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math/9911144 | Suppose MATH and MATH is weakly null and that MATH for all MATH where MATH . Choose MATH so that MATH . By passing to a subsequence we can suppose MATH converges to some MATH . Then MATH . Now MATH and so MATH . This implies that MATH and gives the lemma. |
math/9911144 | Suppose MATH and MATH is weakly null. It is enough to show that MATH under the assumption that both limits exist. MATH . The shrinking case is similar. |
math/9911144 | If MATH is a subspace of MATH then MATH can be identified with MATH and trivially MATH has shrinking unconditional type. Now MATH can be identified with MATH . Let MATH be the canonical quotient map. Suppose MATH and that MATH is weak-MATH-null in MATH. Suppose that MATH where MATH . We may pick (by the NAME theorem) MATH so that MATH and MATH . Passing to a subsequence we can suppose that MATH converges weak-MATH to MATH . Now MATH and MATH by the weak-MATH-continuity of MATH . Hence MATH which yields a contradiction, so that MATH is of shrinking unconditional type. |
math/9911144 | Suppose MATH and that MATH and that MATH is any weakly null sequence; assume that MATH and MATH for all MATH where MATH . By results of CITE we may suppose MATH is isometric to a subspace of a space MATH with a MATH(UFDD). Indeed suppose MATH are finite rank projections defining a MATH(UFDD). Let MATH . Let MATH be the isometric embedding. If MATH we have MATH unconditionally in the weak-MATH-topology. Since MATH does not contain MATH (or equivalently MATH) this series converges in norm so that MATH . Now by the NAME theorem we can find MATH so that MATH and MATH . By passing to a subsequence we can suppose that MATH converges weak-MATH to some MATH . Clearly MATH . Then MATH . This contradiction establishes the lemma. |
math/9911144 | (Compare REF.) Define for MATH to be the infimum of all MATH so that for every weakly null tree-map MATH with MATH there is a full tree MATH with MATH for every branch MATH . Note that MATH, MATH, MATH and that MATH . In particular we have MATH . We now argue exactly as in CITE that MATH is convex. For the convenience of the reader we repeat the argument. Let MATH where MATH . Suppose MATH and MATH . Let MATH be any weakly null tree-map of height MATH with MATH for all MATH . Then we can find a full subtree MATH so that for every branch MATH we have MATH and then a full subtree MATH so that for every branch MATH . Obviously for every branch MATH so that MATH . Next we note that if MATH and MATH weakly then MATH . Assume that MATH . By passing to a subsequence we can suppose MATH for every MATH . Then for each MATH there is a weakly null tree-map MATH of height MATH so that MATH for all MATH and MATH for every branch MATH . Now let MATH be the tree consisting of all sets MATH where MATH such that if MATH then MATH . We define a weakly null tree-map by MATH . Then for every branch MATH we have MATH so that MATH. This implies our claim. Let MATH be the NAME functional of the set MATH. Then MATH is a norm on MATH satisfying MATH . Suppose MATH and MATH is weakly null sequence with MATH . Then MATH and so MATH . Hence MATH . Now by REF we have that MATH is separable. We can then apply REF to deduce that if MATH and MATH is a weak-MATH-null sequence in MATH with MATH then MATH . Thus MATH has a Lipschitz-MATH norm and by the results of CITE (see also CITE) this implies that MATH embeds into MATH . |
math/9911144 | We show that MATH contains no subspace isomorphic to MATH and that MATH has the bounded tree property. To show MATH contains no copy of MATH it suffices to show that MATH does not embed in MATH by CITE. Suppose then MATH is a weak-MATH-null sequence in MATH so that MATH and MATH for any finite subset of MATH of MATH where MATH is an absolute constant. Then define, for any MATH the tree-map on MATH by MATH and MATH otherwise. It is clear that any full subtree MATH has a branch MATH with MATH while MATH, and therefore if MATH is large enough we obtain a contradiction to the weak-MATH summable tree property. Next we need a duality argument. We assume that MATH has the weak-MATH summable tree property with constant MATH . We show that MATH has the bounded tree property with constant MATH for any MATH . Indeed if not there is by REF a weakly null tree-map MATH with the properties that MATH for all MATH and MATH for every branch MATH . For each branch MATH pick MATH with MATH and MATH . Let MATH be the height function of the given tree-map. For each MATH we define MATH by transfinite induction on MATH . If MATH let MATH where MATH is any branch to which MATH belongs. Then if MATH has been defined for MATH and if MATH we define MATH to be any weak-MATH-cluster point of MATH . (Note that according to our definition MATH is a tree-assignment but not necessarily a tree-map because it is not supported on a well-founded tree and we may have MATH) Let us now make a tree-map by defining MATH and then if MATH we define MATH . If MATH we define MATH . This is clearly a tree-map which also satisfies REF and we have that for each MATH zero is a weak-MATH-cluster point of MATH . It is then easy to see that we can pass to a full subtree MATH so that MATH is weak-MATH-null. Let MATH . Now pick MATH so that MATH . We can use the definition of the weak-MATH-summable tree property and also REF to pass to a further full subtree (still labeled MATH) so that we have MATH when MATH and for any branch MATH . For any branch MATH let MATH be the first point where MATH. Then MATH . It follows that MATH . Now we have MATH . This gives a contradiction and so we deduce that MATH has the bounded tree property and we can apply REF to obtain the result. |
math/9911144 | We verify REF with MATH . Let MATH be a normalized sequence with MATH . Form a tree-assignment MATH by putting MATH and then MATH if MATH . Then MATH is a bounded MATH-separated tree-assignment and so there is a branch MATH and MATH with MATH . This leads to a subsequence MATH where MATH and REF holds with either MATH or MATH . |
math/9911144 | We may suppose that MATH is a MATH-L-ideal where MATH . Suppose MATH is the associated MATH-projection. We also use REF to deduce there is a constant MATH so that if MATH is any bounded sequence in MATH with MATH then MATH has a subsequence MATH such that we have an estimate MATH for all finitely nonzero sequences MATH . Now suppose MATH is a MATH-separated tree-assignment. Let MATH. We shall show by an inductive construction that there is a branch MATH and for each MATH with MATH so that MATH if MATH . This will complete the proof since then we can take MATH as any weak-MATH cluster point of MATH . We start the branch with MATH . Now suppose MATH; we must choose a successor MATH and a corresponding MATH . First let MATH be any norm-preserving extension of MATH to MATH . Next we pick a subsequence MATH of MATH satisfying REF . Let MATH be any weak-MATH-cluster point of MATH . Suppose MATH . Let MATH belongs to the weak-MATH-closed convex hull MATH of MATH and hence MATH is in the norm-closure of the set MATH . We deduce that for any MATH we can find convex combinations MATH and MATH of norm at most MATH . Hence MATH . Thus MATH . In particular, MATH . Let MATH be the linear span of MATH . We define a linear functional MATH on MATH by MATH if MATH and MATH if MATH and MATH if MATH . For any MATH we have MATH where MATH and MATH. Then MATH . Hence MATH . It follows that MATH has a weak-MATH-continuous extension MATH with MATH . Now MATH and hence we can pick MATH so that MATH . We thus select MATH and set MATH . This inductive process establishes our result. |
math/9911144 | Note first that MATH cannot contain MATH by results of CITE since MATH has the NAME property. Therefore we can apply REF to deduce that MATH can be given an equivalent norm so that it has shrinking unconditional type. We complete the proof by showing that MATH has the weak-MATH-summable tree property and applying REF . Assume that MATH has the very strong NAME property with constant MATH . We will show that MATH has the weak-MATH-summable tree property with constant MATH . Suppose MATH is a weak-MATH-null tree such that MATH . Assume that MATH fails to have a full subtree such that MATH . Then by considering the tree-map MATH in MATH and using REF we can find a full subtree MATH so that for every branch we have: MATH . Next we can pass to a full subtree MATH so that if for each MATH either MATH or MATH . We then define a tree assignment MATH as follows. Put MATH . If MATH is such that MATH then let MATH be assigned to be any fixed weak-MATH-null normalized sequence. If MATH and MATH we let MATH if MATH . Then MATH is weak-MATH-null and, using the weak-MATH-lower-semicontinuity of the norm we may pass to a full subtree MATH so that for any MATH we have MATH . Next we use the fact that MATH has shrinking unconditional type. For each MATH there is a closed absolutely convex weak-MATH-neighborhood of the origin MATH so that if MATH and MATH then MATH for every choice of signs MATH for MATH . Let MATH . Then MATH is a full subtree of MATH . Let MATH be any branch in MATH . We write MATH where MATH . Let MATH . Then MATH . Notice that if MATH where MATH then since MATH and MATH . Thus MATH . It follows that MATH . Thus we conclude that for any branch and any choice of signs MATH we have MATH . Next we can use the very strong NAME property and the fact that MATH is MATH-separated to find a branch MATH and MATH with MATH for MATH . By the construction of MATH we have MATH so that MATH . Choose MATH so that MATH . Then MATH . Hence MATH . This is a contradiction and shows that MATH has the weak-MATH summable tree property. The proof is complete. |
