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math/0010204 | For a given subset MATH of MATH, the elements of MATH are the vectors in MATH for which the support mod MATH is exactly MATH. In particular, an element MATH of MATH, with MATH and MATH, is in MATH if and only if MATH for MATH and MATH for MATH. As all matrix entries of an element MATH of MATH are nonnegative mod MATH, ... |
math/0010204 | For the proof of the first statement, it suffices to consider MATH as MATH is generated by these elements. As for the description of MATH, only the action of MATH on MATH with MATH is relevant. A computation shows MATH . The set MATH is the set of positive roots for which MATH has coefficient MATH in MATH for any eleme... |
math/0010204 | Since MATH, the subset MATH of MATH coincides with MATH. Hence MATH if and only if MATH, which, by definition of MATH, is equivalent to MATH. By REF , this is the same as MATH, while, since the left hand side equals MATH, this in turn amounts to MATH. Hence the lemma. |
math/0010204 | Let MATH. We show that MATH. We distinguish cases according to MATH. If MATH, then MATH by REF . If MATH, then MATH with MATH. By REF , this implies MATH. If MATH, then MATH with MATH. By REF , this implies MATH. Finally, suppose MATH. Then MATH with MATH. Moreover, since MATH and MATH is closed, we have MATH, which, b... |
math/0010204 | It suffices to prove the assertion for MATH with MATH. Write MATH, so MATH, and MATH is the maximal subset of MATH belonging to MATH. Recall from REF that MATH. It implies MATH, whence MATH, so MATH. In other words, MATH. Since MATH, we obtain MATH. Put MATH. By REF , MATH is the maximal element of MATH contained in MA... |
math/0010204 | CASE: Clearly, MATH, so MATH is nonempty. CASE: This follows immediately from the definition of MATH. CASE: Given MATH, let MATH be such that MATH. Then, by respectively, the definition of MATH, the definition of MATH, and REF , MATH whence MATH. |
math/0010204 | (This is the proof appearing in CITE.) Suppose that the elements MATH and MATH of MATH act identically on MATH. If MATH, then MATH and MATH are both the identity and there is nothing to prove. Suppose therefore, that MATH. Pick MATH. Then MATH, which implies by REF that MATH for some nontrivial MATH. This means that th... |
math/0010204 | The proof is as in CITE and so we do not include it. The use of REF and MATH in CITE is replaced by the following corresponding results for MATH. There is a linear transformation MATH whose matrix with respect to MATH have entries in MATH such that MATH for each MATH, where MATH is the matrix MATH with MATH and MATH re... |
math/0010207 | We see them in CITE. MATH is used in the proof of REF. |
math/0010207 | Most of the results come from CITE. In the case MATH, we classify MATH by REF according to the value of MATH. In REF , we have MATH or MATH from MATH and MATH, which are obtained by REF. In REF is NAME and thus CITE induces the result. In REF , we know all the possible values of MATH and the corresponding values of MAT... |
math/0010207 | CASE: Putting MATH into REF , we have MATH. CASE: We use REF . Because MATH we have MATH. Hence MATH. CASE: We will prove REF with the same idea. Let MATH be a general line on MATH or MATH through MATH, and let MATH be its strict transform on MATH. Then, MATH . CASE: Let MATH be a general conic on MATH through MATH, an... |
math/0010207 | CASE: This comes from REF . CASE: Since MATH in REF , we have MATH and In both cases a general hyperplane section on MATH through MATH has multiplicity MATH along MATH, which means that MATH. This is a contradiction. REF , and REF imply this. CASE: If neither REF nor REF occurs, then from REF we have MATH . This is a c... |
math/0010207 | CASE: This is trivial since MATH. CASE: This comes from REF . CASE: MATH comes from MATH and REF . We know the shape of MATH from the equation below. MATH where the second equality comes from MATH. |
math/0010207 | Though analytic functions seem to emerge in this proof, we stay in the algebraic category by adding higher terms to them if necessary, as we have said in the first paragraph in REF. First we prove a claim on an analytic description. There exists an identification MATH satisfying the following conditions. CASE: MATH. CA... |
math/0010207 | Though MATH or MATH in REF is impossible because MATH. Hence MATH. By REF , it is trivial that the values of MATH in REF cover all the possibility for MATH and MATH. Now we calculate the value of MATH in each case usuig REF . Because MATH and MATH, REF implies that MATH . Thus we have only to the next claim. CASE: In R... |
