collapsed then either MATH or MATH. In the first case MATH holds. So suppose that MATH. Then MATH, and MATH. Then, assuming that MATH, the edge MATH is not collapsed; the exception MATH, MATH, is ruled out because MATH. Thus after the collapse MATH is present, with stabilizer MATH, and MATH by monotonicity. Hence MATH. Thus we have shown that MATH for all MATH, and for MATH if MATH. If MATH and MATH then MATH by equivariance. |
math/0107008 | Recall that collapsibility of MATH means that MATH and MATH. Suppose that REF do not hold, so that MATH meets both MATH and MATH. Relabeling if necessary, we have MATH and MATH for some MATH. Note that by REF the endpoints of MATH are in MATH, and hence MATH. For every MATH in MATH, MATH does not meet MATH and therefore MATH. If MATH then MATH is the endpoint MATH. The edge MATH is in MATH, and since MATH we must have MATH. If MATH and MATH do not meet the interior of MATH then REF now holds. The case MATH is similar. Otherwise, we can assume that MATH for all MATH. Then REF implies that MATH is in another MATH - telescope, which has a common endpoint with MATH by REF . There are now two cases. Suppose first that the MATH - telescope containing MATH has the form MATH. Then MATH is in the MATH - telescope MATH, so REF implies that MATH. Note that MATH separates MATH from MATH, and so MATH separates MATH from MATH. Hence MATH and so MATH. Thus MATH or MATH. The MATH - telescope MATH is separated by MATH into two segments, MATH and MATH. The first is contained in the MATH - telescope MATH. The second is mapped by MATH either to MATH or to MATH, and hence MATH is contained in the MATH - telescope MATH. Therefore MATH is the only geometric edge of MATH that is not contained in another MATH - telescope. It follows, from REF , that MATH is a MATH - minimal edge for MATH. But the vertex MATH is in two MATH - telescopes, contradicting REF . The second case, when the MATH - telescope containing MATH has the form MATH, yields a similar contradiction. |
math/0107008 | First note that elementary moves do not change the set of maximal stabilizers, and that MATH possesses maximal stabilizers, because it is proper. Then MATH and MATH. To avoid confusion, we will decorate segments with subscripts to indicate which tree they are in. REF under collapsesLet MATH be a MATH - telescope. Suppose the edge MATH is collapsed, where MATH. First we arrange that the MATH - minimal edge MATH is not contained in MATH. If MATH, then we have MATH for some MATH (replacing MATH by its inverse if necessary). Then MATH, and so MATH is not an endpoint of MATH. Let MATH be the unique edge in MATH. Since MATH is a telescope, MATH, and so MATH is a MATH - minimal edge. If MATH then MATH is an endpoint of MATH, which implies that this telescope is not proper. However, MATH is a proper telescoping of MATH by REF , so we conclude that MATH. Note also that MATH is not contained in any other MATH - telescope of MATH, for then MATH would be in two MATH - telescopes, contradicting REF . Finally we note that MATH. If MATH then MATH, but this set is empty as MATH is collapsible. If MATH then MATH for some MATH, but then MATH. Thus, replacing MATH with MATH if necessary, there is a MATH - minimal edge MATH. Then MATH is present after the collapse move. We can orient MATH so that MATH separates MATH from MATH. Applying REF to the monotone segments MATH and MATH, we find that MATH is a telescope with MATH - minimal edge MATH, unless MATH and MATH is in the interior of MATH for some MATH, and MATH. If this occurs, then note that MATH by REF , and so MATH by REF . Also, MATH for all MATH, since no MATH meets the interior of MATH. Hence, by REF , MATH is contained in a MATH - telescope. REF implies that this telescope has a common endpoint with MATH. Suppose the telescope is MATH. Note that the subsegment MATH cannot contain a MATH - minimal edge, because MATH. The segment MATH is the intersection of MATH - telescopes MATH and MATH, and hence it does not contain a MATH - minimal edge. Thus the MATH - minimal edge promised by REF can only be MATH or MATH. Hence, MATH is a MATH - minimal edge for MATH (and so is MATH). Now MATH is in two MATH - telescopes, contradicting REF . A similar contradiction ensues if the MATH - telescope containing MATH has the form MATH. Therefore MATH is a telescope with MATH - minimal edge MATH. It is clear that MATH is not contained in any other MATH - telescope of MATH, since every MATH is the image of MATH, and this property holds in MATH. Lastly, we need to know that MATH is contained in the injective sub-segment MATH given by REF of MATH. In REF we show that MATH indeed satisfies REF , and that MATH is the image of MATH. Therefore MATH. REF under