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math/0101148 | Let MATH, MATH be a pair of MATH - essential disks in MATH, MATH respectively, which gives a weak MATH - reducibility of MATH. If MATH and MATH, then we have conclusion REF. Hence in the rest of the proof, we may suppose that MATH. We have the following two cases. CASE: In this case, we first show the following. CASE: MATH is separating in MATH, then we have conclusion REF. Let MATH, MATH be the closures of the components of MATH such that MATH. Then MATH is a MATH - handlebody which is not a MATH - ball. Hence there exists a MATH - essential disk MATH in MATH such that MATH, and MATH (hence, MATH is properly embedded in MATH). See REF . It is clear that MATH is a MATH - meridian disk of MATH. Hence, by regarding, MATH, MATH, we have conclusion REF. By REF , we may suppose that MATH is non-separating in MATH. Let MATH be the surface obtained from MATH by MATH - compressing along MATH, and MATH. We note that MATH separates MATH into two components, say MATH and MATH, where MATH is obtained from MATH by cutting along MATH. Let MATH. Hence, by REF, we see that MATH is a compression body with MATH. Let MATH be a union of maximal mutually disjoint, non parallel compressing disks for MATH such that MATH. Note that since MATH is a compressing disk for MATH such that MATH, there actually exists such MATH. Let MATH. Note that MATH is homeomorphic to MATH, hence, MATH is irreducible REF . Hence REF - sphere components MATH (possibly MATH) of MATH bounds mutually disjoint REF - balls in MATH. Let MATH. Then MATH is a compression body such that MATH. Let MATH. CASE: MATH is compressible in MATH, then we have conclusion REF. Suppose that there exists a compressing disk MATH of MATH in MATH. Let MATH. By the maximality of MATH, we see that MATH is contained in MATH. Note that MATH is a NAME splitting of MATH, and MATH. Hence, by CITE, we see that MATH is weakly reducible, and this implies conclusion REF. By REF , we may suppose that MATH is incompressible in MATH. Note that each component of MATH is a meridian loop of MATH. Since MATH is small and there does not exist a meridional essential surface in MATH, we see that each component of MATH is a boundary parallel annulus properly embedded in MATH. Recall that MATH is the union of REF - sphere components of MATH. Note that we can assign labels MATH and MATH to the components of MATH alternately so that the MATH region are contained in MATH, and that MATH is recovered from MATH by adding tubes along mutually disjoint arcs in MATH - regions. Recall that MATH is connected. Since each component of MATH is separating in MATH, this shows that exactly one component of MATH is a MATH - region. Let MATH be a surface in MATH obtained from MATH by capping off the boundary components by mutually disjoint MATH - disks in MATH (hence, via isotopy, MATH is recovered from MATH by adding the tubes used for recovering MATH from MATH). Then each component of MATH is a MATH - sphere. Since MATH is MATH - irreducible, the components of MATH bounds MATH - balls, say MATH, in MATH. CASE: MATH - balls MATH are mutually disjoint. Suppose not. By exchanging the subscript if necessary, we may suppose that MATH. Since there exists exactly one MATH - region, this implies that the MATH - balls MATH are included in MATH in a non-nested configuration. Hence MATH is contained is the MATH - ball MATH. See REF . Note that MATH is recovered from MATH by adding a tube along the component of MATH, which intersects MATH. Hence we see that MATH is contained in a regular neighborhood of MATH, say MATH. Note that MATH is contained in MATH. Since MATH is a MATH - compression body, there exists a MATH - compressing disk MATH for MATH in MATH. Suppose that MATH. Since MATH is a regular neighborhood of MATH, we see that MATH is a MATH - ball. Since MATH is MATH - irreducible, MATH is also a MATH - ball. These show that MATH is REF - sphere, and MATH is a trivial knot, contradicting REF of REF . Suppose that MATH. Since MATH is MATH - irreducible, we see that MATH is irreducible. This shows that we obtain a REF - ball by cutting MATH along MATH. This shows that MATH is a solid torus. Hence MATH is a genus one NAME splitting of MATH. Hence MATH is a core knot, contradicting REF of REF . This completes the proof of REF . Recall that MATH is the surface obtained from MATH by MATH - compressing along MATH, and MATH, MATH the closures of the components of MATH, where MATH is obtained from MATH by cutting along MATH. By REF , MATH is a MATH - handlebody. By REF , we see that the MATH - region is MATH. Hence MATH is recovered from MATH by adding tubes along arcs properly embedded in MATH. Hence, we see that MATH is obtained from the MATH - balls MATH by adding REF - handles disjoint from MATH. Hence MATH is also a MATH - handlebody. These show that MATH is a NAME surface for MATH. It is clear that MATH gives a genus MATH, MATH bridge position of MATH, and MATH is obtained from MATH by a tubing along a component of MATH. Hence, by regarding MATH, MATH, we have conclusion REF . CASE: In this case, we first show the following. CASE: MATH or MATH is separating in MATH, then we have conclusion of REF . Since the argument is symmetric, we may suppose that MATH is separating in MATH. This implies that MATH is separating in MATH. Let MATH, MATH be the closures of the components of MATH such that MATH. Then MATH is a MATH - handlebody which is not a MATH - ball. Hence there exists a MATH - essential disk MATH in MATH such that MATH, and MATH (hence, MATH is properly embedded in MATH). See REF . It is clear that MATH is a MATH - meridian disk of MATH. Hence by applying the arguments of REF to the pair MATH, MATH, we have conclusion of REF . Let MATH be the surface obtained from MATH by MATH - compressing along MATH, and MATH. Let MATH, MATH be the closures of the components of MATH such that MATH is obtained from MATH by cutting along MATH and attaching MATH, and MATH is obtained from MATH by cutting along MATH and attaching MATH. Then let MATH. CASE: If MATH is compressible in MATH, then we have conclusion of REF . Suppose that there is a compressing disk MATH for MATH in MATH. Since the argument is symmetric, we may suppose that MATH. We may regard that MATH is a compressing disk for MATH. Since MATH is recovered from MATH by adding two tubes along a component of MATH and a component of MATH, we may suppose that MATH. Hence MATH is a MATH - meridian disk of MATH such that MATH. Hence, by applying the arguments of REF to the pair MATH, MATH, we have the conclusion of REF . By REF , we see that, for the proof of REF , it is enough to show that either REF MATH or MATH is separating in MATH, or REF MATH is compressible in MATH. Suppose that MATH REF is non-separating in MATH, and that MATH is incompressible in MATH. Then, by the argument preceding REF of REF , we see that each component of MATH is a MATH - sphere, and MATH is recovered from MATH by adding tubes along two arcs MATH, MATH such that MATH is a component of MATH, and that MATH. Note that MATH is connected. Since MATH, MATH are non-separating in MATH, we see that MATH consists of one MATH - sphere, or two MATH - spheres, and this shows that MATH consists of one arc, or two arcs. But since MATH is a knot, we have MATH in either case, a contradiction. Hence we have the conclusion of REF . This completes the proof of REF . |
math/0101148 | Let MATH-be a NAME splitting of MATH, which gives a genus MATH, MATH - bridge position of MATH with MATH. Suppose, for a contradiction, that MATH is strongly MATH - irreducible. Then, by REF , we may suppose that MATH consists of non-empty collection of transverse simple closed curves which are MATH - essential in both MATH and MATH. By REF , we see that MATH is weakly MATH - reducible, a contradiction. |
math/0101148 | First of all, we would like to remark that the proof given below is just an orbifold version of the proof of CITE. We suppose that MATH is minimal among all surfaces MATH such that MATH gives a two bridge position of MATH, and that MATH consists of non-empty collection of simple closed curves which are MATH - essential in both MATH and MATH. Note that the closure of each component of MATH is either an annulus which is disjoint from MATH, or a disk intersecting MATH in two points. We divide the proof into several cases. Case REFEach component of MATH is not MATH - boundary parallel in MATH, and each component of MATH is not MATH - boundary parallel in MATH. REF is divided into the following subcases. CASE:MATH contains a component which is MATH - compressible in MATH, and MATH contains a component which is MATH - compressible in MATH . In this case, by MATH - compressing the components in MATH and MATH, we obtain MATH - meridian disks MATH, MATH in MATH, MATH respectively. By applying a slight MATH - isotopy if necessary, we may suppose that MATH, and this shows that MATH is weakly MATH - reducible. CASE:REFEither MATH or MATH, say MATH, contains a component which is MATH - compressible in MATH, and each component of MATH is MATH - incompressible in MATH. Let MATH be a MATH - meridian disk obtained by MATH - compressing the component of MATH. Note that MATH is MATH - boundary compressible in MATH REF . Let MATH be a disk in MATH obtained by MATH - boundary compressing MATH. By the minimality of MATH, we see that MATH is a MATH - meridian disk in MATH. Since MATH is a component of MATH, we may suppose that MATH by applying a slight MATH - isotopy if necessary. This shows that MATH is weakly MATH - reducible. CASE:REFEach component of MATH is MATH - incompressible in MATH, and each component of MATH is MATH - incompressible in MATH. Let MATH (MATH respectively) be a MATH - meridian disk obtained by MATH - boundary compressing MATH (MATH respectively). By applying slight isotopies, we may suppose that MATH, MATH (hence, MATH or MATH, MATH or MATH). If one of MATH or MATH is contained in MATH, and the other in MATH, then MATH, and this shows that MATH is weakly MATH - reducible. Suppose that MATH is contained in MATH or MATH, say MATH. Let MATH be a MATH - meridian disk in MATH (that is, MATH is a disk properly embedded in MATH such that MATH, and MATH separates the components of MATH). Note that since each component of MATH, MATH is MATH - incompressible, MATH. We take MATH so that MATH is minimal among all MATH - essential disks MATH in MATH such that each component of MATH (MATH respectively) is a MATH - essential arc properly embedded in MATH (MATH respectively). Suppose that MATH contains a simple closed curve component. Let MATH be an innermost disk. Since the argument is symmetric, we may suppose that MATH. By the minimality of MATH, we see that MATH is a MATH - meridian disk in MATH. Since MATH, MATH. Hence the pair MATH, MATH gives a weak MATH - reducibility of MATH. Suppose that each component of MATH is an arc. Let MATH be an outermost disk. Since the argument is symmetric, we may suppose that MATH. Recall that MATH is a MATH - essential arc in MATH. By the minimality of MATH, we see that at least one component, say MATH, of the surface obtained from MATH by MATH - boundary compressing along MATH is a MATH - meridian disk in MATH. Since MATH, MATH. Hence the pair MATH, MATH gives a weak MATH - reducibility of MATH. Case REFA component of MATH or MATH, say MATH, is MATH - boundary parallel in MATH. By the minimality of MATH, we have either MATH (and MATH (MATH respectively) is a disk intersecting MATH in two points) or, MATH (and MATH is an annulus disjoint from MATH). CASE: Let MATH and MATH. Let MATH be the closure of the component of MATH such that MATH and MATH are MATH - parallel in MATH. Since the argument is symmetric, we may suppose that MATH. We have the following subcases. CASE:MATH is MATH - boundary parallel in MATH . Since MATH, MATH is parallel to MATH in MATH, and cannot be parallel to MATH. Let MATH be a MATH - meridian disk in MATH. Since MATH is not a trivial knot, MATH and MATH are not isotopic in MATH. Hence MATH. We suppose that MATH is minimal among all MATH - meridian disks MATH in MATH such that each component of MATH (MATH respectively) is a MATH - essential arc in MATH (MATH respectively). Suppose that MATH