math/9911144 | We suppose that MATH has REF . Let MATH and denote by MATH the quotient map of MATH onto MATH. We start by supposing that MATH is a bounded MATH-separated tree assignment in MATH . Let MATH be an increasing sequence of finite-dimensional subspaces of MATH whose union is dense. We start by observing that for each MATH and each MATH there is an infinite number of MATH so that MATH . Indeed, if not there are infinitely many MATH so that MATH and for each such MATH we can find MATH with MATH . The set of such MATH is bounded and so by compactness arguments we obtain MATH with MATH . Now we may pass to a full subtree MATH so that there exists a map MATH with the properties that MATH and if MATH where MATH then we have MATH . Now for each MATH we can choose MATH so that MATH . Note that the set MATH forms a weak-MATH-null sequence. For convenience let MATH . Consider the closed unit interval MATH and let MATH be the set of dyadic rationals MATH where MATH and MATH . Let MATH be the space of all real-valued functions MATH on MATH which are continuous on MATH and such that on MATH both left- and right-limits MATH and MATH exist MATH . It is easy to see that MATH equipped with the sup-norm is isometric to MATH where MATH is the NAME set. Then MATH is a closed subspace of MATH and MATH with the quotient map being given by MATH . Now we can define a one-one map MATH with the property that MATH and MATH for MATH . Next define an operator MATH by putting MATH if MATH and MATH otherwise; then MATH . Then MATH can be lifted to an operator MATH so that MATH and MATH . Then MATH maps MATH into MATH and by assumption this restriction MATH has an extension MATH with MATH . Now MATH factors to an operator MATH where MATH satisfies MATH and MATH . We can then write MATH in the form MATH where MATH is weak-MATH-continuous except on points of MATH and has left- and right- weak-MATH-limits MATH and MATH on MATH with MATH . Note that MATH . Finally we build a branch MATH so that for each MATH there exists MATH or MATH so that MATH for MATH . This is done by induction. Let MATH and MATH be any element of MATH with MATH . Then since MATH we can choose an appropriate sign so that the inductive hypothesis holds when MATH . Now suppose MATH have been chosen and that MATH for MATH . Let us assume that MATH; the other case is similar. Then there exists MATH so that if MATH we have for some MATH for MATH . Then we can choose MATH so that MATH . Then MATH for MATH . Now MATH so that we can choose MATH to satisfy the inductive hypothesis. This completes the inductive construction of the branch MATH . Finally we let MATH be any weak-MATH-cluster point of the sequence MATH so that MATH and MATH for all MATH . This shows that MATH has the very strong NAME property with constant MATH . |
math/9911144 | First note that REF implies REF . In fact by REF MATH is a NAME space and hence any (UFDD) is boundedly complete so that MATH is a dual of a space with (UFDD). If we assume REF then REF together yield the result. |
math/9911146 | By REF we have to get an upper bound for MATH on each sphere MATH constant. Using REF with MATH, and going to a unitary gauge we get MATH and MATH. The connection MATH splits into a radially independent MATH connection and an exponentially decaying connection. More precisely, put MATH where MATH is the NAME algebra of MATH and MATH, the NAME algebra of MATH. Alternatively, we can characterise the sub-spaces by MATH . Decompose MATH accordingly as MATH. The connection MATH decomposes as MATH for MATH with the property that MATH is a MATH connection independent of MATH, and MATH is a REF-form that decays exponentially as MATH. Then MATH and MATH is independent of MATH whilst the rest decays exponentially, so MATH. Define MATH to be the smallest eigenvalue of the action of MATH on MATH. Then each time we say that a term decays exponentially, it decays at least as fast as MATH. Notice that MATH. We have MATH for some constant MATH and thus MATH where MATH is the one-dimensional NAME 's function. Now MATH and the second term of the right hand side converges to MATH as MATH. Since MATH, the first term is dominated by MATH for a constant MATH. Since MATH the proposition follows. The constant MATH in the statement of the proposition does depend on the holomorphic map MATH, and is bounded below by a constant independent of MATH. |
math/9911146 | In the proof of the previous proposition we saw that MATH has a radially independent part MATH and an exponentially decaying part. Since MATH vanishes at MATH we get an identity relating MATH with the term that cancels it. In the notation of the previous proof, MATH so MATH where the last equality follows from the fact that MATH is orthogonal to MATH. But at MATH, MATH so MATH . |
math/9911153 | Let MATH be a MATH function with the formal NAME series MATH, supplied by NAME 's theorem. Denote MATH . Let the (nonzero by assumption) series MATH starts with a term MATH, MATH, MATH. Then we have MATH as MATH. On the other hand, it is clear that MATH are flat. We will be looking for MATH of the form MATH where MATH is an unknown continuous MATH-valued function-germ such that MATH as MATH for any MATH. By NAME 's formula, the equation MATH can be written as MATH . For small MATH, this is equivalent to the equation MATH for the function MATH given by MATH . Note that if MATH, and MATH is flat at the origin, while MATH is not flat, then MATH is MATH near the origin and is flat. It follows that MATH. On the complex circle MATH, MATH, the term MATH will dominate the other terms in MATH if MATH is sufficiently small. By Rouche's theorem it follows that the equation MATH has for small fixed MATH exactly MATH roots in the disc MATH, which we denote MATH, MATH. We can arrange so that MATH are continuous in MATH, and the previous argument shows that MATH for any MATH. We now prove REF . Since the functions MATH enter the product in REF in a symmetric way, it is sufficient to prove that the elementary symmetric polynomials MATH in MATH are in MATH. By the NAME relations (see CITE, A. IV. REF), it is sufficient to prove the same for the functions MATH . However, by NAME 's formula we have that for small MATH from where it is clear that MATH. To prove REF , we notice that under the additional assumption made we can take MATH to be real. Then MATH, and therefore non-real roots MATH will appear in conjugate pairs. Then all MATH will be real, which implies REF . |
math/9911153 | By REF , the NAME series MATH of MATH has a factorization REF . Consider the function MATH. Its NAME series has the form MATH, and so factorizes as MATH . Let MATH be one of the series MATH, and assume that among all the MATH there are exactly MATH series coinciding with MATH. Then MATH is a root of multiplicity MATH of the polynomial MATH, and by REF we conclude that there exist MATH functions MATH, MATH, such that REF - REF from the formulation of the lemma are true. In view of REF , we can divide MATH by MATH, and the result is again a polynomial MATH from MATH. The NAME polynomial of MATH will be MATH divided by MATH. Now we can apply REF to MATH choosing a different MATH etc. By repeating this operation several times, we get a complete factorization of MATH. The required factorization of MATH is then obtained by the inverse substitution MATH. REF is ensured by splitting off all real series MATH before non-real ones in the above argument. |
math/9911153 | Notice that the NAME polygon is invariant with respect to multiplication by a nonzero MATH function (see CITE, p. REF). Therefore, for the functions MATH such that MATH is not flat (which is equivalent to having MATH) the proposition follows immediately from REF . Assume now that MATH. In this case we must somehow separate the roots infinitely tangent to the MATH-axis. This can be done as follows. Since MATH is not flat at the origin, there exists a rotated orthogonal system of coordinates MATH such that the restriction of MATH to the MATH-axis is not flat. So we can apply REF to MATH written in coordinates MATH. Let MATH be the arizing polynomial. If MATH is the equation of the old MATH-axis in the new coordinates, then MATH will be a root of multiplicity MATH of MATH. So we can apply REF and obtain MATH roots MATH, MATH of MATH, such that MATH. Moreover, by REF we will have that MATH is in MATH. So we can divide MATH by MATH, and the quotient will be a MATH function, which is no longer flat on the old MATH-axis. Let MATH be this last quotient written in the old system of coordinates. Then MATH is just MATH shifted MATH units to the left. So we can factorize MATH as in the case MATH described above. It remains to get a factorization of MATH in the old coordinates. It is clear that the NAME series of MATH written in the coordinates MATH consists of one term MATH. Interchanging the roles of MATH and MATH brings us back to the case MATH, and the required factorization of the form MATH can be obtained as described above. |
math/9911153 | Put MATH. Consider the convex set MATH given as the intersection of the half-planes lying above the lines MATH, MATH. Let MATH, MATH be all the corner points of the boundary of MATH with the MATH-coordinates MATH. It is clear that MATH. We claim that for any MATH . It is not difficult to see that this implies REF - REF . Let MATH, and assume that the boundary points MATH and MATH belong to the line MATH. Since MATH and MATH lie above all the other lines MATH, we have for all MATH . From these two estimates it follows that MATH . All in all, MATH . This estimate clearly implies MATH, provided that MATH is large enough. |
math/9911156 | A MATH-biorthogonal system in MATH is enough to force MATH to fail the NAME property. |
math/9911156 | Find an (isometric) embedding MATH of MATH into a MATH-space. Assume that there is a MATH-biorthogonal system MATH in MATH with MATH bounded, which would be the case if REF held. If MATH had a weakly NAME subsequence MATH, then MATH would be weakly NAME and NAME would be weakly null, which cannot be since a MATH space has the NAME property (compare CITE). So, by NAME 's MATH theorem, MATH admits a subsequence that is equivalent to the unit vector basis of MATH. |
math/9911156 | Fix a sequence MATH of positive numbers satisfying MATH . Without loss of generality (pass to a subsequence), MATH is a basic sequence with biorthogonal functional MATH satisfying MATH. These MATH's will be used to perturb functionals as needed. Without loss of generality (pass to a subsequence), there is a system MATH in MATH satisfying MATH and MATH . To see how to find such a system by induction, consider a subsequence MATH in MATH given at the beginning of the MATH step (for the base step, let MATH). Let MATH and find MATH in MATH satisfying MATH. Find a subsequence MATH of MATH satisfying MATH for each MATH and let MATH . Without loss of generality (pass to a subsequence), MATH . To accomplish this, iterate REF to produce a sequence MATH of sequences and a sequence MATH so that MATH is a subsequence of MATH and MATH . Then the subsequence MATH works. Clearly, the functionals MATH are biorthogonal to MATH and are of norm at most MATH. |
math/9911156 | Without loss of generality (pass to a subsequence), there is a biorthogonal system MATH in MATH with MATH . For just let MATH be a separable subspace of MATH that REF-norms MATH and take a MATH-basic subsequence of MATH (CITE, compare CITE). For each MATH let MATH . Use the Principle of Local Reflexivity to find a sequence MATH in MATH satisfying MATH and MATH for some MATH. Fix a sequence MATH of positive numbers satisfying MATH . Without loss of generality (pass to a subsequence), MATH . Clearly the vectors MATH are biorthogonal to MATH and are of norm at most MATH. |
