paper
stringlengths 9
16
| proof
stringlengths 0
131k
|
|---|---|
math-ph/0006021
|
CASE: If MATH there is nothing to prove. So, let MATH. We show that we can assume MATH for the proof of MATH- compactness: Let MATH . Then MATH. We set MATH so that MATH . Since MATH is continuous and bounded MATH. If MATH is MATH-elliptic, that is, MATH, then we get MATH, that is . MATH is MATH-elliptic. Denote the spectral projections of MATH with MATH. Then obviously MATH, so that it suffices to test MATH for MATH-compactness. Finally we set MATH. Then MATH is MATH-elliptic by definition, positive by construction, and strictly positive because MATH. If we denote the spectral projections of MATH by MATH then MATH . By REF and therefore MATH, so that MATH if MATH. If MATH we write MATH and apply REF . Hence it suffices to test MATH for MATH-compactness. CASE: We show that every spectral projection MATH for MATH can be produce by continuous functional calculus from MATH, so that it belongs to MATH. For this we note that MATH is densely defined (MATH is regular), self-adjoint and bounded below by MATH. Thus MATH exists, is positive and self-adjoint. By the spectral mapping theorem we have MATH . Therefore, the operator MATH exists for all MATH in the resolvent set of MATH. Since the function MATH is continuous and bounded on every closed set not containing MATH, MATH belongs to the MATH-algebra generated by MATH for every MATH and therefore belongs to MATH, that is, it is MATH-compact. Since MATH for a suitable closed path MATH in MATH with winding number MATH fulfilling MATH, MATH belongs to the MATH-algebra generated by all MATH. CASE: Let MATH be MATH-trace class. Since MATH the spectral projections inherit the trace class property from MATH.
|
math-ph/0006021
|
MATH is the spectral projection of a self-adjoint operator and therefore self-adjoint, and MATH-compact by REF . Thus the eigenspace MATH is the image of a closed adjointable projection MATH and therefore a closed complementable MATH- module. Since the projection MATH is MATH-compact MATH is algebraically finitely generated and projective, because algebraically finitely generated MATH-modules MATH are just the ones with unital MATH and automatically projective CITE. If MATH then so is MATH by REF , and under the same hypotheses we can apply REF .
|
math-ph/0006021
|
Let MATH, so that MATH for MATH, that is . MATH has at least MATH components in MATH. Then MATH since all projections occurring in this sum are MATH-trace class by REF .
|
math-ph/0006021
|
CASE: Let MATH and MATH. By assumption there is MATH such that MATH . For all MATH we have MATH, and MATH. We set MATH. Then MATH CASE: by definition. CASE: because MATH for all MATH. CASE: Let MATH, MATH. Then there is MATH with finite spectrum so that MATH. If MATH is invertible then there is nothing to prove, otherwise we choose MATH so that MATH. Then MATH is invertible, and MATH . CASE: Let MATH. We show first that we can approximate MATH by a self-adjoint element with finitely many gaps. For this we choose a sequence MATH of pair-wise distinct real numbers that are dense in the interval MATH. We set MATH and choose inductively MATH such that MATH . Furthermore MATH so that MATH is in the resolvent set of MATH. The rate of approximation is chosen just so that these gaps remain open (although possibly become smaller) in every step. Now we construct an approximation with finite spectrum for each MATH. For this we arrange, for fixed MATH, the MATH into increasing order, say MATH, and set MATH. We define a continuous monotonically increasing function MATH by MATH such that MATH (see REF ). MATH compresses the spectrum between two gaps into one point. The spectral projections MATH belong to intervals with endpoints in the resolvent set so that they are in MATH. Therefore MATH . Thus MATH, and with the spectral family MATH of MATH we get MATH . Since MATH for MATH and MATH is bounded, we can make the approximation arbitrarily good.
|
math-ph/0006021
|
CASE: As is well known, the spectrum of MATH is MATH, all irreducible representations MATH have dimension MATH. The canonical trace of MATH is MATH with the canonical trace MATH on MATH. Minimal projections have rank MATH in the fiber, and so the NAME constant is MATH. CASE: MATH has real rank zero by CITE. Since MATH is simple and non-elementary also we get MATH from CITE.
|
math-ph/0006021
|
REF .
|
math-ph/0006021
|
Let MATH. Since MATH is closed there is a neigborhood MATH of MATH and MATH with MATH, MATH. Then MATH and therefore MATH . Since MATH is smooth also, it leaves MATH invariant.
|
math-ph/0006021
|
This follows from associativity MATH and the projectivity REF : MATH .
|
math-ph/0006021
|
Let MATH. We choose MATH with MATH and set - a priori depending on MATH - MATH. If MATH, MATH, we get MATH that is, MATH depends on the value of MATH at the point MATH only; hence we omit MATH in the notation. The morphism property follows from the corresponding property of MATH, and from MATH. MATH induces MATH by construction. If there is a proper lift MATH then MATH defines the strict morphism we look for: MATH .
|
math-ph/0006021
|
Easy consequences of the cocycle property.
|
math-ph/0006021
|
For MATH we have by REF since a left Hilbert-MATH-module is a NAME space with conjugated scalar product (complex linear in the first argument, anti-linear in the second). For MATH we get after identifying MATH with MATH . This shows that the structures coincide.
|
math-ph/0006021
|
By REF we have for MATH . Therefore, the completion of MATH with respect to MATH is contained in the one with respect to MATH, that is, in MATH. But MATH is dense in MATH. We get the scalar product of the GNS representation with respect to MATH for MATH from MATH . Since MATH is dense the GNS representation space is exactly MATH.
|
math-ph/0006021
|
For tensor products of left NAME modules we have in general MATH . The statement about MATH is well known in the untwisted case since the opposite of left multiplication is right multiplication. It is easy to check that this holds in the twisted case also.
|
math-ph/0006021
|
Set MATH. Then MATH dense, we set MATH as operators on vector spaces. MATH is adjointable since MATH is symmetric and gauge-periodic: For MATH we have MATH . Finally, MATH is dense in MATH because MATH is essentially self-adjoint; therefore, MATH is regular.
|
math-ph/0006021
|
As a set MATH. We choose the topology so that the natural projection MATH is continuous and open: The topology is generated by the tubular neighborhoods MATH for open sets MATH, continuous fields MATH and MATH . It is easy to check that the tubular neighborhoods generate a topology on MATH with the desired properties. On the fibers MATH it induces the strong topology since the intersections MATH of the fibers with the tubular neighborhoods are just norm balls in the fiber.
|
math-ph/0006021
|
It is well known that right MATH-modules MATH and left MATH-modules MATH are in one-to-one correspondence. So we just have to verify the corresponding NAME module structures: Let MATH. We denote by MATH and MATH corresponding elements in MATH respectively, MATH. then MATH . Since MATH as NAME space we have MATH. Furthermore, for MATH so that MATH and MATH are isomorphic, and so are the corresponding closures and multiplier algebras.
|
math-ph/0006032
|
Diagonality of MATH was stated in REF . For MATH we first calculate MATH . Here we use the notation of REF. In the third line, the summation index was shifted and REF was used. It remains to show that writing coefficients MATH the coproduct MATH has got the following form: MATH . In the last line, the summation index was shifted and, using REF several times, the coefficient MATH was cast in the same form as in REF. Both REF are equal if and only if for all MATH and for all MATH . The first condition holds since the recursion relations REF are satisfied. The second simplifies to MATH which can be verified in a direct computation. The procedure to prove the assertion for MATH is similar. In order to understand in detail how cocommutativity arises from the choice of the coefficients MATH in REF, it is important to contrast the calculation for MATH with that for MATH. Thus we give the main steps in detail again. Firstly, MATH again using the notation of REF. In the third line, the summation index was shifted and REF used. Writing MATH it remains to show that MATH . In the last line, the summation index was shifted and, using REF several times, the coefficient MATH was cast in the same form as in REF. Both REF are equal if and only if for all MATH and for all MATH . The first condition holds since the recursion relations REF are satisfied. The second simplifies to MATH which can be verified in a direct computation.
|
math-ph/0006032
|
The proof is completely analogous to that of REF . Here REF is used to deal with the coefficients MATH and with the shifts of the summation index which are necessary here.
|
math-ph/0006032
|
First we show that all factors in the denominator of the triangular part of MATH cancel with the numerator of MATH. Poles can be present only if MATH, but this implies MATH. The denominator of the triangular part, see REF, applied to MATH is MATH . Since MATH, it cancels with the numerator of MATH as given in REF. The denominator of MATH has a pole if and only if MATH where MATH. But since MATH and MATH, this implies MATH for which MATH is reducible REF .
|
math-ph/0006032
|
The denominator of the triangular part of MATH acting on MATH is given by MATH where MATH. Some factors cancel with the numerator of MATH (see the expression for MATH in REF). The remaining denominator of the triangular part is MATH since MATH and MATH. The denominator of the diagonal part MATH is MATH . Poles in MATH can arise from both REF. If MATH or MATH, then MATH, and both products REF are empty. If MATH and MATH, there is a MATH contribution to the product REF which extends in this case from MATH to MATH. The second product REF runs from MATH to MATH. For arbitrary MATH, these bounds are not exceeded.
|
math/0006001
|
Consider the product MATH and observe the condition of vanishing even-even block, which gives MATH, and other conditions follows obviously.
|
math/0006001
|
It simply follows from supermatrix multiplication law and general previous considerations.
|
math/0006001
|
In our case MATH .
|
math/0006001
|
Follows from REF .
|
math/0006001
|
For the difference using the band REF we have MATH. Then we expand in NAME series around MATH and obtain MATH, where MATH denotes MATH-th derivative which is a finite series for the power-type REF .