math/0010207 | We note that MATH for any MATH. Using REF , for any MATH we have MATH . Hence MATH, and thus MATH. But on the other hand MATH's satisfy the relations in REF . Using them we know that there is no possibility for such MATH's in REF , and that REF is the only possibility in REF . |
math/0010207 | CASE: Let MATH be the multiplicity of MATH along MATH, and let MATH be a general line on MATH. Then, MATH . On the other hand, MATH . By these two inequalities, we obtain MATH and MATH. This shows REF . CASE: Because REF tells that MATH is a line on MATH, we know that MATH induces an isomorphism MATH. Let MATH be the l... |
math/0010207 | We use the same idea as that in the proof of REF . First we remark that MATH by REF . By REF , we have an identification MATH satisfying that MATH. Moreover by REF we may assume that MATH. We know that MATH. Under the above identification, MATH equals, as valuations, an exceptional divisor obtained by the weighted blow... |
math/0010207 | It is sufficient to show that MATH is normal and that MATH, because these imply the last part of the statement. We will prove them simultaneously. Let MATH be the normalization of MATH. First we calculate the dualizing sheaf MATH on MATH. Let MATH be the NAME locus of MATH. We remark that MATH is a finite set. By the a... |
math/0010207 | First we give easy statements about a NAME singularity of type MATH. Let MATH be an algebraic germ (respectively,an analytic germ) of a NAME singularity of type MATH, let MATH be a non-isomorphic partial resolution factored through by the minimal resolution of MATH, and let MATH be a general hyperplane section through ... |
math/0010207 | REF induces that MATH. Thus we have only to show that MATH because of REF . MATH (respectively,MATH, MATH) when the center of MATH on MATH is not a non-Gorenstein point (respectively,is the non-Gorenstein point of index MATH, is the non-Gorenstein point of index MATH). Like the proof of REF , we obtain MATH (respective... |
math/0010207 | The following claim is inevitable. Let MATH, which is the number of elements in the set MATH. Then for MATH, MATH . We note that MATH by REF . Thus by calculation using REF as in CITE, we have, for MATH, MATH . CITE shows that the above dimension equals MATH, which implies the claim. We express MATH's explicitly using ... |
math/0010207 | CASE: Take a surface MATH for a general MATH. Then MATH, which is a NAME singularity of type MATH, where MATH. Here MATH. We remark that MATH if MATH or MATH. Because MATH and MATH, the multiplicity of MATH along MATH equals MATH. Thus MATH is special of type MATH. CASE: We may assume that MATH. Since MATH, MATH is a p... |
math/0010210 | The basic tool needed to construct the natural transformations to other cohomology theories is the theory of NAME classes MATH constructed by CITE and CITE for a very large set of cohomology theories MATH that includes all NAME cohomology theories. These give rise to the NAME character maps MATH . The degree MATH part ... |
math/0010210 | First suppose that MATH is empty. CITE showed that each MATH-group MATH is a finitely generated abelian group. It follows that each of the groups MATH is finite dimensional. The rank of MATH is REF and the rank of MATH is MATH by the NAME Unit Theorem. The ranks of the remaining MATH were computed by CITE. It is zero w... |
math/0010210 | Let MATH denote the tannakian fundamental group of the category of finite dimensional MATH-adic MATH-modules. By CITE, the conditions in Postulate REF are equivalent to the surjectivity of MATH. It is a general fact that the image of MATH is NAME dense. |
math/0010210 | Denote the inverse limit by MATH and by MATH the category of weighted MATH-modules with respect to MATH and MATH. We will show that MATH is the category of finite dimensional MATH-modules, from which the result follows. Suppose that MATH is an object of MATH. Then the NAME closure of MATH in MATH is an extension MATH o... |
math/0010210 | In view of REF , it suffices to prove that MATH is an isomorphism, and that MATH is injective. Since the functor from the category of weighted MATH-modules to the category of MATH-modules is fully faithful, a REF-step extension of weighted MATH-modules splits if it splits as an extension of MATH-modules. This establish... |
