paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0101248 | Let MATH be any totally geodesic hyperplane in MATH. If MATH has a space-like isometric embedding in MATH, then the projection from the image to MATH is conformal by REF . Therefore MATH is conformal to MATH. Conversely, if MATH is conformal to MATH then there exists a function MATH such that MATH; then the graph of MATH above MATH is, by REF , isometric to MATH. |
math/0101248 | Let MATH be an orthonormal frame for MATH which diagonalizes MATH, and let MATH be the associated eigenvalues of MATH. Call MATH the sectional curvature of MATH on REF-plane generated by MATH and MATH. Then, by REF : MATH so that: MATH and MATH so that: MATH and both results follow. |
math/0101248 | We have to prove the existence and uniqueness of MATH. MATH is conformal to MATH, so there exists a function MATH such that MATH. Choose a totally geodesic plane MATH, and let MATH be the graph of MATH above MATH. Then, by REF , the metric induced on MATH is MATH. Now let MATH. By REF , there exists a unique totally geodesic plane MATH in MATH which is tangent to MATH at MATH. MATH is the graph above MATH of a function MATH on MATH. Then MATH is the metric induced on MATH, and is isometric to MATH, so MATH satisfies the conditions set on MATH. Conversely, if MATH satisfies those conditions, then the graph MATH of MATH above MATH has as induced metric MATH, so it is a totally geodesic plane, and moreover it is tangent to MATH at MATH. Thus, by REF , MATH, and MATH. |
math/0101248 | Since MATH is conformal to MATH, so there exists a function MATH such that MATH is isometric to MATH. If MATH is any totally geodesic plane in MATH, the graph MATH of MATH above MATH has MATH as its induced metric. Moreover, by the previous lemma, MATH is NAME if and only if MATH has MATH positive definite, so if and only if MATH is convex. And MATH is NAME if and only if MATH has its eigenvalues in MATH, so if and only if MATH is tamely convex. |
math/0101248 | REF shows that any NAME metric is induced on a convex surface in MATH, and REF then indicates that it has curvature strictly below MATH. Similarly NAME metrics are induced on tamely convex surfaces, which have curvature MATH by REF . |
math/0101248 | We only have to prove the existence and uniqueness of MATH. It is well known that there exists a unique hyperbolic metric in the conformal class of MATH, that is, a unique function MATH such that MATH has constant curvature MATH. We also call MATH the induced function on MATH. Then MATH is isometric to MATH, and this defines a function MATH on MATH which is invariant under an action of MATH by isometries. Now choose a dual plane MATH. Its induced metric is isometric to that of MATH; choose an isometry between MATH and MATH. This defines a function MATH on MATH, and by construction and REF , the graph of MATH above MATH is isometric to MATH. We identify MATH with this graph. Now choose MATH, and let MATH be the totally geodesic plane tangent to MATH at MATH. MATH is a graph above an hemisphere MATH of MATH, thus MATH is also the graph above MATH of a function MATH; by construction, MATH satisfies the conditions on MATH. Conversely, if MATH is a function satisfying those conditions, then the graph of MATH above MATH is a hemisphere of a totally geodesic plane which is tangent to MATH at MATH, so MATH. |
math/0101248 | First note that any metric MATH on MATH has an equivariant isometric embedding into MATH whose representation fixes a dual plane. Indeed, there is a unique function MATH such that MATH is hyperbolic; MATH can then be identified with an equivariant function defined on a dual plane MATH, and then MATH is isometric to the graph of MATH aove MATH. The previous proof then indicates that MATH is convex if and only if MATH is NAME, and tamely convex if and only if MATH is NAME. Therefore, the dual immersion in MATH is NAME if and only if MATH is NAME, and convex if and only if MATH is NAME. In this last case, the convexity implies that the immersion is an embedding. |
math/0101248 | Again we only have to prove the existence and uniqueness of MATH. We know that there exists a function MATH on the universal cover of MATH such that MATH is isometric to an open dense subset of MATH. This defines MATH as the graph of MATH above an open dense subset of a totally geodesic plane MATH (with the induced metric). The rest of the proof is just like for REF , and uses the uniqueness of the conformal change of metric. |
math/0101248 | We already know from the proof of REF that MATH is isometric to the graph of a unique (up to global isometries) graph above a totally geodesic plane MATH. Taking the dual surface in MATH gives an immersion MATH of MATH in MATH which is NAME if MATH is NAME, and convex if MATH is NAME. Moreover, MATH acts by conformal transformations on MATH, so, by REF , by isometries on MATH. By construction, MATH is invariant under those isometries. Thus MATH is isometric to the quotient by MATH of the image of MATH with its horospheric metric. |
math/0101249 | Suppose REF holds for all MATH, and REF for MATH. From REF we deduce that MATH for MATH. Comparing this with REF shows that MATH should satisfy MATH. Therefore, setting MATH shows that REF holds for all MATH. Similarly, if REF holds for all MATH and REF holds for MATH, then it holds for all MATH. This proves the first part of the theorem. For the second part, we must show that MATH is special Lagrangian wherever it is nonsingular, that is, wherever MATH is an immersion. Now MATH is an immersion at MATH when MATH are linearly independent, and then MATH. Thus we must show that MATH is an SL REF-plane MATH in MATH for all MATH for which MATH are linearly independent. By REF , this holds if and only if MATH . Using REF we find that MATH . From REF we deduce that MATH. Thus from REF we have MATH using REF in the second line and REF in the third. This proves the first equation of REF . The second follows in the same way, and the third from MATH using REF . To prove REF , observe that MATH where in the first line the terms MATH are determinants of complex MATH matrices, and MATH are regarded are complex column matrices. As every term in the second line is real, MATH is real. Thus MATH, proving REF . |
math/0101249 | For the first part, by REF - REF we have MATH . The equation MATH follows in the same way. To prove MATH are orthogonal we use REF - REF and the formula MATH. Thus we have MATH using REF in the second line and REF in the third. In the same way we show that MATH, and so MATH are orthogonal. Using REF we obtain MATH . By REF , the terms in MATH and MATH vanish. Also, using REF we have MATH so the term in MATH vanishes. Thus, replacing MATH by MATH, etc., we get MATH . In the same way, we find that MATH . Now using REF one can show that MATH . The last five equations prove REF - REF , as we want. Finally, it follows from what we have proved so far that MATH is a conformal map, and as its image is minimal, it is also harmonic. As MATH has Legendrian image, MATH is also conformal and harmonic, in the usual way. |
math/0101249 | As MATH and MATH are orthogonal, REF implies that the area form on MATH is MATH. Also, as the period lattice is generated by MATH and MATH, we can divide MATH into MATH copies of the basic rectangle MATH, each of which has area MATH. REF follows immediately. |
math/0101249 | Since MATH where MATH by REF , if MATH then the sequence MATH exists for all MATH and is given by MATH . Since MATH, REF - REF follow from REF - REF . Thus MATH is nonisotropic, as MATH exists for all MATH. But any conformal map MATH is isotropic or superconformal from REF, so MATH is superconformal. If on the other hand MATH then MATH, so MATH does not exist. Thus MATH is isotropic. By REF - REF , MATH and MATH exist and are given by REF - REF with MATH. But the harmonic sequence of an isotropic map MATH has length at most MATH, so this is the whole of the harmonic sequence. |
math/0101249 | Let MATH be the subalgebra of MATH generated by MATH, MATH and MATH, and suppose for a contradiction that MATH. Let MATH, and take MATH to be real, and of lowest degree MATH. That is, MATH with MATH for MATH, and every polynomial Killing field of degree less than MATH lies in MATH. As MATH, REF with MATH and REF with MATH show that MATH satisfies MATH . Divide into the three REF MATH, REF MATH, and REF MATH for some MATH. We will prove a contradiction in each case in turn. In REF implies that MATH is diagonal, and then as MATH is nonzero, the first equations of REF show that MATH is a multiple of the identity. So write MATH for some MATH. Taking the trace of REF for MATH gives MATH, as the trace of any commutator is zero. Thus MATH is constant, and MATH, MATH. For MATH, consider MATH. This is a polynomial Killing field of degree less than MATH, as we have cancelled the terms in MATH. Therefore MATH. But MATH, so MATH, a contradiction. Also, when MATH we have MATH. This eliminates REF . Similarly, in REF and the first equations of REF imply that MATH for some function MATH. The second equation of REF is equivalent to MATH, so that MATH is holomorphic. Using the fact that MATH is independent of MATH one can show that MATH must be constant. This determines MATH and MATH. By REF , the leading term of MATH is MATH . Suppose for the moment that MATH, so that MATH. Consider MATH . We have cancelled the terms in MATH, so MATH is a polynomial Killing field of degree less than MATH, and lies in MATH. So MATH lies in MATH, a contradiction. The cases MATH and MATH must be dealt with separately. By explicit calculation we prove that MATH is a multiple of MATH when MATH, and a linear combination of MATH, MATH and MATH when MATH. So MATH, finishing REF . In the same way, in REF we find that MATH for some constant MATH. When MATH we define MATH and deduce that MATH, so that MATH. The case MATH we deal with separately, by showing that MATH is a linear combination of MATH and MATH, and so lies in MATH. This completes the proof. |
