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math/9909141 | First, the vector MATH constructed in REF can be found in polynomial time in the size of the coefficients of MATH. In fact, investigation of CITE shows that the rational numbers MATH in REF in the proof of REF can also be constructed in polynomial time. This implies that MATH and the MATH can be found in time polynomial in MATH. The proof of REF then shows that the final expression can be computed in polynomial time. |
math/9909141 | This follows from CITE by writing the fundamental cycle of MATH in terms of the MATH. |
math/9909142 | The uniqueness of MATH follows from the NAME principle for adjoint groups, thus it remains to show its existence. By the proof of REF, MATH contains a maximal elliptic torus MATH, defined over MATH, and there exists an element MATH which factors through MATH. Let MATH be the corresponding cocharacter of MATH. Since the element MATH is of order two, it defines a certain cohomology class MATH. Moreover, it follows from the construction that the image of MATH in MATH is MATH see REF. Hence it will suffice to show the existence of a cohomology class MATH, whose image in MATH is MATH and whose image in MATH is trivial for each finite prime MATH. By the local and the global NAME duality (see CITE), this would follow if we show that MATH belongs to the kernel of the natural composition map MATH where by MATH we denote the dual of a finite abelian group. Since MATH is anisotropic, MATH splits over a certain NAME MATH. Then we have an action of the NAME group MATH on MATH. Since MATH is a NAME, this action commutes with the complex conjugation, and, therefore, induces an action on the cohomology group MATH. Using the definition of the cohomology groups, one can check that the map MATH is constant on the orbits of MATH. By the construction, the cohomology class MATH is equal to MATH, where we denote by MATH the cohomology class in MATH, induced by the element MATH. Hence MATH has to belong to the kernel of MATH, implying the assertion. |
math/9909142 | The assumption implies that MATH and MATH are birationally isomorphic, therefore they contain open dense subsets MATH and MATH, isomorphic over MATH and containing the generic fiber of MATH and MATH respectively. Since MATH and MATH are of finite type over MATH, the projections of MATH and MATH to MATH are not NAME dense. Therefore there exists an open dense subset MATH, whose inverse images in MATH and MATH are contained in MATH and MATH respectively. Then MATH is the required subscheme of MATH. |
math/9909146 | Clearly, both spectral curves already have the points MATH in common. So let MATH and suppose that MATH is an eigenvector of MATH with eigenvalue MATH. In particular, the point MATH belongs to the NAME spectral curve MATH. By definition, we have: MATH . Clearly, MATH is a holomorphic section in MATH vanishing at MATH. Therefore it induces the following exact sequence: MATH which in turn induces the cohomology sequence: MATH . Thus MATH. In particular, MATH is non-empty, hence MATH also belongs to the instanton spectral curve MATH. The same argument clearly provides the converse statement. Thus the curves MATH and MATH must coincide pointwise. It also follows from the cohomology sequence REF that the dual of the MATH-eigenspace of MATH is exactly MATH. In other words, there are canonical identifications between the fibres MATH and MATH, and the line bundles are isomorphic. Finally, let us check that the connection MATH and MATH also coincide. Noting that the projection MATH is just the restriction map MATH on each MATH, it is easy to see that MATH . Therefore, we have: MATH . |
math/9909148 | Let MATH be any local section of MATH near our given point. Immediately we have MATH, and hence restricting our domain if necessary MATH. As MATH, the function MATH serves as a coordinate function. Define MATH to be the exterior ideal generated by the one-forms MATH and MATH. The curvature conditions MATH and MATH show that MATH is integrable. Since MATH each integral manifold lies in a MATH hypersurface. Again restricting our domain if necessary, we may choose independent functions MATH so that the integral manifolds of MATH are uniquely described by MATH and MATH. Note that the integral manifolds of MATH are also the integral manifolds of MATH. Therefore, at each point of our domain we may express MATH for some invertible matrix MATH and vector MATH. We claim that MATH serve as coordinates of the integral manifolds, and hence completes our local coordinate system. To show this, it suffices to show that MATH forms a local coframe near our given point. To this end, we compute: MATH . Yet, as MATH, we have the structure equation MATH. Therefore, MATH . Every two-form in the left side of this equation contains a MATH. Thus, MATH for some matrix MATH and vector MATH. Yet, MATH are linearly independent. Therefore, the MATH are linearly independent, and also independent from MATH, and thus independent from MATH as well. Finally, define a new section by MATH, where MATH abbreviates MATH. Then we have MATH, MATH, and MATH. In particular, MATH, MATH, and MATH (mod MATH), as desired. |
math/9909148 | We have already seen that the straight lines in MATH are characterized by the equations MATH and MATH. Yet, MATH and MATH are the MATH and MATH terms, respectively, in the NAME forms of the Galilean group. Thus by the definition of development, a curve MATH is a geodesic if and only if it is non-degenerate and MATH. Yet, in normal coordinates MATH and MATH; the integral curves to these forms are precisely the solutions to REF . |
math/9909148 | Let MATH be as given; on the portion of MATH over this coordinate system we have natural coordinates MATH. From the construction of the coordinate MATH, it is clear that MATH. Similarly, if there is a section MATH so that MATH, then we must have MATH (explicitly, the desired section is then MATH). This implies that MATH (mod MATH). The geodesics of the geometry are given by MATH, hence MATH (mod MATH), say MATH where MATH is an arbitrary matrix. If the structure equation MATH is to be satisfied, we must have MATH, where MATH but is otherwise arbitrary. Hence REF is shown. Moreover, the most general Galilean geometry satisfying REF is MATH . We leave it as an easy exercise to the reader to show that for these forms, MATH has no MATH term. To show REF , we compute MATH . Therefore, for any choice of MATH, we have MATH. Finally, suppose MATH, and hence MATH, is now fixed. Note that we may write MATH. One easily computes the MATH term in MATH to be MATH . Only the symmetric part of this involves MATH, hence specifying MATH determines MATH uniquely, and REF is proven. |
math/9909148 | Choosing MATH and MATH, we see that MATH (mod MATH). |
math/9909150 | Move each MATH, MATH slightly along the edge MATH to obtain a new point MATH. Let us show that a generic hyperplane MATH through MATH is transverse to MATH. This will imply the lemma because MATH has at least MATH intersections with MATH. It suffices to show that MATH contains no vertices of MATH. Note first that since MATH is in general position, a generic hyperplane MATH through MATH does not contain any other vertex. The same holds true for every hyperplane sufficiently close to MATH. It remains to show that the chosen MATH does not contain either of MATH. Let MATH contain MATH; then MATH also contains the edge MATH and therefore MATH. If MATH we obtain a contradiction with the previous paragraph. On the other hand, if MATH we can continue in the same way. However, we cannot continue indefinitely since MATH. |
math/9909150 | Connect MATH and MATH by a segment to obtain a closed polygon MATH and consider a hyperplane MATH close to MATH, transverse to MATH and intersecting MATH in MATH points. Since MATH is contractible, MATH intersects MATH in an even number of points. Therefore, MATH intersects the segment MATH for odd MATH and does not intersect it for even MATH. |
math/9909150 | We need to prove that through every MATH-tuple MATH there passes a hyperplane MATH intersecting MATH with multiplicity MATH. Choose a point MATH on the line MATH so that MATH lies on the segment MATH if MATH is even and does not lie on this segment if MATH is odd. Define MATH as the linear span of MATH. We claim that its projection MATH intersects MATH with multiplicity MATH. Let MATH be a hyperplane close to MATH and transverse to MATH; assume, moreover, that MATH contains no vertices. It suffices to show that MATH cannot intersect MATH in more than MATH points. On the one hand, MATH cannot intersect all the edges of MATH. If it were the case, then MATH would separate all pairs of consecutive vertices which contradicts the choice of MATH. On the other hand, if the number of intersections of MATH and MATH were greater than MATH, it would be equal to MATH since, for topological reasons, the parity of this number of intersections is that of MATH. We obtain a contradiction which proves the claim. Finally, by REF , the multiplicity of the intersection of MATH with MATH is not less than MATH. |
math/9909150 | The determinant REF involves all MATH vectors MATH. If MATH is odd then, according to REF , MATH, and REF concerns the cyclic permutation of the vectors that changes sign of the determinant. On the other hand, if MATH is even then MATH which also leads to the change of sign in REF . |