math/9911156 | Let MATH be a biorthogonal system such that MATH is a semi-normalized weak-MATH-null sequence in MATH. First, assume that the above operator MATH factors through an injective space and let MATH be an isomorphic embedding. Consider the diagram MATH where MATH is an injective space and MATH. Since MATH is injective, there exists MATH such that the following diagram (totally) commutes. MATH . Note that the operator MATH is given by MATH similarly, the operator MATH has the form MATH . It is easy to check that MATH is indeed a MATH-stable biorthogonal system: the commutativity of the diagram gives that MATH, the weak-nullness of MATH follows from the fact that MATH is a NAME space (MATH is weak-MATH-null and thus weakly-null), and MATH. Next assume that MATH is a MATH-stable biorthogonal system. Find an embedding MATH from MATH into the injective space MATH for some index set MATH. By the stability of the system, there exists a weakly-null sequence MATH in MATH such that MATH. Define MATH by MATH, for then MATH. |
math/9911156 | Consider the following commutative diagram MATH where MATH is an isomorphic embedding, MATH is the formal injection, and MATH for the NAME functions MATH. Since MATH is MATH-injective, there exists an operator MATH such that the following diagram commutes MATH and MATH. The operator MATH takes the form MATH . It is easy to check that MATH is a MATH-stable biorthogonal system. NAME follows from the commutativity of the diagram. Since MATH is weak-MATH-null, so is MATH. MATH is an injective space through which MATH factors. Furthermore, since MATH and MATH both have norm one and MATH, MATH and so MATH is semi-normalized. If MATH embeds into MATH, then it MATH-embeds into MATH, thus one can arrange that MATH . |
math/9911156 | Let MATH be the annihilator of any finite dimensional subspace of MATH that MATH-norms the annihilator of MATH. For then if MATH then MATH . |
math/9911156 | Without loss of generality, MATH and MATH are each infinite dimensional. Fix a sequence MATH of positive numbers decreasing to zero. Since MATH fails the NAME property, there is a weakly-null sequence MATH in MATH. It suffices to find a system MATH in MATH along with a blocking MATH of MATH, a sequence MATH from MATH, and an increasing sequence MATH from MATH, satisfying CASE: MATH CASE: MATH CASE: MATH CASE: for each MATH, if MATH then MATH CASE: MATH CASE: MATH . The construction will inductively produce blocks MATH. Let MATH and MATH be the zero vectors and MATH. Fix MATH. Assume that MATH along with MATH and MATH and MATH have been constructed to satisfy REF through REF . Now to construct MATH along with MATH and MATH and MATH. Let MATH and MATH . The idea is to find a biorthogonal system MATH in MATH by first finding MATH which helps guarantee REF if MATH is odd and REF if MATH even; however, MATH would not necessarily satisfy REF through REF and so MATH and MATH along with MATH and MATH are constructed and then the NAME matrix is applied to MATH to produce MATH so that MATH with MATH and MATH satisfy REF through REF . MATH is constructed by a standard NAME biorthogonal procedure. If MATH is odd, start in MATH. Let MATH . Set MATH . If MATH is even, start in MATH. Let MATH . Set MATH . Clearly MATH and MATH . Find a natural number MATH larger than one so that MATH and let MATH . Let MATH . The next step is to find a biorthogonal system MATH along with MATH and MATH satisfying MATH and MATH for each MATH. Towards this, fix MATH and assume that a biorthogonal system MATH along with MATH and MATH have been constructed so that REF hold for MATH. Let MATH . By REF , there is a weak-MATH-closed finite co-dimensional subspace MATH of MATH such that MATH is MATH-normed by MATH. Find MATH and MATH such that MATH . Find MATH such that MATH and normalize MATH . This completes the inductive construction of MATH along with the sets MATH and MATH. Now apply the NAME matrix to MATH to produce MATH. With help from the observations in REF , note that MATH is biorthogonal and is in MATH. Furthermore, for each MATH in MATH, MATH and MATH and for each MATH . Thus MATH with MATH and MATH satisfy REF through REF . If MATH is odd, then MATH while if MATH is even, then MATH . Clearly the constructed system MATH, with the blocking MATH of MATH and the increasing sequence MATH from MATH, and the sequence MATH from MATH, satisfy REF through REF . |
math/9911156 | By CITE there is a MATH-isomorphic copy of MATH in MATH and so there is an embedding MATH satisfying, for each MATH, MATH . Moreover, the image MATH of the unit vector basis of MATH can be assumed to be weak-MATH-null (since MATH is separable, MATH has a weak-MATH-convergent subsequence MATH, so just replace MATH by MATH ). Find a sequence MATH in MATH such that MATH norms MATH for each MATH. Fix MATH and let MATH . Note that for each MATH, MATH . The operator MATH defined by MATH illustrates that MATH is isomorphic to MATH. Indeed, fix MATH . Then MATH . On the other hand, MATH . Thus MATH is MATH-isomorphic to MATH, which is MATH-isomorphic to MATH. To see that MATH is MATH-norming for MATH, fix MATH. Let MATH be such that MATH and find MATH such that MATH for then by REF MATH . Thus MATH . So, for MATH sufficiently close to one, MATH is MATH-norming for MATH. |
math/9911156 | The proof of REF is similar to the proof of REF ; thus, notation from the proof of REF will be retained. Fix a strictly decreasing sequence MATH converging to zero with MATH. It suffices to construct a system MATH in MATH along with a blocking MATH of MATH and a sequence MATH from MATH satisfying CASE: MATH CASE: MATH CASE: MATH CASE: MATH if MATH CASE: MATH is orthogonal to MATH for MATH CASE: MATH CASE: MATH . Note that REF through REF imply that MATH is weakly-null in MATH. Clearly all that remains at this point is to show that the MATH-biorthogonal system MATH is indeed stable, which is done in the last step by using the condition that MATH stays inside of MATH. Let MATH . The construction will inductively produce blocks MATH . Fix MATH. Assume that MATH along with MATH and MATH have been constructed to satisfy REF through REF . Now to construct MATH along with MATH and MATH. The idea is to find a biorthogonal system MATH in MATH by first finding MATH that helps guarantee REF if MATH is odd and REF if MATH even; however, MATH would not necessarily satisfy REF and MATH may be far from MATH and so MATH and MATH are then constructed and the NAME matrix is applied to MATH to produce MATH and MATH so that MATH satisfy REF through REF . Find MATH just as in the proof of REF : in the case that MATH is odd, be sure to choose MATH in MATH, which is possible since MATH is total. Find a natural number MATH larger than one so that MATH . Let MATH . Use REF to find a subspace MATH of MATH, a projection MATH with kernel MATH, and a norm one isomorphism MATH so that MATH and MATH . Let MATH be an orthonormal basis for MATH. For each MATH, let MATH and, using Local Reflexivity, find MATH that agrees, on MATH and MATH, with a norm-preserving NAME extension of MATH to MATH and satisfies MATH . Then MATH is a biorthogonal system in MATH and MATH . Now apply the NAME matrix to MATH to produce MATH and let MATH for each MATH in MATH. With help from the observations in REF , note that for each MATH in MATH and MATH and since MATH, MATH . Thus, MATH clearly satisfy REF through REF . This completes the inductive construction of the system MATH in MATH, along with the blocking MATH of MATH and the sequence MATH from MATH, that satisfy REF through REF . The last step is to verify that MATH is indeed stable, which, by REF , is equivalent to verifying that the operator MATH given by MATH factors through an injective space. Towards this, consider the following commutative diagram MATH where MATH is an isomorphic embedding with range MATH and MATH and MATH are given by MATH . Since MATH and MATH are of the form MATH their ranges are contained in MATH; let MATH and MATH be the corresponding maps with their ranges restricted to MATH. Thus the following diagram commutes. MATH . An appeal to REF finishes the proof. |
math/9911156 | Fix a normalized weakly-null sequence MATH in MATH, a dense sequence MATH in MATH, a sequence MATH decreasing to zero, and a sequence MATH such that MATH and MATH. It is sufficient to construct CASE: a sequence MATH in MATH CASE: a sequence MATH in MATH CASE: finite sets MATH in MATH with MATH CASE: an increasing sequence MATH of integers that satisfy CASE: MATH CASE: MATH CASE: MATH CASE: MATH CASE: if MATH, then there is MATH with MATH. For then just take MATH. Note that REF imply REF while REF and biorthogonality imply REF since each MATH has the form MATH . The construction is by induction on MATH. To start, let MATH. Find MATH in MATH that satisfies REF and MATH that satisfies REF . Fix MATH and assume that the items in REF through REF have been constructed up through the MATH-level. From this it is possible to find MATH and MATH. By REF , there is a finite co-dimensional subspace MATH of MATH that is MATH-normed by MATH. Find MATH along with MATH such that REF holds. Since MATH is MATH-normed by MATH, there is MATH such that MATH . Now find MATH satisfying REF . |
math/9911156 | First find the biorthogonal system MATH given by REF . The next step is to perturb this system to produce the desired system. Begin by finding a bijection MATH satisfying CASE: MATH is an increasing sequence for each MATH in MATH. Take a dense set MATH in MATH. The underlying idea is to use MATH to capture MATH, along with MATH, in the span of a small perturbation of MATH. Towards this, with the help of REF and the fact that MATH is not equivalent to the unit vector basis of MATH, for each MATH find a sequence MATH such that CASE: MATH CASE: MATH. Let MATH . Clearly MATH is a MATH-bounded MATH-biorthogonal system. Fix MATH and consider MATH. For each MATH, MATH . Combined with REF , this gives that MATH, which in turn implies that MATH. Thus MATH . Combined with REF , it follows that MATH is fundamental. |