math/0101249 | The first part of the theorem, that if REF - REF hold for all MATH and REF - REF for some MATH, then REF - REF hold for all MATH, follows as in REF . For the second part, we must prove that MATH is special Lagrangian wherever MATH is an immersion. As in REF , this holds if and only if MATH . Using REF - REF we find that MATH . REF give MATH using REF in the second line and REF in the third. This proves the first equation of REF . The second and third follow in a similar way. To prove REF , observe that MATH . Thus MATH is real, and so MATH. |
math/0101249 | REF implies that MATH and MATH are orthogonal to MATH. As MATH, it follows that MATH and MATH are linearly dependent. Therefore MATH. This may be rewritten in matrix form as MATH . Similar equations hold between the MATH and MATH, and between the MATH and MATH. Now the MATH matrix appearing in REF has trace zero and determinant MATH. As by REF are distinct and nonzero, this determinant is nonzero. Hence MATH is a trace-free, nondegenerate quadratic form on MATH. Therefore, MATH must be a unit vector in the intersection of the plane MATH and the quadric cone REF in MATH. Let MATH be the plane perpendicular to MATH, and consider the restriction MATH of MATH to MATH. As MATH is a unit vector, we have MATH . But MATH by REF , so MATH is trace-free. Thus, by the classification of quadratic forms on MATH, there exists an orthonormal basis MATH for MATH such that MATH for some MATH and all MATH in MATH. If MATH then MATH, so MATH is degenerate, a contradiction. So MATH, and therefore MATH are unique up to sign and order, with MATH. As MATH are orthonormal they automatically satisfy the first three equations of REF . But by construction we have arranged that MATH and MATH are linearly dependent, so MATH for MATH. The fourth equation of REF then follows from the second and third. |
math/0101251 | Consider a product of matrices MATH with MATH. The first statement of the lemma can be reformulated that if MATH, MATH for MATH and MATH, with at least one of these inequalities strict, then MATH are all positive. We will show, in fact, that MATH are all positive with the largest given by the following table: MATH . The cases MATH, MATH, or MATH can be quickly verified by hand; we leave this to the reader. Next suppose MATH and MATH. A simple induction on MATH shows that the entries of the product then satisfy MATH, MATH, and MATH. This proves the first case of the above table. Moreover, the same induction shows that MATH unless MATH. By post- and/or pre-multiplying by MATH one now easily completes the proofs of the other three cases of the table. For the converse, consider MATH with MATH and MATH. Assume first that MATH is the largest of MATH. Then we must have MATH and MATH. Put MATH. If we write MATH then MATH, MATH, MATH and MATH. Thus, if MATH we have MATH. Also, MATH, and MATH and MATH. It follows that MATH are all positive and MATH is the largest. Thus we can proceed inductively. The value of MATH is easily seen to decrease at each inductive step, and when MATH we have MATH, so we can write MATH . Thus, our matrix MATH is a product of the desired type with all MATH. The other three cases of the above table now follow easily as before by first post- and/or pre-multiplying MATH by the inverse of MATH. Finally, the uniqueness of the representation follows by noting that the value we chose for MATH in the above argument is forced by the fact that MATH by the calculation of the first part of the proof. |
math/0101251 | As described in CITE (see ``REF " on p. REF), MATH is obtained by plumbing from the plumbing diagram MATH . The two end vertices of this plumbing diagram give the pieces MATH that are pasted. As computed on page REF, the pasting map is given by the matrix product MATH, where MATH and MATH with MATH or MATH according as MATH or MATH. This gives the claimed gluing matrix after left-multiplying by MATH to make the gluing map unimodular (this reverses the orientation of the second MATH). The final part of the lemma says quotient-cusps are taut, in the sense that a singularity with the same topology as a quotient-cusp is analytically that quotient-cusp. This was proved by CITE. The fact that the fundamental group suffices to determine the topology is part of REF. |
math/0101251 | See CITE. The second part of the last sentence is again NAME 's tautness of cusp singularities CITE. Our orientation convention here is different from CITE (where it was not made explicit), so the matrix given there was MATH rather than MATH. |
math/0101251 | Denote by MATH the link of the cusp singularity that canonically double-covers our quotient-cusp. Since this is an abelian cover, MATH is covered by the universal abelian cover MATH of MATH. MATH fibers over the circle MATH with fiber the torus MATH, giving a semi-direct product representation MATH where MATH is free abelian of rank MATH and MATH is infinite cyclic. MATH is the abelian subgroup of MATH generated by MATH and MATH. The elements MATH and MATH are related to these by the unimodular relations MATH . Thus MATH and MATH also generate MATH. We can take MATH to be generated by the element MATH so the subgroup of MATH generated by MATH and MATH is normal of index MATH. Now the subgroup MATH of MATH generated by MATH and MATH clearly has index MATH in MATH. Since MATH equals MATH modulo MATH, the subgroup generated by MATH and MATH has index MATH in MATH, hence index MATH in MATH. On the other hand, it is evident from the presentation of MATH that MATH, and MATH are in MATH. It follows that the subgroup they generate is MATH as claimed. Finally, the above relations show that MATH is also generated by MATH and MATH, giving the second description of MATH in the lemma. |
math/0101251 | Using the notation just introduced, consider the following element MATH of order MATH for a quotient-cusp of the Introduction (coefficients of MATH are in parentheses and omitted if zero - with vertices numbered MATH in the obvious way, MATH) MATH . The double cover is MATH which blows down to MATH . Similarly, the element MATH gives a double cover by a quotient-cusp, while the element MATH gives the canonical double cover by the cusp of the Introduction. These elements MATH and MATH thus generate a NAME four-group MATH whose dual covering is the canonical covering by a cusp followed by the canonical double covering that doubles the resolution cycle. Denote this MATH-fold covering of the links MATH. Now this MATH is self-orthogonal with respect to the linking form, so MATH, so the dual covering MATH determined by MATH factors through the covering just constructed. It is not hard to check that MATH is a MATH-fold cover, but we do not need this. The universal abelian cover MATH factors through MATH and the covering transformation group for MATH is MATH. Since this is MATH acting in the fibers of MATH, the quotient is isomorphic to MATH again. This completes the proof of the first point of the above proposition. |
math/0101251 | The link MATH of the cusp is a torus bundle with monodromy MATH say. We can consider MATH to be MATH with the two boundary components identified by MATH. Replacing MATH by a tubular neighborhood and excising the interior of this neighborhood, we see MATH. The long exact homology sequence for the pair MATH thus leads to a long exact sequence MATH . This shows that the torsion MATH of MATH is the image of MATH and is isomorphic to MATH. If MATH is a REF-cycle in MATH then REF-chain MATH has boundary MATH. It follows that the torsion linking form on MATH is given by MATH (mod MATH), where MATH and MATH is the standard skew-symmetric form on MATH (MATH intersection form on MATH). Suppose now that MATH and write MATH. Then REF below implies MATH, where MATH. The cover and the dual cover for MATH of the given cusp have monodromy given by the action of MATH on MATH and MATH respectively. Let MATH be an oriented basis of MATH and put MATH. Then MATH is a reverse-oriented basis for MATH, since MATH has negative determinant (namely MATH). But the matrix of MATH with respect to MATH equals the matrix of MATH with respect to MATH since MATH and MATH commute. Thus the two cusps are mutual duals. |