math/9909150 | Induction in the number of vertices MATH. Induction starts with MATH. Up to projective transformations, the unique strictly convex MATH-gon is the simplex MATH. Indeed, every generic MATH-tuple of points in MATH can be taken to any other by a projective transformation. Therefore, all generic broken lines with MATH edges are projectively equivalent. It remains to connect the last point with the first one, and there are exactly two ways to do it. One of them gives a contractible polygon and the other non-contractible one. One of these polygons is MATH, while the other cannot be strictly convex since the parity of its intersections with a hyperplane is opposite to MATH. The base of induction is then provided by REF . Let MATH be a strictly convex MATH-gon with vertices MATH. Delete MATH and connect MATH with MATH in such a way that the new edge MATH together with the two deleted ones, MATH and MATH, forms a contractible triangle. Denote the new polygon by MATH. Let us show that MATH is strictly convex. MATH is strictly convex, therefore through every MATH vertices of MATH there passes a hyperplane MATH intersecting MATH with multiplicity MATH. We want to show that the multiplicity of the intersection of MATH with MATH is also MATH. Let MATH be a hyperplane close to MATH and transverse to MATH and MATH. The number of intersections of MATH with MATH does not exceed that with MATH. Indeed, if MATH intersects the new edge, then it intersects one of the deleted ones since the triangle in contractible. By the induction assumption, MATH has at least MATH flattenings. To prove the theorem, it remains to show that MATH cannot have more flattenings than MATH. Consider the sequence of determinants REF MATH. Replacing MATH by MATH we remove MATH consecutive determinants MATH and add in their stead MATH new determinants MATH where MATH with MATH. The transition from REF to REF is done in two steps. First, we add REF to REF so that the two sequences alternate, that is, we put MATH between MATH and MATH. Second, we delete the ``old" determinants REF . We will prove that the first step preserves the number of sign changes while the second step obviously cannot increase this number. If MATH and MATH are of the same sign, then MATH is of the same sign too. Proof of the lemma. Since MATH is in general position, the removed vector MATH is a linear combination of MATH vectors MATH: MATH where dots is a linear combination of the rest of the vectors. It follows from REF that MATH . It is time to make use of strict convexity of MATH. Let MATH be a hyperplane in MATH through MATH vertices MATH that intersects MATH with multiplicity MATH, and let MATH be its lifting to MATH. Choose a linear function MATH in MATH vanishing on MATH and such that MATH. We claim that MATH . Indeed, by REF , the multiplicity of the intersection of MATH with the polygonal lines MATH and MATH are at least MATH and MATH, respectively. Since MATH intersects MATH with multiplicity MATH, the two above multiplicities are indeed equal to MATH and MATH. REF now readily follow from REF . Finally, evaluate MATH on REF : MATH . It follows from REF and MATH that at least one of the numbers MATH or MATH is positive. In view of REF follows. |
math/9909150 | The NAME map establishes a one-to-one correspondence between conics in MATH and hyperplanes in MATH: the image of the conic is the intersection of a hyperplane with the quadratic surface MATH. Since MATH is an embedding, the points MATH and MATH lie on one side of the conic through MATH if and only if the points MATH and MATH lie on one side of the corresponding hyperplane. |
math/9909151 | The symmetry in MATH and MATH follows from the fact that the MATH-torus knot and the MATH-torus knots are of the same knot-type (see CITE). Because the NAME integral is universal for NAME invariants it suffices to prove that in the NAME integral the coefficient of a basis symmetrized diagram has the properties described in the theorem. Fix a basis of symmetrized diagrams of degree MATH. Pick any element of this basis. By the linearity of change of basis from round diagrams to symmetrized diagrams, the coefficient of this element in MATH is a polynomial in MATH of degree at most MATH. The action of MATH will just be to multiply by some power of MATH. By the symmetry in MATH and MATH the result must be a symmetric polynomial of degree at most MATH in MATH and MATH. |
math/9909152 | If three spheres have a common tangency, the three lines all meet at that point; otherwise, each sphere either contains all of or none of the other three spheres. Let the four given spheres MATH REF have centers MATH and radii MATH. If MATH contains none of the other spheres, let MATH, else let MATH. Then the point of tangency MATH between spheres MATH and MATH can be expressed in terms of these values as MATH . In other words, it is a certain weighted average of the two sphere centers, with weights inversely proportional to the (signed) radii. Now consider the point MATH formed by taking a similar weighted average of all four sphere centers. Then MATH that is, MATH is a certain weighted average of the two tangencies MATH and MATH, and therefore lies on the line MATH. By a symmetric argument, MATH also lies on line MATH and line MATH, so these three lines are coincident. |
math/9909152 | By using ideas from our proof of REF , we can express MATH as a weighted average of MATH and MATH: MATH . Hence, MATH is collinear with MATH and MATH. Collinearity with MATH and MATH follows from the known collinearity of MATH with MATH and MATH. A symmetric argument applies to MATH. We omit the proof of harmonicity, which we obtained by manipulating trilinear coordinates of the new centers in NAME. See REF .pdf for the detailed calculations. |
math/9909152 | Let MATH be a circle centered at MATH, such that MATH and MATH cross at right angles. Then inverting through MATH produces a figure in which MATH and MATH have been transformed into lines parallel to MATH, while MATH is unchanged. Since the image of MATH is tangent to MATH and to the two parallel lines, it is a circle congruent to MATH and centered on MATH. Therefore, the inverted image of MATH is a point MATH where MATH and MATH cross. Points MATH, MATH, and MATH are collinear since one is the center of an inversion swapping the other two. |
math/9909153 | This is REF. |
math/9909153 | This is REF. |
math/9909153 | For large MATH, denote MATH constant, where MATH, MATH . We build our proof based on following two Observations. In the first place, we look at instances when fraction MATH is far away, in terms of MATH, from the corresponding MATH. By our instinct, if there is a restraint for such spot, we would have an easy job to evaluate the sum in REF. In the second place, we try to set factor MATH apart from the rest of terms so that we could get a desired outcome. For a fixed denominator MATH, we assume that there are MATH fractions MATH corresponding to MATH, where MATH, fall into a MATH-interval MATH, MATH; there are MATH . NAME fractions MATH such that MATH where MATH; there are MATH fractions MATH corresponding to MATH such that MATH, MATH, MATH. Since there is a one on one correspondence between MATH and NAME fraction MATH, we have MATH . Note that on average there are MATH . NAME fractions MATH fall in a MATH-interval MATH, MATH, the assumption tells us that MATH-intervals between MATH and MATH are void, while MATH-interval MATH is abundant in terms of MATH. Since all MATH, MATH a prime, are equally distributed over MATH, the void and the abundant are caused by MATH, MATH a composite number. If MATH, MATH falls into MATH-interval MATH, but not previous MATH-intervals, we have MATH . From REF , we know that MATH . According to REF , there are at most MATH such NAME fractions. Since MATH, we have MATH . Hence, MATH, MATH by REF; By REF , we know that MATH . There are at most MATH such NAME fractions. However, if MATH, we have MATH . Hence, MATH; If MATH and MATH, we use REF . Since MATH we would get a contradiction if MATH, where MATH. Therefore, MATH. For a fixed MATH, let each MATH-interval has equal number of NAME fractions, we see that fraction MATH would coincide with MATH. Hence, only uneven distribution of MATH might cause MATH deviates from MATH. We study following cases with consideration of REF , and There should be no MATH-interval MATH, MATH, MATH, with negative number of MATH, MATH. If MATH, then MATH. On average, fraction MATH might deviate from the corresponding NAME fraction by MATH. By REF , we know that MATH . If MATH, then on average fraction MATH might deviate from the corresponding NAME fraction by MATH. Applying REF , we know that MATH . There exist a number MATH such that MATH, that is, MATH. Substituting it into REF , we have MATH . In an extreme case, if fraction MATH deviates from MATH by MATH, we have MATH by Condition N. Therefore, MATH, and MATH . There exist a number MATH such that MATH, that is, MATH. Substituting it into REF , we have MATH . In general, if there is a function MATH such that MATH deviates from the corresponding NAME fraction by MATH, through Condition N we have MATH. Subsequently, we could find a corresponding MATH such that MATH, and get an inequality analogous to REF. Summing up these type of inequalities over MATH, we would have the desired consequence by REF . |
math/9909153 | By CITE, CITE and REF . See also CITE. |
math/9909153 | By REF . See CITE. |
math/9909153 | This is Theorem A. REF. |
math/9909153 | For a principal character MATH, we use REF ; for a nonprincipal character MATH, we use REF . |
math/9909153 | Through the NAME product for MATH where MATH prime, MATH, we have the first part of REF . For the second part, using REF we have MATH . Let MATH, we get the conclusion. |
math/9909153 | By REF and analytic continuation, we know that MATH converges for all MATH. Hence, MATH is analytic for the half-plane MATH. Through functional REF , we see that MATH is symmetric with respect to MATH over MATH. Therefore, all nontrivial zeros of MATH lie on line MATH. |
math/9909154 | This is REF. |
math/9909154 | This is REF. |
math/9909154 | This is REF. |
math/9909154 | This is REF. |
math/9909154 | Applying REF , this is REF for REF. |
math/9909154 | Applying REF , this is REF for REF. |
math/9909154 | By REF, we have MATH . Applying REF, we have our result. |
math/9909156 | By definition of the crossed product MATH we have the following equalities between subalgebras of MATH: MATH . On the other hand since the coactions on finite dimensional algebras are non-degenerate we have MATH (see for example, REF). Thus MATH . Let us compute the restriction of MATH to MATH, to MATH and to MATH. REF the restriction of MATH to MATH is MATH. In particular MATH has no fixed points in MATH. REF the algebra MATH is fixed by MATH. REF we prove that the algebra MATH is also fixed by MATH. For, let MATH be an orthonormal basis of MATH consisting of coefficients of irreducible corepresentations of MATH. Since MATH, for any MATH we may use the notation MATH (finite sum). See for example, CITE. From the coassociativity of MATH we get MATH so MATH for any MATH. Thus MATH . Also MATH, so we get that MATH is equal to MATH . By summing over MATH the last term is replaced by MATH. Thus MATH . From MATH we get that MATH, which is the fixed point algebra of MATH under MATH, is MATH. This finishes the proof of the first assertion and proves the non-degeneracy of the diagram in the statement. For the commuting square condition, remark that this diagram is the dual of the square on the left in the following diagram MATH where MATH and MATH. Since MATH is dual, REF applies and shows that the square on the right is a non-degenerate commuting square. We also know that the rectangle is a non-degenerate commuting square. Thus if we denote by MATH the conditional expectation onto MATH, for any MATH, then for any MATH we have MATH. This proves the commuting square condition. |
