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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, the image by MATH acting on two nonzero elements of MATH will have exactly the same support mod MATH. If this is MATH, the images of nonzero vectors of MATH are all in 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 element MATH in MATH. The description of MATH follows directly from this formula. For instance, for MATH with MATH to belong to MATH, we need to have MATH, which is equivalent to MATH, whence MATH. It remains to show that MATH is closed. So suppose that MATH and MATH are in MATH and that MATH is in MATH. Assume MATH. We always have MATH in MATH. As MATH, the inner product MATH equals MATH. By the above, this implies that both MATH and MATH are in MATH. But then MATH satisfies MATH and MATH, so MATH. From now on, we assume that neither MATH nor MATH are equal to MATH. Suppose that both MATH and MATH are orthogonal to MATH. We saw above that being in MATH means that both MATH and MATH are in MATH and because MATH is closed, MATH is also in MATH. But then MATH, being orthogonal to MATH, also belongs to MATH. The case remains where at least one of MATH and MATH is not orthogonal to MATH. Suppose first that MATH. As MATH, by the above, both MATH and MATH are in MATH. If MATH is orthogonal to MATH we know from above and from MATH that MATH is in MATH. Now, as MATH and MATH is closed, also MATH. As MATH and MATH is closed, also MATH is in MATH. Now MATH and so by the above MATH is in MATH. We still need to consider the other possibilities for MATH. As MATH is a root, MATH. Now MATH and we need only show that MATH. But this follows as MATH is closed and MATH. The only case remaining is MATH and MATH. However, the latter inner product cannot be REF, for otherwise MATH, contradicting the fact that MATH is a positive root. This means MATH and as MATH, we find MATH. As MATH and MATH, we have MATH. Since MATH, the vector MATH is a positive root. As both MATH and MATH are in MATH and MATH is closed, the root MATH belongs to MATH. Now as MATH and MATH, we conclude MATH. |
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, by analysis of the case of inner product MATH above, gives MATH. In particular, MATH, so MATH with MATH. Since MATH, REF implies MATH. Hence the lemma. |
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 MATH and, by REF , so is MATH. Therefore, by REF , MATH, proving MATH. |
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 there are MATH, MATH in MATH such that MATH and MATH. But then, as MATH acts injectively, MATH and MATH act identically on MATH, whereas MATH, so we can finish by induction on MATH. |
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 replaced by MATH and MATH, respectively. The matrix MATH is determined by the following rules involving an index MATH such that MATH. MATH . This matrix replaces the matrix MATH in REF. In the representation of REF , MATH is the multiple of the permutation matrix MATH by the scalar MATH. Here MATH permutes MATH according to the action of MATH on MATH and MATH is the number of positive roots that are not orthogonal to a given root. In particular MATH for MATH, MATH for MATH, MATH for MATH, MATH for MATH, and MATH for MATH. Note that this is in accordance with the theorem we are proving as MATH. The matrix MATH replaces the matrix of REF. |
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 MATH as follows by CITE. Considering that MATH, MATH, and that MATH are coprime, we can restrict the possibility again to the three cases below. Thus we have only to exclude the case MATH. In the case MATH, we have MATH, MATH modulo MATH, MATH, and MATH. From the proof of CITE we know that, for MATH, MATH where MATH. In our case we have MATH, MATH, and so on. Putting MATH we obtain MATH . These two equations contradict each other. |
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, and let MATH be its strict transform on MATH. Then, MATH . CASE: Let MATH be a general line on MATH in REF and be a general fiber of MATH in REF , and let MATH be its strict transform on MATH. Then, MATH . CASE: Let MATH be a general conic on MATH, and let MATH be its strict transform on MATH. Then, MATH . CASE: The following proof is a generalization of the proof of CITE using NAME 's connectedness lemma CITE. Let MATH be a general hyperplane section on MATH through MATH, and let MATH be a general hyperplane section on MATH through MATH such that MATH, and that MATH consists of two lines MATH on MATH, which are fibers of two rulings of MATH. Then MATH where we omit the term MATH if MATH. Because MATH we have MATH where LLC denotes the locus of log canonical singularities for a log pair, that is, the union of centers of all algebraic valuations with discrepancies MATH. Moreover, MATH . Since MATH, using the connectedness lemma for two small contractions contracting MATH respectively, we obtain MATH that is, MATH. |
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 contradiction. CASE: We obtain them considering the following inequalities as in the proof of REF . MATH . MATH comes from MATH and REF . |
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. CASE: MATH. CASE: MATH. It is trivial that we can choose an identification satisfying REF . Then by REF , MATH for some MATH. Because MATH for MATH and MATH, by replacing MATH with MATH we may assume REF moreover. Then MATH for some MATH by REF . Because MATH for MATH and MATH, by replacing MATH with MATH we may assume REF moreover. Second we prove that MATH equals, as valuations, an exceptional divisor obtained by a weighted blow-up of MATH. Under the identification in REF , MATH equals, as valuations, an exceptional divisor obtained by the weighted blow-up of MATH with its weights MATH. First we remark that MATH generate local coordinates of MATH at MATH, that MATH, and that MATH equals, as valuations, the exceptional divisor obtained by the weighted blow-up of MATH with its weights MATH. Thus it is trivial that MATH. Because any MATH has an expansion of a formal series MATH, it is sufficient to prove that for any MATH, MATH implies MATH, where MATH . However, by replacing MATH with MATH for a sufficiently large MATH, we have only to show that for any MATH, MATH implies MATH, where MATH. Take any MATH. Then MATH. Because MATH equals, as valuations, the exceptional divisor obtained by the weighted blow-up of MATH with its weights MATH, it is enough to show that the weight of any monomial MATH REF with respect to its weights MATH equals MATH. But this is trivial by a direct calculation MATH. Only the proof of MATH remains. Because of REF and properties of toric geometry, we have only to show the following claim. Consider an analytic germ of a MATH point MATH and blow-up this with its weights MATH. Then the exceptional locus of this weighted blow-up is irreducible, and the weighted blown-up analytic space is normal and has a non-terminal singularity. Direct calculation shows that its exceptional locus is isomorphic to MATH with weighted homogeneous coordinates MATH, which is irreducible, and that all singularities on the obtained analytic space are one terminal quotient singularity of type MATH and one non-terminal singularity isomorphic to MATH. |
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 REF . CASE: In REF . CASE: In REF . we will express MATH's in each case under a suitable identification MATH. CASE: As in REF , we may assume that MATH and MATH. Then MATH where MATH for some MATH such that MATH or MATH is non-zero. This implies REF . CASE: We may assume that MATH. Then MATH . This implies REF . CASE: We may assume that MATH. Then MATH . This implies REF . |
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 linear system on MATH obtained by the total pull-back of MATH with the inclusion map MATH. It is enough to prove that MATH. Let MATH be a general line on MATH, and let MATH be the strict transform of MATH on MATH. Then MATH which shows that MATH. |
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-up of MATH with its weights MATH. First we remark that MATH generate local coordinates of MATH at the generic point of MATH, that MATH, that MATH, and that MATH equals, as valuations, the exceptional divisor dominating MATH which is obtained by the weighted blow-up of MATH along MATH with its weights MATH. Thus we obtain MATH, considering REF . Because any MATH has an expansion of a formal series MATH, it is sufficient to prove that for any MATH, MATH implies MATH, where MATH . However, by replacing MATH with MATH for a sufficiently large MATH, we have only to show that for any MATH, MATH implies MATH, where MATH. Take any MATH. Then MATH. We note that MATH. Because MATH equals, as valuations, the exceptional divisor dominating MATH which is obtained by the weighted blow-up of MATH along MATH with its weights MATH, it is enough to show that the weight of any monomial MATH REF with respect to its weights MATH equals MATH. But this is trivial by a direct calculation MATH. There remains only proving that MATH are coprime. Because of REF and properties of toric geometry, we have only to show the following claim. Consider an analytic germ of a MATH point MATH and blow-up this with its weights MATH, where MATH are positive integers with MATH and are not coprime. Then the exceptional locus of this weighted blow-up is irreducible, and the weighted blown-up analytic space is normal and has a non-terminal singularity. Direct calculation shows that its exceptional locus is isomorphic to MATH or MATH with weighted homogeneous coordinates MATH, which is irreducible, and that all singularities on the obtained analytic space are two non-terminal quotient singularities of types MATH and MATH, and in the case MATH furthermore one terminal NAME singularity isomorphic to MATH. |
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 adjunction formula, we obtain that MATH . On the other hand, we know that MATH is MATH, that MATH is of codimension MATH, and that MATH is invertible. Thus we obtain MATH, and our problem is reduced to only proving that MATH is isomorphism. Second we calculate the dualizing sheaf MATH on MATH. NAME duality tells that MATH where the remark that MATH is invertible induces the third equality. Because MATH is canonical, the above equation shows that the conductor ideal sheaf MATH has to equal MATH. Hence MATH is isomorphism. |