math/0101251 | Suppose we have a cusp given by the monodromy matrix MATH. Assume MATH (if MATH take MATH in the following). Then MATH takes the subspace MATH of MATH generated by MATH to itself by the matrix MATH, where MATH. Thus the cover defined by the subgroup MATH is either the cusp with resolution graph consisting of a cycle with one vertex weighted MATH or the dual cusp of this, according as the above basis is oriented correctly or not, that is, whether MATH or MATH. In any case, by going to the discriminant cover if necessary (see REF ), we get the cusp with resolution graph dual to the above length REF cycle, so it is a cycle of MATH's and one MATH. This is a hypersurface cusp, proving the second statement of the proposition. Since this is a fiberwise cover, this hypersurface cusp not only covers, but is also covered by the original cusp. (In fact any two cusps with the same trace of their monodromy are mutual fiberwise covering spaces, see CITE.) For REF, a necessary condition that a cusp MATH have no finite abelian cover by a complete intersection is that CASE: neither MATH nor the dual cusp MATH is a complete intersection. We shall see that if CASE: the discriminant group MATH has prime order MATH say, then this necessary condition is also sufficient. Indeed, suppose that MATH and MATH are not complete intersections. If MATH has prime order then any abelian cover is either a cover in the circle direction or a cover in the circle direction of the discriminant cover. The resolution graph of the abelian cover is therefore a cyclic covering of the resolution graph or the dual resolution graph. By REF this cannot be the resolution graph of a complete intersection. An explicit example of this is the cusp with resolution cycle MATH with discriminant group of order MATH. This cusp is self-dual and since the cycle is of length MATH it is not a complete intersection. |
math/0101251 | First note that MATH contains MATH, equal to MATH for MATH odd, and MATH for MATH even. Since MATH every element of MATH may be written as desired. For uniqueness consider MATH-with MATH as in the Lemma. The quotient of the last two entries shows MATH is determined. The quotient of the second term by the MATH-th power of the last then shows MATH is determined. This implies the uniqueness. |
math/0101251 | The proof is straightforward and left to the reader. Note that in the second case MATH. |
math/0101251 | MATH acts on MATH because the generators MATH each transform each defining equation by a character. The origin is the only point of MATH with at least MATH coordinates equal to MATH. A non-trivial element of MATH with MATH eigenvalues equal to MATH must have fixed locus a coordinate plane given by setting two coordinates equal to MATH. Such a plane intersects MATH in only finitely many points. |
math/0101252 | We write MATH, where MATH and MATH are the lower and, respectively, the upper parts of MATH, such that MATH for MATH and MATH for MATH, while MATH for MATH and MATH for MATH. Since, formally (that is, computing the entries of the arrays that are involved according to the usual rule of multiplication of operators), MATH and MATH for all MATH, we deduce that MATH . But MATH, the orthogonal projection on MATH, so we can deduce that MATH . |
math/0101252 | This is an immediate consequence of REF . |
math/0101252 | If MATH then we use the fact that for any MATH and MATH, MATH, and deduce REF as in the proof of REF . Conversely, if MATH satisfies REF, then we use the fact that MATH for MATH in order to deduce that MATH . Hence, if MATH for some MATH, then MATH. Since MATH is positive, it follows that there exists a contraction MATH such that MATH. We notice that we can assume, without loss of generality, that MATH and then we deduce that MATH for MATH, which implies that MATH for some MATH. |
math/0101252 | Assume MATH and let MATH be a factorization of MATH with MATH for some NAME space MATH. From REF we deduce that MATH . In matrix form, MATH . Define MATH and MATH, then we deduce from REF that there exists an unitary operator MATH such that MATH . It follows that there exist NAME spaces MATH, MATH, and an unitary extension MATH of MATH, hence this extension satisfies the relation MATH . Let MATH, MATH, MATH, be the matrix coefficients of MATH. It is convenient to rename some of these coefficients. Thus, we set MATH . From REF we deduce that MATH and MATH . By induction we deduce that MATH where MATH are monomials of length MATH in the variables MATH, MATH,MATH, MATH, MATH, MATH. Since MATH is unitary it follows that all MATH are contractions. We define MATH and for MATH, MATH . Then we define MATH, MATH, MATH, by REF and MATH for MATH. We show that MATH belongs to MATH. To that end, we introduce the notation: for MATH, MATH . We check by induction that MATH, and for MATH, MATH . Consequently, MATH is the transfer map of the linear time variant system MATH . Since each matrix MATH, MATH, is a contraction, it follows that MATH is a contraction, hence MATH is a contraction and MATH. Also, since MATH for all MATH, we deduce from REF that MATH. |
math/0101252 | A simple proof can be obtained as an application of the NAME parametrization of upper triangular contractions in CITE. Let MATH and for each MATH, MATH is the row matrix MATH. We use REF in order to associate to MATH a family MATH of NAME parameters. The NAME parameters of the contraction MATH are given by MATH, MATH and MATH. For the next step we associate NAME parameters to the contraction MATH . We notice that MATH . By REF, the entries of the matrix MATH are polynomials in the NAME parameters, their adjoints and their defect operators. Consequently, the contraction associated to the NAME parameters MATH, MATH and MATH is precisely MATH . From the uniqueness of the NAME parameters, we deduce that the NAME parameters of MATH are precisely MATH, MATH and MATH. By induction, we deduce that we can associate to MATH the family of NAME parameters MATH such that each MATH satisfies REF. We notice that MATH for MATH. Also, each MATH is a row contraction, MATH. By REF, MATH where MATH, MATH. Also, there exist unitary operators MATH and MATH. Next, we have that MATH and we can define the row contraction MATH from MATH into MATH by the formula MATH. As above, MATH where MATH, MATH, MATH. We proceed by induction in order to construct the whole family of NAME parameters MATH with MATH and obeying REF. |
math/0101252 | This is a consequence of REF . A direct proof can be obtained as an application of REF. The parametrization given here is a refinement of one given in CITE. |
math/0101252 | We use the construction from the proof of REF. Thus, let MATH be the NAME parameters associated to MATH in the proof of REF . For each MATH we consider the row contraction MATH . Let MATH be the isometry associated to MATH by REF, that is, MATH . MATH is defined on the space MATH with values in in MATH. We set MATH and let MATH be the unitary operator defined by the formula MATH. Then we define MATH . By REF, MATH and for MATH, MATH . Since the elements of the family MATH satisfy REF and the matrix coefficients of MATH are monomials in MATH, their adjoints and defect operators, it follows that the system associated to the families MATH, MATH, MATH and MATH belongs to MATH and MATH is the transfer map of this system. |
math/0101252 | Let MATH be the NAME parameters associated to MATH as in the proof of REF . Then MATH. If MATH, then we must have MATH for MATH, so that MATH and MATH. Assume MATH. Then MATH is the transfer map of the system associated to the operators MATH defined by REF. Thus, we have MATH . We assume first that MATH. Then REF gives MATH . Since MATH, we also have MATH so that we can use a well-known formula for the inversion of a MATH matrix in order to deduce that MATH where MATH . Since MATH, MATH is invertible and the previous calculation makes sense. Also, we define MATH . Since MATH is a contraction (in fact, it is an isometry), we can easily check that MATH. Actually, we can notice that MATH is obtained from the NAME parameters MATH, where MATH and MATH for MATH. We obtain that MATH . Since MATH, the previous formula makes sense without our assumption that MATH. In fact, an approximation argument shows that REF holds without that assumption. Thus, let MATH and consider MATH the transfer map of the system MATH (not necessarely in MATH). All the previous calculations go through and we obtain a contraction MATH such that REF holds for MATH. Letting MATH we obtain REF. |
math/0101254 | It suffices to show that MATH satisfies REF pREF. The proof is exactly same as the proof of REF pREF. |
math/0101254 | See pREF in CITE for details. Since MATH is subanalytic, there is a stratification of MATH for which MATH is a stratified map REF. By NAME 's transversality, any cohomology class MATH in MATH can be represented by a chain MATH which is dimensionally transverse to any stratum in MATH and the cycle MATH lies in MATH because of REF . The class MATH is represented by the cycle MATH. |
math/0101254 | Choose an injective resolution MATH of MATH. (See for instance CITE pREF.) It is elementary to find an injective resolution MATH of MATH such that MATH for MATH. Then MATH is equal to MATH up to degree MATH. So we proved the lemma. |