contains a simple closed curve. Let MATH be an innermost disk. Since the argument is symmetric, we may suppose that MATH. By the minimality of MATH, we see that MATH is a MATH - meridian disk in MATH. Then by pushing MATH into MATH along the parallelism through MATH, we can MATH - isotope MATH to MATH such that MATH, and MATH. Hence, by REF , we see that MATH is weakly MATH - reducible. Suppose that each component of MATH is an arc. Let MATH be an outermost disk. Since the argument is symmetric, we may suppose that MATH. See REF . NAME least one component of the disks obtained from MATH by MATH - boundary compressing along MATH is a MATH - meridian disk. Let MATH, MATH be the disks obtained from MATH by MATH - boundary compressing along MATH. Suppose that MATH is MATH - boundary parallel, that is, there exists a MATH - disk MATH in MATH such that MATH. Note that since MATH is obtained from MATH by MATH - boundary compressing along MATH, there is an annulus MATH is MATH such that MATH, and that MATH: two points. Note also that MATH consists of one point. Hence MATH is not contained in MATH, and this implies that MATH. Then MATH is a disk intersecting MATH in three points, whose boundary is MATH. Since MATH, MATH is not a MATH - disk. Hence MATH is a MATH - meridian disk in MATH. Let MATH be a MATH - meridian disk in MATH obtained as in Claim. By applying a slight isotopy, we may suppose that MATH. Then by pushing MATH into MATH along the parallelism through MATH, we can MATH - isotope MATH to MATH such that MATH, and MATH. Hence, by REF , we see that MATH is weakly MATH - reducible. CASE:MATH is not MATH - boundary parallel in MATH, and MATH is MATH - incompressible in MATH, that is, MATH is MATH - essential in MATH. Since MATH is MATH - incompressible, there is a MATH - boundary compressing disk MATH for MATH in MATH. Claim-MATH. Suppose that MATH. Note that MATH consists of two points in MATH, and MATH is an arc properly embedded in MATH, which separates the points. Then, by MATH - boundary compressing MATH along MATH, we obtain two MATH - disks. Since MATH is MATH - irreducible, these MATH - disks are MATH - boundary parallel in MATH. This shows that MATH is MATH - boundary parallel in MATH, contradicting the condition of Case REFa. REF. Then, by using the argument of the proof of Claim of Case REFa. REF, we see that at least one component, say MATH, of the MATH - disks obtained from MATH by MATH - boundary compressing along MATH is a MATH - meridian disk in MATH. By applying a slight isotopy, we may suppose that MATH. By Claim, we see that MATH. Then by pushing MATH into MATH along the parallelism through MATH, we can MATH - isotope MATH to MATH such that MATH, and MATH. Hence, by REF , we see that MATH is weakly MATH - reducible. CASE:MATH is not MATH - boundary parallel in MATH, and MATH is MATH - compressible in MATH. Let MATH be the MATH - compressing disk for MATH. Since there does not exist a REF - sphere MATH intersecting MATH in three points, MATH. Let MATH be the disk component of a surface obtained from MATH by MATH - compressing along MATH. Since MATH, we see that MATH is a MATH - meridian disk of MATH. By applying a slight isotopy, we may suppose that MATH. Suppose that MATH. Then by pushing MATH into MATH along the parallelism through MATH, we can MATH - isotope MATH to MATH such that MATH, and MATH. Hence, by REF , we see that MATH is weakly MATH - reducible. Hence, in the rest of this subcase, we suppose that MATH REF . Let MATH be a MATH - meridian disk in MATH. Since MATH is not a trivial two component link, MATH and MATH are not isotopic in MATH. Hence MATH. We suppose that MATH is minimal among all MATH - meridian disks MATH in MATH such that each component of MATH (MATH respectively) is a MATH - essential arc in MATH (MATH respectively). Suppose that MATH contains a simple closed curve. Let MATH be an innermost disk. By the minimality of MATH, we see that MATH is MATH - essential in MATH. Note that MATH. If MATH, then the pair MATH, MATH gives a weak MATH - reducibility of MATH. If MATH, then by pushing MATH into MATH along the parallelism through MATH, we can MATH - isotope MATH to MATH such that MATH, and MATH. Hence, by REF , we see that MATH is weakly MATH - reducible. Suppose that each component of MATH is an arc. Let MATH be an outermost disk. If MATH, then by using the argument as in the proof of Case REFa. REF, we see that MATH is weakly MATH - reducible. Suppose that MATH. Then, by using the argument as in the proof of Claim of Case REFa. REF, we can show that at least one component , say MATH, of the MATH - disks obtained from MATH by MATH - boundary compressing along MATH is a MATH - meridian disk in MATH. By applying slight MATH - isotopy, we may suppose that MATH. Hence the pair MATH, MATH gives a weak MATH - reducibility of MATH. CASE: Let MATH, MATH be the components of MATH, and MATH. Recall that MATH is a MATH - boundary parallel annulus in MATH such that MATH, and that MATH, MATH are not MATH - boundary parallel. We also note that MATH bounds an annulus MATH in MATH such that MATH and MATH are MATH - parallel in MATH. Without loss of generality, we may suppose that MATH is contained in REF - ball MATH. CASE:MATH is MATH - incompressible in MATH. Since MATH is MATH - incompressible, there is a MATH - boundary compressing disk MATH for MATH. Without loss of generality, we may suppose that MATH, MATH. Since MATH is not MATH - boundary parallel, at least one component, say MATH, of the MATH - disks obtained from MATH by MATH - boundary compressing along MATH is a MATH - meridian disk in MATH. By applying a slight MATH - isotopy, we may suppose that MATH. Claim-MATH. Suppose, for a contradiction, that MATH. Then MATH is contained in the annulus MATH bounded by MATH. We note that MATH intersects MATH in one point. Hence MATH is not contractible in MATH. This shows that MATH is a core curve of MATH. Let MATH be the annulus in MATH bounded by MATH. Then REF - sphere MATH intersects MATH in three points, a contradiction. By Claim we see that, by pushing MATH into MATH along the parallelism through MATH, we can MATH - isotope MATH to MATH such that MATH. By the above claim, we may suppose that MATH. Hence, by REF , we see that MATH is weakly MATH - reducible. CASE:MATH is MATH - compressible. Let MATH be the MATH - compressing disk for MATH. Without loss of generality, we may suppose that MATH, MATH. Let MATH be a MATH - meridian disk of MATH obtained from MATH by MATH - compressing along MATH. By applying slight isotopy, we may suppose that MATH. Suppose that MATH. By pushing MATH into MATH along the parallelism through MATH, we can MATH - isotope MATH to MATH such that MATH, and MATH. Hence, by REF , we see that MATH is weakly MATH - reducible. Suppose that MATH REF . Let MATH be a MATH - meridian disk in MATH. Since MATH is not a trivial two component link, MATH and MATH are not isotopic in MATH. Hence MATH. We suppose that MATH is minimal among all MATH - essential disks MATH in MATH such that each component of MATH (MATH, MATH respectively) is a MATH - essential arc in MATH (MATH, MATH respectively). Suppose that MATH contains a simple closed curve component. Let MATH be an innermost disk. By the minimality of MATH, we see that MATH is MATH - essential in MATH. Note that MATH. If MATH, then the pair MATH, MATH gives a weak MATH - reducibility of MATH. If MATH, then by pushing MATH into MATH along the parallelism through MATH, we can MATH - isotope MATH to MATH such that MATH, and MATH. Hence, by REF , we see that MATH is weakly MATH - reducible. Suppose that each component of MATH is an arc. Let MATH be an outermost disk. If MATH, then by using the argument as in the proof of Case REFb. REF, we see that MATH is weakly MATH - reducible. Suppose that MATH. Let MATH be the disk obtained from MATH by MATH - boundary compressing along MATH. Claim-MATH is a MATH - meridian disk of MATH. Suppose that MATH is not a MATH - meridian disk of MATH, that is, MATH is MATH - parallel to a disk, say MATH, in MATH. Since MATH, we see that MATH. Note that MATH is recovered from MATH by adding a band along an arc intersecting MATH in one point. This shows that MATH is an annulus not intersecting MATH. Hence MATH is a torus, and MATH is a solid torus such that MATH. However, since MATH is a meridian disk of MATH, this implies that MATH is MATH - reducible, contradicting REF . By Claim, we see that, by applying a slight isotopy, we may suppose that MATH, and MATH. Hence the pair MATH, MATH gives a weak MATH - reducibility of MATH. This completes the proof of REF . |
math/0101148 | We note that MATH is isotopic to either one of the six unknotting tunnels MATH, MATH, MATH, MATH, MATH, or MATH in REF (see CITE or CITE). Suppose that MATH is isotopic to MATH, MATH or MATH, say MATH. Then we may regard that MATH REF . Here MATH, where MATH is a disk properly embedded in MATH, such that MATH separates the components of MATH, and MATH (hence, MATH is properly embedded in MATH). Then we can take a pair MATH, MATH satisfying the conclusion of REF as in REF . Suppose that MATH is isotopic to MATH, MATH, MATH, MATH, or MATH, say MATH. Then we may regard that MATH is obtained from the NAME splitting MATH of MATH as follows. Let MATH be the component of MATH, which is disjoint from MATH, and MATH, MATH. Let MATH. Note that MATH is an arc properly embedded in MATH. Then MATH. See REF . That is, MATH is obtained from MATH by successively tubing along MATH, and MATH. We can take a pair MATH, MATH satisfying the conclusion of REF , as in REF . |
math/0101148 | Let MATH, and MATH. Then from the pair MATH, MATH we can obtain, as in REF NAME splitting, say MATH, of the product region between MATH and MATH. Since MATH, we see, by CITE, that MATH. Hence by REF , MATH is reducible. Hence, by REF , MATH is reducible. |
math/0101148 | Suppose not. Then, by REF , we see that MATH. Recall that MATH. This implies that MATH. Note that MATH is a NAME splitting of type REF. These show that MATH. However, this is impossible since MATH. |
math/0101158 | See REF . (NAME actually works with NAME groups, but his argument can be used with an arbitrary finite geometric reflection group). |
math/0101158 | Let MATH be an essential realization of MATH. We have MATH (this is a consequence of REF , p. REF). The result then follows from REF . |
math/0101158 | Write MATH, MATH, and MATH. Let MATH. Assume MATH, for some integer MATH. Then we have MATH. Indeed, assume that MATH; we will find a contradiction. According to REF (p REF), MATH is a reduced MATH-decomposition of MATH, and MATH is a reduced MATH-decomposition of MATH. Assume that MATH (MATH). Then MATH and MATH are reduced MATH-decompositions of MATH. But MATH is a left descent of only one of these two words, which gives a contradiction. Now assume that MATH is right. Then MATH is MATH-reduced. Since MATH, the word MATH is also MATH-reduced, and MATH. The word MATH must also be MATH-reduced (view it as a subword of the reduced MATH-decomposition of MATH given above). Since MATH, we have a contradiction. This implies that for each orbit MATH, we have MATH. Since MATH has order MATH, we also have MATH. Using the well-known relation MATH, we obtain the claimed results. |
math/0101158 | Let MATH be a chromatic pair such that MATH. Let MATH. By REF , MATH is of the form MATH, with MATH. CASE: Assume MATH. Then MATH is as required. CASE: If MATH, we note that MATH, so, by modifying the chromatic pair, we are back to the case already discussed. |
math/0101158 | Let MATH be a NAME element in MATH. Let MATH. By the previous lemma, we can find MATH such that MATH and MATH is a NAME system. Thus MATH is a NAME element in the parabolic subgroup generated by MATH. By induction, this proves MATH. The converse is easy. |