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. CASE: MATH has its multiplicity MATH along every prime component of MATH, that is, MATH. CASE: The set MATH consists of MATH points, say MATH. These MATH are NAME singularities of types MATH with MATH. Here we define a NAME singularity of type MATH as a smooth point. CASE: For MATH, the local intersection number MATH equals MATH. Let MATH be the minimal resolution of MATH, and let MATH be the induced morphism. MATH is a chain of MATH-curves MATH's. We order the indices MATH's so that they are compatible with the order of MATH's in this chain. It is fundamental to see that MATH and that MATH intersects MATH exactly at a point, say MATH, on MATH and at a point, say MATH, on MATH transversally, where we omit MATH and MATH if MATH. Let MATH (respectively,MATH) be the smallest non-negative integer such that MATH (respectively,MATH) is not contracted by MATH. Then MATH REF is a NAME singularity of type MATH, and MATH consists of MATH. Because MATH (respectively,MATH), we have MATH (respectively,MATH). We begin to prove REF . We keep the notation MATH in REF . Let MATH be a general hyperplane section on MATH through MATH. Then MATH is also a general hyperplane section on MATH through MATH. Because MATH, we have MATH and MATH. The support of MATH is exactly the exceptional locus of MATH, and MATH is factored through by the minimal resolution of MATH by REF . Thus by REF , we obtain that MATH is reduced and that MATH, the strict transform of MATH on MATH. We calculate the intersection number of MATH and MATH around MATH. MATH where the last equality comes from REF . By REF , the set MATH consists of MATH points, say MATH, and thus MATH. We may assume that MATH. Considering the set MATH and REF , we know that MATH and MATH, and that the local NAME indices of MATH are MATH. Therefore by REF, we obtain that MATH are NAME singularities of types MATH with MATH, that is, MATH. |
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 (respectively,MATH, MATH). By this and REF we have MATH (respectively,MATH, MATH) MATH, which implies that MATH. |
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 the above claim. CASE: Take an identification MATH satisfing that MATH. Then for MATH, MATH for some MATH. CASE: Assume MATH. CASE: In REF , MATH, MATH or MATH. CASE: In REF , assume that MATH. Under this situation, for MATH, MATH for some MATH, where MATH is the set in REF . CASE: We will construct MATH inductively starting with MATH. Assume that we have constructed MATH (MATH). There exists a surjective map MATH, MATH . By MATH and REF , MATH. On the other hand because MATH, we know that MATH generates MATH, and that MATH for some MATH. Hence it is enough to put MATH. CASE: As in the above proof, using MATH in REF we have a surjective map MATH, MATH . Dividing by MATH, we have another surjective map MATH, MATH . By REF and MATH, MATH. Hence MATH, which shows REF . CASE: We will prove REF as in the proof of REF , constructing MATH inductively starting with MATH in REF . Assume that we have constructed MATH (MATH). There exists a surjective map MATH, MATH . We know that MATH generate MATH because of the proof of REF . Thus any non-zero element in MATH, which always decomposes into a product of MATH and MATH linear combinations of MATH, has exactly its multiplicity MATH along MATH. This and REF imply that MATH generate MATH, and that MATH for some MATH. Hence it is enough to put MATH. We will construct an identification in REF using REF . It is easy that we can take an identification in REF . REF is trivial if MATH by REF . If MATH, by REF and an equation MATH, we may assume that MATH in the construction of MATH in REF . Then by REF , we obtain that MATH. We express MATH as MATH (MATH, MATH). Thus it is sufficient to replace MATH with MATH, MATH because MATH. |
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 point on MATH except the vertex point of MATH. Thus MATH for some MATH. We note that MATH because MATH. Take a surface MATH for a general MATH. Then MATH, which is a NAME singularity of type MATH. Because MATH, MATH, and MATH, the multiplicity of MATH along MATH equals MATH. Thus MATH is special of type MATH. |
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 of this MATH is a homogeneous polynomial of degree MATH in the NAME classes, just as in the topological case. The key point is the compatibility with the NAME operations which implies that the restriction of MATH to MATH vanishes unless MATH. It follows that MATH factors through the projection onto MATH: MATH . Thus the NAME character induces a natural transformation MATH . It is a ring homomorphism as the NAME character is. Define MATH to be the projection MATH. From this, one can inductively construct NAME classes MATH. Compatibility with the NAME classes MATH is automatic and guarantees the uniqueness of natural transformation MATH. |
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 when MATH is even and MATH, and MATH when MATH. It is easy to see that MATH . CITE constructed regulator mappings MATH and showed that each is injective mod torsion. CITE showed that NAME 's regulator is a non-zero rational multiple of the regulator mapping MATH to NAME cohomology. The properties of the NAME character and NAME 's injectivity together imply that MATH and that MATH vanishes when MATH, and when MATH and MATH. The result when MATH is non-empty follows from this using the localization sequence CITE, and the fact, due to CITE, that the MATH-groups of finite fields are torsion groups in positive degree. Together these imply that each prime removed adds one to the rank of MATH and does not change the rank of any other MATH-group. |
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 of a quotient of MATH by a unipotent group. Here MATH is the NAME closure of the image of MATH in MATH. Because the action of MATH on each weight graded quotient factors through MATH, and because MATH acts on the MATH-th weight graded quotient of MATH with weight MATH, it follows that this is a negatively weighted extension of MATH. Pulling back this extension along the projection MATH, we obtain a negatively weighted extension MATH of MATH and a continuous homomorphism MATH that lifts both MATH and the homomorphism MATH. By the universal mapping property of MATH, there is a natural homomorphism MATH compatible with the projections to MATH and the homomorphisms from MATH to MATH and MATH. Thus every object of MATH is naturally a MATH-module. It is also easy to see that every morphism of MATH is MATH-equivariant. Conversely, suppose that MATH is a finite dimensional MATH-module. Composing with the natural homomorphism MATH gives MATH the structure of a MATH-module. In CITE, it is proven that every MATH-module has a natural weight filtration with the property that the action of MATH on each weight graded quotient factors through the projection MATH and that MATH acts with weight MATH on the MATH-th weight graded quotient. It follows that MATH is naturally an object of MATH. Since MATH-equivariant mappings are naturally MATH-equivariant, this proves that MATH is naturally the category of finite dimensional MATH-modules, which completes the proof. |
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 establishes the injectivity of MATH. To prove surjectivity of MATH, we define a natural weight filtration on each MATH-module extension MATH of MATH by MATH. Simply set MATH and MATH. Since MATH, this makes MATH a weighted MATH-module. To prove that MATH is injective, we need to show that if a REF-step extension MATH lies in the trivial class of extensions of MATH-modules, then it also lies in the trivial class of extensions of weighted MATH-modules. If MATH, then MATH, and there is nothing to prove. Thus we may assume MATH. Since MATH is an exact functor, we may apply MATH to REF to obtain another REF-step extension, without changing the extension class. Then, taking MATH, we have a short exact sequence MATH of MATH-modules. Since MATH is reductive, this has a splitting MATH. Taking the inverse images of this copy of MATH along MATH in MATH and in MATH, we obtain a REF-step extension MATH equivialent to REF satisfying MATH and MATH. Using the dual argument, we may assume that REF satisfies MATH, MATH, MATH, and MATH. By NAME 's characterization CITE of trivial MATH-step extensions, the extension REF represents the trivial REF-step extension class as MATH-modules if and only if there is a MATH-module MATH and exact sequences MATH which are compatible with the existing mappings MATH and MATH. To establish the injectivity of MATH, it suffices to prove that MATH is a weighted MATH-module. But MATH has the weight structure MATH and MATH. This completes the proof of REF . |
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 MATH-modules. |
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 restriction mapping MATH and the transfer mapping CITE MATH . A direct computation on cocycles shows that MATH and MATH are both multiplication by the order of MATH, and are thus isomorphisms. So MATH vanishes if MATH is non-trivial, and is MATH if MATH is trivial. This shows that the unipotent radical of the completion MATH is isomorphic to that of MATH. By functoriality of weighted completion, we have a homomorphism MATH which induces the isomorphism on the unipotent radical. The statement follows. |
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 crystalline, which is well-known. So the first assertion follows. We now consider the case of MATH. Set MATH. We may assume MATH. It suffices to show that MATH in the proof of REF is crystalline provided MATH and MATH are crystalline. But this follows from the next result, which will be proved below. Let MATH be a short exact sequence of crystalline MATH-adic representations of MATH. Assume that MATH is a successive extension of direct sums of a finite number of copies of MATH with MATH. Then, for any extension MATH of MATH-adic representations of MATH, MATH is crystalline if and only if its pushout by the surjection MATH is crystalline. Let MATH be MATH as in the proof of REF . Since MATH, MATH is an extension of MATH for some MATH. Since MATH, the pushout of MATH along MATH is a quotient of MATH, and hence is crystalline. Thus the proposition says that MATH is crystalline, which completes the proof of REF . |
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 have the following commutative diagram whose two rows are exact: MATH . The right vertical arrow is an isomorphism and the left one is also an isomorphism by the induction hypothesis. Hence the middle one is also an isomorphism. |
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 claim, we first define a MATH-linear mapping MATH as follows. For MATH, choose a REF-fold extension MATH that represents it. By CITE, MATH is the image under the connecting homomorphism MATH of the class MATH of the extension MATH. We shall construct MATH so that the diagram MATH commutes, where the rows are parts of the standard long exact sequences constructed in CITE and CITE. Define MATH to be MATH. To prove MATH is well-defined, it suffices to show that two REF-fold extensions that fit into a commutative diagram MATH give a same element of MATH. But this follows from the functoriality of the connecting homomorphism for MATH, that is, the commutativity of MATH . The MATH-linearity of MATH is easily checked. Finally, the injectivity of MATH follows from the fact that for each extension as above, MATH is injective on the image of the connecting homomorphism MATH. |