math/0101254 | Let MATH and MATH. Recall from REF that NAME 's construction applied to MATH with perversity MATH gives us the intersection cohomology sheaf MATH. Suppose we have constructed MATH . This gives rise to MATH by REF where MATH since MATH. We claim MATH . For simplicity, suppose MATH consists of only one connected stratum MATH. Then MATH where MATH is the ``truncation over a closed subset functor" (see REF). In particular, MATH because MATH. When there are more than one strata in MATH we simply repeat the argument for each stratum in the order of increasing codimension. Hence we get a morphism MATH . Uniqueness is an elementary exercise. |
math/0101254 | Recall from REF that MATH and MATH are represented by intersection cycles MATH and MATH that intersect only at finitely many points in the smooth part. In this case, MATH represents MATH. By NAME 's transversality result we can further assume that MATH and MATH are dimensionally transverse to each stratum in MATH so that MATH and MATH are represented by MATH and MATH. Because MATH is smooth, the complex MATH of geometric chains is isomorphic to the constant sheaf MATH and the cup product is just the intersection of chains. Hence the cup product MATH is represented by MATH which also represents MATH. So we proved REF . For the equivariant case, observe that the statement is true for MATH for any MATH since MATH is a finite dimensional manifold. If we take a sufficiently large MATH, then MATH and MATH where MATH. So we are done. |
math/0101254 | The result follows from REF since the intersection pairing is non-degenerate for MATH. |
math/0101254 | Certainly the GIT quotient map is subanalytic and the stratification of MATH in REF is MATH-invariant. The almost balanced condition is equivalent to the MATH-placid condition and hence we have MATH by REF . For injectivity and intersection pairing, it suffices to show that MATH is nonzero in MATH by REF . Let MATH denote the set of points in MATH whose stabilizer is not finite. According to REF, there is an equivariant proper map MATH such that MATH is a homeomorphism and the stabilizer of every point in MATH is a finite group. Thus MATH. Let MATH and consider the commutative diagram of natural maps MATH where the subscript MATH denotes compact support. The class MATH is the image of a nonzero class in MATH and hence it follows from the above diagram that MATH is nonzero. |
math/0101254 | We know both MATH and MATH come from morphisms in the derived category MATH. If we compose them, we get a morphism MATH . On the set of stable points MATH in MATH, the action of MATH is locally free (that is, the stabilizers are finite groups). Let MATH which is an orbifold. Then MATH is the adjunction morphism MATH which is an isomorphism by REF again, and MATH is its inverse by definition since MATH is untouched by the blow-ups in the partial desingularization process. Therefore, MATH is the identity. It is well-known (CITE, V REF) that a morphism MATH which restricts to the identity over the smooth part (that is obviously contained in MATH) is unique. Therefore, MATH. |
math/0101254 | Let MATH and MATH. Let MATH be the identity component of MATH. By REF , a neighborhood of MATH is equivariantly diffeomorphic to MATH for some symplectic MATH-vector space MATH. Let us call MATH the normal slice at MATH. Since MATH is discrete, MATH and MATH acts on the MATH-fixed subspace MATH of MATH. By direct computation, one can check that MATH in this neighborhood is MATH and hence MATH is MATH where MATH (respectively, MATH) is the NAME algebra of MATH (respectively, MATH). Therefore the normal space MATH to MATH is the orthogonal complement of MATH in MATH. If MATH is a generic point in MATH such that MATH is minimal among those containing MATH, then MATH. Suppose the MATH-action on MATH is almost balanced, that is, the action of MATH on the normal slice MATH at MATH is weakly linearly balanced for all MATH. Then by choosing a generic MATH for each MATH, we deduce that the action of MATH on the normal space MATH to MATH is weakly linearly balanced. In general, we only have MATH. But MATH acts trivially on MATH and hence the weights of the maximal torus action on MATH are all zero. By examining REF it is easy to see that if the MATH-action on MATH is weakly linearly balanced then so is the MATH action on MATH. Therefore, if for each MATH the action of MATH on the normal space to MATH at a generic point MATH with MATH is weakly linearly balanced, then the action of MATH on the normal slice MATH at MATH is weakly linearly balanced for all MATH, that is, the MATH-action on MATH is almost balanced. Now let MATH and suppose MATH. Then MATH. Using REF which is purely a group theoretic result, it is direct to check that in the neighborhood of MATH, MATH is MATH . See the proof of REF for a similar computation. If MATH is minimal with MATH so that MATH, then the MATH-fixed sets MATH and MATH are isomorphic. In general, we only have MATH. But by the arguments in the previous paragraphs, we deduce that the action of MATH on MATH is almost balanced if and only if the action of MATH on MATH is weakly linearly balanced for MATH with MATH for all MATH. So we proved the lemma. |
math/0101254 | Let MATH be a semistable vector bundle such that MATH where MATH's are non-isomorphic stable bundles with the same slope. Then the identity component of MATH in MATH is MATH where MATH denotes the subset of elements whose determinant is REF. The normal space to MATH at MATH is CITE MATH . More precisely, MATH . Because MATH is not isomorphic to MATH for MATH, MATH and thus MATH where MATH denotes the NAME number. Therefore, the weights of the representation of MATH on MATH are symmetric with respect to the origin. This implies that the action is weakly linearly balanced. As each subgroup MATH as in REF is conjugate to MATH for a ``subdivision" MATH of MATH, it is easy to check that such MATH action on the MATH-fixed point set is also weakly linearly balanced. |
math/0101254 | See the proof of CITE pREF. |
math/0101254 | This follows from CITE, REF |
math/0101254 | Recall that MATH denotes a contractible free MATH-space for a NAME group MATH and MATH. Let MATH. From the obvious commutative diagram MATH we get a morphism MATH . This induces a morphism by adjunction MATH . The fiber of MATH is homotopically equivalent to MATH and thus this morphism induces the truncation homomorphism MATH where MATH is the identity component of MATH. By composing REF with MATH, we get a morphism MATH whose hypercohomology gives us MATH . Therefore it suffices to show that MATH is equal to zero in view of REF . The sheaf complex MATH is trivial on the complement of the closed subset MATH. Hence MATH is zero on this open dense subset. Hence by adding stratum by stratum in the order of increasing codimension, it suffices to show the following: Let MATH be a subgroup of MATH and consider the stratum MATH defined in REF. Suppose MATH is an open subset of MATH containing MATH such that MATH is open and MATH is equal to zero. Then MATH is also zero. Let MATH and put MATH. We claim that MATH . This claim enables us to deduce that MATH is zero from MATH being zero because MATH is the composition MATH which is zero. Let us now prove REF . If MATH is not conjugate to a subgroup of MATH, then MATH does not intersect with MATH and thus we have nothing to prove. So we may assume MATH after conjugation if necessary. Consider the commutative diagram MATH where MATH is the unique map defined by the universal property of the categorical quotient MATH of MATH. We compute the stalk cohomology of both sides of REF . By REF , the preimage of a contractible neighborhood MATH of a point in MATH by MATH is equivariantly homeomorphic to MATH for some symplectic MATH-vector space MATH. By REF , it is direct to check that MATH in this neighborhood is MATH where MATH and MATH is the MATH-fixed subspace of MATH. Also MATH denotes the NAME algebra of MATH. If we delete MATH from the neighborhood MATH, then the preimage by MATH is MATH where MATH is the GIT quotient map and MATH is the vertex of the cone MATH. The intersection of this with MATH is homeomorphic to MATH . Hence the stalk cohomology of the left hand side of REF is MATH while the right hand side has MATH . Thus it suffices to show that MATH . Without loss of generality, we may assume MATH. By definition, we have MATH . From REF , it is easy to deduce that MATH in the preimage of MATH is MATH where MATH is the complexification of MATH in MATH and MATH is assigned a complex structure compatible with the symplectic structure. Hence, we have MATH . From the surjectivity of the morphism MATH we see that MATH . Comparing REF , we get MATH . Since the action of MATH on MATH is almost balanced REF , by REF we have MATH where MATH is the GIT quotient map. Finally, we observe that MATH . This is because we know the following from CITE: CASE: For MATH, MATH where MATH is the gradient flow for MATH with MATH (MATH is the moment map for MATH). CASE: For MATH, MATH where MATH is the gradient flow for MATH with MATH (MATH is the moment map for MATH). CASE: For a moment map MATH on a symplectic manifold, the gradient vector at MATH for MATH is MATH if MATH is identified with MATH by the Killing form. CASE: MATH if MATH and hence MATH. REF follows from REF . |
