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math/0106228 | If MATH is a regular function on MATH with zero set MATH, then MATH must be homogeneous, say of degree MATH. So MATH then defines a section of MATH with divisor MATH, with MATH a positive integer. This means that MATH. Conversely, an identity of this type implies the existence of such a MATH. |
math/0106228 | We do this in two steps. The set MATH of MATH such that MATH is nonempty and not equal to MATH is a union of double cosets of MATH in MATH distinct from MATH. Each such double coset defines an ordered pair of distinct irreducible hypersurfaces in MATH with the same image in MATH and vice versa. Since MATH is finite and birational, there are only finitely many such pairs of irreducible hypersurfaces and hence MATH for some finite set MATH. Let MATH be the union of the sets MATH, for which MATH runs over a system of representatives of MATH in MATH. This is a finite subset of MATH. If we now pass from MATH to a normal arithmetic subgroup of MATH of finite index which avoids MATH, then the natural maps MATH are injective and local embeddings. They are also proper, and so they are closed embeddings. It is still possible that MATH fails to be injective for some for some MATH (although this can only happen when MATH). We avoid this situation essentially by proceeding as before: given isotropic line MATH defining a cusp then the set MATH such that for some MATH and MATH is a union of double cosets of MATH in MATH. The number of such double cosets is at most the number of branches of MATH at this cusp and hence finite. If MATH is a system of representatives, then let MATH be a of union MATH', where MATH runs over a (finite) system of representatives of MATH in the set MATH. Choose a normal arithmetic subgroup of MATH of finite index which avoids MATH. Then this subgroup is as desired. |
math/0106228 | Let MATH be a MATH-isotropic line. Then MATH contains MATH, is contained in MATH and is defined over MATH since it is an intersection of hyperplanes defined over MATH. Let MATH contain MATH. Then MATH defines a hypersurface MATH in MATH and an abelian subtorsor MATH of MATH with quotient MATH. The latter is a genus one curve which apparently comes with an origin MATH (it is an elliptic curve) and the complement of this origin of is affine. Any regular function on the affine curve MATH with given pole order at MATH lifts to a regular function on MATH with the same pole order along MATH (and zero divisor disjoint from the polar set). This proves that the effect of the normalized blowing up of the fractional ideal MATH is to replace the vertex MATH of MATH by MATH. The strict transform of MATH in this blow-up is clearly a NAME divisor. If we do this for all the MATH that contain MATH, then the vertex gets replaced by MATH as asserted. |
math/0106228 | Since MATH is an orbit of the arithmetic group MATH in MATH, the MATH-submodule MATH spanned by this orbit is discrete in MATH. Now MATH is also contained in the real hypersurface MATH defined by MATH for some MATH. The map MATH defined by MATH is proper. A standard argument shows that the cardinality of MATH is proportional to the MATH-volume of MATH and hence is MATH for some MATH. So MATH is finite and for MATH we have the uniform estimate MATH . The righthand side converges since MATH. |
math/0106228 | Given the MATH-isotropic line MATH, choose a MATH-generator MATH and consider the action of the right half plane on MATH defined by MATH (see Subsection REF): MATH. In view of the characterization in Subsection REF of the analytic structure on MATH it is enough to show that MATH exists as a meromorphic function and this limit is equal to the subseries defined above. We claim that with MATH, MATH and MATH as in the proof of REF above, the map MATH is bounded from below, say by MATH. Once we establish this we are done, for then MATH is monotone increasing for MATH for any for MATH and MATH, implying that the limit is as stated. In order to prove our claim, we first note that without any loss of generality we may assume that MATH for all MATH (just replace MATH by MATH). Write MATH for MATH and choose an isotropic MATH such that MATH. Since the orthogonal complement of MATH is negative definite, we denote the restriction of MATH to this complement by MATH. We write any MATH as MATH with MATH. Notice that the function MATH has a positive minimum MATH. Then for MATH and MATH we have the estimate MATH . The claim follows and with it, the lemma. |
math/0106228 | It suffices to prove this statement for MATH instead of MATH for some MATH. We first show that for MATH the sections of MATH generate this sheaf over MATH. Let MATH and MATH. Choose MATH spanning the orthogonal complement of MATH and let MATH denote the MATH-orbit of MATH. Then according to REF MATH defines a section of MATH. It is in fact a section of the invertible subsheaf MATH on MATH and nonzero as such at MATH. These subsheafs generate MATH. So it remains to verify the generating property over a cusp. But this follows from the conjunction of Lemma's REF. |
