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math/0010020 | We begin with noting that a line MATH in MATH has REF if and only if it is the graph MATH of a nonzero homomorphism MATH. It is clear that there are MATH such lines. They come in MATH pairs: we have MATH if and only if MATH. In that case MATH is isotropic and so MATH correspond to MATH-orbits of MATH-vectors that are perpendicular. There are unique MATH-vectors MATH in these orbits with sum MATH. Let MATH be a MATH-vector with the same mod MATH-reduction as MATH. Since MATH is isotropic, MATH is also orthogonal to MATH. We noted that MATH spans the orthogonal complement of MATH in MATH, and so the second assertion of the corollary follows. |
math/0010020 | We only prove the statements involving a MATH-vector, the proof of the one about a MATH-vector is similar and easier. We begin with the last clause. Let MATH be as in the proposition. By NAME 's result, a unitary transformation will map this into a pair with second component MATH and so we may assume that MATH. Then MATH can be written MATH with MATH, MATH and MATH. We must have MATH. Since the two terms must be nonnegative multiples of three they are MATH, MATH or MATH. The stabilizer of MATH contains the interchange of MATH and MATH as well as the unitary group of each of these summands. So we can eliminate the last case and by REF assume that MATH or MATH. In either case, there exists a MATH-vector MATH with MATH and so by the discussion REF there exists a unitary transformation fixing MATH that sends MATH to MATH. The last assertion follows. We next show that any MATH-vector MATH is perpendicular to a primitive null vector. The orthogonal complement MATH of MATH is a free MATH-module of signature MATH. So its complexification MATH represents zero. Its real dimension is MATH and REF then implies that MATH also represents zero. In other words, there exists a null vector perpendicular to MATH. It remains to see that MATH and MATH are in the same MATH-orbit. This is left to the reader. |
math/0010020 | We apply this to the case at hand: MATH, MATH and MATH spanned by the MATH-vector MATH. Then MATH, where in the latter case the skew form takes the value MATH on a generator. One easily verifies that the group of unitary transformations of MATH is MATH. As this is also the group of unitary transformations of MATH, it follows that every unitary transformation of MATH extends uniquely to unitary transformation of MATH. |
math/0010020 | Suppose first MATH spanned by the MATH-vector MATH and a MATH-vector MATH. We have MATH for some MATH. Since MATH is positive definite, we must have MATH. Since MATH, this implies that up to a unit MATH equals MATH, MATH, MATH or MATH. By multiplying MATH with a unit we may assume that MATH acually equals one of these values. For MATH we get case MATH, and for MATH we get case MATH. For MATH we get case MATH by taking MATH and for MATH we get case MATH by taking MATH. For the last part of the lemma, we observe that for an overlattice MATH we must have that the quotient of the discriminant of MATH by the discriminant of MATH must be the norm of an element of MATH. Since the discriminant of MATH is also divisible by MATH, this implies that MATH is of type MATH or MATH. The case MATH has as underlying integral lattice a root lattice of type MATH. This admits no even overlattice and hence cannot occur. There remains the case that MATH is of type MATH with MATH of discriminant MATH. It is then not hard to see that MATH is as asserted. |
math/0010020 | By REF we may assume that MATH is spanned by MATH and that MATH in case MATH and MATH in case MATH. This identifies MATH with MATH. A MATH-vector in MATH maps to a MATH-vector in MATH of the same MATH-invariant and the MATH-vector of MATH lies in MATH or in MATH. In case MATH it is clear that any MATH-vector in MATH has MATH-invariant MATH, so if we are after the MATH-vectors of MATH-invariant MATH or MATH, then we only have to deal with MATH. Assertion REF then follows from REF . Case MATH follows from the simple observation that MATH cannot be written in any other way as a sum of two MATH-vectors in MATH. |
math/0010020 | Let us first assume that MATH is isomorphic to an orthogonal product MATH with MATH. Since MATH and MATH, we see that it is enough to investigate the corresponding issue in MATH. The MATH-vectors in MATH are all unitary equivalent, and so we can assume that MATH. The assertions regarding the classification now follow from REF . Assume now that MATH is a positive definite sublattice and spanned by MATH and MATH-vectors of MATH-invariant MATH and MATH. Assume also that its rank is MATH. Then the orthogonal complement of the lattice MATH is hyperbolic of sufficiently high rank and so by NAME 's theorem contains a primitive null vector. We may assume that this null vector is MATH and that MATH is either MATH or MATH. So MATH projects isomorphically to a sublattice MATH spanned by MATH-vectors. Since the MATH-vectors helping to span MATH are of MATH-invariant MATH or MATH, REF implies that MATH when MATH and MATH when MATH. In particular, MATH is of rank MATH. All the assertions now follow in a straightforward manner from REF. |
math/0010020 | For MATH or MATH this is part of the statement of the previous proposition. The cases MATH and MATH are handled in a similar way. |
math/0010025 | To prove bijectivity, we show that the inverse assignment is given by taking derived toric manifolds; to each MATH-translation MATH we associate the MATH-equivariant diffeomorphism MATH, where MATH if and only if MATH. It follows directly from the definitions that MATH descends to the original MATH-translation MATH of characteristic pairs. If, on the other hand, we start with a MATH-equivariant diffeomorphism MATH (or its equivalent), then MATH is derived from the corresponding MATH-translation of characteristic pairs. But the preferred section MATH for MATH automatically extends to an equivariant diffeomorphism MATH, and the section MATH extends to an eqivariant diffeomorphism MATH; thus MATH, whence MATH and MATH are equivalent, as required. |
math/0010025 | Given MATH and MATH, we consider tangents to the orbit MATH at MATH. One such has direction vector MATH, whose inner product with MATH is given by MATH. Since this quantity is strictly positive the orbit meets MATH only at MATH, (and the intersection is transverse). On the other hand, given any point MATH in the open cone MATH, the orbit MATH meets MATH in an interior point of MATH, for each face MATH; this follows by taking logarithms and considering the decomposition of MATH into MATH. The required diffeomorphism is therefore given by the map MATH. |
math/0010025 | Since exponentiation is a diffeomorphism, MATH is a tubular neighbourhood of the embedding; each choice of basis for MATH therefore trivialises the normal bundle. |
math/0010025 | Following REF, we must identify MATH with MATH. Any choice of basis for the NAME algebra of MATH will have this effect; since the space of bases has two connected components, it actually suffices to give an orientation. But MATH is the kernel of an epimorphism MATH and MATH is canonically oriented, so it remains to orient the domain; since the latter is isomorphic to MATH, we may simply apply the standard orientation of MATH. The resulting isomorphism MATH provides the structure we seek. |
math/0010025 | The existence of MATH is assured by the fact that MATH and MATH agree on MATH; we write their common restriction as MATH. Our data implies that MATH maps diffeomorphically onto MATH under the projection of MATH onto MATH, so that MATH is isomorphic to MATH in MATH. Since MATH is given by MATH in MATH, it is equivariantly diffeomorphic to MATH with respect to the splitting of MATH; we write MATH for the subspace MATH. We also note that MATH splits as MATH. The stably complex structure on MATH given by factoring out the action of MATH on MATH therefore differs from that given by factoring out the action of MATH on MATH only by the trivial summand MATH, so that we may consider the latter as obtained from MATH by restriction. Similar remarks apply to MATH and MATH. To show that MATH respects the restricted complex structures, we then choose an isomorphism MATH; this immediately extends to an equivariant diffeomorphism MATH, and therefore to an equivariant diffeomorphism MATH which preserves the action of MATH. The differential of the second diffeomorphism is complex linear, and reduces to MATH on the quotient tangent bundles, as required. |
math/0010025 | The data yields a diffeomorphism MATH, with MATH; we then apply REF to MATH with MATH, and to MATH with MATH. |
math/0010025 | Considering MATH as a face of MATH, we note that MATH in MATH, that MATH in MATH, and that MATH in MATH by REF. Thus MATH, and MATH acts as MATH. It follows that the normal bundle MATH of MATH in MATH has total space MATH, and therefore that restriction and induction commute for MATH. Considering MATH as a face of MATH, we may no longer appeal to REF. Instead, we apply REF with MATH and MATH; we let MATH have complement an open tubular neighbourhood MATH of MATH, and MATH be any diffeomorphism extending the identity on MATH. The Lemma provides an equivariant diffeomorphism MATH between MATH and an open tubular neighbourhood of MATH in MATH. By construction, MATH is compatible with the stable complex structures induced by MATH and MATH respectively, and defines an isomorphism between MATH and the normal bundle MATH of MATH in MATH. This isomorphism is well-defined up to homotopy, and therefore confirms that restriction and induction commute for MATH. |
math/0010025 | The first statement follows from the proof of REF by choosing MATH, and letting MATH be a tubular neighbourhood of MATH; then MATH lies in MATH, and MATH is the corresponding coordinate line bundle in the trivial summand MATH. For the second statement, we note that the proof of REF identifies the total space of MATH with MATH. |
math/0010025 | Since MATH restricts to a preferred section for MATH, so MATH pulls MATH back to MATH. From REF , we deduce that MATH is isomorphic to MATH over MATH, and is trivial over the complement. But these properties characterise MATH. |