math/0101254 | Let MATH be a nonzero element in MATH. Then MATH for each MATH . Its image in MATH satisfies MATH for each MATH . This follows from the commutative diagram MATH . Therefore, MATH is mapped to an element in MATH. Recall that MATH is the identity component of a stabilizer which has the maximal dimension MATH and the blowup center is the submanifold MATH. Since MATH is an injection by the well-known argument in CITE pREF, we may think of MATH as an element of MATH. By REF , if MATH that is, MATH, then MATH . Let us now consider the case when MATH. Since MATH is maximal, we have an isomorphism MATH . By the definition of MATH, MATH lies in MATH and we have MATH from REF where MATH is the normal space to MATH. Because the MATH-action is almost balanced, the codimensions of the unstable strata in MATH are greater than MATH by REF . Therefore, by the equivariant NAME theory CITE, we deduce that the restriction homomorphism MATH is an isomorphism. In particular, MATH injects into MATH. By REF , the image of MATH in MATH is not zero and thus MATH injects into MATH. It is obvious from our proof that the statement is true for any open set MATH in MATH. |
math/0101254 | For an open subset MATH of MATH, let MATH be the kernel of the NAME map restricted to MATH. Then by REF , we can write MATH . We have to show that MATH is zero. Let MATH be the preimage of MATH by the blow-up map MATH. By our induction hypothesis, the pull-back MATH is an isomorphism of MATH onto MATH and the NAME map is its inverse. Recall that we have the decomposition REF which induces an isomorphism MATH where MATH is the hypercohomology of MATH over MATH. Since MATH is mapped to zero by MATH we see that MATH injects into MATH. As MATH, MATH and REF all came from sheaf complexes, we have the following commutative diagram by restriction MATH where MATH is the preimage of MATH in MATH. Since MATH is supported over MATH, the vertical map for MATH is the identity map. Now let MATH be a nonzero element in MATH. We know MATH is mapped to a nonzero element, say MATH in MATH. In the above diagram, MATH is mapped to zero in MATH. Then by REF , MATH and thus MATH. This is a contradiction! So we proved that MATH. |
math/0101254 | Let MATH. If MATH is not conjugate to a subgroup of MATH, there is no MATH-fixed point in MATH. Hence, after conjugation if necessary, we may assume that MATH. Let MATH be the MATH-fixed subset of MATH. For MATH we have to consider the map MATH . It is obvious that MATH is a vector bundle over MATH. Using REF , it is easy to check that there are MATH in MATH so that MATH where MATH. Since MATH is maximal in MATH we have isomorphisms MATH by CITE and hence we have MATH . If we apply REF with MATH as subgroups of MATH, we deduce that there exist MATH in MATH such that MATH . But for MATH, MATH and hence we have MATH . This implies that the natural embedding MATH is of finite index. In particular, the identity component of MATH is naturally isomorphic to the identity component MATH of MATH. Let MATH be the identity component of MATH and put MATH. Then MATH. Therefore, MATH is the MATH-invariant part of MATH . The first isomorphism in REF came from REF . Our interest lies in finding the kernel of REF . Combining REF , we see that MATH is the intersection of the kernels of MATH for all MATH. Now observe that the spaces that appear in REF lie over MATH. Applying spectral sequence, we get a homomorphism of spectral sequences whose MATH-terms give us MATH . The right side of REF is isomorphic to MATH by conjugation, where MATH. Note that MATH and MATH share the same identity component say MATH. Thus MATH is the invariant subspace of MATH with respect to a finite group action. With the isomorphism REF , the fiber direction in REF is exactly the truncation map MATH for MATH. Therefore taking the kernel of REF for all MATH gives us MATH . So we are done. |
math/0101254 | We identify the hyperplane at infinity MATH with MATH. The equation for the hyperplane at infinity is MATH-invariant and hence MATH contains MATH. Therefore MATH. Let MATH and suppose be the corresponding point MATH in MATH is semistable. By REF , MATH is fixed by MATH if and only if MATH is fixed by MATH. Hence we may consider only points in MATH. Let MATH. Since we could use any MATH as long as the infinitesimal stabilizer is MATH, we assume that MATH is minimal among the stabilizers whose NAME algebra is MATH so that MATH in the notation of REF . First since the action of MATH on MATH is weakly balanced, the action of MATH on MATH is weakly linearly balanced and so is the action of MATH on the MATH-fixed subspace MATH. Hence we checked the weakly balanced condition for MATH. Now let MATH and let MATH be the identity component of the stabilizer of MATH. Then from REF , we see that the normal space to MATH in MATH is the same as the normal space to MATH in MATH at a generic point. But MATH is the identity component of MATH and MATH. Hence the normal space to MATH in MATH at a generic point is isomorphic to the normal space to MATH in MATH at a generic point. Moreover since the diffeomorphism in REF is equivariant, the actions of MATH on the normal spaces are identical. According to REF , the weakly balanced condition is purely about the action of MATH on the normal spaces to MATH for all MATH. Because the action of MATH on MATH is weakly balanced, we deduce that the action of MATH on MATH is weakly balanced. So we proved that the action of MATH on MATH is weakly balanced. Note that MATH is the first blow-up in the partial desingularization for MATH and hence MATH by CITE By REF , the action on MATH is also weakly balanced. |
math/0101254 | As in the proof of REF , we may assume that MATH is minimal among the stabilizers of points in MATH whose NAME algebra is MATH. So we may use REF with MATH. As MATH is a cone with vertex point MATH, it is well-known that MATH for MATH and if MATH, we have MATH where MATH is the GIT quotient map and the following diagram commutes: MATH . As MATH is the union of the complex cones over the unstable strata of MATH and the real codimension of each unstable stratum is greater than MATH by the weakly balanced condition, the real codimension of MATH is greater than MATH. Hence, the bottom horizontal map is an isomorphism. We claim that the the right vertical in REF is injective and the image is MATH. Consider the following commutative diagram MATH we see that MATH . For the last isomorphism, observe that MATH does not intersect with the blow-up center. Moreover, if MATH, the closure of MATH does not meet MATH (if it does, the point belongs to MATH) and thus MATH is semistable as a point in MATH by REF . Because the action of MATH on MATH is weakly balanced and MATH, REF is true for MATH. In particular, if we apply the theorem for the preimage MATH of the open set MATH by MATH we obtain the isomorphism MATH . Therefore, from REF , it suffices to show that MATH for MATH by restriction. To see this, we note once again that for MATH, MATH is an isomorphism and the same is true for MATH if MATH, MATH, because the action of MATH on MATH is weakly linearly balanced. |
math/0101254 | By REF , we have only to subtract out the NAME series of MATH where MATH is in this case MATH. By the lemma below, which is essentially combinatorial, the image contains MATH and thus the intersection is MATH, whose NAME series is precisely MATH . So we are done. MATH . |
math/0101254 | It is equivalent to show that MATH is surjective. Let MATH and consider, for each MATH, MATH . Then since MATH for all MATH, MATH if MATH and MATH otherwise, where MATH, MATH. Therefore, the images of those MATH span MATH for any MATH and thus the restriction is surjective. MATH . |
math/0101255 | Suppose MATH is in the preimage in MATH of a point MATH in a connected component MATH of a stratum, and that MATH. There is a long exact sequence MATH . We see that MATH. By the definition of MATH we see that MATH lies in this subcategory if, and only if, MATH which is precisely MATH. |
math/0101255 | This follows from the triangle associated to MATH and the above lemma. |
math/0101255 | It is easy to check that the composition MATH is non-zero, but then it must then be a quasi-isomorphism by REF. |
math/0101255 | It follows immediately from the weakly-balanced assumption that the morphism MATH can be factored through MATH since MATH. It is easy to check that the resulting composition MATH is not zero and so must then be a quasi-isomorphism by REF . |
math/0101255 | This follows immediately from REF of MATH. |
math/0101255 | For each MATH, let MATH denote the MATH fixed point set in MATH just like MATH in MATH. Then since the map MATH is equivariant, we have the commutative diagram MATH . This induces the following commutative diagram MATH by assigning zero for MATH. The claim now follows from the definitions. |