|
math/0006001
|
To find the difference MATH we use the expansion REF and the band conditions for components REF - REF .
|
math/0006003
|
The proof is standard. Consider a REF-complex MATH associated to a presentation of MATH with the minimal number MATH of generators. By general position there exists an embedding MATH inducing an isomorphism of fundamental groups. Then a regular neighborhood of MATH in MATH has a handlebody decomposition with MATH REF-handles. Since the complement is REF-connected then by CITE it is g.s.c. for MATH and this yields the claim.
|
math/0006003
|
Reversing the handlebody decomposition of MATH one finds a decomposition from MATH without REF-handles. One slides the handles to be attached in increasing order of their indices. Using NAME Theorem it follows that MATH, where MATH is the number of REF, and thus MATH.
|
math/0006003
|
For MATH this is clear. Thus we suppose MATH. For any compact MATH choose some MATH such that MATH. Consider a loop MATH. Then MATH bounds an immersed (for MATH embedded) REF-disk MATH in MATH. We can assume that MATH is transversal to MATH. Thus it intersects MATH along a collection of circles MATH. Since MATH one is able to cap off the loops MATH by some immersed REF-disks MATH. Excising the subsurface MATH and replacing it by the disks MATH one obtains an immersed REF-disk bounding MATH in MATH.
|
math/0006003
|
Actually the following more general engulfing result holds: If MATH is a REF-dimensional polyhedron whose boundary MATH is contained in MATH then there exists an isotopy of MATH (with compact support), fixing MATH and moving MATH into MATH. Suppose that MATH. One reverses the handlebody decomposition of MATH and obtains a decomposition from MATH without REF- or REF-handles. Assume that we can move MATH such that it misses the last MATH handles. By general position there exists an isotopy (fixing the last MATH handles) making MATH disjoint of the co-core ball of the MATH-th handle, since the co-core disk has dimension at most MATH. The uniqueness of the regular neighborhood implies that we can move MATH out of the MATH-th handle (see for example, CITE), by an isotopy which is identity on the last MATH handles. This proves REF . This yields the result of REF by taking for MATH any REF-disk parameterizing a null homotopy of MATH.
|
math/0006003
|
For MATH it is well-known that g.s.c. is equivalent to w.g.s.c. which is also equivalent to s.c.i. if the manifold is irreducible. For MATH this is a consequence of NAME 's result stating the equivalence of g.s.c. and simple connectivity in the compact case (see CITE). If MATH is w.g.s.c. then it has an exhaustion by compact simply connected sub-manifolds MATH (by taking suitable regular neighborhoods of the polyhedra). One can also refine the exhaustion such that the boundaries are disjoint. Then the pairs MATH are REF-connected, hence CITE they have a handlebody decomposition without REF-handles. Gluing together these intermediary decompositions we obtain a proper handlebody decomposition as claimed.
|
math/0006003
|
There exists at least one group MATH, for instance MATH. Further if MATH and MATH verify the condition MATH, then their product MATH does. Thus, NAME 's Lemma says that a maximal element for the lattice of subgroups verifying this property (the order relation is the inclusion) exists.
|
math/0006003
|
First, MATH satisfies the condition MATH otherwise the minimality will be contradicted. Pick an arbitrary MATH satisfying this condition. If MATH it follows that MATH, hence by a transfinite induction we derive our claim.
|
math/0006003
|
We set MATH in the sequel. We establish first: If the pair MATH is stably compressible then it is MATH-compressible. We will use a transfinite recurrence with the inductive steps provided by the next two lemmas. Set MATH for the morphism making the pair MATH strongly compressible. If MATH and MATH then MATH. By REF. Alternatively, for any MATH there exists some MATH such that MATH and MATH. One can write uniquely MATH in normal form see REF ., REF as MATH where MATH are non-trivial except maybe MATH. Then MATH. Since the normal form is unique in MATH one derives that MATH has the following property. There exists a sequence MATH of integers for which MATH and MATH . Furthermore MATH implies that (recall that MATH) MATH . However each partial product starting at the MATH-th term and ending at the MATH-th term is a product of conjugates of MATH by elements from the image of MATH: MATH . We used above the inclusions MATH and MATH. Therefore MATH for any MATH such that MATH. This implies that MATH and hence MATH. If MATH and MATH then MATH. One can use the symmetry of the algebraic compressibility and then the argument from the previous lemma. Alternatively, choose MATH and some MATH such that MATH and MATH. Using the normal form as above we find this time MATH hence MATH. Using in an alternate way the two previous lemmas one gets REF . Let MATH be a homomorphism such that MATH is strongly compressible. Set MATH. Then MATH is full in MATH. In particular if MATH are finitely generated and MATH is finitely presented then the subgroup MATH is finitely generated. We already saw that MATH. Set MATH for two subgroups MATH. The proof we used to show that MATH and MATH actually yields MATH and respectively MATH. We remark now that MATH. The left inclusion is obvious. The other inclusion consists in writing any element MATH with MATH, MATH as a product of conjugates by elements of MATH. This might be done by recurrence on the length of MATH, by using the following trick. If MATH, MATH then MATH. Consequently the fact that MATH implies MATH hence all inclusions are equalities. Also MATH since both components MATH and MATH are contained in MATH. This shows that MATH hence MATH is full in MATH. We take therefore MATH. It suffices to show now that each of the groups MATH and MATH are finitely generated. MATH is finitely generated since it is the image of MATH. Furthermore MATH is finitely generated since MATH is finitely presented and MATH is finitely generated. REF shows that MATH must be normally finitely generated. This proves the claim. Assume that MATH is full in MATH, where MATH is finitely generated. If the pair MATH is MATH-compressible then it is stably compressible. Consider MATH big enough and a surjective homomorphism MATH. This implies that MATH. We have to show that any MATH is in MATH. Recall that MATH. Set MATH. Then MATH can be written as a product of conjugates of elements MATH by elements MATH. Choose MATH so that MATH and MATH so that MATH. Then MATH and MATH, since MATH. Then REF follows.
|
math/0006003
|
Since MATH is w.g.s.c. there exists a compact REF-connected submanifold MATH of MATH such that MATH. We can suppose MATH, without loss of generality. From now on we will focus on the pair MATH and suppress the index MATH, and denote it MATH, for the sake of notational simplicity. The pair MATH is stably compressible. Let MATH and MATH be the homomorphisms induced by the inclusions MATH, and MATH. If MATH has several components then we choose base points in each component and set MATH for notational simplicity. Let us consider a handlebody decomposition of MATH (respectively a connected component) from MATH, MATH where MATH is a handle of index MATH. We suppose the handles are attached in increasing order of their index. Since the distinct components of MATH are not connected outside MATH (by REF ) REF-handles which are added have the extremities in the same connected component of MATH. Set MATH (respectively MATH) for the submanifold obtained by attaching to MATH only the handles MATH of index MATH (respectively those of index MATH). Then MATH, because adding higher index handles does not affect the fundamental group and we know that MATH. The pair MATH is strongly compressible. Let MATH be the set of attaching circles for REF-handles of MATH and MATH be the corresponding core of REF-handles MATH (MATH). Since MATH is a REF-disk embedded in MATH it follows that the homotopy class MATH vanishes in MATH. Let MATH be the normal subgroup generated by the homotopy classes of the curves MATH which are contained in MATH. Notice that this amounts to picking base points which are joined to the loops. Therefore the image of MATH under the map MATH, induced by the inclusion, is zero. In particular its image in MATH is zero. On the other hand the images of the classes MATH in MATH normally generate all of the group MATH because MATH. These two properties are equivalent to the strong compressibility of the pair MATH which in turn implies that of MATH. Rest of the proof of REF : Assume now that the number of REF MATH is MATH. Notice that MATH, because it can be obtained from MATH by adding REF-handles and REF-handles we added do not join distinct boundary components, so that each one contributes with a free factor. In particular the inclusion MATH induces a monomorphism MATH. Observe that MATH can also be obtained from MATH by adding MATH-handles hence the inclusion MATH induces the isomorphism MATH. The same reasoning gives the isomorphism MATH. In particular we can view the subgroup MATH as a subgroup of MATH. The previous lemma tells us that MATH lies in the kernel of MATH and also projects epimorphically onto MATH. The identification of the respective maps with the morphisms induced by inclusions yields our claim. This finishes the proof of REF .
|
math/0006003
|
One can realize the homomorphism MATH by a disjoint union of bouquets of circles MATH. There is one bouquet in each connected component of MATH. One joins each wedge point to the unique connected component of MATH for which that is possible by an arc, and set MATH for the manifold obtained from MATH by adding a regular neighborhood of the bouquets MATH in MATH (plus the extra arcs). This is equivalent to adding REF-handles with the induced framing. The kernel MATH is normally generated by a finite number of elements MATH. Consider a finite presentation MATH. We know that MATH. Furthermore the composition map MATH is surjective (since MATH is). The first map is the free product of the natural projection with the identity. Therefore MATH is a presentation of the group MATH. REF (see CITE, p. REF) states that any presentation on finitely many generators of a finitely presented group has a presentation on these generators with only finitely many of the given relations. Applying this to MATH one derives that there exist finitely many elements which normally generate MATH in MATH. Then the images of these elements in MATH normally generate MATH (the projection MATH is surjective). This yields the claim. The elements MATH are also in the kernel of MATH. The map MATH induced by the inclusion is injective because MATH is obtained from MATH by adding MATH-handles (dual to REF-handles from which one gets MATH starting from MATH), and MATH. Thus the map MATH factors through MATH and any element in the kernel must be in the kernel of the first map, as stated. The dimension restriction MATH implies that we can assume MATH are represented by embedded loops having only the base point in common. Then MATH bound singular REF-disks MATH. By a general position argument, one can arrange such that REF-disks MATH are embedded in MATH and have disjoint interiors. As a consequence the manifold MATH obtained from MATH by attaching REF-handles along the MATH's (with the induced framing) can be embedded in MATH. Moreover MATH is a compact manifold whose fundamental group is the quotient of MATH by the subgroup normally generated by the elements MATH's. The group MATH is normally generated by the elements MATH. By hypothesis the pair MATH is compressible hence MATH contains MATH. Next MATH since MATH has been supposed surjective. Therefore the quotient of MATH by the subgroup normally generated by the elements MATH is trivial.