math/0010210 | NAME 's conjecture REF implies that MATH, the tannakian fundamental group, is an extension MATH where MATH is a free prounipotent group generated by MATH. This and REF show that the natural map MATH is an isomorphism, and it follows that MATH is fully faithful and its image is equivalent to the category of weighted MAT... |
math/0010210 | If MATH is a set of representatives of the isomorphism classes of irreducible representations of MATH, then MATH is a set of representatives of the isomorphism classes of irreducible representations of MATH, where MATH denotes the exterior tensor product of a representation MATH of MATH and MATH of MATH. Consider the r... |
math/0010210 | It suffices to show that the natural mapping MATH is an isomorphism when MATH and that the natural mapping MATH is injective when MATH-S. The proof is similar to that of REF . To show that MATH is an isomorphism, it suffices to show that an extension MATH of MATH by MATH corresponding to an element of MATH is crystalli... |
math/0010210 | The proof is by induction on the dimension of MATH. In the case dim-MATH, this is well-known REF . Assume MATH and the claim is true for MATH. By assumption, there exists an exact sequence of MATH-adic representations of MATH: MATH for some integer MATH such that MATH satisfies the assumption of the lemma. By REF , we ... |
math/0010210 | By REF , we have the following commutative diagram whose two rows are exact: MATH and the left vertical arrow is an isomorphism by REF . Hence the right square is cartesian. |
math/0010210 | It is well known that an extension MATH in MATH gives a continuous cocycle MATH by choosing a lift MATH of MATH and defining MATH. Conversely, for a given continuous cocycle MATH, we may define continuous MATH-action on MATH by MATH. These are mutually inverse, which establishes the first claim. To prove the second cla... |
math/0010212 | If MATH then MATH. In a genus two handlebody, any pair of separating disks is parallel, so MATH would be isotopic to MATH. If MATH then MATH and MATH would be two separating circles in MATH that intersect in two points, hence they would be isotopic, a contradiction. Suppose MATH. Then each of the disks described in the... |
math/0010212 | There is an obvious (but obviously not unique) orientation-preserving homeomorphism MATH with the property that MATH and MATH. By the NAME trick, MATH is isotopic to the identity. In CITE NAME shows that any isotopy of MATH that ends in a homeomorphism carrying MATH to MATH is a product of particularly simple such isot... |
math/0010212 | The conclusion is obvious if any splitting sphere MATH intersects MATH in a disk disjoint from MATH, for just use the disk MATH. So we may as well assume that every splitting sphere defines an augmented slope. Suppose MATH and MATH are splitting spheres that give rise to two different augmented slopes. Then there is a ... |
math/0010212 | We can regard MATH as the regular neighborhood of a MATH-vertex figure-MATH graph MATH in MATH, in which MATH are meridians of neighborhoods of to the two edges of the graph. Let MATH denote the subknots of MATH corresponding to the meridian disks MATH. It suffices to show that MATH is a standard unknotted figure-MATH ... |
math/0010212 | Suppose, to begin, that there is an extension of MATH to a set of meridians MATH with respect to which MATH is finite. Because MATH is finite, an outermost disk of MATH cut off by the pair of meridians MATH intersects MATH. Then an outermost subdisk MATH of this subdisk, cut off by MATH, is disjoint from both meridians... |
math/0010212 | Cut MATH open along the copies of MATH, denoted MATH and MATH, corresponding to the points of MATH that are nearest to MATH in MATH. Then MATH and MATH lie on different (disk) components of MATH. Cutting MATH open along these meridians leaves one component that is a solid torus MATH whose core is the cycle MATH and who... |
math/0010212 | We know from REF how to find meridians with respect to which MATH is finite: Begin with an outermost disk MATH of MATH cut off by MATH and choose meridians parallel to MATH in the solid torus MATH. REF tells us precisely what those meridians are: one is MATH, the meridian of MATH. The other is bounded by the union of a... |
math/0010212 | Put MATH in thin position so that MATH can be levelled. If MATH is an eyeglass, the result follows from REF , so assume MATH is a level edge MATH. Following CITE we can assume that MATH is in minimal bridge position and the ends of MATH connect the two highest maxima of MATH. If MATH is a level sphere just below MATH, ... |