math/0101255 | We can decompose MATH into a direct sum MATH using REF . It follows from REF that the composition MATH is a quasi-isomorphism on the first term. Now let us consider the composition MATH . By the above and REF the projection onto the first term of the Right-hand side is, up to a quasi-isomorphism, the same as MATH . In particular if we decompose MATH as in REF into MATH then by REF is zero and the choice of MATH in REF induces a unique lift of MATH to MATH. The inductive assumption that MATH now tells us that the two morphisms MATH induced by MATH and MATH respectively must give the same map on hypercohomology. The almost-balanced assumption guarantees that there is a quasi-isomorphism MATH, where MATH is the inclusion of the open complement to MATH. So MATH decomposes as MATH where MATH is defined (up to quasi-isomorphism) by the distinguished triangle MATH . Furthermore because MATH is supported on MATH it includes into MATH as a subobject of MATH. The final observation we require is that the spectral sequence computing the hypercohomology of MATH degenerates at the MATH-term. This follows from NAME 's criterion and the surjectivity of the restriction MATH. Hence the spectral sequence for MATH must also degenerate and there is an injection MATH where MATH is the `exceptional divisor' in MATH. But now we are done for we have shown that the image of the class MATH pulled back to MATH lies in two subspaces, namely MATH and MATH, whose intersection is zero. |
math/0101255 | It is easy to check that MATH. Then by REF we have MATH . Since on the non-singular stratum the two morphisms are both just the product on the constant sheaf with stalk MATH we are done. |
math/0101255 | Note that the maximal degree in which the hypercohomology of a complex MATH does not vanish is bounded by MATH . Now since MATH we see that this quantity is strictly less than MATH. |
math/0101255 | Since MATH the flow induces an equivariant map MATH which is a homeomorphism between dense open sets and so we have MATH . Let MATH. By REF we also have a commutative diagram MATH where the horizontal maps are isomorphisms. The result follows by composing these two diagrams and noting that MATH preserves evaluations against the fundamental class because MATH is `birational'. |
math/0101255 | Both statements follow almost immediately from the above calculation of the normal fibre. |
math/0101260 | Let MATH be a generic bihomogeneous polynomial of bi-degree MATH . Consider the modified complex which is made by replacing MATH with MATH in REF . Because the polynomials MATH and MATH are bihomogeneous but do not have the same bi-degree, the bihomogeneous resultant cannot be taken. However, there is another elimination operator available: the mixed resultant associated with the sequence MATH (CITE, Chapter MATH). It is an irreducible polynomial in the coefficients of MATH which vanishes if and only if these polynomials have a common root in MATH . In order to compute it, we may apply the NAME method for the study of resultants (see CITE, Chapter MATH). Let MATH denote the line bundle on MATH whose sections are homogeneous polynomials of degree MATH in coordinates MATH and degree MATH in coordinates MATH on each MATH . Let MATH and MATH . Each MATH is very ample, and we may regard polynomials of bi-degree MATH with coefficients in MATH as elements of MATH and polynomials of bidegree MATH as belonging to MATH . Every specialization of MATH defines a section MATH of the vector bundle MATH . If we set MATH and construct the complex MATH (for a definition of this complex, see CITE, Chapter MATH), then we recover the modified complex REF . Moreover, it is not hard to check that this complex is stably twisted (that is, has no higher cohomology), so Proposition MATH and Theorem MATH of (CITE, Chapter MATH) hold, and we have that the complex will be exact if and only if the resultant of the specialized MATH is not zero. Furthermore, the determinant of the complex with respect to the monomial bases is equal to MATH . The original complex is recovered by specializing MATH to MATH . Keeping in mind that MATH we have that the determinant of REF is equal to a power of MATH . Comparing the degrees of both the bihomogeneous and the mixed resultant in the coefficients of MATH and MATH respectively (compare CITE), we get that the determinant of the complex REF equals MATH . |
math/0101260 | The proof will be by induction on MATH . For MATH it is clear than MATH and MATH are the same functions, so the proposition follows straightforwardly. Suppose then MATH . The morphism MATH may be factored as follows: MATH where MATH is the morphism defined in REF and MATH . Denote by MATH the matrix corresponding to MATH in the monomial bases for MATH . These are not square matrices (they have sizes MATH and MATH respectively), but applying the NAME formula (see for instance CITE), there is a relationship between their maximal minors and MATH the summation made over all sequences of integers MATH with MATH and MATH (respectively, MATH) denotes the square submatrix of MATH (respectively, MATH) which is made by choosing the MATH columns (respectively, rows) indexed by MATH . Note that MATH is the matrix corresponding to MATH in the monomial bases and, for each MATH the maximal minor MATH in REF outlined in the previous paragraph is denoted MATH in REF . Using REF , one has MATH . An explicit computation of the MATH matrix MATH reveals the following structure: MATH where MATH and MATH have sizes MATH and MATH respectively, and MATH denotes the identity matrix of size MATH . Gluing MATH and the MATH matrix MATH corresponding to MATH one gets a square matrix MATH of size MATH which has the following structure: MATH where the block MATH denotes the matrix corresponding to the linear map MATH which maps MATH to MATH and the other blocks have the same meaning. It is easy to check that MATH is a square matrix. Moreover, MATH . In the same way, it holds that MATH . Then, the determinant of MATH equals MATH and then the inductive hypothesis yields the following equality: MATH . This determinant may also be computed as a sum of maximal minors of MATH times their complementary minor in MATH . This is exactly the sum which appears in REF , that is, MATH . Replacing REF , the Lemma follows. |
math/0101260 | As in the Lemma, the proof will be by induction on MATH . For MATH it happens that MATH so the statement holds straightforwardly. Take MATH and factor MATH as follows: MATH where MATH . Denote by MATH the matrix corresponding to MATH in the monomial bases. The NAME formula gives the following relationship: MATH where, as before, MATH runs through all sequences of integers MATH satisfying MATH and MATH (respectively, MATH) have the same meaning as in REF . Proceeding as in the proof of the previous Lemma, one gets MATH . Gluing MATH and matrix MATH one gets a square matrix with the following structure: MATH . Here, MATH and MATH are the same blocks which appear in REF , and it is easy to check that the block MATH is the matrix MATH . Then, using inductive hypothesis and the previous Lemma, the following equalities hold: MATH . Computing this determinant as sum of maximal minors of MATH times their complementary minor in MATH it appears the summation in REF . Replacing it with this last expression, the Theorem follows straightforwardly. |
math/0101260 | Set MATH . Computing explicitly the last row of MATH because of the order given to the rows of MATH in REF , one obtains: MATH which coincides with one of the rows in MATH . In the same way, one can check that the row inmediately before the last in MATH is the following MATH so, substracting from it the last row, one gets another row of MATH . A similar situation happens in all rows. This implies that the matrix MATH may be transformed in the matrix MATH applying operations on its rows which do not change the determinant. |
math/0101260 | The fact that the degree of MATH in MATH is MATH can be easily checked in NAME 's matrix (compare CITE). Set MATH . Suppose MATH where MATH . The polynomial MATH is homogeneous in the variables MATH which implies that MATH and MATH are also homogenous in the MATH . Specializing MATH one has that MATH . But the left-hand side is irreducible. This implies that one of the factors must have degree MATH in the variables MATH lets say MATH . The factorization now reads as follows MATH where MATH . But if MATH then the variety MATH in MATH . On the other hand, it is well known that, for a given family of bihomogeneous polynomials MATH with no base points, the equation MATH is a power of the implicit equation of the rational surface defined by MATH (actually, if we use NAME matrices for computing this resultant, we will find moving planes encoded in their rows; see the comment at the end of REF). If MATH had a nonempty zero locus, this would imply that every rational parametric surface of bi-degree MATH without base points will contain the zero locus of MATH which is impossible. |
math/0101260 | To begin, it will be proven that the ratio between MATH and MATH is in MATH . Note that there is a positive integer MATH such that MATH (this is due to the fact that, in the construction of MATH one needs the inverse of MATH . ) Set MATH . Both MATH and MATH are homogeneous in MATH and of the same degree, which is equal to MATH so it will be enough to show that one of them is a polynomial multiple of the other. Suppose that the variables have been specialized: MATH with the following conditions: CASE: MATH where MATH denotes the matrix MATH after specializing MATH . CASE: MATH vanishes. This means that the projective point MATH belongs to the rational surface defined by MATH . Using MATH and the same argument as in the proof of REF, one can verify that, if MATH belongs to that rational surface, the determinant of MATH must vanish. REF implies that MATH must vanish. Hence, MATH . If MATH the inclusion holds trivially and, using the NAME 's NAME, one can conclude that a power of MATH must be a multiple of MATH . As MATH is irreducible, MATH must be a multiple of MATH so MATH where MATH as expected. In order to compute MATH do the following replacement: MATH . Then, it will hold that MATH if the columns of MATH are properly ordered (compare CITE, proof of REF). So, MATH must be MATH . |
