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math/0103134 | For MATH, let MATH be the rotation of angle MATH in the plane of the last MATH coordinates. Let MATH, the diagonal matrix with entries MATH. Let MATH be a MATH-equivariant bundle. Choose MATH, with MATH and let MATH. For MATH, one has MATH and MATH whence MATH, which proves REF . Let MATH be a smooth homomorphism and set MATH. Suppose that MATH is a MATH-equivariant MATH-bundle with MATH such that MATH and MATH for MATH. Then MATH. let MATH be the curve MATH, where MATH. Using that MATH for MATH, on checks, as in REF , that MATH is a MATH-isotropic lifting. By REF , one has MATH in MATH. By REF , this proves that MATH is determined by MATH, which proves the uniqueness statment of REF . It remains to construct, for a smooth homomorphism MATH, a MATH-equivariant MATH-bundle with MATH. Consider the map MATH sending a matrix to its last column. This makes a MATH-equivariant MATH-bundle (the principal MATH-bundle associated to the tangent bundle of MATH). Let MATH be the MATH-bundle obtained by the NAME construction, using the homomorphism MATH . This is a MATH-equivariant MATH-bundle. Choosing MATH and MATH, one sees that MATH and MATH . Therefore, MATH is a MATH-equivariant MATH-bundle with MATH. |
math/0103134 | Suppose that we are given an admissible MATH-isotropy group system over MATH. Let MATH be a special MATH-bundle over MATH realizing the given MATH-isotropy group system. By restricting the bundle to MATH, we get a map MATH classifying the principal MATH-bundle MATH (which completely determines MATH) by REF . We can apply the Tube REF to the MATH action on MATH, since MATH means that MATH and similarly for MATH. This identifies the restriction of our bundle to the part over MATH as MATH, where MATH is the MATH-equivariant projection, and MATH is a principal right MATH-bundle over MATH. The classifying map MATH for MATH therefore extends to a map MATH where the notation means that each boundary component MATH is mapped into the MATH-component of MATH. The restriction of MATH to MATH classifies MATH. This shows that MATH is determined up to equivalence by MATH. Conversely, if we are given a map MATH as above we can reconstruct a special MATH-equivariant bundle over MATH realizing the isotropy group system, up to fine equivalence. It can be checked that this bundle is unique up to equivalence. |
math/0103134 | This is proved using CITE, as for the finiteness of MATH. |
math/0103134 | The exact sequences MATH and MATH of groups induce a map MATH. By REF , we also have a surjective map MATH induced by MATH and our construction of equivariant bundles. |
math/0103134 | The map MATH is given in the diagram above, and we have already observed that it is surjective. Suppose that MATH, MATH with MATH. Then we have a continuous map MATH realizing the homotopy between the classifying maps for MATH and MATH. By the surjectivity of MATH for MATH-bundles over MATH, we get a bundle MATH over MATH which restricts to MATH and MATH at the ends MATH. Since MATH, we get MATH. |
math/0103134 | The map from one set of bundles to the other is defined by regarding a MATH-bundle over MATH as an element of MATH, and this is well-defined since the equivalent relation in MATH is stronger. Moreover, two bundles with isotropy group system MATH over MATH are equivalent if and only if they are in the same orbit of the action of MATH, hence our correspondence is injective. On the other hand, if MATH is a bundle with base space MATH in MATH, then there exists a MATH-equivariant diffeomorphism MATH covering the identity on MATH. Then MATH is an equivalent element in MATH, and is a bundle over MATH, so our correspondence is surjective. |
math/0103134 | This follows directly from REF , since MATH . |
math/0103134 | Since MATH comes from an admissible system, the image of MATH is contained in MATH. On the other hand, let MATH be two smooth homomorphisms such that MATH. The special MATH-manifold constructed as in REF , with isotropy group system MATH (associated to a MATH-meridian) is a MATH-bundle MATH over MATH (the special MATH-manifold with MATH-meridian isotropy group system MATH). One has MATH, which proves REF . If MATH then its isotropy group system is fine equivalent to MATH. In the notation introduced earlier, we have MATH and the result now follows from REF . |
math/0103134 | The pointed maps MATH send MATH into the component of MATH containing the base point MATH. Let MATH. Using our MATH-meridian MATH, one defines a smooth path MATH by the formula MATH . The map MATH being MATH-equivariant, one has, for all MATH . This implies, for all MATH, that MATH. For MATH, one gets in addition that MATH. We check that this defines an anti-homomorphism from MATH to the group MATH. Now, if MATH represents a class in the latter, the formula MATH defines an element MATH of MATH and constitutes an inverse to the above anti-homomorphism. |
math/0103134 | By REF , there is a surjective map MATH. Since MATH and MATH are surjective, we have the fibration REF . From the homotopy exact sequence of this fibration, we see that MATH is in bijection with the quotient of MATH by the action of MATH. But this is just the action of MATH by REF . |
math/0103134 | As MATH, one has, for all MATH, MATH . Therefore, MATH. Recall that our MATH-meridian for MATH is MATH. By REF and its proof there exists a MATH-bundle MATH over MATH, with a MATH-isotropic lifting MATH of MATH such that the isotropy representations associated to MATH are MATH. The curve MATH is the horizontal lifting of MATH for a MATH-invariant connection. There is no problem to extend the definitions of MATH and MATH for MATH. This defines MATH by MATH . For MATH, one has MATH (since this is true for MATH and both side are horizontal). For MATH, this gives MATH . Therefore, MATH. Now, let MATH; one has MATH and the MATH-action on MATH comes from the diagonal action. A MATH-isotropic lifting MATH of MATH is then given by MATH . By REF , one has MATH in MATH iff MATH in MATH. By REF, using the curves MATH and MATH, one has in MATH . This proves that MATH represents the unit element in MATH. |
math/0103139 | MATH, where MATH is the space of symmetric, non-degenerate MATH matrices over MATH. MATH is an open subset of the vector space of all MATH symmetric matrices over MATH. By the fundamental exact sequence for NAME groups (REF below), the NAME groups of any NAME open subset of affine space vanish in codimensions higher than zero and are MATH in codimension zero. |
math/0103139 | As REF states, the NAME ring of MATH is trivial. Therefore, if MATH is a MATH-orbit in MATH whose codimension is larger than zero, then MATH in MATH. By REF , any orbit in MATH which is not fixed point free lifts to a single orbit in MATH and is therefore rationally equivalent to zero. On the other hand, if an orbit MATH is fixed point free, then MATH where MATH and MATH are two disjoint copies of MATH distinguished by the sign of MATH. Again, since the NAME ring of the base space is trivial, MATH, so MATH. Therefore, MATH is generated by elements corresponding to closures of fixed point free orbits in MATH, along with the codimension zero cycle. |
math/0103139 | Fix a quadratic form MATH in MATH. An element MATH is determined by the numbers MATH . These numbers may be chosen freely subject only to the constraint that MATH . |
math/0103139 | Since the map MATH is a fibration with fibers isomorphic to an open subset of affine space, it induces a surjection of NAME rings REF . By our induction hypothesis MATH. More specifically, the NAME ring of MATH is generated by cycles MATH and MATH, where MATH is the codimension zero cycle and MATH is in codimension MATH. Therefore, MATH is generated by the pullbacks of these cycles. |
math/0103139 | We will show by induction on MATH that MATH is a quotient of MATH . For the case MATH, since MATH, REF tells us that MATH is a quotient of MATH. The pullback of the codimension zero orbit MATH is indexed by the permutation MATH so since MATH it is not a fixed point free orbit in MATH. Therefore, by REF , MATH is rationally equivalent to zero in MATH, and the subvariety MATH contributes at most MATH to the NAME ring of the whole symmetric space. This completes the MATH case. As stated in REF , we know that MATH is exact. Therefore, by induction, MATH is a quotient of MATH. Again, we note that MATH is indexed by the permutation MATH so is not fixed point free. For MATH, this orbit has codimension larger than zero, so by REF , this orbit is rationally equivalent to zero in MATH. Finally, if MATH, as we noticed previously (immediately following REF ), the subvariety MATH contributes only the codimension zero cycle MATH. |
math/0103140 | The transpose of MATH with respect to the MATH inner product on the NAME algebra of divergence free vector fields exists: MATH. and the NAME cocycle is of the required form MATH with MATH. Then REF can be applied with initial condition MATH. |
math/0103140 | All the conditions required in REF are fulfilled by this extension. The mapping MATH is MATH and the mapping MATH is MATH. We have again denoted by MATH the orthogonal projection on the space of divergence free vector fields on MATH. The transpose MATH and MATH is skew-adjoint. Now REF can be applied. |
math/0103140 | We observe that MATH, so in the curvature REF applied to MATH, only REF terms are non-zero: MATH . Using the NAME coefficients, we get the following expressions of the cocycle MATH and of the mapping MATH: MATH and MATH . From the hypothesis, MATH is an orthogonal basis of MATH. In the associated orthonormal basis the coordinates of MATH are denoted by MATH. Let MATH (respectively MATH) be the curvature tensor of the central (respectively general) extension of the group of volume preserving diffeomorphisms. Then MATH . So the sectional curvature on the central extension is positive if and only if MATH. An analogous computation for MATH gives the condition MATH. |
math/0103141 | Each MATH is a derivation of the NAME algebra MATH and this implies MATH . Using this and REF , the result follows. |
math/0103142 | It is enough to note that the orbits of MATH in MATH can be identified to the orbits of MATH on MATH. |
math/0103142 | The regular case has been treated in CITE. The fact that MATH projects on a Killing vector field follows from a local computation, so it holds also in the quasi-regular case. As any isometry of an orbifold has to fix (or interchange) the conical points, it follows that MATH vanishes at the conical points. Now, if a Killing vector field on a (even open) surface has a zero, then it is isolated and has index REF; we can smoothen the conical points and deform a little MATH such as the resulting vector field on MATH has the same number of zeros, each with the same index as before. They must be thus exactly REF, as the NAME characteristic of MATH; we will call these degenerate orbits of MATH poles. The other orbits of MATH are smooth submanifolds, because they coincide with the level sets of MATH (recall that MATH), and that MATH is non-degenerate away from REF poles. |
math/0103142 | We know that there exists such a MATH CITE. Suppose MATH is another regular MATH . NAME vector field such that MATH. Consider the closure of an orbit of MATH: this is a torus MATH. From the standpoint of MATH, MATH is a commuting MATH . NAME vector field: as such, MATH, MATH is a Killing vector field on MATH, the space of the orbits of MATH, and we know from REF that the orbits of MATH - as an irregular MATH . NAME vector field commuting with MATH - are dense in tori to which MATH is tangent. So MATH, as well as MATH, is tangent to the closures of MATH's orbits. On a dense set of such a torus MATH, MATH is determined by its value at a point MATH (as MATH commutes with MATH, we transport, through the flow of MATH, this vector to all the points situated on the orbit of MATH passing through MATH, and this set is dense). The same holds for MATH, so if MATH coincides with MATH at a point of MATH, it coincides everywhere. In any case we get that MATH on MATH, and, as MATH was arbitrarily chosen, MATH and MATH commute on MATH. As they are both regular, they coincide (REF , see also the remark below). |