math/9909156 | CASE: For MATH this follows from REF. We prove it for MATH. Let MATH and consider the following diagram MATH . By REF, the square on the left and the rectangle are non-degenerate commuting squares. We want to prove that the square on the right is a non-degenerate commuting square. The non-degeneracy condition follows from MATH . Let MATH and write MATH with MATH and MATH. Then MATH . This proves the commuting square condition. CASE: A similar proof shows that the proposition holds in the case MATH. CASE: General case. We will use many times the following diagram MATH in which all the rectangles and all the squares, except possibly for the square in the statement, are non-degenerate commuting squares (compare REF, II, III). The non-degeneracy condition follows from MATH . Let MATH and write MATH with MATH and MATH. Then MATH . This proves the commuting square condition. |
math/9909156 | We prove only that the rows are NAME towers (same proof for columns). By restricting attention to a pair of consecutive inclusions, it is enough to prove the following statement: if MATH is a basic construction in MATH and MATH is an object of MATH then MATH is a basic construction in MATH. For, we will use many times the following diagram MATH in which all squares and rectangles are non-degenerate commuting squares (compare REF ). We use the abstract characterisation of the basic construction: MATH is a basic construction with NAME projection MATH if and only if: CASE: MATH; CASE: MATH; CASE: MATH for any MATH and REF MATH for any MATH, where MATH is the inverse of the index of MATH. Let MATH be the NAME projection for the basic construction MATH. With MATH, MATH and MATH the verification of REF is as follows. REF follows from the following computation: REF follows from MATH, from MATH and from MATH. CASE: Let MATH and write MATH with MATH and MATH. Then MATH REF With the above notations, we have that MATH and that MATH for every MATH, so MATH and this proves REF . |
math/9909156 | Consider the MATH-morphism MATH . Since both squares in the statement are non-degenerate commuting squares, all the assertions are consequences of the formulae MATH for any MATH and MATH for any MATH, to be proved now. The second one follows from the following computation MATH . For the first formula, what we have to prove is that MATH . By moving the unitaries to the left and to the right we have to prove that MATH . Call this unitary MATH. Since MATH we have MATH . By CITE the comultiplication of MATH is given by MATH. On the other hand since MATH is a corepresentation of MATH, its adjoint MATH is a corepresentation of MATH, that is, we have MATH. We get that MATH . Thus MATH and we are done. |
math/9909156 | CASE: Denote by MATH the multiplication and by MATH the unit. We recall that the universal MATH-algebra MATH generated by the entries of a unitary matrix MATH such that MATH and MATH has a unique structure of NAME MATH-algebra which makes MATH a corepresentation of it. The element MATH is a coaction. See CITE, CITE. Since MATH is a MATH-morphism we get that MATH and that MATH with the notations in CITE (same proof as for REF in there). Since MATH is a commuting square we have that MATH, so the same computation as in the first part of the proof of REF shows that MATH, that is, that MATH is a unitary. It follows that there exists a representation MATH such that MATH. Thus MATH is biunitary and MATH is a model for it (see CITE). CASE: By using the universal property of the minimal model we get a morphism MATH such that MATH. It follows that MATH is equal to MATH, which is a coaction. CASE: Since MATH is a corepresentation we have MATH in MATH, for any MATH. By applying MATH we get that MATH in MATH, for any MATH. The term on the right is MATH and the term on the left is MATH. CASE: Consider the following diagram MATH associated to the objects MATH and MATH of MATH and MATH of MATH. By REF this is a basic construction for non-degenerate commuting squares. By applying REF with MATH we get an isomorphism between the big square in the above diagram and the big square in the diagram below MATH which sends MATH for any MATH and MATH for any MATH. It follows that this diagram is also a basic construction for non-degenerate commuting squares. Thus MATH. |
math/9909156 | CASE: Since MATH is the minimal model for MATH, MATH is the minimal model for MATH. Thus REF shows that MATH is a coaction. CASE: We prove that MATH is a coaction. We use terminology and notations from REF below. By using the construction MATH in the beginning of this section we can define an embedding MATH by MATH for any coefficient MATH. We have MATH for any representation MATH. Let us apply now the result in REF with the dual of MATH instead of MATH and with MATH as the choice of the biunitary. We get that if MATH is the minimal model for MATH then MATH is a coaction of MATH. By applying REF below we get an embedding MATH such that MATH. It follows that MATH is a coaction of MATH, so MATH is a coaction of MATH. CASE: It remains to prove the isomorphism part. For, consider the diagram MATH associated to the objects MATH and MATH of MATH and MATH and MATH of MATH. By REF this is a lattice of basic construction for non-degenerate commuting squares. On the other hand, by applying REF with MATH we get an isomorphism from the big square in this diagram, say MATH, to MATH, which sends MATH for any MATH and MATH for any MATH. By composing it with the isomorphism MATH we get an isomorphism MATH. Since MATH and MATH, MATH implements an isomorphism between the lower left square in MATH and MATH. REF shows that inclusion in REF is the vertical subfactor associated to the commuting square in REF . Let us prove that MATH is outer. For, we apply REF . Since MATH is inner faithful (compare CITE) it remains to prove that MATH contains the counit. For, we come back to REF . With the notations in there, we have that MATH, so in particular MATH contains a copy of the trivial corepresentation MATH. Since MATH, we have that MATH. By arguing as in REF in the above proof of REF , with the notations in there, we have MATH. Since MATH, it follows that MATH and by dualising we get that MATH contains the counit. Finally, the ergodicity condition MATH follows from factoriality and from REF. |
math/9909156 | We prove first that MATH is subreflexive (same proof for MATH). We have MATH . Thus if MATH then MATH is a model for MATH. By using this twice we get that MATH is a model for MATH. Since MATH is the minimal model for MATH, we get a morphism MATH satisfying MATH. Consider now the diagram MATH . The condition MATH shows that this diagram commutes on the coefficients of MATH. Since these coefficients generate MATH as a NAME algebra, this diagram commutes. It remains to prove that MATH is injective. By using the above diagram it is enough to prove that the composition MATH is injective. The kernel of this map is by REF. Since the coefficients of MATH generate a dense subalgebra of MATH (see CITE), we have in particular that MATH is dense in MATH and this gives MATH. We construct now the duality. Since MATH is a model for MATH, the triple MATH is a model for MATH. By definition of MATH its space of coefficients MATH is the image of the space MATH by the embedding MATH. On the other hand for any subreflexive NAME algebra MATH we have from definitions that MATH, so MATH is dense in MATH. Together with the fact that MATH generates MATH as a NAME algebra, we get that MATH generates in MATH a NAME subalgebra which is dense in MATH. Thus if we define MATH to be the space of coefficients of MATH and denote by MATH the NAME subalgebra generated by it, then MATH is a bi-faithful model for MATH. By CITE we get that this triple is isomorphic to the minimal model MATH. We get in this way an isomorphism MATH and by composing it with the embedding MATH we get an embedding MATH. The same method gives also an embedding MATH and all the assertions are clear with these MATH and MATH. |
math/9909156 | Note first that the equality MATH gives an embedding of MATH in the relative commutant of in the statement. The ``only if" part is clear from REF (because MATH is a NAME subfactor) and will not be used here. For the converse, let MATH be a complete system of non-equivalent unitary irreducible corepresentations of MATH, each of them co-acting on some MATH (see for example, CITE). For any MATH we write MATH and we use the formula MATH, which follows from the coassociativity of MATH. For any MATH we define its spectral projection MATH by MATH. With the above notation, we have MATH, so it follows that MATH are orthogonal projections with respect to the trace of MATH, that their images MATH are in MATH and that MATH decomposes as a direct sum MATH (see CITE for details). CASE: We prove that an element MATH commutes with MATH if and only if all its spectral projections commute with MATH. The ``if" part is clear. Conversely, for any MATH we have MATH, so MATH commutes with MATH. Thus MATH for any unitary corepresentation MATH. CASE: Assume that MATH is not minimal and choose a non-scalar element MATH. By REF all its spectral projections commute with MATH. The condition in the statement gives for MATH and MATH that MATH is a factor, so the spectral projection MATH is a scalar. Thus there exists a non-trivial irreducible corepresentation, say MATH and a nonzero element MATH. If we denote MATH then we have MATH. This shows that the matrix MATH is a (nonzero) MATH-eigenmatrix, that is, that MATH satisfies MATH. It follows that MATH . Consider now the unitary corepresentation MATH . The above equality shows that the matrix MATH is in the fixed point algebra MATH (note that this is the algebra appearing in the quantum extension of NAME 's extension of NAME 's MATH matrix trick, REF). Thus MATH where we use the hypothesis with MATH. Thus MATH, contradiction. |
math/9909156 | Let MATH be our representation and MATH be its product type coaction. By applying REF we have to prove the following statement: for any corepresentation MATH the relative commutant of the inclusion MATH is MATH. We may and will assume that the coefficients of MATH generate MATH. It is easy to see that this inclusion is isomorphic to MATH, where MATH is the corepresentation equal to the transpose of MATH and where we use notations from REF (see CITE for details). By using REF , this inclusion is isomorphic to the vertical subfactor associated to the vertex model corresponding to the biunitary MATH. REF applies and shows that the relative commutant is MATH, where MATH is the minimal model for MATH where MATH. Since MATH contains the counit, the spaces of coefficients of both MATH and MATH are contained in the space of coefficients of MATH. It follows that MATH is inner faithful, so MATH is the minimal model for MATH. Thus MATH, so our commutant is MATH. |