math/0010025 | Pulling REF back along MATH yields the complex structure on MATH, which REF converts to an isomorphism MATH. Different choices of preferred section MATH yield homotopic isomorphisms, and therefore homotopic stably complex structures, because the corresponding diffeomorphisms MATH are isotopic. By REF , the complex structure on MATH is given by an isomorphism MATH, which is defined uniquely up to homotopy. The structures restrict to MATH as claimed, by appeal to REF . |
math/0010025 | By definition, there is a canonical diffeomorphism between MATH and MATH, and a canonical isomorphism between MATH and MATH which preserves their respective actions on MATH and MATH. We obtain a diffeomorphism MATH which repects the induced stably complex structures, and the result therefore follows by pullback along inverse preferred sections, as in REF . |
math/0010025 | The first statement follows from our introductory remarks, in view of the cell decomposition defined above. Since there are MATH submanifolds MATH, but only MATH two-cells in the decomposition, NAME duality shows that there are MATH linear relations amongst the cobordism classes of the inclusions. |
math/0010025 | We first extend REF to MATH-fold products by induction. For MATH, the inductive hypothesis provides a combinatorial equivalence MATH where the MATH are products of simplices; iterating REF replaces the right-hand expression by MATH and applying REF yields MATH, where the MATH are products of simplices as required. As in REF, we may ensure that MATH corresponds to one of the original vertices of MATH (for which we retain the label MATH) under the resulting equivalence. Our connected sum is therefore combinatorially equivalent to MATH . Since none of the faces MATH contains the vertex MATH, they may be identified with the corresponding faces of MATH; in other words, REF is combinatorially equivalent to MATH, and the result follows by final appeal to REF. |
math/0010025 | Since the subtori MATH generate MATH, we may define an automorphism MATH of MATH by mapping MATH onto MATH, preserving orientation, for each MATH. We conclude by replacing MATH with MATH, and appealing to REF . |
math/0010025 | We consider the halfspace MATH of MATH given by MATH for some small MATH, and note that its inverse image under the projective transformation MATH is a halfspace MATH, which intersects MATH in a closed neighbourhood MATH of the vertex MATH. We then apply REF by choosing MATH, MATH, MATH, and MATH; we take MATH to be MATH. We obtain an equivariant diffeomorphism MATH of MATH into MATH (for some invariant neighbourhood MATH of MATH), which identifies the restrictions of the stably complex structures induced by MATH and MATH respectively. We then repeat the process for MATH, using the halfspace MATH, and obtain a corresponding diffeomorphism MATH. The images of MATH and MATH overlap in a collared MATH-sphere, and the proof is complete. |
math/0010025 | This is a direct consequence of REF , since the connected sum of any two stably complex structures is cobordant to their disjoint union. |
math/0010029 | Let MATH be a cycle representing the class MATH. There is a smooth (see CITE) curve MATH on MATH that contains the support of MATH. Hence it is enough to show that there is a homologically trivial cycle MATH on MATH so that MATH for every cycle MATH on MATH such that the image under MATH lies in MATH. Let MATH denote the graph of the inclusion MATH. Then clearly MATH but MATH is not homologically trivial. Choose a point MATH on MATH. Now, by a result of NAME (see CITE), for some positive integer MATH we have an expresssion in MATH where MATH is the diagonal and MATH is a cycle so that its cohomology class has non-zero NAME component only in MATH. In particular, MATH gives a map MATH which induces multiplication by MATH on MATH. Since MATH and MATH induce MATH on MATH it follows that the correspondence MATH induces multiplication by MATH on MATH. Now, MATH so we have an expression MATH . Let MATH. Then the cohomology class of MATH is REF. Moreover, the map MATH induced by MATH is zero. Thus, by the above propery of MATH we see that MATH for any MATH in MATH whose image lies in MATH. By NAME 's theorem (see CITE) the group MATH is divisible. Hence, we conclude the result. |
math/0010029 | By the functoriality of the NAME map we have a commutative diagram MATH . By projection we can replace the bottom right corner with MATH. Now we tensor this with the NAME map for MATH to obtain, MATH . The required commutative diagram now follows from the definition of MATH. |
math/0010029 | The following composite map is multiplication by REF MATH . By the divisibility of MATH for a surface MATH we see that the lemma follows from the following result. |
math/0010029 | Let MATH, MATH be points on MATH; we get points MATH and MATH of MATH. The image of MATH in MATH is MATH. Thus, we have an expression MATH . Now, any cycle MATH in MATH can be written as MATH. The NAME variety of MATH is MATH and the image of MATH under the NAME map is MATH. Thus, if the cycle is in MATH, then MATH. Now we combine this with the above expression to obtain MATH which proves the result. |
math/0010029 | By NAME 's result there are non-trivial classes in MATH. The Jacobian variety MATH is spanned by decomposable elements. Moreover MATH. Thus there is a pair of elements of MATH of the form MATH, MATH such that the image of their tensor product in MATH is non-zero. By a result of NAME, for any fixed class MATH in MATH, the collection MATH forms a countable union of abelian subvarieties of MATH. Since MATH does not lie in MATH, the latter is a proper subgroup of MATH. Since MATH is assumed to be simple this is forced to be a countable set. In particular, there is an element of the form MATH which is not in MATH; so that the product of this with MATH is non-zero in MATH. But we just saw that all such elements are mapped to REF in MATH. |
math/0010030 | Consider the first identity. By definition both sides are degree MATH super-derivations on MATH so it suffices to check that they agree on generators. Clearly, both sides give MATH when evaluated on MATH and for MATH we have MATH A similar argument proves the second identity. |
math/0010030 | Clearly MATH when MATH or MATH would be a vertex-idempotent whence in MATH. Let MATH be the starting point of MATH and MATH the end point of MATH and assume that MATH, then MATH from which the assertion follows. |
math/0010030 | Define the NAME derivation MATH on MATH by the rules that MATH . By induction on the length MATH of an oriented path MATH in the quiver MATH one easily verifies that MATH. By induction one can also proof that MATH. This implies that MATH is a bijection on each MATH, where MATH and on MATH, MATH has V as its kernel. By applying the NAME homotopy formula for MATH, we obtain that the complex is acyclic. |
math/0010030 | Let MATH be the MATH-space spanned by all necklace words MATH in MATH and define a linear map MATH for all oriented paths MATH in the quiver MATH, where MATH is the necklace word in MATH determined by the oriented cycle MATH. Because MATH it follows that the commutator subspace MATH belongs to the kernel of this map. Conversely, let MATH be in the kernel where MATH is a linear combination of non-cyclic paths and MATH for MATH is a linear combination of cyclic paths mapping to the same necklace word MATH, then MATH for all MATH. Clearly, MATH as we can write every noncyclic path MATH as a commutator. If MATH with MATH, then MATH and MATH for some paths MATH whence MATH is a commutator. But then, MATH is a sum of a commutator and a linear combination of strictly fewer elements. By induction, this shows that MATH. |
math/0010030 | If MATH is not a cycle, then MATH and so vanishes in MATH so we only have to consider terms MATH with MATH an oriented cycle in MATH. For any three paths MATH and MATH in MATH we have the equality MATH whence in MATH we have relations allowing to reduce the length of the differential part MATH so MATH is spanned by terms of the form MATH with MATH and MATH an oriented cycle in MATH. Therefore, we have a surjection MATH . By construction, it is clear that MATH lies in the kernel of this map and using an argument as in the lemma above one shows also the converse inclusion. |
math/0010030 | We know that MATH. By a result of CITE one knows that the geometric points of the quotient scheme MATH are the isomorphism classes of MATH-dimensional semi-simple representations of MATH. Moreover, the MATH-stabilizer of a point in MATH is trivial if and only if it determines a simple MATH-dimensional representation of MATH. The result follows from this and the results recalled above. |
math/0010030 | Assume that MATH is a point of semi-simple representation type MATH, that is, MATH and MATH a simple MATH-representation. We have MATH . But then, the dimension of MATH is equal to MATH from which it follows immediately that MATH . On the other hand, as MATH we know that MATH . But then, equality occurs if and only if MATH, that is, MATH or MATH is a simple MATH-dimensional representation of MATH. |
math/0010030 | As MATH we know that MATH has dimension MATH which is equal to MATH. By the hyper-Kähler correspondence so is the dimension of MATH, whence the open subset of MATH consisting of MATH-semistable representations has dimension MATH as there are MATH-stable representations in it (again via the hyper-Kähler correspondence). Take a MATH-closed orbit MATH in this open set. That is, MATH is the direct sum of MATH-stable subrepresentations MATH with MATH a MATH-stable representation of MATH of dimension vector MATH occurring in MATH with multiplicity MATH whence MATH. Again, the normal space in MATH to MATH can be identified with MATH. As all MATH are MATH-representations we can determine this space by the knowledge of all MATH. MATH . But then the dimension of the normal space to the orbit is MATH . By the Luna slice theorem CITE, the étale local structure in the smooth point MATH is of the form MATH where MATH and is therefore of dimension MATH . This number must be equal to the dimension of the subvariety of MATH-semistable representations of MATH which has dimension MATH if and only if MATH and MATH, that is if and only if MATH is MATH-stable. Hence, if MATH is smooth, then MATH must be a minimal non-zero vector in the set of dimension vectors of MATH-stable representations of MATH and hence by the hyper-Kähler correspondence, MATH is a minimal non-zero vector in MATH. Conversely, if MATH is a minimal vector in MATH, then MATH is a coadjoint orbit, whence smooth and hence so is MATH by the correspondence. Moreover, all MATH-dimensional MATH-semistable representations must be MATH-stable by the minimality assumption and so MATH is a principal MATH-fibration over MATH whence smooth. Therefore, MATH is a sheaf of MATH-Cayley smooth algebras. |