math/0103142 | In the first case, MATH and MATH coincide in a neighborhood of one of their common generic orbits. If they are irregular, we consider the uniquely (see REF ) associated regular MATH . NAME vector fields MATH, respectively, MATH, and we easily get, as in REF , that they commute in a neighborhood of a torus MATH, which is the closure of a common orbit of MATH and MATH. Then, for a suitable constant MATH, MATH is a CR NAME vector field, and it coincides with MATH on a neighborhood of an orbit of MATH. If we prove that they coincide everywhere, it follows that MATH and MATH are two regular CR NAME vector fields that commute, so (REF and the Remark thereafter) they coincide. We can suppose thus that MATH and MATH are both (quasi-) regular, and they coincide on a neighborhood of some orbit, which we may suppose generic. Then MATH (the function defining MATH starting from MATH) is constant in such a neighborhood, and MATH is constant on an open set of MATH, the orbit space of MATH. But MATH, so MATH is a Killing vector field, vanishing in an open set, thus everywhere. So MATH is constant. At this point we can apply REF , or the method below, to conclude that MATH is constant. The method we apply to prove the second claim in the Lemma is the same we used - in the regular case - in REF ; we recall it briefly: Using the NAME inequality MATH we get the implication MATH with equality if and only if MATH is constant. Here, MATH is the contact form such that MATH, MATH is the volume of MATH with respect to the NAME metric defined by MATH, and MATH is the length of the orbits of MATH. The hypothesis of the Lemma ensures that the parenthesis of the above implication, as of the following one, are true, so we get MATH and MATH . : MATH . If MATH is quasi-regular, we denote by MATH the length of the generic orbits of MATH ; besides, all the integrals above will be taken over the complement of the exceptional orbits in MATH. Note that the exceptional orbits, as well as the wrapping numbers of MATH and MATH coincide (otherwise a linear combination of them would be irregular - see the Remark at the end of REF - , and this implies that MATH is even-dimensional (see REF below), but if the commutator of MATH is REF-dimensional, then MATH is odd-dimensional, contradiction). The proof follows as in the regular case (we get equality in the NAME inequality, thus MATH is a constant). |
math/0103142 | The exponential of MATH is the adjoint action of the flow of MATH on MATH; as this flow is periodic, it follows that the eigenvalues of MATH are imaginary. The image of MATH lies in MATH, because, for an infinitesimal MATH automorphism MATH, MATH, where MATH, and, of course, the integral of MATH along the orbits of MATH is zero. It remains to prove that if MATH, then MATH. MATH implies that MATH is constant on the orbits of MATH, and MATH implies that this constant is REF. |
math/0103142 | It follows from the previous Lemma that MATH is always even. The second claim readily follows. For the first claim, we take MATH to be the regular MATH . NAME vector field which commutes with an irregular one, and then MATH is REF or REF (see above). |
math/0103142 | If the MATH . NAME vector field is quasi-regular, this follows from the theorems above; if it is irregular, then any second type deformation of it necessarily commutes with the first type deformation that replaces the irregular MATH . NAME vector field with the unique (commuting) regular one, and we can apply the conclusion above. |
math/0103142 | We know, from CITE, that, if MATH admits a normal MATH structure, then it is a finite quotient of a circle bundle MATH over a NAME surface MATH with non-zero NAME class, and, in most of the cases (see the exceptions below), this MATH structure is MATH-invariant, where the MATH-action is free and transverse to the underlying contact structure. The latter is thus given by a connection on the MATH-bundle MATH (where MATH is the orbit space of MATH), such that its curvature is a given volume form on MATH (from here follows that the NAME class should be non-zero). On MATH, we have thus a MATH-invariant NAME structure, where MATH is the covering group of the finite covering MATH. On MATH, all (MATH-invariant) contact structures which are invariant and transverse to the free action of MATH are homotopic within (MATH-invariant) contact structures CITE, and this (MATH-equivariant) homotopy is followed by a (MATH-equivariant) isotopy CITE, CITE. On the other hand, all free MATH actions on MATH (and, implicitly, the diffeomorphism type of the quotient space MATH) are isomorphic, because the only invariants are MATH and the NAME class of the fibration MATH, and these are determined by the cohomology of MATH. The exceptional cases are the normal MATH structures on MATH (or finite quotients) which are MATH-invariant, but the MATH action is not free; it is given by a (unique up to a constant factor) quasi-regular MATH . NAME vector field. We can, nevertheless, deform the metric on the orbifold MATH such that it admits a Killing vector field; the corresponding (second type) deformation of the MATH structures is such that the underlying contact structures are all isotopic CITE; but the new MATH structure admits at least REF linearly independent MATH . NAME vector fields, therefore it is invariant to a free MATH action, which is the case treated above. On the other hand, the MATH invariant and transversal contact structures on the total space of a circle fiber bundle as above are tight CITE. |
math/0103146 | Set MATH and MATH. With this an explicit coloring is given by MATH . This rule assigns to each MATH-set MATH an integer between MATH and MATH. If MATH is a MATH-disjoint MATH-family, then every minimal element MATH can appear at most MATH times in the family; thus if MATH, then this value is assigned to at most MATH sets in a MATH-disjoint MATH-family, where MATH. On the other hand, if all the MATH-sets MATH get the color MATH, then they are contained in the set MATH, of cardinality MATH. But the pigeonhole principle demands that a MATH-disjoint MATH-family of MATH-sets uses at least MATH elements. |
math/0103146 | No set MATH of MATH contiguous elements mod MATH contains a MATH-stable MATH-set. (Note MATH.) Furthermore, there is a MATH-disjoint packing of (at most MATH elements from) MATH such contiguous subsets into MATH: Such a packing can be written down as MATH . This proves that MATH. To prove that MATH, it suffices to verify that every set MATH of cardinality MATH contains a MATH-stable MATH-subset. Take MATH, and let MATH be an arbitrary set of size MATH; this exists since MATH. Now MATH has cardinality MATH, and we can partition it into MATH disjoint MATH-stable MATH-subsets, by taking ``every MATH-th element" to go into the same MATH-subset. At least one of these MATH-subsets contains no element from MATH, since MATH. |
math/0103146 | Let MATH be a proper MATH-coloring, and assume that MATH, that is, if any subset of MATH of size at least MATH is colored by two colors, then it contains a monochromatic subset from MATH. Fix an arbitrary linear ordering MATH on the subsets of MATH. Then define a map MATH, as follows: CASE: If MATH, then define MATH as MATH, where MATH is the smallest set (according to MATH) from MATH that is contained either in MATH, or in MATH. Take the sign to indicate which of MATH or MATH you took MATH from. Thus we obtain a value MATH in the set MATH. CASE: If MATH, then define MATH as MATH, where the sign indicates which of MATH or MATH is nonempty, and if they both are, then it indicates which is smaller (according to MATH). Thus we obtain a value MATH in the set MATH. This map MATH is antipodal. Thus by the Octahedral NAME REF , there are signed sets MATH, MATH with MATH, where MATH, MATH, not equality in both cases, and so MATH. This is possible only if both signed sets are labeled according to the first case. But then (assume without loss of generality that above we have MATH) there are sets MATH with MATH and MATH, MATH, where MATH and MATH are disjoint: so also MATH and MATH are disjoint, but they get the same color from MATH, contradiction. |
math/0103146 | For this, we watch what happens if we increase the ground set, by extending MATH to MATH, with MATH, where MATH is not changed. Since MATH is not changed, the NAME hypergraph MATH and its chromatic number don't change, either. On the other hand, with this operation MATH increases by MATH, and MATH also increases by MATH, since we may extend exactly one of the MATH-s by an extra element MATH. Thus, in summary, extending the ground set with MATH changes neither the chromatic number of the NAME hypergraph, nor the colorability defect, so validity of the theorem is unchanged. By applying this operation, which increases MATH by MATH, at most MATH times, we get the required divisibility. |
math/0103146 | Let MATH, and let MATH be a coloring such that no MATH-disjoint sets from MATH get the same color. At the same time we assume that MATH, that is, if MATH elements of the multiset MATH are colored by MATH colors (which we take from MATH), then some set from MATH is monochromatic. We define a labeling MATH using an arbitrary linear ordering MATH on the subsets of MATH, as follows: CASE: If MATH, then define MATH as MATH, where MATH is the smallest set (according to MATH) from MATH that is contained in one of the MATH's; take the sign MATH to indicate which MATH you took MATH from. Thus we obtain a value MATH in the set MATH. CASE: If MATH, then define MATH, where the sign MATH indicates which of the nonempty sets MATH is the smallest one according to MATH. In this case we obtain a value MATH in the set MATH. This labeling is MATH-equivariant. By REF is an integer, that is, MATH. Thus we can apply the MATH-Tucker REF : there is a chain of MATH-matrices MATH such that MATH, for some permutation MATH. Since at most MATH of the matrices MATH can have the same MATH, and thus the same ``color" MATH according to the second case, the chain consists of MATH matrices that fall into the first case in the definition of MATH. Thus there are sets MATH that satisfy MATH, with the same MATH. The MATH sets MATH are MATH-disjoint, since they are contained in distinct parts of MATH, which is itself MATH-disjoint, but they all get the same color MATH: contradiction. |
math/0103146 | We proceed by induction on MATH, where we assume that the result is true when MATH is prime. Thus let MATH, let MATH with MATH, let MATH, and assume that MATH . We construct an auxiliary hypergraph MATH (on the same ground set as MATH) by MATH . Note that for this we use ``disjoint" colorability defect, corresponding to MATH. Using induction and the definition of MATH, we now get MATH and thus MATH . Claim: MATH. Proof of the Claim. Otherwise we could find a MATH-disjoint MATH-family MATH such that no MATH contains a set from MATH and such that MATH. In particular, none of the sets MATH lies in MATH, so by definition of MATH we have MATH for all MATH. Thus for each MATH we can find MATH disjoint sets MATH, such that no MATH contains a set from MATH, with MATH. Taking all the sets MATH together, we have MATH subsets of MATH, none of which contains a set from MATH, and they are MATH-disjoint: they form a MATH-disjoint union of disjoint families. We compute MATH which contradicts (MATH). Thus we have established the Claim. Using induction, together with the Claim, we get MATH and thus MATH . Now consider a coloring MATH of MATH by MATH colors. By (MATH), in every set MATH we find MATH disjoint sets from MATH which from MATH get the same color MATH. Using this, we construct a new coloring MATH which assigns to MATH one of the (possibly several) colors MATH which MATH assigns to MATH disjoint sets in MATH. By (MATH), there are MATH sets MATH, which are MATH-disjoint, and which from MATH get the same color MATH. Thus we have MATH sets MATH with MATH, also MATH-disjoint, that get from MATH the same color MATH. This contradicts the definition of MATH and MATH. |