|
math/0006003
|
We first consider a simpler case: The result holds in the case of a manifold MATH of dimension at least MATH with one end. In this case MATH has an exhaustion MATH with MATH connected for all MATH. MATH has an exhaustion such that the map MATH induced by inclusion is a surjection for each MATH. As MATH is simply connected, by taking a refinement we can assume that each inclusion map MATH is the zero map. As usual we denote MATH and MATH by MATH and MATH respectively. Now, take a handle-decomposition of MATH starting with the boundary MATH. Suppose the core of each MATH-handle of this decomposition is homotopically trivial in MATH, then it is immediate that MATH is a surjection. We will enlarge MATH by adding some MATH-handles and MATH-handles (that are embedded in MATH) at MATH in order to achieve this. Namely, let MATH be the core of a MATH-handle. By hypothesis, there is a disc MATH in MATH bounding the core of each of the MATH-handles, which we take to be transversal to MATH. As the dimension of MATH is at least MATH, MATH can be taken to be embedded. Notice that REF-disks corresponding to all REF-handles can also be made disjoint, by general position. Thus MATH intersects MATH in a collection of embedded disjoint planar surfaces. The neighborhood of each disc component of this intersection can be regarded as a MATH-handle (embedded in MATH) which we add to MATH at MATH. For components of MATH with more than one boundary component, we take embedded arcs joining distinct boundary components. We add to MATH a neighborhood of each arc, which can be regarded as a MATH-handle. After doing this for a finite collection of arcs, MATH becomes a union of discs. Now we add MATH-handles as before. The disc MATH that MATH bounds is now in MATH. Further, the dual handles to the handles added are of dimension at least MATH. In particular we can extend the previous handle-decomposition to a new one for (the new) MATH starting at (the new) MATH with no new MATH-handles. Thus, after performing the above operation for the core of each MATH-handle of the original handle decomposition, the core of each MATH-handle of the resulting handle-decomposition of MATH starting at MATH bounds a disc in MATH. Thus MATH is a surjection. The above exhaustion is in fact MATH-compressible by the following algebraic lemma. Suppose that we have a square of maps verifying the NAME theorem: MATH . Let MATH. Suppose also that MATH is surjective and MATH is the zero map. Then MATH. Observe that MATH, hence MATH. Hence we can define another diagram with MATH, MATH, MATH, MATH: MATH . Notice that the map MATH is well-defined. Again this diagram verifies the NAME theorem. For this diagram, the induced MATH is an isomorphism. It is immediate then that the universal (freest) MATH must be MATH. In fact MATH. Consider the map MATH, where the second arrow consists in replacing any occurrence of MATH by the element MATH and taking the product in MATH. This composition is the identity and the first map is a surjection, hence the map MATH is an isomorphism. The map induced by MATH is zero since MATH is the zero. But the map MATH is the map MATH followed by an isomorphism, hence MATH. This is equivalent to MATH. Note that the above lemma is purely algebraic, and in particular independent of dimension. The two lemmas immediately give us the proposition for one-ended manifolds MATH with MATH. The general case. We now consider the general case of a simply-connected open manifold MATH of dimension at least MATH, with possibly more than one end. We shall choose the exhaustion MATH with more care in this case. We will make use of the following construction several times. Start with a compact submanifold MATH of codimension MATH, with possibly more than one boundary component. Assume for simplicity (by enlarging MATH if necessary) that no complementary component of MATH is pre-compact. As MATH is simply connected, we can find a compact submanifold MATH containing MATH in its interior such that the inclusion map on fundamental groups is the zero map. Further, we can do this by thickening and then adding the neighborhood of a MATH-complex, that is, a collection of MATH-handles and MATH-handles. Namely, for each generator MATH of MATH, we can find a disc MATH that MATH bounds, and then add MATH-handles and MATH-handles as in REF . Thus, as MATH, the boundary components of MATH correspond to those of MATH. We repeat this with MATH in place of MATH to get another submanifold MATH. Observe that as a consequence of this and the simple-connectivity of MATH, the inclusion map MATH is the zero map for any component MATH of MATH. More generally if MATH, then MATH. Similar results hold with MATH and MATH in place of MATH and MATH. Now start with some MATH as above and construct MATH and MATH. Thicken MATH slightly to get MATH. We will eventually choose a MATH, but for now we merely note that it can (and so it will) be chosen in such a manner that the inclusion map MATH is the zero map. Let MATH be the boundary components of MATH. Let MATH be the union of the component of MATH containing MATH and MATH, and define MATH analogously with MATH in place of MATH. Denote the image of MATH in MATH by MATH. We then have a natural map MATH. Let MATH be the component of MATH containing MATH. By adding MATH-handles and MATH-handles to MATH, we can ensure that MATH surjects onto MATH. The proof is essentially the same as that of REF . We start with a handle-decomposition for MATH starting from MATH. We shall ensure that the image in MATH of the core of each MATH-handles is trivial. Namely, for each core, we take a disc MATH that it bounds. By the above remarks, we can, and do, ensure that the disc lies in MATH, and in particular does not intersect any boundary component of MATH except MATH. As in REF we may now add MATH-handles and MATH-handles to MATH to achieve the desired result. Notice that the changes made to MATH in the above lemma do not affect MATH, MATH and MATH for MATH. Hence, by repeated application of the above lemma, we can ensure that all the maps MATH are surjections. Also notice that the preceding remarks show that MATH. Now take MATH, MATH and MATH, and let MATH be the image of MATH in MATH. Then, by the preceding remarks and REF , the diagram MATH satisfies the hypothesis of REF . REF-compressibility for the pair MATH follows. Now we continue the process inductively. Suppose MATH has been defined, choose MATH so that it contains MATH and also in such a manner as to ensure that MATH's exhaust MATH. Then find MATH, MATH and MATH as above. The rest follows as above.
|
math/0006003
|
There is a natural projection map on the spine MATH, which is the fiber bundle projection of MATH (with fiber a REF-disk). When both MATH and MATH are fixed then such a projection map is also defined only up to isotopy. Set therefore MATH . Since MATH does not depend on the particular projection map (in the fixed isotopy class) this number represent a topological invariant of the pair MATH. Hence the claim follows from the following result: MATH . Consider a position of MATH for which the minimum value MATH is attained. A small isotopy make MATH transversal to MATH. Then, for this precise position of MATH there exists some number MATH such that MATH . Denote by MATH the NAME measure on MATH. For any MATH one can move MATH in MATH by an ambient isotopy such that the following conditions are fulfilled: MATH . MATH . The set MATH is an open subset of positive measure. Consider then a flow MATH on the torus MATH which expands a small ball contained in MATH into the complement of a measure MATH set (for example, a small tubular neighborhood of a spine of REF-holed torus). Extend this flow as MATH all over MATH and consider its action on MATH. MATH . The map MATH is the projection of the metric tube around MATH on its spine, hence the Jacobian MATH has bounded norm MATH. It follows that MATH for any MATH, hence the claim. MATH . Set MATH for the map given in coordinates by MATH, MATH. Here the projection MATH provides a global trivialisation of MATH. Then MATH . Therefore for MATH close enough to REF one derives MATH . Since the position of MATH was chosen arbitrary, this inequality survives after passing to the infimum and the claim follows.
|
math/0006003
|
The proof here follows the same pattern as that given by NAME CITE for REF-dimensional case. Let us establish first the following useful property of the wrapping number: If MATH then MATH. This is a consequence of the two lemmas below: MATH. Consider MATH, where MATH is a very thin tube around MATH, and the two projections to MATH and MATH respectively. Using REF one can assume that the conditions MATH hold. For small enough MATH one derives that MATH . This proves that the minimal cardinal of the MATH is not greater than MATH, hence the claim. MATH. We can assume that MATH. Consider an embedding of the MATH-torus MATH for which the value of MATH (as function of MATH) is closed to the infimum in the isotopy class. We will assume that in all formulas below the tori lay in their respective isotopy classes. Then MATH . The last inequality follows from the fact that MATH. In fact MATH implies that MATH intersects any REF-disk MATH (that is, any fiber of the projection MATH) of MATH in at least one point. Then the transversal disk MATH of radius MATH is therefore contained in the tube MATH of radius MATH around MATH, establishing the claimed inclusion. On the other hand the following holds MATH due to the topological invariance of the wrapping number. Letting MATH go to REF in the previous inequality yields the claim. There exist NAME links whose wrapping number has the form MATH for any natural number MATH. The claim is well-known for MATH. One uses NAME 's construction CITE of NAME links by induction on the dimension. If MATH is a NAME link then set MATH. Consider the projection MATH of the solid torus MATH onto MATH (the first and the last factors). Choose some NAME link MATH, and set then MATH. The pair MATH is a NAME link of dimension MATH. The Proposition then is an immediate consequence of: MATH. From the multiplicativity of MATH and the triviality of the projection MATH it is sufficient to prove that MATH. This formula can be checked directly using MATH instead of MATH. For any sequence MATH of positive integers consider a NAME manifold MATH, where MATH. If the sequences MATH and MATH have infinitely many non-overlapping prime factors then the manifolds MATH and MATH are not PL homeomorphic. The proof is similar to that of (CITE, p. REF). Set MATH, MATH, where MATH,MATH, are tori, as above. If MATH is a PL homeomorphism, there exist integers MATH such that MATH, MATH has a prime factor which occurs in MATH but not in MATH, MATH and MATH. We have therefore MATH . We have obtained a contradiction because MATH divides MATH but not the left hand side (which is non-zero also).