math/0106228 | The codimension assumption implies that any section of MATH extends to MATH. Since MATH, the first assertion follows. This induces an isomorphism between the underlying proj's and so we obtain an isomorphism MATH, as stated. |
math/0106229 | The lemma is equivalent to the following equality: MATH . It follows from the definition of MATH that MATH . On the other hand, we shall rewrite MATH. It follows from the definition of MATH that MATH where MATH denotes the projection image of MATH on the quotient vector space of MATH by the subspace MATH spanned by the cone MATH. Note that MATH lies in MATH if and only if MATH lies in MATH modulo MATH, and that MATH by definition. Therefore, writing MATH as MATH, the equality above turns into MATH where MATH runs over elements in MATH such that MATH and MATH modulo MATH. Putting this in the defining equation of MATH, we have MATH where the sum is taken over elements MATH and MATH such that MATH and MATH modulo MATH. Fix MATH with MATH, and observe how many times MATH appears in the above sum. It is equal to the number of MATH such that MATH and MATH modulo MATH. But the number of such MATH is MATH. To see this, we note that MATH means that MATH by definition, where MATH, and that the condition that MATH modulo MATH is equivalent to saying that MATH contains the complement of the set MATH in MATH. Therefore, any such MATH is obtained as the complement of a subset of MATH with cardinality MATH, so that the number of such MATH is MATH. This together with REF proves the equality REF . |
math/0106229 | CASE: This immediately follows from REF because MATH's are independent of MATH. CASE: If we take MATH instead of MATH, then MATH turns into MATH, so that MATH turns into MATH. Since MATH's are independent of MATH as shown in REF above, this proves MATH. |
math/0106229 | For simplicity we assume that there is only one MATH such that MATH. We may assume that MATH is positive without loss of generality. The situation is as in REF . It follows from the definition of MATH that the difference between MATH and MATH arises from the cones MATH's for MATH such that MATH and MATH. In fact, one sees that MATH where MATH runs over the elements as above. Since MATH and MATH, the equality above turns into MATH . Here MATH may be viewed as the value at MATH of the characteristic function of the cones in MATH with apex MATH spanned by MATH's MATH. This shows that the right-hand side at the equality above agrees with MATH, proving the lemma. |
math/0106229 | Induction on the dimension of simple multi-polytopes MATH. We have observed the lemma in REF when MATH. Suppose MATH and suppose that the lemma is true for simple multi-polytopes of dimension MATH. Then the support of MATH is bounded by the induction assumption. This together with REF implies that MATH takes the same constant on unbounded regions in MATH. On the other hand, it follows from the definition of MATH that MATH vanishes on a half space MATH for a sufficiently small real number MATH, because for each MATH the cone MATH is contained in the complement of MATH if MATH is sufficiently small. Therefore the constant which MATH takes on the unbounded regions in MATH is zero, proving the former assertion in the lemma. As for the latter assertion in the lemma, it follows from the induction assumption that the right-hand side of the wall crossing formula in REF is independent of MATH, and we have seen above that MATH vanishes on unbounded regions regardless of the choice of MATH. Thus, it follows from REF that MATH is independent of MATH on any regions of MATH. |
math/0106229 | Through MATH and the inverse of the excision isomorphism, the cycle MATH maps to MATH. We express MATH as MATH where MATH or MATH and define an oriented MATH-simplex MATH in MATH by MATH. It follows that MATH . Here MATH by the definition of MATH, and MATH if and only of MATH. Therefore, the right-hand side above reduces to MATH, that is MATH in MATH. |
math/0106229 | Existence. We will construct MATH inductively using decending induction on MATH. If MATH, then we map MATH to any point in MATH. Thus MATH is defined on MATH with the image in MATH. Let MATH be a nonnegative integer less than MATH and MATH. Suppose that MATH is defined on MATH with the image in MATH. Then MATH extends to a continuous map from MATH to MATH because MATH is contractible. Thus MATH is defined on MATH with the image in MATH. This completes the induction step, so that we obtain the desired map MATH defined on MATH. Uniqueness. We construct a homotopy MATH of given two maps MATH and MATH in the lemma. The argument is almost same as above. Since MATH is contractible, MATH can be defined on MATH with MATH as the image. Let MATH be as above and MATH. Suppose that MATH is defined on MATH with the image in MATH and that MATH agrees with MATH on MATH for MATH. Then a map MATH extends to a continuous map from MATH to MATH because MATH is contractible. Thus MATH is defined on MATH with the image in MATH. This completes the induction step, so that we obtain the desired homotopy MATH defined on MATH. |