expansionsSuppose that an expansion move is performed, creating MATH with MATH. By REF , MATH is a telescope, unless the following conditions hold: MATH for some MATH with MATH equal to one of the endpoints of MATH, and MATH. But then MATH, which contradicts maximality of MATH or MATH (one of which is equal to MATH). REF under collapsesLet MATH be a MATH - telescope, and suppose that the edge MATH is collapsed, where MATH. The image of MATH in MATH is MATH, which is a MATH - telescope by REF . Let MATH and MATH be the images of MATH and MATH respectively. If MATH then also MATH. Note that MATH and MATH are telescopes, since they are nontrivial sub-segments of the telescope MATH. Now, to establish REF for MATH, it remains to show that MATH and MATH are injective (we are not required to show that MATH is proper). If MATH is not injective then two of its interior vertices are in the same orbit. The preimages in MATH of these vertices must be incident to opposite ends of the geometric edge orbit MATH. By relabeling if necessary, we can arrange that MATH and MATH are in the interior of MATH for some MATH. Injectivity of MATH implies that if MATH for some MATH, then the endpoints of MATH are among the vertices MATH, MATH, MATH, and the endpoint of MATH in MATH. But MATH does not join MATH to MATH because these vertices have separate images in MATH. Also MATH does not join an endpoint of MATH to one of MATH, MATH, because the images of these vertices are in the interior of MATH. Therefore MATH. By REF we have MATH, and so MATH. We now obtain a contradiction by applying REF , since none of its three conclusions holds. Thus, MATH is injective. Similarly, MATH is injective. REF under expansions Suppose that an expansion move is performed at MATH, creating the edge MATH with MATH. Consider the intersection of MATH with MATH. First we observe that no two adjacent edges of MATH are in MATH. For if this occurs, then the two edges cannot be coherently oriented. Their common endpoint would be in MATH, but this intersection is empty because MATH is collapsible. Also, if the two edges were incoherently oriented, then they would be related by an element of MATH fixing their common endpoint. This element would violate the telescope property of MATH, which holds by REF . Thus, no two geometric edges in MATH collapse to the same vertex of MATH. The telescope MATH is now obtained from MATH by replacing each of various vertices in MATH by a single geometric edge in MATH (vertices are omitted if the corresponding elements of MATH are not in MATH). Note that REF implies that MATH contains at most two vertices of MATH, and if there are two, then they are in MATH and MATH respectively, and are related by MATH. Suppose that MATH is in the interior of MATH, and suppose that MATH expands to MATH. Let MATH be the two neighboring edges of MATH. Then MATH and MATH are separated by MATH in MATH, and MATH and MATH are in MATH, separated by MATH. Hence MATH. Similarly if MATH is the endpoint MATH and MATH, then MATH. In each of these cases, MATH does not meet MATH by REF , and so MATH is unchanged during the move, and hence MATH and MATH are injective. Also, the relation MATH holds. Next suppose that MATH for some MATH. If MATH then MATH and MATH is injective. Otherwise, MATH for some MATH, and MATH. Recall that REF implies that MATH meets MATH in MATH and MATH only. Then MATH meets MATH in zero, one, or two geometric edges. If MATH, let MATH and MATH be the vertices of MATH that are incident to MATH, and which map to MATH and MATH respectively under the collapse move from MATH to MATH. Then MATH and MATH are injective and MATH. If MATH meets MATH in one edge MATH which collapses to MATH, let MATH be the endpoint of MATH closest to MATH. Let MATH be the vertex corresponding to MATH. Then MATH, and MATH and MATH are injective. If MATH contains two edges of MATH, then call the edges MATH and MATH, where MATH collapses to MATH and MATH collapses to MATH. Let MATH and MATH be the endpoints of MATH. Then MATH and MATH map bijectively to MATH and MATH respectively under the collapse from MATH to MATH. Note that MATH. We have that MATH for MATH. Then MATH implies MATH. Now if MATH, then MATH, contradicting the fact that MATH is collapsible. Hence MATH. This implies that MATH fixes MATH, because MATH. Since MATH is monotone with MATH minimal, MATH also fixes all of MATH. Therefore MATH. Setting MATH and MATH, we now have that the injective segment MATH is the sub-telescope required by REF , where the element MATH has taken the place of MATH. Lastly suppose that MATH meets the interior of MATH. Then REF implies that MATH meets MATH in no other point. Then MATH and MATH are injective and so REF is satisfied by MATH. REF under expansionsSuppose that an expansion move is performed. Let MATH be a MATH - minimal edge for MATH. In REF we showed that MATH is MATH - minimal for MATH. It is contained in the segment MATH, according to the description of MATH obtained in REF . Also MATH is not contained in another MATH - telescope after the move, and so it is MATH - minimal. |