math/0103146 | Let us assume that, for some MATH, a MATH-equivariant coloring MATH exists that does not produce a ``fully colored chain of MATH signed faces," as promised by the MATH-Tucker lemma. CASE: In the first half of the proof we will construct, under the assumption MATH, a square of MATH-equivariant chain maps MATH . Here the vertices of MATH correspond to MATH-matrices of size MATH with row sums at most MATH, as discussed above. The faces of the complex correspond to chains of such matrices, with respect to the entrywise MATH-partial order. The MATH-action is free for prime MATH, if MATH. The faces of MATH are MATH-matrices of size MATH with no full row of ones: we interpret them as admissible color sets. Again MATH acts cyclically on the columns; this is free for prime MATH. The map MATH of the MATH-Tucker lemma yields a simplicial map, and thus the chain map MATH used here. The map is equivariant, by assumption. The barycentric subdivision operators MATH, which yield the vertical arrows in the square above, have explicit combinatorial descriptions that we do not have to work out here. They are MATH-equivariant. Finally, MATH is a simplicial map that we construct orbitwise, as follows. The space to be mapped, MATH, is the barycentric subdivision of a simplicial complex of dimension MATH, so it is the order complex of a graded poset MATH with MATH rank levels, MATH. The free MATH-action on it respects the grading, so it decomposes the rank levels of the poset MATH into disjoint orbits of size MATH. The target space is a simplicial complex whose vertex set is identified with the positions in a MATH matrix; its faces are the MATH-matrices with at most MATH ones in the MATH-th row, for all MATH; the MATH-orbits of its vertices are exactly the rows of the matrix. An equivariant simplicial map can now be defined orbitwise, where the image of any element of an orbit determines the images for all others. We construct MATH such that the lowest MATH rank levels are mapped to the lowest row of the MATH matrix. The next MATH rank levels are mapped to the second lowest row of the matrix, etc. Thus the orbits in the MATH-th rank level of MATH are mapped to the row number MATH. (The following figure illustrates this for MATH, MATH, MATH.) MATH . This is well-defined if the target matrix has enough rows, that is, if MATH. The definition on the vertices indeed yields a simplicial map into the target space: any chain in MATH contains at most MATH elements in the (at most MATH) adjacent rank levels that are mapped to the MATH-th row of the MATH matrix. In summary, for MATH, we can combine the four MATH-equivariant chain maps of the square into a chain self-map MATH . Furthermore, the chain maps involved are induced either by simplicial maps, or by barycentric subdivision. Thus all four of them, and thus in particular MATH, are augmentation preserving in the sense that they preserve the sum of the coefficients of the vertices. CASE: In the second half of the proof, we compute the NAME number of MATH in two ways. First, the MATH-actions are free and the chain maps are MATH-equivariant, hence the NAME number of MATH satifies MATH . However, we will show that if MATH, then MATH restricts to an augmentation preserving chain map of the chain complex of a cone, and thus necessarily has MATH which yields a contradiction for MATH. If MATH, then we can extend MATH to a simplicial map MATH, where MATH denotes a new top element that is added to the poset MATH. Indeed, just map this new element into the ``top row" of the matrix; this gives a well-defined simplicial map (not MATH-equivariant, of course). The order complex MATH is a cone, and thus so is its image MATH: the image of a cone under a simplicial map is always a cone. We conclude that the image of MATH is contained in MATH, where the restriction of MATH to MATH has the same NAME number as MATH itself. The following lemma thus completes the proof. |
math/0103146 | Assume that we have a coloring MATH and set MATH, so that MATH. For MATH we define MATH the simplicial complex of all chains of sign vectors that have an alternating subsequence with more than MATH components. It has a free MATH-action, given by MATH. The coloring MATH of MATH now yields a simplicial, MATH-equivariant map MATH . If MATH is a correct coloring, then the color sets MATH and MATH are disjoint. Moreover, MATH has MATH, so both MATH and MATH contain at least one stable MATH-set, so MATH and MATH cannot be empty, thus MATH, hence MATH is well-defined for every vertex. Furthermore, any chain of sign vectors is mapped by MATH to a (weak) chain of signed color sets, so we obtain a simplicial map. This map is equivariant. Topologically, MATH is a simplicial (MATH)-sphere, namely the barycentric subdivision of the topological representation CITE of the alternating oriented matroid of rank MATH on MATH elements, see CITE CITE CITE. Similarly, MATH is a simplicial (MATH)-sphere. Both spheres have natural antipodal actions, and the map MATH respects these. Thus, the NAME theorem completes a topological proof at this point, but we keep going on the combinatorial track. Our next step is a quite trivial simplicial map, MATH which deletes the last component of each sign vector MATH. This map is well-defined: deleting the last component reduces MATH at most by MATH, the operation is compatible with the partial order, and it is MATH-equivariant. Now we use the canonical simplicial embedding maps of MATH into a cone MATH and then into a suspension MATH . Thus we have maps MATH where the composition MATH is MATH-equivariant (with the natural MATH-action on the suspension that interchanges MATH and MATH). Finally, we will construct an equivariant chain map MATH and sequence of equivariant chain maps MATH for MATH. Once these maps have been constructed, the proof will be complete, since then we have a square of equivariant, augmentation preserving chain maps MATH . Indeed, as in REF we can then argue that the composition MATH is MATH-equivariant, so it has even NAME number, but it also restricts to a cone (the image of MATH is contained in MATH), and thus its NAME number is MATH REF . The chain maps MATH and MATH (for MATH) can be written down combinatorially, by giving a formula for the image of an arbitrary MATH-simplex as a sum of MATH-simplices with MATH-coefficients. However, for the exposition we prefer to give a geometric description, from which the combinatorial one can then be derived. For MATH, note that MATH can be interpreted as the boundary complex of MATH, whose ``equator" subsphere naturally corresponds to MATH. MATH . This suggests a natural subdivision chain map MATH, as indicated in the figure, where every simplex in MATH is mapped ``to itself" on the equator, the vertices MATH are mapped to the north/south poles MATH, and any MATH-simplex MATH, say, is mapped to a signed sum of MATH-simplices, MATH of which triangulate the prism MATH, while the last one is MATH. The construction of the chain maps MATH is quite similar: MATH is the face poset of the dual cyclic oriented matroid of rank MATH on MATH elements, as described and analyzed in CITE. The following figure illustrates the sign vectors associated to various faces (vertices and edges) of MATH, for MATH and MATH. MATH . The vertex set of MATH is MATH. Thus MATH decomposes into a positive hemisphere MATH, the induced subcomplex given by all vertices with MATH, and the negative hemisphere MATH of all vertices with MATH. Every simplex is contained in one of these two ``hemispheres," whose intersection is the equator MATH, induced on all vertices MATH with MATH. The equator is naturally isomorphic to MATH. Similarly, the simplicial complex MATH decomposes into a positive hemisphere MATH, on the vertices MATH with last component MATH, and a negative hemisphere MATH, on the vertices MATH with last component MATH, whose intersection is the equator MATH, given by the vertices MATH with MATH, which is again isomorphic to MATH. For the construction of MATH, let MATH be an (oriented) simplex of MATH; since we are dealing with the order complex of a poset, all simplices have a natural ordering of their vertices, and thus they have natural orientations. We distinguish three cases. CASE: If MATH lies in the equator MATH, then MATH is the corresponding oriented simplex in MATH. CASE: If MATH has more than one vertex that does not lie in the equator, then it is mapped to the corresponding simplex in the positive or negative hemisphere. CASE: Assume that MATH has a facet in the equator, but does not itself lie in the equator; that is, all its vertices lie in the equator, except for MATH, which lies in the interior of the positive hemisphere, say. Then MATH is mapped to a sum of MATH oriented MATH-simplices in MATH, which again correspond to MATH simplices that triangulate a prism over MATH, plus one extra simplex in the interior of the positive hemisphere, and which in the following formal description comes first: MATH . Our figure tries to illustrate this for MATH and MATH. In the left figure, which represents MATH, the equator arises as a subcomplex of the barycentric subdivision. The right figure, depicting MATH, has the equator as a regular subsphere. MATH is a chain map from the left simplicial complex to the right one; it maps the shaded triangle MATH on the left to the sum of the three triangles shaded on the right: Now one verifies either geometrically (depending on the geometric realization of the spheres in question as barycentric subdivisions of arrangement spheres), or combinatorially, that these rules satisfy MATH for all MATH, that is, they provide equivariant simplicial chain maps MATH, as required. |
math/0103150 | Again, let MATH. NAME will be denoted MATH. Note that REF is a piece-wise linear function of MATH, where MATH is the cone defined by MATH. This is because it is defined as the minimum among a finite set of linear functions, namely the functions MATH for MATH. There is a decomposition of MATH into a finite number of subcones MATH and REF is linear on each cone MATH. Choose one vector MATH in each edge of each cone MATH. Multiply all these vectors by MATH, so that all their coordinates are divisible by MATH, and call this set of vectors MATH. All vectors in MATH come from a collection of weights MATH, MATH, given by the formula MATH. Hence to obtain the finite set MATH of vectors it is enough to consider a finite set of values for MATH, and hence there is a maximum value MATH. Finally, it is easy to see that it is enough to check REF for the weights associated to the vectors in MATH. Indeed, first note that since the first term in REF is linear on MATH, then it is also linear on each MATH. Then the left hand side of REF is linear on each MATH, and hence it is enough to check it on all the edges of all the cones MATH. |
math/0103150 | Let MATH be the set of all multi-indexes MATH. Given a multi-index MATH, we have MATH where MATH is the number of elements MATH of the multi-index MATH such that MATH. If MATH is the multi-index giving minimum in REF , we will denote MATH (or just MATH if the rest of the data is clear from the context). Then MATH . We index the filtration MATH with MATH. Let MATH be the multi-index giving minimum for the filtration MATH. In particular, we have MATH. Then MATH . |
math/0103150 | Follows from REF and an easy calculation. |