math/0010030 | As MATH we know that the local quiver MATH in a simple representation MATH corresponding to MATH is a one vertex quiver having MATH loops. That is, MATH . But then, for MATH a point corresponding to MATH, the local quiver is still MATH but this time the local dimension vector MATH. If MATH lies in the smooth locus, then by the Luna slice theorem we must have MATH . The left hand side is MATH whereas the right hand side is equal to (because MATH) MATH, a contradiction. |
math/0010031 | For a fixed REF-PS of MATH, there are two possibilities in the previous proposition: either all the MATH's vanish for MATH and we are done, or it is not so. Assume that all MATH for MATH. Because the sum MATH, it follows that there must exist a MATH. We know that MATH and therefore the restriction MATH. Because MATH is a basis of MATH, one can write MATH as a non-zero linear combination of these vectors. Now all we have to do is to replace a section from the set MATH which appears in this linear combination with MATH. |
math/0010031 | The line bundle MATH is again very ample and its associated linear system gives an embedding MATH . The MATH-action on MATH induces one on MATH. For a stable map MATH which is MATH-semi-stable and MATH a REF-PS of MATH, REF ensures the existence of sections MATH whose restrictions to MATH give a basis of MATH; in particular, they are linearly independent. Because of the direct sum decomposition MATH these sections can be completed with sections MATH to a basis of MATH. This basis defines coordinates on MATH in which the MATH-action is diagonal. Claim There exists an irreducible component MATH of MATH having the property that among MATH there are two sections MATH and MATH such that MATH and MATH and their restriction to MATH is non-zero. We know already that there are two sections MATH and MATH such that MATH and MATH, and their restriction to MATH is nonzero. Let MATH and MATH respectively two irreducible components of MATH on which these two sections do not vanish. Because MATH is connected, there is a chain of irreducible components MATH connecting these two components. Since MATH is a basis of MATH and MATH is very ample, it follows that there are sections MATH with the property that: MATH does not vanish at a (certain) point in MATH, MATH does not vanish at a (certain) point in MATH does not vanish at a (certain) point in MATH. Notice that MATH does not vanish on MATH and MATH, MATH does not vanish on MATH and MATH and so on. Let MATH denote the weights of the sections MATH respectively. If MATH, then the component MATH satisfies the requirement of the claim. If it is not the case, we look at the chain MATH whose length is one less than the length of MATH. Because at the end MATH the weight MATH is positive, an induction argument on the length of the connecting chain shows that it must exist an irreducible component MATH of the chain MATH having the property of the claim. Let us look now at the image of a point MATH inside MATH: a representative MATH of it will have non-zero coordinates with both positive and negative weights (for the MATH action), so MATH is in the MATH-semi-stable locus of MATH. Since obviously MATH, we deduce that MATH is in the MATH-semi-stable locus of MATH. |
math/0010031 | By the NAME criterion, MATH . Since MATH is assumed irreducible, REF implies that for any REF-PS MATH of MATH, the image of MATH intersects the MATH-unstable locus of MATH in finitely many points; denote by MATH the NAME open subset of MATH consisting of points which are mapped by MATH into the MATH-semi-stable locus of MATH. Because in a torus there are countably many one-parameter subgroups, MATH . |
math/0010031 | The geometric invariant quotient MATH is a projective subvariety of MATH. This last geometric quotient can be described as MATH . Let's denote by MATH the quotient map. There is an invertible sheaf MATH such that MATH for some MATH. It has the additional property that for large enough values of MATH, MATH . The assumption that the image of the stable map is contained in the semi-stable locus of MATH implies the existence of the commutative diagram MATH . The map MATH is still stable. Indeed, problems appear only if MATH contracts some MATH-components, without enough special points on them, which are not contracted by MATH. If MATH denotes such a MATH-component of MATH, MATH so that MATH must be constant on MATH. This contradicts the stability of MATH. The group MATH acts on MATH trivially on the first factor and consequently its invariant quotient is MATH. The quotient map MATH is just MATH. Let us define the line bundle MATH . It has the property that MATH and it can be easily checked that MATH for large MATH. There is again a commutative diagram MATH . Because MATH is an embedding, MATH is also. The MATH-dimensional subvariety MATH of MATH has NAME polynomial: MATH where MATH is the NAME polynomial of MATH. It is independent of MATH satisfying the hypothesis of the proposition. NAME proves in CITE that there is an integer MATH such that for all MATH, MATH is generated by its global sections and moreover, for any closed subscheme MATH of MATH whose NAME polynomial is MATH there is an epimorphism MATH . Recall that for obtaining a projective embedding of MATH we had to chose a high enough power MATH. Since MATH, we can chose from the very beginning an integer MATH large enough such that MATH with MATH (MATH). The following three relations MATH prove that there are MATH-invariant sections MATH such that the restrictions MATH form a basis of MATH. The problem with the marked points is easy: by REF and we may consider their images MATH. The number MATH was chosen large enough to ensure that MATH is globally generated by its sections. Consequently, we find MATH such that MATH for MATH. Since MATH, there are MATH-invariant sections MATH such that MATH for MATH. The MATH sections MATH now obviously satisfy the conditions of REF . |
math/0010031 | There are two things to prove in this statement: the first one is that the stabilizer of MATH in MATH is finite, and the second one is that this point is indeed stable. When MATH, for any MATH we have MATH, and therefore the stabilizer of the map is indeed finite. Let us prove that it is so in general. Consider a representative MATH of the point MATH and define MATH . Let's assume that MATH is not finite. By definition, for any MATH, there is an automorphism MATH having the property that MATH. In particular, for all MATH, Image-MATH. For MATH an arbitrary point, MATH has finite stabilizer in MATH by assumption and therefore dim-MATH. Since MATH which is one dimensional, we deduce that dim-MATH. Let us look at the connected component of the identity MATH of MATH: it is a connected MATH-dimensional group and therefore isomorphic either to the multiplicative group MATH or to the additive group MATH. In both cases MATH will be fixed. This contradicts the assumption that Image-MATH. We will show that the point MATH is MATH-stable using REF . For a REF-PS MATH, there are sections MATH and MATH with MATH, and whose restriction to MATH is non-zero (this is because Image MATH). Using now REF, we complete MATH and MATH with MATH-invariant sections MATH in order to fulfill the requirements of REF . |
math/0010031 | We may assume as usual that MATH. Recall that a map MATH is stable if and only if MATH is ample. Also, there is an integer MATH with the property that MATH is very ample. In this way, any stable map MATH gave rise to an embedding MATH into a space isomorphic to MATH, where MATH. The ambiguity in the choice of this isomorphism is given by elements in MATH. In order to define the space MATH we shall use a fixed but otherwise arbitrary stable map MATH. Let us consider the fixed embedding MATH defined by the very ample line bundle MATH and let MATH be its degree on MATH. For another stable map MATH, the NAME polynomial of its graph MATH is MATH . So each stable map MATH defines a point MATH. Using the embedding MATH, the graph MATH can be viewed as a subvariety of MATH and its NAME polynomial with respect to the very ample line bundle MATH is MATH also. Clearly, the same is true for any closed subscheme MATH of MATH: the NAME polynomial inside MATH with respect to MATH is the same as the NAME polynomial of its image MATH inside MATH with respect to MATH. So we obtain a closed immersion MATH . The ample line on MATH is det-MATH, where MATH is the universal quotient bundle on some NAME variety (see REF). Remember that there is a subscheme MATH of MATH corresponding to the locus of MATH-marked, genus MATH stable maps to MATH which represent MATH times the generator of MATH. Let consider the commuting diagram MATH . Since MATH is an immersion, MATH is a quasi-projective scheme. Its ample line bundle is determined by the restriction to MATH of det-MATH. The space MATH is quasi-projective, being an open subset of MATH. |
math/0010031 | Because MATH are all MATH-invariant, it follows that MATH is also. For proving that MATH is a moment map, we need to show that its differential is the same as the contraction of the NAME form MATH on MATH with the vector field MATH. It what follows, the symbol MATH will always denote the contraction of a differential form with a vector field. The contraction MATH because the MATH-component of MATH is zero. MATH . MATH . At a point MATH, MATH and consequently MATH . MATH . For writing equality MATH, we have used that MATH . MATH . All together, these equalities show that MATH is indeed a moment map. |
