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math/9910107 | We consider the variety MATH with coordinate MATH on the first factor and MATH on the second factor. There is a morphism MATH sending a point MATH to the first MATH coefficients of the series MATH and MATH. Let MATH be the image of MATH by MATH. It is a locally closed subvariety of MATH. It follows from REF that MATH is a NAME cover with group MATH acting by multiplication on MATH. By construction, MATH is equal to MATH, with MATH the central function on MATH which takes value REF at the identity and zero elsewhere. But now remark that MATH for MATH a non trivial irreducible character of MATH and that MATH when MATH is the trivial character (compare REF). The second equality follows, since, when MATH is expressed as a MATH-linear combination of irreducible characters, the coefficient of the trivial character is MATH. For the first equality it is enough to remark that MATH and that MATH. |
math/9910108 | Let us show, that the compact set MATH complies with the conditions of REF . We will divide our argument into several steps. CASE: Since MATH then for any MATH we can find MATH such that MATH. REF states that the point MATH is accessible from MATH. Hence MATH is accessible from MATH because MATH. CASE: Let us show, that any point MATH is accessible from MATH. Without loss of generality we can assume that the origin of coordinates lies in MATH. We fix MATH. The set of all points accessible from MATH is dense in MATH CITE, therefore there exists a point MATH accessible from MATH which do not coinside with MATH. All compact subsets of MATH, MATH, are known to be limited. Therefore there exists MATH such that MATH . We fix a point MATH which meets an equality MATH. It is known (see CITE) that there exists a cut MATH, MATH, MATH. Let MATH . According to the conditions of REF. Let MATH. Denote a polar angle of MATH by MATH. Consider continuous injective mapping MATH . This map is an imbedding of MATH into MATH, moreover MATH, MATH. CASE: Let us show, that the open set MATH is connected. Consider an involution MATH . This map is known to be a homeomorphism. Under the action of MATH the area MATH will pass to an open connected set MATH. NAME that the origin of the coordinates is an isolated point of the boundary MATH because MATH . Therefore, MATH appeares to be the open connected set and the map MATH is a cut of the set MATH. Moreover MATH. So, for a proof of connectivity of the set MATH it is sufficient to check the validity of the following statement. Let MATH be an open connected set, point MATH be accessible from MATH, MATH be a cut of MATH with the end in MATH (a continuous injective mapping such that MATH and MATH). Then the set MATH is connected. Let us prove this statement. Let MATH for some MATH. According to REF from CITE there exists a homeomorphism MATH of MATH onto MATH, such that the map MATH complies the relation MATH . Since MATH, there exists a MATH such that MATH . Obviously, MATH is a neighbourhood of a point MATH in MATH and the set MATH is connected. Besides, a set MATH is not empty for any MATH, MATH. Therefore MATH is a connected open neighbourhood of a set MATH in MATH, hence MATH is a connected set. MATH . So, the set MATH is connected. CASE: Select a point MATH, MATH for each MATH. The set MATH is connected by the condition of theorem and the point MATH is accessible from MATH for any MATH. Therefore we can find a continuous map MATH which meets the following conditions MATH . Consider a continuous map MATH . Since the relations MATH hold and a compact set MATH does not contain a point MATH on a construction, there exists MATH which complies the inequality MATH . Now we are ready for proof of local linear connectivity of the area MATH in the point MATH. CASE: Let MATH be a curtain neighbourhood of the point MATH. Find MATH which meets the conditions MATH . Fix imbeddings MATH . Here MATH. The metric on MATH we shall define as follows: MATH . NAME that maps MATH, MATH are uniformly continuous. Fix MATH such that an inequality MATH implies MATH for any MATH and MATH, MATH. Find also MATH, such that MATH has as a consequence an inequality MATH for every MATH and any MATH, MATH. Assume MATH. CASE: Let us show, that for any MATH, MATH there exists a continuous map MATH such that MATH, MATH. The inequality MATH is fulfilled for all MATH, MATH, hence MATH for every MATH and in the case MATH the circle MATH could be decomposed into two not intersecting intervals MATH and MATH with common endpoints in such a way that the following relations are fulfilled MATH . In the case MATH set MATH, MATH. Therefore, MATH and MATH. Let MATH be a closed disk satisfying the following conditions: MATH . Then MATH, MATH. On the condition of REF for every MATH. Therefore, MATH or MATH. By a construction MATH and MATH, hence MATH, MATH. MATH . Let MATH, MATH. Since MATH, then MATH, MATH. From connectivity of the set MATH follows, that there exists an injective continuous map MATH complying the equalities MATH, MATH (the concepts of connectivity and linear connectivity coincide for open subsets of MATH). Find smooth imbeddings MATH such that the points MATH, MATH lie inside disks bounded by curves MATH, MATH. It is known that an imbedding of a segment or circle into MATH could be as much as desired precisely approximated by a smooth imbedding. It is known as well that any two one-dimensional smooth compact submanifolds of MATH could be reduced to the general position by a small perturbation fixed on their boundary. Therefore, there exists smooth imbedding MATH such that the sets MATH and MATH consist of final number of points. For every MATH there exist MATH, MATH, MATH, which comply with the following conditions MATH . We receive a finite family of nonintersecting intervals MATH satisfying to relations MATH . Now for each MATH we fix an arc MATH with the endpoints MATH and MATH. A set MATH is homeomorphic to a circle, therefore it bounds a closed disk MATH such that MATH . By REF these relations has as a consequence following inclusion MATH . Since MATH, then MATH . Here MATH is a final family of nonintersecting arcs of the circle MATH. In addition MATH, MATH. A set MATH represents a final union of connected components homeomorphic to the two-dimensional disk, lying either inside or outside the closed disk limited by a circle MATH. Select that from components, which bounds with an arc MATH. Designate by MATH a closure of this component. Obviously, MATH . Let MATH be an arc of a circle MATH with the endpoints MATH, MATH. As we already have shown, it complies with the relation MATH . A continuous curve MATH represents a continuous path in MATH, connecting points MATH and MATH. Therefore, the point MATH is accessible from MATH and all the more it is accessible from MATH. Then each point of MATH is accessible from MATH because of the arbitrary rule we selected the point MATH. CASE: The set MATH is connected on a condition of theorem. CASE: Let us show that the set MATH is connected. The set MATH is connected since any point of MATH is accessible from a connected set MATH, therefore it is sufficient to show that the boundary MATH does not lie in the set MATH for any connected component MATH of the set MATH, different from MATH. Assume that MATH. The set MATH divides MATH, consequently it has dimension not less than one (see CITE). Therefore, we can find three different points MATH, MATH, MATH. Each of these points is accessible from the connected sets MATH and MATH. There exists a continuous injective mapping (see CITE) MATH which satisfies the conditions MATH . Let MATH. There exists a continuous injective mapping MATH . Let MATH. We have MATH since MATH. Denote MATH. Then MATH is uniquely defined, such that MATH. Consider continuous injective mappings MATH which comply with the relations MATH . Similarly, there exists a point MATH and continuous injective mappings MATH, MATH, such that MATH . Since MATH the equality MATH is valid. Therefore, everyone from the sets MATH is homeomorphic to a circle. The set MATH falls into the three connected components MATH, MATH, MATH, two of which are homeomorphic to the open two-dimensional disk and third is not limited. As MATH then there exists MATH such that MATH. But it is impossible because everyone from the sets MATH, MATH contains exactly two from the points MATH, MATH, MATH. So, we have proved that the set MATH consists not more than from two points. Therefore, MATH and MATH could be found to comply the inclusion MATH. The set MATH is connected since MATH and the sets MATH, MATH, MATH are connected. By virtue of arbitrariness in a choice of MATH, the set MATH is connected. Applying to MATH REF we conclude that this set is homeomorphic to the closed two-dimensional disk. |
math/9910108 | Let MATH, MATH be the inclusion maps, MATH. Without loss of a generality, it is possible to assume that a NAME Pole MATH of MATH lies in MATH. Consider a stereographic projection MATH . As is known, this map is a homeomorphism. Since MATH, MATH and the set MATH is open in MATH, the compositions MATH are continuous and are one-to-one. The set MATH is compact, therefore maps MATH, MATH are imbeddings. Sign MATH, MATH. From a mutual uniqueness of map MATH follows that MATH . The set MATH is connected as an image of a connected set at a continuous map. So, family MATH satisfies to conditions of REF . Consider an open set MATH. It is easy to see that MATH and MATH is an isolated point of the boundary of MATH. Denote MATH. Obviously, MATH is the unique unlimited connected component of a set MATH. Applying REF , we conclude that a set MATH is homeomorphic to the closed two-dimensional disk, and it's boundary MATH is homeomorphic to a circle MATH. From this immediately follows, that the set MATH of the limit points of MATH is homeomorphic to a circle. From REF it immediately follows that the set MATH divides MATH into two opened connected components and for each of these components it's closure is homeomorphic to the closed two-dimensional disk. Consequently, the set MATH is homeomorphic to the closed two-dimensional disk. |