math/0106229 | Consider the following commutative diagram: MATH where MATH runs over the indices of MATH's which intersect MATH. The element MATH maps to MATH through the upper horizontal sequence by REF and down to MATH by REF . Now we trace the lower horizontal sequence from the right to the left. Through the inverse of MATH, MATH maps to the fundamental class MATH, and further maps to MATH, where the sign arises from the choice of the orientation on MATH. These together with the commutativity of the diagram above show that MATH . On the other hand, MATH by definition. The lemma follows by comparing these two equalities. |
math/0106229 | Subtract the equality in REF for MATH from that for MATH. Since one can take MATH to be MATH, the lemma follows. |
math/0106229 | The equality is established in REF when MATH. Suppose MATH and suppose that the equality holds for simple multi-polytopes of dimension MATH. Both MATH and MATH are locally constant, satisfy the same wall crossing formula REF and MATH by induction assumption. Therefore, it suffices to see that MATH and MATH agree on one region. But we know that they vanish on unbounded regions (REF and the remark after the definition of MATH), hence they agree on the whole domain. This completes the induction step, proving the theorem. |
math/0106229 | The NAME expansion of MATH is given by MATH . Taking this into account, we expand the sum MATH into NAME series and get MATH where the summation MATH runs over the collection of such MATH that MATH (see REF for MATH). Since MATH by REF , the NAME expansion of the left-hand side of the equality in REF has the form MATH where MATH . One easily checks that MATH agrees with MATH, proving the lemma. |
math/0106229 | We shall prove REF first. It suffices to prove MATH. Since MATH by definition, it suffices to prove that MATH . Let MATH and MATH be the maps introduced in REF which are associated with multi-polytopes MATH and MATH respectively. We note that MATH and MATH considered as maps from MATH to MATH for MATH are homotopic. Since the multiplication by MATH on MATH sends the fundamental class MATH to MATH, we obtain REF . We shall prove REF . Because of REF , it suffices to prove REF for MATH. We apply REF to MATH in place of MATH (so that MATH is replaced by MATH), and approach MATH to MATH in the equality. Since the right-hand side approaches MATH, it suffices to show that the left-hand side approaches a polynomial in MATH of degree at most MATH. When MATH is the identity element, MATH. Therefore, the term in the summand MATH in the left-hand side has a pole at MATH of degree exactly MATH when MATH is the identity element, and of degree at most MATH otherwise. Thus the left-hand side of the equality in REF applied to MATH can be written as MATH where MATH and MATH are polynomials in MATH and MATH. Then the repeated use of NAME 's Theorem implies that when MATH approaches MATH, the limit of the above rational function is a polynomial in MATH of degree at most MATH. Finally we prove REF . Since MATH it follows from the definition of definite integral that MATH proving that the coefficient of MATH in MATH is MATH. We apply REF to MATH, that is MATH with MATH. Then the MATH in the lemma is zero, and MATH unless MATH because the origin is the only vertex of MATH so that the vertices of MATH are very close to the origin. Thus the right-hand side of the equality in the lemma applied to MATH is a constant, say MATH, which is nothing but the constant term in MATH. Now we approach MATH to MATH. Then the equality reduces to MATH because MATH for all MATH if and only if MATH with MATH for all MATH, and the latter is equivalent to saying that MATH belongs to the cone MATH spanned by MATH's MATH. Since MATH by definition, the constant term in MATH, that is MATH, agrees with MATH. |
math/0106229 | Since MATH and MATH, it follows from REF of MATH that MATH, which is equal to MATH by definition. |
math/0106229 | Let MATH denote a finite set which consists of elements in MATH taken with multiplicity, that is, elements in MATH may appear in MATH repeatedly. Set MATH and denote by MATH the subset of MATH consisting of elements appearing in MATH. It follows from the definition that MATH is additively generated by MATH's such that MATH, so it suffices to prove the lemma for such MATH. We shall prove it by induction on MATH. If MATH, then MATH; so MATH is obviously of the form in the lemma in this case. Suppose MATH. Then there is a MATH which appears in MATH at least twice. Set MATH. Then MATH and MATH. Multiplying the both sides at REF by MATH, we obtain MATH for any MATH. We choose MATH such that MATH and MATH for all MATH different from MATH. (Such MATH exists because MATH is a subset of a basis of MATH.) Then the equality above reduces to MATH . Here MATH for MATH, so the right-hand side above are of the form in the lemma by the induction assumption, showing that so is MATH. This completes the induction step and proves the lemma. |