math/0101158 | Thanks to REF , it is enough to deal with the case of NAME elements: indeed, if MATH is a parabolic NAME in MATH, it is a NAME element in some MATH, and MATH. We prove the proposition, for NAME elements, by induction on the rank MATH of MATH. It is obvious when MATH is MATH or MATH. Assume MATH, and assume the proposition is known for NAME elements in parabolic subgroups of rank MATH. Let MATH be a NAME element in MATH. Our goal is to prove that MATH forms a single orbit under the action of MATH. Denote by MATH the concatenation of finite sequences; we write MATH for MATH. We have MATH . For all MATH, MATH is a parabolic NAME element. The induction assumption ensures that the action of MATH is transitive on MATH. In particular, since the MATH-action on the last MATH terms is a restriction of the MATH-action, for any MATH, the MATH-orbit of MATH contains the whole MATH. To conclude, it is enough to exhibit a particular element of MATH such that its orbit under the action of MATH contains at least one element in each of the MATH. This may be done as follows: Let MATH be a chromatic pair such that MATH. Write MATH, MATH. Clearly, MATH. A direct computation shows that, for all MATH, the word MATH starts with MATH. Another straightforward computation yields the following: MATH . As a consequence, for all MATH and for all positive integer MATH, MATH is an element in the MATH-orbit of MATH starting by MATH. REF ensures that all elements of MATH are of the form MATH. |
math/0101158 | The monoid MATH is generated by its atoms, which are exactly the elements of MATH. A presentation for MATH, with respect to this generating set, is obtained by taking all relations of the form MATH, where MATH and MATH are reduced MATH-decompositions of the same MATH, with MATH. Let us prove that such a relation MATH is a consequence of the dual braid relations with respect to MATH. By REF , the corresponding MATH is a NAME element in a parabolic subgroup MATH. Of course, MATH and MATH are reduced MATH-decompositions of MATH. By REF , any two reduced MATH-decompositions of MATH are in the same orbit for the braid group action. By the very definition of this braid group action, this ensures that the relation MATH is a consequence of the dual braid relations with respect to MATH and MATH. Since MATH, these ``parabolic" dual braid relations constitute a subset of the set of dual braid relations with respect to MATH and MATH. |
math/0101158 | The implication MATH is obvious, since if MATH and MATH are crossing, no dual braid relation can be applied to MATH. Let us now prove MATH. We set MATH. Without loss of generality, we may assume MATH. Let MATH, MATH and, for MATH, MATH. We have, for all MATH, MATH, and MATH is a sequence of dual braid relations. Assume MATH is even. By multiple use of the above relations, we have MATH . Assume MATH is odd. We have MATH . |
math/0101158 | It is enough to prove the result when MATH is irreducible. The implication MATH is obvious. Let MATH, let MATH such that MATH in MATH. To obtain the converse implication, we have to prove that MATH in MATH. It suffices to deal with the case MATH. From now on, we assume MATH. By REF , this implies that MATH is a multiple of MATH (of course, if MATH is odd, then MATH must be a multiple of MATH). According to REF , p REF, we have MATH when MATH is even, and MATH with no restriction on MATH. In any case, we have MATH. The conjugation by MATH is a diagram automorphism of the classical diagram for MATH; the relation MATH follows immediately. Since the natural map MATH is MATH-equivariant (for the respective conjugacy actions by powers of MATH and MATH), the description of MATH given in REF and the definition of MATH prove that MATH is a bijection. |
math/0101158 | An obvious induction reduces the lemma to the case MATH. Let MATH. Let MATH. Let MATH such that MATH REF . The relation MATH is a consequence of the dual braid relations. Thus, in MATH, we have MATH. |
math/0101158 | We only have a case-by-case proof. It is enough to deal with the irreducible case. The exceptional types are dealt with by computer, using the package CHEVIE of GAP. The dihedral case is obvious. For the infinite families MATH, MATH and MATH, see REF. Note however that the geometric interpretation of the next section allows a reformulation of this fact which, we hope, could lead to a general proof. |
math/0101158 | Here again, we only have a case-by-case proof (the reduction to the exceptional case is obvious). We hope that the geometric approach of the next section will eventually provide a general proof. The exceptional types are dealt with by computer, using GAP. The type MATH is trivial (the poset has height MATH, with only one maximal element). For type MATH, MATH and MATH, see REF. Note that, using REF , we only have to check that pairs of reflections have a right lcm. |
math/0101158 | NAME monoids satisfy the embedding property: the natural monoid morphism MATH is injective. We conclude using REF . |
math/0101158 | CASE: Let MATH. Since all hyperplanes have real equations, we have for all MATH . The chamber MATH is separated from MATH by only one wall, MATH. Thus the segment MATH intersects only one real hyperplane, MATH. This proves that MATH. The identity MATH is easy: choose a path representing MATH and use REF . If MATH, then for any MATH, we have MATH . If MATH, then the path MATH, which exits the chamber MATH, must cross at least a wall of MATH, and MATH. The second part of the statement follows immediately. |
math/0101158 | REF . A priori, the isomorphism MATH only depends on the homotopy class of MATH. To prove that it does not depend on MATH, it is enough to check it when MATH, that is, to prove that the conjugacy action of MATH on MATH is trivial. But MATH is cyclic, generated by an element which is well-known to be central in MATH (see for example REF ). The transitivity of the system of isomorphisms follows from the independence of the choice of MATH. CASE: The visibility condition is invariant by scalar multiplication. The rest is an easy computation. |
math/0101158 | By REF , if MATH, then ``MATH is generated (as a group) by the local generators at MATH" is equivalent ``MATH is generated (as a group) by the local generators at MATH". Since MATH is finite, it is always possible to find MATH such that MATH or, in other words, MATH is in a chamber MATH. By REF , the set of local generators at MATH contain a classical NAME generating subset for MATH. |
math/0101158 | By assumption, we have MATH, where MATH is a primitive MATH-th root of unity. Write MATH. Applying REF to the path MATH, we obtain an isomorphism MATH such that, whenever MATH, MATH . In particular, MATH restricts to an automorphism of MATH. The order of MATH is the same as the order of the action MATH on MATH; this action is isomorphic to the conjugation action of MATH on MATH. By REF , the set MATH generates MATH. Thus the order of MATH is the smallest MATH such that MATH is central in MATH. If MATH, then MATH. Otherwise, the only non-trivial central element is the (unique) regular element of order MATH. The conclusion follows. |
math/0101158 | Since MATH has complex dimension MATH, and since the claimed properties are invariant under multiplication of MATH by a non-zero complex number, we only have to prove the proposition for a particular MATH. It is easy to build one from the information provided by NAME. Let us summarize various results from pages REF. According to NAME, it is possible to find MATH such that: CASE: For any left wall MATH and any right wall MATH, we have MATH . CASE: The MATH-plane MATH generated by MATH and MATH is stable by MATH and MATH. CASE: The element MATH (respectively, MATH) acts on MATH as a reflection with hyperplane MATH (respectively, MATH). Note that there is a unique (up to scalar multiplication) scalar product on MATH invariant by MATH and MATH and therefore there is a well-defined notion of angle in MATH. We have MATH. Since the conditions specifying MATH and MATH are stable by multiplication by an element of MATH, we may assume that both of their norms are MATH. The vector MATH is such that MATH is an orthonormal basis. A direct computation shows that the element MATH defined by MATH is a MATH - eigenvector for MATH. Assume MATH is a left wall and MATH is a right wall. We have MATH . This proves REF . Let MATH. For any MATH, the linear form MATH has kernel MATH, thus MATH, with MATH. In particular, for any MATH, we have MATH . By REF , MATH is the closure of MATH for the conjugacy action of MATH; rephrased in terms of hyperplanes, this says that MATH is the closure of MATH for the multiplication action of MATH. Using REF , we see that MATH is the regular MATH-gon containing MATH. Since all MATH are linear combinations of the MATH with real positive coefficients, MATH must consist of the MATH consecutive points from MATH to MATH. |
math/0101158 | REF follows from REF by an immediate computation, done with the technique described in Subsection REF; we leave the details to the reader. From REF and the proof of REF , it follows that the automorphism MATH from REF is the conjugation by MATH. REF follows. |
math/0101158 | Set MATH. By REF , pp REF, we have MATH. Using the commutation relations within MATH, we see that, when MATH, MATH. When MATH, the commutation relations within MATH yield MATH . We have proved MATH . For all MATH, we have MATH. From this and the above description of MATH, we deduce MATH. From the NAME exercice quoted above, we also get MATH. Since MATH, we have MATH. |
math/0101158 | This can be checked, case-by-case, on the list given in CITE . |
math/0101158 | Let MATH be the right lcm of the elements of MATH. Since MATH, one has MATH. More precisely, for all MATH, one has MATH. Let MATH be the element uniquely defined by MATH. As MATH is non-empty, one has MATH, and NAME 's proof's induction hypothesis implies that MATH is contractible. The map MATH is increasing and induces an isomorphism between MATH and an initial segment of MATH. The map MATH is increasing and one has, for all MATH and all MATH, MATH and one concludes as in NAME 's proof. |
math/0101158 | Since the defining relations are valid in MATH, the monoid MATH comes equipped with a natural morphism MATH. If for example MATH, then MATH in MATH. Since MATH is cancellative, MATH. To obtain the first claim, observe that the composition of the natural pre-monoid morphism MATH with MATH is the restriction of identity map of MATH. |
math/0101158 | The pre-monoid MATH is unitary REF ; the length function MATH satisfies REF . With the assumption MATH, MATH is the set of atoms of MATH. The existence of left and right lcm's for elements of MATH are exactly REF : let MATH, MATH, such that MATH, MATH and MATH; then MATH, MATH, so MATH and MATH. REF is MATH-cancellativity, which we have proved in REF . The element MATH is a common multiple of all elements of MATH. We conclude using REF. |
math/0101159 | First we show that MATH is closed. Let MATH be closed. We need to show that MATH is closed. Let MATH be a sequence in MATH converging to MATH. Let MATH be the face containing MATH. After passing to a subsequence we may assume that all MATH are in the same face, say MATH. Then MATH, so MATH and MATH. Since MATH is compact, MATH is closed, so MATH. Hence MATH that is, MATH is closed. The fact that MATH has compact fibres now implies that it is proper. The stated properties of MATH follow easily, and the local closedness of MATH follows from the observation that MATH is equal to MATH. |
math/0101159 | Let MATH be the face containing MATH. Assume first that all points in MATH are of the same orbit type for the action of MATH, so that MATH is a smooth symplectic manifold. By reduction in stages, it is naturally symplectomorphic to the iterated quotient MATH . Since MATH and the restriction of MATH to MATH is the moment map for the MATH-action, it is clear that REF is naturally symplectomorphic to MATH. If MATH consists of more than one stratum, the same argument, using stratified reduction in stages (see CITE), shows that the quotient map MATH induces a homeomorphism MATH, which maps strata symplectically onto strata. |
math/0101159 | The second equality is immediate from the first. Because of MATH-invariance, the first equality need only be checked at points of the form MATH. Let MATH. Then MATH by the definition of MATH. (We identify MATH with the annihilator of MATH in MATH, so that MATH is well-defined.) Moreover, MATH which is equal to MATH because MATH and MATH is equal to the projection of MATH onto MATH. |