math/9910111 | CASE: Let MATH be given, and assume MATH satisfies MATH. Then using REF , MATH . By the uniqueness assumption, MATH, and thus by REF , MATH. Conversely, set MATH. Then using REF , MATH . Therefore MATH is a right quasigroup and hence a quasigroup. CASE: Obvious. |
math/9910111 | For all MATH, MATH. Cancelling MATH, we have MATH, and thus MATH is idempotent. If MATH, then clearly MATH. |
math/9910111 | REF is immediate from REF since MATH. To simplify further calculations, set MATH. For REF , the restriction on MATH implies MATH, and the result follows immediately. For REF , we compute MATH and MATH . Using these calculations in REF gives the desired result. |
math/9910111 | All three assertions are trivial for MATH, and thus we may assume MATH. REF follows from the computation MATH using REF and the fact that MATH is an orthogonal projection. Now MATH and thus MATH using REF (with MATH). This establishes REF . Finally, REF follows from REF , and the symmetry of MATH: MATH . This completes the proof. |
math/9910111 | Since MATH is a group, MATH by REF . We have MATH. In view of REF , this establishes the result. |
math/9910111 | The left inverse property follows from REF . The automorphic inverse property follows from REF . The MATH property follows from REF . |
math/9910111 | By the preceding discussion, MATH. Thus REF becomes MATH, that is, MATH. This is equivalent to MATH, and thus to MATH, or MATH, with the constraint MATH. An obvious particular solution is MATH. Thus the solution set in MATH is the affine subspace MATH, and hence the solution set in MATH is MATH, as claimed. The nonuniqueness assertion follows from geometric considerations: MATH is a punctured sphere of codimension MATH in MATH with center MATH and containing MATH. The antipode of MATH is the removed point MATH. If MATH is a singleton, then either MATH, in which case MATH, or MATH is tangent to MATH at MATH, in which case MATH or MATH, a contradiction. |
math/9910111 | Fix MATH with MATH. To simplify notation, set MATH and MATH. For MATH such that MATH, we have MATH using REF . Now MATH . Using REF , we may expand REF into an expression of the form MATH where the coefficients MATH are to be determined. We find, for instance, MATH . Similar calculations give MATH . Thus MATH . This and REF give the desired result. The remaining assertion is clear. |
math/9910111 | Fix MATH with MATH, and assume MATH. Then MATH. Choose MATH. Then MATH. Suppressing details, we have MATH, a contradiction. |
math/9910111 | We have MATH by REF , and MATH . Now choose MATH. We have MATH and MATH. Thus MATH if and only if MATH, that is, if and only if MATH. Now assume MATH and MATH (that is, MATH is given by REF ). Then MATH, and thus MATH. If the NAME identity holds, then MATH. Applying both sides to MATH, we have MATH. |
math/9910111 | Note that REF is immediate since MATH is the restriction of a bounded linear operator to MATH. For REF , it is clear from REF that MATH REF has an analytic extension to MATH. Similarly, REF shows that MATH has an analytic extension to MATH, and this implies REF . |
math/9910111 | REF , and REF are easy consequences of the definitions. For REF , we use REF and the NAME inequality to compute MATH . By REF , we have MATH since MATH. If equality holds, then the NAME equality implies that MATH and MATH are parallel, which is equivalent to the desired result. The remaining inequality of REF is clear. |
math/9910111 | In any LIP left loop, we have the identity MATH for all MATH (see, for example, CITE). In any NAME left loop, we have the identities MATH for all MATH (see, for example, CITE and REF ). We compute MATH . Now MATH using REF and LIP. Thus MATH using the MATH property. Finally by REF , and REF again, MATH . This establishes the result. |
math/9910111 | REF follows from REF follows from REF . For REF , we have MATH . Taking norms of both sides, and then using REF , we obtain the desired result. For REF , we use REF in MATH, take norms, and apply REF : MATH . This completes the proof. |
math/9910112 | Let MATH be a positive deformation of MATH and let MATH be a non-zero Hermitian element with respect to MATH with first non-vanishing order MATH. Then MATH is Hermitian with respect to the classical MATH-involution and thus we find a positive MATH-linear functional MATH of MATH with MATH. Moreover, we find MATH such that MATH is a positive MATH-linear functional of MATH with respect to the deformed MATH-algebra structure. But then clearly MATH, proving the proposition. |
math/9910112 | Thanks to CITE it is sufficient to check the positivity of a MATH-linear functional MATH on elements in MATH alone. But if MATH is a positive linear functional then clearly MATH. |
math/9910112 | First we note that the MATH-series is well-defined thanks MATH. As already observed , see REF , the MATH-linear maps MATH are again derivations and thus all occuring derivations commute which implies that REF is associative. Clearly MATH is Hermitian and one has MATH for all MATH whence we can apply the above lemma. |
math/9910112 | Let MATH be a classically positive linear functional. Choose a locally finite open cover MATH of MATH by contractable charts and let MATH be a subordinate `quadratic' partition of unity, that is, MATH. Endow each MATH with a local NAME star product MATH and let MATH be a local and real equivalence transformation between MATH and MATH, that is, MATH on MATH. Such an equivalence transformation exists since MATH is contractable, see for example, CITE. Then it is easy to check that the MATH-linear functional MATH defined by MATH is well-defined, positive with respect to MATH, and a deformation of MATH. |
math/9910112 | Let MATH be Hermitian then we may assume that MATH. Thus we find a point MATH and an open neighborhood MATH of MATH such that MATH is either strictly positive or strictly negative. Choose now a smooth density MATH with MATH, MATH, and MATH. Then it follows by the same argument as in CITE that the functional MATH is positive with respect to any star product on MATH and clearly we have MATH. |
math/9910119 | The existence and the uniqueness of the solution follows immediately from REF . From the homogeneity of the symbols and from the uniqueness of the solution we see that MATH holds for every MATH. If we set MATH and MATH we obtain MATH . The theorem will be proved if we show that for MATH we have MATH for MATH and that the left-hand side is bounded by a constant for MATH. The boundedness for MATH follows easily from REF . We have to consider the case of large MATH. To find an estimate in this case, we represent the solution in a form suggested in a paper of CITE. This representation is different from the formula used for the NAME boundary value problem in CITE and allows us to separate the two parts of the zeros of the polynomial MATH in a more adequate form. Due to REF , the roots of this polynomial consist of two groups, the first group, denoted by MATH, being bounded for MATH, the other group, denoted by MATH, being of order MATH for MATH. We define MATH . Let MATH be a contour in the upper half of the complex plane enclosing the zeros MATH. From REF we see that MATH can be chosen independently of MATH and MATH for all MATH and MATH. From the same lemma we see that MATH as MATH. Therefore we obtain from REF that for all MATH and MATH the polynomials MATH are independent modulo MATH. Thus there exist polynomials (with respect to MATH) MATH, depending continuously on MATH, such that MATH . For the construction of MATH compare , for example, CITE, p. CASE: Analogously, we define MATH . Let MATH be a contour in the upper half of the complex plane enclosing the zeros MATH. From REF we know that this contour is of order MATH for MATH. Therefore we may fix a contour MATH, independent of MATH and MATH such that MATH encloses all values MATH with MATH. We also remark that due to the regular degeneration we may choose MATH with a positive distance to the real axis (compare also REF ). From REF we know that MATH is linearly independent modulo MATH. From REF we know that MATH . Due to continuity, the polynomials MATH are for sufficiently large MATH linearly independent modulo MATH. Therefore there exist polynomials (in MATH) MATH for MATH, depending continuously on MATH and MATH, such that MATH . Now we need a lemma which will be proved below. The solution MATH of REF - REF can be represented in the form MATH where for MATH and MATH we have MATH and MATH . As a direct corallary of the lemma we obtain MATH . The estimate REF trivially follows from these relations. |
math/9910119 | Let MATH be a solution of REF - REF with MATH replaced by MATH. We seek the solution in the form MATH where the functions MATH still have to be found. Applying the boundary operator MATH to both sides of REF and taking MATH we obtain the following system for the unknown functions MATH: MATH . Here we have set MATH . We remark that we have used MATH. Now we write MATH, where MATH consists of the first MATH components of the vector MATH, and MATH consists of the other MATH components. In the same way we write MATH. In these notations the system REF - REF can be rewritten in the form MATH where we use the notation MATH and MATH . If me multiply the second equation by the matrix MATH from the left and subtract it from the first equation we obtain MATH . In a similar way we obtain MATH . The matrices in brackets in the left-hand sides of above relations differ from the identity by matrices whose elements can be estimated by a constant times MATH. According to REF , their norms tend to zero as MATH. From this it follows that the matrices in brackets for large MATH have inverses which we denote by MATH and MATH, respectively. Then we obtain MATH . If we take MATH, where MATH stands for the MATH-th unit vector, and denote by MATH the first MATH components of MATH, we obtain MATH . In the same way if MATH and MATH denotes the components MATH of MATH, we obtain MATH . The statement of the lemma directly follows from these relations. |