math/0010031 | We claim that for MATH there is a point MATH with the property that MATH that is, MATH. Suppose that it is not the case, so either Image-MATH or Image-MATH. Let assume that we are in the first case. At a point MATH, MATH . Recall that the smooth real-valued function MATH defined on MATH is the ``quotient" MATH, where MATH denotes the curvature of the tangent bundle of MATH with respect to the NAME form MATH. This last form was defined in terms of a fixed projective embedding of MATH; in particular, it does not depend on MATH. For obtaining the projective embedding of the NAME scheme we had to take large positive integral values for MATH and therefore we may assume that MATH is large enough for MATH to be a strictly positive function on MATH. Notice also that since MATH is holomorphic and MATH is a positive MATH-form, the MATH-form MATH on MATH is still positive. It becomes now clear that if Image-MATH, MATH will be negative also. This contradicts the choice of MATH in the zero locus of MATH. The lemma follows now because, by assumption, MATH acts with finite stabilizers on MATH. |
math/0010031 | According to the NAME criterion, is sufficient to prove the statement for every REF-PS MATH. For a fixed REF-PS MATH of MATH, the point MATH is MATH-stable, if its MATH-orbit meets the zero-level set of the moment map MATH on MATH. Assume, for instance, that MATH. By hypothesis, Image-MATH, so that under the MATH-action all the points MATH meet the MATH-level; consequently MATH for all MATH and MATH. For such a large MATH, the translated map MATH will have the property that MATH. A continuity argument proves that there is (a unique) MATH such that MATH. Now, by REF , MATH has also finite stabilizer. |
math/0010031 | REF follow immediately from the irreducibility of MATH and from the initial assumption that the class MATH is can be represented by a morphism MATH. CASE: If both MATH and MATH are convex, MATH and MATH are open and dense (see REF , page REF); convexity implies also that we are working in the expected dimension. The only thing to check is REF : this follows from the fact that the evaluation map at the MATH marked point on MATH is submersive, and therefore any map MATH can be `pulled away' from the unstable locus of MATH. For both (iia) and (iib) we must prove that MATH is still convex. In the first case, we use the exact sequence MATH on MATH associated to the principal bundle MATH. In the second case we notice that for a morphism MATH, MATH is a principal MATH-bundle which is topologically trivial since MATH is simply connected; a result due to NAME says that in this case MATH is in fact holomorphically trivial. This implies that the morphism MATH can be lifted to a morphism MATH. The conclusion follows now from the convexity of MATH and the exact sequence MATH on MATH, where MATH denoted the trivial sub-bundle generated by the infinitesimal action of MATH. |
math/0010031 | I start noticing that MATH is generically injective on its image. Indeed, by REF , we may restrict our attention to maps whose image is contained in MATH: if MATH are morphisms such that MATH, it follows from REF that there is a morphism MATH such that MATH for all MATH. Since MATH is affine, the morphism MATH must be constant and therefore MATH and MATH represent the same point in MATH. Write MATH as the union of its irreducible components. REF says that each component MATH contains a non-empty open subset MATH having the property that its points represent stable maps defined on MATH, with image completely contained in the MATH-stable locus of MATH. We know already that the maps MATH are birational on their image; let us denote MATH the closures of these images. They are distinct irreducible components of MATH, each of them having expected dimension. In fact, for MATH the composite MATH is a smooth point of MATH; this can be seen pulling back by MATH the exact sequence REF. For proving that MATH is dominant, we have to show that if MATH is an irreducible component of MATH, then MATH is one of the MATH's comming from MATH. This is the place where we are using REF which says that we may restrict ourselves to morphisms MATH. The topological type of MATH is determined by the class MATH, and we assumed that MATH can be represented by a morphism MATH. Consequently, for any morphism MATH representing the class MATH, the principal bundle MATH is topologically trivial and therefore holomorphically trivial by the same result of NAME. |
math/0010031 | We have seen in REF that MATH and MATH . Moreover, MATH are birational for all MATH. Let MATH and MATH be respectively the images of the MATH-point evaluation maps on MATH and MATH. Then MATH is the closure of the image of the evaluation map on MATH and there is a natural map MATH compatible with the other arrows in REF which is dominant. The class MATH which appears in the statement of the theorem is just the class of a ``rational section" of the quotient MATH, that is MATH for a general complete intersection MATH which transversally intersects, in MATH points, the closures of the general MATH-orbits in MATH. With this choice for MATH, the rational map MATH is generically finite of degree MATH. Consider MATH and let MATH for MATH. The discussion preceding this proof applied to the composite MATH shows that MATH . We can write therefore, MATH . This finishes the case when MATH. When MATH (that is MATH is not generically finite), both sides are zero. Summing these equalities after MATH we get the conclusion. |
math/0010031 | It is clear that REF just extends the action of the real gauges by composition on the right. We shall prove the formula for the action of MATH on MATH searching a connection MATH on MATH which makes MATH-holomorphic the map MATH. For doing computations we use a local trivialization of MATH, so that MATH itself may be assumed trivial (as long as the objects found in the end are globally defined). In what follows, MATH denotes a point on MATH. By REF and MATH, for some MATH. I want to find a connection MATH on MATH such that MATH. Since MATH is MATH-equivariant, it is enough to check this condition at points MATH. Because MATH is MATH-holomorphic, MATH for MATH, or equivalently MATH . REF implies that MATH and MATH . For MATH to be MATH-holomorphic it is necessary and sufficient that the difference of these two quantities is zero. Imposing this condition, we find MATH . It remains to separate the MATH and MATH components of the last line. MATH . Inserting this into the previous relation, we obtain MATH . For MATH defined by MATH the last equality is satisfied. Notice that in general this is the only possible choice for MATH since the vectors MATH and MATH are linearly independent in most cases. Using local normal coordinates on MATH, it follows that for any MATH-form MATH, MATH. |
math/0010031 | Clearly, we may assume that MATH. The uniqueness part is immediate. For the existence part, notice that if MATH is exact, that is, MATH for MATH, then MATH does the job. Homotopy classes of maps MATH are parameterized by MATH, so for MATH there exists MATH such MATH and MATH. By the discussion above, there exists MATH such that MATH and MATH. Now MATH will be convenient. |
math/0010031 | The curvature of MATH is MATH. For MATH and MATH, MATH, and MATH evaluates zero on other pairs of vectors. It is easy to see that for any MATH-form MATH on MATH and any tangent vector MATH to MATH, MATH. This implies that the connection MATH defines indeed a holomorphic structure in MATH because its curvature is a MATH-form on MATH. |
math/0010031 | REF implies that the formula above is indeed an action. It is also holomorphic because MATH preserves the connection MATH; indeed, MATH . CASE: The statement is a direct consequence of REF. |
math/0010031 | CASE: MATH has a holomorphic structure because MATH is a holomorphic bundle, according to REF . That it is also a projective variety follows from the fact that the NAME torus is projective. REF implies that given MATH there is a unique MATH (depending on MATH) such that MATH . The composed map MATH is MATH-equivariant and therefore defines MATH . Since MATH is MATH-holomorphic, it follows that this map is holomorphic. If MATH (or in MATH) denotes a point lying over MATH, the explicit formula for MATH is MATH where the square brackets denote obvious equivalence classes. Suppose that MATH for a map MATH. Then from the commutative diagram MATH we can see that the homology class MATH depends only on the MATH-equivariant class MATH and also that MATH. CASE: Consider now MATH and MATH, MATH with MATH and MATH. According to REF, MATH, so MATH . Notice that MATH is actually an integral MATH-valued harmonic form and according to REF there is a unique MATH such that MATH. I claim that MATH, that is, that MATH. Indeed, MATH . We are going to check now that MATH and MATH induce the same (holomorphic) map MATH. Using REF , MATH reads MATH . This finishes the proof of the lemma. |
math/0010031 | Consider a MATH-holomorphic, MATH-equivariant map MATH together with MATH marked points MATH and let MATH. REF says that this data induces a morphism MATH and moreover, it does not depend on the MATH-orbit of MATH. So we get a map MATH . This map is clearly MATH-invariant and therefore descends to the quotient MATH . Because the composition MATH, for some MATH, the map MATH is in fact a representative for the corresponding stable map (see REF ). The map MATH is clearly surjective: given a point MATH, consider the diagram MATH . The composed map MATH will be a MATH-equivariant, MATH-holomorphic map, for MATH (see REF for the definition of MATH). As marked points in MATH, one may take any MATH lying over MATH. We have to prove that MATH is injective. Consider MATH and MATH which induce the same morphism MATH. It follows from REF that necessarily MATH, so that MATH. REF implies that MATH and MATH are in the same MATH-orbit and consequently MATH and MATH are also in the same MATH-orbit. Since MATH is gauge invariant, we may assume that MATH and even that MATH is a harmonic form. The problem is reduced to the following: two maps MATH and MATH which define the same MATH must be equal. REF says that MATH . A moment's thought shows that this imply MATH and MATH. |
math/0010031 | Notice that in the diagram MATH. Indeed, for MATH there is a unique MATH such that MATH, so we may identify MATH. Consequently, MATH and we obtain a MATH-equivariant map MATH which covers the identity of MATH. This one must be an isomorphism. |