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 lemma has boundary a bigon in MATH that can be made disjoint from MATH and MATH. If any boundary bigon were inessential it could be used to reduce MATH. So each bigon in MATH (hence each disk in MATH) is essential. Furthermore, any essential circle in MATH that is disjoint from MATH and MATH is non-separating. See REF . |
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 isotopies, whose effect on a fixed separating sphere MATH is simple to describe. In each case, either MATH is preserved or the intersection number of MATH with its image is MATH. The upshot is this: the homeomorphism MATH is the composition of homeomorphisms MATH where each MATH is the MATH-preserving homeomorphism of MATH obtained by one of the simple isotopies. To obtain a sequence of splitting spheres we take MATH. Then notice that MATH can be understood by viewing it as the image under the homeomorphism MATH of MATH, so MATH or MATH. |
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 sequence of splitting spheres, beginning with MATH and ending with MATH, such that each has intersection number MATH with the previous splitting sphere. Since the first and last terms have different augmented slopes, somewhere there is a pair in sequence with different augmented slopes. So we may as well assume that MATH or MATH. Let MATH and MATH be the complete pair of arcs associated to the waves of MATH and MATH respectively. First notice that MATH. For if not, then MATH. If we double each of the four arcs, the total number of intersection points is MATH, and converting a doubled arc into a wave can never remove intersection points, only add them. Suppose next that MATH. Then MATH and MATH can be made disjoint, but not the waves that define them. Indeed, if one pair of waves has its ends on MATH and the other on MATH then each wave from MATH intersects each wave from MATH in at least two points, a total of at least MATH points. See REF . On the other hand, if MATH and both pairs of waves have their ends on MATH (or both on MATH) then each wave from MATH intersects each pair of waves from MATH in at least MATH points, a total intersection of just the waves of MATH points. Any other arc of MATH with the same slope will intersect a wave of MATH, and vice versa, so MATH and MATH must both be disjoint from MATH. Following REF we can say more: the bigons determined by the intersection points bound non-separating disks, two in MATH and two in MATH. In our case each relevant bigon in MATH are made up of a unions of arcs, each with one end on MATH and one end on MATH. In particular, the bigon intersects MATH twice or more, always with the same orientation. But no such curve can bound a disk in MATH, for the union of the solid torus MATH with the disk would define a punctured lens space in MATH. See REF . Finally, if MATH but MATH has waves on MATH and MATH on MATH, then each wave of MATH intersects each wave of MATH in at least (hence exactly) two points. Because of the waves, any further component of MATH would intersect MATH and vice versa, so there can be no further components: MATH and MATH each intersects MATH in exactly the two waves. Then the bigons cut out by MATH and MATH that bound disks in MATH each intersect each of MATH and MATH in exactly one point. |
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 graph in MATH since then the boundary of a regular neighborhood of the vertex of MATH serves as a bridge sphere for a MATH-bridge presentation of the knot core of MATH. Following the unpublished CITE (see CITE) it suffices then to show that each of the knots MATH is the unknot. Suppose that MATH is an essential disk, given by REF , that intersects each of MATH and MATH at most once. If MATH is disjoint from exactly one of the meridians, say MATH, then MATH is an unknotting disk for MATH, and MATH is an unknotted solid torus whose core is MATH. Similarly, if MATH is disjoint from both meridians, then MATH divides MATH into two solid tori, each of whose meridians is an unknotting disk for one of MATH. If MATH intersects both meridians, then MATH is an unknotted solid torus in which both MATH and MATH can be viewed (individually) as core curves. |
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. Furthermore, MATH intersects MATH in a single arc dividing MATH into two subdisks. The union of each of those subdisks with MATH gives meridian disks for MATH parallel to MATH. (See REF ). Now MATH is an essential arc MATH in the twice punctured torus MATH, and the arc has both its ends on a single puncture. Let MATH be closed curves in MATH parallel to MATH, lying on either side of MATH. If MATH is any other arc of MATH which has both its ends on a single puncture, then MATH is disjoint from MATH; this is obvious if the ends of MATH lie on the other puncture, and follows from a counting argument on ends of arcs in MATH if the ends of MATH lie on the same puncture as those of MATH. Any non-parallel separating pair of closed curves, e. CASE: MATH, in MATH must be parallel to MATH. So we see that MATH are determined precisely by taking closed essential curves in MATH that are parallel to MATH. (See REF ) It's now easy to see that there is always some such pair. Consider an outermost disk MATH of MATH cut off by MATH in MATH. Then the union of MATH with each of the two subdisks of MATH into which MATH splits MATH produces two natural meridian disks MATH for the solid torus MATH. These, together with MATH comprise a complete collection of meridian disks for MATH. Moreover, MATH is an essential arc that is disjoint from both meridians MATH of the solid torus MATH bounded by MATH. Since some arc of MATH lying in the pairs of pants MATH has both ends on MATH, no arc can have both ends on the same component of MATH. Hence MATH is finite with respect to the meridian set MATH. |
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 whose boundary contains disks corresponding to MATH and MATH. The meridian MATH of MATH and the curves MATH (parallel in MATH) naturally define respectively a meridian curve (which we continue to call MATH) and horizontal longitudes in the twice-punctured torus MATH. We want to understand the pattern of arcs MATH. See REF . We begin by examining how MATH intersects the twice punctured annulus MATH obtained by cutting open MATH along the horizontal longitude MATH at the top of MATH. Note that MATH intersects MATH in a single spanning arc. The boundary of MATH can be thought of as two copies of MATH which we denote MATH. Claim: Among the arcs in MATH that intersect MATH, either there is one that has both ends on the same component of MATH and separates the punctures MATH and MATH, or there is one that has one end on MATH and the other end on a puncture MATH or MATH. Proof of Claim: Let MATH be the exterior disk, and consider an outermost disk MATH cut off from MATH by an outermost arc MATH of MATH. (It is easy to remove all closed components of MATH, since MATH is thin.) Since MATH obviously lies in a single component of MATH, it follows easily that MATH must lie below MATH (all arcs of MATH can be assumed to be essential and so each spans the annulus MATH) and so MATH lies in MATH. The arc MATH can't have one end on each of MATH and MATH since these lie in distinct components of MATH. If the ends of MATH both lie on the meridian MATH, say, then MATH is a longitudinal arc in the punctured annulus MATH. That is, MATH is a core curve of MATH. On the other hand, the outermost disk of MATH cut off by MATH must be a meridinal wave in MATH, so it too must also be based at MATH. The complement of the two arcs, one meridinal and the other longitudinal, is then a disk in MATH containing just the puncture MATH. But then no wave could be based at MATH, and there would be more ends of MATH on MATH than on MATH, an impossibility. We deduce that MATH has one or both ends on MATH. Notice that if both ends of MATH lie on MATH then they must lie on the same component (MATH, say) of MATH (since they are connected by an arc in MATH) and then the subdisk MATH of MATH cut off by MATH contains at least one puncture (or it would be inessential) but not both else MATH would intersect MATH more often than it intersects MATH. It remains to show that MATH intersects MATH. We will show that if it doesn't, it can be used to make MATH thinner. Suppose that MATH is disjoint from MATH and consider first the case in which MATH has one end on MATH (say) and other end on MATH. Then there is an arc MATH so that MATH and one end of MATH lies on each of MATH and MATH. Since MATH is disjoint from MATH it is isotopic in the ball MATH to MATH. Then MATH can be used to pull a minimum of MATH past MATH to MATH, thinning MATH, a contradiction. Similarly, if both ends of MATH lie on (necessarily the same component of) MATH, then MATH is a lower cap, separating one component MATH of MATH below MATH from all the others; MATH is parallel to an arc MATH in the punctured plane MATH that lies entirely in the twice punctured disk in MATH bounded by MATH. There is then an arc MATH in MATH whose interior is disjoint from MATH and intersects MATH once and whose ends lie on the two punctures. Then MATH can be pushed up to MATH and MATH pushed down to MATH, thinning MATH. From this contradiction, we conclude that MATH intersects MATH, establishing the Claim. Following the Claim, we have two cases to consider, corresponding to the two types of arcs given by the Claim. Both arguments will use the MATH-punctured sphere MATH bounded by MATH and two copies MATH of MATH. We briefly recount some of its properties. An outermost disk MATH of MATH cut off by MATH is a wave of MATH in MATH and if it's based at MATH an end count shows that there is a wave based at MATH and vice versa. Similarly if a wave is based at one of MATH there is a wave based at the other. Furthermore, any two waves in MATH must have the same slope. The arc MATH is a path joining the two copies MATH; we can use it to establish slope MATH in MATH. In these terms, a restatement of the Lemma is the claim that the wave determined by MATH is based at one of MATH and is disjoint from MATH. Suppose first that the arc MATH given by the Claim has both ends on a boundary component MATH of MATH and cuts off from MATH a disk MATH containing a single puncture MATH (say). Once MATH is minimized by isotopy, an outermost arc of MATH in MATH cuts off a bigon MATH containing MATH and bounded by subarcs of MATH and MATH. (See REF .) An outermost arc of MATH in MATH (possibly the subarc of MATH in MATH) is a wave of MATH in MATH based at MATH that is disjoint from MATH, since MATH is. It follows that all the waves in MATH, including that determined by MATH, have the same property. Suppose finally that only one end of MATH lies on MATH and the other end lies on MATH (say). Since MATH intersects MATH, the end segments of MATH in MATH have these properties: One end, MATH, connects one of MATH to MATH and is disjoint from MATH. The other end connects MATH to one of MATH essentially; let MATH denote the segment of MATH that contains this end. Since one of MATH intersects MATH and one doesn't, they have different slopes in MATH. (See REF .) Observe that if waves are based on two boundary components of a MATH-punctured sphere, the only disjoint arc that can have a different slope than the waves is an arc that connects the bases of the waves. Since only one end of MATH can be the base of a wave, it follows that MATH must connect the two bases of the waves. Since one end of MATH lies on one of MATH this means that the waves must be based at MATH, and MATH runs between MATH. Finally, the slope of the wave must be that of MATH, so the wave is disjoint from MATH. |