|
math/0006003
|
It is sufficient to consider the case of the NAME manifolds since the pair of groups appearing in the product exhaustions are the same as this case. We start with REF-dimensional case, and take for MATH the classical NAME manifold. Recall that MATH is an increasing union of solid tori MATH, with MATH embedded in MATH as a neighborhood of a NAME link. We shall first show that the pair MATH is not MATH-weakly compressible, and hence not stably compressible. We then extend this argument to show that any pair of the form MATH is not MATH-weakly compressible, and hence not stably compressible. By REF , it follows that MATH is not MATH-compressible, and hence not w.g.s.c. Let MATH and MATH be as usual and let MATH, and fix a base point MATH. Then MATH in our usual notation. Note that MATH is normally generated by the meridian of MATH and hence MATH. Thus MATH consists of the homologically trivial elements in MATH. Consider now the cover MATH of MATH with fundamental group MATH. This is MATH with the neighborhood of an infinite component link, say indexed by the integers, deleted. Further each pair of adjacent components has linking number MATH. Pick a lift MATH of the base point MATH, which we use for all the fundamental groups we consider. In this cover, MATH is the image of the bounding torus MATH of the component of this link containing MATH, and MATH is generated by the meridian of this component. Thus, MATH is the fundamental group of MATH, that is, of MATH with a solid torus glued along MATH to kill the meridian. But, because of the linking, the longitude MATH is not trivial in this group, that is, MATH. Since MATH, we see that the NAME link is not MATH-compressible. We shall now consider a pair MATH in some refinement of the given exhaustion. This is homeomorphic to a pair of the form MATH for some MATH. As before pick a base point MATH. Let MATH and let MATH. Note that MATH. In terms of earlier notation, MATH and MATH. Further MATH is normally generated by the meridian of MATH. We have a sequence of subgroups MATH and hence covers MATH of MATH corresponding to these subgroups. Pick lifts MATH of the base point MATH to these covers. Then MATH is generated by the meridian of the component of the inverse image of MATH containing MATH. As the meridian is in MATH, and each MATH is the normal subgroup generated by MATH in MATH, we see inductively that the lift of the meridian is a closed curve in MATH so that the previous sentence makes sense. Let MATH be the result of gluing a solid torus or cylinder to MATH along the component containing MATH so that the meridian is killed. Then by the above MATH. We shall prove by induction the following lemma. MATH lifts to MATH, or equivalently, MATH lifts to MATH. Furthermore the longitude of the lift of MATH is a non-trivial element in MATH. The case when MATH is the above special case. Suppose now that the statement is true for MATH. As the longitude of the lift of MATH is a non-trivial element in MATH, in MATH the inverse image of MATH is a cylinder with a sequence of linked lifts of MATH deleted. Thus, MATH lifts to MATH, and its longitude is linked with other lifts. It follows that the longitude of the lift of MATH is non-trivial in MATH. As a subgroup of MATH, MATH is the image of the lift of MATH containing the base point. As in the special case, as the longitude of this torus is a non-trivial element of MATH, it follows that MATH. Thus MATH is not MATH-compressible. This ends the proof of the claim for the NAME manifold. Observe however that the same proof works for uncountably many similar manifolds - namely we may embed MATH in MATH as a link similar to the NAME link that winds around the solid torus several times. We will use now a recurrence on the dimension and the results of the previous section in order to settle the higher dimensional situation. Consider for simplicity MATH and a NAME manifold MATH which is the ascending union of solid tori as in NAME 's construction. We use the notations from REF below. Then the pair of tori MATH is constructed out of the two NAME links in one dimension less MATH and MATH. As above MATH is normally generated by the meridian and MATH consists of homologically trivial elements of MATH, by using NAME and the fact that MATH is abelian. The cover MATH of MATH with fundamental group MATH is MATH with a thick infinite link deleted. There is an obvious MATH action on the components of this link, and so we can label the boundary tori as MATH, for integer MATH. Let MATH be the longitude curve having the parameters MATH on the torus MATH. Then one can compute the linking numbers MATH and MATH. This follows because both links used in the construction were the standard NAME link. Variations which yield non-zero linking numbers are also convenient for our purposes. Consequently for non-zero MATH we obtained an element which is non-trivial in MATH hence the pair of solid tori is not MATH-compressible. A similar argument goes through the higher compressibility as well. Using suitable variations in choosing the links and mixing the pairs of solid tori as in previous section yields uncountably many examples as in the theorem.
|
math/0006003
|
The proof given in CITE for REF-dimensional statement extends without any essential modification, and we skip the details.
|
math/0006003
|
Consider a connected compact submanifold MATH. Assume that there exists a compact polyhedron MATH with MATH and an immersion MATH such that MATH. One can suppose that MATH is a manifold. The polyhedron MATH is endowed with an immersion MATH into the manifold MATH. Among all abstract regular neighborhoods (that is, thickenings) of MATH there is a MATH-dimensional one MATH, which is called the regular neighborhood determined by the immersion, such that the following conditions are fulfilled: CASE: MATH extends to an immersion MATH. CASE: The image of MATH is the regular neighborhood of the polyhedron MATH in MATH. The construction of the PL regular neighborhood determined by an immersion of polyhedra is given in CITE. The authors were building on the case of an immersion of manifolds, considered previously in CITE. Moreover, if one replaces MATH by the manifold MATH and MATH by MATH we are in the conditions required by the NAME lemma. Consider now a handlebody decomposition of MATH and let MATH be the union of MATH with the handles of index REF. Then MATH. Let MATH be the cores of these extra REF- and REF-handles. By using a small homotopy of MATH one can replace MATH by some embedded REF-disks MATH with the same boundary. Also by general position these REF-disks can be chosen to have disjoint interiors. Both assertions follow from the assumptions MATH, and MATH. This implies that the restriction of the new map MATH, obtained by perturbing MATH, to MATH (and MATH) is an embedding into MATH. Using the uniqueness of the regular neighborhood it follows that MATH can be chosen to be an embedding on MATH. In particular MATH is engulfed in REF-connected compact MATH.
|
math/0006003
|
We have to reconsider the proof of REF . Everything works as above except that the disks MATH cannot be anymore embedded, but only (generically) immersed. They may have finitely many double points in their interior. Then the manifold MATH obtained by adding REF-handles along the MATH has a generic immersion MATH, whose double points MATH are outside of MATH. This implies that MATH is NAME exhaustible. Conversely assume that MATH is NAME exhaustible. Let MATH be a compact submanifold of MATH and MATH be the immersible simply connected polyhedron provided by the NAME exhaustibility property. REF allow us to assume that MATH and MATH are REF-manifolds. Consider now to the proof of the first claim from REF . It is sufficient to consider the case when MATH that is, MATH is obtained from MATH by adding REF- and REF-handles. If MATH is the normal subgroup generated by the attaching curves of REF-handles of MATH then the same argument yields: MATH . Since MATH is a generic immersion we can suppose that MATH is an embedding of the cores of REF-handles and so MATH is an embedding. We have MATH because the double points of MATH are outside MATH. Now the homomorphism induced by MATH on the left side of the diagram MATH is an isomorphism and we derive that MATH . Meantime MATH surjects onto MATH under the map MATH. But MATH is homeomorphic to MATH, hence it is obtained from MATH by adding REF-handles. This shows that the pair MATH is stably compressible, from which one obtains the end compressibility of MATH as in the proof of REF .
|
math/0006003
|
We use the fact that degree-one maps are surjective on fundamental group. Given an exhaustion MATH of MATH, pull it back to MATH of MATH. Notice that MATH where MATH stands for the frontier. One needs then the following approximation by manifolds result. Given two MATH-complexes MATH there exist regular neighborhoods MATH such that MATH, MATH is homotopy equivalent to MATH, MATH is homotopy equivalent to MATH, and moreover MATH is homotopy equivalent to MATH. This uses essentially the fact that MATH are of codimension zero in MATH. Now, the hypothesis applied to the approximating exhaustion consisting of submanifolds implies the existence of a stably-compressible refinement of MATH. Since degree-one maps are surjective on fundamental groups the lemma below permits to descend to MATH. Suppose that the triple MATH of groups surjects onto MATH that is, we have three surjections with diagrams commuting. Then if MATH is strongly (or stably) compressible then so is MATH. The proof is straightforward. The only subtlety above is to make sure the inverse image of boundary components is connected (else we can connect them up in the one-ended case).
|
math/0006003
|
Let MATH be the quotient map. Assume that MATH satisfies the NAME conjecture. Suppose that G violates the NAME conjecture. Then we have generators MATH and relations such that MATH is the trivial group. Let MATH be the map extending MATH by mapping MATH to MATH. This is clearly a surjection, and induces a surjective map MATH. But since the domain of the surjection MATH is trivial, so is the codomain. But this means that MATH is trivial, and so MATH violates the NAME conjecture, a contradiction.