math/0101159 | The map MATH sends MATH to the Lagrangian MATH, so MATH is a symplectic embedding. The moment map for the diagonal MATH-action on MATH is given by MATH, so MATH maps MATH into MATH. One checks readily that the induced map MATH is a diffeomorphism. It is a symplectomorphism because MATH is a symplectic embedding. Moreover, MATH and MATH, so MATH is MATH-equivariant and intertwines the MATH-moment maps on MATH and MATH. This proves REF is proved in exactly the same way. |
math/0101159 | The MATH-moment map on MATH is given by MATH, so MATH maps MATH into MATH. Moreover, MATH implies MATH, so MATH induces a continuous map MATH. Upon quotienting by MATH we obtain the map MATH. As in the proof of REF one checks that MATH is a homeomorphism which is MATH-equivariant and intertwines the MATH-moment maps on MATH and MATH. Note that MATH restricts to a map MATH . This is none other than the map induced, upon reduction with respect to MATH, by the map MATH defined in REF. Because MATH is an isomorphism of Hamiltonian MATH-manifolds, it preserves the MATH-orbit types and therefore REF maps strata onto strata and is a symplectomorphism on each stratum. |
math/0101159 | By REF and hence MATH so REF holds. It follows that MATH, so any MATH is a regular value and MATH is transverse to MATH. It is now easy to see that MATH is a contact form on MATH; see for example, CITE. Furthermore, on MATH we have MATH and MATH, so MATH is the NAME vector field, which proves REF. |
math/0101159 | The cone spanned by the positive coroots is the dual of the cone MATH. Since MATH is in its interior and MATH is semisimple, the linear function MATH is proper on MATH. It follows that MATH is proper on MATH. The other assertions in REF are now obvious. To prove REF, apply REF to each stratum MATH. REF follows from the observation that the simplex MATH is a global section of the MATH-action MATH on the punctured NAME chamber MATH. The equality MATH is readily checked on the tangent space MATH at any MATH and therefore holds globally by homogeneity. To prove REF, observe that MATH defines a map MATH which sends MATH to MATH and therefore descends to a homeomorphism MATH. |
math/0101159 | Let us denote the element MATH of MATH by MATH and its image in the orbit space MATH by MATH. To prove REF we need to show that for every MATH in the zero level set of the MATH-moment map, the infinitesimal stabilizers MATH and MATH are equal. There exists an infinitesimal character MATH such that for all MATH . Let MATH be the face containing MATH. Then REF boils down to MATH . Differentiating at MATH yields MATH . Since MATH is in the zero fibre of the MATH-moment map, MATH . Combined with REF this gives MATH . Applying MATH to both sides of REF we then obtain MATH because MATH and MATH. Since MATH on MATH, MATH and hence MATH. From REF we conclude that MATH, and therefore MATH is contained in MATH. The reverse inclusion is obvious. This proves REF follows immediately from REF. |
math/0101159 | REF follows immediately from MATH . From REF it follows that MATH is open in MATH. For MATH, MATH, so for MATH, MATH, MATH is equivalent to MATH. Hence MATH. |
math/0101159 | Everything has been proved except the frontier condition. Let MATH be the stratum of MATH and suppose that MATH is in the closure of a stratum MATH. Put MATH. Then MATH and MATH is closed in MATH. In the local model MATH around MATH, the stratum MATH is of the form MATH, whereas MATH is of the form MATH for some MATH. Here the notation is as in REF. Every stratum MATH in MATH is stable under the MATH-action and therefore has the vertex MATH as a limit point. It follows that MATH, where MATH is an appropriate open neighbourhood of MATH in MATH. Hence MATH is open, and therefore MATH. We have shown that MATH. |
math/0101159 | Let MATH denote the equivalence class of MATH in the product of projective spaces MATH. Then MATH and MATH . (Here we identify MATH with the character of MATH that it exponentiates to.) Let MATH and let MATH be the unipotent radical of MATH. The corresponding NAME algebras are MATH and we have NAME decompositions MATH . Since characters of MATH vanish on MATH, we have MATH with MATH. Thus both MATH and MATH are semisimple groups with root system MATH. To finish the proof it suffices to show that MATH is connected. The elements of MATH being MATH-invariant, MATH is stable under conjugation by the connected group MATH. Therefore we need only show that the intersection of MATH with the maximal torus MATH is connected. Observe that MATH . If MATH is any top-dimensional polyhedral cone in a vector space MATH and MATH is a lattice in MATH, then the set MATH contains a MATH-basis for MATH. Therefore MATH contains a MATH-basis for the lattice MATH. This basis can be extended to a basis of MATH, and this implies that MATH is connected. |
math/0101159 | It is plain from REF that MATH is continuous and closed. Furthermore, by REF the stabilizer of MATH for the MATH-action is equal to MATH, where MATH is the face containing MATH. This implies REF. It is clear from REF that MATH is smooth on MATH for every MATH, and therefore MATH restricted to MATH is a smooth embedding. We check that it preserves the symplectic form by showing that MATH, where MATH is the one-form on MATH considered in REF . Because MATH is MATH-equivariant we need only show this at the points MATH, where MATH denotes the coset MATH. By REF , MATH for all MATH and MATH. On the other hand, MATH . Here MATH . Together with REF this yields MATH where we have used MATH. Comparing with REF we conclude that MATH. This proves REF is a consequence of the NAME decomposition. Put MATH, MATH and, for each face MATH, MATH and MATH. Then MATH and MATH. Recall also that MATH. Here MATH and MATH are as in REF. Let MATH so that MATH. Writing MATH and MATH we find MATH and, using REF, MATH . Hence MATH as smooth MATH-manifolds. To finish the proof it suffices to show that MATH is equal to the MATH-orbit through MATH for all MATH. For MATH put MATH where the sum is over all MATH such that MATH. For each face MATH, MATH defines a diffeomorphism from MATH to MATH, and therefore MATH is also a diffeomorphism. Moreover, for MATH . Hence MATH. |
math/0101159 | Let MATH and let MATH be the orthogonal projection. Then MATH is MATH-equivariant and MATH is the closure of MATH. For any face MATH, MATH, where MATH is the interior of MATH. Hence MATH . Consider the subsets of MATH given by MATH . Then MATH is MATH-stable and REF implies that MATH. Hence MATH is NAME. Similarly, MATH is equal to MATH, and it is a MATH-stable NAME neighbourhood of MATH. If MATH and MATH are in MATH, then MATH, so that MATH by REF . It follows that the multiplication map MATH induces a MATH-equivariant isomorphism MATH. The affine MATH-variety MATH is the union of all orbits MATH with MATH. The groups MATH are exactly the parabolic subgroups of MATH that contain MATH, and therefore it follows from the corollary to CITE that MATH. |
math/0101159 | It follows from REF that MATH is smooth at MATH if and only if MATH is smooth at the vertex. CITE proved that this is the case if and only if MATH and MATH. This proves REF. Likewise, MATH has an orbifold singularity at MATH if and only if MATH has an orbifold singularity at the vertex. This is the case precisely when MATH has an orbifold singularity at the vertex, where MATH is the universal cover of MATH and MATH is the preimage of MATH in MATH. Since MATH is simply connected and MATH is connected, MATH is simply connected. The complement of MATH inside MATH has complex codimension at least MATH, and therefore MATH is also simply connected. Moreover, MATH is a union of MATH-orbits and therefore homogeneous, that is, stable under dilations on MATH. We conclude that the vertex has a basis of open neighbourhoods MATH such that the nonsingular part MATH is simply connected. A result of Prill CITE implies that if a complex analytic space MATH has a finite-quotient singularity at MATH and if MATH has a basis of neighbourhoods MATH such that MATH is simply connected, then MATH is smooth at MATH. We conclude that MATH has an orbifold singularity at the vertex if and only if it is smooth. Therefore REF follows from REF. |
math/0101159 | Embed MATH into a finite-dimensional unitary MATH-module MATH as in REF . The MATH-action on MATH extends uniquely to an algebraic MATH-action, which preserves the stratification of MATH. To examine MATH let us assume that MATH is semisimple and simply connected. (This is justified by REF .) Then MATH is a closed MATH-stable affine subvariety of MATH, and REF give us isomorphisms MATH . A well-known result of CITE (see also CITE) says that the symplectic quotient on the right is homeomorphic to the invariant-theoretic quotient of MATH by MATH, that is, the affine variety (associated to the scheme) MATH. The homeomorphism is induced by the inclusion of the zero fibre of the moment map for the MATH-action on MATH and is therefore MATH-equivariant. According to CITE the ring MATH is isomorphic to MATH. |
math/0101159 | That the strata are NAME follows from REF, which shows that every stratum is a symplectic quotient of a NAME manifold. The other assertions follow from REF. |
math/0101159 | The first assertion follows immediately from CITE and REF . Taking the MATH-invariant parts of both sides we get MATH as virtual characters of MATH, where MATH. By construction the moment polytope of MATH lies within the fundamental chamber MATH, so, by the quantization commutes with reduction theorem CITE, the weights occurring in MATH are all dominant. Let MATH be such a weight and MATH the representation of MATH with weight MATH. By the NAME Theorem MATH, the irreducible representation with highest weight MATH. Hence MATH, since only the highest weight vector MATH is invariant under MATH. We conclude that MATH. Together with REF this proves the second assertion. |
math/0101159 | Let MATH be the NAME space MATH, which is equipped with a Hamiltonian MATH-action. Denote the MATH-moment map on MATH by MATH. We will show that for all sufficiently small MATH there exists a bimeromorphic map MATH. (Properness and surjectivity are then immediate from the compactness of MATH and the irreducibility of MATH.) This is well-known in the algebraic category. Let us briefly indicate how the argument carries over to the analytic category thanks to results of CITE. Let MATH. The set of MATH-semistable points is MATH . It is open and dense, if nonempty CITE. Two points in MATH are equivalent under the MATH-action if their orbit closures intersect in MATH. For every MATH there is a unique MATH such that MATH is closed in MATH and MATH is in the closure of MATH. This implies that the inclusion MATH induces a homeomorphism MATH. (These assertions follow from the holomorphic slice theorem, CITE or CITE.) A MATH-semistable point is MATH-stable if MATH is closed in MATH and MATH is finite. The set of stable points is denoted MATH. It too is open and dense, if nonempty. A point MATH is stable if and only if MATH is finite. (These facts follow also from the holomorphic slice theorem.) The last fact we need is a generalization of NAME 's result CITE that for every MATH the image MATH is the convex hull in MATH of the MATH-fixed points contained in MATH. Furthermore, MATH is equal to the full image MATH for all MATH in an open dense subset MATH. The convexity is proved in CITE. For the set MATH we can take MATH, where MATH, MATH, MATH are the vertices of MATH. Take MATH so small that MATH is contained in MATH. Then MATH, and this inclusion induces an analytic map MATH. To see that this map is bimeromorphic, observe that the stable set MATH is nonempty, since MATH is a regular value of the MATH-moment map on the manifold MATH. Let MATH. Then the image of MATH in MATH is open and dense and for MATH we have MATH, that is, MATH. Thus we obtain an analytic map MATH which inverts the previously defined map MATH over an open dense set. |