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 an arc disjoint from the highest horizontal longitude MATH of the circuit MATH and an arc that intersects MATH once. Hence the boundary of the second meridian has a single maximum, so it can be viewed as simply the vertical meridian of MATH at MATH, separating the ends of MATH. |
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, then the part of MATH lying above MATH is just a collar MATH, so we may as well assume that the exterior disk MATH intersects this product region in bigons whose boundaries each consist of an arc in MATH and an arc in MATH. Let MATH and MATH be the meridians of the two arcs MATH; we know that MATH, say, cuts off an outermost disk MATH of MATH. If MATH intersects the meridian MATH of the tunnel MATH we are done, so suppose that MATH is disjoint from MATH. Somewhere below MATH and above the highest minimum of MATH, there is a generic level sphere MATH which cuts off both an upper disk MATH and a lower disk MATH from MATH. Since MATH and MATH can be made disjoint, it follows that MATH crosses the meridian MATH of the tunnel at most once. We are now in a position to apply the argument of CITE, though now in the context that the edge MATH disjoint from MATH is MATH, not a subarc of MATH: There cannot be simultaneously an upper cap and a lower cap that have disjoint boundaries in MATH, or MATH could be thinned. If there is an upper cap and a disjoint lower disk or a lower cap and a disjoint upper disk, then, as in CITE we can find such a pair for which the interior of the disk is disjoint from MATH. Then thin position implies that there cannot be simultaneously an upper cap and a lower disk and, if there is a lower cap which is disjoint from an upper disk (whose interior is now disjoint from MATH) then we can ensure that the boundary of the upper disk runs across the tunnel, hence exactly once across the tunnel. Similarly, if there is an upper disk which is disjoint from a lower disk then we can ensure that the interior of the upper disk is disjoint from MATH and is disjoint either from a lower cap or a lower disk whose interior is also disjoint from MATH. Furthermore, the boundary of the upper disk must be incident to the tunnel, hence run exactly once across the tunnel. But then the upper and lower disks describe how to push the tunnel down to, or even below, the level of at least one minimum. Following CITE this implies (almost, see next paragraph) that the tunnel can be pushed down to connect two minima, which we may make to be the lowest minima. Furthermore, this operation lowers by two the number of extrema of MATH (in our case, the number of minima) found in the interior of the component of MATH containing MATH, the base of the waves. The argument then continues; it finally fails when there is only one extremum in the component of MATH containing MATH. That is, in the argument above, it may finally happen that the two maxima to which the ends of MATH are attached have only one minimum lying between them, and it's the component on which MATH is found. In this case, the upper disk pushes MATH down to MATH, and then MATH together with the segment MATH containing the minimum form an unknotted cycle. If any other minimum could be pushed up above this cycle then, again following CITE, all the other minima could, and the cycle would bound a disk disjoint from MATH, a contradiction. We conclude that the lower disk or cap is in fact a lower disk that pushes MATH up to MATH, thereby forming a level cycle. It can be used to slide MATH to form a level eyeglass, at which point we appeal to REF . |
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, with three (preferred) longitudinal boundaries on the boundary of MATH. If MATH is disconnected (corresponding to the case in which MATH is separating) then one of the components of MATH is a NAME surface for MATH. Since it can't be of lower genus than MATH, the other component must be an annulus, defining a parallelism between MATH and MATH in MATH, as required. Suppose MATH is non-separating. Then zero-framed surgery on MATH yields a manifold MATH and ``caps off"' MATH; call the capped-off surface MATH. A capped-off version MATH of the NAME surface MATH also imbeds in MATH and MATH and MATH represent the same homology class in MATH. Since MATH is less than MATH, it follows from work of CITE that genus MATH is less than genus MATH, a contradiction. |
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 to show that MATH . Suppose to the contrary that after the slides and isotopies MATH is non-empty. Let MATH be an essential disk in the handlebody MATH chosen to minimize the number MATH of components in MATH. MATH for otherwise the incompressible MATH would lie in a solid torus, namely (a component of) MATH, and so be a disk. Furthermore, since MATH is incompressible, we can assume that MATH consists entirely of arcs. Let MATH be an outermost arc of MATH in MATH, cutting off a subdisk MATH of MATH. The arc MATH is essential in MATH, for otherwise we could find a different essential disk intersecting MATH in fewer components. Let MATH, an arc in MATH with each end either on the longitude MATH or a meridian disk of MATH corresponding to a point of MATH. CASE: If any meridian of MATH is incident to exactly one end of MATH, then we can use MATH to describe a simple isotopy of MATH which reduces the number of intersections between MATH and MATH. CASE: If no meridian of MATH is incident to and end of MATH, then both ends of MATH lie on MATH. If the interior of MATH runs over MATH we are done, for MATH is disjoint from MATH. If the interior of MATH lies entirely in MATH then MATH would be a boundary compressing disk for MATH (since MATH is essential), contradicting the minimality of MATH. The only remaining possibility is that both ends of MATH lie on the same meridian of MATH. In this case, MATH forms a loop in MATH and the ends of MATH adjacent to MATH both run along the same subarc MATH of MATH. Since MATH is disjoint from MATH, MATH either terminates in MATH or MATH is an eyeglass and MATH terminates in the interior vertex of MATH. If MATH terminates in an end of MATH in MATH then, since the interor of MATH is disjoint from MATH, MATH must intersect MATH in either an inessential arc in the torus or in a longitudinal arc. The former case is impossible, since if the disk bounded by the inessential arc did not contain the other end of MATH then it could be isotoped away and MATH reduced. If the disk did contain the other end of MATH, then MATH would cross one end of MATH more often than the other, an impossibility. It follows that MATH intersects the torus MATH in a longitudinal arc. Then MATH is a thickened annulus MATH, defining a parallelism in MATH between MATH and the loop MATH on MATH. By REF that means the loop MATH is parallel to MATH. Substituting MATH for the annulus between MATH and MATH in MATH would create a NAME surface for MATH with fewer intersections with MATH, a contradiction. So MATH is an eyeglass and MATH terminates in the interior vertex of MATH. If, nonetheless, the interior of MATH intersects MATH this means that MATH traverses the edge MATH so MATH is disjoint from MATH. In that case, we can just repeat the argument above, absorbing MATH into MATH. So we can assume that MATH lies entirely on MATH. Now the component MATH of MATH on which MATH lies is either a punctured torus (if MATH is disjoint from MATH) or a pair of pants. In the former case, consider the NAME surface MATH obtained from MATH by removing the meridian disk MATH of MATH on which the ends of MATH lie and substituting MATH. MATH is of one higher genus than MATH, and intersects MATH in one fewer point. Surgery to MATH using MATH reclaims the minimal genus without introducing another point. Thus we get a contradiction to our choice of MATH. If MATH is a pair of pants a similar argument works: Since MATH is incompressible, it follows that the loop MATH bounds a disk in MATH. Since MATH is essential, that disk contains exactly one of the other two meridian disks (call it MATH) of MATH in MATH that correspond to boundary components of MATH. Remove the meridian disks MATH and MATH from MATH and attach instead an annulus that runs parallel to the subarc of MATH (containing the interior vertex) that has ends at MATH and MATH. This creates a NAME surface MATH of genus one greater than MATH, but having one fewer intersection point with MATH. Now do surgery on MATH using MATH, deriving the same contradiction as above. |
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 the ends of MATH. Using a collar MATH of MATH in MATH isotope MATH until it is a standard longitude lying on the boundary of the smaller tubular neighborhood MATH of the core MATH. Extend the isotopy to an ambient isotopy of MATH, i. e. a self-homeomorphism of MATH. This extends the ends of MATH as a MATH-braid through MATH; call the extended arc MATH. The construction shows that MATH is isotopic in MATH to MATH. Now make the braid trivial by absorbing it into MATH. (This translates into slides of the ends of MATH on MATH). Afterwards, MATH is just a collar of MATH. |