math/9910119 | To see this, we only have to remark that the right-hand side of REF is equivalent to MATH . The first and fourth lines above coincide with the corresponding lines in the right-hand side of REF . The ratio of the second line in REF and the second line above is equal to MATH . Respectively, the ratio of the third line in REF and the third line above is equal to MATH . Now our statement follows from REF . |
math/9910119 | The proof of this theorem is similar to the proof of REF; therefore we only indicate the main steps. By the localization method (``freezing the coefficients"), it is possible to reduce the proof to the proof of the corresponding results for model problems in MATH and MATH. The case of the whole space MATH is quite elementary and needs only slight changes in comparison with REF . The key result is the a priori estimate in the half space MATH. So we assume that MATH is a solution of MATH . Let MATH be a cut-off function, that is, MATH for MATH and MATH for MATH. As in CITE we write MATH in the form MATH . Here we fixed an extension operator MATH from MATH to MATH, being continuous from MATH to MATH and from MATH to MATH; we define the distribution MATH as MATH. By MATH we denote the operator of restriction of functions on MATH onto MATH. In REF , the pseudodifferential operator (ps.d.o.) MATH in MATH is defined as MATH . It is easily seen that we have MATH . We still have to estimate MATH defined in REF . By definition, MATH is a solution of MATH where we set MATH. Now we use the fact that MATH . As MATH it is sufficient to estimate MATH. It can be seen, using the binomial formula, that MATH and therefore the semi-norm MATH is equivalent to MATH . Taking the partial NAME transform MATH with respect to MATH, we obtain from REF - REF that for MATH the function MATH is a solution of MATH . Due to REF , this solution is unique and given by MATH with MATH being the solution of REF - REF . Now we can apply REF to obtain MATH . Integrating this inequality with respect to MATH and using the norm REF we get the desired estimate for MATH, which finishes the proof of the theorem. |
math/9910119 | Let MATH and set MATH. Using the equivalent norm MATH in MATH, we see that we have to show that MATH for some function MATH whose MATH-norm is bounded by a constant independent of MATH. For this it is sufficient to show that for MATH we have MATH . As we have for MATH the equivalence MATH for all MATH (with MATH defined by the right-hand side of REF ), the case MATH is already covered by REF . Here we take into account that, due to REF , the constant MATH in REF applied to the symbols REF may be chosen independently of MATH. The case MATH follows after differentiation of REF with respect to MATH along the same lines as in the proof of REF . |
math/9910119 | The symbol of the ps.d.o. MATH in MATH with parameter MATH is given by MATH with MATH . Consider the famliy MATH of polynomials in MATH. As the degree of the polynomial MATH is equal to MATH for all MATH, the family MATH is a subset of the finite-dimensional vector space of all polynomials in MATH of degree not greater than MATH. Therefore, there exists a finite set MATH such that every MATH may be represented in the form MATH with smooth coefficients MATH. Taking into account that the operators of multiplication by MATH are bounded in MATH, we reduce our problem to the proof of the boundedness of operators of the form MATH where MATH . Literally repeating the proof of REF we establish the boundedness of the operator MATH . According to REF the operator MATH is bounded. As MATH is the product of of the above operators this operator is also bounded. |
math/9910119 | We define MATH with MATH from REF and MATH given by REF . The continuity of MATH follows from REF . In order to see that the operator MATH is continuous with respect to the spaces given in the theorem, we denote the components of MATH by MATH. The operator MATH is given by MATH . We see from REF that MATH maps the space REF continuously into MATH. Turning to the other components MATH, we remark that for MATH the operator MATH equals MATH up to operators of lower order. More precisely, the operator MATH is a ps.d.o. in MATH which is continuous from MATH . This is due to the fact that MATH satisfies REF - REF ; the estimates for the lower order terms of the ps.d.o. MATH can be found in the same way as it was done for MATH in the proof of REF . From the continuity of MATH the continuity of MATH in the spaces given in the theorem immediately follows. |
math/9910119 | Direct calculation shows that MATH . If we substitute the last expression in the left-hand side of REF and change MATH to MATH we obtain the right-hand side of REF . |
math/9910119 | Applying the a priori estimate REF to the function MATH (compare REF ), we obtain, according to the lemma, MATH . The MATH norm of this expression tends to the left-hand side of REF , as MATH tends to MATH. Now we turn to the right-hand side of REF . We have MATH where MATH as MATH. It is easy to check that the limit of the second term of the right-hand side of REF is equal to zero. |
math/9910119 | We apply REF with MATH replaced by MATH to the function MATH defined in REF , noting that MATH is again defined in MATH because of MATH and MATH. From REF and the fact that for any function MATH we have MATH we see that the MATH-th term in the first sum in REF is equal to MATH which tends to the corresponding term in REF for MATH. The remaining expressions in REF can be treated analogously; the term MATH tends to zero for MATH. For the terms involving the boundary operators we remark that MATH where MATH stands for the trace operator. Therefore we may apply REF to the function MATH defined in MATH. |
math/9910119 | This can be seen in exactly the same way as REF , now applying the a priori estimate REF with MATH instead of MATH to the function MATH where MATH is fixed and MATH tends to infinity. |
math/9910119 | We apply REF with MATH replaced by MATH to the function MATH with MATH fixed. Now we use MATH where MATH indicates that the operator MATH acts on the first MATH variables (and analogously that MATH acts on the last variable), and apply REF twice. For the MATH-th term in the first sum of REF we obtain the expression MATH . For MATH this expression tends to zero for MATH, for MATH its limit equals MATH. The remaining terms can be treated analogously; to finish the proof we use MATH and MATH as MATH. |
math/9910121 | Denote points in MATH by MATH, where MATH and MATH satisfies MATH for each MATH. Similarly, denote points in MATH by MATH, where MATH and MATH. Then MATH and MATH. Let MATH be the total space of the pullback of MATH along MATH. It is then readily checked that the map MATH defined by MATH is an equivalence of bundles. |
math/9910121 | From REF , we have MATH. For MATH, let MATH. Then MATH is normal in MATH and MATH is free, so REF are satisfied. We show that REF holds by induction on the cohomological dimension of MATH, which we may assume without loss of generality is equal to MATH. In the case MATH, MATH is a (single) finitely generated free group acting on itself by conjugation, and REF clearly holds. In general, from REF we have a commuting diagram MATH where MATH is induced by the map MATH from REF, and MATH is induced by MATH defined by MATH. From REF , we have MATH, where MATH and MATH is the NAME representation. Using the semidirect product structure, every element MATH may be expressed as MATH, where MATH and MATH. For MATH, consider the conjugation action of MATH on MATH. In the case MATH, for MATH and MATH as above, we have MATH . Thus in this instance, conjugation by MATH coincides with conjugation by the pure braid MATH. So for MATH, REF holds by the result of NAME, NAME, and NAME stated in REF above. For the case MATH, let MATH and consider MATH. In this instance, we have MATH . Now MATH is in MATH, since MATH and MATH is normal in MATH. Consequently, the conjugation action of MATH on MATH coincides with that of MATH for MATH. So REF holds by induction in this case. |
math/9910121 | Write MATH, where MATH and MATH, and identify a pure braid with its image under the NAME representation. Then MATH. The action of MATH is by conjugation, so MATH, where MATH. A calculation with the NAME representation reveals that MATH for MATH. So we have MATH, where MATH, and the result follows. |
math/9910121 | Denote points in MATH by MATH, where MATH and MATH satisfies MATH for each MATH. Similarly, denote points in MATH by MATH, where MATH and MATH satisfies MATH for each MATH. Then MATH and MATH. Let MATH be the total space of the pullback of MATH along MATH. It is then readily checked that the map MATH defined by MATH is an equivalence of bundles. |
math/9910121 | It suffices to show that the braids MATH, MATH generate the free group MATH. Identify the pure monomial braid group with the fundamental group of the space MATH. The free group MATH is the fundamental group of the fiber of the projection MATH. Via the imbeddings MATH, this projection corresponds to the map MATH defined by forgetting the last MATH coordinates. In terms of geometric braids, this map is given by forgetting the last, or outermost, MATH strands. From this and the definitions of the braids MATH and MATH, it is clear that these braids are elements of MATH. Recall from REF that the strands of monomial braids are indexed by the points MATH from which these braids emanate. Checking that the braid MATH performs a full twist on the MATH strands emanating from MATH, and that MATH simultaneously performs full twists on the pairs of strands emanating from MATH and MATH, MATH, we see that the homology classes of these braids are independent. It follows that the braids MATH and MATH generate the free group MATH. |