math/0103150 | Let MATH be as in REF . Choose MATH large enough so that MATH and the leading coefficient of MATH is negative, where MATH . Choose MATH large enough so that for MATH, MATH . Since the filtration is assumed to be saturated, and since MATH is torsion free, we have MATH. CASE: Suppose MATH. For each MATH, consider the one step filtration MATH. The leading coefficient of the semistability condition applied to this filtration, together with MATH, implies REF . Let MATH be the term in the NAME filtration with maximal slope. Then the same argument applied to MATH gives MATH . Now assume that the first alternative does not hold, that is, MATH . This gives MATH . Combining REF with REF , we have MATH . CASE: Suppose MATH for some MATH. For each MATH, consider the quotient MATH. Let MATH be the last factor of the NAME filtration of MATH (that is, MATH). Let MATH be the kernel MATH and consider the one step filtration MATH. REF imply that MATH. Then a short calculation using REF , the fact that MATH, REF shows MATH . It can be seen that if this inequality of polynomials holds for some MATH, then it holds for all larger values of MATH, hence choosing MATH large enough and looking at the coefficients, we have MATH . A short calculation using this, MATH and MATH (hence MATH), yields REF . Now assume that the first alternative does not hold, that is, MATH . It follows that MATH, and hence MATH . |
math/0103150 | Let MATH. Let MATH be a subsheaf of MATH, and MATH the saturated subsheaf of MATH generated by MATH. Using REF MATH . Then by NAME 's REF , the set MATH is bounded. |
math/0103150 | Let MATH. The sheaf MATH is torsion free, and MATH is bounded because the set MATH is bounded and MATH . Then by NAME 's REF , the set of sheaves MATH obtained in this way is bounded, and hence also MATH. |
math/0103150 | Since MATH is bounded, the set that consists of the polynomials MATH, MATH, MATH and MATH for MATH is finite. On the other hand, REF implies that we only need to consider a finite number of values for MATH, hence the result follows. |
math/0103150 | Let MATH and such that all sheaves in MATH and MATH are MATH-regular, and MATH is MATH-regular for all MATH in MATH. MATH. Let MATH. Consider a weighted filtration MATH. Then MATH. Let MATH and consider a saturated weighted filtration as in REF . Since MATH is MATH-regular, MATH. If MATH, then MATH. If MATH, then the second alternative of REF holds, and then MATH . Let MATH be the subset of those MATH for which MATH. Let MATH be the corresponding subfiltration. REF and a short calculation shows that MATH . The condition that MATH is saturated can be dropped, since MATH and MATH, where MATH is the saturated subsheaf generated by MATH in MATH. MATH. Let MATH and consider a saturated weighted filtration MATH. Since MATH is MATH-regular, MATH. If MATH, then MATH. Hence REF applied to the subfiltration MATH obtained by those terms such that MATH implies MATH . This is equivalent to MATH and by REF , this is in turn equivalent to MATH . If MATH, then the second alternative of REF holds, and then MATH . Using REF MATH . Again, we can drop the condition that the filtration is saturated, and this finishes the proof of REF . |
math/0103150 | By the proof of the part MATH of REF , if we have this equality then all inequalities in REF are equalities, hence MATH, MATH for all MATH, and the result follows. |
math/0103150 | The family is given by a tuple MATH as in REF , where MATH is the constant map from MATH to MATH with constant value MATH. Shrinking MATH, we can assume that MATH is trivial. Let MATH. Let MATH. Since it has no MATH-torsion, MATH is flat over MATH. The natural map MATH is an isomorphism on MATH, hence we have a homomorphism MATH on MATH, and this extends to a homomorphism MATH on MATH because MATH is locally free. Finally define MATH, and let MATH be the homomorphism induced by MATH. |
math/0103150 | Uniqueness is clear. To show existence, assume that MATH is very ample (taking a multiple if necessary) and let MATH be the projection to the second factor. Since MATH is MATH-flat, taking MATH large enough, MATH is locally free (recall MATH). The question is local on MATH, so we can assume, shrinking MATH if necessary, that MATH and MATH is given by a free MATH-module. Now, since MATH is affine, the homomorphism MATH of sheaves on MATH is equivalent to a homomorphism of MATH-modules MATH . The zero locus of MATH is defined by the ideal MATH image of MATH, thus the zero scheme of MATH is given by the ideal MATH, hence MATH is a closed subscheme. Since MATH is very ample, if MATH we have an injection MATH (and analogously for MATH), hence MATH, and since MATH is noetherian, there exists MATH such that, if MATH, we get a scheme MATH independent of MATH. To check the universal property first we will show that if MATH then MATH factors through MATH. Since the question is local on MATH, we can take MATH, MATH, and the morphism MATH is locally given by a ring homomorphism MATH. Since MATH is flat over MATH, for MATH large enough the natural homomorphism MATH (defined as in CITE) is an isomorphism. Indeed, for MATH sufficiently large, MATH and MATH for all points MATH, MATH and MATH, and since MATH is flat, this implies that MATH and MATH are locally free. Then, to prove that the homomorphism MATH is an isomorphism, it is enough to check it at the fiber of every MATH, but this follows from CITE or CITE. Hence the commutativity of the diagram MATH implies that MATH. This means that for all MATH, in the diagram MATH it is MATH. Hence the image MATH of MATH is in the kernel MATH of MATH. Therefore MATH, hence MATH factors through MATH, which means that MATH factors through MATH. Now we show that if we take MATH and MATH the inclusion, then MATH. By definition of MATH we have MATH for any MATH with MATH. Showing that MATH is equivalent to showing that MATH is zero for some MATH. Take MATH large enough so that MATH is surjective. By the right exactness of MATH the homomorphism MATH is still surjective. The commutative diagram MATH implies MATH, hence MATH. |
math/0103150 | Given MATH, the sheaves MATH for MATH form a bounded family, so if MATH is large enough, we will have MATH for all subspaces MATH. By the NAME criterion, a point is GIT-(semi)stable if and only if for all one-parameter subgroups MATH of MATH, MATH A one-parameter subgroup of MATH is equivalent to a basis MATH of MATH and a vector MATH with MATH. This defines a weighted filtration MATH of MATH as follows: let MATH be the different values of MATH, let MATH be the vector space generated by all MATH such that MATH, and let MATH. Denote MATH. We have (CITE or CITE) MATH and MATH . The last statement follows from an argument similar to the proof of REF , with MATH replaced by MATH. |
math/0103150 | First we prove that if MATH is NAME, then the induced linear map MATH is injective. Let MATH be its kernel and consider the filtration MATH. We have MATH and MATH. Applying REF we have MATH and hence MATH. Using REF , the inequality of REF becomes MATH . An argument similar to REF (using MATH instead of MATH) shows that we can take MATH large enough (depending only on MATH and MATH), so that this inequality holds for MATH if and only if it holds as an inequality of polynomials. Now assume that MATH is GIT-(semi)stable. Take a weighted filtration MATH of MATH. Then REF applied to the associated weighted filtration MATH of MATH give REF . On the other hand, assume that REF holds. Take a weighted filtration MATH of MATH. Then REF applied to the associated weighted filtration MATH of MATH give REF , and it follows that MATH is GIT-(semi)stable. |
math/0103150 | We prove this in two steps: CASE: MATH NAME MATH MATH-semistable and MATH induces an isomorphism MATH. The leading coefficient of REF gives MATH . Note that even if MATH is NAME, here we only get weak inequality. This implies MATH . To be able to apply REF , we still need to show that MATH is torsion free. By REF , there exists a tensor MATH with MATH torsion free such that MATH and an exact sequence MATH . Consider a weighted filtration MATH of MATH. Let MATH, and let MATH be the image of MATH in MATH. Let MATH be the kernel of MATH. Then MATH, MATH, and MATH. Using this and applying REF to MATH we get MATH and hence REF implies that MATH is MATH-semistable. Next we will show that MATH, and hence, since MATH, we will conclude that MATH is isomorphic to MATH. Define MATH to be the image of MATH in MATH. Then MATH where the last inequality follows from REF applied to the one step filtration MATH. Hence equality holds at all places and MATH. Since MATH is globally generated, MATH, and hence MATH. Finally, we have seen that MATH is injective, and since MATH is MATH-semistable, MATH, hence MATH is an isomorphism. CASE: MATH-stable (respectively strictly MATH-semistable) and MATH induces an isomorphism MATH NAME (respectively strictly semistable). Since MATH is an isomorphism, we have MATH for any subsheaf MATH. Then REF implies that for all weighted filtrations MATH . If the inequality is strict, then MATH . If MATH is strictly MATH-semistable, by REF there is a filtration giving equality in REF , then REF implies that MATH, and by REF MATH and a short calculation using this and REF gives MATH . So we finish by using REF . |
math/0103150 | Let MATH be a one-parameter subgroup of MATH with MATH. The proof of REF then gives equality in REF . Using REF (relationship between MATH and MATH), REF (relationship between MATH and MATH) and REF (relationship between MATH and MATH), this equality becomes MATH . We have chosen MATH so large that this holds if and only if it holds as a polynomial in MATH, hence taking the leading coefficient in MATH we obtain MATH where MATH and MATH. Using REF , this is MATH . By REF , MATH, hence MATH but by REF this must be nonpositive, hence MATH, and the last inequality is an equality. By REF , MATH, and hence MATH for all MATH, and then REF gives MATH . Conversely, let MATH be a filtration with MATH together with a splitting of the filtration MATH of MATH, and let MATH be the associated one-parameter subgroup of MATH. REF gives in particular MATH . By the proof of implication MATH in REF , since we get an equality, it is MATH for all MATH, hence MATH for all MATH, and the previous equality becomes MATH . Furthermore, the strong version of REF gives MATH. Using REF , together with REF and the strong form of REF , we obtain MATH . Hence, we also get MATH after evaluating this polynomial in MATH, but by the proofs of REF , this is equal to MATH. We have seen that MATH and MATH, and it is easy to check that this gives a bijection. |
math/0103150 | The main ingredient of the proof is REF , showing that GIT-(semi)stable points correspond to MATH-(semi)stable tensors. Using the notation of REF, let MATH (respectively MATH) be the GIT quotient of MATH (respectively MATH) by MATH. Since MATH is projective, MATH is also projective. GIT gives that MATH is an open subset of the projective scheme MATH. The restriction MATH to the stable part is a geometric quotient, that is, the fibers are MATH-orbits, and hence the points of MATH correspond to isomorphism classes of MATH-stable tensors. It only remains to show that MATH corepresents the functor MATH. We will follow closely CITE Let MATH be a family of MATH-semistable tensors (compare REF ) parametrized by a scheme MATH. Then MATH is locally free on MATH. The family MATH gives a map MATH, sending MATH to MATH. NAME MATH with small open sets MATH. For each MATH we can find an isomorphism MATH (where MATH is the NAME bundle in the definition of MATH at the beginning of REF), and a trivialization MATH . Using this trivialization we obtain a family of quotients parametrized by MATH giving a map MATH. And using the quotient MATH and isomorphism MATH we have another family of quotients parametrized by MATH . Then, using the representability properties of MATH and MATH, we obtain a morphism to MATH, and by REF this morphism factors through MATH and since a MATH-semistable tensor gives a NAME point REF , the image is in MATH. Composing with the geometric quotient to MATH we obtain maps MATH . The morphism MATH is independent of the choice of isomorphism MATH. A different choice of isomorphism MATH will change MATH to MATH, where MATH, so MATH is independent of the choice of MATH. Then the morphisms MATH glue to give a morphism MATH and hence we have a natural transformation MATH . Recall there is a tautological family REF of tensors parametrized by MATH. By restriction to MATH, we obtain a tautological family of MATH-semistable tensors parametrized by MATH. If MATH is another natural transformation, this tautological family defines a map MATH, this factors through the quotient MATH, and it is easy to see that this proves that MATH corepresents the functor MATH. Note that in CITE, the moduli space of stable framed modules is a fine moduli space. In our situation this is not true in general, because the analog of the uniqueness result of CITE does not hold in general for tensors. |