math/0010212 | Let MATH be the annulus giving the parallelism between MATH and MATH. Let MATH be a neighborhood of MATH containing a neighborhood of MATH. Since MATH is an annulus, we can think of MATH as being a ribbon-like neighborhood of MATH itself. In the complement of MATH, the remnant of MATH is a possibly disconnected surface... |
math/0010212 | First choose a minimal genus NAME surface MATH and slide and isotope MATH, doing both so as to minimize the number of points of intersection between MATH and MATH. The slides and isotopies may leave MATH as either an edge or an eyeglass. (In the latter case, let MATH be the edge in MATH and MATH be the circuit.) We aim... |
math/0010212 | Since MATH traverses MATH once, we can shrink MATH, dragging along its end points in MATH until MATH is just a spanning arc of the annulus MATH. (At this point, we can identify MATH with MATH but we cannot yet identify MATH with MATH.) MATH is a possibly complicated curve lying on MATH and MATH is incident to MATH at t... |
math/0010212 | Let MATH be the copy MATH of MATH in MATH. Consider the hemispheres MATH and MATH. By definition of MATH there are meridians MATH and MATH for MATH that realize the slope MATH. Then, in particular, there are subarcs of MATH that are waves based at one of these meridians. (Warning: we know little about how these meridia... |
math/0010212 | By REF we may assume that MATH is disjoint from some genus one NAME surface MATH. REF shows that we can then isotope MATH onto MATH, necessarily as an essential arc. Then MATH is an incompressible annulus MATH whose ends comprise a non-simple (because of MATH) tunnel number one link MATH. (The core of MATH's unknotting... |
math/0010215 | Clearly, the isotropy group of MATH in MATH equals MATH, as a set. Thus, it suffices to check that the isotropy NAME algebra of MATH is MATH. To see this, recall that the NAME tangent space to MATH at MATH is MATH . Moreover, the differential of the morphism MATH at the identity element of MATH, is the map MATH . |
math/0010215 | Choose opposite NAME subgroups MATH, MATH of MATH, with common torus MATH and unipotent radicals MATH, MATH. By the NAME decomposition, the product map MATH is an open immersion. By CITE and CITE, this map extends to an open immersion MATH where MATH is an affine open subset of the closure of MATH in MATH. Moreover, MA... |
math/0010215 | Since the approach of CITE does not extend to arbitrary characteristics in a straightforward way, we provide an alternative argument. First we review some results from CITE. Let MATH be a parabolic subgroup of MATH such that MATH and MATH are opposite, and that MATH is minimal for this property. Choose a maximal MATH-s... |
math/0010215 | As a first step, we show the equality in MATH: MATH (decomposition into irreducible components). For this, note that we have MATH . Moreover, by REF , the irreducible components of MATH are the closures MATH where MATH satisfies MATH . Thus, we obtain the following decomposition into irreducible components: MATH the un... |
math/0010215 | The first assertion is a direct consequence of REF . If MATH is irreducible, then MATH acts transitively on MATH, and hence its unipotent radical acts trivially. Since MATH acts faithfully on MATH, it follows that MATH, that is, MATH. |
math/0010215 | We begin with the following observation. With the preceding notation and assumptions, every connected component of MATH contains a unique closed MATH-orbit, and admits a MATH-equivariant embedding into the projectivization of a MATH-module. Let MATH be a connected component of MATH. Then all points of MATH, viewed as c... |
math/0010216 | CASE: MATH : CASE: MATH. Let MATH be the maximal root. For MATH we have MATH which proves that MATH is a root. In the same way it is seen that MATH is also a root. We have MATH, thus MATH. CASE: MATH. Reasoning as before, it follows that MATH and MATH, as MATH is a particular root. CASE: MATH : For MATH we have MATH so... |
math/0010216 | We have, for any MATH, MATH. If MATH we have the brackets MATH . Applying the adjoint operator MATH we obtain the condition : MATH . Now MATH, so that MATH for all MATH. On the other hand, the previous condition implies MATH, so MATH, contradiction with the assumption. |