|
math/0006003
|
Suppose that MATH (with some REF-handle or REF-handle added if one boundary component is empty). It is well-known (see CITE) that the homology groups MATH are the same as those of a differential complex MATH, whose component MATH is the free module generated by the MATH-handles. Therefore this complex has the form: MATH . Thus MATH implies that MATH holds. Consider now the handlebody decomposition is turned up-side down: MATH (plus possibly one REF-handle or REF-handle if the respective boundary component is empty). By the NAME theorem it follows that MATH is obtained from MATH by adding one generator for each REF-handle and one relation for each REF-handle. Therefore MATH where MATH is the free group on MATH generators MATH and MATH is a normal subgroup of the free product generated also by MATH words MATH. Consider a virtually geometric REF-manifold MATH such that MATH is surjective. If MATH is a geometric REF-manifold then its fundamental group is residually finite (see for example, REF ). Let MATH be the degree of the letter MATH in the word representing MATH. The result of NAME REF states that for any locally residually finite group MATH and choice of words MATH such that MATH is of (maximal) rank MATH, the natural morphism MATH is an injection. We have therefore a commutative diagram MATH whose vertical arrows are surjections. The kernel of the map induced by inclusion, MATH is contained in MATH. This contradicts our hypothesis. On the other hand if the rank of MATH is not maximal then by considering the abelianisations one derives MATH, hence MATH, which is also false.
|
math/0006003
|
Assume now that MATH admitted a proper handlebody decomposition without REF-handles. One identifies MATH with MATH. We can truncate the handle decomposition at a finite order to obtain a manifold MATH such that MATH, because the decomposition is proper. We can suppose that MATH is connected since MATH has one end. Then MATH is g.s.c. hence MATH. By hypothesis, we can choose MATH to be a homology sphere. Then MATH separates the cylinder MATH into two manifolds with boundary which, by NAME, have the homology of MATH. This implies that MATH (again by NAME). Let us consider now the map MATH, the composition of the inclusion with the obvious projection. The map MATH has degree one hence induces a surjection on the fundamental groups. REF-manifold MATH separates the two components of the boundary. In particular the generic arc joining MATH to MATH intersects transversally MATH in a number of points, which counted with the sign sum up to REF (or -REF). If properly interpreted this is the same as claiming the degree of MATH is one. It is well-known that a degree one map between orientable REF-manifolds induce a surjective map on the fundamental group (more generally, the image of the homomorphism induced by a degree MATH is a subgroup whose index is bounded by MATH). This shows that MATH is surjective. On the other hand MATH surjects onto a non-trivial residually finite group. Since MATH is the trivial map, the argument we used previously (from NAME 's theorem) gives us a contradiction. This settles our claim.
|
math/0006003
|
Since attaching MATH-handles and MATH-handles correspond to surgery and cutting along MATH-spheres respectively, we merely have to show that the surgery is REF-frame about a homologically trivial curve and the spheres along which one cuts are non-separating. First note that the absence of MATH-handles implies MATH, for all MATH. Further, each MATH is connected because MATH has one end. Thus the MATH-spheres along which any MATH is split have to be non-separating. Using NAME, the fact that MATH, and the long exact sequence in homology we derive that MATH. Also adding a MATH-handle decreases the rank of MATH by one hence every surgery increases the rank of MATH by one unit. But this means that the surgery must be a zero-frame surgery about a homologically trivial curve.
|
math/0006003
|
If a surgery is performed along a curve MATH, this means that a REF-handle is attached along the curve in the MATH-manifold MATH. Hence MATH bounds a disk in MATH, which projects to a disk bounded by MATH in MATH.
|
math/0006003
|
We can find MATH so that MATH is entirely after MATH, and MATH so that MATH, because the handlebody decomposition is proper. We then define MATH by repeating this process once, that is, MATH, for some MATH for which MATH is entirely after MATH. Consider MATH which persists till MATH. This means that there is an annulus properly embedded in MATH, whose boundary curves are MATH and MATH. Since MATH it bounds a disc in MATH. This disc together with the above annulus ensure that MATH dies by MATH, as they bound together a disc entirely in MATH, and MATH-handles do not affect the fundamental group.
|
math/0006003
|
It follows from a transversality argument that the image of MATH intersects only REF-handles, along REF-disks. Further it is sufficient to see that the circles MATH are homotopic to REF-framing since in MATH homotopy implies isotopy for circles. If one circle is null-homotopic then it can be removed by means of an ambient isotopy. If a circle turns MATH-times around the longitude, then it cannot bound a disk in MATH unless MATH.
|
math/0006003
|
Consider the diagram MATH where MATH is surjective, and the subscript MATH means abelianisation. Then it is automatically that MATH. Since MATH, and the strong compressibility is symmetric, the result follows.
|
math/0006003
|
The pull-backs in MATH of spheres MATH are planar surfaces with boundary components being the loci of future surgeries. Further, after compressing the spheres MATH of MATH (hence arriving into MATH) we have a surjection MATH, thus the map MATH is also surjective. This means that there exist curves in the complement of the planar surfaces in MATH mapping to every element of MATH. Moreover, by the above lemma, we have such curves that are homologically trivial in MATH, and hence in MATH as all surgery curves are null-homologous.
|
math/0006003
|
If not then there exists a curve MATH that represents a non-trivial element of MATH. Modifying by a homologically trivial element if necessary, we may assume that MATH. By the previous lemma MATH persists. The group MATH is the quotient of MATH by the relations generated by the surgery curves, which are homologically trivial. In particular MATH. Then the class of MATH is non-zero since its image in MATH is non-zero by hypothesis. This gives the required contradiction.
|
math/0006003
|
Let MATH be the rank of MATH and MATH be the fundamental groups of the surfaces. Since the surfaces are disjoint, MATH is obtained by HNN extensions from the fundamental group MATH of the complement of the surfaces. Thus, if MATH are the gluing maps, we have MATH . Now, since MATH and MATH, MATH surjects onto MATH, the group obtained by adding MATH generators to MATH. But, MATH is obtained by using MATH REF-handles and MATH REF-handles. Thus, as in NAME 's theorem, MATH is killed by adding MATH generators and MATH relations. This implies that MATH is killed by adding MATH generators and MATH relations.
|
math/0006003
|
By construction the images of the seams (which are roughly speaking half the generators of the fundamental group) are null-homotopic. If the fundamental groups of the generalized NAME surfaces from the previous remark map to the trivial group, then after doing surgery on the seams we obtain surfaces MATH representing homology with trivial MATH images by MATH. Thus, it suffices to show that we obtain this condition for a choice of NAME surfaces for all seams, at some time in the future. Fix MATH large enough so that MATH is in the collar MATH. There exists some MATH such that, whenever a REF-sphere immersed in MATH bound a REF-ball immersed in the collar, it actually bounds a REF-ball immersed in MATH. Choose MATH large enough so that a small collar MATH. Then use the horizontal flow to send MATH into MATH by preserving the right side boundary. This yields a ball in the complement of MATH. We need the following analogue, for MATH instead of MATH, of REF . There exists MATH such that any immersed REF-sphere in MATH (respectively in the intersection of MATH with the collar) which bounds a REF-ball in MATH (respectively in the collar) does so in MATH (respectively in the collar). The same trick we used in the proof of the previous lemma applies. Choose now MATH large enough such that MATH contains a non-trivial collar MATH, and MATH provided by REF . Consider a surgery curve MATH for some MATH. There exists a generalized NAME surface for MATH in MATH, so that the other boundary components are surgery curves from MATH, with MATH, and whose fundamental group maps to the trivial group under MATH. The curve MATH bounds a disc in the MATH-handle attached to it. Further, as it can be pulled back, say along an annulus, to time MATH, and then dies by MATH, it bounds another disc consisting of the annulus and the disc by which it dies. These discs together form an immersed MATH-sphere. Consider the class MATH of this REF-sphere by using the projection of the collar on MATH. We can realize the element MATH by an immersed REF-sphere in a small collar MATH. Therefore by modifying the initial REF-sphere by this sphere (which is far from MATH) in the small collar one finds an immersed REF-sphere whose image in MATH is trivial. Since MATH was large enough MATH is a subset of a larger collar MATH. Then REF-sphere we constructed bounds a REF-ball in MATH, and so by REF it also does so in MATH. Let MATH denote this immersion realizing a trivial element of MATH. Then MATH lifts to a map MATH, where MATH is the universal covering space of MATH. Since MATH is null-homotopic the homology class of MATH is trivial, when interpreting MATH as a REF-cycle in MATH. The homology of MATH is computed from the MATH-equivariant complex associated to the handle decomposition, whose generators in degree MATH are the MATH-handles attached to MATH in order to get MATH. Therefore one has then the following relation in this differential complex: MATH . The action of the algebraic boundary operator MATH on the element MATH can be described geometrically as the class of REF-cycle which represents the attachment REF-sphere MATH of REF-handle MATH. Consequently the previous formula can be rewritten as MATH where MATH are closed surfaces (actually these are closed REF-cycles, but they can be represented by surfaces by the well-known results of NAME) with the property that MATH (MATH denotes the core of REF-handle MATH). Let us compute explicitly the boundary operator on REF-handles, in terms of the surfaces we have in REF-complex MATH. Set MATH . Then the coefficient MATH is the number of times the boundary MATH runs over the core of MATH. But REF-sphere MATH, when pulled back in MATH, is a planar surface in MATH whose boundary circles (that is, at seams) are capped-off by the core disks MATH of REF-handles MATH. Therefore the number MATH is the number of times the seam MATH appears in the planar surface which is a pull-back of MATH. In particular the coefficient of a MATH-handle vanishes in a MATH-cycle only if the boundaries of the planar surface glue together to close up at the corresponding surgery locus. Thus the pull-backs of the surfaces MATH give a surface MATH in MATH with boundary the curve MATH with which we started, plus some other curves along which surgery is performed by time MATH. As this is in fact a closed cycle in the universal cover, the surface MATH lifts to a surface in MATH, with a single boundary component, corresponding to a curve which is not surgered by the time MATH. Therefore the map MATH factors through MATH, hence the image of MATH in MATH is trivial. Thus, after surgering along the curves up to the MATH we do have the required NAME surfaces to compress to get embedded surfaces with trivial MATH image.