math/0101159 | The image of the map MATH is the subvariety MATH of MATH and we can therefore regard it as a proper morphism MATH. The MATH-orbits in MATH are in natural one-to-one correspondence with the MATH-orbits in MATH, which are identical to the MATH-orbits in MATH. Each MATH-orbit in MATH passes through a unique point of the form MATH, so each MATH-orbit in MATH passes through a unique point of the form MATH. (Here points in MATH are written as MATH with MATH and MATH.) The stabilizer of MATH for the MATH-action is MATH, where the second equality follows from REF . Thus the fibre MATH is the flag variety MATH. In particular, MATH contains a NAME orbit of type MATH, namely the orbit through MATH, where MATH is the top face of MATH. Hence MATH is birational, which proves REF. If MATH is integral, then MATH is integral on MATH. Since MATH is exact, this implies that MATH is integral. Furthermore, being a sum of pullbacks of two NAME forms, MATH is positive semidefinite. To prove that it is NAME, it is therefore enough to show that it is nondegenerate. We shall do this by showing that MATH pulls back to the symplectic form on MATH under a suitable diffeomorphism. The principal face MATH of MATH is the top face of MATH and its principal cut (for the right MATH-action) is MATH. We noted in REF that the toric manifold associated to the polyhedral cone MATH is the symplectic vector space MATH, so by REF the partial desingularization of MATH is MATH. To see that this space is actually a manifold rather than an orbifold, observe that the moment map for the MATH-action on the product MATH is given by MATH, where MATH is the MATH-moment map on MATH. The map MATH which sends MATH to MATH is a MATH-equivariant diffeomorphism onto MATH. It therefore descends to a MATH-equivariant diffeomorphism MATH which shows that MATH is smooth. Moreover, the inclusion map MATH induces a diffeomorphism MATH. Composing with the inverse of the map REF we obtain a diffeomorphism MATH . To finish the proof of REF we must show that MATH is the symplectic form on MATH. Recall that the symplectic cut MATH contains a copy of MATH as an open dense submanifold. The symplectic form on this subset is the form MATH of REF and the embedding MATH is given by MATH . Here MATH is any section of the map MATH, such as for example MATH . Let us denote the open embedding MATH by MATH. We need to show that MATH . It suffices to check this identity at points of the form MATH with MATH. For MATH and MATH in MATH one readily checks that MATH . On the other hand, MATH. Now MATH, where MATH denotes the coset of MATH, so MATH . A computation as in the proof of REF yields MATH where MATH . Combining this with REF gives MATH which together with REF proves REF. |
math/0101162 | First assume this canonical model category exists. Then since Reedy cofibrations are cofibrations and level equivalences are weak equivalences, a Reedy cylinder object (CITE, CITE) for a Reedy cofibrant object is also a cylinder object in the canonical model category. This shows using CITE that for MATH . Reedy cofibrant and MATH homotopically constant and Reedy fibrant the homotopy classes of maps coincide in the homotopy category of the Reedy model category and the homotopy category of the canonical model category, MATH. Since level equivalences are weak equivalences in both cases this means that for arbitrary MATH and homotopically constant MATH, MATH. A map MATH is a weak equivalence in the canonical model category if and only if for each homotopically constant MATH, MATH is a bijection. Or, equivalently, MATH is a bijection. Since MATH is level equivalent to MATH, this is equivalent to MATH being a bijection for each MATH in MATH. So the weak equivalences are the realization weak equivalences. Since the cofibrations and weak equivalences are determined, the fibrations are determined by the right lifting property. Hence there is at most one model category on MATH with the above properties. |
math/0101162 | If the Realization Axiom holds, then REF gives the characterization of the fibrations as equifibered Reedy fibrations. For the other implication, an equifibered Reedy fibration that is also a realization weak equivalence is a trivial fibration in the canonical model structure by assumption. But a trivial fibration has the right lifting property with respect to the Reedy cofibrations, and hence is a level equivalence. Thus the Realization Axiom holds. |
math/0101162 | For the second statement, note that MATH preserves cofibrations and trivial cofibrations. By adjointness MATH preserves fibrations and trivial fibrations, and hence also weak equivalences between fibrant objects. But, if MATH is a weak equivalence then MATH is a level equivalence since fibrant objects are homotopically constant. To show that the adjoint functor pair MATH induces a NAME equivalence, we use the criterion in CITE since MATH preserves and detects weak equivalences between fibrant objects. So we must show for any cofibrant object MATH in MATH that MATH is a weak equivalence where MATH is a fibrant replacement of MATH in MATH. Take MATH to be the Reedy fibrant replacement of MATH, it is homotopically constant and hence also a fibrant replacement in the canonical model category. Then MATH and MATH are level equivalent so MATH is indeed a weak equivalence in MATH. Since fibrations have the right lifting property with respect to level trivial Reedy cofibrations, a fibration is a Reedy fibration. So we assume MATH is a Reedy fibration between two fibrant objects and show that it is a fibration. Factor MATH with MATH a trivial cofibration and MATH a fibration. Then MATH is a weak equivalence between fibrant objects, hence a level equivalence by part two. Thus MATH is a trivial Reedy cofibration so it has the left lifting property with respect to MATH. This implies that MATH is a retract of MATH, and hence a fibration in MATH. |
math/0101162 | In the proposition take MATH, the simplicial indexing category. An equifibered Reedy fibration MATH, viewed as a map of MATH-diagrams, satisfies the hypotheses of REF , and MATH is a realization weak equivalence precisely when MATH is a weak equivalence by REF . Therefore, for such MATH and for every MATH the map MATH is a weak equivalence, that is, MATH is a level weak equivalence. |
math/0101162 | Let MATH be an equifibered Reedy fibration and a realization weak equivalence in MATH. Since MATH is a right proper model category, the condition for an equifibered Reedy fibration is invariant under level equivalences. By definition level equivalences are realization equivalences. Hence, we can assume that MATH and MATH are Reedy cofibrant. For simplicial model categories, the realization, MATH is weakly equivalent to the homotopy colimit on Reedy cofibrant objects. This follows from the generalization of CITE to general simplicial model categories, see CITE. So MATH is a weak equivalence in MATH by REF , since MATH is a realization weak equivalence. By REF , this means that MATH and MATH are weak equivalences. Thus, MATH is a realization weak equivalence of bisimplicial sets, by REF and the fact that all bisimplicial sets are Reedy cofibrant. Since MATH preserves fibrations and weak equivalences, it preserves homotopy pullback squares, and hence MATH preserves equifibered Reedy fibrations. So, by REF , MATH is a level equivalence. Thus MATH is a level equivalence. |
math/0101162 | First note that since MATH is stable the Reedy model category on MATH is also stable. This follows since Reedy cofibrations and fibrations are level cofibrations and fibrations and colimits and limits are taken levelwise. So the suspension and loop functors in the Reedy model category are level equivalent to the levelwise suspension and loop in MATH. Now given a realization weak equivalence MATH in MATH that is an equifibered Reedy fibration, we must show that MATH is a level equivalence. Since MATH is right proper, the level homotopy fiber of MATH is weakly equivalent to MATH, the fiber of MATH. In a stable model category fiber sequences induce long exact sequences after applying MATH. So MATH is trivial for any MATH in MATH. Since MATH is equifibered, MATH is homotopically constant and hence level equivalent to MATH. Thus MATH is trivial in MATH. This implies that MATH is level trivial, and hence that MATH is a level equivalence since MATH is stable. |
math/0101162 | Apply REF to both MATH and MATH. Then they are both simplicially NAME equivalent to MATH. |
math/0101162 | To show that MATH is a simplicial functor we show that MATH is isomorphic to MATH. Here MATH and MATH are the simplicial actions in the respective categories. These are not to be confused with the coends, see CITE, MATH and MATH which follow. Since the left adjoint MATH is a strong simplicial functor, that is, the natural transformation is an isomorphism, it follows that the right adjoint MATH is also a simplicial functor. Let MATH be the functor such that MATH, the simplicial MATH-simplex. Then MATH is isomorphic to the coend MATH and for any simplicial set MATH, MATH. Because MATH commutes with colimits, MATH. Here MATH. The functor MATH is the left NAME extension of MATH across the diagonal functor MATH. So MATH. But MATH is the functor describing MATH, so this gives an isomorphism of the last step with MATH. This produces the required isomorphism. |
math/0101162 | First note that MATH where MATH denotes the boundary of the simplicial MATH-simplex. So if MATH is a Reedy (trivial) fibration then MATH is a Reedy (trivial) fibration because the induced map MATH is equivalent to the map MATH which is a (trivial) fibration by the adjoint form of REF, see SMREF. The trivial fibrations in MATH are the Reedy trivial fibrations. Since the fibrations in MATH between fibrant objects are Reedy fibrations by REF , this shows that MATH preserves trivial fibrations and fibrations between fibrant objects. Hence, by CITE, MATH also preserves fibrations. By adjointness, MATH preserves cofibrations and trivial cofibrations. Since MATH preserves trivial cofibrations it preserves weak equivalences between cofibrant objects. It also detects weak equivalences between cofibrant objects by REF since MATH is weakly equivalent to the homotopy colimit on Reedy cofibrant objects, by CITE and CITE. Hence by the dual of the criterion for NAME equivalences in CITE, we only need to check that for fibrant objects MATH in MATH, MATH is a weak equivalence where MATH is a trivial fibration from a cofibrant object in MATH. By the simplicial model category structure on MATH, MATH is homotopically constant. Since MATH is level equivalent to MATH, it is also homotopically constant. Consider the following commuting square MATH . The left vertical map is a weak equivalence since MATH. The top map is a weak equivalence since MATH is homotopically constant. Finally, the bottom composite is the identity map. Hence the right hand map is a weak equivalence as required. |
math/0101162 | First MATH is a simplicial functor. The necessary natural transformation, MATH is given on each level by the canonical maps MATH. Since MATH preserves fibrations, trivial fibrations, and limits, MATH preserves Reedy fibrations and Reedy trivial fibrations. So MATH preserves trivial fibrations and fibrations between fibrant objects. By CITE this implies MATH also preserves fibrations. Hence MATH are a NAME adjoint pair. The last statement follows from REF and the two out of three property for equivalences of categories, since NAME equivalences are NAME adjoint functors that induce equivalences of homotopy categories CITE. |
math/0101162 | By REF , MATH and MATH are simplicially NAME equivalent respectively to MATH and MATH. By REF , the NAME equivalence between MATH and MATH can be lifted to a simplicial NAME equivalence between MATH and MATH. |
math/0101162 | MATH is a simplicial functor because each of its composites is simplicial by REF , and REF. The first step in the zig-zag between MATH and MATH uses the natural transformation MATH. This induces MATH . Note that for MATH fibrant MATH is a level equivalence between level fibrant objects by the simplicial model category structure on MATH. Since MATH preserves trivial cofibrations by REF , MATH preserves weak equivalences between cofibrant objects. So, since MATH preserves weak equivalences between fibrant objects, MATH is an equivalence for MATH fibrant. To relate this to MATH, note that MATH. Since MATH is a level equivalence, MATH is homotopically constant. Thus, MATH is a level equivalence between cofibrant objects. Hence MATH is also a weak equivalence for any MATH. MATH is an isomorphism. Since MATH is a level equivalence, MATH is also an equivalence. Combining this with the first step finishes the proof. |