math/0106229 | Both MATH and MATH are ring homomorphisms and MATH is a polynomial ring generated by elements in MATH, so it suffices to check the lemma on MATH. Let MATH. It follows from the definitions of MATH and MATH that MATH which agrees with MATH because MATH is the dual basis of MATH. Since MATH is arbitrary, this proves that MATH is the identity on MATH. |
math/0106229 | Since MATH is a linear combination of MATH for MATH, it suffices to show that MATH . As remarked above, MATH is realized as the equivariant first NAME class of a lattice multi-polytope, so it follows from REF that MATH . The NAME character MATH mapping MATH to MATH extends to a map from MATH and it further extends to a map between the quotient fields. Sending the element above by this extended NAME character, we obtain REF . |
math/0106229 | Since MATH is a MATH-module map, it suffices to check the lemma for elements MATH's MATH by REF . We distinguish two cases. CASE: The case where MATH, that is, MATH. In this case MATH . Therefore MATH . CASE: The case where MATH. In this case we will show that MATH. Since MATH for any MATH if and only if MATH is the identity, and MATH the term of lowest degree in REF (up to a nonzero constant multiple) is MATH that is, MATH, and REF tells us that it is an element of MATH. This means that MATH because the degree of MATH is equal to MATH. |
math/0106229 | We shall compute MATH. For that, we need to see MATH. Let MATH. If MATH, then there is a MATH such that MATH; so MATH for such MATH because the NAME expansion of MATH has no constant term and MATH. Therefore, only elements MATH in MATH contribute to MATH. Now suppose MATH. Then MATH for MATH, so MATH for such MATH because the NAME expansion of MATH has the constant term MATH and MATH. Finally, since MATH for MATH, we thus have MATH . This together with the definition of MATH and REF shows that MATH . This implies that MATH . |
math/0106229 | The latter equality is obvious because only elements of degree MATH in MATH survive through the map MATH. We shall prove the former equality. CASE: If MATH is a lattice multi-polytope, then REF applied to MATH for any integer MATH implies MATH . We compare the coefficients of MATH at the both sides above. Since MATH, the coefficient of MATH at the left-hand side is MATH, while the one at the right-hand side is MATH by REF . Therefore the lemma is proven for a lattice multi-polytope MATH. CASE: If MATH is rational, by which we mean that there is a nonzero integer MATH such that MATH is a lattice multi-polytope, then MATH by REF . Since MATH and MATH, the lemma is proven for a rational multi-polytope MATH. CASE: The functions MATH and MATH are defined on the vector space MATH through the equivariant first NAME class, and they are obviously continuous. By REF they agree on all rational multi-polytopes which form a dense subset of the vector space, so they must agree on the entire vector space by continuity. This completes the proof of the lemma. |
math/0106229 | An elementary computation shows that MATH . Therefore, it follows from REF that MATH proving the theorem. |
math/0106229 | The proof is given in CITE, but we shall give a simple proof for the reader's convenience when MATH is as in REF . Since MATH is additively generated by MATH's, one can express MATH with a unique integer MATH depending on MATH for each MATH. We view MATH as a function of MATH. Since it is linear, it defines an element MATH of MATH such that MATH. |
math/0106229 | As we remarked in REF after the definition of the completeness of a multi-fan, it suffices to prove the pre-completeness of MATH for any MATH. Choose a generic vector MATH from MATH. The sign MATH for MATH is defined as in REF with respect to the projection image of MATH on the quotient lattice of MATH by the sublattice generated by MATH. The pre-completeness of MATH is equivalent to the equality: MATH which we will verify in the following. Since MATH, a connected component of MATH containing a MATH-fixed point is a MATH-dimensional sphere on which MATH acts effectively. We denote those connected components by MATH's. They are torus manifolds equipped with the orientations discussed before this lemma. Since MATH has two MATH-fixed points, MATH consists of two elements, denoted by MATH, corresponding to the MATH-fixed points. One easily checks that the multi-fan MATH of MATH is complete, which is equivalent to the equality: MATH . As discussed before this lemma, we have a natural map MATH. Note that if MATH where MATH stands for MATH or MATH, then MATH. On the other hand, we have MATH . Therefore MATH which vanishes by REF , proving the lemma. |
math/0106229 | Recall that the cycle which defines MATH is MATH. We reverse the orientation on MATH. Obviously, MATH and MATH remain unchanged unless MATH. Suppose MATH. Then, since the orientation on MATH is reversed, MATH and MATH will be interchanged, so that MATH turns into MATH. As for MATH, MATH turns into MATH as remarked before and hence so does MATH by REF . Thus, MATH turns into MATH if MATH. After all, MATH does not depend on the orientations on MATH's for any MATH. |