math/0103150 | We start with a general observation about GIT quotients. Let MATH be a projective variety with an action of a group MATH linearized on an ample line bundle MATH. Two points in the open subset MATH of semistable points are NAME (they are mapped to the same point in the moduli space) if there is a common closed orbit in the closures (in MATH) of their orbits. Let MATH. Let MATH be the unique closed orbit in the closure MATH in MATH of its orbit MATH. Assume that MATH is not in MATH. There exists a one-parameter subgroup MATH such that the limit MATH is in MATH (for instance, we can take the one-parameter subgroup given by CITE) . Note that we must have MATH (otherwise MATH would be unstable). Conversely, if MATH is a one-parameter subgroup with MATH, then the limit is NAME CITE. Note that MATH, and then MATH. Repeating this process with MATH we then get a sequence of points that eventually stops and gives MATH. Two points MATH and MATH will then be NAME if and only if MATH. Let MATH be a MATH-semistable tensor with an isomorphism MATH, and let MATH be the corresponding NAME point. Recall from REF that there is a bijection between one-parameter subgroups of MATH with MATH on the one hand, and weighted filtrations MATH of MATH with MATH together with a splitting of the filtration MATH of MATH on the other hand. The action of MATH on the point MATH defines a morphism MATH that extends to MATH with MATH for MATH and MATH. Pulling back the universal family parametrized by MATH by MATH we obtain the family MATH . Then, as in CITE, MATH corresponds to MATH (in particular, if MATH, then MATH is canonically isomorphic to MATH), and MATH is the admissible deformation associated to MATH. |
math/0103150 | The category of principal MATH-bundles is equivalent to the category whose objects are pairs MATH (where MATH is a principal MATH-bundle, MATH is a section of the associated fiber bundle MATH) and whose isomorphisms are isomorphisms MATH of principal bundles respecting MATH (that is, MATH and MATH). Note that this notion of isomorphism is not the same as isomorphism of reductions. The category of principal MATH-bundles is equivalent to the category of vector bundles of rank MATH. The quotient MATH is the set of invertible symmetric matrices (send MATH to MATH). Hence, a section MATH is the same thing as a homomorphism MATH as in REF . Now it is easy to check that there is a bijection betweeen these sets of isomorphisms classes. |
math/0103150 | Since MATH is locally free, the last term in the following exact sequence is zero MATH and hence MATH is surjective. Combining this with REF we get the exact sequence MATH . If MATH is torsion free, then MATH and we can use this sequence to obtain MATH. To prove REF , first we show that MATH is saturated. The composition MATH factors as MATH . The sheaf MATH is torsion free, and hence also MATH is torsion free. We conclude by showing that the stalk MATH is torsion free for all points MATH. Let MATH and let MATH be a nonzero element in the maximal ideal of the local ring of MATH, such that MATH. Since MATH, and MATH is saturated, then MATH. The same argument applies to MATH, and hence MATH. REF are easy to check. To show REF , if MATH, use the exact sequence REF , together with REF . If MATH, then MATH is a subsheaf of MATH of rank MATH, then MATH, and hence MATH. |
math/0103150 | First we will show that if MATH is a bilinear nondegenerate form on a vector space MATH, then MATH is NAME under the natural action of MATH (with the natural linearization induced on MATH). The point MATH is unstable if and only if there is a one-parameter subgroup MATH of MATH such that MATH. But this is impossible because MATH, and then MATH hence MATH is semistable. Then, using this and REF , it follows that MATH for all weighted filtrations. |
math/0103150 | To see that MATH-(semi)stable as a tensor implies (semi)stable as an orthogonal sheaf, we apply the stability condition to the weighted filtration MATH with weights MATH. By REF , MATH. Since MATH is isotropic, MATH, hence the stability REF gives the result: MATH . Now we will show that if MATH is (semi)stable as an orthogonal sheaf, then it is MATH-(semi)stable as a tensor. We start with a vector space MATH and a nondegenerate bilinear form MATH. Let MATH be a weighted filtration with MATH . Denote MATH. Take a basis of MATH adapted to the filtration, and let MATH be the one-parameter subgroup of MATH associated to this basis and weights MATH. Let MATH as in REF . Since MATH, the limit MATH exists, and MATH. Furthermore, we also have MATH . Write MATH and MATH as block matrices MATH . Note that if MATH, then MATH because of REF . We have MATH . The weights MATH strictly increase with both MATH and MATH. Assume MATH. Then, if MATH, and either MATH, MATH, or MATH, MATH, we have MATH. In matrix form: MATH . Since MATH, in each row of MATH there must be at least one nonzero block (and the same for columns). This, together with REF implies MATH with nonzero blocks in the second diagonal, and zero everywhere else. Since MATH is nondegenerate, these blocks give isomorphisms for all MATH and a short calculation then gives MATH. This, together with REF , implies that MATH . Finally REF imply that MATH for all MATH. Then, using this and the definition of MATH, MATH . Let MATH be a weighted filtration with MATH. We can assume that all subsheaves MATH are saturated. Apply the previous argument to MATH, the fiber over a point where MATH is locally free, and MATH the bilinear form induced by MATH on the fiber. We have MATH, hence it follows that MATH. Furthermore, as we have just seen MATH and MATH for all MATH. Hence we can write MATH where the last inequality is given by REF . Let MATH be a weighted filtration with MATH . By REF , it is strictly negative. We claim that MATH for all MATH. Assume that this is not true. Then there is a saturated subsheaf MATH with MATH. By REF , MATH, and MATH. By REF , MATH is saturated, and by REF , MATH. Consider the weighted filtration MATH with weights MATH. Since MATH is isotropic, MATH, and since MATH, this weighted filtration contradicts REF . Hence, using MATH together with MATH, MATH . |
math/0103150 | The proof of REF , replacing the NAME polynomials MATH, MATH, MATH,. by the degrees MATH, MATH, MATH,. proves that MATH is slope-MATH-(semi)stable if and only if for all isotropic subsheaves MATH, MATH . We can assume that MATH is saturated, hence MATH by REF , and since MATH, the result follows. |
math/0103150 | The proof that MATH corepresents the functor MATH is completely analogous to the proof of REF (see REF), so we will not repeat it. The subscheme MATH is open because being locally free is an open condition. Now we will prove that this moduli space is projective. REF are closed conditions, so they define a projective subscheme MATH of the moduli space of MATH-semistable tensors. The lemma will be proved by showing that MATH. If MATH is MATH-semistable then MATH is torsion free, so it only remains to check that if REF does not hold, then MATH is MATH-unstable. Assume that the homomorphism MATH induced by MATH is zero. Then the sheaf MATH defined as MATH is nonzero. Let MATH be the cokernel of MATH . Taking the dual of this sequence and restricting to the open subset MATH of MATH where MATH is locally free, we get MATH . By REF we have MATH, hence MATH, and then MATH, and since MATH, MATH. The exact sequence on MATH implies that MATH. Consider the weighted filtration MATH, MATH. We have MATH . Recall that MATH. Then MATH and hence MATH is slope-MATH-unstable REF , and in particular, MATH-unstable. |
math/0103150 | The first claim is proved by considering the filtration MATH. If MATH, then this filtration does not satisfy REF , hence contradicts semistability. Now we will prove that if MATH, then MATH is MATH-stable. Using REF , we only have to study filtrations of the form MATH with MATH, MATH and MATH isotropic and saturated. Using the NAME formula we have MATH so we need to estimate the second NAME classes of MATH and MATH. The sheaf MATH is saturated (see the proof of REF ). Define the torsion free rank one subsheaf MATH where MATH, MATH and MATH are respectively elements of MATH, MATH and MATH. We have MATH, so MATH is the ideal sheaf of a zero-dimensional subscheme of MATH. We distinguish several cases: CASE: If MATH, then MATH, and MATH. CASE: If MATH, MATH, then again MATH, and MATH. CASE: If MATH, MATH and MATH, then MATH does not contain a subsheaf MATH with MATH, hence this cannot happen. CASE: If MATH, MATH and MATH, then again MATH does not contain a subsheaf MATH with MATH, hence this cannot happen. So we conclude that MATH. The sheaf MATH is a rank one subsheaf of MATH, hence MATH unless MATH is the third summand MATH, but this is not possible because the third summand is not isotropic. Putting everything together, MATH hence MATH is MATH-stable by REF . |
math/0103150 | Let MATH act by multiplication on the right on MATH, and consider the quotient MATH. Let MATH, and let MATH be the class in MATH. To this class we associate the pair MATH. This gives a bijection between the set MATH and the set of pairs MATH, where MATH is a symmetric invertible matrix and MATH is a nonzero complex number such that MATH . Given a principal MATH-bundle MATH (or equivalently a vector bundle MATH), a reduction of structure group to MATH is a section MATH of the associated bundle MATH, and then this is equivalent to a pair of homomorphisms MATH as in REF . The rest of the proof is analogous to the proof of REF . |
math/0103150 | REF implies that to obtain the isomorphism MATH we have to extract a square root, so we obtain two special orthogonal sheaves MATH and MATH mapping to the given orthogonal sheaf. It only remains to check if theses two objects are isomorphic or not. If there is an automorphism MATH with the above properties, then MATH is an isomorphism between MATH and MATH. Conversely, assume that there is an isomorphism MATH between MATH and MATH. Then MATH is an automorphism of MATH with MATH and MATH. |
math/0103150 | Apply REF to MATH (multiplication by MATH). |
math/0103150 | The parameter space MATH for orthogonal sheaves is a subscheme of MATH, where MATH (this is a particular case of the parameter space defined in REF). Let MATH be the polarization defined in REF , and consider the natural linearization of the action of MATH on this polarization. There is a morphism MATH with MATH. This morphism is equivariant with respect to MATH, and the linearizations are compatible. Property REF implies that MATH is finite étale (because MATH is given locally by the equation MATH), and then it follows that a point in MATH is GIT-(semi)stable if and only if its image in MATH is GIT-(semi)stable. The result follows from REF , and REF . |
math/0103153 | Given MATH, define MATH by MATH . Also, for MATH, say MATH, define MATH by MATH for all MATH. So MATH. Given MATH, a sequence of predictors for the space MATH, and MATH, say MATH, put MATH . For MATH, define MATH. So, if MATH, MATH. MATH for all MATH. Assume that, for some MATH, we have MATH. List MATH and list MATH. Fix MATH and MATH such that MATH. Define MATH and consider MATH. Then MATH. This is a contradiction to the definition of MATH for it would mean MATH cannot predict correctly all MATH somewhere in the interval MATH. For MATH define MATH as follows. First let MATH be minimal such that MATH. Such MATH exists by the claim. Then let MATH be any MATH such that MATH is of maximal size. To see that this works, let MATH. Let MATH be predictors such that for all MATH and almost all MATH, there is MATH such that MATH. Fix MATH such that for all MATH and all MATH, there is MATH such that MATH. Let MATH with MATH. Thus MATH for all MATH. We need to find MATH such that MATH. To this end simply note that if MATH is such that MATH, then, by definition of MATH, MATH where MATH is minimal with MATH. This means in particular MATH. A fortiori, MATH. Since MATH, this entails that if we had MATH for all MATH, we would get MATH for some MATH. Thus MATH. So MATH. However MATH, a contradiction. This completes the proof of the theorem. |