math/0101260 | In order to make the method work correctly, one can suppose without loss of generality that MATH . The method computes MATH which is equal to a constant times MATH because of REF . But MATH vanishes if and only if the projective point MATH belongs to the implicit surface. This, combined with the fact that the implicit equation of the parametric surface is always irreducible, completes the proof. |
math/0101260 | The proof follows applying mutatis mutandis all the tools developed in REF. Consider the following NAME Complex: MATH . Here, MATH and MATH are defined by REF respectively. A similar version of REF holds, applying the same trick used there to the formulation of the multivariate resultant as the determinant of a NAME Complex (compare CITE). The complex REF is exact, and after a specialization of the coefficients MATH it will be exact if and only if MATH does not vanish. The determinant of the complex with respect to the monomial bases is equal to MATH . The linear map MATH may be factored as follows: MATH where MATH is defined as in REF . In order to apply to this situation the proof of REF , the ``triangular" version of REF is needed. Let MATH be the linear map defined as in REF , and set MATH the matrix in the monomial bases corresponding to the restriction of MATH to MATH . It is a square matrix of the same size as MATH . The following equality follows straightforwardly: MATH . Now, the same proof performed in REF for the tensor product case, may be applied for triangular polynomials, giving the desired result. |
math/0101260 | The fact that there are MATH linearly independent moving planes of degree MATH implies that MATH has maximal rank. REF implies that there exists in index set MATH such that MATH is non-singular. |
math/0101262 | This is a more or less straightforward chase though the definition. |
math/0101262 | Let MATH be a convex subset of a real vector space and let MATH, MATH and MATH. Let MATH be defined by MATH . As MATH there is a MATH so that MATH. Then as MATH is MATH-almost convex MATH . Thus MATH. |
math/0101262 | Let MATH with MATH. Then there is a MATH so that if MATH and MATH, then MATH, MATH and by REF MATH . We assume that MATH, the case of MATH having a similar proof. As any compact subset of MATH is contained in a bounded convex open subset of MATH we can also assume, without loss of generality, that MATH is bounded. Let MATH be compact and let MATH. For any MATH let MATH be the open ball of radius MATH about MATH. Then for any MATH we have MATH. For MATH define MATH by MATH . Then it is easy to check that MATH for all MATH and MATH for all MATH. Also MATH is a dilation in the sense that MATH for all MATH. As MATH and MATH this implies MATH. Let MATH be NAME measure on MATH. Then for any measurable subset MATH of MATH . Choose a positive real number MATH so that MATH where MATH is the open ball of radius MATH about the origin. Because MATH is measurable and MATH there is a positive MATH so large that MATH . Therefore if MATH then MATH. Let MATH. We now claim that MATH has positive measure. For if not then MATH and MATH would be essentially disjoint subsets of MATH and therefore, using that MATH, MATH which can be rearranged as MATH contradicting REF. Therefore MATH as claimed. Let MATH. Then MATH and MATH are both in MATH and therefore MATH. Thus MATH which shows that MATH is bounded above on MATH. To show that MATH has a lower bound on compact subsets of MATH, let MATH and let MATH be small enough that the closed ball MATH is contained in MATH. Then MATH-is compact so by what we have just done there is a constant MATH so that MATH for all MATH. Let MATH. Then, again as above, MATH, and therefore MATH which can be solved for MATH to give MATH . Therefore MATH is bounded below on MATH. But any compact subset of MATH can be covered by a finite number of such open balls and thus MATH is bounded below on all compact subsets of MATH. |
math/0101262 | By doing a linear change of variable (which preserves MATH-almost convexity) we can assume that MATH. Also by replacing MATH by MATH we can assume that MATH. Let MATH. Then by REF there is a constant MATH such that MATH on MATH. Let MATH . We now show that MATH on MATH. If MATH, MATH, or MATH this is clear. Let MATH then the choice of MATH ensures that there is a MATH such that MATH for some positive integer MATH. Also, as MATH, MATH. Therefore MATH . If MATH a similar calculation shows that MATH (or this can be reduced to the case MATH by the change of variable MATH). This completes the proof. |
math/0101262 | If MATH is a vertex of MATH, which without lost of generality we can take to be MATH, then let MATH be any MATH-ranked probability measure and let MATH and MATH for MATH. Partition MATH as MATH and MATH an arbitrary partition of MATH. Then MATH and MATH for MATH and therefore MATH . Thus MATH. Now assume that MATH is not a vertex and let MATH be a MATH-ranked probability measure and MATH with MATH and MATH. Then as MATH is not a vertex we have that MATH for MATH and therefore MATH. Thus MATH and therefore MATH. This gives MATH . Taking an infimum then gives that MATH. Now assume that MATH contains a point that is not a vertex and note that if MATH then MATH for all MATH. Thus it suffices to show that MATH is bounded when MATH is a single point MATH with MATH. Suppose MATH. We let MATH be the product measure as in REF and we let MATH as in REF and use the alternative definition of MATH given in REF . For each MATH, MATH, we select inductively a set MATH with MATH pairwise disjoint such that MATH . Note that if MATH, then MATH. We carry out the the inductive selection as follows: Let MATH . Then MATH . If MATH, let MATH. If MATH let MATH be the first integer such that MATH . Since MATH, MATH . Let MATH. Continue choosing from MATH to obtain MATH. Note that by REF, the supply of atoms in MATH is sufficient to choose the sets MATH. For MATH we have MATH which implies MATH for MATH. As MATH we can use REF (with MATH) to compute MATH . Thus, in the notation of REF , MATH which bounds MATH as required. |
math/0101262 | Let MATH and MATH. For MATH, let MATH be a MATH-ranked probability measure. We let MATH be a disjoint sequence such that MATH. Now let MATH be the MATH-ranked probability measure on MATH as in REF , that is, MATH. It is easily checked that MATH (and MATH). Thus MATH . Taking the infimum over all MATH on the right hand side of this gives MATH which completes the proof. |
math/0101262 | Let MATH. Also let MATH be a MATH-ranked probability measure and MATH a disjoint sequence in MATH such that MATH where MATH is partitioned by MATH and MATH. If MATH (the MATH-algebra generated by MATH), that is, MATH, we define (for MATH, so that MATH) MATH . Then the map MATH is a vector measure on MATH. Note that MATH (as MATH except for a set of MATH-measure zero so that MATH). For each MATH let MATH . Note that if MATH, then MATH for some MATH. Since MATH each MATH is (except for a set of MATH-measure zero) a disjoint union of countable many sets MATH with MATH so that MATH. We require the following lemma to complete the proof. With MATH as in the statement of REF MATH . Before proving the lemma we show that it implies the theorem. As MATH is bounded there is a MATH so that MATH for all MATH. Therefore by the lemma MATH . Since MATH this yields MATH. Taking the infimum over MATH gives MATH and completes the proof of REF . |
math/0101262 | The proof is by induction on MATH. The base case is MATH which amounts to MATH, which is in fact an equality. Now assume for some MATH that REF holds. Consider MATH. Then MATH divides into sets MATH such that MATH . Since MATH the MATH-almost convexity of MATH implies MATH . Multiplying this by MATH . If we let MATH and apply the above to each MATH . If MATH and MATH for some MATH then MATH, where MATH. Thus the term for MATH satisfies MATH since MATH and MATH. Now MATH. Thus MATH . This closes the induction and completes the proof of the lemma. |
math/0101262 | By replacing MATH by MATH, which will still be MATH-almost convex, we may assume that MATH for MATH. If MATH is bounded then MATH by REF . So to finish the proof it is enough to show that MATH is bounded. In doing this we can use REF and note that there are MATH with MATH so that if MATH then MATH is MATH-almost convex. (To be a bit more precise let MATH with MATH and then the choice MATH and MATH works.) With this choice of MATH we now prove by induction on MATH that if MATH is MATH-almost convex and vanishes on the vertices of MATH then MATH. The base case is MATH. Then as a NAME measurable function is NAME measurable REF implies MATH is bounded. But then REF implies MATH. For the induction step let MATH be MATH-almost convex and suppose MATH vanishes on the vertices of MATH. Let MATH be the function MATH. Then MATH is MATH-almost convex, vanishes on the vertices of MATH and is NAME measurable. Therefore by the induction hypothesis MATH. Let MATH and consider the function MATH given by MATH . Then this is MATH-almost convex on MATH and is NAME measurable. Therefore another application of REF implies that MATH bounded and as MATH vanishes at the endpoints of MATH we have that MATH. This implies MATH . But every MATH can be expressed as MATH for some MATH and some MATH. Therefore MATH is bounded on MATH. Then REF implies MATH. This closes the induction and completes the proof. |