math/0010031 | Let me start explaining the guiding idea: we have learned in REF that we should transfer the integrals on MATH used for defining NAME invariants of MATH to integrals on MATH because MATH. All the evaluation maps involved for making these computations live on the right-hand-side of REF. On the other hand, the invariant MATH is defined using equivariant cohomology classes which are pulled-back to MATH by maps which live on the left-hand-side of REF. So, loosely speaking, what we have to do is to pass from the left to the right in REF. For making this passage, we need to understand the relationship between the various group actions and evaluation maps which appear in the context. On MATH there are two commuting actions. First, MATH acts by constant complex gauges MATH . The map MATH is MATH-invariant for this action, while MATH is MATH-equivariant for the diagonal right action of MATH on the MATH-factor of MATH. Secondly, MATH acts on MATH with quotient MATH: MATH . The evaluation map MATH is MATH-equivariant for the MATH-action on MATH on both terms. Convention In what follows, the symbol MATH will denote homotopy equivalence and the letter MATH obvious inclusions. For understanding better the forthcoming calculations, we should keep in mind that for integration purposes homotopy equivalent spaces are equal. MATH . The map MATH being MATH-equivariant and MATH-invariant, induces MATH . On the other hand, MATH . The reason for the last equality is that MATH acts on MATH only and therefore the MATH-action on MATH is trivial. The class MATH we start with lives in MATH; pulling it back using MATH, we get the class MATH. Relation REF implies that MATH and we deduce from REF that MATH. The invariant MATH can therefore be defined as MATH where the relevant maps fit in the diagram MATH . From the diagram MATH we see that MATH, where MATH is by definition the cohomology class on MATH determined by MATH. REF says that the subset MATH of morphisms with image contained in MATH is dense. We deduce, going the other way round in REF, that MATH . At this point we have finally moved from the left-hand-side to the right-hand-side of REF as we wished. Notice that since the maps involved in the computations are rational, the pull-backs are defined as in REF. The composition of morphisms MATH representing the class MATH with the projection MATH is of the form MATH because MATH induces the class MATH. Consequently, a map MATH defines a map MATH which is the identity on the first component. The map MATH is birational. REF implies that MATH is dominant if every morphism MATH representing MATH is in its image. From the diagram MATH we deduce that the pull-back MATH is a topologically trivial, holomorphic principal bundle (this follows from the assumption that MATH is induced by a class MATH); it is therefore isomorphic to MATH for a certain MATH. This data induces the map MATH which represents the class MATH and also MATH. Now let us prove that MATH is generically injective. Using REF , we may restrict ourselves to the (dense) open subset MATH representing morphisms whose image is contained in MATH. Let us assume that MATH are such that MATH. Then MATH and REF implies that MATH. Then MATH induce the same map to MATH and consequently for any MATH, MATH for a unique MATH. The morphism MATH must be constant, so MATH and they define the same point in MATH. Since MATH has trivial automorphism group, MATH . Finally, from the diagram MATH we conclude that the invariant MATH does coincide with a NAME invariant of MATH. Indeed, MATH . |
math/0010033 | The first statement is well known. To prove the second part of the lemma we have to show that there exists a ray with infinite diameter in any graph MATH that does not contain a star ball. If there is no star ball in MATH then the complement of any vertex MATH in MATH must contain a connected component MATH with infinite diameter. At least one of the components in the ball MATH which are contained in MATH must have infinite diameter. Otherwise MATH would be a star ball. We call this component MATH. By induction we obtain a strictly decreasing sequence MATH of components of MATH with infinite diameter. In the proof of REF we will see that there must exist a ray that has infinitely many vertices in common with every set MATH. Since the ray is not contained in any ball it must have infinite diameter. |
math/0010033 | For two sets MATH and MATH in MATH we have MATH . MATH is a vertex-cut and MATH the set of ends whose rays are completely contained in MATH and MATH from some index onwards. Thus MATH, and finally MATH . |
math/0010033 | As MATH is connected and a superior set of MATH any vertex MATH in MATH can be connected by a path MATH of vertices in MATH with a vertex MATH in MATH. By induction we get a path MATH such that the initial vertex of MATH is adjacent to the last vertex of MATH and MATH is contained in MATH for every natural MATH greater than one. The union MATH of these paths must have finite intersection with every ball with center MATH (eq. every bounded subset of MATH). Thus MATH is infinite and therefore it constitutes a ray. The corresponding end lies in all cuts MATH. |
math/0010033 | Under the given preliminaries there exists a vertex MATH in MATH with infinite degree, which is adjacent to infinitely many connected components of MATH containing elements of MATH. Every element MATH of the base MATH which contains MATH must contain almost all of these components, since the vertex-boundary of MATH is finite. The same holds for every neighbourhood of MATH. Thus MATH is an accumulation point of MATH. |
math/0010033 | Let MATH be a sequence of pairwise different elements of MATH. If MATH is the only vertex-cut, then every element of MATH is an accumulation point of the sequence. Otherwise let MATH be a vertex in the complement of a vertex-cut MATH, in which lie infinitely many elements of MATH. If the sequence has no accumulation point in MATH, then by REF there exists a connected component MATH of MATH, such that infinitely many elements of MATH lie in MATH. In the case that MATH has no accumulation point in MATH, we construct a connected vertex-cut MATH, that is a subset of MATH and contains infinitely many elements of MATH: For a vertex MATH in MATH with infinite degree, which is no accumulation point of MATH, there exists a vertex-cut MATH, that contains MATH and for which MATH contains only finitely many elements of MATH. Almost all vertices in MATH, that are neighbours of MATH, lie in MATH. Thus MATH is a cut, such that infinitely many elements of the sequence MATH lie in it. This and the finiteness of MATH imply that MATH is a vertex-cut, which is a subset of MATH and contains almost all elements of MATH. Thus MATH is also a vertex-cut. It is a subset of MATH. If MATH has no accumulation point in MATH, once more by REF , there exists a connected component MATH of MATH, that has again the properties requested before. By induction we obtain a strictly decreasing sequence MATH of connected vertex-cuts, in all of which lie infinitely many elements of the sequence MATH. Since MATH is a subset of MATH and the vertex MATH lies in the complement of MATH we have MATH and thus MATH . Following REF we obtain a ray MATH, whose end MATH lies in every cut MATH. Every neighbourhood MATH of MATH contains a base element MATH with MATH, for which MATH is a subset of one of the balls MATH. This implies, that almost all cuts MATH and thus infinitely many elements of the sequence MATH lie in any neighbourhood of MATH. |
math/0010033 | We may assume that an open cover MATH consists of base elements. Given a set MATH of the cover, we choose a vertex MATH in MATH. Since MATH is finite there exists a finite subcover MATH of MATH in MATH. The set MATH of vertices in MATH is a vertex-cut. Every ray in MATH must lie in one of those vertex-cuts MATH for which MATH is an element of the cover MATH. The same holds for any end containing this ray and therefore MATH . By induction we obtain a sequence of vertex-cuts MATH and a corresponding finite subcover MATH of MATH, such that MATH is a subset of MATH. In other words: MATH . For an end MATH and a vertex MATH we define MATH, the distance of MATH to MATH with respect to the cover MATH, as the minimal radius MATH for which there exists a base element MATH in MATH such that MATH is a subset of MATH. Note that this radius exists for all vertex-ends. Let MATH be a base element in MATH. If the vertex boundary of MATH is contained in MATH, then for a MATH in MATH all connected components of MATH which are adjacent to MATH are completely contained in MATH. Let MATH denote the set of elements of MATH whose vertex-boundaries are subsets of MATH. We now choose a finite subcover MATH of MATH which covers MATH. Now MATH covers all ends MATH with MATH and the union MATH is a countable covering of both MATH and MATH which meets the statement of the lemma. |
math/0010033 | A sequentially compact NAME space is compact. Thus we can conclude from REF , that MATH is compact. It remains to show that at least one of two elements MATH and MATH of MATH has an open environment, in which the other element is not contained. We distinguish: CASE: One of the two elements is a vertex. If MATH is a vertex, then MATH is the requested neighbourhood of MATH. CASE: MATH and MATH are ends. In the complement of a finite set of vertices that separates MATH and MATH there exist two disjoint base elements, of which one contains the end MATH and the other MATH. |
math/0010033 | We define the MATH-image of an end in MATH as the end in MATH that contains it. The preimage of every base element of the edge-topology is open in the vertex-topology. |
math/0010033 | The statement is a consequence of REF . |
math/0010033 | Let MATH and MATH be open and closed in their corresponding topologies. They can be represented as a finite union of elements of the base, because they are open and compact. Hence MATH is an edge-cut and MATH is a vertex-cut. The same argument holds for MATH and therefore MATH is a vertex-cut, too. Thus MATH is an edge-cut. An end in MATH or MATH must lie in MATH or MATH, respectively, because MATH and MATH are open. On the other hand every end lying in MATH or MATH must be an element of MATH or MATH, respectively. Otherwise the complements of MATH and MATH would not be open. |