math/9911156 | Find MATH in MATH which is MATH - equivalent to the standard unit vector basis of MATH and a surjective contractive projection MATH . Without loss of generality, MATH is weak-MATH-null (similar to before, just replace MATH with a weak-MATH-convergent subsequence MATH, which will be MATH - equivalent to MATH and contractively complemented in MATH). Find a sequence MATH in MATH such that MATH norms MATH for each MATH. Fix MATH and let MATH . Note that for each MATH, MATH . Each element in MATH has a unique expression as MATH where MATH; the operator MATH defined by MATH illustrates that MATH is isomorphic to MATH. Indeed, fix MATH . Then MATH . Thus MATH for each MATH; thus, MATH is MATH-isomorphic to MATH. To see that MATH is MATH-norming for MATH, fix MATH. Let MATH be such that MATH and find MATH such that MATH for then by REF MATH . Thus MATH . So MATH is MATH-norming for MATH. |
math/9911156 | Let MATH and MATH satisfy REF. Set MATH . Without loss of generality each MATH. Find MATH so that MATH and let MATH . The condition MATH guarantees that MATH is non-empty. Since MATH REF holds. For each MATH, MATH and so MATH by uniform convexity. |
math/9911156 | The underlying idea behind the proof is to use REF to find a small perturbation MATH of MATH that are disjointly supported on the standard unit vector basis of MATH. For then, MATH is equivalent to the standard unit vector basis of MATH and so, for a small enough perturbation, MATH is also equivalent to the standard unit vector basis of MATH. Find MATH in MATH so that MATH . Thus MATH and so by the first inequality in REF MATH . Write MATH where MATH and let MATH; thus, MATH . So by REF there is a sequence MATH of finite subsets in MATH so that MATH and for each MATH and MATH. Let MATH . To see that the MATH's are disjoint, suppose that there is MATH for some distinct MATH. Pick MATH so that MATH. Then MATH . On the other hand, from REF and the third inequality of REF it follows that MATH . A contradiction, thus the finite subsets MATH of MATH are indeed disjoint. For each MATH in MATH, MATH and so REF holds. |
math/9911156 | The underlying idea behind the proof is that for sufficiently small MATH . REF gives that MATH is equivalent to the standard unit vector basis of MATH: indeed, REF will hold and REF implies REF. Then MATH is equivalent to the standard unit vector basis of MATH. But if MATH is small enough, then condition (REF d) cannot hold since MATH has finite cotype. Let the hypotheses of REF hold. Since MATH there are constants MATH and MATH so that MATH . Find MATH sufficiently small enough so that MATH and so that there exists MATH satisfying MATH where MATH. To see that REF is easily accomplished, note that if MATH and MATH then REF holds. Let conditions (REF a) through (REF d) hold. Assume that MATH. By (REF c), without loss of generality, for each MATH, MATH . Keeping with the notation from REF , let MATH . From (REF a), (REF d), and REF it follows that there is MATH such that MATH . Thus REF from REF holds. Furthermore REF also holds since MATH . So by REF , since MATH is fundamental, MATH is equivalent to the standard unit vector basis of MATH with MATH for each MATH in MATH. For each finite subset MATH of MATH . The right-hand side of REF mimics MATH-growth in that for each choice MATH of signs MATH . The left-hand side of REF mimics MATH-growth since MATH for each MATH. Thus MATH . This contradicts REF. Thus MATH. |
math/9911156 | Let MATH. Thus MATH. It is straight forward to verify that MATH satisfies REF . Note that MATH. Let MATH be the predual of MATH. There is an operator MATH such that MATH is the formal pointwise embedding; for indeed, since MATH is reflexive, this formal pointwise embedding is weak-MATH-to-weak-MATH continuous. Similarly, by reflexivity, there is MATH such that MATH is a MATH - isomorphism. Let MATH. Thus MATH and MATH is a MATH - isomorphic embedding. Thus REF follows from REF . |
math/9911157 | By REF , the complex REF is chain homotopy equivalent to a free chain complex MATH over the ring MATH, such that the rank of each module MATH equals MATH for any MATH. Hence the NAME numbers MATH and MATH can be computed starting from the complex MATH. REF is now standard for complexes over principal ideal domains. This completes the proof. |
math/9911157 | Suppose first that MATH is not an algebraic integer. Let MATH be the monodromy representation of the flat bundle MATH, where MATH is the fiber over the base point. Note that holds MATH that is, MATH defines a right action of MATH on MATH. Let MATH be the representation defined in REF. We obtain a ring homomorphism MATH and MATH becomes a MATH-bimodule via MATH, satisfying the conditions of REF (arguments proving that are the same as in REF). Hence by REF for any closed REF-form MATH on MATH lying in class MATH and having only NAME zeros, there exists a chain complex MATH of complex vector spaces having MATH generators in any dimension MATH and computing the homology MATH. REF follow from existence of MATH via the standard well-known argument, compare CITE. Suppose now that MATH is not an algebraic integer. Consider the dual vector bundle MATH. The above arguments applied to the form MATH prove the inequalities MATH where MATH and MATH is the orientation bundle of MATH (that is, a flat line bundle such the monodromy along any loop equals MATH depending on whether the orientation of MATH is preserved or reversed along the loop). Note, that both flat bundles MATH and MATH have integral lattices. REF follows now using MATH and the NAME duality MATH. REF is obtained similarly. By the above arguments applied to the form MATH and the flat vector bundle MATH we obtain the inequalities MATH for MATH (a variant of the usual inequality with increasing indices) and then one uses the NAME duality and MATH to obtain REF . |
math/9911157 | The proof is similar to proof of REF . Let MATH be the monodromy representation of MATH, where MATH denotes the fiber of MATH over the base point. For a flat line bundle MATH, the representation MATH defines on MATH a structure of MATH-bimodule. Assuming that the flat line bundle MATH is not a MATH-algebraic integer (compare above) the bimodule MATH satisfies the condition of REF . Indeed, let MATH denote an integral lattice which is preserved under the monodromy transformation along loops in MATH. Consider MATH as a right MATH-module, where each MATH acts as follows MATH where MATH is the abelinization homomorphism and MATH is the monodromy along MATH, viewed as a map MATH. Note that MATH, that is, we indeed have a right action. MATH has also the obvious left MATH action. Any MATH matrix MATH with entries in MATH yields a MATH-linear map MATH (by acting on the right), which may be represented by a square matrix MATH of size MATH with entries in MATH. If MATH is MATH-negative then MATH is also MATH-negative. Hence the determinant MATH of matrix MATH is an integer polynomial with MATH-top coefficient REF. Therefore, the determinant MATH of the MATH-linear map MATH, given by action of MATH, is a nonzero complex number, since this determinant MATH equals MATH and hence is nonzero, since otherwise we would have MATH, contradicting the assumption that MATH is not a MATH-algebraic integer. Using REF we obtain that for any closed REF-form MATH on MATH having NAME zeros and lying in the class MATH, there exists a chain complex MATH over MATH with MATH and with MATH. REF now follow from the standard argument CITE. In case when MATH is not a MATH-algebraic integer, we apply the previous arguments to the class MATH and the flat bundle MATH instead of MATH and use the NAME duality as in the proof of REF . |
math/9911157 | We may view MATH as a homomorphism MATH and let MATH denote the kernel. Let MATH denote MATH. We have the monodromy representation MATH of the flat bundle MATH, where MATH is the fiber over the base point. Consider the field MATH of rational functions on MATH and the following right action of MATH on MATH: MATH and MATH is the natural projection. MATH also has an obvious MATH-action from the left. Note that MATH yields a MATH-bimodule satisfying the condition of REF (because of the argument similar to REF). Hence REF applied to bimodule MATH implies MATH . The last inequality gives REF , since the assumption, that the flat bundle MATH is MATH-generic is in fact equivalents to MATH . Indeed, consider the MATH-bimodule MATH, defined similarly to MATH but replacing the field of rational functions MATH by the ring MATH of NAME polynomilas. Let MATH denote the chain complex of the universal covering of MATH. For any flat line bundle MATH we have the monodromy homomorphism MATH. We will view the dimension of the homology MATH as a function of MATH. It is well know from the algebraic geometry that there exists a proper algebraic subset MATH, so that for MATH (that is, when MATH is a generic point) MATH and for MATH . Hence, if MATH is MATH-generic, then MATH for all MATH and therefore REF follows. |
math/9911157 | First note that MATH implies MATH and also MATH . Using REF we obtain MATH . Now we define two chain maps: MATH . One checks that MATH and MATH . Hence MATH and MATH are mutually inverse homotopy equivalences. This completes the proof. |
math/9911157 | Fix on MATH a Riemannian metric and let MATH be the gradient field of MATH with respect to this metric. If MATH denotes the union of open MATH-balls (MATH is small) around the zeros MATH of MATH, then on MATH holds MATH for some MATH. Hence for any closed REF-form MATH, which sufficiently closely approximates MATH on the compact subset MATH, holds MATH on MATH. It is clear that we may find representation REF with the forms MATH approximating MATH on MATH and such that on each of the MATH-balls the forms MATH coincide with MATH. Then the gradient MATH of MATH will belong to the intersection of all the half-spaces above for all MATH, distinct from MATH. |