math/0103153 | Let MATH be a family of predictors in MATH weakly MATH - constantly predicting all functions. Put MATH. By the theorem, every MATH is MATH - constantly predicted by a member of MATH. This shows MATH. Next let MATH be a family of functions such that no predictor MATH - constantly predicts all of MATH. Let MATH and MATH. Assume MATH weakly MATH - constantly predicts all members of MATH. Then MATH - constantly predicts all members of MATH, where we put MATH with MATH for all MATH, a contradiction. |
math/0103153 | We force with the countable support product MATH where CASE: MATH is NAME forcing MATH for MATH, CASE: MATH is MATH - ary NAME forcing MATH for MATH and MATH, and CASE: MATH is MATH where MATH for all MATH, for MATH. By MATH, MATH preserves cardinals and cofinalities. MATH is also immediate. Note that if MATH and MATH, then there is MATH of size MATH such that MATH, the generic extension by conditions with support contained in MATH, that is, via the ordering MATH. So there is MATH. Clearly the generic real added by MATH is not MATH - constantly predicted by any predictor from MATH. This shows MATH. A similar argument shows MATH. So it remains to see that MATH for MATH. Put MATH. Let MATH be a MATH - name for a function in MATH. By a standard fusion argument we can recursively construct CASE: a strictly increasing sequence MATH, MATH, CASE: MATH countable, CASE: MATH, a partition of MATH into countable sets, CASE: a condition MATH, and CASE: a tree MATH such that CASE: if MATH, MATH, and MATH (MATH), then MATH where we put MATH, CASE: MATH, and CASE: whenever MATH where MATH with MATH, MATH, and MATH are such that MATH, then there are MATH and MATH with MATH, such that MATH where MATH with MATH for MATH. Now let MATH be MATH - generic with MATH. By REF above, there is, in MATH, a tree MATH such that for all MATH, MATH and MATH, there is a unique MATH extending MATH, and such that MATH is forced to be a branch of MATH by the remainder of the forcing below MATH. By REF, we also have that for all MATH, MATH and MATH, there are at most MATH many MATH extending MATH. This means we can recursively construct a predictor MATH which MATH - constantly predicts all branches of MATH. A fortiori, MATH is forced to be predicted by MATH by the remainder of the forcing below MATH. On the other hand, MATH satisfies MATH so that there are a total number of MATH many predictors in MATH, and they MATH - constantly predict all reals of the final extension. This completes the argument. |
math/0103153 | Let MATH be a sequence of natural numbers MATH in which each MATH appears MATH often and such that in each limit ordinal, the set of MATH with MATH is cofinal. We perform a countable support iteration MATH such that MATH . By MATH, MATH preserves cardinals and cofinalities. As in the previous proof, we see MATH for all MATH. We are left with showing that MATH. Let MATH be a MATH - name for a function in MATH. Notice given any MATH, we can find MATH and MATH such that MATH . First consider the case MATH is a successor ordinal, say MATH. Let MATH be such that MATH. The following is the main point. There are MATH and a predictor MATH such that MATH . We construct recursively CASE: MATH countable, CASE: MATH, a partition of MATH into countable sets, CASE: finite partial functions MATH, MATH, CASE: conditions MATH, MATH, CASE: a strictly increasing sequence MATH, MATH, CASE: a tree MATH, and CASE: a predictor MATH such that CASE: MATH, CASE: MATH, CASE: if MATH, then MATH; in case MATH, we have MATH, otherwise MATH; MATH for MATH, CASE: MATH, CASE: MATH; furthermore for all MATH, MATH, CASE: MATH, CASE: if MATH, MATH, then MATH, CASE: for each MATH, there is MATH which forces MATH; furthermore MATH, and CASE: MATH - constantly predicts all branches of MATH. Most of this is standard. There is, however, one trick involved, and we describe the construction. For MATH, there is nothing to do. So assume we arrived at stage MATH, and we are supposed to produce the required objects for MATH. This proceeds by recursion on MATH. Since the recursion is straightforward, we confine ourselves to describing a single step. Fix MATH. Let MATH be such that MATH. Without loss MATH (the case MATH being easier). Consider MATH. Step momentarily into MATH with MATH. Then MATH. Since MATH is forced not to be in MATH, we can find MATH, pairwise incompatible MATH, and distinct MATH where MATH extending MATH such that MATH. As MATH is MATH - ary NAME forcing, we may do this in such a way that the predictor MATH can be extended to MATH - constantly predict all MATH. NAME in MATH, by extending the condition MATH if necessary, we may without loss assume that it decides MATH and the MATH. We therefore have the extension of MATH which MATH - constantly predicts all MATH already in the ground model MATH. We may also suppose that MATH decides the stem of MATH, say MATH. For MATH define MATH such that CASE: MATH, MATH, CASE: MATH, CASE: MATH. Doing this (in a recursive construction) for all MATH and increasing MATH if necessary, we may assume there is MATH with MATH for all MATH. Finally MATH is the least upper bound of all the MATH. This completes the construction. By REF, and REF, the sequence of MATH's has a lower bound MATH. By REF, MATH. By REF, MATH which means that REF entails MATH is MATH - constantly predicted by MATH," as required. Now let MATH be a limit ordinal. Using a similar argument and the fact that below MATH, MATH is cofinally often NAME forcing, we see There are MATH and a predictor MATH such that MATH . This completes the proof of the theorem. |
math/0103153 | Assume MATH where each MATH is MATH - linked. Define MATH such that, for each MATH, MATH is a MATH such that no MATH forces MATH. (Such a MATH clearly exists. For otherwise, for each MATH we could find MATH forcing MATH. Since MATH is MATH - linked, the MATH would have a common extension which would force MATH, a contradiction.) Let MATH. Now choose MATH such that for all MATH there are infinitely many MATH with MATH. Fix MATH and MATH. There is MATH such that MATH. We can find MATH such that MATH. By definition of MATH, there is MATH such that MATH. Thus MATH, as required. |
math/0103153 | This is a standard argument which we leave to the reader. |
math/0103153 | This is well - known and trivial. |
math/0103153 | Simply note MATH is a witness for MATH. For given a predictor MATH, define MATH by MATH the unique MATH such that MATH predicts MATH incorrectly on the whole interval MATH where MATH. If MATH is such that MATH, then MATH does not MATH - constantly predict MATH. |
math/0103153 | Simply adapt the argument from CITE, or see CITE. |
math/0103153 | Let MATH be a finite support iteration of ccc forcing such that each factor MATH is forced to be a MATH - linked forcing notion of size less than MATH for some MATH. Also guarantee we take care of all such forcing notions by a book - keeping argument. Then MATH is straightforward. In view of REF it suffices to prove MATH for all MATH. So fix MATH. Note that in stage MATH of the iteration we adjoined a family MATH of size MATH satisfying MATH above with countable replaced by less than MATH. Show by induction on the remainder of the iteration that MATH continues to satisfy this version of MATH. The limit step is taken care of by REF . For the successor step, in case MATH is MATH - linked for some MATH, use REF , and in case it is not MATH - linked (and thus of size less than MATH), use REF . By REF , MATH follows. |
math/0103153 | CASE: Let MATH be a finite support iteration of ccc forcing such that CASE: for even MATH, MATH is amoeba forcing, CASE: for odd MATH, MATH is a subforcing of some MATH of size less than MATH. Guarantee that we go through all such subforcings by a book - keeping argument. Then MATH is straightforward, as is MATH. Now note that amoeba forcing is MATH - linked (like random forcing). Therefore we can apply REF , and REF for all MATH simultaneously, and see that there is a family MATH of size MATH which satisfies the appropriate modified version of MATH (such a family is adjoined after the first MATH stages of the iteration). CASE: First add MATH many NAME reals. Then make a MATH - stage finite support iteration of amoeba forcing. Again, MATH is clear. MATH follows from REF using standard arguments. |
math/0103153 | List all predictors as MATH. Choose reals MATH such that MATH does not MATH - constantly predict MATH for MATH. Let MATH. Let MATH be property MATH. Also let MATH be a MATH - name for a predictor. Assume there are conditions MATH such that MATH - constantly predicts MATH from MATH onwards." Without loss MATH for all MATH, and any MATH many MATH have a common extension. Let MATH be the tree of initial segments of members of MATH. Given MATH with MATH, let MATH and MATH. Note that if MATH for all such MATH, then we could construct a predictor MATH - constantly predicting all of MATH past MATH as in the proof of REF . So there is MATH with MATH. Find MATH such that MATH and notice that a common extension of the MATH forces a contradiction. |
math/0103153 | Use the lemma and the folklore fact that the iteration of REF p.o.'s is REF. |
math/0103158 | The proof of this Proposition follows from REF. The input we need for applying these Theorems is the linking matrix between the link components MATH and the virtual link components corresponding to the regular fiber MATH and the singular fiber MATH of each NAME fibration appearing in the minimal splice diagram of MATH, see REF . As we chose carefully all the NAME invariants to be equal, the linking matrix with the MATH-th splice component has the relatively simple form MATH the reader familiar with the content of CITE can read this result from the diagram in REF above, as linking numbers coincide with the product of the edge weights adjacent to the nodes contained in the path from MATH (or MATH) to MATH but not on the path itself. To prove REF , we recall that REF claims that MATH is fibered if and only if its coefficients satisfy MATH . Substituting with the result of REF we obtain the statement in REF above. All the MATH (top dimensional) faces of the NAME unit ball are consequently fibered, that is, all the integral point laying in the cone over these faces are represented by fibrations. It is well known (see for example, CITE) that in this case the NAME and NAME norm coincide. Concerning REF , in REF it is proven the NAME norm of any element MATH of MATH is given by MATH where MATH is the number of incident edges of the virtual component (in our case, MATH for the nodes, MATH for the boundary vertices). Plugging in this formula the coordinates of MATH, and using the values of the linking matrix of REF we obtain REF . The rest of REF follows from these results. REF gives the example of the NAME unit ball for MATH. Finally, REF computes the NAME polynomial of the link as MATH by using REF again and taking the symmetrized polynomial we obtain REF . |