math/0010033 | For any closed set MATH not containing some point MATH we can find a neighbourhood MATH of MATH, which is element of the base and contained in the complement of MATH. As the base elements in MATH are open and closed the indicator function on MATH is continuous. Thus the proper metric end topology is MATH or completely regular. NAME is also an immediate consequence of the fact that elements of the base are open and closed. It is easy to see that this topology is NAME which now implies that it is NAME and totally disconnected. Let MATH be a locally countable graph. For some given vertex MATH let MATH denote the set of all base elements MATH such that MATH is a connected component of MATH for some natural MATH. MATH is countable. Any element of an open cover MATH containing a proper metric end must contain an element of MATH. Thus we can find a countable subcover of MATH of the set of proper metric ends. The set of vertices is countable anyway and hence the proper metric end topology is NAME. Every regular NAME space is normal. For any vertex MATH with infinite degree MATH is an open cover which is not locally finite in MATH and therefore the proper metric end topology is not paracompact. |
math/0010033 | Let MATH be an open cover of base elements. It must contain sets MATH and MATH such that MATH is a locally complete set and MATH is a global star-set. After removing these base elements from MATH we obtain a set MATH such that MATH and MATH. MATH is a set of vertices, that contains only finitely many elements of any ball in MATH. Thus MATH is countable and we can find a countable subcover MATH of MATH in MATH. We now copy the ideas in the second part of the proof of REF to construct a countable subset of MATH which covers MATH. Let MATH again denote the distance of the end MATH to the vertex MATH with respect to the cover MATH. As MATH contains no star-boundary, it has the property that for every metric cut MATH we can find a union MATH of finitely many components of MATH, such that MATH has a finite diameter. Thus there are only finitely many connected components in MATH for any natural number MATH, we can find a finite subcover MATH of MATH covering all ends with MATH, that lie in MATH. Thus MATH is a countably finite subcover of MATH covering MATH. |
math/0010033 | To show that every sequence in MATH has an accumulation point we can assume that it contains an infinite partial sequence MATH consisting of pairwise different elements. Otherwise the existence of an accumulation point is immediate. If MATH contains a bounded subsequence of vertices, the local end is an accumulation point. In the other case we choose a vertex MATH and distinguish between two cases. CASE: For every natural MATH there exists a sequence MATH of components of MATH such that MATH contains infinitely many elements of the sequence MATH and is a superset of MATH. Following the idea of REF we now can construct a metric ray, that lies in all cuts MATH. Its end is an accumulation point of MATH. CASE: There exists a ball MATH such that only finitely many elements of MATH lie in every component of MATH. Let MATH be a base element that contains the star-end MATH, which means that MATH must be a global star-set. By the definition of a global star-set there exists a star-annulus MATH of MATH such that MATH has a finite diameter. The set MATH must contain all but finitely many elements of MATH, but as MATH contains no bounded subsequence, only finitely many elements of the sequence MATH lie in MATH. Now MATH must contain all but finitely many elements of MATH, which means that MATH in an accumulation point of MATH. We have shown that MATH is sequentially compact. By REF it is also a NAME space and hence it is compact. The other statements of REF now follow easily by the above considerations. |
math/0010033 | MATH implies MATH . Let MATH be a quasi-inverse to MATH. If MATH is infinite, then, by REF , this also holds for MATH. Now MATH implies that MATH has no upper bound. |
math/0010033 | By connecting the MATH-images of adjacent vertices of a metric ray MATH in MATH with geodesic paths with lengths, that are smaller or equal MATH, we obtain a path MATH in MATH. Its diameter is infinite by REF . If it had infinite subset MATH with finite diameter, we also could find infinitely many elements in MATH that lie in MATH, contradicting REF and the assumption that MATH is a metric ray. As a graph is locally finite if and only if every bounded set of vertices is finite, MATH is a locally finite subgraph of MATH. REF implies that it must contain a ray which we denote with MATH. By the above consideration it is also a metric ray in MATH. Thus MATH maps metric rays in MATH onto metric rays in MATH. Let MATH and MATH be metric equivalent rays in MATH. If MATH and MATH were not metrically equivalent, we could find disjoint metric cuts MATH and MATH, such that MATH lies in MATH and MATH lies in MATH. Again by REF , MATH must have a finite diameter and hence the rays MATH and MATH cannot lie in both cuts MATH and MATH in contradiction to the assumption that they are metrically equivalent. Thus we can say that metric equivalence is an invariance under MATH on sets of metric rays. Now for every end MATH in MATH we define MATH as the unique end in MATH which contains the MATH-images of the elements of MATH and set MATH for every vertex MATH. By the above invariance of the metric equivalence and the construction of MATH we obtain MATH for every metric cut MATH in MATH, and therefore by REF MATH and MATH for every metric cut MATH in MATH. Now let MATH be a proper metric end in MATH. Then of course MATH. If MATH did not lie in MATH then there would exist infinitely many vertices MATH in a ray MATH of MATH which are not elements of MATH. Their MATH-images would not be elements of MATH, MATH would not lie in MATH and MATH would not element of MATH. Hence MATH and MATH . For every base element MATH in MATH we obtain MATH and MATH is continuous. To prove the third property of MATH, it now suffices to show that the restriction MATH is a bijection of the sets of ends in MATH and MATH. To a given end MATH in MATH we can find a decreasing sequence MATH of connected metric cuts in MATH such that MATH and MATH. Now we choose an increasing sequence MATH of concentric balls in MATH such that MATH is contained in MATH. To every natural MATH there exists a connected component MATH of MATH which contains all but finitely many vertices of the MATH-preimage of a ray MATH of MATH. Copying the idea of REF we can find a metric ray MATH that lies in all cuts MATH. The end MATH of MATH is mapped onto MATH under MATH. Thus MATH is surjective. Let MATH and MATH be two different ends of the graph MATH and MATH and MATH two disjoint metric cuts such that MATH lies in MATH and MATH lies in MATH. Now MATH and MATH must be disjoint too, and therefore MATH is injective. To complete the proof of the theorem, it remains to show that MATH is the unique extension of MATH with the requested properties. We assume that there exists another extension MATH which has these three properties but does not equal MATH. Let MATH be an end in MATH and MATH and MATH for two different ends MATH and MATH in MATH. Now we choose again two disjoint metric cuts MATH and MATH such that MATH lies in MATH and MATH lies in MATH. As MATH and MATH, the preimages MATH and MATH are disjoint metric cuts in MATH. We know that MATH must lie in MATH and therefore MATH must lie in MATH. This implies that every open neighbourhood of MATH has a nonempty intersection with MATH. In other words, no open neighbourhood of MATH is completely contained in MATH and MATH is not continuous. |
math/0010033 | Let MATH be an element of MATH. As MATH the vertex MATH must not be contained in MATH, so MATH and MATH. Hence MATH . Now let MATH be an element of MATH which is adjacent to MATH and choose an element MATH of MATH such that MATH . For a MATH in MATH with MATH we now have MATH . By the boundedness of MATH we have MATH and by the quasi-injectivity of MATH we get MATH which implies MATH . In other words MATH . As MATH is a subset of MATH we now obtain MATH . Consequently MATH has a finite diameter and by REF this also holds for MATH and MATH is a metric cut in MATH. By the definition of MATH and to prove that this set is open in MATH we show that MATH is contained in MATH. For every metric cut MATH in MATH REF implies MATH . As MATH is a subset of MATH and MATH is a metric cut we now obtain MATH by replacing MATH with MATH. |
math/0010033 | As the support of MATH generates MATH we can choose four distinct elements MATH, MATH, MATH and MATH in MATH such that MATH and MATH are positive. We define a random walk MATH on the free group MATH with generating system MATH by choosing MATH, MATH, MATH and MATH as the probabilities for the right multiplication with the generating elements, respectively. The probability of not making a move is set MATH. We know that MATH is transient (compare for example CITE). The corresponding probability measure on the set of trajectories is denoted by MATH. As the corresponding NAME graph MATH is locally finite, the random walk MATH does not enter any ball from an index on with probability one. Thus MATH . Now MATH for all elements MATH and MATH of MATH and therefore MATH . This is equivalent to the almost sure convergence of MATH to a proper metric end in the metric end topology of MATH. |
math/0010036 | The construction above gives sets of linear evolution data MATH and MATH from MATH and MATH, where MATH lie in vector spaces MATH. It is easy to see that MATH is a vector subspace of MATH, since the NAME algebra MATH of MATH is a vector subspace of MATH. Let MATH be the annihilator MATH of MATH in MATH. Then MATH. Now the construction of SL REF-folds using MATH in REF involves a REF-parameter family of linear maps MATH in MATH satisfying the o.d.e. CASE: Let MATH be the pull-back of MATH from MATH to MATH. It is easy to show that this family MATH also lie in MATH and satisfy REF , and that the SL REF-fold MATH constructed using the MATH is a subset of the SL REF-fold MATH constructed using the MATH. Thus the SL REF-folds constructed using MATH are included in those constructed using MATH. |