math/0010216 | The characteristic sequence imposes the existence of a basis MATH such that MATH thus we have MATH . The central descending sequence induces the following relations for the associated graduation MATH . If MATH is satisfied, there exist two nonzero vectors MATH such that MATH. Without loss of generality we can suppose M... |
math/0010216 | Let MATH be a basis of MATH over MATH and MATH its dual basis. Any central extension is specified by the adjunction of a differential form MATH . The graduation forces the depth of MATH to be MATH . So MATH is of the following type MATH where MATH for MATH and MATH. The structure of MATH implies MATH . Moreover, the fo... |
math/0010216 | Let MATH . We define the cocycle MATH by MATH . The action of the adjoint operator MATH implies the conditions MATH where MATH is the last value for which MATH is nonzero. Moreover, MATH where MATH . The NAME condition MATH implies MATH so that the cocycle MATH is identically zero. Thus the vector MATH is central and t... |
math/0010216 | An extension MATH is determined by the cocycles of the space MATH . Then the differential form MATH is of type MATH where the indexes MATH satisfy MATH . As MATH is even, let MATH . The the form MATH can be rewritten as MATH . It is trivial to verify that the equations MATH are satisfied. This allows us to take a commo... |
math/0010216 | It is not difficult to see that if MATH, then MATH . For MATH and any of the nongiven MATH's the nonexistence of naturally graded extensions with the required nilindex is routine. The remaining cases are a direct consequence of the previous results. |
math/0010216 | For MATH the cocycles MATH must satisfy the relation MATH . It is immediate to verify that this space is generated by the cocycles MATH subjected to the relations MATH . If MATH is the dual base of MATH, we have MATH and MATH-Thus there is, for any MATH only one extension, which is isomorphic to MATH . For the remainin... |
math/0010216 | Similarly to the previous case we have MATH and MATH so that there exists a unique extension, isomorphic to MATH . |
math/0010216 | MATH . We can suppose MATH as we have studied the case MATH before. We know that if the depth of the vector MATH is MATH the factor algebra MATH is naturally graded and filiform, thus MATH is a central extension of either MATH or MATH . Let also be MATH . If MATH is central, we obtain again a central extension of MATH ... |
math/0010216 | Bot the graduation and MATH imply that the only cocycles that must be considered are those belonging to the space MATH. Thus the only cohomology classes that give central extensions with the prescribed conditions are MATH and MATH. The differential form MATH associated to the koined vector MATH has the form : MATH wher... |
math/0010216 | Again, the main idea of the proof is the same as in REF . We only comment few aspects : for the exceptional ( nine dimensional ) case MATH a central extension satisfying MATH is determined by the cocycles MATH and MATH subjected to the relations MATH . It is clear that they define a unique extension. Any central extens... |
math/0010216 | For any case the reasoning is similar to previous ones. CASE: The cocycle MATH makes reference to the differential form MATH. To this we have to add, by the characteristic sequence and the closure of the forms system, the cocycle MATH, subjected to the condition MATH. A second class of extensions is defined by the cocy... |
math/0010216 | If MATH, it is trivial to verify that this vector must be in the center. Then the factor algebra MATH has characteristic sequence MATH and MATH. We know that for this depth there does not exist any nonsplit model. Thus MATH and MATH must be isomorphic to either MATH or MATH. If MATH the characteristic sequence and the ... |
math/0010216 | We prove the assertion for MATH. For the remaining cases the reasoning is similar. Recall that for MATH the last differential form is given by MATH A central extension of MATH by MATH which is a MATH-algebra will be determined by the adjunction of a differential form MATH, whose structure is MATH where the cocycles MAT... |
math/0010216 | Any extensions which satisfies the centralizer property, preserves the graduation and has characteristic sequence MATH is determined by MATH subjected to the relations MATH where MATH the class is unique, and by an elementary change of basis the extended algebra is easily seen to be isomorphic to MATH. The centralizer ... |