math/9911157 | For any proper cell MATH we will denote by MATH the smalles index MATH where MATH. For any MATH we will denote by MATH the submodule of MATH generated by all the basis elements MATH, such that the subset MATH satisfies MATH . Also, we will denote by MATH the submodule of MATH generated by all the basis elements MATH, such that the subset MATH satisfies MATH . Here MATH runs over all the cells of MATH, compare above. MATH and MATH are considered as graded modules with the grading given by the dimension of the corresponding cells, compare REF . Hence we obtain MATH where MATH is a graded free MATH-module generated by the cells of MATH corresponding to the internal critical points of MATH, that is, to zeros of the closed REF-form MATH; the grading of each generator of MATH equals the index of the corresponding zero of MATH. We will show that one may perform MATH subsequent collapses in the decomposition REF . First we may perform a collapse with respect to MATH. Indeed, from REF for the boundary operator we see that only elements in MATH have their boundary components lying in MATH; in other words, we have two zeros in the first row of the matrix REF . Also, the map MATH (in the notations of REF ), that is, the corresponding part of the differential of the chain complex MATH, is invertible. To see this, we may label the generators MATH and MATH (for a given MATH there is at most one element of this form) by the same letter, and then we see from REF that the matrix representing MATH has the form MATH where MATH is a diagonal matrix with the diagonal entries MATH and MATH is a matrix with all entries MATH-negative. Hence the matrix MATH is invertible over MATH according to our assumptions about MATH. Note that, assuming that MATH, this collapse is simple (compare REF), that is, the map MATH (compare matrix REF ) vanishes. Indeed, any basis element, having in the decomposition of its boundary a nontrivial summand in MATH, belongs to MATH. Hence we will have the collapsed complex supported on the graded module MATH and having the differential given by the same REF , in which we simply ignore the basis elements lying in MATH. Now we may perform the second collapse. The basis elements in MATH have the maximal length among the survived generators. Indeed, MATH contains all generators MATH with MATH, except the ones with MATH, which were killed during the first collapse. Also, the corresponding map MATH acting MATH, is again an isomorphism, because it can be again represented by a matrix of the form MATH, where MATH is diagonal with entries MATH and where MATH is MATH-negative. Hence, we may implement the second collapse. Let us show that, assuming MATH, the second collapse is again simple. A basis element having in its boundary a non-trivial MATH-component must have the form MATH, where MATH. Such MATH must contain MATH (and hence to belong to MATH), since all subsets MATH with MATH and MATH were killed on the first collapse. Suppose now that we have performed a sequence of simple collapses and have arrived to a chain complex MATH with the differential given by REF , in which we ignore all the cells of the form MATH with MATH and also all the cells MATH with MATH and MATH, which were killed on the previous collapses. Repeating the previous arguments we observe that we may perform the next collapse, killing MATH. Indeed, from REF we see that no element lying in MATH has a nontrivial MATH-component in its boundary (since MATH contains all survived generators with maximal MATH). In other words, we have two zeros in the first row of matrix REF . The generators of MATH and MATH are naturally split into pairs (by removing index MATH from MATH), and if one labels the corresponding basis elements of MATH and MATH by the same name, we will find that the corresponding matrix MATH (compare REF ) has the form MATH, where MATH is a diagonal matrix with MATH entries, and where MATH is MATH-negative matrix. This holds since the first sum in REF contains precisely one non-zero term in MATH. If MATH this collapse is simple, since a basis element, having in its boundary a non-trivial MATH-component must have the form MATH, where MATH. Such MATH must contain MATH, since all subsets MATH with MATH and MATH were killed on the previous collapse; hence the corresponding element MATH belongs to MATH. These arguments show that we may successfully collapse all the chain complex MATH to a chain complex supported on graded module MATH. Note that the last collapse MATH to MATH is not simple, since the boundaries of the cells lying in the interior of MATH (which form the basis of MATH) have components in MATH. This completes the proof. |
math/9911158 | Note that for a vector bundle MATH, specifying a classifying map MATH together with a covering map MATH is equivalent to specifying a map for MATH which is linear and injective on each fiber. (Here MATH.) If MATH has a basis MATH, then MATH is tame if and only if the function MATH is constant on the interior of cells. We prove something slightly more general than REF . Suppose that MATH has a basis MATH and that MATH are two tame classifying maps for MATH with covering maps MATH. Suppose also that for every MATH and for every MATH, MATH and MATH do not have opposite signs. Then MATH defines a classifying map MATH for the bundle MATH which is tame with respect to the product cell structure on MATH and hence a tame homotopy between MATH and MATH. Applying this to MATH gives REF . For REF , it suffices to prove it for the universal case. We will find a triangulation of MATH so that the identity map is tame, that is, MATH is constant on the interior of simplices. Then given a classifying map MATH where MATH has dimension MATH, by the cellular approximation theorem and the NAME cell decomposition of the NAME, there is a homotopic map MATH where MATH is a vector space spanned by MATH elements of the basis. Finally, apply the Simplicial Approximation Theorem to map from the barycentric subdivision of MATH to the tame triangulation of MATH. Thus the following lemma applied to the coordinate oriented matroid (where MATH) completes the proof of the combinatorialization theorem. (compare CITE) Let MATH be a realizable rank MATH oriented matroid with a fixed realization. Then there is a semi-algebraic triangulation MATH of MATH and a simplicial map with respect to its barycentric subdivision MATH such that for every vertex MATH in the barycentric subdivision, one has MATH. Furthermore, the homotopy class of MATH is independent of the choice of semi-algebraic triangulation. This is an application of REF, theorems on existence and uniqueness of semi-algebraic triangulations, and the fact that (compare REFEF) MATH is upper semi-continuous. A key tool is CITE which proves that for any finite partition MATH of a bounded, semi-algebraic set MATH into semi-algebraic sets there exists a semi-algebraic triangulation of MATH such that each MATH is a union of the interiors of simplices. Furthermore, by CITE, for any two such semi-algebraic triangulations, there is a semi-algebraic triangulation which is a common refinement. Thus MATH is a MATH space. In the case at hand MATH is a semi-algebraic partition of MATH. In the language of REF , the corresponding triangulation refines the stratification given by the upper semi-continuous map MATH, so the result follows. |
math/9911158 | Let MATH be a MATH-dimensional vector bundle over MATH. CASE: Existence: The combinatorialization theorem shows there is a tame classifying map MATH . Define MATH to be the corresponding combinatorial vector bundle MATH. CASE: Uniqueness: Suppose MATH are two tame classifying maps. Applying the first part of the combinatorialization theorem gives a tame classifying map MATH for MATH restricting to MATH and MATH at either end. The resulting combinatorial vector bundle over MATH gives a MATH-isomorphism between MATH and MATH. CASE: Naturality: Clear. |
math/9911158 | This follows from REF and results on existence of semi-algebraic triangulations, by the same argument as the proof of REF . |
math/9911158 | By replacing MATH by MATH, we may assume that MATH is the geometric realization of a simplicial complex. Let MATH be a classifying map for MATH. Apply the simplicial approximation theorem to MATH to find a subdivision MATH of MATH and a map MATH, where MATH is homotopic to MATH. Then set MATH. |
math/9911158 | This conversion process has two nice properties. The first is that it sends a fiber-preserving homotopy equivalence to a fiber-preserving homotopy equivalence. The second is that given a map MATH where MATH and MATH have the homotopy type of a NAME, then MATH also has the homotopy type of a CITE. We need to see that the map MATH is well-defined. If MATH is a MATH-isomorphism between quasifibrations MATH and MATH, then as above, there is a a f.p.w.h.e. from MATH (or MATH) to MATH, which is a f.p.h.e by the NAME. Now this conversion process takes a f.p.h.e to a f.p.h.e, and hence the corresponding fibrations are MATH-CW-isomorphic. If one first converts and then forgets, one obtains a quasifibration which is f.p.h.e. to the original one, and hence equivalent. Conversely, if one has a fibration and converts it, the result is a fibration equivalent to the original one. |
math/9911158 | Let MATH. Since MATH is nonzero in MATH, there are covectors MATH and MATH of MATH so that MATH and MATH. Since MATH, there are covectors MATH and MATH of MATH so that MATH and MATH. Since MATH and MATH, we can apply the elimination axiom in the definition of an oriented matroid. What results is a covector MATH of MATH so that MATH. |
math/9911158 | We induct on MATH and on the number of elements of MATH. When MATH, the existence of MATH is easy. If MATH, it suffices to consider the case when MATH is not maximal, since if MATH is maximal, there is a nonzero covector MATH of MATH so that MATH and we replace MATH by MATH. So assume that there is some nonzero element MATH of MATH such that MATH. We have two cases: CASE: If MATH is zero in MATH, then MATH. Then by induction on the number of elements we get MATH such that MATH. CASE: If MATH is nonzero in MATH, then by REF we have MATH. Since MATH, induction on rank gives a nonzero MATH such that MATH. |
math/9911158 | Enumerate the simplices MATH of MATH so that the dimension is monotone decreasing. Let MATH . Note MATH and MATH. We show MATH can be obtained from MATH by a sequence of elementary collapses and expansions. Define an elementary collapse to be inward if it collapses out a pair of simplices MATH and MATH with MATH. (We will use the term inward to apply to collapses of any complex, not just MATH.) We show by induction on dimension and induction on MATH that MATH can be obtained from MATH by a sequence of elementary collapses and expansions. Both initial cases hold because MATH is full. We will use MATH to denote equivalent via a sequence of elementary collapses inwards and elementary expansions. MATH . We leave for the reader to verify that such a sequence of collapses inward and expansions from MATH to a point gives a sequence of collapses inward and expansions from MATH to MATH, and hence from MATH to MATH. |