math/0106229 | Since MATH's generate MATH additively modulo MATH-torsions, MATH modulo MATH-torsions with some integers MATH's. We define a map MATH by MATH . The pair MATH is a lattice multi-polytope, and MATH for MATH which follows from REF . Since MATH is non-singular by REF and complete by REF , the lemma follows from REF applied to the multi-polytope MATH. |
math/0106229 | Look at the expansion of the right-hand side of REF with respect to MATH. It follows from REF that all coefficients of powers of MATH in REF are elements of MATH. Take a generic vector MATH and evaluate the right-hand side of REF on MATH. Then we get the following polynomial in MATH whose coefficients are NAME polynomials in MATH: MATH . It is easily seen that REF approaches to a polynomial in MATH with constant coefficents if MATH tends either to MATH or to MATH. This means that REF itself is a polynomial with constant coefficients. Since MATH is generic, this implies that REF , that is MATH, is actually a polynomial with constant coefficients equal to REF . Then, by letting MATH tend to MATH, we obtain MATH where MATH. This MATH agrees with the MATH in REF because MATH is the dual basis of MATH. Hence MATH, proving the former equality in the theorem. The latter follows from REF . As noted in the definition of MATH in REF, MATH. Since MATH, the last statement in the theorem follows. |
math/0106229 | Since MATH equals MATH, the former statement follows from REF . The latter statement is noted in the definition of MATH in REF. |
math/0106229 | We prove the lemma by induction on the dimension MATH. When MATH, MATH is MATH with a nontrivial smooth MATH-action. In this case, it is not difficult to check the lemma, which we leave to the reader. Assume that MATH. Since a characteristic submanifold of MATH is a connected component of MATH for some MATH and such MATH is uniquely determined by the characteristic submanifold of MATH, there is a natural map MATH. This map is an isomorphism if MATH is connected for any MATH, but otherwise it is only surjective. As we did in REF , we identify the realization of MATH with MATH. One sees that MATH . Since MATH is itself a torus manifold, the spaces MATH and MATH have stratifications like for MATH, and hence we have a map MATH preserving the stratifications. By the induction assumption MATH . On the other hand, MATH has a stratification induced from MATH and each stratum is contractible. Since MATH restricted to MATH is a map from MATH to MATH preserving the stratifications and so is MATH as well, they are homotopic preserving the stratifications by REF . Therefore, we have the following commutative diagram: MATH where the left horizontal maps are natural ones. Tracing the upper horizontal sequence from the left to the right, MATH maps to MATH, and down to MATH by REF , while MATH maps through the lower horizontal sequence to the same element as observed in REF . Since the horizontal sequences above are injective, the lemma follows. |
math/0106229 | By REF we have a map MATH associated with the multi-polytope MATH. We denote the map by MATH. The composition MATH is a map from MATH to MATH sending MATH to MATH for any MATH, and so is MATH as well. Therefore, MATH and MATH are homotopic preserving the stratifications by REF . It follows from REF that MATH . This together with REF and the definition of MATH (that is, REF ) proves the theorem. |
math/0106229 | To MATH we correspond the degree of the minimal finite covering MATH of MATH such that there is a lifting of the action to MATH. The lifted action necessarily preserves MATH. It is not difficult to see that the correspondence is locally constant. Since MATH is connected the correspondence must be constant. |
math/0106229 | The argument is almost similar to the case of torus manifolds. One has only to observe that the characteristic suborbifolds and their intersections are torus orbifolds and a REF-dimensional torus orbifold is topologically a REF-sphere acted on by a circle group with exactly two fixed points. |
math/0106229 | Let MATH be an orbifold chart around MATH. We may regard MATH as a MATH dimensional MATH-module as before. As such, MATH is decomposed as a direct sum of MATH-modules MATH where MATH is projected into MATH by MATH. MATH can be regarded as embedded in the general linear group of MATH. Since MATH acts on each MATH preserving its orientation, there is a homomorphism MATH. The action of MATH on MATH is trivial. Moreover the action of MATH on MATH is effective because MATH. It follows that the homomorphism above embeds MATH into MATH. Since the kernel of MATH is equal to the intersection of MATH with the image of MATH, it is isomorphic to MATH. |