math/9910121 | It follows from the above discussion that the group MATH admits a presentation of this form. We sketch how one may use REF and the monomial braid relations REF to show that the conjugating words appearing in the relation families REF - REF are as asserted. Using REF, one can show that MATH commutes with MATH, MATH, and MATH for each MATH. Relations REF with MATH follow. Let MATH and MATH for MATH. Then MATH and MATH. Conjugating REF with MATH by MATH yields REF for any MATH and MATH. Conjugating these last relations by MATH yields REF in general. Similarly, using REF, one can show that the following relations hold in MATH: MATH . Repeated conjugation these relations by MATH, MATH, and use of REF yields the relations REF. Finally, it is clear from REF that the pure monomial braids MATH satisfy the classical pure braid relations REF. These relations may be rewritten as MATH . Repeated conjugation these relations by MATH, MATH, and use of REF yields the relations REF - REF. |
math/9910121 | Rewriting MATH as MATH, the result follows from the commutator identity MATH in MATH. |
math/9910121 | The isomorphism, MATH, of graded abelian groups was noted in REF above. The NAME bracket relations in MATH may be obtained by applying REF to the relations REF. |
math/9910122 | Denote by MATH the semigroup generated by MATH. For any trigonometric polynomial MATH define MATH by: MATH . Since MATH has a negative exponential type, the definition of MATH makes sense. Moreover there is a constant MATH independent of MATH so that MATH . Now we solve MATH . Elementary calculations show that MATH on MATH. Now, the fact that MATH has REF yields the result. |
math/9910122 | Assume MATH and that MATH is a sectorial operator of type MATH on MATH which fails REF . Then the operator MATH, defined on MATH by MATH and MATH when MATH with MATH provides a counterexample to REF on MATH. |
math/9910122 | Let MATH and MATH be two sequences defined by MATH . We let MATH and MATH . It is easy to see that MATH while MATH . Hence MATH . If we assume that MATH has (MRP) then both MATH and MATH satisfy REF and so we can apply REF to each in turn for the polynomial MATH . Subtracting gives us the first estimate. The second estimate follows by duality. More precisely the operators MATH and MATH can be extended to bounded linear operators on MATH . Taking adjoints and restricting to the subspace MATH one easily obtains similar estimates in the dual. |
math/9910122 | Denote by MATH the canonical basis of MATH and let MATH. MATH is a NAME basis of MATH which is usually called the summing basis of MATH. We now apply REF with the sequence of projections MATH associated with the NAME basis MATH and MATH. Then we obtain that there is MATH so that for every MATH: MATH . The right-hand side is equal to MATH but, considering only the first co-ordinate of the left-hand side with respect to the canonical basis, we have MATH . This is a contradiction. Assume now that MATH has the (MRP). Let MATH be the coordinate functionals associated with the summing basis MATH of MATH. The closed linear space MATH spanned in MATH by the sequence MATH is of codimension REF in MATH and hence is isomorphic to MATH . The bi-orthogonal functionals MATH in MATH are equivalent to the summing basis of MATH . Hence using the same calculation as above and the second inequality of REF we again get a contradiction. |
math/9910122 | The idea is to show that if MATH has the (MRP) and an unconditional basis MATH, then for every permutation MATH of the integers and for every block basis MATH of MATH the closed subspace of MATH spanned by the MATH's is complemented in MATH. Once we have shown this the proof is completed by using REF and NAME (CITE, see also REF ) which asserts that MATH must be equivalent to the canonical basis of MATH or MATH for some MATH in MATH. Then, by REF , MATH is equivalent to the canonical basis of MATH for some MATH in MATH. Now, if MATH, MATH admits an unconditional basis which is not equivalent to any of the canonical bases of MATH or MATH where MATH. Indeed CITE showed that, for MATH, MATH is isomorphic to MATH. So assume, as we may, that MATH is a normalized REF-unconditional basis of MATH and that MATH is a normalized block basis of MATH, with MATH where MATH and MATH. For MATH, let MATH and MATH. Then MATH is an unconditional NAME decomposition of MATH with associated projections MATH say. Now, by the NAME theorem there is a norm-one projection MATH . Let MATH . Then MATH is a NAME decomposition of MATH with associated projections MATH and MATH . We now apply REF and exploit the unconditionality of the NAME decomposition MATH . There is a constant MATH so that if MATH is in the linear span of the MATH then MATH . This implies that MATH is complemented in MATH . Clearly the same reasoning can be applied to any permutation of the basis MATH so that the proof is complete. |
math/9910122 | For MATH, the NAME system is known to be an unconditional basis of MATH (CITE, see also CITE). So the result follows from the preceding Theorem. The fact that MATH fails (MRP) was proved by NAME in CITE. Notice that MATH contains a complemented copy of MATH, so this result can be derived from REF . |
math/9910122 | It suffices to show that if MATH with MATH then MATH converges if and only if MATH . As above in the proof of REF , let MATH be a norm-one projection of MATH onto MATH . Then let MATH and MATH . Reasoning exactly as in REF gives that MATH is complemented in MATH . But this subspace has an unconditional basis and so REF yields that MATH is equivalent to the canonical basis of MATH . |
math/9910122 | We first note that MATH is complemented in MATH so that if the latter has (MRP) then MATH by REF . Assume that MATH has the (MRP); we will show that MATH is isomorphic to a NAME space (the opposite implication is due to CITE). By CITE, MATH must have the UMD property. In particular, MATH does not contain the MATH's uniformly. Hence by NAME 's theorem CITE, the space MATH is complemented in MATH (here MATH is a standard NAME function). Therefore, by REF , MATH has the (MRP). Now, MATH is an unconditional NAME decomposition of MATH. So, by REF , MATH must be isomorphic to MATH. Finally, it follows from NAME 's theorem CITE that MATH is isomorphic to a NAME space. |
math/9910122 | Let MATH be an order continuous NAME lattice. By REF , it is enough to show that every normalized sequence of disjoint elements of MATH is equivalent to the canonical basis of MATH. So let MATH be such a sequence in MATH. Then MATH admits an unconditional NAME decomposition MATH such that the MATH's are ideals of MATH and for all MATH, MATH. Now, by REF MATH is isomorphic to MATH and MATH is equivalent to the canonical basis of MATH . |
math/9910122 | Let MATH be a separable NAME lattice which is not order continuous. Then MATH is not MATH-complete (see REF ) and by a result of NAME (CITE, see also REF ) MATH contains a subspace isomorphic to MATH. Since MATH is separable, it follows from NAME 's Theorem CITE that this subspace is complemented in MATH. So, by REF , MATH does not have the (MRP). Then, the preceding Theorem concludes our proof. |
math/9910123 | Assume that MATH is exceptional. Let MATH be a lc blow-up. Since MATH is MATH-ample, the linear system MATH is base point free over MATH for MATH. Let MATH be a general member and let MATH. By NAME Theorem, MATH is lc. Since MATH is MATH-linearly trivial, MATH is lc and MATH for any component MATH of MATH. Hence MATH is irreducible and MATH. If MATH is another lc blow-up, then similarly MATH is irreducible and we have a boundary MATH such that MATH is lc and MATH. We claim that MATH and MATH define different discrete valuations of the function field MATH. Indeed, otherwise MATH is an isomorphism in codimension one and MATH. Since both MATH and MATH are ample over MATH, we have MATH. We may assume also that MATH and MATH have no common components. For MATH, define the linear function MATH by MATH and put MATH. Clearly, MATH and MATH (because MATH). We claim that MATH is lc for all MATH. Assume the opposite. Then MATH . Fix some log resolution of MATH factoring through MATH and let MATH be the new exceptional divisors. The value MATH can be computed from a finite number of linear inequalities MATH. Therefore MATH is rational and MATH is lc. Since MATH is not lc for any MATH, some inequality MATH is an equality (see CITE). Hence MATH. This contradicts the exceptionality of MATH. Thus MATH is lc for all MATH. In particular, MATH is lc. Since MATH is exceptional, MATH and therefore MATH. On the other hand, MATH and MATH, a contradiction. |
math/9910123 | Let MATH be a toric morphism which is a log-resolution of MATH. Denote MATH where MATH is the proper transform of MATH on MATH and by abuse of notation the divisors corresponding to MATH on MATH and on MATH are both denoted by MATH. Here we may assume that MATH are all primitive. Then if MATH, we obtain that MATH . This is because the discrepancy of MATH at MATH is MATH by CITE and the coefficient of MATH in MATH is MATH. Now consider the discrepancy MATH. Since MATH by REF , it follows that MATH, if MATH. In case MATH, define MATH from the proportion MATH. Then MATH and MATH for MATH. Since MATH, it follows that MATH . Here if one assumes that MATH, the right hand side of the inequality above is MATH which yields that MATH. If one assumes that MATH, then the inequality above is strict, since MATH. So the first assertions of REF are proved. For the second assertion of REF , consider the discrepancy MATH. This value is MATH which is greater than MATH, if MATH and non-negative if MATH by the argument above. Hence MATH is plt. Then, by the inversion of adjunction (CITE or CITE), MATH is normal and MATH is klt. For the second assertion of REF , note that MATH is dlt and MATH is smooth outside of MATH. Then MATH is smooth at a general point of MATH by the classification of MATH-dimensional log-canonical pairs CITE. Therefore MATH is smooth in codimension two along MATH, which yields MATH. |