|
math/0006003
|
We will express each surgery locus MATH as a product of commutators of the form MATH, with each MATH being conjugate to a surgery locus (possibly MATH itself). It then follows readily that MATH, as now if each MATH, then each MATH. Suppose now MATH is a surgery locus. Then the MATH-frame surgery along MATH creates homology in MATH which by our structure theorem is represented by a surface MATH. The pullback of MATH to MATH gives a surface with boundary along seams, and being compressed to a sphere by the seams, so that the algebraic multiplicity of MATH is MATH while that of all other seams is MATH. In terms of the fundamental group, this translates to the relation that was claimed.
|
math/0006003
|
For the first stage, take two surfaces of genus MATH, and let them intersect transversely along two curves (which we call seams) that are disjoint and homologically independent in each surface. Next, take as NAME surfaces for these curves once punctured surfaces of genus MATH intersecting in a similar manner, and glue their boundary to the above-mentioned curves of intersection. Repeat this process to obtain the complex. At the first stage, we cannot have embedded, disjoint surfaces representing the homology as the cup product of the surfaces is non-trivial. As the surfaces are compact, we must terminate at some finite stage. We will prove that if we can have disjoint surfaces at the stage MATH, then we do at stage MATH. This will suffice to give the contradiction. Now, we know the complex cannot be embedded in the first stage. Suppose we did have disjoint embedded surface MATH and MATH at stage MATH. Since these form a basis for the homology, they contain curves on them that are the seams at the first stage with algebraically non-zero multiplicity, that is, the collection of curves representing the seam is not homologically trivial in the intersection of the first stage with the surface. Further, some copy of the first seam must bound a subsurface MATH in each of the surfaces, for otherwise the surface contains a curve dual to the seam. For, the cup product of such a dual curve with the homology class of the other surface is non-trivial, hence it must intersect the other surface, contradicting the hypothesis that the surfaces are disjoint. Similarly, at the other seam we get surfaces MATH. By deleting the first stage surfaces and capping off the first stage seams by attaching discs, we get a complex exactly as before with the MATH-th stage having become the MATH-th stage. Further, the MATH and MATH now give disjoint, embedded surfaces representing the homology that are supported by stages up to MATH. This suffices as above to complete the induction argument. It is easy to construct a handle-decomposition corresponding to this complex. REF shows a construction of tori with one curve of intersection. Here we have used the notation of NAME calculus, with the thickened curves being an unlink along each component of which MATH-frame surgery has been performed. It is easy to see that the same construction can give surfaces of genus MATH intersecting in MATH curves. On attaching the first two MATH-handles, the boundary is MATH. Since the curves of intersection are unknots, after surgery they bound spheres. Further, it is easy to see by cutting along these that the boundary is MATH after attaching the MATH-handles and MATH-handles as well. Repeating this process, we obtain our embedding. Thus we have an infinite handle-decomposition satisfying our hypothesis for which this MATH-complex is MATH.
|
math/0006003
|
We will take a variant of the example in the last section. Namely, we construct an explicit handle-decomposition according to a canonical form. Start with a MATH-handle and attach to its boundary three MATH-handles along an unlink. the resulting manifold has boundary MATH obtained by MATH-frame-surgery about each component of an unlink with MATH components. We now take as NAME surfaces for these components surfaces of genus MATH, so that each pair intersects in a single curve, so that the curves of intersection form an unlink and are unlinked with the original curves. Now, attach MATH-handles along the curves of intersections, and then MATH-handles along the NAME surfaces compressed to spheres by adding discs in the MATH-handles just attached. It is easy to see that the resulting manifold once more has boundary MATH. Thus, we may iterate this process. Further, the generators of the fundamental group at any stage are the commutators of the generators at the previous stage. Suppose MATH is in fact tame. Then, we may use the results of the previous sections. Now, by construction no curve dies as only trivial relations have been added. Thus every element in kernel REF must fail to persist by some uniform time. In particular, the image of the group after that time in the present (curves that persist beyond that time) must inject under MATH. But we know that it also surjects. Thus, we must have an isomorphism. Thus, there is a unique element mapping onto each element of MATH. Hence this element must persist till infinity as we have a surjection at all times. On the other hand, since the limit of the lower central series of the free group is trivial, no non-trivial element persists. This gives a contradiction unless MATH is trivial. But there are non-trivial elements that do persist beyond any give time. As no element dies, we again get a contradiction.
|
math/0006003
|
If we did have such a sequence of surgeries, then MATH bounds a MATH-manifold MATH with MATH , with a half-basis formed by embedded spheres. Now glue this to a manifold with form MATH which is bounded by MATH to get MATH. We can surger out the disjoint family of MATH's from MATH to get a MATH-manifold with form MATH and trivial MATH. This contradicts NAME 's theorem.
|
math/0006014
|
By REF , we know that MATH is a free MATH-module for all MATH, hence, MATH. Recall that MATH . Now, since MATH is an isomorphism, we conclude that, for all MATH, MATH is also a free MATH-module. Therefore, one has: MATH and MATH . Every NAME invariant MATH can then be seen as a linear map from MATH to MATH which vanishes on MATH. Via MATH, this means that MATH is a linear map from MATH to MATH, which vanishes on MATH. Therefore, if MATH is a NAME invariant of type MATH, it can be lifted in a unique way to a linear map MATH, which verifies MATH.
|
math/0006014
|
Since MATH is a normal subgroup of MATH, it is straightforward to prove that MATH. So, it suffices to prove that MATH. The inclusion MATH is obvious, once we notice that MATH and that MATH. For the other inclusion, we must prove that for all MATH one has MATH. Suppose that MATH, with MATH; then MATH, so it suffices to show it for a set of generators of MATH. As we said before, MATH is the normal closure of MATH in MATH, so a set of generators of MATH consists on the elements of the form MATH, where MATH and MATH. Take an element MATH as above. One has: MATH, so we only have to show that MATH for MATH. It is known REF that MATH belongs to the ideal of MATH generated by MATH. But the extension of this ideal to MATH is precisely MATH, so MATH.
|
math/0006014
|
Consider the NAME graph of MATH, which is defined as follows. Its vertices are the elements of MATH, and its edges are labeled by MATH. For every vertex MATH and for every MATH, there is exactly one edge labeled by MATH, with source MATH and target MATH. In this graph, the normal form of an element MATH corresponds to a unique path going from MATH to MATH. By the prefix-closed condition mentioned above, the set of normal forms of MATH defines a maximal tree MATH of the NAME graph. The NAME graph of MATH can be seen as the one-skeleton of a tiling of the plane. For every vertex MATH, the path which starts at MATH and which is labeled by MATH bounds a fundamental region MATH of this tiling, and all fundamental regions are obtained in this way. Hence, there is a one to one correspondence between the vertices of the NAME graph and its fundamental regions. Therefore, the fundamental group of the NAME graph of MATH is the free group with free system of generators MATH . We now define a graph MATH as follows. Take the NAME graph of MATH and replace the labels MATH by MATH. Then, for every vertex MATH, add MATH edges with source and target MATH, labeled by MATH, respectively. Notice that the fundamental group of MATH is the free group with free system of generators MATH, where MATH. Recall the exact sequence MATH . One can easily verify that MATH is freely generated by the set MATH and that MATH sends MATH to MATH for all MATH, and sends MATH to MATH for all MATH. It follows from classical geometric methods (see CITE) that the group MATH is the fundamental group of the graph MATH, hence MATH is a free system of generators for MATH, as we wanted to prove.
|
math/0006014
|
The case MATH is trivial, since MATH. Hence MATH is a free group of infinite rank. Suppose now that MATH. It is well known that the kernel of the homomorphism MATH is MATH (see CITE). Moreover, one can easily see that MATH lies in this kernel, namely MATH. Similarly, one has MATH. The homomorphism MATH which sends MATH to MATH is the restriction of MATH to MATH. In particular, it sends MATH onto MATH. We consider an embedding MATH satisfying: CASE: MATH for MATH; CASE: MATH does not lie in the image of MATH; CASE: MATH is a homotopy equivalence relatively to MATH. Then, the map MATH induces a homomorphism MATH, which sends MATH to MATH. By the third condition, this homomorphism is a section of MATH. It obviously sends MATH to MATH.
|
math/0006014
|
We only need to verify that the action of the generators of MATH on the generators of MATH is trivial after abelianization. Moreover, let us see that it suffices to show the result for the action defined by any set-map section MATH of MATH. Indeed, if MATH is a section of MATH, then for every MATH, there exists an element MATH such that MATH. Therefore, if MATH acts trivially on MATH via MATH, we obtain, for every MATH, MATH . As we said in Subsection REF, a set of generators for MATH consists on the elements of the form MATH, where MATH and MATH. On the other hand, it is known that MATH is a set of generators for MATH, where MATH denotes the braid defined in Subsection REF. Therefore, the following is a set of generators for MATH: MATH . We take MATH such that MATH. In other words, we just add a trivial string based at MATH for any element of the set of generators of MATH. We remark that MATH is a set map section of MATH, but it is not a homomorphism. Now, MATH is by definition a normal subgroup of MATH. Therefore, if we show that each MATH acts trivially on MATH by conjugation, then we will have finished the proof, since in that case: MATH . Recall from REF that MATH is a free system of generators for MATH, where MATH is a word over MATH for all MATH. One can verify (just drawing the corresponding braids) that in MATH one has the following relations: MATH where MATH, MATH and MATH. Therefore, in MATH, we have: MATH as we wanted to show.