math/0103158 | We start by observing that homotopic forms must lie in the cone over the same fibered face of the NAME unit ball. We will prove the Theorem by showing that the action of Diff-MATH on MATH partitions the MATH . NAME classes in MATH different orbits. The diffeomorphisms of MATH act over MATH through isometries of the NAME REF norm, preserving the unit ball; they preserve moreover the lattice of integral points (and consequently, divisibility). We will leave aside the case of MATH, where the statement of the Theorem holds true by direct check of the divisibility of the vertices dual to the fibered faces (MATH and MATH are two inequivalent vertices, that we can easily determine from the NAME polynomial). Let's assume then that MATH; any primitive element MATH, lying in one non-fibered ray must be sent to a primitive element lying in a non-fibered ray and having the same norm, that is, Diff-MATH acts on the set of MATH elements MATH preserving the norm. REF implies that there are at least MATH different orbits for this action: the orbit of MATH can contain only MATH, as these are the only primitive non-fibered integral homology classes having the same NAME norm, and the same holds for the other MATH elements. As MATH, the knowledge of the action of Diff-MATH on a single primitive element determines completely the action on the others (for example, if MATH goes to MATH, then MATH must go to MATH and so on). This implies a constraint on the possible orbit of fibered faces, and it is easy to check that there must be at least MATH orbits of the action of Diff-MATH on the set of fibered faces, the possible identifications corresponding to reflections through the diagonals of MATH (the `+REF' comes from the fact that the fibered faces hit by the diagonals have orbits containing at most MATH elements, while the orbits of the remaining faces contain at most MATH elements). REF illustrates the case of MATH: the fibered faces marked with a different symbol in Figure are inequivalent. This result implies that the vertices dual to inequivalent fibered faces (and consequently the NAME classes) lie in different orbits of Diff-MATH, and completes the proof. |
math/0103158 | This is a consequence, as in REF, of the property of diffeomorphism extension of MATH; first, the fiber sum in REF above is well defined, and independent of the choice of the framing for MATH; then, because of the diffeomorphism of REF , for all choice of MATH the fiber sum manifold is defined as one of those in REF , where we determine the pair of MATH according to REF and we complete it to a homology basis of MATH respecting the framing of MATH. Up to diffeomorphism, the smooth manifold defined this way is the same for all MATH. |
math/0103158 | The NAME - NAME polynomial of MATH is determined by the NAME polynomial of MATH through REF ; therefore, the first two coefficients of MATH determine its SW norm, and that coincides with the NAME norm of MATH. |
math/0103158 | CASE: For all choice of a fibered class MATH we have a symplectic manifold MATH, with a pair of embedded, framed symplectic tori MATH. MATH, with its standard NAME form, is as well symplectic, with MATH an embedded, framed symplectic torus. NAME 's theory of symplectic fiber sum (see for example, CITE) guarantees that the manifolds defined as in REF admit a symplectic structure. REF tells that the smooth manifold carrying the symplectic structure is the same for any MATH. Once MATH is fixed, REF allows to identify the canonical class of the symplectic structure on MATH as the image, with reversed sign, of the NAME class of the fibration under the injective map MATH described in REF . In fact, in CITE it is shown that the anticanonical bundle has a section whose zero set is homologous to the image of the NAME class associated to MATH cross the class of MATH. This section is obtained in the following way. The anticanonical bundle restricts on MATH to the pull-back of the plane field defined by the fibration MATH, which has a preferred section which is inward pointing on the boundary and whose zero set is the NAME class of MATH. This section obviously induces a section of the anticanonical restricted to MATH. On MATH the anticanonical is a trivial bundle (recall that MATH) and we can construct a nowhere vanishing section by restricting a section of MATH with a simple zero along MATH: on the boundary MATH, the section can be assumed to depend only on the MATH coordinate, (as the tangent bundle to the elliptic fiber is canonically trivial), and outward pointing, because of the simple zero at MATH. These two sections glue on the boundary to give a section, on MATH, whose zero set is the image of the NAME class under the map of REF . Note that the canonical classes arising through this construction satisfy (as they should) all NAME 's constraints for MATH, as they are image of a vertex of the polyhedron dual to the NAME norm ball and therefore, by REF , they are the vertices of the convex hull of basic classes. To prove REF we note that, as symplectic forms with different canonical bundles are not homotopic, we can construct at least MATH non-homotopic symplectic forms on MATH by choosing, in the symplectic fiber sum definition of REF , elements MATH laying in different fibered cones of MATH (and having thus different NAME classes). We need to prove now that the MATH canonical bundles arising this way partition in at least MATH orbits under the action of the diffeomorphism group of MATH. NAME preserve the convex hull of NAME - NAME basic classes, as well as the unit ball of the NAME - NAME norm, which is its dual polyhedron. We will study the orbits of the vertices of the convex hull (and therefore of the canonical bundles) by studying the orbits of their dual (top dimensional) face. Again, we leave aside the case of MATH, where we can easily apply divisibility to prove the result. Let's assume then that MATH. We will study the orbits of the classes of MATH laying in the plane spanned by MATH (which can be identified with MATH) with same coordinates as MATH or MATH. For sake of notation, we denote these elements with the same symbol. By invariance of the unit ball, diffeomorphisms exchange the cones MATH: the image of an element MATH must be therefore a primitive element laying in one of the MATH and having the same NAME - NAME norm. Note that there is no reason why diffeomorphism should preserve the span of MATH. We need to modify a bit the approach to the proof of REF to keep track of this. Consider one of the four primitive classes MATH whose coefficients MATH lie on non-fibered rays of MATH, and which have the smallest SW norm. According to REF , this must be one of the MATH. Take MATH for example. Under the action of a diffeomorphism MATH, MATH is sent to a primitive class MATH where, in principle, MATH can be divisible, but must be integer multiple of the coordinates of one of the MATH. Moreover we must have MATH: consequently, by REF MATH . As the norm of MATH is minimal, we deduce that in fact MATH are the coordinates of one of the MATH. This condition constrains the orbit of the entire cone MATH; MATH must be one of the four cones MATH. We can repeat this argument for the other three elements of minimal SW norm, and then pass to the other four of second minimal norm and so on. Or, more simply, we observe that as MATH when we know MATH then we can unambiguously determine the image of the other cones (take MATH and proceed as above to determine MATH, and then use the invariance of the unit ball). It is easy to check that this result implies that there are at least MATH orbits of the action of the diffeomorphism group of MATH on the set of MATH top dimensional faces of the unit ball, which play the role of the fibered faces for the link exterior. Again, the possible identifications correspond to reflection of these faces about the diagonals MATH, MATH. The dual vertices to these faces, and so the canonical bundles, lie in at least MATH distinguished orbits of the group of diffeomorphism. This completes the proof of REF . |
math/0103159 | The proof repeats the computation in the proof of REF with REF replaced with their generalizations, REF. |
math/0103159 | We can assume that MATH is a neighborhood of MATH in MATH such that MATH and MATH . Define MATH by MATH . Consider the following commutative diagram: MATH . Here MATH is the connecting homomorphism, MATH the inclusion, MATH the radial projection. Let MATH . Then MATH is given in the first column of the diagram. Now we apply the Extension Theorem, REF. Suppose MATH . Then from the commutativity of the diagram, MATH. Thus the primary obstruction to extending MATH to MATH vanishes. By REF the other obstructions MATH also vanish-MATH . Next, MATH has the form MATH . Define a map MATH by MATH where MATH satisfies the following: CASE: MATH is small enough so that MATH for all REF MATH for all MATH . Then MATH since MATH . To complete the proof observe that MATH is homotopic to MATH relative MATH because MATH is convex. |
math/0103159 | By the NAME Isomorphism Theorem CITE, we have MATH . By REF and the Universal Coefficient Theorem, REF, we have also MATH . |
math/0103159 | CASE: MATH and MATH for all MATH . CASE: MATH for all MATH . CASE: Either MATH or MATH for all MATH . Thus the two conditions of REF are satisfied. Now the conclusion follows from REF . |
math/0103159 | Suppose MATH is zero. Choose MATH to be identically equal to MATH . Then MATH. Recall MATH . Then for all MATH we have the following. MATH . Hence MATH . Therefore by REF the coincidence set can be removed. Thus MATH is not weakly coincidence-producing . |
math/0103166 | CASE: Consider the (direct limit) exact sequence MATH . The coefficient of MATH in MATH lies in MATH. CASE: Suppose first that we restrict the elements MATH to those for which MATH. Then the symbols of the coefficients of MATH lie in the ``subspace of vertical symbols" MATH . Letting the subscript MATH denote the coefficient of MATH, we have by REF that we can vary MATH and MATH to obtain elements MATH whose symbols generate MATH . Furthermore these elements MATH have the property that MATH . Thus for MATH we can consider MATH as a map MATH . CASE: Notice that our restriction on the support of the MATH insures that, for MATH, the coefficient of MATH in MATH has the same symbol in MATH as MATH does. So it suffices to show that the symbols of the coefficients of MATH in elements of the form MATH generate MATH . But MATH generates MATH, so it suffices to show that elements of the form MATH generate MATH. But this is again the content of REF. |
math/0103166 | By REF therefore, it suffices to show that, for each expression MATH, there are operators MATH such that REF MATH has the same symbol in MATH as the coefficient of MATH in MATH REF MATH . The proof of this fact is identical to the proof given in REF. The notational dictionary is MATH . In the proof we consider the line bundle MATH on the product MATH . Just as in REF we use REF of CITE to twist the NAME data MATH above to obtain MATH for which the condition that the tautological section extends under deformation implies that the map MATH given by MATH is MATH. |
math/0103166 | Referring to MATH, consider the chain MATH of subsheaves and the image of MATH of each in MATH. By MATH, REF , and the surjectivity of MATH, the induced maps on MATH of the graded quotients of this chain into respective quotients of MATH are all zero from MATH on. |
math/0103167 | NAME map is not defined for the NAME examples, see CITE or REF. The indeterminacy locus for every rational map is closed, therefore degenerations of the NAME examples are also in the indeterminacy locus. |
math/0103167 | We will construct two disjoint equivariant subgraphs MATH and MATH such that together they contain all vertices of the graph MATH and every edge of MATH either belongs to one of the subgraphs MATH or MATH or is an ordinary edge which begins at one subgraph and ends at the other. To find these MATH and MATH we will take MATH and MATH and start adding to them vertices and edges until we will get required MATH and MATH. First let us consider connected components of the bold subgraph MATH. Each bold component cannot intersect with both subgraphs MATH and MATH, otherwise we have a bold path connecting these subgraphs. We will change each MATH by the union of MATH and all components of MATH which has non empty intersection with MATH. It is easy to see that the new MATH and MATH still satisfy all the conditions from the statement of the Lemma. In addition, the compliment MATH is attached to MATH through the ordinary edges only. Next let us consider the following topological space MATH. Connected components of this space can be of three different types: REF components attached to MATH only, REF components attached to MATH only, REF components attached to both MATH and MATH. We will change MATH by the union MATH and MATH by MATH. Again, MATH and MATH still satisfy all the conditions from the statement of the Lemma. Now, to get desired MATH and MATH, we can take MATH to be equal MATH. For vertices of MATH we will take all vertices of MATH which are not from subgraph MATH and for edges of MATH we will take all edges whose beginning and ends are vertices of MATH. Note that the number of edges connecting MATH and MATH is at least the same as the number of edges connecting original MATH and MATH, which is not less then REF. Consider decomposition of the curve MATH where each MATH correspond to subgraph MATH. Each part MATH is invariant with respect to involution and can be obtained as a degeneration of smooth curves with involution. Then the whole curve MATH is a degeneration of the NAME example and as we have noticed the number of nodes is at least REF. |