math/9910123 | For the first assertions of REF , by REF , it is sufficient to prove that MATH is smooth in codimension one under the conditions MATH and MATH. If MATH has a singular locus in codimension one, then MATH has a singular locus in codimension one. Indeed, the restriction MATH of the canonical projection MATH is an open map. Then we obtain that MATH is a smooth morphism on the smooth locus of MATH, as MATH is a NAME variety and each fiber of MATH is smooth. Here the singular locus of MATH is contained in the invariant divisor MATH, because MATH is defined by MATH in the weighted projective space MATH and MATH is non-degenerate. Let MATH be a MATH-codimensional component of the singular locus of MATH and contained in an invariant divisor MATH of MATH. For simplicity, let MATH. Let MATH be a log-resolution of MATH which factors through the blow-up of MATH and MATH the proper transform of MATH. Since MATH, MATH is lc by the previous lemma and the adjunction. From the classification of MATH-dimensional log-canonical pairs CITE there exists an exceptional divisor MATH mapped onto MATH such that the discrepancy MATH. If MATH is mapped onto MATH, it follows that MATH, where MATH, MATH are positive rational numbers. Since MATH for MATH, it follows that MATH, which shows that the order of MATH is MATH, a contradiction. Now the first assertions of REF are proved. In particular MATH is irreducible, therefore MATH is irreducible and reduced. Noting that MATH is MATH-ample, one obtains that MATH is a lc blow-up. If MATH, then, by REF , MATH is dlt, therefore it is plt. |
math/9910123 | In this case, MATH, therefore the condition of REF holds. |
math/9910123 | For REF , it is sufficient to show the existence of a compact face MATH of MATH such that MATH, because for every compact face MATH there exists a weight MATH such that MATH. Since MATH there exists a compact face MATH such that MATH. Assume that MATH is a boundary point of MATH for every such MATH as above. Then MATH belongs to the non-compact face of MATH. So for any such MATH there exists MATH such that MATH contains MATH and MATH is on the boundary of MATH. Since MATH is on the boundary of MATH, it is on the boundary of MATH, therefore on the boundary of MATH, a contradiction. For the first statement of REF , note that one can take a weight MATH such that MATH by the argument above and apply REF . For such MATH, denote the proper transform of MATH under the MATH-blow-up MATH by MATH and MATH by MATH. Then in MATH, MATH and MATH coincide and MATH and MATH coincide, because the both are equal to MATH. Therefore the conditions for MATH and MATH to be non-exceptional are the same REF . |
math/9910123 | Let MATH be the toric morphism corresponding to the star-shaped decomposition by adding a MATH-dimensional cone MATH and MATH be the proper transform of MATH by MATH. First we show that MATH. Assume the contrary. If MATH, then MATH and MATH are MATH-divisors with different supports, a contradiction to MATH. If MATH, then both sides coincide, because the left hand side is of dimension MATH, while the right hand side is irreducible and of dimension MATH. Therefore the support of MATH is an invariant divisor on MATH, which yields that MATH is a power of a single coordinate, a contradiction. Now we obtain MATH. It implies that the coefficient of MATH in MATH is REF. Let MATH and MATH. If we put MATH, then MATH and the coefficient of MATH in MATH is MATH by CITE. Hence MATH for all MATH. Now we obtain that MATH for all MATH. By making the same procedure with exchanging the role of MATH and MATH, we obtain the opposite inequality MATH for all MATH, which yields MATH. |
math/9910123 | By REF , plt bow-up for MATH is unique. If a weight MATH gives a plt blow-up, then by REF , MATH is not a power of a single coordinate. Then apply REF . |
math/9910123 | By REF , it is sufficient to prove that MATH. Assume that MATH. Then a MATH-codimensional irreducible component of the singular locus of MATH is contained in MATH, which implies that MATH in MATH is contained in MATH in MATH for some MATH. Therefore MATH. By MATH, it follows that MATH. For the simplicity, let MATH. Here MATH is singular along MATH in MATH, so the weight MATH is represented by MATH for some integer MATH with MATH. Now, by the assumption on MATH, it follows that MATH, where MATH and MATH does not appear in MATH, as MATH is irreducible. Then MATH and MATH, which is a contradiction. |
math/9910123 | Let MATH a non-degenerate weighted homogeneous power series defining an exceptional canonical singularity in MATH and MATH . Assume that the set MATH is infinite and induce a contradiction. For MATH, define MATH. CASE: For MATH, define MATH. Then MATH is bounded by MATH from above, since MATH. Hence there exist a subset MATH and MATH such that MATH is a infinite set and MATH, MATH for any MATH, because MATH is finite and MATH is infinite. Therefore every MATH is written as MATH, MATH. CASE: For MATH, define MATH. If MATH is bounded, then, by taking infinite sets MATH and MATH smaller, we can write MATH for every MATH in the same way as in REF . In particular, if MATH is a power of a single coordinate, then MATH is bounded. Indeed if MATH is unbounded, then MATH, where MATH belongs to the left hand side and not to the right hand side, a contradiction. CASE: By the successive procedures, one obtains infinite sets MATH, MATH such that for every MATH, MATH and MATH is unbounded, where MATH. Indeed, if these procedures do not terminate, one obtains an infinite series MATH in MATH. Let MATH be the linear subvariety spanned by MATH. Then MATH for all MATH, because MATH appear in the weighted homogeneous polynomials in MATH. Since MATH, there exists MATH such that MATH for all MATH. Let MATH be the weight of an element MATH. Then MATH is contained in a hyperplane MATH for some MATH. Hence the infinite set MATH is contained in MATH which is a finite set because of MATH, (i=REF,,n), a contradiction. CASE: Now fix an element MATH and let MATH be a weight of MATH. As MATH is unbounded, one can take MATH such that MATH for MATH. Then MATH. So if MATH, then MATH. On the other hand, from the unboundedness of MATH, it follows that MATH. Now by REF , MATH-blow-up of MATH is an lc blow-up. On the other hand, for the weight MATH of MATH, MATH-blow-up of MATH is a plt blow-up by REF . Since neither MATH nor MATH is a power of a single coordinate, by REF MATH, which is a contradiction to that MATH is exceptional. |
math/9910123 | By REF we may assume (up to finite numbers of cases) that MATH is a weighted blow-up of fixed weight MATH. Then the exceptional divisor MATH is defined in the weighted projective space MATH by MATH. Thus we may assume that MATH is contained in some algebraic family. Now let MATH be a very ample divisor on MATH. Write MATH, where MATH and MATH's are prime divisors. Since MATH is ample, MATH. Thus the degree of components of MATH under the embedding MATH given by MATH is bounded. Then MATH belongs to a finite number of families (see, for example, CITE) and we may assume that MATH is fixed. Now we need to show only that MATH for all MATH. Indeed, in the opposite case we can take an infinite sequence MATH. Let MATH. Then MATH is nef and MATH. This contradicts REF . |
math/9910123 | By REF , MATH is lc. Note that MATH is plt (by Inversion of Adjunction). Applying CITE twice we obtain that both MATH and MATH are lc. |
math/9910123 | REF is obvious. Indeed, codimension two singularities of MATH are contained in MATH and MATH does not contain its components. To prove REF we consider the finite map MATH given by MATH . The ramification divisor is MATH, where MATH is MATH-th coordinate hyperplane on MATH. Let MATH be the preimage of MATH. The restriction MATH is also a finite morphism of degree MATH. Put MATH. By the ramification formula we have MATH . Since MATH is a normal crossing divisor, MATH is lc. Then MATH is lc by CITE or CITE. REF can be proved in a similar way. |
math/9910123 | Restricting MATH to MATH we obtain MATH . By REF , MATH and MATH. Hence, MATH. Taking into account that MATH, we get the assertion. |
math/9910123 | REF follows, for example, from the discussion REF REF follows easily from the fact that MATH is a finite abelian quotient of the projective space. |
math/9910123 | Let MATH be a finite étale in codimension one cover such that MATH is smooth. Set MATH and MATH. By CITE, MATH is plt for all MATH. Hence, all MATH's are smooth irreducible curves. Again, by CITE it is sufficient to show that MATH is klt. By our assumption, MATH, where MATH and MATH. Let MATH be the blow-up of MATH and let MATH be the crepant pull-back of MATH (that is, MATH). By CITE, MATH is klt if and only if so is MATH. Clearly, all the irreducible components MATH of MATH are smooth. Write MATH so that MATH is the exceptional divisor and MATH's are proper transforms of MATH's for MATH. It is easy to see that MATH for MATH and MATH. So we again have MATH. Thus, it is sufficient to prove our statement on MATH. We replace MATH with MATH and continue the process. At the end, we get the situation when MATH is a normal crossing divisor. In this situation, the inequalities MATH and MATH gives us that MATH is klt (and even canonical). |