math/0106229 | We may assume that MATH. Take an equivariant NAME form MATH for the equivariant orbifold bundle MATH (we refer to CITE for NAME form and NAME form, compare also CITE). Let MATH be a point in the principal stratum of MATH, and MATH a reduced orbifold chart around MATH. The restriction of MATH to MATH is invariant under the action of MATH and its support is contained in a tubular neighborhood MATH of MATH, where MATH. Moreover, with respect to the fibering MATH, we have MATH, where MATH is the integration along the fiber of MATH. Note that the fiber is MATH, and that the action of MATH preserves the orientation of MATH. The equivariant NAME class MATH is the restriction to MATH of the cohomology class MATH of MATH. Here MATH is considered as a relative class in MATH where MATH is a tubular neighborhood of MATH. On the other hand, MATH is the restriction of a cohomology class MATH such that MATH where MATH denotes the projection of the fibration. Note that the fiber of MATH is MATH, where MATH acts effectively on MATH. We have MATH . But MATH is an isomorphism (NAME isomorphism). Hence we have MATH, and consequently MATH . |
math/0106229 | By REF we have MATH . But MATH viewed as MATH-module is MATH. It follows that MATH. Hence MATH . If MATH, then MATH. Therefore MATH. |
math/0106229 | Let MATH be a fixed point of the MATH-action. We restrict both sides of the equality in REF to MATH. On the left hand side we get MATH. On the right hand side the result is MATH by virtue of REF . But this is equal to MATH by the definition of the MATH. Thus both sides coincide after the restriction to each MATH. Since the restriction homomorphism MATH is injective modulo MATH-torsion, the equality is confirmed. |
math/0106229 | We have already shown that MATH is isomorphic to the kernel of MATH in REF . For the second part it suffices to note that MATH and MATH can be identified with the fundamental group of MATH and MATH. Therefore the kernel of MATH is isomorphic to MATH. |
math/0106229 | Let MATH be a torus orbifold of dimension MATH and let MATH be a point in the principal stratum of MATH. We may suppose that MATH. A closed tubular neighborhood MATH of the orbit of MATH is equivariantly diffeomorphic to MATH and the complement of MATH is connected because MATH is connected and the orbit has codimension MATH. Let MATH be another torus orbifold of dimension MATH with MATH, and let MATH be a closed subset in MATH corresponding to MATH in MATH. Since both MATH and MATH are equivariantly diffeomorphic to MATH and MATH has an orientation reversing diffeomorphim, there is an orientation reversing equivariant diffeomorphism between MATH and MATH. We remove the interior of MATH and MATH from MATH and MATH respectively and glue their boundaries through the diffeomorphism restricted to the boundaries and obtain a new torus orbifold MATH. The multi-fan MATH is the disjoint union of MATH and MATH. (Precisely speaking, MATH is the disjoint union of MATH and MATH with the empty sets in them identified.) If all connected components of MATH are geometrically realized, then we connect torus orbifolds that realize the connected components of MATH by the above method. Then the resulting torus orbifold realizes MATH. |
math/0106229 | Let MATH be a non-zero vector in the cone MATH. One may choose it to be MATH if MATH and MATH if MATH. Then one has a relation MATH with positive numbers MATH. This implies REF in the lemma. We shall prove REF in the lemma. Let MATH. Since the cardinality of MATH is MATH, there are exactly two elements MATH not contained in MATH, and MATH and MATH are in MATH, in other words, the MATH-dimensional cone MATH is a facet of only two MATH-dimensional cones MATH and MATH. We project them on MATH (the quotient space of MATH by the subspace generated by MATH). Then the vectors projected from MATH and MATH are toward opposite directions if and only if MATH. It follows from the completeness of MATH that MATH. This together with the definition of MATH shows that MATH. Since MATH is arbitrary, this proves REF . It also proves the completeness of MATH, so that MATH is minimal. The procedure from MATH to MATH corresponds to reversing orientations on characteristic suborbifolds MATH with MATH, so the latter statement in the lemma is obvious. |
math/0106229 | By REF , we may assume that the union of cones MATH over MATH covers the entire space MATH and the pair MATH, which we denote by MATH, is independent of MATH. When MATH, MATH can be realized by a weighted projective space, say MATH. There is an orbifold structure on a weighted projective space such that the edge vectors are all primitive. We admit these facts for a moment; the proof will be give in the appendix at the end of this section. Then MATH realizes the case when MATH. This completes the proof when MATH. Suppose MATH. For a general value of MATH, we prepare MATH copies of MATH and MATH copies of MATH and do equivariant connected sum along all MATH's and MATH's for each MATH. Then the resulting torus orbifold realizes MATH. The edge vectors are all primitive in this construction since it is so for MATH. |