math/9911158 | We apply NAME 's Criterion, first with MATH and MATH, then with MATH and MATH. In the first case, let MATH and MATH. Then MATH and MATH. If MATH, then MATH is isomorphic to the poset of all covectors MATH of MATH such that MATH. This is the poset of all covectors in MATH corresponding to cells of MATH given as a subcomplex of the pseudosphere picture of MATH. By the Topological Representation Theorem, this is either empty, a MATH-sphere, or a MATH-ball. It can't be a sphere since MATH and it is non-empty by REF . Thus REF 's Criterion is fulfilled. If MATH, then MATH is isomorphic to the poset of all covectors MATH of MATH such that MATH. This is the poset of all covectors in MATH corresponding to cells in the interior of the cell complex MATH given as a subcomplex of the pseudosphere picture of MATH. Now MATH must be empty or a MATH-ball, but is in fact a MATH-ball of full rank since this is a non-empty intersection (containing cells corresponding to covectors MATH). By the previous lemma, the realization of the poset of cells in the interior of this ball is contractible. Thus REF 's Criterion is fulfilled, and so the realization of the combinatorial sphere bundle is a quasifibration. In the case of the disk bundle, the first condition to check is trivial, since the poset in question will have a unique minimal element. The second condition follows immediately from the proof of the second condition for sphere bundles. |
math/9911158 | For any quasifibration MATH and point MATH, there is a NAME spectral sequence MATH given by the NAME spectral sequence of the associated fibration MATH. If MATH were a fibration to begin with, there are, a priori, two different NAME spectral sequences, since MATH can be considered as a quasifibration or as a fibration. They coincide, since if MATH is a fibration then MATH and MATH have the same fiber homotopy type. The collapsing of the relative NAME spectral sequence MATH gives the NAME isomorphism, and the NAME class MATH is the image of REF under the NAME isomorphism. With integer coefficients, the same argument applies except that the MATH-term might have twisted coefficients. However the existence of an integral NAME class in REF guarantees that the coefficients are untwisted (look at MATH). |
math/9911158 | It suffices to prove the axioms hold for spherical fibrations. REF holds since MATH and MATH is zero on MATH for MATH. REF is clear by construction. The NAME sum formula REF holds since the NAME class MATH for the fiberwise join is the external product of the NAME classes of the two summands and there is a sum formula for the NAME squares. As for REF , one may compute MATH by restricting the canonical line bundle to the circle. Here the bundle is the NAME strip, which has non-trivial MATH by direct computation. |
math/9911158 | Any topological lifting is simplicial, and any simplicial lifting to a poset covering space is the realization of a poset lifting. (This is clear from looking at the lifting on individual simplices.) |
math/9911158 | Suppose first that MATH is a matroid bundle. From the double cover result, MATH, and is generated by MATH (where MATH is the universal bundle) since the first NAME class is a non-trivial characteristic class. Note that the map MATH is a MATH bundle, and hence has a classifying map into MATH. Thus we have maps MATH . Let MATH be the composition of the lower two maps, MATH be the generator of MATH. Then by covering space theory MATH has a lifting if and only if MATH. One can see directly that the combinatorial vector bundle corresponding to the NAME strip mapping to the circle is non-orientable, and thus MATH. Hence MATH, and the result follows. In the other cases, it suffices to consider a spherical fibration. One could either use the classifying spaces MATH and MATH for (oriented) spherical fibrations and proceed as above, or use the fact the MATH is the mod REF NAME to show that MATH if and only if the coefficients in the spectral sequence used in the NAME isomorphism theorem are untwisted. |
math/9911158 | We again use REF and the semi-algebraic triangulation theorem of CITE. As in the proof of REF , there is a semi-algebraic triangulation MATH refining the stratification of MATH and a map MATH so that MATH is simplicial. Now consider the semi-algebraic stratification of MATH given by the intersections of the preimages of elements under MATH and the preimages of simplices under MATH. By CITE, there is a triangulation MATH refining this stratification, and so by REF there is a map MATH so that MATH is simplicial. To see that these maps make the above diagram commute up to homotopy, let MATH. Then MATH lies in a simplex MATH of MATH and there is a simplex MATH of MATH so that MATH. By construction MATH is contained in the simplex spanned by the totally ordered set MATH and MATH is contained in the simplex spanned by the totally ordered set MATH. Then MATH and MATH both lie in the closed simplex spanned by MATH, so there is a straight-line homotopy from MATH to MATH. Finally, note that for every vertex MATH of MATH, the topological realization theorem gives a homotopy equivalence from the fiber over MATH to the fiber over MATH. |
math/9911158 | The proof uses REF and the techniques of the proof of the previous lemma. |
math/9911158 | Let MATH be a vector bundle. Convert the map MATH to a fibration MATH and consider the diagram MATH . By the homotopy lifting property, there is a map MATH so that MATH. Furthermore, MATH gives a homotopy equivalence on fibers (since MATH is a quasifibration) and MATH gives a homotopy equivalence on a fiber, so MATH gives a homotopy equivalence on fibers. Recall MATH is defined by applying the Simplicial Approximation Theorem to find a subdivision MATH of MATH and a map MATH, where MATH is homotopic to MATH. Then MATH. We then have the following equations in MATH: MATH . |
math/9911158 | Let MATH denote the monoid of self-homotopy equivalences of MATH, given the compact-open topology. Its classifying space MATH classifies rank MATH spherical (quasi)-fibrations CITE. MATH denotes the monoid of self-homotopy equivalences of MATH which fix a base point MATH. The result follows from commutativity up to homotopy of the diagram MATH the surjectivity of MATH when MATH, and the identification of MATH with MATH. The map MATH is given by inclusion, the map MATH by classifying the canonical bundle minus its zero section, and the map MATH is MATH applied to the injection of monoids MATH given by suspension. The map MATH, the map MATH, and the homotopy MATH, are given by our three main theorems: the NAME Theorem, Spherical NAME Theorem, and the Comparison Theorem (see also REF ). The surjectivity of MATH when MATH follows either from the Cellular Approximation Theorem applied to the NAME cell decomposition, or by viewing MATH as MATH and using the homotopy long exact sequence of a fibration. Now MATH since there is a fibration MATH and the NAME manifold is contractible. Also MATH, which is in turn MATH by the adjoint property of smash and mapping spaces in the category of based CW complexes. Thus we have identified the domain and range of MATH with that of MATH, and the identification of the two maps consists of tracing through these identifications. |
math/9911158 | CASE: By CITE, the combinatorialization map MATH is surjective, so REF follows from REF . An alternate route to REF is to use characteristic classes to detect elements of MATH, by applying the NAME class and second NAME class of the combinatorialization of the complex NAME bundle over REF-sphere. CASE: There is a characteristic class version and a homotopy theoretic version of the proof. The characteristic class proof is as follows. Consider the oriented combinatorial NAME MATH, defined analogously to MATH. The forgetful map MATH is a double cover, so MATH for each lifting MATH of MATH. The tangent bundle of the MATH-sphere combinatorializes to a map MATH. The evaluation of the NAME class of the tangent bundle of a manifold on the fundamental class of that manifold is the NAME characteristic of the manifold (compare REF). In particular, the NAME class of the tangent bundle of an even-dimensional sphere represents twice the generator of MATH. The NAME class applied to oriented vector bundles over MATH-spheres can be viewed as a homomorphism from MATH to MATH, where MATH classifies rank MATH oriented spherical fibrations. Thus the combinatorialization of this tangent bundle generates an infinite subgroup of MATH. The homotopy theoretic version is to consider the NAME invariant MATH . The domain of MATH is MATH, which classifies MATH-bundles over MATH. Now MATH is the NAME class, so MATH, using the tangent bundle of the MATH-sphere again. (For the reader's edification, we note the NAME invariant is onto if and only if MATH or REF, as can be seen by NAME periodicity or by NAME 's work on NAME invariant one.) REF.,REF. This follows from REF and the deep homotopy theoretic computation of MATH due to CITE. The result is that MATH is MATH for MATH (mod REF), MATH for MATH, and is zero otherwise. |
math/9911158 | CASE: This follows immediately from REF The proof is by induction on MATH, and relies on the Order Homotopy Lemma. If MATH, first note that MATH for all MATH, so MATH. (Here MATH means MATH-isomorphic. All homotopies occur in MATH; there is no need to go to MATH.) Thus, it suffices to show that MATH for all MATH. Any covector MATH is built from covectors MATH and MATH. Since MATH is independent in MATH, there is a MATH, so that MATH. Thus MATH is a covector of MATH which is greater than or equal to MATH. For MATH, let MATH be the set of MATH independent vector fields obtained from MATH by deleting MATH from each oriented matroid MATH, and let MATH be the resulting quotient bundle. Note that the vector field MATH on MATH given by MATH is non-vanishing and MATH is the quotient bundle of MATH by MATH. Hence we have MATH . |
math/9911158 | If MATH is a set of MATH independent vector fields in MATH and MATH is the resulting quotient bundle, then by the above lemma, MATH. Thus by the NAME sum formula, MATH where MATH is the total NAME class. But MATH since MATH is trivial, and MATH since MATH is a rank MATH bundle. |
math/9911158 | For a simplex MATH of MATH, let MATH denote both the barycenter of the simplex MATH and the corresponding vertex of MATH. Let MATH be a chain of simplices of MATH. Note that MATH hence inductively, MATH . Since the triangulation refines the stratification and MATH is upper semi-continuous, MATH for all MATH. |
math/9911158 | REF follows from the last lemma by defining MATH on vertices and extending by linearity with respect to MATH. REF follows from a straight-line homotopy MATH across the simplices of MATH. |