math/9910123 | Let MATH be the minimal resolution, let MATH be the (irreducible) exceptional divisor and let MATH be the proper transform of MATH. Write MATH. Then MATH is a log-resolution and it is sufficient to show that MATH. By Adjunction, MATH. Thus MATH. |
math/9910123 | Let MATH be the set of points where MATH is not plt. Let MATH be a birational morphism which is a log resolution over MATH and an isomorphism outside of MATH. Write MATH where MATH is the proper transform of MATH and MATH is the exceptional divisor. Then MATH is smooth and by Adjunction MATH . We can write MATH . If MATH, then by construction, MATH is an isomorphism over MATH and by CITE, MATH, MATH and MATH. If MATH then by Connectedness Lemma CITE, MATH is connected near MATH. Thus the coefficient MATH of MATH at MATH is MATH. Now we have MATH, where MATH. Combining this with REF and MATH, we obtain MATH and MATH. In particular, MATH and MATH. |
math/9910123 | We have an exact sequence MATH . By NAME vanishing MATH, MATH. Hence, MATH. Similarly, MATH. Since MATH is an analytic germ near MATH, MATH. If MATH is analytically MATH-factorial, then MATH and MATH. |
math/9910124 | Let MATH be the characteristic of MATH, and choose a prime MATH. Let MATH denote the NAME characteristic. Since MATH is isomorphic to the blowup of MATH at REF points (CITE, for example), MATH . Since MATH is the blowup of MATH at REF points, MATH . On the other hand, combining the NAME spectral sequence MATH with the NAME formula (CITE or CITE) yields MATH where MATH is the generic fiber, MATH is the fiber above MATH, and MATH is the alternating sum of the NAME conductors of MATH considered as a representation of the inertia group at MATH of the base MATH. Since MATH is a smooth curve of genus MATH, MATH. If MATH is such that MATH is smooth, then all terms within the brackets on the right side of REF are MATH, so the sum is finite. The NAME conductor of MATH is trivial. Hence REF becomes MATH . Since MATH is a dimension, it is nonnegative, so the lemma will follow from the following claim: if MATH is singular, MATH with equality if and only if MATH is a nodal cubic. To prove this, we enumerate the combinatorial possibilities for a plane cubic, corresponding to the degrees of the factors of the cubic polynomial: see REF . The NAME characteristic for each, which is unchanged if we pass to the associated reduced scheme MATH, is computed using the formula MATH where MATH is the normalization of MATH, MATH is the genus of the MATH-th component of MATH, and MATH is the set of singular points of MATH. For example, for the ``conic + tangent," REF gives MATH . |
math/9910124 | The polynomial MATH must be squarefree, since otherwise MATH would factor completely. Hence MATH is reduced. If MATH is a smooth cubic curve, then it is of genus REF, and MATH by the NAME bound. Otherwise, enumerating possibilities as in REF shows that MATH is a nodal or cuspidal cubic, or a union of a line and a conic. The NAME action on components is trivial, because when there is more than one, the components have different degrees. There is an open subset of MATH isomorphic over MATH to MATH with at most two geometric points deleted. But MATH, so there remains a smooth MATH-point on MATH. |
math/9910124 | Existence of real points is automatic, since MATH is a plane curve of odd degree. Existence of MATH-points for MATH was proved just above the statement of REF . Consider MATH. NAME basis calculation shows that there do not exist MATH, MATH, MATH, MATH, MATH, MATH, MATH such that MATH and MATH are identical. Hence REF applies to show that for any MATH, the plane cubic defined by REF over MATH has a smooth MATH-point, and NAME 's Lemma implies that MATH has a MATH-point at least when MATH. When MATH, the curve reduced modulo REF, MATH consists of three lines through MATH, so it does not satisfy the conditions of REF , but one of the lines, namely MATH, is defined over MATH, and every MATH-point on this line except MATH is smooth on MATH. Hence MATH has a MATH-point. The same argument shows that MATH has a MATH-point whenever MATH, since the curve reduced modulo REF is MATH, which contains MATH. Finally, when MATH, the point MATH satisfies REF modulo MATH, and NAME 's Lemma gives a point MATH with MATH. This completes the proof. |
math/9910127 | It suffices, by a NAME argument, to extend the contact structure on MATH above to MATH, such that the characteristic foliation on MATH has a half-elliptic or a half-hyperbolic singularity. It therefore also suffices to treat the neighborhood of a Legendrian divide. Without loss of generality, let the Legendrian divide be MATH, with contact REF - form MATH. Now extend to MATH for a half-elliptic singularity and MATH for a half-hyperbolic singularity, on MATH, where MATH on MATH and MATH on MATH. |
math/9910127 | We will take an easy way out by using the following theorem, due to CITE: Let MATH be a smooth REF - manifold with boundary. Suppose MATH, MATH are two NAME structures with boundary on MATH. If the induced contact structures MATH, MATH on MATH are isotopic, then MATH. Let MATH be the NAME surface obtained from MATH by attaching REF - handles MATH corresponding to Legendrian surgeries with coefficients MATH along the link in REF . If MATH is the canonical class and MATH is a REF - dimensional class supported on MATH, then MATH. For the various MATH, the MATH are distinct. |
math/9910127 | CASE: Let MATH. Consider its one-sided components MATH (they are all along MATH). If MATH, not all the one-sided components have the same sign. We have two possibilities: REF there exists a positive one-sided MATH and negative one-sided MATH which lie further toward MATH as in the left-hand side of REF (or signs reversed), or REF there is a positive boundary-parallel MATH as well as a negative boundary-parallel MATH. Let MATH be the dividing curve on MATH which is `farthest' from MATH (that is, the half-disk cut off by MATH contains the other dividing curves which bound MATH). For both cases, pass to the cover MATH. Let us first consider REF . There exist lifts MATH and MATH, MATH, for which MATH, after rounding the edges, has a dividing curve MATH which bounds a disk. In fact, a lift of MATH on MATH will connect up to a lift of MATH on MATH for suitably chosen MATH. The existence of a null-homotopic dividing curve then implies that MATH is overtwisted. For REF , take MATH, MATH as above, as well as lifts MATH on MATH and MATH on MATH. Pick MATH so that MATH connects the left endpoint of MATH to the left endpoint of MATH, MATH connects the right endpoint of MATH to the left endpoint of MATH, and so on. What we still lack is a dividing curve connecting the right endpoint of MATH to the right endpoint of MATH. Take MATH, MATH, as well as MATH, after rounding the edges. If we pick MATH appropriately, we can make the desired connection along MATH. Now, the dividing curve sits on the branched surface MATH, and there exists an overtwisted disk on this branched surface. If the tight contact structure on MATH has a horizontal convex annulus MATH, all of whose one-side components are boundary-parallel with the same sign, then MATH can be embedded into, and is universally tight because MATH is. REF are left for the reader. |
math/9910127 | Consider the factorization MATH, where MATH is a nonrotative outer layer and MATH is its horizontal annulus. We prove that MATH corresponding to another nonrotative outer layer MATH is disk-equivalent to MATH. Write MATH. Let MATH (respectively, MATH) be the set of isotopy classes of nonrotative tight contact structures MATH with a fixed boundary characteristic foliation and MATH, which glue to MATH to yield a tight contact structure on MATH which is MATH-invariant (respectively, a tight contact structure on MATH). Here, the MATH-invariant tight contact manifold is isomorphic to an invariant neighborhood of a convex surface MATH (or MATH). By REF , a nonrotative MATH is characterized by the dividing set of its horizontal annulus MATH. Any MATH will have exactly two endpoints along MATH and exactly two nonseparating arcs. Associate to MATH (respectively, MATH) the corresponding set of isotopy classes MATH (respectively, MATH) of MATH. Let MATH be the horizontal annulus for MATH, where we assume that MATH and MATH have common boundary MATH. Now, MATH if and only if MATH consists of exactly two parallel nonseparating arcs. Clearly, MATH, since the MATH-invariance of MATH implies there is a contact diffeomorphism MATH. Of course, MATH, unlike MATH, depends on the ambient MATH, and MATH may or may not contain certain MATH for which MATH contains (necessarily homotopically essential) closed curves. See REF for various possibilities of MATH. The induction is done by fixing MATH and inducting on MATH over the space of all nonrotative outer layers MATH with MATH. Note that all nonrotative MATH with MATH can be embedded inside a MATH-invariant neighborhood of MATH by folding (see REF), so all contact structures on MATH constructed this way are tight. When MATH, the nonrotative outer layer is clearly unique. Therefore, assume the theorem is true for MATH, and we prove it for MATH. There are two cases: either MATH has at least two MATH-parallel curves or there is only one MATH-parallel curve. Suppose first that there are at least two MATH-parallel curves on MATH. Let MATH be an arc on MATH whose endpoints are consecutive points of MATH, that is, MATH is MATH-parallel. If the endpoints of MATH coincide with the endpoints of a MATH-parallel curve of MATH, then, for any completion of MATH to a dividing set MATH, the gluing MATH corresponds to an overtwisted contact structure. On the other hand, if the endpoints of MATH are not