math/0010036 | Since MATH by REF , one can show from MATH that MATH, MATH and MATH are linearly independent if and only if MATH, or equivalently if and only if MATH and MATH are linearly independent. But MATH is an immersion near MATH if and only if MATH, MATH and MATH are linearly independent. |
math/0010036 | As MATH span a vector space of dimension REF, they satisfy MATH for some MATH not all zero. We may think of the MATH as giving a linear map MATH with kernel MATH, and so MATH defines a point in the kernel in MATH. Under the action of MATH on MATH defined in REF - REF , the vector MATH transforms under the natural action of MATH on MATH. It can be shown that MATH acts on MATH as the group preserving the Lorentzian conformal structure with null cone MATH. Thus there are three MATH-orbits of nonzero vectors in MATH: the `time-like' vectors with MATH, the `space-like' vectors with MATH, and the `null' vectors with MATH. Therefore every nonzero vector MATH is equivalent under the MATH-action to MATH or MATH or MATH. Hence we can transform MATH under the MATH action to MATH satisfying either REF MATH, REF MATH, or REF MATH. We shall show that REF contradict REF, which leaves REF . In REF , as MATH are linearly independent and MATH by REF , we see that MATH and MATH are linearly independent. But REF gives MATH for small MATH. Therefore MATH and MATH are linearly independent for small, nonzero MATH. This contradicts REF . Similarly, REF leads to a contradiction. Thus REF holds, and we can transform the MATH to MATH with MATH. Clearly MATH must be nonzero for REF to hold. We can then use a dilation in MATH to rescale the MATH to get MATH. This completes the proof. |
math/0010036 | Suppose for simplicity that MATH and MATH are linearly independent in MATH. Consider the action of MATH in MATH upon MATH and MATH given in REF - REF , where MATH. We can think of this as an action of MATH on MATH, where MATH in MATH is mapped to MATH. This action preserves the quadratic form MATH upon MATH. Now there is a second, positive definite quadratic form on MATH given by MATH . By standard results on simultaneous diagonalization of quadratic forms, there is a basis of MATH in which both quadratic forms are diagonal. Choose an element of MATH which takes this basis to vectors proportional to MATH, MATH and MATH. Let MATH be the transforms of MATH under this element of MATH. Then the quadratic form MATH is diagonal with respect to the standard basis of MATH. That is, MATH and MATH are orthogonal. Now from REF we see that MATH span a Lagrangian plane in MATH. But any three nonzero orthogonal elements in a Lagrangian plane in MATH are conjugate under a matrix in MATH to MATH, MATH and MATH for some MATH. Applying this MATH matrix to MATH gives MATH, where REF holds for MATH. But it is easy to see from REF that if REF holds at MATH then it holds for all MATH. This completes the proof. |
math/0010036 | REF follows immediately. One can verify that REF - REF are solutions to REF by subtituting them in, and using MATH. The six solutions of REF this gives, with coefficients MATH, are easily seen to be linearly independent. But as REF is a well-behaved first-order o.d.e., its solutions are determined by the initial data MATH, which comprise REF complex or REF real numbers. Thus REF can have no more than REF linearly independent solutions, so REF - REF are the full solutions to REF . |
math/0010036 | When MATH, we have MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH. Thus REF holds at MATH if and only if REF holds, so by REF - REF hold when MATH if and only if REF holds. But REF - REF hold at MATH if and only if they hold for all MATH. |
math/0010036 | Substitute REF into the first equation of REF , and use the values for MATH in REF . We get MATH . But MATH as MATH. So dividing by MATH gives MATH which is equivalent to the first line of the equation we have to prove. The other lines follow in the same way. |
math/0010036 | From REF we see that MATH, where MATH . Thus the eigenvalues of MATH are the same as those of MATH, which are the roots of the polynomial MATH. Direct calculation shows that MATH . Now MATH and MATH with MATH. Thus we may substitute MATH, MATH, for MATH in MATH. Then MATH becomes MATH which factorizes as MATH . Thus the roots of MATH are zero (twice), MATH and MATH, where MATH . This proves the first equation of REF . Note also that as MATH and MATH, we have MATH, so we can take MATH to be real and positive. The second two equations of REF follow from MATH, as for instance MATH . The first equation of REF follows from REF . The second is equivalent to MATH which has a unique solution MATH, as the MATH matrix appearing here has determinant MATH, and so is invertible. To prove REF , observe that as MATH is a real eigenvalue of the real matrix MATH, there exists a real eigenvector MATH of MATH with eigenvalue MATH. This gives the first equation of REF , and the second then holds because of the form of MATH. One can also easily show that MATH and MATH are both nonzero. REF is proved in the same way, as MATH is an eigenvalue of MATH. |
math/0010036 | The MATH matrix MATH studied above is not symmetric, because the signs of the MATH terms are not right. However, if we define MATH then MATH is a symmetric complex matrix, and REF shows that MATH are eigenvectors of MATH, with eigenvalues MATH and MATH respectively. Now if MATH are eigenvectors of a symmetric matrix with different eigenvalues, then MATH. Taking inner products between the five eigenvectors above proves REF . Also, MATH is essentially an eigenvector with eigenvalue REF, modulo nilpotent behaviour, and taking inner products of this with the MATH and MATH eigenvectors gives REF . To prove the last part, as in the proof of REF put MATH, MATH and MATH for some MATH. Then MATH when MATH, that is, when MATH or REF, which contradicts MATH. Similarly, MATH when MATH, that is, when MATH, contradicting MATH, and MATH is ruled out in the same way. To eliminate MATH for MATH, note that MATH, but MATH for MATH implies that MATH. |
math/0010037 | CASE: Use the surjectivity of the map MATH. CASE: It follows from the fact that, by MATH-invariance, MATH contains the vertical part of the tangent space to the orbit of MATH under the action of MATH. |
math/0010037 | CASE: We start observing that the sheaf MATH is globally generated. Indeed, by NAME 's vanishing REF we have MATH . Therefore, by NAME 's m-regularity theorem ( REF , page REF) the maps MATH are surjective for MATH. The result now follows immediately recalling the isomorphism MATH and the surjection MATH . CASE: Again it suffices to check the case MATH. Using the irreducibility of the action of MATH on the NAME, it suffices to construct a single non-trivial meromorphic section of MATH with simple pole along the zero set of a NAME coordinate. To do this, for all lines MATH not meeting MATH the NAME coordinate MATH so there is, by NAME 's rule, a unique MATH containing MATH. |
math/0010037 | By REF , we have MATH . Let MATH be a smooth point of MATH. Since the bundle MATH is generated by its global sections, there exists a section MATH such that MATH . Since MATH is generically an immersion, we obtain from the above a non zero element in MATH coming from MATH. |
math/0010037 | Recall that MATH can be naturally interpreted as the kernel of the NAME map MATH. Hence one easily verifies that MATH contains MATH, for all MATH and MATH. Then MATH contains elements of the form MATH for all MATH and MATH, coming from the wedge product of elements in MATH. Since we are supposing that MATH is in the base locus of MATH, the previous fact implies in particular that the dimension of the subspace MATH is at most MATH, that is, the multiplication map MATH cannot be surjective, for any MATH. Recall that if MATH and MATH are vector spaces, and MATH, then MATH . If, for generic MATH, the map MATH has rank one, from REF we obtain that MATH, for any MATH, that is, MATH, where MATH is the line determined by MATH. Then MATH contains a hyperplane of MATH and the Lemma is proved. Thus, we can assume that, for generic MATH. the map MATH has rank at least two. Let MATH such that MATH is of codimension MATH in MATH. For generic MATH, consider the map MATH whose rank is then equal to MATH or MATH. In the former case MATH would then contain MATH, which is absurd since MATH has codimension MATH. Hence we can suppose MATH. Then by REF , Lemma MATH, MATH contains the degree d part of the ideal of a line MATH passing through MATH, and hence MATH contains at least a codimension two subspace of MATH. Assume first that the line does not vary with MATH, and denote it by MATH. If MATH then the image MATH of MATH in MATH has codimension REF in MATH. On the other hand, since MATH contains MATH, its reduction MATH modulo MATH has also codimension REF in MATH. Hence MATH. By the genericity of the choice MATH in MATH, this fact would imply that MATH thus leading to a contradiction. Assume now that MATH, for generic MATH. Since MATH contains a codimension two subspace of MATH, from the exact sequence MATH and the fact that MATH has codimension MATH, it follows that MATH. Let MATH be the span of MATH and MATH: we study the variation of this plane with MATH. If for generic MATH the intersection MATH is equal to the line MATH, then MATH and we are done. If otherwise MATH, then it is immediate to see that MATH contains MATH because MATH and MATH will vary in this plane. But this is absurd since MATH is of codimension MATH. |