math/0101162 | Given MATH in MATH we construct a simplicial functorial factorization, MATH, as a cofibration followed by a trivial fibration. The other factorization is similar. Let MATH be a set of generating cofibrations in MATH. Define the first stage, MATH, as the pushout in the following square. MATH . By CITE, any object that is small with respect to the regular MATH-cofibrations is small with respect to all cofibrations. So each MATH is small relative to the cofibrations. Let MATH be the regular cardinal such that each MATH is MATH-small with respect to the cofibrations. Let MATH and for any limit ordinal MATH let MATH. Then we claim that MATH is a cofibrant replacement functor which is also a simplicial functor. We need to show that MATH is a cofibration and that MATH is a trivial fibration. Since MATH is a simplicial model category the left map in the square above is a cofibration. Since pushouts and colimits preserve cofibrations this shows that MATH is a cofibration. To show that MATH is a trivial fibration we need to show that it has the right lifting property with respect to any map MATH. Because MATH is MATH-small with respect to cofibrations, the map MATH factors through some stage, MATH. Then, by construction, there is a lift MATH. We now show that MATH is simplicial. The colimit of a diagram of simplicial functors is again a simplicial functor. Since the composition of simplicial functors is again simplicial, we only need to show that MATH is a simplicial functor. But MATH itself is a colimit of functors which are simplicial, so we are done. |
math/0101162 | We need to consider the matching maps of MATH, that is MATH in MATH by REF . Since MATH is a cofibration in MATH, it is enough to show that MATH is a (trivial) fibration in MATH. In fact it is enough to show this for each MATH since they generate the cofibrations in MATH by CITE. But MATH is a (trivial) fibration by REF since MATH is a Reedy (trivial) fibration. |
math/0101162 | The first two statements follow by adjointness from REF . For all three statements, see also CITE and compare with CITE. |
math/0101162 | We prove the statement for MATH, the statement for MATH follows similarly. A finite colimit of small objects is small since finite limits commute with small filtered colimits, CITE. The domains of MATH can be built by finite colimits from objects MATH for MATH a domain or range of a map in MATH or MATH. Since MATH and MATH is small relative to regular MATH-cofibrations by REF , MATH is small relative to maps in MATH that are regular MATH-cofibrations on each level. But each level of a regular MATH-cofibration is a regular MATH-cofibration. This is because each level of a map in MATH is just a direct sum of copies of maps in MATH or identity maps. Identity maps and coproducts of regular cofibrations are regular cofibrations. So each level of each map in MATH is a regular MATH-cofibration. Hence this is also true of the regular MATH-cofibrations. |
math/0101162 | A Reedy fibration is a map MATH whose matching maps are fibrations. These matching maps are MATH with MATH by REF . That is, MATH has the right lifting property with respect to each map in MATH. By adjointness, this is equivalent to MATH having the right lifting property with respect to the maps in MATH. Given a Reedy fibration MATH, then MATH is a fibration by REF . So a Reedy fibration MATH is equifibered if and only if MATH is a trivial fibration. By adjunction MATH is a trivial fibration if and only if MATH has the right lifting property with respect to MATH. So MATH is an equifibered Reedy fibration if and only if MATH has the right lifting property with respect to MATH. The last statement of the proposition follows since MATH is an equifibered Reedy fibration if and only if MATH is Reedy fibrant and for each MATH and MATH the map MATH is a trivial fibration. |
math/0101162 | Much as in the previous proof, a map MATH is a Reedy trivial fibration if the matching maps MATH are trivial fibrations. That is MATH has the right lifting property with respect to each map in MATH. By adjointness, this is equivalent to MATH having the right lifting property with respect to the maps in MATH. By the Realization REF , an equifibered Reedy fibration that is also a realization weak equivalence is a level equivalence, and hence a Reedy trivial fibration. Conversely, for MATH a Reedy trivial fibration, the maps MATH are trivial fibrations. Since MATH factors as MATH and the second map here is the pull back of a trivial fibration, the map MATH is a weak equivalence. So a Reedy trivial fibration is equifibered. Then, since level equivalences are realization weak equivalences, this shows that a Reedy trivial fibration is a realization trivial fibration, that is, an equifibered Reedy fibration that is also a realization weak equivalence. |
math/0101162 | A MATH-cofibration has the left lifting property with respect to the MATH-injective maps, the equifibered Reedy fibrations. Since any Reedy fibration that is also a level equivalence is equifibered, a MATH-cofibration has the left lifting property with respect to the Reedy trivial fibrations. Hence a MATH-cofibration is a Reedy cofibration. Each MATH-cofibration is a retract of a directed colimit of pushouts of maps in MATH by REF . The maps in MATH are level equivalences, hence the maps built from MATH are Reedy trivial cofibrations. These level equivalences are realization weak equivalences. So we only need to consider MATH-cofibrations. Since the maps in MATH are trivial cofibrations of simplicial sets, they are MATH-cofibrations where MATH is the set of inclusions of the horns into simplices. Hence MATH-cofibrations are MATH-cofibrations. Below, in REF , we show that any MATH-cofibration is a realization weak equivalence. |
math/0101162 | MATH is a fibration, by REF . Since MATH, MATH is the map MATH, which is a trivial fibration since MATH is a homotopically constant, Reedy fibrant object. This proves the lemma for MATH. We proceed by induction. MATH is the pullback of a punctured MATH-cube where each arrow is of the form MATH, that is, MATH for MATH. These maps are fibrations by REF and they are weak equivalences because MATH is homotopically constant. By induction the map from the object at the puncture of each contained punctured MATH-cube, for MATH, to the pullback is a trivial fibration. For any such punctured MATH-cube, the added maps from the pullback are trivial fibrations. That is, the maps from the pullback, MATH, to each MATH are trivial fibrations. Since each MATH factors as MATH, this proves the lemma holds for MATH by the two out of three property for weak equivalences. |
math/0101162 | First note that by an adjoint of REF , which is verified for the Reedy model category in REF , MATH is Reedy fibrant. Hence by REF , MATH is MATH-injective and we only need to show that MATH is MATH-injective to finish the proof. Here we say ``MATH has the lifting property," as short hand for MATH has the left lifting property with respect to MATH. This also extends to sets of maps. By REF , MATH has the lifting property for MATH in MATH, MATH in MATH, and MATH any homotopically constant, Reedy fibrant object. Let MATH be the class of maps MATH for such MATH. Then, by adjointness MATH has the lifting property. But then pushouts, colimits and retracts of maps in MATH also have the left lifting property with respect to MATH. That is, MATH-cofibrations, MATH has the lifting property. For MATH a cofibration and MATH a trivial cofibration of simplicial sets , the pushout product MATH is a MATH-cofibration. So MATH is a MATH-cofibration for MATH in MATH. Hence MATH has the lifting property. Consider the cofibration MATH. By adjointness this shows that MATH has the lifting property. Hence MATH is MATH-injective. |
math/0101162 | Our first claim is that MATH is naturally isomorphic to MATH for MATH . Reedy cofibrant and MATH homotopically constant and Reedy fibrant. Indeed the maps MATH produce MATH as a path object for MATH. Here MATH is a level equivalence by REF since it is a map between homotopically constant objects whose zeroth level is given by the equivalence MATH and REF shows that MATH is a Reedy fibration. This implies the claim. Since MATH is a realization weak equivalence if and only if its cofibrant replacement is, we can restrict to the case when MATH is its own cofibrant replacement. Then requiring that MATH is a weak equivalence for all homotopically constant, Reedy fibrant objects MATH is equivalent to requiring that for all simplicial sets MATH, MATH is a bijection for all such MATH. By REF and the above, this is equivalent to MATH being a bijection for all such MATH and MATH. As MATH runs through all homotopically constant, Reedy fibrant objects and MATH runs through all simplicial sets, MATH runs through all fibrant objects in MATH. Since MATH is a level equivalence, this is equivalent to MATH being a bijection for all MATH in MATH. |
math/0101162 | By the proof above of REF , MATH has the lifting property for MATH homotopically constant and Reedy fibrant. Then by adjointness, MATH also has the lifting property for any such MATH. That is, any map in MATH is a trivial fibration. Since the maps in MATH are assumed to be between cofibrant objects, the maps in MATH are Reedy cofibrations between Reedy cofibrant objects. So they are their own cofibrant replacements. Hence the maps in MATH are realization weak equivalences by REF . Since the maps in MATH are Reedy cofibrations, to finish this proof it is enough to show that Reedy cofibrations that are realization weak equivalences are preserved under pushouts, directed colimits, and retracts. Since MATH is left proper, if MATH is a pushout of a Reedy cofibration MATH then one can choose a cofibrant replacement MATH for MATH as a pushout of the cofibrant replacement MATH of MATH. Hence MATH is a pullback of MATH. We show in the next paragraph that if MATH is a Reedy cofibration then MATH is a fibration. So if MATH is a Reedy cofibration and realization weak equivalence then MATH and hence also MATH is a trivial fibration. Thus, MATH is a realization weak equivalence. Since retracts and directed limits of trivial fibrations are also trivial fibrations, it follows that retracts and directed colimits also preserve Reedy cofibrations that are realization weak equivalences. Since MATH has the lifting property, so does MATH-cofibrations, MATH for MATH any homotopically constant, Reedy fibrant object. By adjointness this shows that for any Reedy cofibration MATH, MATH is a fibration since it has the right lifting property with respect to MATH. |
math/0101162 | As always, we assume that MATH is a left proper, cofibrantly generated model category that satisfies Realization REF . The category MATH has all limits and colimits since MATH does. The two out of three axiom for weak equivalences and the retract axiom for the cofibrations and weak equivalences are easily checked. The retract axiom for fibrations follows from REF . The two factorizations follow from REF by REF . One lifting property follows from REF since the realization trivial fibrations are the Reedy trivial fibrations. So only the lifting of a realization trivial cofibration with respect to an equifibered Reedy fibration is left. Assume MATH is a Reedy cofibration and a realization weak equivalence. Factor MATH where MATH is a MATH-cofibration and MATH is MATH-injective. Since MATH and MATH are realization weak equivalences, MATH is also a realization weak equivalence. Since MATH is a Reedy cofibration, REF show that it has the left lifting property with respect to MATH. Thus, MATH is a retract of MATH. Hence MATH is a MATH-cofibration and so it has the left lifting property with respect to any equifibered Reedy fibration. This finishes the proof that the realization model structure on MATH is a model category. |