REF the two endpoints of the nonseparating curves of MATH and not REF the two endpoints of a MATH-parallel curve of MATH, then MATH can be completed into some MATH. We now summarize the completability of MATH to an element in MATH: unknown if endpoints are REF , no if endpoints are REF , and yes otherwise. (Here ``unknown" means that it depends on whether adding an extra MATH-twisting MATH layer to MATH yields a tight contact structure or an overtwisted contact structure.) Now, since there are at least two MATH-parallel curves of MATH, there are at least two MATH-parallel MATH which cannot be completed to an element of MATH, and at least one of them must have the same endpoints as a MATH-parallel curve of MATH. (This follows from repeating the same argument for MATH instead of MATH.) Thus, there is a common MATH-parallel position for both MATH and MATH. Now, attach a horizontal annulus with MATH nonseparating curves and one MATH-parallel dividing curve MATH right next to the common MATH-parallel position of MATH and MATH as in REF , and use the inductive step. Suppose now that there exists only one MATH-parallel arc of MATH. Then the two nonseparating curves must be consecutive (that is, one of the regions of MATH divided by these two curves does not contain any other dividing curves), and all the separating curves must be nested concentrically around the one MATH-parallel dividing curve. See REF . The MATH-parallel arc MATH on MATH satisfying REF is at the center (solid line), and MATH satisfying REF is given by dotted lines. Then MATH satisfies one of the following: CASE: MATH. CASE: The positions of REF are reversed. CASE: Positions REF for MATH are both REF for MATH. In each case, MATH and MATH are disk-equivalent. |
math/9910127 | The actual isomorphism is not an arbitrary isomorphism, but an isotopy in the following sense. Let MATH be an element of MATH as in the proof of REF . Then there exists a contact isotopy of MATH to MATH inside MATH. This is clear from the MATH-invariance of MATH. Now we claim that the disk-equivalence of MATH and MATH implies that MATH is MATH-invariant, thus proving the contact isotopy of MATH to MATH inside MATH. Write MATH, MATH, and MATH. We then complete MATH (respectively, MATH) by attaching a disk MATH and (respectively, MATH) along MATH (respectively, MATH). By the disk-equivalence, the dividing sets on MATH and MATH are identical and consist of exactly one MATH-parallel arc along MATH. This in turn implies that, after removing MATH from MATH, MATH consists of exactly two nonseparating arcs. This completes the proof. |
math/9910127 | The smallest state transition unit MATH consists of attaching a bypass along MATH to obtain MATH. Hence, every pair MATH, MATH of isotopic tori is related by a sequence of bypass attachments. Suppose that MATH is tight for every convex MATH with MATH, obtained from MATH via a sequence of bypass moves which do not change MATH. Observe that if MATH is the sequence of bypass moves which extricates the original MATH from a candidate overtwisted disk, then there will exist intervals MATH where MATH and MATH inbetween, or half-intervals MATH, where MATH and MATH thereafter. We will prove that the state transitions when MATH are rather superficial, and that MATH. We inductively assume the following: CASE: MATH is one of the MATH between MATH and MATH (or MATH). CASE: MATH is tight. CASE: There exist nonrotative layers MATH, MATH with MATH and MATH, and such that MATH is MATH-invariant. CASE: There is an isomorphism MATH . Let MATH and MATH be the horizontal annuli corresponding to MATH and MATH. Let MATH be the layer between MATH and MATH. It is nonrotative because MATH and we are considering a single bypass move from MATH to MATH. The hypotheses of the theorem guarantee an extension to MATH, a nonrotative outer layer of MATH. There also exists a nonrotative outer MATH on the other side of MATH. Call the corresponding new horizontal annuli MATH and MATH. (Also let MATH.) The key is to prove that the new layer MATH containing MATH is MATH-invariant. This is done by completing MATH to a disk MATH, MATH to a disk MATH, and likewise forming MATH and MATH from MATH and MATH. If we put MATH and MATH together to form MATH so the dividing curves match up, then there is exactly one dividing curve, since MATH consists of two parallel nonseparating curves. (The corresponding toric annulus is MATH-invariant.) Now use REF to see that MATH must also consist of exactly one dividing curve, due to disk-equivalence. Now, MATH is obtained by removing two small disks from MATH, each containing a short arc of the dividing set. Therefore, MATH must consist of parallel nonseparating curves. This proves that Condition C of the inductive step also holds for MATH. Next, Condition D is satisfied, since MATH and MATH due to REF . Condition B is now obvious, since MATH is obtained from MATH by folding. |
math/9910127 | Let MATH. Then MATH is rotative and any pair of nonrotative layers MATH can be extended to a pair of nonrotative outer layers MATH using bypasses and the Imbalance Principle CITE. Moreover, for each state transition MATH, if MATH is rotative, then so is MATH. |
math/9910127 | We take MATH, MATH, and MATH. As in the proof of REF , consider the set MATH of nonrotative tight contact structures MATH with MATH, which glue to MATH to yield a tight contact structure on MATH. The key difference between this case and REF is that it is possible to determine MATH and its corresponding MATH precisely. That is, MATH consists of all MATH for which MATH does not have any homotopically trivial dividing curves. - in other words, the ``unknown" gluings which produced the middle configuration in REF are now known to be tight gluings. Elements of MATH correspond to MATH, whose attachment makes MATH either into a basic slice or adds extra twisting by a multiple of MATH. Now, we want to prove that if MATH and MATH are two horizontal annuli for MATH, then MATH modulo parallel closed essential curves. This is proved by induction on MATH. If MATH, then there are two possibilities for MATH modulo parallel closed essential curves, corresponding to the two possible positions for MATH-parallel dividing curves. In this step only, we attach templates which are basic slices MATH (not nonrotative layers) with MATH and MATH, and corresponding ``horizontal" annuli MATH. The two basic slices are also distinguished by the positions of the MATH-parallel dividing curves along MATH. (As before, we are assuming that MATH and MATH. Note they have a common boundary MATH.) Since the gluing is tight if and only if a closed homotopically trivial curve does not appear on MATH, the two possible MATH can be distinguished using templates. Next, assume inductively that the claim holds for MATH. Let MATH. Now any arc MATH on MATH with consecutive endpoints on MATH can be extended to some MATH, if and only if the endponts of MATH are not the endpoints of a MATH-parallel dividing curve of MATH. This implies that the set of MATH-parallel curves must be the same for MATH and MATH. We then reduce to the case MATH in the same manner as in the proof of REF . |
math/9910127 | In this case, we can apply the same template matching as in REF . Let MATH be an outermost rotative layer with MATH, and MATH the corresponding horizontal annulus. Let MATH be the set of configurations on MATH, corresponding to nonrotative MATH for which MATH remains tight. We claim that MATH once again is the set of MATH for which no homotopically trivial dividing curves appear after merging with MATH. Note that there might be some attachments of MATH for which the twisting increases by a multiple of MATH when we compare MATH and MATH. This happens when homotopically nontrivial closed curves are created on MATH. The tightness is guaranteed by NAME 's gluing theorem for universally tight contact structures along incompressible tori (see CITE). Finally, MATH is sufficient to recover MATH. This proves that any two rotative outermost layers are contact diffeomorphic. |
math/9910133 | Any saturated torsion free rank REF subsheaf of MATH is invertible of the form MATH and gives an exact triple MATH where MATH is a subscheme of MATH of codimension REF. The triple breaks the NAME stability of MATH if and only if MATH. If we assume that MATH has global sections, then there exists a triple REF with MATH, hence MATH is not stable. If we assume that MATH has no global sections, then in any triple REF for MATH, we have MATH, because MATH. Hence MATH is stable. |
math/9910133 | This is NAME 's Theorem, proved by him for MATH-(semi)stable sheaves CITE. For another approach to the proof and for relations between different notions of (semi)stability, see e. CASE: CITE. |
math/9910133 | The case of odd MATH in REF follows trivially from NAME - NAME - NAME: we have MATH. For the remaining cases, the proof goes exactly as in CITE. The first step is to show that if MATH (MATH), then MATH for all MATH. The second step is to verify that there are no curves on MATH that might be zero loci of sections of MATH. So, let MATH, MATH, MATH. Assume that MATH. By NAME duality, MATH. Hence MATH. We have MATH if MATH and MATH if MATH. Hence MATH, and there exists a non-trivial extension of vector bundles MATH . We have MATH if MATH and MATH if MATH, so MATH is unstable by REF . To fix ideas, restrict ourselves to the case MATH, the other case being completely similar. The unstability of MATH can manifest itself in two ways: either MATH contains a rank REF saturated subsheaf MATH with MATH, or there exists a non-trivial morphism of sheaves MATH with MATH. In the first case, MATH, hence MATH, hence MATH, and this contradicts the stability of MATH. In the second case, MATH. If MATH, then MATH and MATH descends to a non-trivial morphism MATH, which contradicts the stability of MATH. Hence MATH or MATH. It cannot be MATH, because otherwise the extension REF would be split. So MATH and we obtain the exact triple MATH in which MATH is a reflexive rank REF sheaf. We have MATH, MATH, hence MATH is unstable by NAME 's inequality. By REF , MATH, which implies MATH and hence MATH. This contradicts the stability of MATH. Thus we have proved MATH. As MATH, the scheme MATH of zeros of a non-trivial section MATH of MATH is a l. c. i. of pure codimension REF. Hence MATH fits into the following exact triple MATH . We have MATH, MATH, MATH. It remains only to verify the case of MATH. We have MATH and MATH is embedded into MATH by a subsystem of the canonical system. The exact triple REF , twisted by MATH, implies that MATH is not contained in a hyperplane. Hence MATH is not connected. It cannot be a union of more than one connected components either, because at least one of them should be of degree MATH and hence MATH cannot be ample. The proof is completed in a similar way in the case MATH. |