math/0107008 | REF together imply that the property of being a proper telescoping of MATH, when MATH is strongly slide-free, is invariant under elementary deformations. Then since MATH is a proper telescoping of itself, MATH is a proper telescoping of MATH. If MATH is proper then MATH as remarked earlier. Uniqueness of the isomorphism follows from the fact that the only equivariant self map of a proper MATH - tree is the identity. To prove this fact, let MATH be equivariant. If MATH is proper then the stabilizer of a vertex does not fix any other vertex. Since equivariance requires MATH for all MATH, the map is the identity. For the last conclusion of the theorem, let MATH be obtained from MATH by a maximal sequence of collapse moves. Then MATH is related to MATH by a deformation, so it is a proper telescoping of MATH by REF. Since MATH is reduced, every MATH - telescope in MATH has length MATH, by REF . Thus MATH. The case when MATH is cocompact and reduced follows from the statement just proved. |
math/0107008 | Call a subgroup of MATH weakly elliptic if its elements are all elliptic. The vertex stabilizers of MATH and MATH are maximal weakly elliptic subgroups, by properness (and the Hyperbolic Segment Condition). Let MATH be a vertex stabilizer of MATH and consider its action on MATH. By REF , MATH fixes either a vertex or an end of MATH. Suppose MATH fixes MATH. Let MATH be the set of elliptic elements of MATH; then MATH. Note that MATH is a subgroup (and is weakly elliptic), and hence MATH by maximality. Suppose first that MATH is rational, meaning that MATH contains a hyperbolic element MATH (that is, MATH is an endpoint of some hyperbolic axis). Then MATH is maximal weakly elliptic and is also contained in MATH, hence is equal to MATH. Since MATH is hyperbolic, the subgroups MATH and MATH are the stabilizers of different vertices of MATH, contradicting properness. Next assume that MATH is irrational, so that MATH, and hence MATH. Let MATH be any sequence of vertices of MATH tending monotonically to MATH. Then MATH is the increasing union MATH. We can assume that the inclusions MATH are strict infinitely often, for otherwise MATH would fix a vertex of MATH. Each MATH is equal to the intersection MATH, and therefore MATH . Now we ask whether a vertex stabilizer of MATH can have this structure. Each stabilizer MATH fixes either a vertex or an end of MATH, and in the latter case the end must be irrational as above. Then, by maximality of MATH, it must be equal to the stabilizer of that vertex or end. Thus we have MATH for some MATH. Note that MATH is infinite, for otherwise only finitely many of the inclusions MATH would be strict. Let MATH denote the ball of radius MATH at MATH. As MATH is locally finite, MATH separates MATH into finitely many connected components, one of which meets MATH in an infinite set. Thus, by passing to a subsequence we can arrange that MATH is contained in a connected component of MATH. Note that REF remains valid after passing to any subsequence. Now consider the subtree spanned by MATH. MATH is disjoint from this subtree, so there is a vertex MATH separating MATH from every MATH. Then MATH for all MATH, hence MATH for all MATH. But MATH, so REF yields MATH. This contradicts properness of MATH as MATH. |
math/0107008 | In fact, we show directly that MATH and MATH are related by a deformation, which implies the conclusion by REF . The proof is nearly the same as the proof of REF . We can assume without loss of generality that MATH is reduced, by performing collapse moves. By REF there exist a MATH - tree MATH obtained from MATH by subdivision and an equivariant morphism MATH. This morphism is surjective because MATH is proper and hence minimal. Thus, by REF , it factors as MATH where MATH is a finite composition of folds and MATH induces an isomorphism of quotient graphs. Since MATH is proper, REF implies that MATH is an isomorphism. Thus, MATH is a finite composition of folds. Each individual fold is equivariant and therefore it preserves ellipticity of elements of MATH. Since their composition MATH preserves hyperbolicity, each fold must preserve hyperbolicity as well. Therefore REF applies to each fold and MATH and MATH are related by an elementary deformation. Then MATH and MATH are related by a deformation. |
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