math/9909158 | The proof uses the edge concept, in particular the result that an achronal set is a closed MATH hypersurface if and only if it is edgeless; see for example, REF , p. CASE: Let MATH be a neighborhood of MATH in MATH in which MATH is achronal and edgeless. Then MATH is a neighborhood of MATH in MATH in which MATH is achronal and edgeless in MATH. Hence, MATH is a closed achronal MATH hypersurface in MATH, and it remains to show that it is actually acausal in a perhaps smaller neighborhood. Suppose there exists a sequence of neighborhoods MATH of p, which shrink to MATH, such that MATH is not acausal in MATH for each MATH. Then, for each MATH, there exists a pair of points MATH such that MATH and MATH. Hence for each MATH, there exists a MATH-null geodesic MATH from MATH to MATH. Now, MATH is a causal curve in MATH, and, in fact, must be a null geodesic in MATH, since otherwise we would have MATH, which would violate the achronality of MATH in MATH. Hence MATH, and the initial tangent MATH to MATH, when suitably normalized, is a semi-tangent of MATH at MATH. By the second part of REF , MATH, where MATH is tangent to MATH at MATH. But MATH is tangent to MATH, since each MATH is, which contradicts the assumption that MATH is transverse to MATH at MATH. |
math/9909158 | Let MATH be a neighborhood of MATH such that MATH is closed and achronal in MATH and MATH. For simplicity, we may assume that MATH and MATH. Let MATH be an initial segment in MATH of the future directed null generator of MATH starting at MATH with initial tangent MATH. Then MATH. By the remark before REF , if MATH leaves MATH at some point, it will enter either MATH or MATH. Either case leads to an achronality violation. Hence, MATH must be a null generator of MATH, which implies that MATH is a semi-tangent of MATH. |
math/9909158 | The basic idea for constructing past support hypersurfaces for MATH is to consider the ``past null cones" of points on generators of MATH formed by past null geodesics emanating from these points. For the purpose of establishing certain properties about these null hypersurfaces, it is useful to relate them to achronal boundaries of the form MATH, MATH, which are defined purely in terms of the causal structure of MATH. For this reason we assume initially that MATH is globally hyperbolic. At the end we will indicate how to remove this assumption. For each MATH and normalized semi-tangent MATH at MATH, let MATH, MATH, be the affinely parameterized null geodesic generator of MATH starting at MATH with initial tangent MATH. Since MATH is achronal each such generator is a null ray, that is, a maximal null half-geodesic. Since MATH is maximal, no point on MATH is conjugate to MATH. For each MATH, consider the achronal boundary MATH, which is a MATH hypersurface containing MATH. By standard properties of the null cut locus (see especially, REF, which assume global hyperbolicity) there is a neighborhood MATH of MATH such that MATH is a smooth null hypersurface diffeomorphic under the exponential map based at MATH to a neighborhood of the line segment MATH in the past null cone MATH. From the achronality of MATH we observe, MATH. This implies that MATH is a past support hypersurface for MATH at MATH. Let MATH denote the null mean curvature of MATH at MATH with respect to MATH. We use REF to obtain the lower bound MATH . The argument is standard. In the notation of REF, let MATH, MATH, be the null mean curvature of MATH at MATH with respect to MATH. REF and the energy condition imply, MATH . Without loss of generality we may assume MATH. Then MATH is strictly negative on MATH, and we can devide REF by MATH to obtain, MATH . Integrating REF from MATH to MATH, and letting MATH we obtain the lower bound REF. Thus, since MATH can be taken arbitrarily large, we have shown, in the globally hyperbolic case, that MATH has null mean curvature MATH in the sense of support hypersurfaces with respect to the collection MATH of smooth null hypersurfaces. By REF , MATH is tangent to MATH at MATH. Let MATH denote the null second fundamental form of MATH at MATH with respect to MATH. We now argue that the collection of null second fundamental forms MATH, for some MATH, is locally bounded from below. Recall, ``locally", means ``locally in the point MATH"; the lower bound must hold for all MATH, compare , REF . This lower bound follows from a continuity argument and an elementary monotonicity result, as we now discuss. Fix MATH. Let MATH be a convex normal neighborhood of MATH. Thus, for each MATH, MATH is the diffeomorphic image under the exponential map based at MATH of a neighborhood of the origin in MATH. Provided MATH is small enough, we have that for each MATH, MATH, where MATH is the past null cone in MATH. In particular, for each MATH, MATH is a smooth null hypersurface in MATH such that if MATH in MATH, MATH converges smoothly to MATH. There exists a neighborhood MATH of MATH, MATH, and MATH such that for each MATH, and each normalized null vector MATH, the null geodesic segment MATH, MATH is contained in MATH. Let MATH be the null second fundamental form of MATH at MATH with respect to MATH. Since, as MATH, MATH converges smoothly to MATH, we have that MATH smoothly. Returning to the original family of support hypersurfaces, MATH, with associated family of null second fundamental forms MATH, observe that when MATH, MATH agrees with MATH near MATH. Hence, if MATH in MATH, MATH smoothly. It follows that the collection of null second fundamental forms MATH is locally bounded from below. Consider as in the beginning, for MATH and MATH a normalized semi-tangent at MATH, the null geodesic generator, MATH, MATH, MATH. For MATH, MATH. Then, since MATH is achronal, MATH cannot enter MATH. It follows that for MATH, MATH lies to the past side of MATH near MATH. By an elementary comparison of null second fundamental forms at MATH we obtain the monotonicity property, MATH . This monotonicity now implies that the entire family of null second fundamental forms MATH is locally bounded from below. This concludes the proof of REF under the assumption that MATH is globally hyperbolic. We now describe how to handle the general case. In general, MATH may have bad causal properties. In particular the generators of MATH could be closed. Nevertheless, the past support hypersurfaces are formed in the same manner, as the ``past null cones" of points on generators of MATH formed by past null geodesics emanating from these points. But as an intermediary step, to take advantage of the arguments in the globally hyperbolic case, we pull back each generator to a spacetime having good causal properties. Again, consider, for MATH and MATH a normalized semi-tangent at MATH, the null geodesic generator, MATH, MATH, MATH. Restrict attention to the finite segment MATH. Roughly speaking, we introduce NAME type coordinates near MATH. Let MATH be an orthonormal frame of spacelike vectors in MATH. Parallel translate these vectors along MATH to obtain spacelike orthonormal vector fields MATH, MATH along MATH. Consider the map MATH defined by MATH. Here MATH is an open set containing the line segment MATH. By the inverse function theorem we can choose MATH so that MATH is a local diffeomorphism. Equip MATH with the pullback metric MATH, where MATH is the Lorentzian metric on MATH, thereby making MATH . Lorentzian and MATH a local isometry. Let MATH be the MATH-th coordinate function, MATH. Since the slices MATH are spacelike, MATH is timelike, and hence MATH is a time function on MATH. Thus, MATH is a strongly causal spacetime. The curve MATH, MATH, defined for MATH sufficiently small, is a maximal null geodesic in MATH. Let MATH be a compact neighborhood of MATH in MATH. Then by REF, any two causally related points in MATH can be joined by a longest causal curve MATH in MATH, and each segment of MATH contained in the interior of MATH is a maximal causal geodesic. This property is sufficient to push through, with only minor modifications, all relevant results concerning the null cut locus of MATH on MATH. In view of the above discussion, there is a neighborhood MATH of MATH such that MATH is a smooth null hypersurface diffeomorphic under the exponential map based at MATH to a neighborhood of the line segment MATH in the past null cone MATH. Let MATH be a neighborhood of MATH on which MATH is an isometry onto its image, and let MATH. Then MATH is the desired collection of past support hypersurfaces for MATH, having all the requisite properties. In particular, the mean curvature REF and the monotonicity REF hold for the family MATH, and hence the family MATH, by just the same arguments as in the globally hyperbolic case. |
math/9909158 | The first and main step of the proof is to show that MATH is flat (that is, has vanishing NAME curvature). Then certain global arguments will show that MATH is isometric to NAME space. For technical reasons, it is useful to extend MATH slightly beyond its boundary. In fact, by smoothly attaching a collar to MATH and to MATH, we can extend MATH to a spacetime MATH without boundary such that MATH is a retract of MATH and both MATH and MATH separate MATH. It follows that MATH and MATH are globally achronal null hypersurfaces in MATH. Let MATH. A straight forward limit curve argument, using the asymptotic simplicity of MATH, shows that MATH is causally simple. This means that the sets of the form MATH, MATH, are closed subsets of MATH. (The limit curve lemma, in the form of REF, for example, is valid in MATH.) Let MATH be a null line in MATH which has past end point MATH and future end point MATH. Consider the ``future null cone" at MATH, MATH. From the causal simplicity of MATH it follows that MATH. Hence each point in MATH can be joined to MATH by a null geodesic segment in MATH. In particular MATH is connected. From the simple equality MATH, it follows that, MATH where MATH and MATH is the future directed null generator of MATH starting at MATH. Since asymptotically simple spacetimes are null geodesically complete, REF implies that MATH is a smooth null hypersurface in MATH which is totally geodesic with respect to MATH. Since it is smooth and closed the generators of MATH never cross and never leave MATH in MATH to the future. Moreover, since MATH is achronal, there are no conjugate points to MATH along the generators of MATH. It follows that MATH is the diffeomorphic image under the exponential map MATH of MATH, where MATH is the future null cone in MATH. We are now fully justified in referring to MATH as the future null cone in MATH at MATH. Given a smooth null hypersurface, with smooth future pointing null normal vector field MATH, the shear tensor MATH is the trace free part of the null second fundamental form MATH, MATH. Since MATH is totally geodesic in the physical