math/0010037 | Since the distribution MATH is integrable, the brackets induce a map MATH which is given at the point MATH by MATH . Since we are supposing that MATH contains MATH but not the whole ideal MATH, there is a canonical isomorphism MATH and hence MATH identifies with a map MATH which we also denote by MATH. To prove the integrability of MATH, by NAME 's theorem it will suffice to show that MATH is zero. In what follows we will denote MATH and MATH respectively by MATH and MATH. Now, choose coordinates on MATH such that MATH and MATH. Recall that MATH identifies naturally with the set of MATH-graded homomorphisms from MATH to MATH. Hence there is a natural bilinear map, denoted by MATH : MATH which is explicitely given by MATH . Remark that, since MATH and MATH, any deformation of MATH belonging to MATH passes through MATH, that is, MATH. A verification in local coordinates shows that MATH . Note that MATH for every MATH and MATH, and therefore MATH, for every MATH. We first show that MATH vanishes on MATH: if we had MATH with MATH, then MATH would contain the elements MATH for any MATH, hence at least a hyperplane of MATH. Thus MATH vanishes on MATH, giving a map MATH, which we still call MATH. Identify MATH with MATH, and recall again the natural isomorphism MATH . Then MATH corresponds to a subspace MATH with MATH, and the dot map is simply given by the contraction MATH . Hence, by REF the map MATH identifies with MATH . To conclude we need the following linear algebra result: Let MATH and MATH be two vector spaces, MATH a codimension MATH subspace of in MATH and MATH a linear map. If MATH, then the image of the map MATH contains at least a codimension MATH subspace of MATH. Let MATH. Pick a complementary subspace MATH to MATH in MATH and a basis MATH for MATH which is compatible with the decomposition MATH . Let MATH be a basis of MATH and MATH the dual one. Pick a complementary space MATH to MATH in MATH which will be generated by a monomial MATH and extend MATH to the whole MATH by setting MATH . The map MATH extends naturally to MATH. Since MATH we are reduced to proving that if the extended map MATH is not zero, then the codimension of the image of MATH is at least MATH. This has already been checked in REF , Lemma MATH, and so we are done. Take MATH and MATH, and apply REF . to our situation. If we had MATH, then the image of the map MATH would contain at least a codimension MATH subspace of MATH. But the image of MATH is contained in MATH, and MATH contains a hyperplane in MATH. Hence the codimension of MATH in MATH would be at most MATH, which is in contradiction with REF . Thus MATH, hence MATH is zero and so is MATH. By NAME 's theorem the distribution MATH is integrable. |
math/0010037 | CASE: Assume MATH . Then the assertion follows from the fact that the contact condition MATH for a line MATH through MATH with respect to MATH becomes MATH . REF A dimension count shows that all hypersurfaces MATH in MATH of degree MATH admit a point through which passes a line having contact with MATH of maximal order. |
math/0010041 | First, we will show that MATH is generated as a MATH-module by MATH . Let MATH be a homogeneous endomorphism in the MATH-centre of MATH. Then, MATH for some MATH and all MATH in MATH. Since MATH is homogeneous, MATH is some homogeneous MATH of the same degree as MATH. Then MATH. Since MATH has the same value on MATH as MATH, we have MATH. To prove MATH contains all of the prescribed operators, let MATH be any homogeneous element and MATH any element of MATH. Since left multiplication and right multiplication commute, for any MATH we have MATH. Furthermore, by REF we have MATH. Hence, MATH is in the MATH-centre of MATH. Now, by its definition MATH is the MATH-subspace of MATH spanned by MATH. However, MATH commutes with MATH up to multiplication by a scalar, and commutes with MATH without incident, so MATH can be generated by the given operators as required. |
math/0010041 | It is clear that MATH is contained in MATH so we only need to show that every MATH is in the MATH span of MATH. To this end, suppose for some MATH and all MATH that MATH. Then we have for any MATH, MATH . Since MATH is an automorphism of MATH, MATH. Hence MATH. Since MATH is a group, we may put MATH. Then MATH for any MATH. Hence MATH. |
math/0010041 | Since MATH is a bracket, MATH is a derivation. Hence, for any MATH, we have MATH . Inductively, we have that if MATH then MATH. Hence MATH is in the MATH-span of MATH, and so it is in MATH. It follows that for any MATH, MATH. Hence MATH. |
math/0010041 | Indeed, if MATH, then MATH for homogeneous MATH. Corresponding to each MATH, there exists MATH such that MATH. Hence, MATH . Now induction completes the lemma. |
math/0010041 | Let MATH. By REF , it is enough to show MATH. For any MATH, MATH . Thus MATH for any MATH. Hence MATH, as required. |
math/0010041 | This follows from the fact that MATH. |
math/0010041 | It is enough to prove the lemma when MATH is a monomial, for if the lemma holds for monomials MATH, MATH, and MATH, then we can find differential operators MATH of the prescribed form such that MATH. Putting MATH yields the desired result. Now, suppose MATH where each MATH. We will show by induction on MATH that there is a homogeneous MATH-differential operator MATH of degree MATH, comprised of monomials in the MATH's such that MATH. When MATH we have only one possibility: MATH. Put MATH. Since MATH, this proves the base case. Before proceeding, we must address one technical point. We shall assume that if MATH then every MATH. To see that no generality is lost, suppose that MATH and MATH. Put MATH. By REF MATH. Thus we have MATH for some nonzero constant MATH. If MATH then MATH. However, since MATH, we have MATH. Now suppose the statement holds for monomials of degree MATH. Let MATH . That is MATH is the monomial obtained from MATH by a cyclic permutation of its factors. Then MATH, and MATH when MATH. Let MATH, MATH, MATH, MATH be arbitrary integers. Then the series MATH is a telescoping series which reduces to MATH . By the induction hypothesis, there are MATH homogeneous in the MATH's of degree MATH such that MATH. For each MATH, put MATH. The homogeneity of the MATH ensures that for each MATH, MATH . Thus we get MATH . We can use this and REF to get MATH . Thus, applying MATH to MATH will yield MATH . We use the identity MATH on REF to get MATH which can be reduced to MATH . We chose the MATH so that MATH. Thus MATH. By our assumption on MATH, either all MATH or MATH is nonzero. If some MATH, then MATH as required. However, if all MATH, then MATH. In this case, MATH is the required operator. |
math/0010041 | Let MATH be the MATH-module generated by MATH . Since MATH commutes with MATH up to multiplication by a scalar in MATH, we have MATH for any MATH. We will show by induction on MATH that MATH. Since MATH, we have MATH. This proves the base case. Now suppose that MATH. By REF above, we know that MATH. Thus, we must show MATH. Since MATH, and MATH, it suffices to show MATH. Let MATH. Then MATH by the induction hypothesis. Thus MATH where MATH, MATH with MATH, and MATH. By REF , we can find MATH such that MATH. Put MATH. Then MATH is in MATH. Since MATH and MATH commute with MATH, we have MATH. Hence MATH. That is, MATH. Hence, MATH, and so MATH. We have shown that MATH generates MATH over MATH. Moreover, it is clear from the construction of MATH that MATH is generated as a MATH-algebra by MATH and MATH. Furthermore, by Note REF, each MATH is in the MATH-span of MATH. Hence, it is enough to show that each MATH is generated over MATH by MATH, MATH, and MATH. To this end, note that MATH, MATH, and MATH. For any positive integer MATH, MATH and MATH. This proves the theorem. |
math/0010041 | Let MATH be an ideal in MATH. Let MATH. Then MATH can be written as MATH where MATH are polynomials in MATH, MATH and the multi indices MATH have entries in MATH. We induct on MATH, to claim that the ideal containing MATH should contain REF. Assume that MATH. Then MATH where MATH and MATH. Now MATH have fewer monomials than MATH. Continuing thus, we can assume that MATH for some MATH. Since MATH the base case is proved. Assume that the claim has been proved for all positive integers less than MATH. We can write MATH as MATH where MATH and MATH are polynomials in MATH where MATH are multi indices of length MATH, and these multi indices can repeat in the above sum. This implies that, MATH where MATH and MATH. This shows that MATH has fewer monomials which correspond to multi indices of length MATH. This completes the theorem. |
math/0010041 | It is clear that MATH, MATH, and MATH are homogeneous of degree REF. We must show that they generate all of MATH, that they commute, and that they have no other relations among them. Suppose that MATH. Then by REF , MATH can be written as MATH where each MATH is a monomial in MATH, MATH, MATH, and MATH. Moreover, each MATH has degree MATH, so each MATH has exactly one MATH for each MATH, MATH, and MATH. Since MATH for MATH, we can rewrite MATH so that each MATH has the form MATH. If MATH, then MATH. Since MATH, we have that each MATH is in the span of MATH, MATH, and MATH. Hence MATH has the required generators. Since MATH and MATH are multiplicative inverses, they commute. Comparing the values of MATH and MATH on MATH for any MATH shows that MATH and MATH also commute. All we need to show now is that there are no remaining relations. Suppose that we have a relation MATH for some MATH. Then we can multiply this expression by MATH and MATH an appropriate number of times to ensure that all all powers of MATH in the MATH and all MATH are nonnegative. Then for every MATH, MATH. Fix a MATH such that for every MATH and MATH, all the powers of MATH appearing in MATH are less than all the powers of MATH appearing in MATH. Then the polynomial MATH is a polynomial in MATH with no terms canceling. Since it is zero, all its coefficients are zero. But if MATH is the coefficient of MATH in MATH, then MATH is the coefficient of MATH in MATH. Hence MATH. It follows that MATH completing the proof. |
math/0010041 | First we show that MATH cannot be a zero divisor. It is obvious that MATH if MATH. Suppose MATH, then MATH for all positive integers MATH. Also note that MATH. Hence MATH does not satisfy REF . Suppose MATH. Let MATH and MATH be the highest degree parts of MATH and MATH of degrees MATH and MATH respectively. Then MATH. If MATH is positive, then MATH where MATH is of degree REF. Note also that MATH is a non-zero homomorphism. If MATH is negative, then MATH and MATH is of degree REF. Hence, we can assume that MATH. Now, if MATH is positive, then MATH implies MATH, where MATH is of degree REF. If MATH is negative then MATH where MATH is of degree REF. Hence, we can assume that MATH also. Now the corollary follows from REF . |
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