math/0010219 | The proof is made by using induction over MATH. First we prove the result for MATH, in this case the tournament MATH is isomorphic to the first tournament in REF . Calculating MATH using REF we obtain MATH and MATH. Then MATH is MATH - symplectic if and only if MATH . Suppose that the result is true to MATH. For MATH w... |
math/0010220 | Straightforward using the definition of MATH and concatenation. |
math/0010220 | CITE proved that for functions satisfying the SAC, the nonlinearity satisfies MATH . In CITE the following inequality is obtained: MATH . Using REF we obtain easily the right inequality of REF , that is MATH . From the proof of REF we get that MATH satisfies MATH where MATH. Using the trivial identity MATH and the fact... |
math/0010220 | The corollary follows from the proof of the theorem. For a Boolean balanced function, MATH. Therefore for any MATH, such that MATH is a linear structure of MATH of even NAME weight, we have MATH. Thus MATH . |
math/0010220 | This follows from the fact that any nonzero element of MATH is a linear structure for an affine function. |
math/0010220 | We will prove the theorem for the case of MATH even, that is MATH, pointing out, whenever necessary, the differences for the case of odd MATH. The function MATH can be written as MATH . The fact that MATH is balanced can be seen by pairing the functions MATH with MATH and MATH with MATH in the two segments MATH and MAT... |
math/0010220 | We know that MATH. Therefore, MATH and the result follows. |
math/0010220 | We proved that, if MATH is even, then MATH. If there is a MATH not equal to the four displayed vectors in the proof of REF , for which MATH is not PC, then MATH. If so, then by the same argument we would get MATH, which is not true. So MATH is PC with respect to all but four vectors. In CITE, NAME and NAME proved that,... |
math/0010221 | We have MATH . By taking the transformation MATH we see that MATH is affinely equivalent to a bent function in the NAME class (see CITE), therefore it is also bent. |
math/0010221 | Using the above algorithm, we deduce that the RotS function on MATH of degree REF can be be evaluated in MATH steps, which requires MATH operations, since at each step MATH we complement MATH bits. First, we take an example, say MATH on MATH. We see that MATH, therefore it is semi-bent. It is very easy to see that MATH... |
math/0010221 | We recall that MATH. We show that for any MATH, MATH . Since MATH, MATH . Now, from MATH, we get MATH . The above equation, for MATH, produces MATH . Now, we add REF plus twice REF, and we get MATH . But MATH. By adding the two previous equations we get MATH . Replacing REF into REF we obtain MATH . This together with ... |
math/0010221 | Straightforward using the truth table. |
math/0010222 | We may assume that MATH. Let MATH be as in REF . CASE: Let MATH preserves (reverses) orientation-MATH. If MATH, then MATH and by REF we obtain MATH. If MATH in MATH, then MATH converges to MATH in the sense MATH of REF . Hence the map MATH is continuous. Let MATH be the cone extension map and let MATH be the reflection... |
math/0010222 | Suppose MATH in MATH. It suffices to show that the sequence MATH has a subsequence MATH such that MATH and MATH converges uniformly to MATH. Let MATH the radius of MATH and MATH. CASE: Passing to a subsequence we may assume MATH for some MATH. First we will show that MATH. CASE: Suppose MATH. Take MATH, MATH, such that... |
math/0010222 | CASE: Let MATH. Comparing two maps MATH, MATH, we obtain a unique map MATH such that MATH. Extend MATH radially to MATH by MATH. The required map MATH is defined as the unique map MATH with MATH. In Claim below we will show that the map MATH is continuous. This implies the continuity of the map MATH. CASE: Since MATH, ... |
math/0010222 | Under the notations of REF , let MATH and MATH. For the inclusion MATH, we abbreviate as MATH and MATH. Let MATH (the circular arc in MATH). Also let MATH. Note that MATH is continuous in MATH REF , MATH, MATH and that MATH maps MATH homeomorphically onto MATH. CASE: First we will show the following statement: CASE: Su... |
math/0010222 | CASE: Let MATH, MATH and let MATH be the unique point such that MATH. Under REF , MATH corresponds to MATH, where MATH. Thus MATH and the conclusion follows from MATH . CASE: Since MATH corresponds to MATH and MATH corresponds to MATH, it follows that MATH. The conclusion follows from MATH . |