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 meridians intersect MATH.) If the exterior disk MATH were disjoint from MATH, then MATH would lie in a solid torus obtained by compressing MATH to the outside along MATH, contradicting the assumption that MATH is incompressible. So MATH. We can isotope MATH and MATH to have minimal intersection and then remove any closed components of MATH since MATH is incompressible. Let MATH be an outermost disk of MATH cut off by MATH. Then MATH consists of two arcs, MATH lying on MATH and MATH lying on MATH. We may assume that MATH is essential in MATH, for otherwise the subdisk of MATH it cuts off, together with MATH, would again give an essential disk in MATH that is disjoint from MATH. An important observation is that the ends of MATH lie on MATH and are incident to the same side of MATH. That is, if MATH is normally oriented, the orientation points into (say) MATH at both ends of MATH. It's also true that MATH must cross the meridian MATH of MATH at least once. For otherwise, MATH would give a MATH-compression of MATH to MATH, contradicting the fact that MATH is an incompressible (hence MATH-incompressible) NAME surface for MATH. If MATH crosses MATH exactly once, then, following REF , MATH is equivalent to MATH, so MATH provides a way of isotoping MATH to MATH, completing the proof. Hence it suffices to show: Claim: Any subarc of MATH whose interior is disjoint from MATH and whose ends lie on the same side of MATH crosses MATH at most once. Proof of claim: Continue to denote this subarc by MATH. Let MATH denote the four-punctured sphere obtained by cutting open MATH along the meridians MATH and MATH of MATH. CASE: MATH intersects MATH in loops as well as arcs. Say the loops are based at MATH and MATH (see REF ). We first note that we may as well assume that MATH lies entirely in MATH. Indeed, since MATH is finite, there are waves of MATH based at MATH that lie in MATH and are on opposite sides of MATH (see REF ). It follows that MATH can't cross MATH. For the same reason, we can assume that any component of MATH with an end on MATH must lie in the component MATH of MATH that contains MATH, for any other component can be isotoped out through MATH. But the segments MATH and MATH are disjoint, for otherwise the two adjacent loops of MATH based at MATH and MATH would form a simple closed curve, which is impossible. So now assume that MATH. (We say MATH is short.) Since MATH intersects MATH in loops as well as arcs, then to intersect MATH at all, MATH must lie in the annulus lying between the two outermost loops. (See REF .) This annulus has MATH as its core (since MATH is disjoint from MATH) and any essential path in the annulus intersects MATH at most once. CASE: MATH intersects MATH only in arcs. If MATH intersects MATH only in arcs, then (considering the torus MATH) it must be in precisely two arcs, both of infinite slope (connecting MATH to MATH and MATH to MATH.) If MATH, the argument is the same as above, using the annulus MATH (see REF ). So finally suppose MATH is not contained in MATH and suppose with no loss that the waves of MATH are based at MATH and MATH. Any arc of MATH with an end on either MATH or MATH will then have a fixed slope, and since the slope MATH is not MATH it will intersect MATH in its interior. Moreover, if the normal orientation induced by that of MATH points towards MATH on an arc with an end on MATH it will point away from MATH on an arc with an end on MATH (see REF ). Hence we conclude that MATH cannot cross MATH. It is as easy to rule out the possibility that MATH crosses MATH. An arc of MATH with one end on MATH may have slope MATH or have slope MATH with MATH. That is, fixing a meridian MATH so that MATH, p odd, we could have that MATH with MATH. The arc couldn't have its other end on MATH so MATH is even, hence MATH is odd and MATH. Note that MATH whereas both MATH and MATH. In words, MATH is at least as close to MATH as it is to MATH or MATH. Hence MATH and either MATH (e. g. when MATH) or MATH has the same sign as MATH. So any subarc of MATH with an end on MATH or MATH is either disjoint from MATH (if MATH) or it first intersects MATH on the same side as an arc with slope MATH. In particular, a subarc of MATH that intersects MATH and intersects MATH precisely in its endpoints, necessarily ends on opposite sides of MATH, and so cannot be MATH. |
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 tunnel is the dual arc to MATH in the rectangle MATH.) This implies, via CITE, that each component of MATH is unknotted. This then implies via CITE or CITE that REF graph obtained from MATH by crushing MATH to a point MATH can be isotoped to lie in a plane. This finally implies that MATH is a MATH-bridge knot, with MATH the bridge sphere. |
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, MATH is isomorphic to an affine space, and it meets every MATH-orbit in MATH along a unique MATH-orbit. As a consequence, MATH is a MATH-invariant open affine subset of MATH, and MATH meets the closed MATH-orbit (isomorphic to MATH) at the unique point MATH. It suffices to show that MATH is bijective, and restricts to an isomorphism over MATH. For the latter assertion (the main point of the proof), we shall show that every subscheme MATH intersects transversally certain NAME varieties in MATH. Mapping MATH to its common points with these NAME varieties, and to the NAME tangent spaces of MATH at these points, will yield appropriate morphisms from MATH to MATH, MATH and MATH, and hence an inverse to MATH. We need more notation, and results from CITE. Choose for MATH a NAME subgroup of MATH; let MATH be the NAME subgroup of MATH containing the maximal torus MATH. Let MATH be the NAME group of MATH and let MATH be the root system of MATH, with the subset MATH of positive roots defined by MATH and the corresponding basis MATH. Every MATH defines a simple reflection MATH. The MATH generate MATH; this defines the length function MATH on that group. Let MATH be the unique element of maximal length; then MATH. Let MATH be the subset of MATH consisting of all simple roots of MATH, and let MATH be the corresponding root subsystem of MATH. The NAME group of MATH is the NAME group MATH of the root system MATH; it is generated by the MATH, MATH. Let MATH then MATH consists of all elements of MATH of minimal length in their right MATH-coset; it is a system of representatives of the quotient MATH. The latter identifies to the double coset space MATH, via MATH. Recall that the MATH-fixed points in MATH are the MATH for MATH, and the MATH-orbits (the NAME cells) are the MATH (where MATH); the NAME varieties are their closures MATH . We shall also need the opposite NAME cells MATH and the opposite NAME varieties MATH (note that MATH is invariant under MATH). By CITE, we have MATH as sets. Thus, the same holds for the scheme MATH, since it is reduced. The isotropy group of MATH in MATH equals MATH (as sets). Moreover, MATH is bijective. The isotropy group MATH contains MATH; thus, it can be written as MATH for parabolic subgroups MATH containing MATH, and MATH containing MATH. Since MATH is connected, it stabilizes all NAME varieties MATH (MATH). As a consequence, MATH stabilizes all NAME cells. If MATH, then we can choose MATH such that MATH has a representative in MATH. Let MATH, then MATH, so that MATH by the NAME decomposition. Thus, MATH. Since this holds for all MATH in a system of representatives of MATH, it follows that the orbit MATH is contained in MATH. Therefore, MATH contains the intersection of MATH with the linear span of MATH. As a consequence, MATH contains a non-trivial closed normal subgroup of MATH. But this contradicts the assumption that MATH acts faithfully on MATH. Thus, MATH; likewise, MATH. In other words, the restriction of MATH to the closed MATH-orbit is bijective. Using REF, it follows that MATH is the normalization of its image; in particular, MATH is finite. Moreover, by CITE and CITE, the MATH-isotropy group of every point of MATH is connected, and the conjugacy class of that group determines the orbit uniquely an explicit description of these isotropy groups will be recalled in REF. Therefore, MATH is bijective. As a consequence, MATH is an open subset of MATH. We now investigate the structure of this subset, in a succession of lemmas. For all MATH and MATH, the scheme-theoretic intersection of MATH with MATH is supported at the unique point MATH, and this intersection is transversal. Note that MATH contains the MATH-fixed point MATH; therefore, this point belongs to every MATH. We first determine the intersection MATH. It is invariant under MATH, and contains MATH. Let MATH be another fixed point. Then, by the description of MATH, there exists MATH such that: MATH and MATH. Thus, MATH for the NAME ordering on MATH, so that MATH. Moreover, MATH, so that MATH. Likewise, we have MATH. Therefore, we must have MATH, and MATH is the unique MATH-fixed point of the set MATH. It follows that MATH is the unique point of that set. Since MATH and MATH intersect transversally at MATH, we see that MATH and MATH intersect transversally at MATH. To extend this to all MATH, consider the universal family MATH of MATH, and its pullback MATH to MATH. This is a closed subscheme of MATH; the scheme-theoretic intersection MATH is invariant under the natural action of MATH, and the second projection MATH is equivariant and proper. By the preceding discussion, the scheme-theoretic fiber of MATH at is the closed point MATH. Since each MATH-orbit closure in MATH contains MATH, it follows that all fibers of MATH are finite. Thus, MATH is finite, and MATH is affine. Now a version of NAME 's lemma implies that MATH is an isomorphism. By REF and the structure of MATH, every MATH intersects MATH transversally, at a unique point of MATH; let MATH be that point. Every MATH is a MATH-equivariant morphism, and the image of the product map MATH is closed and isomorphic to MATH. Consider again the universal family of MATH, and its restriction MATH to MATH. Then the scheme-theoretic intersection MATH projects isomorphically to MATH, for every fiber is a point. It follows that MATH is a morphism; it is clearly MATH-equivariant. Thus, the product morphism MATH is equivariant as well. To show that its image is isomorphic to MATH, it suffices to check that the map MATH is a closed immersion. For this, identify each NAME cell MATH with the homogeneous space MATH. Arguing as in the proof of REF , we see that the intersection of the subgroups MATH REF is trivial, and that the same holds for the intersection of their NAME algebras. Thus, the orbit map MATH is an immersion. Moreover, its image is closed, as an orbit of an unipotent group acting on an affine variety. By REF , every MATH contains MATH as a non-singular point, and the NAME tangent space MATH is transversal to MATH . Let MATH be the NAME of linear subspaces of MATH, and let MATH be the open affine subset consisting of those subspaces that are transversal to MATH. Every MATH is a morphism, and the image of the product map MATH is isomorphic to MATH. Consider once more the universal family of MATH, and its restriction MATH to MATH. Let MATH be the ideal sheaf of the closed point MATH of MATH and consider the scheme-theoretic intersection MATH with projection map MATH to MATH. Then the fiber of MATH at MATH equals MATH; its length is finite and independent of MATH, since MATH is a non-singular point of that scheme. Thus, MATH is finite and flat; it follows that the assignment MATH is a morphism. Note that MATH maps MATH to the diagonal of MATH. For the second assertion, fix MATH and set MATH . Then MATH is the direct sum of its subspaces MATH and MATH. Thus, every subspace of MATH, transversal to the subspace MATH, can be written as a graph MATH with uniquely defined linear maps MATH. This defines isomorphisms MATH mapping MATH to the identity. Moreover, for every MATH, the point MATH is mapped to the endomorphism MATH . Since each weight space of MATH is one-dimensional, it follows that the eigenvalues of MATH are non-zero regular functions on MATH, eigenvectors of MATH (acting on the left) with weights: the MATH-weights of MATH and the opposites of the MATH-weights of MATH. Let MATH be the set of weights of these functions; for MATH, let MATH be the corresponding function. Then MATH is an eigenvector of MATH (acting on the right) of weight MATH. The set of weights of MATH (respectively, MATH, MATH) equals MATH (respectively, MATH; MATH). Thus, we have MATH . Moreover, for every simple root MATH, there exists MATH such that MATH belongs to MATH, by the proof of REF ; and the functions MATH REF generate the coordinate ring of MATH, by REF. Thus, the composition of MATH with MATH is a closed immersion. REF imply that the restriction MATH is an isomorphism. Since MATH is bijective and MATH meets all MATH-orbits in MATH, it follows that MATH is an isomorphism. |