metric MATH and the shear scalar MATH is a conformal invariant, it follows by continuity that the shear tensor of MATH (with respect to an appropriately chosen MATH- null normal MATH) vanishes in the metric MATH, MATH. The trace free part of REF then implies that the components MATH (with respect to an appropriately chosen pseudo-orthonormal frame in which MATH, compare , REF) of the conformal tensor of MATH vanish on MATH. An argument of NAME in CITE, which makes use of the NAME identities and, in the present case, the vacuum field equations expressed in terms of his regular conformal field equations, then shows that the so-called rescaled conformal tensor, and hence, the conformal tensor of the physical metric MATH must vanish on MATH. Since MATH is NAME flat, we conclude that MATH is flat on MATH. In a precisely time-dual fashion MATH is flat on MATH, where MATH is the past null cone of MATH at MATH. To conclude that MATH is everywhere flat we show that MATH. Consider the set MATH. It is clear from the fact that MATH is an achronal boundary that MATH is open in MATH. As MATH, it follows that MATH, and time-dually, that MATH. Using the fact that MATH and MATH are closed subsets of MATH and MATH, respectively, it follows that MATH is closed in MATH. Hence, MATH. We show that each term in this union is a subset of MATH. Trivially, MATH. Consider MATH. We claim that MATH. If not, then MATH. Choose a point MATH, and let MATH be a null generator of MATH with future end point MATH. Since MATH is edgeless, MATH remains in MATH as it is extended into the past. By asymptotic simplicity, MATH must meet MATH. In fact, since no portion of MATH can coincide with a generator of MATH, MATH must meet MATH transversely and then enter MATH. But this means that MATH has left MATH, which is a contradiction. Since MATH, we have shown that MATH. By the time-dual argument, MATH. Thus, MATH is globally flat. Also, as an asymptotically simple spacetime, MATH is null geodesically complete, simply connected and globally hyperbolic; compare , CITE, CITE. We use these properties to show that MATH is geodesically complete. It then follows from the uniqueness of simply connected space forms that MATH is isometric to NAME space. We first observe that there exists a time orientation preserving local isometry MATH, where MATH is NAME space. To obtain MATH, first construct by a standard procedure a frame MATH of orthonormal parallel vector fields on MATH. Then solve MATH, MATH, for functions MATH, and set MATH. From the fact that MATH is a local isometry and MATH is null geodesically complete, it follows that any null geodesic, or broken null geodesic, in MATH can be lifted via MATH to MATH. In particular it follows that MATH is onto: If MATH is not all of MATH then we can find a null geodesic MATH in MATH that meets MATH but is not entirely contained in MATH. Choose MATH such that MATH is on MATH. Then the lift of MATH through MATH is incomplete, contradicting the null geodesic completeness of MATH. We now show that MATH is timelike geodesically complete. If it is not, then, without loss of generality, there exists a future inextendible unit speed timelike geodesic MATH, MATH, with MATH. Let MATH; MATH can be extended to a complete timelike geodesic in MATH which we still refer to as MATH. Let MATH be a future directed broken null geodesic extending from MATH to MATH. Let MATH be the lift of MATH starting at MATH; MATH extends to a point MATH, with MATH. Since MATH is globally hyperbolic there exists a maximal timelike geodesic segment MATH from MATH to MATH. Then MATH is a timelike geodesic segment in MATH from MATH to MATH. Hence, MATH. It follows that MATH extends MATH to MATH, which is a contradiction. Finally, we show that MATH is spacelike geodesically complete. If it is not, then there exists an inextendible unit speed spacelike geodesic MATH, MATH, with MATH. Let MATH; MATH can be extended to a complete spacelike geodesic in MATH which we still refer to as MATH. We now use the fact that timelike geodesics, and broken timelike geodesics, in MATH can be lifted via MATH to MATH. Let MATH be a two-segment broken timelike geodesic extending from MATH to MATH, with MATH starting at MATH and past pointing. Similarly, let MATH be a two-segment broken timelike geodesic extending from MATH to MATH, with MATH starting at MATH and future pointing. Let MATH be the point at the corner of MATH, and let MATH be the point at the corner of MATH. Note that MATH. Let MATH be the lift of MATH starting at MATH; MATH extends to a point MATH with MATH. Similarly, let MATH be the lift of MATH starting at MATH. Let MATH be the point at the corner of MATH, and let MATH be the point at the corner of MATH. Note that an initial segment of MATH is contained in MATH. We claim that MATH is entirely contained in MATH. If not then MATH either leaves MATH or MATH. Suppose it leaves MATH at MATH. Since MATH is globally hyperbolic, there exists a null geodesic segment MATH from MATH to MATH. Then MATH is a null geodesic in MATH from MATH to MATH. But since, by construction, MATH, no such null geodesic exists. Thus, MATH. We show that MATH extends to MATH, thereby obtaining a contradiction. Consider a sequence MATH, MATH. Since MATH is compact, there exists a subsequence MATH which converges to a point MATH. Since MATH, it follows by continuity that MATH. Let MATH be a causal geodesic segment from MATH to MATH. Then MATH is a causal geodesic from MATH to MATH in MATH. Thus, MATH, and hence MATH. Since MATH has future end point MATH, we conclude that MATH. Hence, since every sequence MATH, MATH, has a subsequence converging to MATH, it follows that MATH, and so MATH extends to MATH. This concludes the proof that MATH is geodesically complete and hence, as noted above, isometric to NAME space. |
math/9909160 | REF follows from the fact that MATH. From the fact that MATH follows that MATH does not admit any nontrivial deformations as an algebra, (see CITE), which proves REF. From the fact that MATH follows that any deformation of the algebra morphism MATH appears as a conjugation of MATH. In particular, the comultiplication in MATH looks like REF with some MATH such that MATH. From the coassociativity of MATH follows that MATH satisfies the equation MATH for some invariant element MATH. The element MATH satisfying REF can be obtained by correction of some MATH only obeying REF , CITE. This procedure also makes use simple cohomological arguments and essentially REF . This proves REF. |
math/9909160 | Let the comultiplication for MATH have the form REF . Let MATH be a commutative algebra with the MATH invariant multiplication MATH. Suppose MATH is a MATH invariant quantization of MATH . This means that the deformed multiplication has the form MATH and satisfies the relation MATH . Substituting REF and collecting the terms by MATH we obtain MATH . Subtracting from this equation the similar one with permuting MATH and MATH and making use that MATH is commutative and MATH is skew-commutative, we derive that the NAME bracket MATH has to satisfy the property MATH . Let us prove that the bracket MATH is MATH invariant. Indeed, from REF we have for MATH, MATH . Using this expression, REF , and the fact that MATH, we obtain MATH which proves the invariantness of MATH. So, we have MATH, as required. It is easy to check that any bracket of the form MATH, for MATH, is compatible with any invariant bracket. In particular, a MATH-matrix bracket is compatible with MATH. In addition, MATH is a NAME bracket, so its NAME bracket with itself is equal to zero. Using this and the fact that the NAME bracket of MATH-matrix bracket with itself is equal to MATH, we obtain from REF that MATH. |
math/9909160 | For the two parameter quantization, the NAME brackets MATH and MATH form a NAME pencil, hence must be compatible. Also, MATH is a MATH invariant bracket, so that MATH is compatible with the MATH-matrix bracket MATH. It follows from REF that MATH has to be compatible with MATH. |
math/9909160 | Apply the above construction to MATH, the deformed basic representation of MATH. In this case MATH, where MATH is a deformed adjoint representation. So, MATH is a deformation of the standard embedding of MATH in MATH. It is easy to see that in this case MATH is a deformation of the usual permutation: MATH, and MATH is a deformation of the NAME bracket on MATH: MATH, MATH. Hence, at MATH, the quadratic-linear relations REF are exactly the defining relations for MATH, therefore the map REF is an isomorphism at MATH. It follows that REF is an embedding. (Actually, REF is essentially an isomorphism, that is, it is an isomorphism after completion of MATH in MATH-adic topology.) So, the kernel of the map MATH is defined by the quadratic-linear relations REF . |
math/9909160 | Since in this case MATH, from REF follows that REF is a monomorphism at MATH. Due to the PBW theorem the algebra MATH at the point MATH is a free MATH-module and is equal to MATH . For MATH this algebra is the symmetric algebra MATH, the algebra of algebraic functions on MATH. For MATH this algebra is isomorphic to MATH. Since MATH is a free MATH-module, it follows that MATH in REF is a monomorphism of algebras over MATH and MATH is a free MATH-module isomorphic to MATH . It is clear that MATH is the standard quantization of the NAME bracket on MATH. |
math/9909160 | That MATH coincides with the NAME bracket is obvious from REF . From REF we have MATH. Since MATH is a quadratic algebra over MATH, MATH must be a quadratic bracket. But the MATH-matrix bracket MATH is quadratic, too. Hence, MATH must be a quadratic invariant bracket. There is only one possibility for such a bracket: it must be a unique (up to a factor) nontrivial invariant map MATH. Now apply REF . |
math/9909160 | Let MATH. The invariance condition for MATH means: MATH . The NAME bracket MATH is: MATH . In the left hand side of this expression, REF-st, REF-th, and REF-th terms are canceled due to REF , and we have MATH . Putting in this equation instead of MATH its expression from REF , that is, MATH, we obtain, since the term MATH is canceled: MATH . Now observe that, due to the NAME rule, REF is valid for any MATH. To prove the proposition, it is sufficient to show that if MATH belongs to the ideal MATH defining the orbit MATH, then MATH also belongs to this ideal. Again, due to the NAME rule, it is sufficient to show this for MATH. Since MATH is semisimple, there are elements MATH such that MATH. We have from REF MATH . But MATH, since the NAME bracket is restricted on any orbit, MATH by hypothesis of the proposition. So, MATH. |
math/9909160 | Follows from REF . |