math/9910133 | We are using the same argument as that of REF . The family of quartic REF-folds in MATH is parametrized by MATH, and that of the Pfaffian representations of quartic REF-folds by an open set in the variety MATH of linear morphisms between the two projective spaces. So, we are going to specify one particular quartic REF-fold MATH which admits a Pfaffian representation MATH, then we will show that the differential of the map MATH at MATH is surjective, and this will imply that MATH is dominant. Let MATH . A point MATH is the proportionality class of a MATH skew-symmetric matrix of linear forms MATH and is given by its MATH homogeneous coordinates MATH such that MATH (MATH). We have MATH, where MATH denotes the Pfaffian of the MATH matrix obtained by deleting the MATH-th and the MATH-th rows and the MATH-th and the MATH-th columns of MATH. Computation by the NAME REF program CITE shows that, for the above matrix MATH, REF quartic forms MATH generate the whole REF-dimensional space of quinary quartic forms, hence MATH is of maximal rank at MATH. One can also easily make NAME REF to verify that MATH is in fact nonsingular, though this is not essential for the above proof. It remains to verify that the generic fiber of the induced map MATH is REF-dimensional. By counting dimensions, one sees that this is equivalent to the fact that the stabilizer of a generic point of the NAME MATH in MATH is REF-dimensional. Take a generic REF-dimensional linear subspace MATH. Then the quartic REF-fold MATH is generic, and hence MATH is trivial. Thus the stabilizer MATH of MATH in MATH acts trivially a MATH, and hence on MATH. This implies the triviality of MATH by REF. |
math/9910133 | If MATH is given by the extension REF , then twisting by MATH and using MATH for MATH REF , we see that MATH. The stability follows from REF . Smoothness and dimension. The stability implies that MATH is simple, that is MATH. Hence the tangent space MATH at MATH is identified with MATH, and if MATH, then MATH is smooth at MATH of local dimension MATH. As MATH, we have MATH. By NAME duality, MATH. By REF , MATH. Together with the ACM property for MATH this gives MATH for all MATH. Now, from REF tensored by MATH, we obtain the isomorphisms MATH . Further, the restriction sequence MATH yields MATH, so to finish the proof, it remains to prove the vanishing of MATH. By REF , the vanishing of MATH follows from the fact that the map MATH, introduced in the proof of REF , is dominant. As MATH, we have MATH, and we are done. |
math/9910133 | CASE: The restriction sequence REF yields MATH. We proved in REF the vanishing of MATH. As MATH is the scheme of zeros of a section of MATH, we have MATH. So, we obtain MATH. By NAME - NAME, MATH and we are done. CASE: We have MATH. We are going to show that this implies MATH. First, by NAME duality, MATH. From the restriction sequence MATH and from the fact that MATH, we deduce that MATH. Now, the exact triple MATH yields MATH. The ACM property for MATH and MATH imply MATH. Now, the triple MATH and the NAME duality give MATH. By NAME - NAME, MATH. CASE: The sections of MATH are naturally identified with elements of MATH via the embedding of MATH into the trivial rank REF vector bundle MATH. Let MATH be the classifying map, sending each MATH to the projectivized kernel of MATH, considered as a point of MATH, and MATH the restriction of MATH to MATH. We can choose the coordinates in MATH in such a way that MATH. Hence MATH, where MATH is the hyperplane MATH in MATH, and MATH is the NAME subvariety of all the lines contained in the hyperplane. We can also write MATH. The closure of the MATH-fold MATH in MATH is defined by the MATH cubic NAME MATH. As cubic forms, the NAME MATH, MATH do not depend on the variables MATH, MATH. Therefore MATH is isomorphic to the cone MATH with vertex (or ridge) MATH and base MATH; here MATH is REF-th principal adjoint matrix of the matrix MATH, that is, MATH is obtained from MATH by deleting its last column and row. It is well known that the vanishing of the principal minors of order MATH of a skew-symmetric MATH matrix is equivalent to the vanishing of all its minors of order MATH, so MATH is the locus of MATH skew-symmetric matrices of rank MATH. The projection MATH with center MATH maps isomorphically (for generic MATH) the intersection MATH to MATH, where MATH. This ends the proof. |
math/9910133 | CASE: Indeed, MATH is the image of MATH. CASE: This follows from the count of dimensions: MATH. CASE: According to CITE, if MATH, then the codimension of any component of MATH in MATH is at most MATH. Applying this to our case, we see that the dimension of every component of MATH is at least REF. Hence the component MATH, containing the image of MATH, is of dimension MATH. The dimension of MATH is REF, so it remains to show that MATH is dominant over MATH. Take a generic MATH from the image of MATH. MATH is a smooth ACM curve in MATH. By the definition of MATH, every small (analytic or étale) deformation of MATH is accompanied by a deformation of the theta-characteristic MATH embedding MATH into MATH. The ACM property being generic, any generic small deformation of MATH is again in the image of MATH, and we are done. |
math/9910133 | It is easily seen that MATH, so, given MATH, the NAME construction determines MATH uniquely. This yields MATH as a set theoretic map. An obvious relativization of the NAME construction shows that it is indeed a morphism. Further, we have MATH by stability of MATH and REF , so the projective space MATH is injected into MATH by sending each section MATH of MATH to its scheme of zeros. Hence the fibers of MATH are set-theoretically REF-dimensional projective spaces. The proof of the last assertion of the proposition is completely similar to that of REF. |
math/9910133 | Twisting REF exact triples in the proof of REF by MATH, one can see that the assertion is equivalent to MATH . The last equalities follow immediately from the resolution for MATH, obtained from REF by restriction to MATH and twisting by MATH: MATH . |
math/9910135 | Let MATH, MATH be a rigidity pair and MATH be the associated NAME transform: MATH and MATH. Note here that MATH is a NAME space because of the simplicity of the unit object MATH. From the inequalities MATH the C*-algebra MATH is continuously isomorphic to the NAME space MATH with the bounded inverse, whence MATH is reflexive as a NAME space, proving MATH. |
math/9910135 | If MATH has the finite NAME index, it is rigid as a consequence of the existence of NAME duality. Conversely, let MATH be a rigid object in MATH. Then by NAME 's finite-dimensionality, we may assume that MATH is simple. In that case, the finiteness of NAME index follows from the non-triviality of MATH and MATH as pointed out in CITE. |
math/9910135 | Recall that MATH is imbedded into MATH by MATH . Thus a central element MATH in MATH is mapped into MATH . If we identify this with a central element MATH in MATH, then we have MATH for any MATH, MATH and MATH (MATH as MATH supports the whole MATH), that is, MATH because MATH for some MATH. |
math/9910135 | Consider the case MATH. By the assumption MATH for any MATH, we can find a finite family of partial isometries MATH in MATH (MATH can be chosen as the positive integer satisfying MATH) such that MATH . Then MATH and we have MATH . The reverse inclusion is trivial. |
math/9910135 | We will only show the density of MATH. Let MATH be orthogonal to MATH and express it as MATH with MATH an orthonormal basis in MATH and MATH. Given any MATH, choose MATH, MATH, MATH and MATH arbitrarily. Then MATH is orthogonal to the vector MATH by assumption, that is, MATH . Since the set MATH is total in MATH by REF , we can simplify the orthogonality condition to MATH for any MATH. On the other hand, NAME isomorphisms MATH show that MATH is total in MATH. Consequently we have MATH for any MATH and any MATH, MATH. |
math/9910135 | By the formula of MATH-valued inner product in MATH REF , MATH that is, the correspondence MATH gives a well-defined isometry MATH, which is in fact a unitary by REF . |
math/9910135 | From the definition of unitaries, it is immediate to check that both of MATH and MATH are mapped to MATH. Thus the unitaries turn out to be the obvious identification MATH on the subspaces MATH and MATH. Since MATH and MATH are orthogonal complements of these, the assertion holds. |
math/9910135 | By REF , MATH where an element MATH in the right end is transfered as MATH . |
math/9910135 | Let MATH be an object in MATH. Then we have the following commuting squares MATH and then we can apply the imbedding theorem of amalgamated free products (see REF for example) to find that MATH is a NAME subalgebra of the amalgamated free product MATH. |
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