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-split subtorus MATH of MATH, a maximal torus MATH of MATH containing MATH, and a NAME subgroup MATH of MATH containing MATH. Then MATH is MATH-stable, so that MATH acts on the character group of MATH, and on the root system MATH; the opposite NAME subgroup MATH is contained in MATH, and contains MATH. Moreover, the natural map MATH extends to an open immersion MATH where MATH is isomorphic to affine space where MATH acts linearly with weights MATH, MATH. The isomorphism MATH restricts to a closed immersion MATH where MATH. Note that MATH is isomorphic to MATH, equivariantly for the action of MATH on MATH by MATH, and for the natural action of MATH on MATH. Moreover, MATH is contained in the image of MATH in MATH (for every MATH can be written as MATH for some MATH). Let MATH be the closure in MATH of the image of MATH, and let MATH. Then MATH induces an isomorphism MATH . It follows that the rational map MATH is defined on MATH and maps it isomorphically to MATH. Since MATH meets all MATH-orbits in MATH, this rational map is an isomorphism. |
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 union over all MATH as above. Since every irreducible component of MATH is invariant under MATH, we can rewrite this as MATH . Set MATH. Then, since MATH is invariant under the map MATH, we have MATH . This in turn amounts to: MATH is the unique element of maximal length in its right MATH-coset. On the other hand, we have MATH which is equivalent to: MATH is the element of maximal length in its left MATH-coset. So we have proved that MATH the union over all MATH of maximal length in their right MATH-coset and in their left MATH-coset. But every irreducible component MATH depends only on the double coset MATH (for the isotropy group of MATH contains MATH). Moreover, this double coset contains a unique element of maximal length (for MATH is the set of MATH-orbits in MATH, and the latter contains a unique open MATH-orbit). Thus, the preceding union is over the set MATH, or, equivalently, over MATH. This completes the first step of the proof. As a second step, we show that MATH (decomposition into irreducible components). For this, given MATH, note that MATH . Note also that MATH and MATH are exchanged by MATH. The assertion follows from these remarks, together with the first step. Now recall that MATH . Moreover, for every MATH in MATH, we have MATH for MATH and MATH. As a consequence, MATH is invariant under left multiplication by the group MATH. Likewise, since MATH, we have MATH and this subset is MATH-invariant. Thus, MATH . Together with the second step, this proves all assertions of the Proposition, except for birationality of MATH. But MATH contains MATH as an open subset, mapped under MATH onto MATH . We shall show that the restriction MATH is an isomorphism. For this, we describe the (set-theoretical) isotropy group of MATH in MATH, that is, MATH. Since MATH this isotropy group is contained in MATH and contains MATH. It follows that MATH . The isotropy NAME algebra of MATH in MATH is described similarly; it follows that MATH and that MATH is bijective and separable, hence an isomorphism. |
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 cycles in MATH, are algebraically equivalent. But algebraic equivalence in MATH coincides with rational equivalence, and the NAME group of MATH is freely generated by the classes of the NAME varieties (see REF ). Therefore, every cycle in MATH is algebraically equivalent to a unique cycle with MATH-stable support. Thus, MATH contains a unique fixed point of MATH, and hence a unique closed MATH-orbit. For the second assertion, choose an equivariant embedding of MATH into the projectivization MATH of a MATH-module. Then MATH is contained in some MATH. The latter is contained in the projectivization of a MATH-module, by the construction of the NAME variety. Applying REF to MATH, we see that MATH contains a unique closed MATH-orbit. The latter is isomorphic to MATH, by REF . Thus, REF is a consequence of Let MATH be a MATH-module. Let MATH satisfy the following conditions: CASE: The isotropy group MATH equals MATH. CASE: The orbit closure MATH contains a unique closed MATH-orbit, and the latter is isomorphic to MATH. Then the map MATH extends to an isomorphism MATH. By REF, the map MATH extends to an equivariant birational morphism MATH . We shall construct an inverse to that morphism. By REF , there exists a unique line MATH in MATH such that the corresponding point of MATH belongs to MATH and has isotropy group MATH. Thus, MATH consists of eigenvectors of MATH of weight MATH, where MATH and MATH are regular dominant weights of MATH. Let MATH be the MATH-submodule of MATH generated by MATH. By complete reductibility, we may choose a MATH-equivariant projection MATH. Then the corresponding rational map MATH is MATH-equivariant and defined at MATH, and hence defined everywhere on MATH. The image of MATH under MATH is a fixed point of MATH in MATH. In particular, the MATH-module MATH contains an eigenvector of MATH. Thus, this module is the space of endomorphisms of the simple MATH-module MATH with highest weight MATH; moreover, the image of MATH in MATH is the line spanned by the identity map. By CITE, the MATH-orbit closure of that line is isomorphic to MATH. Thus, MATH restricts to an equivariant birational morphism from MATH onto MATH. |
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 that MATH. Now MATH and MATH, thus MATH . Considering the MATH-string through MATH we obtain that MATH and MATH. CASE: MATH : Consider the maximal root MATH and the particular root MATH. Now MATH and MATH. CASE: MATH : As before, we have MATH. Considering the MATH-string through MATH and the MATH-sting through MATH we obtain MATH and MATH. CASE: MATH : MATH where MATH. CASE: MATH : MATH . CASE: MATH : MATH. CASE: MATH : MATH. |
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 MATH for MATH. Then MATH . This shows that the unique admissible graduation block for MATH is MATH. Suppose therefore that MATH. Then MATH implies MATH for a nonzero value MATH; moreover, MATH must be even, MATH. As MATH belongs to the commutator algebra, there exist two indexes MATH with MATH and a pair of vectors MATH, MATH such that MATH, where MATH is nonzero. Let MATH be the minimal pair with this property; it is not difficult to see that it is MATH. Then the associated differential form to the vector MATH is of the following type : MATH . On the other hand MATH . It is immediate to verify the nonexistence of nonzero coefficients MATH such that the previous forms satisfy simultaneously MATH . |
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 following relations hold MATH from which we deduce, by the structure of MATH that MATH. Observe in particular that the nullity of this cocycle implies the existence of a unique extension. Through an elementary change of basis it follows that this extension is isomorphic to MATH for MATH. An algebra MATH is determined by the adjunction of a differential form MATH. As the nilindex MATH is fixed, this implies that MATH . Then this form must be of the following type : MATH where MATH and MATH . The determinant cocycle is MATH : if it is nonzero we obtain MATH and otherwise MATH . Thus there are two nonequivalent extensions, the first being isomorphic to MATH and the second to MATH. |
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 the factor algebra MATH is naturally graded and filiform, isomorphic to MATH if MATH and isomorphic to MATH if MATH . |
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 common factor, so that MATH where this form is easily proven to be nonclosed. So we deduce the nonexistence of naturally graded with the required nilindex in MATH . |
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 remaining values of MATH it is easy to see that MATH does not admit naturally graded extensions with the prescribed characteritic sequence. |
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 . If not, then the differential form MATH has a nonzero coefficient associated to the summand MATH . In this case, the central element to be taken is MATH and it is not difficult to see that MATH is a naturally graded NAME algebra isomorphic to MATH . As the central graded extensions of this algebra which increment the nilindex in one unity are unique, this algebra must be isomorphic to MATH. Finally, for the fractionary depths we have seen the nonexistence of extensions of this type. MATH . It is a trivial verification that the models satisfy the requirements. |
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 where MATH. Clearly MATH must be nonzero, and by a change of basis we can suppose MATH. If MATH we obtain MATH, while for nonzero MATH we obtain MATH. |
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 extension of degree one of MATH is determined by the adjunction of a differential form, which we will call MATH . The graduation and the characteristic sequence imply that this differential form is of the type MATH where MATH, as we have MATH. The following relations hold MATH . This implies the existence of a unique extension having characteristic sequence MATH, and given by the equations MATH . We denote this algebra with MATH. |