math/0010222 | We may assume that MATH, since if MATH satisfies the above condition in the case where MATH then we have MATH for any MATH. CASE: The case when MATH: We fix a triangulation of MATH and let MATH denote the set of MATH-simplices of this triangulation and MATH denote REF-skeleton of MATH. For each MATH with ends MATH, MAT... |
math/0010222 | Let MATH, MATH, denote the restriction map. By NAME REF (with MATH) there exists an open neighborhood MATH of MATH in MATH and a map MATH such that MATH. Then MATH, MATH, is a homeomorphism with the inverse MATH. Since MATH is an ANR CITE and MATH is open in MATH, MATH is also an ANR. |
math/0010222 | CASE: MATH is completely metrizable by the sup-metric, and MATH is MATH in MATH. CASE: For MATH let MATH denote the subspace of MATH-Lipschitz embeddings. If we write MATH (MATH is compact and MATH), then MATH. Since MATH is equicontinuous and closed in MATH, it is compact by NAME Theorem CITE. Hence MATH is MATH-compa... |
math/0010222 | CASE: MATH is MATH in MATH. CASE: We may assume that MATH. It suffices to show that each MATH has a MATH-fd-compact neighborhood. Since MATH, we may assume that MATH. Choose a sequence of small collars MATH of MATH in MATH pinched at MATH such that MATH becomes thinner and thinner and also the angle between MATH and MA... |
math/0010222 | CASE: For every MATH, take a compact PL REF-submanifold neighborhood MATH of MATH in MATH and consider the map MATH, MATH. By REF there exists an open neighborhood MATH of MATH in MATH and a map MATH such that MATH. Since MATH is an ANR by REF , so is MATH. Hence MATH is an ANR. CASE: By REF it suffices to show that ev... |
math/0010223 | In CITE the NAME groups of compact REF-manifolds was studied in the context of semisimplicial complex. However, using REF and the results in CITE, we can apply the arguments and results in CITE to our setting. CASE: Let MATH be a small regular neighborhood of the union MATH of the nondegenerate components of MATH and l... |
math/0010223 | Attaching MATH to MATH, we may assume that MATH. Take a covering MATH such that MATH. If MATH is noncompact, then by CITE there exists a compact connected REF-submanifold MATH of MATH such that MATH and the inclusion induces an isomorphism MATH. Since MATH, it follows that MATH is a free group, so it is an infinite cyc... |
math/0010223 | By the claim below we have a MATH. If MATH for some MATH, then by REF MATH for some MATH and MATH for some MATH. Since MATH and MATH, by REF MATH. Hence by REF MATH so MATH, a contradiction. |
math/0010223 | First we note that MATH does not contain any handles or NAME bands. In fact if MATH is a handle or a NAME band in MATH, then we can easily construct a retraction MATH which maps MATH homeomorphically onto MATH REF , and we have the contradiction MATH. In particular, if MATH is compact then MATH is a disk or an annulus.... |
math/0010223 | Let MATH be represented by MATH and let MATH. The homotopy MATH implies that MATH. Since MATH, by REF MATH for some MATH. Since MATH does not bound a disk or a NAME band, by REF MATH so MATH for some MATH. Let MATH. By REF there exists a MATH such that MATH for any MATH REF . The homotopy MATH implies that MATH in MATH... |
math/0010223 | Let MATH be any homotopy and let MATH be the components of MATH. By REF the loop MATH in MATH for any MATH. We must find an isotopy MATH rel MATH. Let MATH be any path with MATH, MATH. The homotopy MATH yields a contraction of the loop MATH in MATH. Since MATH, MATH, it follows that MATH in MATH. Since MATH is monomorp... |
math/0010223 | We can assume that MATH is a finite set, since there exists a finite subset MATH of MATH such that MATH satisfies REF . Replacing MATH by MATH, we may assume that MATH. CASE: The case where MATH is compact : Let MATH be the components of MATH and MATH. Let MATH be the components of MATH which are disks or NAME bands an... |
math/0010223 | Let MATH denote the unit path-component of a topological group MATH. REF implies MATH. When MATH is compact, from REF it follows that MATH for any compact subpolyhedron MATH of MATH. Since MATH can be replaced by a finite subset MATH of MATH as in the above proof, we have MATH. The noncompact case follows from the same... |
math/0010223 | REF follows from REF , and REF follows from REF . For REF , note that MATH is path connected REF and each MATH is isotopic to the inclusion MATH in a compact subset of MATH. |
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