math/0101162 | Given MATH a Reedy cofibration in MATH and MATH a cofibration in MATH, MATH is a Reedy cofibration by REF . So we are left with showing that if MATH or MATH is also a weak equivalence then so is MATH. First consider the case where MATH is a trivial cofibration. Since the pushout product of a trivial cofibration and a cofibration of simplicial sets is a trivial cofibration, MATH-cofibrations MATH-cofibrations is contained in MATH-cofibrations. So by REF , MATH is a realization weak equivalence for MATH any Reedy cofibration. Next consider the case where MATH is a realization weak equivalence. Since trivial cofibrations are preserved under pushouts, retracts and colimits, it is enough to show that for MATH in MATH, MATH is a realization weak equivalence. For MATH in MATH this follows from REF . For MATH in MATH this follows from the previous paragraph by associativity, since the maps in MATH are trivial cofibrations. |
math/0101163 | We will give the proof in the smooth category and discuss the necessary modifications for the topological result at each point. It is clear that the composition of the inclusion of MATH into a MATH - cobordism MATH between MATH and MATH and the homotopy inverse of the inclusion from MATH is an orientation preserving homotopy equivalence and thus induces an isometry between the intersection forms. This way one obtains a map from the set of diffeomorphism classes rel. boundary of MATH - cobordisms between MATH and MATH to the set of isometries from MATH. It is known that this map is surjective. Namely, each isometry can be realized by a homotopy equivalence CITE. And each homotopy equivalence can after composition with a self equivalence of MATH which operates trivially on MATH be realized by a smooth MATH - cobordism (CITE and the correction in CITE - the proof of this result implies that not every homotopy equivalence can be realized by a MATH - cobordism). If MATH and MATH are topological manifolds with MATH, then it is known that each isometry can be realized by a homeomorphism CITE. This implies surjectivity in the topological case. A different argument for surjectivity both in the smooth and topological category can be found in the proof of CITE. Thus we only have to show injectivity. Let MATH and MATH be two smooth MATH - cobordisms between MATH and MATH inducing the same isometry between the intersection forms. We will use CITE to show that MATH and MATH are diffeomorphic rel. boundary. For this we first determine the normal REF - type of a MATH - cobordism MATH. By CITE this is the fibration MATH, if MATH, the non-spin case, and MATH, if MATH, the spin case. In the topological case we have to take instead MATH or MATH. If we want to apply CITE we have as a first step to check that normal REF - smoothings of MATH and MATH exist which coincide on the common boundary MATH. A normal REF - smoothing is in the non-spin case equivalent to an orientation and in the spin case to a spin-structure. Thus, since MATH are simply connected, compatible choices exist. The next step is to decide if MATH is MATH - zero-bordant. In the smooth spin case the MATH - bordism group is spin-bordism which vanishes in dimension MATH. In the smooth non-spin case the MATH - bordism group is oriented bordism which is MATH detected by MATH. One has the same answer in the topological case. One can argue that all MATH - manifolds can be made MATH - connected by surgery and then they admit a smooth structure since the NAME - NAME obstruction for the existence of a MATH - structure in REF - th cohomology with MATH - coefficients vanishes, and in dimension MATH the MATH and the smooth categories are equivalent. In the rest of the argument there is no difference between the smooth and topological case. Now and later on we need information about the (co)homology of MATH. For this we choose a fibre homotopy equivalence between MATH and the mapping torus of the homotopy equivalence on MATH given by MATH, where MATH and MATH are the inclusions from MATH to MATH respectively, MATH. If MATH and MATH induce the same isometry between the intersection forms of MATH and MATH, then MATH induces the identity map in second (co)homology. Thus by the NAME sequence for the mapping torus of MATH we obtain, for arbitrary coefficients, isomorphisms MATH, where MATH is the inclusion, and MATH and MATH. By the NAME we have MATH, since MATH in MATH. Thus the characteristic number MATH vanishes and also in the non-spin case MATH bounds. Choose in both cases a zero bordism MATH and use surgery to make the map MATH REF - connected CITE. The next step is to analyze the surgery obstruction MATH. Note that in both cases MATH implying that the obstruction is contained in MATH instead of MATH making life easier since we do not have to consider quadratic refinements. The obstruction is given by the equivalence class MATH where the maps are induced by inclusion and MATH is the intersection pairing between MATH and MATH. We will show that this obstruction is elementary, that is, there is a submodule MATH such that under both maps MATH maps to a half rank direct summand and MATH vanishes on MATH. We first note that since MATH, we can replace MATH by MATH and since MATH is surjective we can work with MATH instead. The situation is here particularly easy since by our homological information both MATH and MATH are isomorphic to MATH. Thus we have to find MATH such that, under inclusion, MATH maps to a half rank direct summand of MATH and MATH vanishes on MATH. Looking at the exact sequence MATH and using that the latter group is free we can pass to rational coefficients. Here we make use of the fact that we do not have to take quadratic refinements into account. Thus the obstruction is elementary if there is MATH such that, under inclusion, MATH maps to a half rank summand of MATH and MATH vanishes on MATH. Namely, for such a MATH choose MATH such that MATH is a direct summand in MATH and MATH. Since MATH is torsion free MATH maps to a direct summand in MATH. If MATH vanishes for MATH the same holds for MATH and thus our obstruction is elementary. Using that MATH and MATH by the homology information above we obtain an exact sequence MATH . By the homological information above we have isomorphisms MATH . Together with the exact sequence MATH this implies MATH . Since the intersection form on MATH is unimodular and skew symmetric there is a submodule MATH of half rank of this cokernel, on which the intersction pairing vanishes. Finally the intersection form on the image MATH of MATH in MATH is contained in the radical and has rank equal to MATH. Thus MATH is the desired submodule in MATH implying that the obstruction MATH is elementary. Then MATH and MATH are diffeomorphic rel. boundary by CITE. |
math/0101164 | First we show that the best constant MATH in REF is at most the right hand side of REF . Clearly, it is sufficient to prove an inequality of the type REF for monic polynomials only. Thus, we assume that MATH, are all monic, so that MATH is monic too. Let MATH be the zeros of MATH and let MATH be the normalized zero counting measure for MATH. Then, we use REF , NAME 's theorem and REF in the following estimate: MATH . This gives that MATH . To show that equality holds in REF , we consider the MATH-th NAME points MATH for MATH and the NAME polynomials MATH . Observe that MATH . Since MATH, the set MATH consists of more than one point and, therefore, MATH is a strictly positive continuous function in MATH. Consequently, MATH is also continuous in MATH, and we obtain by REF that MATH . Finally, we have from the above and REF that MATH . |
math/0101164 | Since MATH is a bounded continuum, we obtain from REF that MATH . Thus, the Corollary follows by combining this estimate with the obvious inequality MATH and by using that MATH . |
math/0101164 | The proof of this result is quite similar to that of REF (also see CITE). For MATH, consider a function MATH . One can immediately see that MATH is a subharmonic function in MATH, which has the following integral representation (see CITE): MATH where MATH is the normalized angular measure on MATH. Let MATH be such that MATH . If MATH are the zeros of MATH, arranged so that the first MATH zeros belong to MATH, then MATH by REF . It follows from REF and NAME 's theorem that MATH . Using the definition of MATH, we obtain from the above estimate that MATH . Hence MATH . In order to prove the inequality opposite to REF , we consider the MATH-th NAME points MATH for the set MATH and the NAME polynomials MATH. Let MATH be a point, where the maximum of the right hand side of REF is attained. Define the factor MATH for MATH, with zeros being the MATH-th NAME points satisfying MATH. Then we have by REF that MATH . Combining the above inequality with REF and the definition of MATH, we obtain that MATH . This shows that REF holds true. Moreover, if MATH is a regular point for MATH, then we obtain by REF that MATH . Hence MATH which implies REF by REF . |
math/0101164 | It is well known CITE that cap-MATH and MATH where MATH is the angular measure on MATH. If MATH then the numerator of REF is equal to REF, so that MATH . Assume that MATH . We set MATH and let MATH be a point where the maximum in REF is attained. On writing MATH we obtain that MATH by the change of variable MATH . |
math/0101164 | Recall that cap-MATH (see CITE) and MATH . It follows from REF that MATH . If MATH then the integral in REF obviously vanishes, so that MATH. For MATH, let MATH . One can easily see from REF that MATH and MATH . However, if MATH then MATH . It is not difficult to verify directly that MATH which implies that MATH for MATH . Hence MATH . Collecting all facts, we obtain that the maximum for MATH on MATH is attained at the endpoints MATH and MATH, and it is equal to MATH . Thus REF follows from REF and the above equation. |
math/0101164 | Note that the numerator of REF is equal to REF, because MATH. Thus REF follows immediately. |
math/0101164 | Observe that MATH for any MATH, so that MATH. Since MATH is regular, we use the representation for MATH in REF . Let MATH be the dilation mapping. Then MATH and MATH. This gives that MATH where MATH. Using the absolute continuity of the integral, we have that MATH which implies REF . |
math/0101166 | Consider the NAME polynomials for MATH, given by MATH . We already observed in REF that MATH is a monic polynomial with integer coefficients, whose roots are given by REF . NAME showed that if MATH is a prime number, then MATH is irreducible over integers (see CITE). Hence the numbers MATH form a complete set of MATH conjugate algebraic integers in MATH, for any prime MATH. It is clear that the corresponding roots MATH of MATH, obtained by shifting the above set by MATH, form a complete set of conjugates on MATH. Let MATH be an integer NAME polynomial of degree MATH for MATH. Note that MATH where MATH . Indeed, if MATH for just one MATH, then this must be true for all MATH . Since the product MATH is a symmetric form in MATH's with integer coefficients, it may be written as a polynomial in the elementary symmetric functions of MATH's, with integer coefficients, by the fundamental theorem on symmetric forms. Thus the above product must be a nonzero integer, so that we have MATH and MATH . Observe that MATH where the last inequality follows from REF . Combining REF with the above estimate and REF , we obtain that MATH . |
math/0101166 | For the sequence of polynomials MATH we have MATH . Thus the upper bound in REF follows. Suppose that MATH has a leading coefficient MATH. Then we estimate MATH by CITE, which gives the lower bound of REF . |
math/0101166 | Since MATH, REF follows at once from REF . Furthermore, if MATH is of exact degree MATH, with the leading coefficient MATH, then MATH where we used CITE. Hence MATH is an integer NAME polynomial of degree MATH on MATH . |
math/0101166 | If MATH is an integer NAME polynomial of degree MATH on MATH, for any MATH then REF is immediate from REF . Therefore, we only need to prove the second statement of the theorem. It is trivial for MATH, so that we assume MATH . Suppose to the contrary that there exists a polynomial MATH, of exact degree MATH, such that MATH . Let MATH be the zeros of MATH. Clearly, all MATH's are inside MATH, so that we have by the maximum principle MATH . Using a known argument based on the fundamental theorem of symmetric forms (see Lemma in CITE), we obtain that MATH . On the other hand, we estimate MATH . Consequently, this integer N is equal to zero, which means that MATH for some MATH. But then the irreducible polynomial MATH must divide MATH. Assume that MATH, where MATH and MATH, of exact degree MATH, does not have MATH as a factor. It follows that MATH and MATH . Hence we can use the same argument for MATH, to conclude that MATH . This implies that MATH divides MATH, contradicting our assumption. |