math/9909160 | If such a quantization exists, then from REF follows that there exists an invariant bracket MATH on MATH such that MATH and MATH. Here MATH is the NAME bracket and MATH is the three-vector field induced by MATH (see REF ). We show that such MATH does not exist. Observe that MATH has type MATH, that is, is a sum of terms of the view MATH, where MATH is a homogeneous polynomial of degree MATH. Observe also that the NAME bracket of two polyvector fields of degrees MATH and MATH is a polyvector field of degree MATH. We shall write MATH for degree MATH when the second number, MATH, is clear from context. It is obvious that on MATH there are no invariant bivector fields of degree MATH and, up to a factor, there is a unique invariant bivector field of degree MATH, the NAME bracket MATH itself. Since MATH, there are no bivector fields of degree MATH (see REF ). Therefore, MATH must be of the form: MATH, where MATH is a bracket of degree MATH. Since MATH is compatible with MATH and MATH, it must be MATH. But it is impossible, because MATH has at least degree MATH. |
math/9909160 | First of all, define a quantum exterior algebra, MATH, an algebra of differential forms with constant coefficients. Let us modify the operator MATH from REF . Since the representation MATH is isomorphic to MATH, there exists a MATH invariant bilinear form on MATH, deformed Killing form. This form can be naturally extended to all tensor degrees MATH. Let MATH be the MATH submodule in MATH orthogonal to MATH. Define an operator MATH on MATH in such a way that it has the eigenvalues MATH on MATH and MATH on MATH. It is clear that MATH and MATH are deformed skew symmetric and symmetric subspaces of MATH. Now observe that the third graded component in the quadratic algebra MATH is the quotient of MATH by the submodule MATH, hence this submodule and, therefore, the submodule MATH are direct submodules in MATH, that is, they have complement submodules. As the complement submodules one can choose the submodules MATH and MATH, respectively, since they are complement at the point MATH and MATH is orthogonal to MATH with respect to the Killing form extended to MATH. Hence, MATH is a direct submodule. Also, the symmetric algebra MATH is NAME. From a result of NAME, CITE (see also CITE), follows that the quadratic algebra MATH is a free MATH module, that is, is a MATH-invariant deformation of the exterior algebra MATH. Call MATH a quantum exterior algebra over MATH. Define a quantum algebra of differential forms over MATH as the tensor product MATH in the tensor category of representations of the quantum group MATH. The multiplication of two elements MATH and MATH looks like MATH, where MATH for MATH being the permutation in that category. So, MATH. As in the classical case, the algebras MATH and MATH can be embedded in MATH as a graded submodules in the following way. Call the submodule MATH of MATH a MATH-th symmetric part of MATH. It is clear that the natural map MATH restricted to MATH is a bijection onto the MATH-degree component MATH of MATH. Denote by MATH the inverse bijection. Similarly we define MATH, the MATH-th skew symmetric part of MATH, and the bijection MATH. Now, define a differential MATH in MATH as a homogeneous operator of degree MATH. It acts on an element, MATH, of degree MATH in the following way. Let MATH be its realization as an element from MATH. Then the formula MATH presents the element MATH through its realization in MATH. It is obvious that MATH. So, the graded differential algebra MATH is constructed. It is easy to see that at the point MATH this algebra coincides with MATH. |
math/9909160 | There exists an isomorphism of MATH modules MATH, where MATH, the direct sum of the MATH-th symmetric parts of MATH (see previous Subsection). Consider the composition MATH, where the last map appears from REF . It is an isomorphism, since it is an isomorphism at the point MATH. It follows that MATH is isomorphic to MATH as a MATH-module, Denote by MATH the submodule of MATH invariant elements in MATH. It is obvious that MATH is isomorphic to MATH, where MATH is the invariant submodule in MATH. Hence, MATH is a direct free MATH submodule in MATH. Moreover, MATH is a central subalgebra in MATH. Indeed, for a generic MATH the algebra MATH can be invariantly embedded in MATH. But MATH invariant elements in MATH form the center of MATH. Also, MATH as an algebra is isomorphic to MATH with the trivial action of MATH, where MATH, the algebra of invariant elements in MATH. This follows from the fact that MATH is a polynomial algebra, CITE, and, therefore, admits no nontrivial commutative deformations. By the NAME theorem, CITE, MATH is a free module over its center. It follows that at the point MATH the module MATH is a free module over the algebra MATH. One can easily derive from this that MATH is a free module over MATH. Now, let the maximal semisimple orbit MATH be defined by invariant elements from MATH. Consider a character defined by MATH, the algebra homomorphism MATH which takes each element from MATH to its value on MATH. Then, MATH may be considered as a MATH-module, and the function algebra MATH on MATH is equal to MATH. Extend the character MATH up to a character MATH in the trivial way and consider MATH as a MATH-module. The tensor product over MATH is a MATH-algebra. It is a free MATH-module, since MATH is a free one over MATH. It is obvious, that MATH, MATH gives a quantization of the KKS bracket on MATH, and MATH is a quotient algebra of MATH. |
math/9909160 | REF are proven in CITE. REF follows from the fact that all the weight subspaces for all MATH have the dimension one (see NAME, NAMEgèbres NAME, REF ). |
math/9909160 | Follows from REF |
math/9909160 | See CITE. |
math/9909160 | See CITE. |
math/9909160 | First, let MATH be a NAME bracket, that is, MATH. Then the complex of polyvector fields on MATH, MATH, with the differential MATH is well defined. Denote by MATH the NAME complex on MATH. Since none of the coefficients MATH of MATH are zero, MATH is a nondegenerate bivector field, and therefore it defines a MATH-linear isomorphism MATH, MATH, which can be extended up to the isomorphism MATH of MATH-forms onto MATH-vector fields for all MATH. Using NAME identity for MATH and invariance of MATH, one can show that MATH gives a MATH invariant isomorphism of these complexes, so their cohomologies are the same. Since MATH is simple, the subcomplex of MATH invariants, MATH, splits off as a subcomplex of MATH. In addition, MATH acts trivially on cohomologies, since for any MATH the map MATH, MATH, is homotopic to the identity map, (MATH is a connected NAME group corresponding to MATH). It follows that cohomologies of complexes MATH and MATH coincide. But MATH gives an isomorphism of complexes MATH and MATH. So, cohomologies of the latter complex coincide with NAME cohomologies, which proves the proposition for MATH being NAME brackets. Now, consider the family of complexes MATH, MATH. It is clear that MATH depends algebraicly on MATH. It follows from the uppersemicontinuity of MATH and the fact that MATH for odd MATH, CITE, that MATH for odd MATH and almost all MATH. Using the uppersemicontinuity again and the fact that the number MATH is the same for all MATH, we conclude that MATH for even MATH and almost all MATH. |
math/9909160 | In this case the system of positive quasiroots consists of MATH, MATH, and MATH, where MATH, MATH are the simple quasiroots. From REF follows that invariant elements in MATH and MATH form subspaces of dimension one, MATH and MATH. Moreover, all the ivariant elements of MATH are lying in MATH. Since MATH takes MATH onto MATH, there is only one-dimansional MATH invariant subspace in MATH, which is necessarily generated by MATH. |
math/9909160 | To begin, consider the multiplication MATH. The corresponding obstruction cocycle is given by MATH considered modulo terms of order MATH. No MATH terms appear because MATH is a biderivation and, therefore, a NAME cocycle. The fact that the presence of MATH does not interfere with the cocyle condition and that this equation defines a NAME MATH-cocycle was proven in CITE. It is well known that if we restrict to the subcomplex of cochains given by differential operators, the differential NAME cohomology of MATH in dimension MATH is the space of MATH-polyvector fields on MATH. Since MATH is reductive, the subspace of MATH invariants splits off as a subcomplex and has cohomology given by MATH. The complete antisymmetrization of a MATH-tensor projects the space of invariant differential MATH-cocycles onto the subspace MATH representing the cohomology. The equation MATH implies that obstruction cocycle is a coboundary, and we can find a MATH-cochain MATH, so that MATH satisfies MATH . Assume we have defined the deformation MATH to order MATH such that MATH associativity holds modulo MATH, then we define the MATH obstruction cocycle by MATH . In CITE REF we showed that the usual proof that the obstruction cochain satisfies the cocycle condition carries through to the MATH associative case. The coboundary of MATH appears as the MATH coefficient of the signed sum of the compositions of MATH with MATH. The fact that MATH mod MATH together with the pentagon identity implies that the sum vanishes identically, and thus all coefficients vanish, including the coboundary in question. Let MATH be the projection of MATH on the totally skew symmetric part, which represents the cohomology class of the obstruction cocycle. The coefficient of MATH in the same signed sum, when projected on the skew symmetric part, is MATH which is the coboundary of MATH in the complex MATH. Thus MATH is a MATH cocycle. By REF , this complex has zero cohomology. Now we modify MATH by adding a term MATH with MATH and consider the MATH obstruction cocycle for MATH. Since the term we added at degree MATH is a NAME cocyle, we do not introduce a MATH term in the calculation of MATH and the totally skew symmetric projection MATH term has been modified by MATH. By choosing MATH appropriately, we can make the MATH obstruction cocycle represent the zero cohomology class, and we are able to continue the recursive construction of the desired deformation. |