|
math/0006014
|
We can suppose, without loss of generality, that MATH. Suppose first that MATH is a single letter. If MATH and MATH is odd, then one can easily show by drawing the braids that the following equality holds in MATH: MATH . Hence MATH where MATH . If MATH and MATH is even, then one has MATH . Therefore, MATH where MATH . The computations for MATH are the same as for MATH interchanging the case MATH odd with the case MATH even. Suppose now that MATH is a word of length MATH, MATH, and that the result is true for words of length less than MATH. We write MATH, with MATH. Consider MATH . The hypothesis MATH implies that MATH. Furthermore, both MATH and MATH are normal subgroups of MATH, thus MATH. Finally, a direct calculation shows that: MATH as we wanted to show.
|
math/0006014
|
Recall the epimorphism MATH. One has MATH, thus MATH, where MATH is a word over MATH. By drawing the braids, one sees that MATH, for MATH. Moreover, since MATH, one has MATH, where MATH, for all MATH. Therefore, since MATH is a word over MATH, one has: MATH, where MATH. Hence, MATH where MATH, as we wanted to show.
|
math/0006014
|
Clearly, it suffices to show the lemma when MATH is a single letter. Besides, we can suppose that MATH. Then the result is a consequence of the following relations in MATH. MATH where MATH. These relations can be easily verified by drawing pictures.
|
math/0006014
|
This lemma follows from the well known congruences MATH (see, for example, CITE), together with the inclusion MATH.
|
math/0006014
|
We only need to prove that MATH is the inverse of MATH as a homomorphism of MATH-modules. For MATH, let MATH be the submodule of MATH consisting of the homogeneous polynomials of degree MATH. Consider as well MATH, where MATH, MATH and MATH. By Relations REF , seen as relations in the enveloping algebra MATH of MATH, one has: MATH . Therefore, a set of generators for MATH as a MATH-module consists on the elements of the form MATH where MATH and MATH for all MATH. But MATH so, by definition of MATH, and since MATH, one has MATH. This is true for all MATH, so it follows that MATH. Hence, since MATH is an isomorphism, MATH is its inverse, as we wanted to show.
|
math/0006014
|
Recall that, by REF , the ideal MATH of MATH is isomorphic to MATH via MATH, for all MATH. Moreover, since MATH is a free MATH-module, one has: MATH . Hence, MATH via MATH. Now, MATH and both MATH and MATH are isomorphisms of MATH-modules, thus MATH is an isomorphism of MATH-modules.
|
math/0006014
|
Let MATH. Write MATH and MATH for MATH. Then MATH . So, in order to prove that MATH, it suffices to show that MATH . But MATH, thus there exists MATH such that MATH with MATH, hence, in MATH, MATH since MATH, as we wanted to show.
|
math/0006014
|
Write MATH and MATH, to simplify notation. Write as well MATH. We know that MATH is a MATH-algebra isomorphism, so MATH . On the other hand: MATH . Therefore, we need to show that, in MATH, MATH . Since the action by conjugation does not depend on MATH, we only need to verify the above formula when MATH is a generator of MATH. In addition, since MATH is a homomorphism of MATH-algebras, it suffices to verify it when MATH is a generator of MATH as a MATH-algebra, that is, when MATH, MATH. Hence, it suffices to prove REF below.
|
math/0006014
|
The first equation is a consequence of the following relations in MATH, which are easily verified. MATH . The second equation comes from REF , and from the following relations, where MATH and we have denoted MATH if MATH is odd, and MATH if MATH is even. MATH . Indeed, in this case, MATH and by REF , this is equivalent to MATH. Finally, the third equation is verified as follows. MATH where MATH, so this is equivalent to MATH.
|
math/0006022
|
For MATH, we compute MATH where in the penultimate equality, we have used the fact that MATH is a homomorphism of NAME algebras and REF . This establishes REF , and REF follows from REF .
|
math/0006022
|
For MATH, we have MATH . Thus MATH is closed under the product MATH if and only if MATH is a homomorphism. The remaining assertions are clear.
|
math/0006022
|
It is straightforward to check that REF give the NAME identity for the bracket REF . That MATH is immediate from REF , and so the decomposition is reductive. From REF we have MATH and MATH for MATH, which proves the remaining assertion.
|
math/0006022
|
Indeed, for MATH, we have MATH using REF . This establishes the equivalence of REF .
|
math/0006029
|
See REF .
|
math/0006029
|
We can decompose MATH where the MATH's are irreducible representations. By what we have said before, there are integers MATH, MATH, with MATH, MATH, such that MATH is a direct summand of MATH. Our assumption on the action of MATH implies that MATH. Let MATH be a positive integer which is so large that MATH for MATH. Then, MATH is the natural representation of MATH on MATH . Now, the MATH-module MATH is a direct summand of MATH and we are done.
|
math/0006029
|
We may assume that MATH is a very ample line bundle. Set MATH. The linearization MATH provides us with a representation MATH and a MATH-equivariant embedding MATH. Since obviously MATH for all points MATH and all one parameter subgroups MATH of MATH, we can assume MATH. Now, there are a basis MATH of MATH and integers MATH with MATH . A point MATH can be thought of as the equivalence class of a linear form MATH . Then, MATH . Therefore, MATH, and this implies the assertion.
|
math/0006029
|
This theorem can be proved with the methods developed in CITE for MATH. A more elementary approach is contained in the note CITE.
|
math/0006029
|
Let MATH be any subbundle. By REF, MATH, so that MATH-semistability gives MATH that is, MATH so that the theorem holds for MATH.
|
math/0006029
|
By our assumptions, the sheaf MATH is locally free of rank MATH. Therefore, we can choose a covering MATH, MATH, of MATH, such that it is free over MATH for all MATH. For each MATH, we can choose a trivialization MATH so that we obtain a surjection MATH on MATH. Therefore, MATH is a quotient family of MATH-pairs of type MATH parameterized by MATH, and we can conclude by REF .
|
math/0006029
|
The two morphisms MATH and MATH provide us with quotient families MATH and MATH of MATH-pairs parameterized by MATH. By hypothesis, MATH, and we have isomorphisms MATH and MATH with MATH . In particular, there is an isomorphism MATH . This yields a morphism MATH and MATH. Let MATH be the fibre product taken with respect to MATH and MATH, MATH. The morphism MATH is then a MATH-sheeted étale covering coming with the projection map MATH. In the following, we set MATH, MATH. One has MATH. By construction, the morphism MATH factorizes over a morphism MATH. The quotient family defined by the morphism MATH is just MATH with MATH . The assertion of the proposition is that this family is equivalent to the quotient family MATH. But this is easily seen, using MATH and MATH.
|
math/0006029
|
First, suppose we are given a weighted filtration MATH, MATH, such that MATH is globally generated and MATH for MATH. Then, for MATH, MATH so that the claimed condition follows from MATH being MATH-(semi)stable. Next, recall that we have found a universal positive constant MATH depending only on MATH, MATH, and MATH, such that for every MATH, every semistable MATH-pair MATH, and every non-trivial subbundle MATH of MATH . If we fix another positive constant MATH, then the set of isomorphy classes of vector bundles MATH such that MATH, MATH, and MATH is bounded. From this, we infer that there is a natural number MATH, such that for every MATH, every semistable MATH-pair MATH of type MATH, and every proper subbundle MATH of MATH CASE: either MATH CASE: or MATH is globally generated and MATH. Moreover, the NAME estimate compare REF and proof of REF, p. REF gives in the first case MATH that is, for large MATH and thus MATH . Our contention is now that for MATH with MATH and MATH, the theorem holds true. So, assume that we are given a weighted filtration MATH with MATH and MATH. Let MATH be the indices such that MATH, for MATH, so that MATH is globally generated and MATH, MATH. We let MATH be the indices in MATH in increasing order. We introduce the weighted filtrations MATH and MATH with MATH, MATH and MATH, MATH. REF yields MATH whence MATH . Since this last expression is positive by assumption, we are done.
|
math/0006029
|
To see the equivalence between REF. and REF., observe that MATH provides an equivariant embedding of MATH into MATH. Via the canonical surjection MATH, the latter space becomes embedded into MATH, so that we have an equivariant embedding MATH. Since for every point MATH and every one parameter subgroup MATH the claimed equivalence is easily seen. For the equivalence between REF. and REF., we have to go into the GIT construction of the moduli space of MATH-semistable MATH-pairs. We choose MATH, such that MATH is a direct summand of MATH. Therefore, MATH is also a direct summand of MATH, MATH, so that we can assume MATH for MATH. For a tuple MATH, positive rational numbers MATH, and MATH as in the statement, we thus find MATH for some MATH. Recall that in our GIT construction of the moduli space of MATH-semistable MATH pairs of type MATH, we had to fix some natural number MATH which was large enough. Being large enough depended on constants MATH, MATH, MATH, and MATH which in turn depended only on MATH, MATH, MATH, and MATH. One now checks that MATH, MATH, MATH, and MATH yield exactly the same constants, so that the construction will work also - for all MATH and all MATH - for MATH-semistable MATH-pairs of type MATH. Fix such a MATH. We can now argue as follows. Set MATH, and let MATH be a complex vector space of dimension MATH. Given a MATH-semistable MATH-pair MATH of type MATH, we can write MATH as a quotient MATH where MATH is an isomorphism. Set MATH and MATH. Then MATH defines a NAME point MATH which is semistable for the linearization of the MATH-action in MATH with MATH. By REF , we find indices MATH and positive rational numbers MATH with MATH, such that MATH, MATH, and the point MATH is semistable with respect to the linearization of the MATH-action in MATH. As before, there is an embedding MATH such that the pullback of MATH is MATH. The point MATH is thus semistable with respect to the linearization in MATH. Now, the second component of MATH is defined by the homomorphism MATH obtained from MATH and the components MATH, MATH. Composing this homomorphism with the natural map MATH, we find a point MATH, MATH. The point MATH is semistable with respect to the linearization of the MATH-action in MATH. By construction, MATH is the NAME point of the quotient MATH-pair MATH. Since MATH, we infer that MATH is MATH-semistable. The converse and the polystable part are similar.