math/9911158 | We will inductively define maps MATH so that CASE: MATH CASE: For a vertex MATH of MATH, MATH CASE: For a point MATH in the interior of a simplex of MATH which is spanned by vertices MATH, MATH . Assume inductively that MATH has been defined satisfying REF , and REF , and also assume inductively that MATH satisfies REF . MATH. We then use REF to define MATH on MATH and REF to define MATH on MATH. We need to verify three things: first that REF continues to hold for MATH and MATH, second that MATH is totally ordered for MATH, and third that REF holds. We leave the verification of the first part to the reader. We now show that MATH is totally ordered for MATH. Suppose MATH is in the interior of a simplex of MATH with vertices MATH with the MATH's in MATH and the MATH's in MATH. Then by choosing a point MATH in the interior of the simplex spanned by the MATH's and by the (omitted) proof that REF holds for MATH and MATH, we see MATH is totally ordered. The proof of REF shows that MATH is totally ordered. Because MATH is a subcomplex of the triangulation MATH, because we have taken a barycentric subdivision, because MATH is upper semi-continuous, and because MATH refines the stratification, MATH for all MATH and MATH. Hence MATH is totally ordered. Finally REF holds since we have taken a barycentric subdivision. Next we use MATH to define MATH inductively. For MATH, let MATH. For a vertex MATH of MATH, define MATH. Then for MATH in the interior of a simplex spanned by MATH, define MATH by linearity, noting inductively that MATH is in the closed simplex spanned by MATH. This completes the proof of REF. For REF , it suffices to consider the case when each MATH subdivides MATH. Then note that MATH, so we can use the straight-line homotopy across the simplex spanned by MATH. |
math/9911158 | By Theorem A applied to MATH and REF , we only need show that the realization of the inclusion MATH is a homotopy equivalence for all MATH. But this follows from REF and by applying Theorem A to MATH, noting that MATH . |
math/9911158 | We wish to apply NAME 's Theorem B to the induced map of posets MATH. For a chain MATH, there is a retraction MATH . By the order homotopy lemma, MATH is actually a deformation retraction since MATH for every MATH. For an inequality MATH in MATH, consider the following commutative diagram: MATH where MATH, MATH is the retraction and the other two arrows are inclusions. To verify REF 's Theorem B applied to MATH, it suffices to prove the following lemma. Let MATH be a poset map satisfying the conditions of REF . Given an inequality MATH in MATH, the geometric realization of the map MATH is a homotopy equivalence. To prove this it suffices to check only those inequalities in MATH given by deleting the smallest or largest element of a chain. Let MATH be a chain in MATH. Let MATH and MATH. We will use REF to prove that MATH and MATH are homotopy equivalences. Note MATH is isomorphic to MATH by the isomorphism MATH, and the geometric realization of MATH is contractible by REF . Now if MATH where MATH and MATH are chains which map to MATH and MATH respectively, MATH has a maximum element, namely MATH, so its geometric realization is contractible. Thus by REF , MATH is a homotopy equivalence. The proof that MATH is a homotopy equivalence is similar and uses REF . Now we return to the proof of REF . Note that for MATH, the geometric realization of the diagram MATH is homotopy cartesian. Thus for all vertices and barycenters of MATH the inclusion of the fiber of MATH in the homotopy fiber is a homotopy equivalence. Since this is true for the barycenters and since MATH is a simplicial map, this is true for all points in the interior of a simplex, so MATH is a quasifibration. |
math/9911159 | By definition, the underlying chain of MATH is the oriented intersection chain in MATH of the marked points of MATH and of MATH respectively. Once the orientation of MATH is fixed, the orientations of intersections behave for calculations like orientations of normal directions. Now, for each MATH, we denote by MATH the chain of MATH of marked points of MATH. Thus, for transversal intersection one gets the familiar MATH where MATH. Since MATH is the underlying set of MATH we get MATH . The chain MATH is the restriction of MATH to MATH . Therefore, the above formula yields the same formula for MATH, MATH . |
math/9911159 | Assume that the three homology classes in question MATH are represented by cycles that are pairwise transversal. The intersection locus of MATH and MATH are literally equal with identical coorientations. The loop product is associative up to homotopy using the same considerations as in the based loop product, now parameterized by the points of the intersection MATH. |
math/9911159 | Let MATH denote the chain of MATH of the marked points of MATH and let MATH denote the chain of MATH with parameter space MATH, given by MATH, where MATH is the usual projection. Suppose that MATH and MATH are appropriately transversal. Since the underlying sets of MATH and MATH coincide, MATH is parametrized by the underlying set of MATH. Hence, for transversal intersection, one obtains MATH . Since MATH, the restriction of MATH to the underlying set of MATH is MATH . One the other hand, the restriction of MATH to the underlying set of MATH is MATH, which completes the proof. |
math/9911159 | These follow directly from the definitions. |
math/9911159 | By REF , MATH is a surjection. With MATH coefficients, the homology of MATH is the homology of MATH and the NAME product gives a polynomial algebra on this generator in degree MATH. |
math/9911159 | Observe that the parameter spaces of MATH and MATH consist in those values where the basepoint of the loops of MATH coincides with the one of the points of the loop of MATH and the basepoint of loops of MATH coincides with another. (see REF ) Symbolically, MATH . Clearly, there is a bijection between the underlying sets of the chains in question, MATH and MATH, which is orientation preserving if and only if MATH. |
math/9911159 | Let us prove MATH. By REF , MATH . |
math/9911159 | The proof of MATH is easier because the equation holds transversally at the chain level: The set of parameters where the marked point of MATH coincides with one of the images of MATH is the union of the set of parameters where the marked point of MATH coincides with one of the image of MATH union the set of parameters where the marked point of MATH coincides with one of the image of MATH. (see REF ) Let us prove MATH. The idea (see REF ) is that the MATH dimensional chain MATH where MATH and MATH attach to MATH at pairs of arbitrary points in such a way relative to the cyclic order that MATH is between the marked point and MATH, provides a chain homotopy between the two sides. More precisely, for each pair of points MATH we will define a chain such that MATH is attached to MATH at MATH and MATH is attached to MATH at MATH. Let MATH . We define a map MATH. MATH for each MATH, MATH. Since the restriction of MATH to CASE: MATH is MATH . CASE: MATH is MATH CASE: MATH is MATH then MATH . MATH is the restriction of MATH to MATH. Thus MATH which completes the proof of MATH. (See also CITE). |
math/9911159 | First, the idea of the proof: consider the chain operation MATH where the loop of MATH is attached to any point of the loop of MATH and one goes around a part of the loop of MATH, starting at any point between the marked points of MATH and MATH and ending where MATH is attached, then goes around the loop of MATH and finally, around the rest of MATH. (See REF ). More precisely, Let MATH, MATH be two cells of MATH. Consider MATH, the parameter space of MATH and set MATH and define a chain MATH as follows MATH . Observe that MATH is attached to MATH at the image of MATH by MATH and the image of MATH by MATH is the marked point of the resultant loop. Proceeding in an analogous way as we did in the proof of REF , we obtain MATH as desired. |
math/9911159 | By REF . |
math/9911159 | By REF , MATH is graded commutative. So, since MATH has degree MATH, MATH . To prove NAME, replace MATH (respectively, MATH, MATH) by MATH (respectively, MATH, MATH) in the NAME property MATH of REF and apply MATH to both sides of the equation to obtain MATH . Now, use REF and the fact that MATH to replace in the above equation each of the brackets MATH by the formula MATH . Since MATH we cancel the terms where MATH appear and so we obtain MATH . Now, replacing in the above formula each occurrence of MATH by MATH yields MATH . Hence, MATH . |
math/9911159 | We only have to show the equations in the case when the range of the commutator being studied lies in monomial degree one. Since the commutator of coderivations is a coderivation, this is enough to show it is identically zero. We illustrate the proof of MATH for MATH, MATH. We have four families of closed curves MATH. Then MATH can be viewed a sum of twelve terms, each of them labeled with MATH where MATH runs over all possible choices of three families from MATH with a preferred element. Analogously, MATH can be viewed as a sum of twelve terms, each of them labeled with MATH where MATH runs over all possible choices of two families with a preferred element. A correspondence of the two sets of labels is given by the map MATH where MATH is the only family not in MATH. Now, we will see that corresponding pair of terms appear in MATH with different sign. Consider, for instance, the corresponding pair of terms labeled by MATH . For simplicity, let us denote by MATH the parameter space of the family MATH. We can assume that each MATH is a cell, and that MATH is a map MATH where MATH . The parameter space of the term labeled MATH is MATH, the preimage of MATH under the map MATH given by MATH . The parameter space of the term labeled with MATH, MATH is the preimage of MATH under the map MATH given by MATH . Over each point of these parameter spaces, the loops of the terms labeled with MATH and MATH are as in REF . Observe that the only difference between MATH and MATH is that MATH and MATH are interchanged. This produces the difference of sign. Now, we prove MATH. We illustrate for MATH. There will be five families, MATH. MATH is the sum of thirty terms, each of them labeled with MATH where MATH runs over all subsets of three elements of MATH. We group these terms in pairs with the following correspondence MATH where MATH . As in the proof of case MATH, we can see that the parameter spaces corresponding to pairs cancel, so MATH holds. |
math/9911159 | By REF each MATH is a coderivation of MATH of square zero and so determines (by definition) a MATH structure on MATH. |
math/9911159 | The loop product is the composition of intersection with image MATH which is represented as a codimension MATH submanifold being the transverse image of the diagonal under the map MATH with the induced transformation of MATH in homology. The first process commutes with cap product with first MATH then with its restriction to MATH. The second process commutes with capping with a class in MATH or with its pull back via MATH to MATH . |