math/0106229 | The proof is essentially the same as that of REF . As remarked in REF, it suffices to show that, when a generic vector MATH gets across a MATH-dimensional cone, the integer MATH in REF remains unchanged. Let MATH be an element of MATH and let MATH and MATH be the two elements in MATH. Then MATH and MATH are the elements in MATH which contain MATH. We project cones MATH and MATH on MATH. Then it follows from REF that the vectors projected from MATH and MATH are toward opposite directions if and only if MATH, where MATH is understood to be MATH. This together with REF of MATH implies that MATH remains unchanged regardless of the sign of MATH when MATH gets across the MATH-dimensional cone MATH. |
math/0106229 | Consider the projection of the cones MATH's on MATH. We define MATH or MATH according as the projection image of MATH disagrees or agrees with that of MATH. Applying REF with MATH and MATH, one sees that MATH . On the other hand, it follows from the completeness of MATH that MATH . These two equalities imply the lemma. |
math/0106229 | By REF we may assume that MATH is connected. We choose a generic (rational) MATH-dimensional cone MATH and decompose MATH using MATH into minimal multi-fans MATH's MATH. By REF MATH is realized by a torus orbifold, say MATH, such that all its edge vectors are primitive. We consider the disjoint union of MATH over MATH and piece them together using equivariant connected sum in the following way. For each MATH we do equivariant connected sum of MATH successively along MATH's, and similarly do equivariant connected sum of all MATH's along MATH as well. The resulting space is connected because MATH is connected, and becomes a torus orbifold. Its multi-fan is close to MATH but contains extra cones which are the cones spanned by MATH and MATH for MATH with MATH. For a fixed MATH, it follows from REF that there are the same number of MATH-fixed points MATH with MATH and MATH with MATH contained in the union of MATH with MATH and corresponding to the cone spanned by MATH and MATH. Hence one can do equivariant connected sum at pairs of MATH-fixed points MATH and MATH so that those MATH-fixed points will be eliminated. Doing this for each MATH, we obtain a torus orbifold, say MATH, realizing MATH. In fact, the characteristic suborbifolds MATH turn into a codimension two suborbifold of MATH, which is fixed by the circle subgroup determined by MATH but has no MATH-fixed point, so it is not a characteristic suborbifold of MATH by definition. This means that all the cones in MATH's containing MATH as an edge do not show up in the mulit-fan of MATH. |
math/0106229 | We already observed the "only if" part, so we prove the "if" part. By REF we may assume that our MATH, which satisfies REF , is connected. Then (the realization of) MATH is either REF . a MATH-simplex, or REF . the boundary of a MATH-gon where MATH, and that MATH . Using the latter statement in REF , the same argument as in the proof of REF shows that MATH in REF is geometrically realized. As for REF , let MATH be the unique simplex. There exist a finite covering MATH whose kernel MATH is isomorphic to MATH where MATH is the sublattice generated by the primitive vectors MATH's for MATH, and a MATH-dimensional MATH-module MATH corresponding to the cone MATH, as was explained in REF. Then the one point compactification of MATH, that is, the orbit space of MATH by an action of MATH, realizes our MATH in REF . |
math/0106229 | We shall define a correspondence MATH which is to be the inverse of MATH. Take a multi-fan MATH in MATH and MATH. It is easy to see there is a unique set MATH of edge vectors of MATH such that MATH and the maximal common divisor of MATH is MATH. Define a homomorphism MATH by requiring MATH. Then there is a unique finite covering map MATH which induces MATH. Let MATH be the kernel of MATH. The homomorphism MATH, hence MATH either, does not depend on the choice of identification MATH, but it depends on the numbering of MATH's. So if we put MATH, it induces a correspondence MATH as above. It is clear that MATH is in fact the inverse of MATH. |
math/0106229 | For each MATH let MATH be the isotropy subgroup at MATH of the MATH-action on MATH where MATH. MATH does not depend on the choice of MATH in MATH. If MATH lies in MATH for MATH, then MATH. We put MATH. We take a family MATH of small MATH-invariant open neighborhoods of MATH such that MATH converges to MATH when MATH tends to infinity. We may assume that MATH is equivariantly diffeomorphic to a MATH-invariant open disk in MATH. It is possible to make MATH's so small that they satisfy the following condition: MATH . Then MATH is an open neighborhood of MATH in MATH, and MATH is an orbifold chart of MATH compatible with MATH. On the other hand the fact that MATH implies that the kernel of MATH contains MATH,which we denote by MATH. Since MATH is the kernel of MATH, MATH is contained in MATH. We put MATH for MATH and define MATH . MATH can be considered as an open disk in MATH as pointed out in Remark above. The projection MATH induces a map MATH which induces a homeomorphism MATH. We shall prove that the family MATH forms a set of orbifold charts of an orbifold structure on MATH. For that purpose it suffices to show that, if MATH, then there are an injective homomorphism MATH and a MATH-equivariant open embedding MATH such that MATH . REF implies that, if MATH and MATH with MATH, and if MATH, then MATH. Therefore MATH . It follows that the inclusion MATH induces an injective homomorphism MATH. If MATH is taken in MATH, then MATH is contained in MATH. The inclusion induces an embedding MATH. MATH is clearly MATH-equivariant. REF follows from REF . If MATH lies in MATH, then the action of MATH lifts to the action of MATH on MATH and the lifting is minimal. Hence the edge vector of MATH corresponding to the orbifold structure defined above must be MATH. |