math/0101166 | We begin by choosing an increasing subsequence MATH such that MATH where MATH is the power of the factor MATH in MATH. Clearly, MATH and MATH . We may assume that MATH . Our goal is to show that the factors with MATH do not have influence on the MATH-th root behavior for the norms of the sequence. If MATH is not a zero of MATH then MATH where convergence in the above equation is uniform on compact subsets of MATH . Hence MATH for any MATH, with finitely many exceptions. But the function on the left of REF is continuous, so that MATH . It is easy to obtain the opposite inequality from MATH . |
math/0101166 | Suppose that this is not the case, and there are infinitely many real solutions of the above equation. Since the function MATH on the left hand side grows indefinitely, as MATH, all these solutions are contained in a bounded open interval MATH. Note that in this case MATH must have infinitely many points of local maximum in MATH with at least one point of accumulation MATH. Let MATH be the sequence of maxima for MATH in MATH, such that MATH and MATH . Observing that MATH, we conclude that there exists a REF-dimensional neighborhood MATH of MATH, free of zeros of MATH . Hence MATH can be defined as a single valued analytic function in MATH, which is real valued on MATH by an appropriate choice of branches for the powers MATH . It follows that MATH where MATH is also analytic in MATH. Thus the zeros of MATH have a point of accumulation in its domain of analyticity, forcing this function to vanish identically in MATH. This implies that MATH, which can be extended to the whole domain MATH of definition for MATH. But that gives an immediate contradiction, as MATH has the zeros of MATH on the boundary. |
math/0101166 | We first note that the actual degrees of integer NAME polynomials on MATH cannot be bounded, for our assumption that MATH has infinitely many points would give at once that MATH, by REF . Suppose to the contrary that there are only finitely many polynomials, with integer coefficients, that can be factors of MATH . Then we have by REF that, for a subsequence MATH where MATH and MATH . Observe that there are only finitely many points MATH, where the function MATH attains its norm on MATH: MATH according to REF . Let MATH . By choosing MATH sufficiently small, we can make the logarithmic capacity of MATH as small as we wish (see REF). This implies that MATH can also be made arbitrarily small by REF . In particular, we can find an integer NAME polynomial MATH for MATH such that MATH . It follows that MATH for any MATH . Note that MATH where MATH . Hence we can estimate for MATH that MATH . It is clear that we can now choose MATH sufficiently small, to insure that MATH . But this immediately implies that the polynomials MATH have smaller sup norms on MATH than those of integer NAME polynomials, as MATH. This is an obvious contradiction. |
math/0101166 | Take MATH and choose MATH such that MATH . Clearly, MATH . On the other hand, MATH is inside the lemniscate MATH by the maximum principle, so that we can set MATH . Hence MATH lies interior to MATH and MATH . The last inequality follows by considering a sequence of polynomials MATH on MATH . |
math/0101166 | We first assume that MATH, and then obtain a contradiction, to prove REF . Let MATH be the closure of all zeros of the integer NAME polynomials MATH for MATH . Since the sets MATH and MATH differ only by countably many isolated points, we have that MATH (compare REF). Hence MATH is a connected open set, which follows from REF. Consider a sequence of functions MATH that are subharmonic in MATH and harmonic in MATH . It follows from NAME lemma (see, for example, REF) that this sequence is bounded on compact subsets of MATH. Therefore, we can select a subsequence MATH, converging to a harmonic function MATH locally uniformly in MATH . Note that MATH which is inherited from the polynomials MATH . Also, MATH . We want to show that all accumulation points for the solutions of the equation MATH belong to MATH . Indeed, if MATH is such a point, then it must also be a point of accumulation for the local maxima of MATH on MATH . Let MATH be those maxima of MATH such that MATH . Consider a REF-dimensional neighborhood MATH of MATH. We can define an analytic completion of MATH in MATH, denoted by MATH such that MATH . It is easy to see from the NAME integral formula that MATH because MATH . Hence MATH which means that MATH where by MATH we understand the complex derivative of MATH. It follows that MATH vanishes identically in MATH, so that MATH and MATH are identically constant in MATH. But then MATH is identically constant in the whole domain MATH, which cannot be true, because MATH contains compact sets MATH of arbitrarily large capacity and MATH, for any MATH (compare REF). Thus the set MATH of solutions for REF in MATH consists of isolated points, that is, MATH is countable and MATH. Furthermore, MATH by REF. Set MATH . We choose a sufficiently small MATH, so that MATH by REF . Hence there exists MATH, such that MATH for integer NAME polynomials MATH on MATH of degree MATH. Recall that for the integer NAME polynomials MATH on MATH, we have MATH where MATH . We now let MATH and consider sequences of polynomials MATH. Using two preceding estimates and NAME 's inequality, we obtain that MATH for all large MATH . Observe that we can find MATH, so that MATH by our construction of the set MATH . Therefore, MATH for all sufficiently large MATH . This gives the following estimate MATH where MATH is selected to be sufficiently large. The last inequality in REF follows because MATH by NAME inequality, and because MATH . Finally, we combine REF to obtain the contradiction: MATH . |
math/0101166 | Observe that if the set MATH is finite, then MATH. Indeed, we can use the regular integer NAME polynomials MATH on MATH, to find that MATH . But MATH in this case, so that MATH by REF . Thus, REF is trivially true when MATH is finite, and we assume that MATH has infinitely many points for the rest of this proof. We need to find a sequence of polynomials MATH with small weighted norms MATH . It is possible to use the NAME interpolation in weighted NAME points for this purpose. The weighted NAME points MATH are defined as a set of points maximizing the absolute value of the ``weighted NAME determinant" (compare CITE) MATH among all MATH-tuples MATH. Note that MATH and we obtain from the NAME interpolation formula that MATH . Since MATH for any MATH we have that MATH . It follows at once from REF that MATH (also see REF). Observe that MATH are linear forms in MATH, with real coefficients. Applying NAME 's theorem (see CITE), we conclude that there exists a set of integers MATH, not all zero, such that MATH . But MATH so that we can find a sequence MATH satisfying MATH . Hence MATH by REF. |
math/0101166 | Let MATH be a sequence polynomials satisfying MATH where MATH is defined in REF . We construct the following new sequence of polynomials with integer coefficients: MATH where MATH are selected so that MATH . Using REF , we obtain that MATH because MATH . It now follows from REF that MATH . |
math/0101166 | We need to find a solution of the weighted energy problem on MATH, corresponding to the weight MATH of REF . It follows from REF that there exists a weighted equilibrium measure MATH, whose support is a compact set MATH . Let MATH be a unit point mass at MATH. Observe that MATH where MATH is the logarithmic potential of the measure MATH . It is clear that MATH is a positive NAME measure of total mass MATH . Let MATH be the balayage of MATH from MATH onto MATH (see, for example, REF). Then MATH is a positive NAME measure of the same mass as MATH, which is supported on MATH. Furthermore, we can express MATH via harmonic measures MATH (compare REF). The potentials of MATH and MATH are related by the equation MATH which holds quasi everywhere on MATH (see REF). Hence the measure MATH is a probability measure on MATH. Using REF , we obtain for quasi every MATH that MATH . Note that MATH has finite logarithmic energy, since it is composed of harmonic measures. Thus we can apply REF to prove that MATH is the weighted equilibrium measure MATH and that the associated modified NAME constant MATH is given by REF . REF expresses MATH through MATH, MATH and MATH (compare REF). We now obtain REF from REF by the following simple manipulation. Recall that MATH (see REF). Substituting this relation into REF , we have that MATH . Hence REF is proved too. |
math/0101166 | Observe that the actual degree of MATH is MATH . It follows from REF that there exists a weighted equilibrium measure MATH, whose support is a compact set MATH . Hence the factors MATH do not vanish on MATH . Estimating MATH and noting that MATH by REF , we obtain REF . |
math/0101166 | Clearly, the degrees of integer NAME polynomials on MATH must be unbounded, because MATH so that the assumptions of this theorem are valid. Since the leading coefficient of MATH is at least REF in absolute value and the degree of MATH is MATH, we obtain from REF that MATH . Thus REF follows by taking the MATH-th root in the above inequality, and using REF together with REF , as MATH . Finally, REF is a direct consequence of REF , for MATH . |
math/0101166 | Recall that the actual degree of MATH is given by MATH . Since MATH, we have that MATH for a positive integer MATH. On the other hand, REF gives the estimate MATH . Consequently, MATH . Taking the MATH-th root in the above inequality, and using REF together with REF , we obtain REF , as MATH . |
math/0101166 | Consider a sequence of integer NAME polynomials MATH satisfying REF . It is not difficult to see that we can assume MATH while preserving the property MATH . Since MATH we only need to show that, for some MATH, MATH . Using the same notations as in the proofs of REF , we have that MATH . We next estimate, multiplying REF by MATH, MATH . Recall that MATH is a compact set, MATH. Therefore, the potential MATH is harmonic and bounded on MATH, if MATH is sufficiently small. On the other hand, we can obviously make MATH smaller than any negative number on MATH, by choosing MATH small. It follows that MATH for some MATH, which further implies that MATH by REF . Using REF , we now obtain that MATH . |
math/0101166 | Note that the weight MATH of REF is just a special case of the NAME weights of REF. Thus our problem on the interval MATH is easily reduced to that on the interval MATH considered in CITE, with the help of the change of variable MATH. We obtain from REF that MATH, with MATH and MATH given by REF (see REF). Similarly, REF gives REF after this change of variable. Consider the following natural extension for MATH of REF : MATH . It follows from REF that MATH for quasi every MATH (that is, with the exception of a set of zero capacity). Denote the right hand side of REF by MATH . Then MATH is a harmonic function in MATH, such that MATH for quasi every MATH, by REF functions (see CITE). Using the uniqueness theorem for the solution of the NAME problem in MATH (compare REF and its Corollary in CITE), we conclude that MATH . Thus MATH can be found from MATH or from REF . We obtain the explicit representation of REF by expressing the NAME functions of REF via the conformal mappings of MATH onto the exterior of the unit disk. Indeed, introducing these conformal mappings by MATH and MATH we observe that MATH . Hence MATH by REF. |
math/0101166 | This theorem is an immediate application of REF to the integer NAME polynomials on MATH. Using REF , we obtain the upper bound MATH . Then we choose MATH and MATH to produce the inequalities, defining the region MATH of REF , by REF . The values of MATH are readily found from REF and the explicit formulas for the NAME functions, obtained in the proof of REF , as well as the bounds for MATH and MATH, is generated by NAME. |
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