math/9911171 | This follows by the above discussion and the sequence MATH (note that the vanishing of MATH is implied by our assumptions). |
math/9911171 | The first part follows directly from the discussion above and REF . The second part follows from the fact that the MATH term in the second stage of the resolution of the ideal may be removed. |
math/9911171 | This follows exactly as above taking into account: CASE: MATH CASE: MATH satisfies condition MATH (CITE,CITE) CASE: MATH for MATH (CITE, CITE) |
math/9911171 | The first two statements follows immediately from the basic sequence MATH . The third is just the statement that a form of degree MATH cannot vanish MATH times on a non-degenerate variety. |
math/9911171 | The first statement follows from the fact that MATH is smooth and MATH has reduced, connected fibers. The second follows from the smoothness of MATH and the description of the fibers MATH above. For the third, MATH is NAME 's Main Theorem. MATH follows from the fact that MATH for MATH (Compare CITE). To show MATH, note that MATH is the composition of birational morphisms to smooth varieties, followed by a projective bundle, followed by a birational morphism to MATH which is normal by the first statement. |
math/9911171 | The case MATH is automatic by REF . The case MATH is contained in REF . For MATH, by REF we need only prove the result for MATH. First note that as MATH is MATH-very ample, we can apply REF to MATH to obtain MATH. Furthermore, as the restriction of MATH to a fiber of the MATH bundle MATH is MATH, we see immediately MATH . Now, beginning anew with MATH, apply REF to MATH. This yields MATH where MATH. We deduce the desired vanishing from the sequence where MATH is arbitrary. The fact that MATH (this is true by REF ) implies that MATH . By REF , if MATH then the cohomology along the fibers vanishes, implying the groups in REF vanish (note the higher direct images vanish by REF ). MATH is shown in REF . MATH is more difficult and is shown in REF . |
math/9911171 | From REF parts MATH and MATH, we have MATH. Using part MATH of REF to check the vanishing of MATH of the rightmost term of sequences of the form MATH and gives MATH. |
math/9911171 | As before, REF imply the result for MATH. Hence we show MATH for MATH. By restricting to MATH and computing direct images (recall MATH is flat), this immediately implies MATH for MATH. Because MATH, it suffices to show MATH. We prove MATH and the result follows by a regularity argument (note that MATH is smooth and MATH globally generated). As before, we have MATH because the restriction of MATH to a fiber of the MATH-bundle is MATH. The fact that MATH follows immediately from projective normality and the first paragraph. The final step is to note MATH (this follows from CITE), hence we have the exact sequence MATH which we push down MATH. As MATH is trivial along the fibers, it is the pull back of a line bundle MATH on MATH. As the restriction of MATH to MATH is flat of degree two, MATH for some line bundle MATH. Therefore, MATH implies MATH. It is, however, not difficult to verify that MATH, and the vanishing follows from the fact that MATH is non-special and very ample. |
math/9911171 | This follows by induction after tensoring the sequence: MATH by MATH. As we work over MATH, REF implies MATH of the left term vanishes. By MATH, the Theorem on Formal Functions CITE implies the completion MATH. As MATH is coherent, the result follows (for example, by CITE). |
math/9911171 | We proceed via REF . As MATH is supported on MATH, we need to check three classes of fibers. First, let MATH. Then MATH where MATH determines the unique MATH-secant MATH containing MATH. The restriction of MATH to such a fiber is simply MATH, hence MATH. The conormal sequence for MATH is MATH . The required vanishings follow after twisting by MATH. Let MATH. Then MATH with the embedding MATH where MATH determines the unique secant line containing MATH. The restriction of MATH to such a fiber is MATH where MATH. Therefore, MATH by projective normality of the above embedding. The conormal sequence for MATH is MATH and as above the required vanishing follows after twisting by MATH. If MATH, then MATH where MATH. The restriction of MATH to such a fiber is MATH where MATH. By REF , MATH for MATH and the vanishing of tensor powers of the conormal bundle follows exactly as above. |
math/9911171 | For the first, we need only note that a line bundle of degree at least MATH satisfies condition MATH CITE. For the second statement, we need only add that under our hypotheses, surjectivity of MATH is equivalent to the vanishing MATH CITE. |
math/9911171 | For MATH the result is again automatic by REF , and MATH is in REF . Recall MATH. To settle the case MATH, let MATH and let MATH be the morphism induced by MATH. We have the diagram: Twisting the entire diagram by MATH, we see from the top row that showing MATH is equivalent to showing MATH. However, from the pictured diagram, it is easy to see that MATH is equivalent to the vanishing MATH. Therefore, twisting the last row by MATH we need MATH. Clearly, it suffices to prove that the higher direct images of the blow down to the curve vanish. As MATH is flat, we only need vanishing along the fibers, which are isomorphic to MATH. However, MATH maps a fiber of MATH isomorphically to a linearly embedded subspace MATH, and the vanishing follows easily. Repeating this argument after tensoring by MATH yields the stated result. |
math/9911171 | REF is REF applied to REF is the earlier statement that MATH is surjective exactly when MATH applied to REF . |
math/9911172 | That there exists a properly embedded oriented norm minimizing surface MATH for any class MATH is explained in CITE. It remains to show that we may arrange to have MATH as indicated. We may assume that MATH is incompressible and boundary incompressible, since compressions do not change the homology class of the surface and do not raise the value of MATH. So, since MATH is a collection of tori, MATH has MATH consisting of some number (possibly zero) of parallel simple closed curves, essential in MATH. If there are two components of MATH with opposite orientations, then there is an innermost pair of such components. That is, there are two components MATH and MATH (without loss of generality) with opposite orientations such that MATH is the boundary of an annulus MATH in MATH with MATH empty. We can alter the surface MATH by attaching a copy of the annulus MATH to MATH and pushing it into the interior of MATH. This operation does not change the homology class of MATH or MATH. By repeatedly applying this procedure, we arrange that every component of MATH has the same orientation. Now homotop MATH so that MATH is minimal. Then the absolute value of the algebraic intersection number of MATH and MATH equals the number of points of intersection of MATH and MATH. That is, MATH. |
math/9911172 | Suppose that MATH is a norm minimizing surface representing MATH. By REF we may assume that MATH is a single simple closed curve isotopic to MATH for each MATH and that MATH is no fewer than MATH meridional curves for each MATH. We can fill in each boundary curve of MATH on MATH with a meridian disk for MATH to obtain a NAME surface MATH for the link MATH in MATH. Then MATH . NAME 's inequality tells us that MATH. So MATH . |
math/9911172 | We first note that MATH where MATH is the algebraic crossing number between components in MATH and components not in MATH. Crossings in MATH occur in the following ways: CASE: For each crossing of MATH with itself, there are MATH crossings with the same sign in MATH. CASE: For each component MATH, there are MATH crossings in MATH (counted with sign) coming from the braid inserted above. CASE: For each crossing of MATH with MATH for MATH, there are MATH crossings with the same sign in MATH. Therefore, MATH . The sums can be taken over all MATH in these equations because each relevant term contains a MATH factor, eliminating those components in MATH. Hence, MATH . The equality in line REF is true because the second term in line REF will produce all terms of the form MATH for all pairs MATH with MATH and neither MATH nor MATH equal to zero. The fourth term in line REF will produce all terms of the form MATH for all pairs MATH with MATH and MATH. Together, they produce all terms of the form MATH for all pairs MATH with MATH and MATH. The sum in line REF can be taken over all MATH because a factor of MATH appears in every term, eliminating those with MATH. |
math/9911172 | The pair of link diagrams MATH and MATH are defined in terms of a norm minimizing surface MATH for the class MATH so that MATH is also a surface spanning MATH in the complement of MATH. Therefore, by REF , MATH. |
math/9911172 | Assume that the basis MATH is chosen so that MATH for MATH and MATH for MATH. The unit ball MATH for MATH is convex. Therefore MATH is an intersection of half-spaces. If a point MATH has norm MATH then there exists a MATH-dimensional affine space MATH through MATH such that MATH is contained in the closure of one of two halfspaces defined by MATH. Let MATH be the point MATH for MATH. Choose MATH, hence MATH. The only MATH-dimensional affine space MATH that contains MATH, and has a closed halfspace that contains all MATH and all MATH is the MATH-dimensional affine space containing the MATH's times MATH. For all MATH it follows MATH. |
math/9911172 | First we prove the theorem for the case that all MATH are equal to either MATH or MATH. By assumption the NAME norm at the point MATH equals MATH where MATH is the word length of the braid and MATH the number of strands. Choose a component MATH with MATH strands of the link. Construct a surface that spans this component as in the generalization of NAME 's algorithm explained in REF. That is, if another component undercrosses an arc of MATH, we get a piercing of the surface. Let MATH be the number of these piercings and MATH be the number of self-crossings of the component MATH. The so constructed surface has NAME characteristic MATH and thus, the NAME norm at the point MATH is less than or equal to MATH. By REF and MATH. Since the NAME norm is a semi-norm, we have MATH . Thus, the NAME norm of MATH equals MATH. REF gives us the additivity for all other points MATH with MATH. |
math/9911172 | By REF these links are fibred and the fibre surface is the one that one gets by the application of NAME 's algorithm to the closed braid. Hence, the NAME norm equals the NAME norm at the point MATH (for example, CITE) and the claim follows by REF . |
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