math/0101262 | Let MATH be given by MATH. Then MATH is MATH-almost convex, bounded (as MATH is bounded on the convex hull of MATH as it is compact) and vanishes on the vertices of MATH. Therefore by REF MATH which implies REF. If MATH is compact and MATH dimensional with extreme points MATH, then MATH is the convex hull of MATH. By NAME 's Theorem for any MATH there are MATH and MATH so that MATH which, along with REF, implies REF. |
math/0101262 | In the special case that MATH a proof, based on ideas of CITE and CITE, can be found in CITE. As the details in the present case are identical we omit the proof. |
math/0101262 | The proofs of REF are similar, with the proof of REF being the simpler of the two, so we will give the details in the proof of REF. The inequality MATH implies for MATH and any MATH there is a MATH and MATH such that MATH . As MATH and MATH are bounded we can assume, by adding appropriate positive constants to MATH and MATH, that MATH for some MATH. Set MATH . We need to show that MATH (as MATH is clear). We may assume that MATH, for if MATH then MATH and there is nothing to prove. Let MATH, but MATH and assume for some MATH that MATH . Then there is a MATH so that MATH . We now prove REF from the lemma. Let MATH. We now choose a finite sequence MATH with MATH as follows. From the definition of MATH there is a MATH with MATH. If MATH we stop. If MATH, then by the lemma, there is a MATH with MATH and MATH. If MATH then stop, otherwise use the lemma (with MATH replacing MATH and MATH replacing MATH) to get a MATH with MATH. If MATH, stop. If MATH then we continue to use the lemma to get MATH with MATH for MATH. This implies that MATH. But as MATH this process must terminate for some MATH with MATH. Then MATH . Letting MATH in this implies MATH which completes the proof. |
math/0101262 | Let MATH be as in the statement of the lemma. From the definition of MATH there is a sequence MATH so that MATH and MATH. By REF there is a sequence MATH and sequences MATH so that (replacing MATH by the appropriate subsequence). MATH for some non-negative real number MATH. By compactness of MATH and MATH we can assume, by possibly going to a subsequence, that MATH and MATH and that MATH for some MATH, MATH and MATH. Then MATH and from the definition of MATH, MATH. Therefore MATH . This is turn implies that MATH . Because MATH is MATH-almost convex, MATH . Combining REF yields MATH . We now claim there is a MATH so that MATH . To see this partition MATH into two sets MATH and MATH where MATH and MATH. Note that as MATH we have MATH so that MATH. For MATH let MATH. Then MATH. Using REF, MATH . We have already seen that MATH and therefore MATH. Thus MATH where we have used that MATH and that MATH is decreasing for MATH. As MATH this implies there is at least one MATH with MATH. For this MATH REF holds. Letting MATH be so that REF holds and using that MATH, and that MATH for all MATH in REF, we have MATH . This implies MATH . As MATH this gives MATH . Letting MATH completes the proof of the lemma. |
math/0101262 | This can be derived from REF in the same way that CITE is derived from CITE. |
math/0101262 | Notice that if MATH, then MATH if and only if for every MATH, there exists MATH such that MATH where MATH and MATH is the disjoint union of MATH. Let MATH and define a function MATH by MATH . Then this is continuous on MATH. Let MATH . Then MATH. As an intersection of closed sets is closed, to finish the proof it is enough to show that each MATH is closed. Let MATH and suppose MATH in MATH. For each MATH there is a MATH such that MATH. Since MATH is compact, by passing to a subsequence, if necessary, we may assume that MATH. Thus MATH . Therefore MATH is closed. |
math/0101262 | First we select a subsequence MATH of MATH for some infinite MATH by first choosing sets MATH and MATH as follows: For each MATH, we can use point REF to partition the terms of MATH into MATH sequences MATH where MATH and MATH. We may assume that for every MATH that MATH. If MATH, let MATH and MATH, otherwise MATH is bounded for some infinite set of MATH. Since for any integer MATH, there are only finitely many sets in MATH of rank MATH, there is a MATH so that MATH on an infinite subset MATH of MATH. Similarly choose MATH infinite in MATH and MATH such that either MATH and MATH or MATH for all MATH. Continue selecting infinite sets MATH of MATH and MATH such that MATH and either MATH and MATH or MATH for all MATH. The inequalities MATH yield MATH and therefore MATH . Also the sets MATH are pairwise disjoint. Now let MATH be an infinite set in MATH such that each MATH is finite. Let MATH. REF implies MATH . Thus for fixed MATH and MATH and therefore MATH . Hence MATH where MATH. It follows that MATH . But since the sets MATH are pairwise disjoint MATH . But this implies that there must be equality for each MATH: MATH . Once again fix MATH. For MATH suitably large in MATH, MATH if MATH. Thus MATH . (All the sums are finite so there is no problem in interchanging the limit with the summation.) Since this holds for all large MATH, MATH . Now splice the sequences MATH into a single sequence MATH. This sequence satisfies the conclusion of the Lemma. |
math/0101262 | We First show the lower semi-continuity of MATH. Suppose that MATH is a sequence in MATH and that MATH. Further suppose that MATH is convergent. For each MATH, select a measure MATH and a sequence MATH in MATH such that MATH, MATH, and MATH. By passing to a subsequence, if necessary, we may assume that MATH. By REF , there is a sequence MATH in MATH so that MATH, MATH and MATH . Thus MATH is lower semi-continuous. We now show the second conclusion of REF . Let MATH. Select MATH a sequence in MATH and for each MATH choose a sequence MATH in MATH such that MATH, MATH and MATH . By passing to a subsequence, if necessary, MATH for some MATH. Let MATH be the sequence obtained by REF . Then for the measure MATH and the sequence MATH the equality REF holds. This completes the proof. |
math/0101262 | Since for MATH we have MATH it is enough to show the existence of a disjoint sequence MATH with MATH for then MATH automatically holds. We select this sequence recursively. Suppose that MATH have been chosen to be pointwise disjoint with MATH. Then MATH . Since each of the sets MATH is a union of atoms from MATH, there is an atom of MATH that is disjoint from MATH. As atoms of MATH have MATH-measure MATH we can use this atom as MATH. |
math/0101262 | Clearly MATH (otherwise MATH). Suppose that for some MATH that MATH. Then let MATH . Then each MATH is nonnegative integer, MATH and MATH . This contradicts the minimality of the sum and completes the proof. |
math/0101262 | From REF we know that if MATH with MATH nonnegative integers is the expansion of MATH so that MATH, then MATH. When MATH is not a MATH-adic rational uniqueness of base MATH expansions implies that MATH and we are done. If MATH is a MATH-adic rational and so has two expansions with MATH then direct calculation shows that MATH is smaller when the finite expansion is used. Thus MATH in this case also. |
math/0101262 | Other than the lower bound MATH, we refer the reader to the proofs of CITE and CITE which cover the case when MATH. Only trivial changes are required for the general case. To prove the lower bound, suppose MATH is the base MATH expansion for MATH with MATH. Then MATH and therefore MATH. Thus MATH as required. |
math/0101262 | Suppose not and let MATH be the least MATH such that MATH. If MATH, then MATH and if MATH, then MATH. In either case MATH. There is a MATH such that for MATH, MATH and MATH. Thus for each MATH there is a MATH with MATH defined so that MATH . (this is possible because MATH) and MATH (this is possible becasue MATH so that MATH). But then for MATH, MATH . But then MATH which is impossible. |
math/0101262 | Suppose that for some MATH that MATH implies MATH. Then there exists MATH such that for MATH and MATH we have MATH. But then for any MATH (recall that for fixed MATH there holds MATH for MATH sufficiently large) MATH which is a contradiction. |
math/0101262 | By REF MATH and by REF , MATH for infinitely many MATH. Thus MATH . Set MATH. Then MATH where the first inequality follows form REF and the second from REF . Thus MATH with MATH and MATH a positive integer. But then MATH which implies MATH. That is MATH. Thus MATH for MATH. |
math/0101262 | Using the results from the last several lemmas: MATH . |
math/0101262 | By REF MATH and MATH so MATH. For the other inequality, use MATH so that MATH . |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.