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 cocycle ( class ) MATH. CASE: The cocycles which define the desired extensions are MATH satisfying MATH CASE: We have to consider the cocycles MATH subjected to the relations MATH . |
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 graduation imply that MATH, thus MATH. In consequence MATH MATH-the NAME conditions give two solutions : if MATH then MATH is isomorphic to MATH, and if MATH then MATH must be a central vector, from which MATH is isomorphic to MATH; in the first case MATH and in the second MATH. For MATH the characteristic sequence and the graduation imply that MATH, thus MATH by the previous reasoning. Finally, for the depth MATH the factor of MATH by the central ideal MATH is either isomorphic to MATH or MATH. |
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 MATH satisfy MATH . We thus obtain a unique extension class which is isomorphic to MATH. This shows the assertion for MATH. Let it be true for MATH. Then the NAME equations of MATH are MATH where MATH . Now we extend this algebra by MATH. Supposing that the extension satisfies the centralizer property and is naturally graded of the prescribed characteristic sequence, the determining cocycles are MATH . We have the relations MATH and by an elementary change of basis, the adjoined differential form MATH is of type MATH . Both the characteristic sequence and centralizer property are obviously satisfied. |
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 property is given by the form MATH. |
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 we must consider two cases: CASE: MATH. Then MATH and MATH. CASE: MATH. Then MATH, MATH and MATH. MATH . Then MATH is MATH -symplectic if and only if MATH satisfies the linear system MATH . This system contains all of equations corresponding to the system for MATH. Then all of elements of MATH to MATH are equal to the matrix for MATH, except MATH. Using the system above we see how to write MATH: MATH . |
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 that MATH is balanced, we get MATH . We note that MATH satisfies the PC with respect to MATH if and only if MATH. Since MATH is balanced, MATH. It follows that MATH . We want to evaluate MATH. In order to do that we have to compute MATH . CASE: MATH. In this case MATH . CASE: MATH. In this case, since MATH satisfies MATH for any element with odd NAME weight, it follows that MATH. Therefore, REF becomes MATH . When MATH is a linear structure of MATH, MATH, where MATH. CASE: MATH. Then MATH and REF becomes MATH . CASE: MATH. In this case, MATH can be evaluated as follows: MATH . CASE: MATH. Since MATH, we get MATH . CASE: MATH. Since MATH, we get MATH . From the above analysis we deduce that: CASE: MATH and if MATH is a linear structure for MATH, MATH or MATH. CASE: MATH. CASE: MATH, and if MATH is a linear structure for MATH, MATH . CASE: MATH. We observe that the only cases where we do not know precisely MATH are when MATH is an element of odd NAME weight with MATH not a linear structure for MATH. We deduce that in REF with MATH a linear structure for MATH, MATH . Now, returning to the computation of MATH, with the new results we get MATH . |
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 MATH. To show that MATH satisfies the SAC we use some results of NAME and NAME, that is REF or relation REF, which says that a function MATH satisfies the SAC if and only if MATH for each MATH, where MATH or equivalently (if MATH), MATH for each MATH, where MATH is equal to the number of REF's minus the number of REF's in MATH . If we associate REF-bit blocks MATH and MATH, we see that, for MATH, the relation REF holds. Obviously, if MATH is balanced, then MATH. Thus, in the sum REF the sum in each parenthesis is zero, except perhaps the ones based entirely on MATH (which are the only unbalanced REF-bit blocks in MATH). However, those terms will have an antidote in another parenthesis. For instance, since MATH, MATH will have the antidote MATH, according to the form of our functions. In order to compute the nonlinearity of MATH we have counted the bits at which our function differ from any linear or affine function. Intuitively, we need to prove that on average an affine function cannot cancel to many blocks in a segment. Precisely, we show that given any two segments MATH in the same half of MATH, based on the same block MATH, then MATH, for any affine function MATH based on the same block MATH. This is shown easily using the folklore lemma, and observing that on the positions of MATH, MATH can have only the following forms: MATH, MATH, MATH, etc. Since all cases are treated similarly, we may assume that MATH (recall the definition of MATH). Without loss of generality we may assume that MATH are in the first half of MATH and MATH, MATH. Thus MATH . Here we used MATH, the fact that MATH is balanced and MATH, if MATH. Next, we compute MATH. One may assume that MATH is based on MATH. From the part of MATH that does not contain MATH we get MATH units for the weight (we recall that only a quarter of all blocks contain MATH). We consider now the part of MATH based on MATH. Using the previous result, we deduce that in the worst case (minimum weight), MATH cancels completely at most four functions from each half, and from the rest of the part of MATH based on MATH, half of the blocks are cancelled. Since there are MATH functions based on MATH and we cancel MATH functions, we gather that there remain MATH functions uncancelled. Since each uncancelled function contributes MATH units to the weight (recall that if two affine functions MATH are not equal or complementary, their sum is balanced), we get MATH units contributed to the weight by the part based on MATH, so the nonlinearity is at least MATH. In the odd case we get MATH (the lengths of the affine functions MATH double, while the number of segments remains the same), by a similar argument. Now, since MATH and from the above analysis MATH we get MATH which will produce our right hand side inequality MATH . In order to evaluate MATH for suitably chosen MATH we apply the same technique as in the proof of REF . For MATH, let MATH . Using the form of our functions and taking MATH, we get MATH . Thus, MATH. Now, we take MATH. Thus, we get MATH . Now, taking MATH and MATH, we obtain MATH for any function MATH. In particular, for the functions in our class, we get MATH . Similarly, MATH. Thus, MATH and MATH . In any of the three cases MATH, we have MATH. Thus, MATH . |
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, if a function satisfies the PC with respect to all but four vectors, then MATH must be even, the nonzero vectors, where the propagation criterion is not satisfied, must be linear structures and MATH. We have the result. |
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 . But MATH therefore MATH is semi-bent. By REF , MATH satisfies the propagation criterion for all weights MATH. Similarly, MATH . Now, we shall use REF to compute the nonlinearity of MATH. First, we observe that MATH . Take MATH and MATH. We see that MATH . We denote the last expression by MATH. Using REF we compute the NAME transform MATH since by REF , MATH is bent. For simplicity we set MATH and MATH. Thus, MATH since MATH and MATH are bent. Therefore, MATH . By REF , MATH or MATH. Let MATH. By REF , MATH . Thus, by the same REF , MATH which implies MATH. Therefore MATH is not bent (any bent function in MATH variables has nonlinearity MATH [REF , p. REF]) and the theorem is proved. |
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 REF will give the following recurrence for the weights of MATH. MATH . A generating function for the above recurrence is MATH . We can linearize the recurrence by using the transformation MATH thus obtaining the recurrence MATH . Using the above simple recurrence with MATH and MATH we get a closed formula for the weights of MATH in dimension MATH, namely MATH and the theorem is proved. |
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. Then the extension map MATH is defined by MATH for MATH and MATH for MATH. CASE: The tree MATH separates the disk MATH into subdisks MATH. By REF each disk MATH admits an extension map MATH. Every MATH can be extended to MATH. The required extension map MATH is defined by MATH. To achieve MATH, replace MATH by MATH. |
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. Choose MATH such that MATH and MATH. We can apply CITE for any point MATH (with MATH) to find MATH, such that MATH, where MATH is one of the two arc components of MATH which connects MATH and MATH. This implies MATH, a contradiction. CASE: Suppose MATH. Take MATH, MATH, such that MATH. Choose MATH such that MATH and MATH. By CITE there exists MATH, MATH such that MATH. Since MATH is contained in the interior of the circle MATH, we have MATH, a contradiction. CASE: Next we will show that the sequence MATH is equicontinuous, that is, for every MATH there exists a MATH such that MATH for any MATH and MATH, MATH with MATH. Let MATH be given. We may assume that MATH for each MATH. Since the sequence MATH, MATH is uniformly locally connected, there exists a MATH, MATH, such that if MATH, MATH (respectively MATH) and MATH, then there exists an arc MATH in MATH (respectively MATH) connecting MATH and MATH and with MATH. Choose MATH, MATH, such that MATH and MATH. Suppose MATH, MATH and MATH. By CITE (with MATH) we have MATH, MATH, such that MATH, where MATH. Since MATH, MATH, it suffices to show that MATH. By the choice of MATH, MATH meet at most one of MATH and MATH. If MATH or MATH, then MATH and MATH is a disk bounded by MATH, so MATH. Otherwise, MATH is an arc connecting two points MATH, MATH with either REF MATH or REF MATH. In both cases MATH, hence by the choice of MATH, we have an arc MATH in MATH (respectively, MATH) connecting MATH and MATH and MATH. In REF MATH separates MATH into the subdisk MATH and another subdisk. Since MATH and MATH, the NAME curve MATH bounds the disk MATH, so MATH. In REF the NAME curve MATH bounds a disk MATH in MATH with MATH. Since MATH is contained in the exterior of MATH and MATH, it follows that MATH so MATH. CASE: Since the sup-metric MATH, the sequence MATH is also equicontinuous. By the NAME theorem, passing to a subsequence, we may assume that MATH converges to a map MATH. Set MATH. Then MATH, MATH and MATH. Since the sequence of univalent analytic maps MATH converges weakly uniformly to the map MATH (that is, for each compact subset MATH of MATH, MATH (MATH large) converges uniformly to MATH) and MATH is not constant, MATH is also a univalent analytic map CITE. It follows that MATH and MATH is a conformal map, so MATH by the uniqueness in REF . This completes the proof. |