math/0103167 | Let curve MATH be a degeneration of the NAME example with MATH edges with MATH. Thus curve MATH is a union MATH where each MATH is a degeneration of a family of smooth curves and intersection MATH consists of MATH pairs of exchanged nodes. Then in the dual graph for MATH there are two equivariant subgraphs MATH and MATH corresponding to MATH and MATH and these subgraphs MATH and MATH are connected by the MATH pairs of exchanging edges. We will denote them MATH. It is clear that these edges are of type REF. In definition of REF all MATH corresponding to MATH equal REF. Now if we have an arbitrary simple cycle MATH then either it does not contain any of the edges MATH or it passes through even number of them. With counting multiplicities the sum MATH passes through MATH the same even number of times. All the elements of the lattice MATH are of the form MATH where MATH and MATH are simple oriented cycles. Therefore for any element of MATH we have MATH is an integer. We can choose some MATH edges MATH with the functions MATH so that the intersection of MATH hyperplanes MATH is REF-dimensional. But the point of intersection MATH cannot be an element of MATH since MATH is not an integer. REF fails. |
math/0103167 | If REF does not hold then REF does not hold either. We only need to show that if REF holds but the graph has an edge MATH with the corresponding function MATH of type REF then the graph is a degeneration of a NAME example with REF edges. Let MATH be the beginning of MATH and MATH be the end of MATH. Let us consider connected components of MATH. The number of components is at most REF. If there is only one component in MATH then there is a path MATH in MATH connecting MATH and MATH. Thus for the simple cycle MATH me have MATH and MATH which is impossible since MATH is an edge of type REF. If there were three components in MATH then edges MATH and MATH would have been of type REF. The only possibility is when MATH has two components MATH and MATH. Since for every simple cycle MATH we have MATH then beginnings MATH and MATH belong to the same component of MATH, say MATH. Therefore MATH is invariant with respect of involution MATH and so is MATH. Now consider decomposition of the curve MATH where each MATH correspond to subgraph MATH. Each part MATH is invariant with respect to involution and can be obtained as a degeneration of smooth curves with involution. Then the whole curve MATH is a degeneration of the NAME example with the number of edges REF. REF implies REF . Assume that REF fails. The curve MATH is a degeneration of a NAME example with REF edges. Thus curve MATH is a union MATH where each MATH is a degeneration of a smooth curve and intersection MATH consists of two exchanged nodes. Then in the dual graph for MATH there are two equivariant subgraphs MATH and MATH corresponding to MATH and MATH and these subgraphs MATH and MATH are connected by the pair of exchanging edges. It is clear that these edges are of type REF. REF implies REF . Since there are no edges of type REF then in definition of REF all MATH equal REF. REF holds, so if the intersection of MATH hyperplanes MATH is REF-dimensional, then it is in the lattice MATH. Thus intersection of MATH is in MATH, but this is the same point as intersection of MATH. Therefore REF holds. REF implies REF . Let us take an arbitrary edge MATH and the corresponding function MATH. If MATH is nontrivial (that is, not of type REF) then we can choose some MATH edges MATH with the functions MATH so that the intersection of MATH hyperplanes MATH is REF-dimensional and by REF is some point MATH. Now MATH and MATH. Thus edge MATH and the corresponding function MATH are of type REF and not of type REF. |
math/0103170 | This follows from REF in the book of CITE; see also CITE. |
math/0103170 | The matrix MATH is positive semidefinite if and only if there exists a real NAME factorization MATH. If this holds then MATH is a sum of squares, and every sum of squares representation arises in this way. |
math/0103171 | Assume that MATH extends to a diffeomorphism MATH as above; then, by virtue of REF , MATH, where MATH. Because MATH is smooth and MATH, the function MATH is smooth and MATH never vanishes. Recall from REF , that the form MATH extends smoothly to all of MATH and restricts to a contact form on MATH. This implies that MATH is smooth on all of MATH and restricts to a contact form on MATH. Conversely, suppose that MATH is regular on the blowup and that the form MATH is extends smoothly to all of MATH and restricts to a contact form on MATH. Then because MATH is non-vanishing, MATH is diffeomorphic to MATH. It also follows that the construction of the vector field MATH given in REF extends to define a vector field MATH on all of MATH. Since MATH is transverse to MATH, the map MATH where MATH is the flow of MATH is a diffeomorphism. By uniqueness of integral curves, MATH agrees with MATH on the interior of MATH. |
math/0103171 | We claim that MATH away from MATH and that MATH and MATH commute everywhere. To verify the first condition, recall that MATH is the extension of MATH to MATH, and that MATH, where MATH is the complex structure tensor; the identity MATH follows from REF . To see that the vector fields MATH and MATH commute, recall that because MATH is a volume form on MATH, we need only prove that the NAME bracket MATH is in the kernel of each of the forms MATH, MATH, and MATH. But MATH and, because the kernel of MATH is an involutive distribution and MATH and MATH are both in the kernel, so is MATH. Because MATH and the level sets of MATH are all compact, MATH is well defined on all of MATH. Because MATH on MATH, MATH is a holomorphic curve in MATH. By construction, the vectors MATH and MATH are non vanishing for all MATH. Consequently, MATH is a non-singular parameterization of a leaf of the NAME foliation. To see that every leaf of MATH is contained in the image of MATH for some MATH, choose a point in MATH. Then MATH for a unique point MATH. Hence, the leaf of MATH through MATH is contained in the image of MATH. Finally, to verify that the leaves of MATH intersect MATH along non-singular curves of the form MATH, recall that MATH. This shows that the curve is contained in MATH. Moreover, by construction, MATH showing that the curve is non-singular. |
math/0103171 | Assume that MATH is regular on the normal blowup. First observe that the flow of MATH induces a continuous deformation retract MATH defined as follows MATH . Hence, MATH and MATH have the same homotopy type. By REF, the NAME manifold MATH has the homotopy type of a MATH-dimensional cell complex. Consequently, MATH can have dimension at most MATH. Let MATH denote the tangent bundle of MATH, and let MATH denote the complex structure tensor of MATH. We claim that the composition MATH is a surjective map onto the normal bundle of MATH in MATH. Because the dimension of MATH is at most MATH, this claim implies, that the map REF is an isomorphism of vector spaces, hence, that MATH is totally real. To prove that the map REF is surjective, first choose a point MATH and a non-zero vector MATH. We need only show that a multiple of MATH is in the image of this map. But the vector MATH defines an oriented ray, which by definition of MATH is a point MATH with MATH. By REF , the oriented rays defined by MATH and MATH coincide. |
math/0103171 | We first claim that the form MATH extends smoothly to all of MATH. To see this, give MATH the NAME metric defined by the NAME potential MATH. One easily verifies that the vector field MATH defined in REF satisfies the identity MATH on MATH. Therefore by REF , the gradient vector field MATH is a scalar multiple of MATH, and each integral curve of MATH is contained in a geodesic of MATH that intersects the level sets of MATH orthogonally. Consequently, these geodesics lift to the blowup and intersect the boundary of MATH transversely. The union of all of these curves forms a one dimensional foliation of MATH with tranversal intersection with the boundary of MATH. Moreover, the leaves of this foliation are (by construction) the closures of the integral curves of MATH. The identities MATH then show that the form MATH extends smoothly to all of MATH and is non-vanishing at all points of MATH. Let MATH denote the pullback of MATH to the boundary MATH. The non-degeneracy condition MATH, implies that MATH is a contact form on MATH. Therefore, to conclude the proof of regularity, we need only show that MATH is smooth on all of MATH and that MATH is non-vanishing near MATH. We do this obtaining explicit formulas for MATH and MATH in blowup coordinates adapted to the complex structure on MATH. By REF and CITE (see also CITE), MATH, the zero set of a smooth strictly plurisubharmonic function, is totally real. Let MATH be the dimension of MATH, and let MATH, and let the indices MATH and MATH range between MATH and MATH and MATH and MATH, respectively. We choose holomorphic coordinates MATH with MATH and a smooth function MATH such that MATH . Because MATH is totally real, we may choose coordinates so that MATH vanishes to arbitrarily high order at MATH. These coordinates are not adapted to MATH, so they must be replaced by the adapted coordinates MATH defined by MATH . Blowup coordinates are then given by the formulas MATH where Greek indices range between MATH and MATH. Since we only have to compute MATH on the set MATH, and since MATH vanishes to high order, we may assume that MATH is identically zero in any finite order computation along MATH. In particular, up to first order along the set MATH, MATH, we have MATH . (To highlight the special role played by the radial parameter MATH, we have written MATH.) A straightforward computation using the chain rule, shows that MATH . We claim that MATH can we written in the form MATH where MATH is an integer, MATH, and MATH is a differentiable function of MATH such that MATH . Assume this claim for the moment. Then, substituting REF into the formula for MATH and simplifying gives MATH . But we have already proved that MATH extends smoothly to the set MATH and is nowhere-vanishing. Inspection of the above formula for MATH shows that this implies that MATH. Thus, MATH, which is smooth on all of MATH. The formula MATH for MATH shows that MATH is regular. That that MATH is totally real and has dimension MATH follows from REF . It remains only to prove that MATH is of the form REF. Because MATH is real analytic, MATH has a series expansion of the form MATH where MATH and MATH is a smooth function such that MATH. Therefore, the equation MATH can be inverted to show that MATH is of the form MATH for MATH a smooth (in fact, analytic) function satisfying the condition MATH. On the other hand, MATH is smooth and vanishes precisely on MATH. This, together with the positivity condition MATH, implies that MATH vanishes precisely to order REF on MATH. Therefore, MATH can be expressed in the form MATH where MATH is a smooth function and MATH. Combining REF, and setting MATH results in REF . Setting MATH in the above formula for MATH and simplifying yields the identity MATH . At this point, we invoke REF to conclude that MATH and MATH, and that MATH is totally real. |
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