math/9909160 | The existence of a multiplication which is MATH associative up to and including MATH terms is nearly identical to the previous proof. Both MATH and MATH are anti-invariant under the NAME involution MATH. We shall look for a multiplication MATH such that MATH is MATH anti-invariant and skew-symmetric for odd MATH and MATH invariant and symmetric for even MATH. So, suppose we have a multiplication defined to order MATH, MATH with mentioned above invariance properties and MATH associative to order MATH. Further we shall suppose that MATH has the properties: It is invariant under the NAME involution MATH and MATH. Such MATH always can be choosen, CITE. Using these properties for MATH, direct computation shows that the obstruction cochain, MATH has the following invariance properties: For odd MATH, MATH is MATH invariant and MATH, and for even MATH, and MATH is MATH anti-invariant and MATH. Hence, the projection of MATH on MATH is equal to zero for even MATH. It follows that all the MATH are NAME coboundaries, and the standard argument implies that the multiplication can be extended up to order MATH with the required properties. For odd MATH, REF shows that the projection on MATH has the form MATH . The KKS bracket is given by the two-vector MATH . Setting MATH gives MATH . Defining MATH the new obstruction cohomology class is MATH . Finally we define MATH and get an obstruction cocycle which is zero in cohomology. Now the standard argument implies that the deformation can be extended to give a MATH associative invariant multiplication with the required properties of order MATH. So, we are able to continue the recursive construction of the desired multiplication. |
math/9909166 | This follows easily from the definition of MATH. |
math/9909166 | Let us choose a point in the relative interior of each face of MATH (we also take all vertices and a point in the interior of the polytope). The resulting set MATH of MATH points will be the vertex set of the cubical complex MATH. Since the polytope MATH is simple, the number of MATH-faces meeting at each vertex is MATH, MATH. Hence, for each vertex MATH there is defined a MATH-element subset MATH of MATH consisting of the points chosen in the interiors of faces containing MATH (including MATH itself and the point in the interior of MATH). This set MATH is said to be the vertex set of the cube MATH corresponding to MATH. The faces of MATH are defined as follows. We take any two faces MATH and MATH of MATH such that MATH, MATH. Then there are MATH faces MATH of dimension MATH such that MATH, MATH. Hence, there are MATH faces ``between" MATH and MATH. The points inside these faces define a MATH-element subset of MATH, which is said to be a vertex set of a MATH-face MATH of the cube MATH. Now, to finish the definition of the cubical complex MATH we need only to check that the intersection of any two cubes MATH, MATH is a face of each. To do this we take the minimal-dimension face MATH that contains both vertices MATH and MATH (clearly, there is only one such face). Then it can be easily seen that MATH is the face of MATH and MATH. Now let us construct an embedding MATH. First, we define the images of the vertices of MATH, that is, the images of the points of MATH. To do this, we fix the numeration of facets: MATH. Now, if a point of MATH lies inside the facet MATH, then we map it to the vertex MATH of the cube MATH, where REF stands on the MATH-th place. If a point of MATH lies inside a face MATH of codimension MATH, then we write MATH, and map this point to the vertex of MATH whose MATH coordinates are zero and all other coordinates are REF. The point of MATH in the interior of MATH maps to the vertex of MATH with coordinates MATH. Hence, we constructed the map from the set MATH to the vertex set of MATH. This map obviously extends to a map from the cubical subdivision MATH of MATH to the standard cubical subdivision of MATH. One of the ways to do this is as follows. Take a simplicial subdivision MATH of MATH with vertex set MATH such that for each vertex MATH there exists a simplicial subcomplex MATH with vertex set MATH that subdivides the cube MATH. The simplest way to construct such a simplicial complex is to view MATH as the cone over the barycentric subdivision of the complex MATH dual to the boundary MATH. Then the subcomplexes MATH are just the cones over the barycentric subdivisions of the MATH-simplices of MATH. Now we can extend the map MATH linearly on each simplex of the triangulation MATH to the embedding MATH (which is therefore a piecewise linear map). REF illustrates this embedding for MATH, MATH. In short, the above constructed embedding MATH is determined by the following property: MATH . Thus, all cubes of MATH map to faces of MATH, which proves the assertion. |
math/9909166 | This follows from the fact that MATH-cubes of MATH are in one-to-one correspondence with pairs MATH of faces of MATH such that MATH (see the proof of REF ). |
math/9909166 | REF shows that MATH is presented as the union of MATH-cubes MATH indexed by the vertices of MATH. Let MATH be the orbit map. It follows easily from the definition of MATH that for each cube MATH we have MATH, where MATH is the polydisc in MATH with the diagonal action of MATH. Hence, MATH is presented as the union of ``blocks" of the form MATH. Gluing these ``blocks" together smoothly along their boundaries we obtain a smooth structure on MATH. Since each MATH is MATH-invariant, the MATH-action on MATH is also smooth. Now, let us prove the second part of the theorem. Recall our numeration of codimension-one faces of MATH: MATH. Take the block MATH corresponding to a vertex MATH. Each factor MATH and MATH above corresponds to a codimension-one face of MATH and therefore acquires a number (index) MATH, MATH. Note that MATH factors MATH acquire the indices of those facets containing MATH, while other indices are assigned to MATH factors MATH. Now we numerate the factors MATH of the polydisc in any way and embed each block MATH into MATH according to the indices of its factors. It can be easily seen the set of embeddings MATH define an equivariant embedding MATH. |
math/9909166 | It can be easily seen that an embedding of a face MATH defined by MATH equations of the type MATH (as in REF ) induces an equivariant embedding of MATH into MATH. Then our assertion follows from the representation of MATH as the union of blocks MATH and from REF . |
math/9909166 | If MATH, then one can find a zero non-trivial linear combination of vectors MATH. But this is impossible: since MATH is simple, the set of normal vectors of facets meeting at the same vertex constitute a basis of MATH. |
math/9909166 | A point from MATH may have the non-trivial isotropy subgroup with respect to the action of MATH on MATH only if at least one its coordinate vanish. As it follows from REF , if a point MATH has some zero coordinates, then all of them correspond to facets of MATH having at least one common vertex MATH. Let MATH. The point MATH has non-trivial isotropy subgroup with respect to the action of MATH only if some linear combination of vectors MATH lies in the coordinate subspace spanned by MATH. But this means that MATH, which contradicts REF . Thus, MATH acts on MATH freely. Now, let us prove the second part of the theorem. Here we use both embeddings MATH from REF and MATH from REF . It is sufficient to prove that each orbit of the action of MATH on MATH intersects the image MATH in a single point. Since the embedding MATH is equivariant, instead of this we may prove that each orbit of the action of MATH on the real part MATH intersects the image MATH in a single point. Let MATH. Then MATH lies in some MATH-face MATH of the unit cube MATH as described in REF . We need to show that the MATH-dimensional subspace spanned by the vectors MATH is in general position with the MATH-face MATH. But this follows directly from REF . |
math/9909166 | This follows directly from the definition of MATH: MATH . |
math/9909166 | A simple polyhedral complex MATH is defined as the cone over the barycentric subdivision of a simplicial complex MATH with MATH vertices. We construct a cell embedding MATH by induction on the dimension of MATH. If MATH, then MATH is a disjoint union of vertices MATH and MATH is the cone on MATH. In this case MATH is a bouquet of MATH copies of MATH and we have the obvious inclusion MATH. In degree zero MATH is just MATH, while in degrees MATH it is isomorphic to MATH. Therefore, MATH, where MATH is the ideal generated by all square free monomials of degree MATH, and MATH is the projection onto the quotient ring. Thus, the theorem holds for MATH. Now let MATH. By the inductive hypothesis, the theorem holds for the simple polyhedral complex MATH corresponding to the MATH-skeleton MATH of MATH, that is, MATH. We add MATH-simplices one at a time. Adding the simplex MATH on vertices MATH results in adding all cells of the subcomplex MATH to MATH. Then MATH, where MATH is generated by MATH and MATH. It is also clear that a map of the cochain (or cohomology) algebras induced by MATH is the projection onto the quotient ring. |
math/9909166 | The identities MATH and MATH follow from the cell decomposition of MATH described in the previous theorem. In order to calculate MATH and MATH we consider the following fragment of the exact homotopy sequence for the bundle MATH with the fibre MATH: MATH . It follows from REF that MATH above is an isomorphism, and hence, MATH. The third assertion of the theorem follows from the fragment MATH in which MATH for MATH. Finally, the cell structure of MATH shows that if MATH is MATH-neighbourly, then the MATH-skeleton of MATH coincides with the MATH-skeleton of MATH. Thus, MATH for MATH. Now, the last assertion of the theorem follows from the third one and from REF . |
math/9909166 | Consider the NAME - NAME spectral sequence of the fibration MATH . REF gives the monomorphism MATH such that MATH. The MATH term of the NAME - NAME spectral sequence is MATH . The right-hand side above is a bigraded MATH-module (see CITE, CITE). The first (``external") grading arises from a projective resolution of MATH as a MATH-module used in the definition of the functor MATH. The second (``internal") grading arises from the gradings of MATH-modules which enter the resolution; we assume that non-zero elements appear only in even internal degrees (remember that MATH). Since MATH is a free MATH-module, we have MATH . Therefore, MATH and MATH for MATH. Thus, MATH and MATH. |
math/9909166 | It follows from CITE that for the NAME - NAME spectral sequence of an arbitrary commutative square REF one has MATH. In our case this gives MATH. Now, the proposition follows from the fact that the map MATH is an epimorphism (see REF ). |