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, if we take a sufficiently small neighborhood MATH of MATH, then MATH is close to MATH for MATH, and we can use a collar of MATH in MATH and a local contraction of a neighborhood of MATH in MATH to modify the map MATH to obtain the desired map 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: Suppose MATH is any open neighborhood of MATH in MATH and MATH is a small compact neighborhood of MATH in MATH such that MATH (hence MATH). If MATH is sufficiently close to MATH, then MATH. In particular, MATH is continuous in MATH. In fact, there exists a MATH such that MATH. If MATH is sufficiently close to MATH then the sup-metric MATH and MATH. Hence, MATH does not meet MATH, so MATH. CASE: To show that MATH is continuous in MATH, let MATH and MATH be given. It suffices to show that for each MATH there exists a small neighborhood MATH of MATH in MATH such that MATH and MATH are MATH-close on MATH for every MATH. Set MATH and MATH, and let MATH and MATH be small circular arc neighborhoods of MATH and MATH in MATH as in REF with respect to MATH and MATH respectively. Set MATH and choose small circular arc neighborhoods MATH and MATH of MATH and MATH in MATH such that MATH meets neither MATH nor MATH. Choose MATH such that MATH meets neither MATH nor MATH. By the compactness argument there exists MATH, MATH, such that for any MATH, MATH. By REF there exists a neighborhood MATH of MATH in MATH such that if MATH, then MATH, MATH, MATH, MATH and MATH, MATH. Since MATH is orientation preserving, MATH and MATH, it follows that MATH. If MATH, then MATH so that MATH. For each MATH we have the same conclusion. Suppose MATH. Since MATH is MATH-close to MATH and MATH, we have MATH. Similarly MATH, and so MATH. Since MATH is MATH-close to MATH and the latter is also MATH-close to MATH, we have MATH. Hence by the choice of MATH, MATH. This completes the proof. |
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, MATH we choose two disjoint subarcs MATH, MATH of MATH with MATH, MATH and a subarc MATH of MATH with MATH. For each MATH set MATH, which is an ad or a single point. We choose two disjoint families of closed disks MATH and MATH in MATH such that REF MATH and REF MATH and MATH that is, MATH is a proper arc of MATH. MATH . By REF for each MATH there exists a neighborhood MATH of MATH in MATH and an extension map MATH. In turn, by REF for each MATH there exists a neighborhood MATH of MATH in MATH and an extension map MATH. If MATH is a sufficiently small neighborhood of MATH in MATH, then for any MATH we have MATH for every MATH and we can define a map MATH by MATH . Since MATH and MATH, if MATH is small enough, then MATH is sufficiently close to MATH so that MATH. Hence we can define a map MATH by MATH . Then MATH and MATH is equal to the identity map on MATH for each MATH. Since MATH, we can define a map MATH by MATH and MATH. Since MATH and MATH, the map MATH, MATH satisfies the desired conditions. MATH . CASE: The case when MATH: We can use the double MATH. Since MATH is a subpolyhedron of MATH, MATH is also a subpolyhedron of MATH. CASE: By REF (where MATH) we have a neighborhood MATH of MATH in MATH and an extension map MATH. We can extend every MATH to a MATH by the identity on MATH. If MATH is a small neighborhood of MATH in MATH, then for every MATH we have MATH, so MATH is defined and MATH. Thus we have an extension map MATH, MATH. CASE: Since MATH is locally contractible, using a collar of MATH in MATH, we have a neighborhood MATH of MATH in MATH and a map MATH such that MATH and MATH. We can easily verify a REF-dimensional version of REF and find a neighborhood MATH of MATH in MATH and an extension map MATH. We may assume that MATH. Hence if MATH is a small neighborhood of MATH in MATH, then we have a map MATH, MATH. Then MATH and MATH. If MATH is small, then we have MATH and the required extension map MATH is defined by MATH. |
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-compact. |
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 MATH at MATH becomes smaller and smaller as MATH. Let MATH. Then MATH and MATH are MATH-fd-compact by CITE. By REF there exists an open neighborhood MATH of MATH in MATH and a map MATH such that MATH and MATH. Let MATH be the restriction map, MATH and let MATH. Then MATH, MATH, is a homeomorphism with the inverse MATH. Hence MATH is also MATH-fd-compact. This implies the conclusion. |
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 every MATH admits a neighborhood MATH and a homotopy MATH such that MATH and MATH. Take a compact PL REF-submanifold MATH of MATH with MATH. Let MATH be given by REF . Since MATH, by REF we have an absorbing homotopy MATH of MATH into MATH. There exists a MATH such that MATH. Define MATH by MATH. CASE: There exists a MATH with MATH. It induces a homeomorphism MATH, MATH. Hence we may assume that MATH. Pushing towards MATH using a collar of MATH pinched on MATH, it follows that MATH has the h.n. complement in MATH. Thus REF follows from REF , and REF follows from REF and MATH. |
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 let MATH. Since MATH deforms into MATH, we may assume that MATH. This case follows from CITE and CITE. CASE: Let MATH denote the union of the components of MATH which meet MATH. Then MATH strongly deformation retracts onto MATH, and the latter is contractible by REF . |
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 cyclic group MATH. By REF MATH, so MATH. Suppose MATH is compact. Since MATH or MATH and MATH, it follows that MATH or MATH and MATH is closed and nonorientable. If MATH so MATH, then MATH and MATH, a contradiction. Therefore, MATH and MATH, so MATH and MATH. We have MATH. |
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. Suppose MATH is noncompact. It follows that MATH contains no circle components other than MATH. In fact if MATH is a circle in MATH, then we can join MATH and MATH by a proper arc MATH in MATH and by REF we have a retraction MATH, a contradiction. We can write MATH, where MATH is a compact connected REF-submanifold of MATH, MATH, MATH and each component of MATH is noncompact. We will show that each MATH is an annulus. This easily implies the conclusion. Let MATH be the components of MATH. By the above remark MATH, so MATH meets a component of MATH. Let MATH be a submanifold of MATH obtained by removing an open color of each MATH from MATH. It follows that MATH, MATH is the union of circles MATH associated with MATH's, each MATH is contained in some component MATH of MATH, MATH, and each MATH is noncompact. Since MATH contains no handles or NAME bands (so no one point union of two circles), it follows that MATH and MATH. By REF MATH is a retract of MATH, so MATH. This implies that MATH is 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. Since MATH is monomorphic by REF , MATH in MATH so that MATH and 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 monomorphic by REF , the loop MATH in MATH, and the desired isotopy is obtained by REF . |
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 and let MATH be the remaining components. For each MATH we can write MATH where MATH's are the circle components of MATH, MATH's are the components of MATH which contain some arc components of MATH and MATH's are the remaining components of MATH. We choose disjoint collars MATH of MATH and MATH of MATH in MATH and set MATH, MATH and MATH . Note that MATH. Since MATH by REF , we can isotope MATH rel MATH to a MATH. By the construction MATH satisfies the following conditions: CASE: MATH is a REF-submanifold of MATH, every component of MATH is a circle and MATH. CASE: MATH and MATH is isotopic to MATH rel MATH. CASE: Suppose MATH is a component of MATH. If MATH bounds a disk MATH then MATH, and if MATH bounds a NAME band MATH then MATH. CASE: If MATH is an annulus component of MATH then MATH. To see REF first note that MATH is the union of compact MATH-manifolds MATH's, MATH's, MATH's, MATH's and MATH's, which have disjoint interiors. Suppose MATH is a compact connected MATH-manifold in MATH with MATH. Since MATH and each MATH meets MATH, it follows that MATH is the union of MATH's, MATH's, MATH's and MATH's contained in MATH. Since MATH is an annulus and MATH is not a disk or a NAME band, from REF it follows that REF if MATH is a disk, then MATH contains a disk which is some MATH or MATH with MATH, so MATH, REF if MATH is a NAME band, then MATH contains a disk or a NAME band which is some MATH or MATH with MATH, so MATH. As for REF , MATH is the union of MATH's MATH's, MATH's and MATH's contained in MATH, and MATH contains at least one MATH, which is a disk with MATH holes. If MATH then by the assumption MATH. If MATH then we can find a disk MATH in MATH such that MATH REF . Since MATH, MATH is a union of MATH's and MATH's and we can conclude that it coincides with some MATH, which meets MATH. These imply REF . It remains to show that MATH is isotopic to MATH rel MATH under REF When MATH, we can apply REF to the triple MATH. To verify REF , note that REF each component of MATH takes of the form MATH for some component of MATH of MATH, and, in particular, REF if MATH, then MATH is a disk and MATH, a contradiction. Therefore MATH is isotopic to MATH rel MATH. Extending this isotopy over MATH by MATH we have the required isotopy MATH rel MATH. CASE: In the case where MATH, let MATH and consider MATH, where MATH is the extension of MATH by MATH. Then REF MATH and MATH satisfies REF - REF , and REF an isotopy of MATH to MATH rel MATH restricts to an isotopy of MATH to MATH rel MATH. (Alternatively, we can modify the isotopy of MATH to MATH rel MATH to an isotopy rel MATH, where MATH is a neighborhood of MATH in MATH. We can replace MATH so that MATH.) This completes the proof of REF . CASE: The case where MATH is noncompact: Choose a compact connected MATH-submanifolds MATH and MATH of MATH such that MATH and MATH. Let MATH. Since MATH is a subpolyhedron of MATH with respect to some triangulation of MATH (compare CITE), by REF we have the principal bundle : MATH. Let MATH, MATH, be any lift (= extension) of the path MATH defined by MATH and MATH on MATH. We can apply REF to MATH, MATH. For REF , when MATH is a component of MATH, REF if MATH, then MATH is a component of MATH and REF if MATH, then MATH contains a component of MATH and MATH (it also follows that MATH is not connected since MATH, so if MATH is a disk or a NAME band, then MATH is a disjoint union of arcs). Therefore we have an isotopy MATH rel MATH such that MATH, MATH. We can extend MATH and MATH to MATH by MATH. The required isotopy MATH is defined by MATH. |
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 argument when we will show that MATH is an ANR REF in the next section. |
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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