|
math/0006029
|
First, assume REF. Let MATH be a homomorphism. Call a sub vector space MATH-superinvariant, if MATH and MATH. Let MATH. Given a basis MATH of MATH and MATH, set MATH. Then REF MATH, if MATH is not MATH-invariant. CASE: MATH, if MATH is MATH-superinvariant and MATH. CASE: MATH in all the other cases. Now, let MATH be a MATH-pair of type MATH. For any subbundle MATH of MATH with MATH, we find MATH. Let MATH and MATH a MATH-(semi)stable MATH-pair of type MATH. Then MATH for every non-trivial proper subbundle MATH of MATH with MATH. This condition implies that for every MATH, every MATH-semistable MATH-pair MATH of type MATH, and every subbundle MATH of MATH . See, for example, CITE. Therefore, the set of isomorphy classes of bundles MATH, such that there exist a positive rational number MATH and a MATH-semistable MATH-pair of type MATH of the form MATH, is bounded. Now, the only thing we still have to show is that for every sufficiently large positive rational number MATH and every MATH-semistable MATH-pair MATH of type MATH, such that MATH, the homomorphism MATH can't be nilpotent. First, let MATH be a MATH-pair of type MATH, such that there exists a positive rational number MATH with respect towhich MATH is semistable and such that MATH is nilpotent. Then, there is a filtration MATH with MATH, MATH. It is clear by the boundedness result that the MATH's occurring in this way live in bounded families, so that we can find a positive constant MATH with MATH for all such filtrations. One checks MATH, so that the semistability assumption yields MATH . This is impossible if MATH. To see the converse, let MATH be a MATH-pair satisfying REF. Let MATH. Then, as before, MATH for every non-trivial proper subbundle MATH of MATH, that is, MATH. First, consider a weighted filtration MATH such that MATH, MATH. Then, the condition that MATH be not nilpotent if MATH implies MATH, so that MATH follows from REF. Second, suppose that we are given a weighted filtration MATH such that, say, MATH are not invariant under MATH, that is, MATH, MATH, and MATH. Let MATH. One readily verifies MATH. We thus find MATH so that MATH will be positive if we choose MATH.
|
math/0006033
|
First note that MATH is a one-dimensional non-commutative MATH complex, as defined in CITE. Hence MATH is semiprojective by CITE and finitely generated by CITE.
|
math/0006033
|
Let MATH denote the MATH identity matrix for any non-negative integer MATH. For each MATH, define an integer matrix of size MATH by MATH . Let MATH denote the MATH matrix MATH and define for MATH, an integer matrix of size MATH by MATH . For MATH, define yet another MATH matrix by MATH . Finally, let MATH be the MATH matrix MATH . Note that for MATH, MATH . Using this, it is easily seen by induction that MATH . It follows that MATH . Since all the matrices on the left-hand side, except MATH, are invertible in MATH, we obtain the desired calculation of MATH. Finally, it is easily verified that MATH . The last part of the lemma follows from this.
|
math/0006033
|
Define a *-homomorphism MATH by MATH . Via the identification MATH we define a *-homomorphism MATH by MATH . The short exact sequence MATH gives rise to a six-term exact sequence MATH where MATH denotes the exponential map. By NAME periodicity MATH is generated by MATH, where MATH is a unitary in MATH. Note that MATH and hence MATH in MATH. Since the map MATH is surjective it follows that MATH is generated by MATH, and that MATH gives rise to an isomorphism between the cokernel of MATH and MATH. Let MATH denote the standard matrix units in MATH. Recall that MATH is generated by MATH. We leave it with the reader to check that MATH and for MATH, MATH . The conclusion follows from REF .
|
math/0006033
|
Let us start with a simple and well-known observation. Let MATH be a unitary in the MATH-algebra MATH such that the winding number of MATH is MATH. Then MATH can be connected to MATH via a continuous path MATH in MATH. If MATH we may assume that MATH for every MATH. Let MATH be the exceptional points of MATH, where MATH are numbers in MATH. Set MATH, MATH and let MATH be the inclusion, MATH. Since the group of unitaries in MATH with determinant MATH is path-connected there exists a continuous function MATH such that MATH, MATH, and MATH. Set MATH . It follows from the above observation that MATH can be connected to MATH via a continuous path of unitaries in MATH. Upon replacing MATH with MATH we may thus assume that MATH for MATH. Set MATH . Then MATH. Again by the above observation, MATH can be connected to MATH within MATH for MATH.
|
math/0006033
|
Let MATH denote the rank of MATH and let MATH be the exceptional points of MATH, where MATH. Since MATH divides MATH for MATH, it follows that MATH also divides MATH. Hence there is a projection MATH with the same trace as MATH. For each MATH there is a unitary MATH such that MATH . We may assume that MATH, MATH, and that MATH. By compactness MATH where MATH, MATH, and MATH . Set MATH for MATH, MATH, where MATH . Then MATH, MATH, is a continuous path of unitaries in MATH, and by CITE MATH . As MATH is path-connected there is for each MATH a continuous map MATH such that MATH . Since MATH for MATH, we can define a unitary MATH by MATH . Then MATH.
|
math/0006033
|
As in the proof of REF we see that there exists a projection MATH of rank MATH if and only if MATH divides MATH. The conclusion follows.
|
math/0006033
|
Let MATH where MATH is a positive integer, MATH are non-negative integers, and MATH are prime numbers. Let MATH be prime numbers, mutually different as well as different from MATH. Define integers MATH and MATH by MATH . Set MATH. Then MATH by REF . MATH is unital projectionless by REF .
|
math/0006033
|
Choose MATH such that MATH. Set MATH . We have a pull-back diagram MATH where MATH are the restriction maps and MATH evaluation at MATH. Apply the NAME sequence CITE to get a six-term exact sequence MATH . Note that MATH and MATH. Thus the exact sequence becomes MATH . Since MATH is homotopic to evaluation at MATH in MATH and MATH is homotopic to evaluation at MATH in MATH we see that MATH . MATH is homotopic to MATH via MATH and hence MATH is generated by MATH. For MATH we have that MATH is generated by MATH and that MATH . Thus MATH is generated by MATH and MATH and MATH . Proceeding by induction, assume that the lemma holds for MATH. By the induction hypothesis MATH is generated by MATH. Note that MATH such that MATH is generated by MATH. It also follows that MATH if and only if there exists MATH such that MATH . By the induction hypothesis this happens if and only if there exist MATH such that MATH and MATH . The desired conclusion follows easily from these equations.
|
math/0006033
|
Choose MATH such that MATH, MATH, are the exceptional points for MATH. Set MATH . Define a *-homomorphism MATH by MATH. Let MATH be evaluation at MATH. Let MATH denote the map MATH. Let MATH be the map MATH. We have a pull-back diagram MATH and hence by CITE a six-term exact sequence of the form MATH . MATH is generated by MATH and MATH where MATH are the coordinate projections. MATH is generated by the class of the identity map MATH on MATH. Note that MATH . As MATH maps onto MATH (because MATH) and as MATH, we see that MATH is surjective. Assume that MATH. Then MATH and hence MATH by the above. Thus MATH is an isomorphism and the conclusion follows from REF .
|
math/0006033
|
By REF , or simply because homotopic *-homomorphisms MATH define the same elements in MATH, we have that MATH . From this the existence follows. To check uniqueness, assume MATH where MATH . Fix some MATH. By REF there exist integers MATH such that MATH, MATH. Therefore MATH for MATH. Finally, to prove the equations above, fix again some MATH. Note that MATH . Hence there exist integers MATH, MATH, such that MATH and MATH . The desired conclusion follows easily from these equations.
|
math/0006033
|
Let MATH. By REF there is an integer MATH, MATH, and integers MATH, MATH, such that MATH . Note that MATH for MATH, and MATH. By REF we see that for MATH, MATH . By REF there exists a unitary MATH such that the matrix MATH belongs to MATH for all MATH. Set MATH . Let MATH denote the exceptional points of MATH and let MATH be those of MATH. Choose continuous functions MATH such that MATH as ordered tuples. Choose a unitary MATH such that MATH. Define a unital *-homomorphism MATH by MATH . By REF we have that MATH for some MATH. Let MATH in MATH. Define MATH by MATH and define MATH by MATH. Note that MATH, MATH. Define MATH by MATH . By REF we see that in MATH, MATH . Since MATH, MATH, MATH, we conclude that MATH .
|
math/0006033
|
By REF there exists an element MATH such that MATH. Let MATH have standard form MATH . By adding an integer-multiple of MATH we may assume that MATH for MATH. Define MATH and MATH, MATH, as in the proof of REF . Let MATH . Choose a positive integer MATH such that MATH. Choose for each MATH, a unitary MATH such that the matrix MATH belongs to MATH for all MATH. As in the proof of REF these matrices can be connected to define a *-homomorphism MATH. We leave it with the reader to check that MATH on MATH. Set MATH.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.