math/9909166 | First, we show that MATH . To prove this inequality we construct an injective map from the union of generator sets of the free MATH-modules MATH (see REF ) to a generator set of MATH (over MATH). Consider the NAME - NAME spectral sequence of the bundle MATH with the fibre MATH. The first column of the MATH term of this spectral sequence is the cohomology of the fibre: MATH. By REF , non-zero elements can appear in the MATH term only in the bottom line; this bottom line is the ring MATH: MATH . Therefore, all elements from the kernel of the map MATH (see REF ) must be killed by the differentials of the spectral sequence. This kernel is just our ideal MATH. Let MATH be a minimal generator set of MATH. Then we claim that the elements MATH can be killed only by the transgression (that is, by differentials from the first column). Indeed, suppose that the converse is true, so that MATH is killed by a non-transgressive differential: MATH for some MATH, where MATH is not from the first column. Then MATH is sent to zero by all differentials up to MATH. This MATH arises from some element MATH in the MATH term, MATH, MATH. Suppose that all the elements MATH are transgressive (that is, MATH for MATH), and let MATH, MATH. Since all MATH are killed by differentials, their inverse images in MATH belong to MATH. Hence, we have MATH, MATH, which contradicts the minimality of the basis MATH. Therefore, there are some non-transgressive elements among MATH, that is, there exists MATH and MATH such that MATH (see REF ). Then MATH survives in MATH and we have MATH - contradiction. This means that all the minimal generators of MATH are killed by the transgression, that is, they correspond to some (different) generators MATH. Since MATH, a free MATH-module generated by the elements MATH is included into the MATH term as a submodule. Therefore, we have MATH and the map MATH is defined by the differentials of the spectral sequence. The kernel of this map, MATH, can not be killed by the already constructed differentials. Using the previous argument, we deduce that the elements of a minimal generator set for MATH can be killed only by some elements from the first column, say MATH. Therefore, a free MATH-module generated by the elements MATH is also included into the MATH term as a submodule, that is, MATH. Proceeding with this procedure, at the end we obtain MATH generators in the first column of the MATH term. Using REF , we deduce the required REF . Now let us consider the NAME - NAME spectral sequence of the bundle MATH with the fibre MATH. This spectral sequence has MATH, and MATH. It follows from REF that MATH, which concludes the proof of the theorem. |
math/9909166 | It follows from REF that MATH . By REF , MATH. Take a basis of MATH consisting of elements MATH of internal degree MATH such that MATH are primitive collections of the vertices of MATH (see the proof of REF ). If MATH is a primitive collection, then the subcomplex of MATH consisting of all simplices with vertices among MATH is a simplicial complex consisting of all faces of a simplex except one of the highest dimension (that is, the boundary of a simplex). In terms of the simple polytope MATH the element MATH corresponds to the set MATH of codimension-one faces such that MATH, though any proper subset of MATH has a non-empty intersection. Note that the element MATH defines, by means of the isomorphism from REF , an element of MATH of dimension MATH. Now, we take one point inside each face MATH, MATH, (MATH is dropped); then we can embed the simplex MATH on these points into the polytope MATH in such a way that the boundary MATH embeds into MATH. (Compare this with the construction of the cubical decomposition of MATH in REF .) Let MATH be the projection onto the orbit space; then it can be easily seen that MATH. In this way we obtain an embedding MATH that realize the element of MATH dual to MATH. |
math/9909166 | Let us consider the introduced above MATH-free NAME resolution MATH of MATH. Then MATH . |
math/9909166 | First, consider the MATH term of the spectral sequence. Since MATH, MATH, we have MATH . It can be easily seen that the differential MATH acts as follows: MATH (see REF ). Now, since MATH, our assertion follows from REF by putting MATH, MATH. |
math/9909166 | Let us consider the bundle MATH with the fibre MATH. It follows from REF that the correspondent cochain algebras are MATH and MATH, and the action of MATH on MATH is defined by the quotient projection. It was shown in CITE that there is an isomorphism of algebras MATH . Now, it follows from above arguments and REF that MATH which concludes the proof. |
math/9909166 | Let MATH be the NAME - NAME spectral sequence of the first commutative square, and let MATH be that of the second one. Then, as it follows from the results of CITE, CITE, the inclusions MATH, MATH, MATH, and MATH define a homomorphism of spectral sequences: MATH. First, we prove that MATH is an isomorphism. The map MATH is the quotient projection MATH with the kernel MATH. By REF , we have MATH. Hence, MATH is the quotient epimorphism. The MATH-terms of our spectral sequences are MATH and MATH. To proceed further we need the following result. Let MATH be an algebra and MATH a subalgebra and set MATH. Suppose that MATH is a free MATH-module and we are given a right MATH-module MATH and a left MATH-module MATH. Then there exists a spectral sequence MATH with MATH . See CITE. The next proposition is a modification of one assertion from CITE. Suppose MATH is an epimorphism of graded algebras, MATH, and MATH is an ideal generated by a length MATH regular sequence of degree-two elements of MATH. Then the following isomorphism holds: MATH . Let MATH, MATH, and MATH is a regular sequence. Let MATH, MATH, be degree-two elements of MATH such that MATH. Hence, MATH and MATH. Let us take elements MATH of degree two such that MATH and put MATH. Then MATH is a free MATH-module, and therefore, by REF , we have a spectral sequence MATH where MATH. Since MATH is a regular sequence, MATH is a free MATH-module. Therefore, MATH which concludes the proof of the proposition. Now, we return to the proof of REF . Setting MATH in REF we deduce that MATH is an isomorphism. The MATH terms of both spectral sequences contain only finite number of non-zero modules. In this situation a homomorphism MATH that defines an isomorphism in the MATH terms is an isomorphism of the spectral sequences (see CITE). Thus, REF is proved. |
math/9909166 | By REF , MATH. Hence, our assertion follows from the isomorphism between the MATH terms of the spectral sequences from REF . |
math/9909166 | The cohomology algebra MATH is exactly the MATH term of the NAME - NAME spectral sequence for the bundle MATH. At the same time, it follows from REF that this cohomology algebra is isomorphic to MATH. REF shows that this is exactly MATH. Since the NAME - NAME spectral sequence converges to MATH, it follows that it collapses in the MATH term. |
math/9909166 | The inclusion of the subgroup MATH defines a map of classifying spaces MATH. Let us consider the bundle pulled back by this map from the bundle MATH with the fibre MATH. It follows directly from the construction of MATH (see subsection REF) that the total space of this bundle has homotopy type MATH (more precisely, it is homeomorphic to MATH). Hence, we have the commutative square MATH . The corresponding NAME - NAME spectral sequence converges to the cohomology of MATH and has the following MATH term: MATH where the action of MATH on MATH is defined by the map MATH, that is, MATH. Using CITE in the similar way as in the proof of REF , we show that the spectral sequence collapses in the MATH term and the following isomorphism of algebras holds: MATH . Now put MATH, MATH, MATH, and MATH in REF . Since MATH here is a free MATH-module and MATH, a spectral sequence MATH arises. Its MATH term is MATH and it converges to MATH. Since MATH is a free module over MATH with one generator REF, we have MATH . Thus, the spectral sequence collapses in the MATH term, and we have the isomorphism of algebras: MATH which together with the isomorphism REF proves the theorem. |
math/9909166 | It follows from REF that the orbits of the action of MATH on MATH corresponding to the vertices MATH of MATH have maximal (rank MATH) isotropy subgroups. These isotropy subgroups are the coordinate subgroups MATH. A subgroup MATH acts freely on MATH if and only if it has only unit in the intersection with each isotropy subgroup. This means that the MATH-matrix obtained by adding MATH columns MATH (REF stands on the place MATH, MATH) to MATH defines a direct summand MATH. (This matrix corresponds to the subgroup MATH.) Obviously, this is equivalent to the requirements of the lemma. |
math/9909166 | The map MATH takes the cohomology ring MATH of MATH to the subring MATH of the cohomology ring of MATH (see REF ). This subring is isomorphic to the quotient ring MATH. Now, the assertion follows from the fact that a polytope MATH is MATH-neighbourly if and only if the ideal MATH (see REF ) does not contain monomials of degree less than MATH. |
math/9909166 | In the similar way as in REF we show that the MATH term of the spectral sequence is MATH . REF shows that this is exactly MATH. |
math/9909166 | The first assertion follows from the fact that the cohomology class under consideration is a generator of the module MATH (see REF ). The second assertion holds since two cohomology classes are NAME dual if and only if their product is the fundamental cohomology class. |
math/9909166 | It is sufficient to prove that any cocycle MATH from MATH that does not lie in MATH is a coboundary. To do this we note that if there is MATH, then MATH contains the summand MATH, hence, MATH - a contradiction. Therefore, MATH. If MATH contains at least one MATH with degree MATH, then since MATH, we have MATH. Now, our assertion follows from the fact that all non-zero elements of MATH of the type MATH correspond to simplices of MATH. |
math/9909166 | Using REF , we calculate MATH . |
math/9909166 | This can be seen either directly from the construction of the minimal resolution REF , or from REF . |
math/9909166 | It follows from REF that MATH . By REF , MATH . Now the theorem follows from the fact that MATH if MATH is a REF-simplex, and MATH if MATH is a pair of disjoint vertices. |
math/9909168 | The monomials of MATH outside MATH form a MATH-basis for MATH. If there were infinitely many initial ideals then REF would give a proper inclusion of MATH-bases. |
math/9909168 | Consider the map MATH given by MATH. For a monomial ideal MATH, we define MATH. Then MATH, so the result follows from applying REF to the set MATH. |
math/9909168 | MATH is isomorphic to MATH for some binomial ideal MATH. Any element of MATH which is homogeneous with respect to the MATH grading can be written as MATH where MATH is some monomial in MATH, so homogeneous ideals of MATH lift to monomial ideals in MATH. Containment in MATH implies containment in MATH, so the result follows. |
math/9909168 | For MATH, let MATH and MATH is a vertex of MATH. Then the fiber over MATH is atomic if and only if MATH is not contained in any MATH for MATH. If there were an infinite number of atomic fibers, then MATH would be an infinite antichain of monomial ideals, contradicting REF . |
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