math/0106233 | Let MATH, MATH, MATH, MATH. Denote: MATH . For MATH, set MATH and let MATH be the set of even and odd supercommuting indeterminates. The correspondence MATH determines the homomorphism desired. |
math/0106233 | We have MATH . |
math/0106233 | Clearly, there exists a linear map MATH such that MATH. Let MATH. Then MATH . On the other hand, MATH . Further, MATH . On the other hand, MATH . This proves the first statement. The second one is obvious. |
math/0106233 | CASE: MATH . REF follows from REF CASE: MATH . On the other hand, MATH . |
math/0106233 | Let MATH. Then MATH . The identity MATH is similarly verified. |
math/0106233 | The functional MATH is uniquely, up ot a scalar multiple, characterized by its invariance with respect to the right coregular representation. Further, MATH is an invariant of the MATH-module MATH. Hence, by REF , MATH is an invariant with respect to the right coregular representation on MATH, that is, MATH is an invariant functional on MATH. Hence, MATH. On the other hand, MATH . Hence, MATH . To find MATH, let us substitute MATH into both parts of the identity. We obtain: MATH . |
math/0106233 | Let MATH be a base (system of simple roots) in the root system MATH. Set MATH . By induction on MATH we prove that MATH for any MATH, MATH, MATH and MATH. Let MATH. Then we obtain an exact sequence MATH with MATH. Therefore, MATH and MATH; moreover, MATH. Let REF hold. Consider the set MATH, where MATH and let MATH be the least element of this set. Clearly, MATH. There exists then a finite number of vectors MATH with weights MATH such that MATH. For all the other weights MATH of MATH we have MATH. The vectors MATH are, obviously, the highest weight ones, so we have an exact sequence MATH . This implies: MATH and if MATH, then MATH. This proves REF . Let MATH be any weight. Set MATH; apply REF to see that MATH. This proves REF . |
math/0106233 | First, observe that MATH . But, as is easy to verify, MATH. Therefore, MATH . Further, it is easy to verify that MATH . Hence, MATH. |
math/0106233 | Let us prove a more general statement; namely, let MATH be the character of an irreducible MATH-module with highest weight MATH. Then MATH . Indeed, by REF MATH . By multiplying both parts of the inequality by MATH we obtain MATH . If MATH, then MATH. Since MATH, it follows that MATH. But by REF MATH . So for MATH, we have MATH . In particular, applying this statement to the trivial module we obtain MATH, implying MATH. |
math/0106233 | MATH . |
math/0106233 | REF are similarly proved. Consider REF: MATH . This proves REF. CASE: We have MATH or MATH. This proves REF. Heading REF is similar. |
math/0106235 | Since MATH has MATH-boundary, there is a MATH-defining function MATH with MATH on MATH, such that MATH. We compute at MATH : MATH with MATH since MATH is a complex tangent vector to MATH at MATH, causing MATH to vanish. Let MATH, MATH tend to zero with MATH. Then the term MATH tends to zero. For every MATH close to zero, there is a small MATH such that MATH, in other words : MATH. Since MATH is not a complex tangent vector (if we choose MATH small enough), MATH. Hence this term has to be negative. Then there is a constant MATH such that MATH implies that MATH in other words : MATH. |
math/0106235 | Observe that MATH (since MATH intersects MATH transversally), such that the Jacobian of MATH does not vanish at MATH. Now apply the inverse function theorem. |
math/0106235 | First we consider the case that MATH. Then we choose MATH such that MATH. Let MATH. Then MATH for all MATH. If MATH is the MATH'th unit vector, we have MATH . We construct a compact MATH such that for all MATH contains the circle in the complex line through MATH and MATH with center MATH and radius MATH. Then we have for all MATH : MATH where MATH is the circle with center MATH and radius MATH. Hence MATH . Now we consider the case that MATH. Choose MATH such that MATH. Let MATH. Take MATH, MATH corresponding to MATH. One can make the appropriate estimate on MATH as above. Let MATH be a complex unit tangent vector to MATH at MATH. We construct MATH such that for all MATH contains the circles in the complex line through MATH and MATH with center MATH and radius MATH. Then MATH where MATH is the circle with center MATH and radius MATH. Hence MATH . We can choose MATH to be compact in MATH. For MATH the corresponding MATH depends continuously on MATH, hence we can choose linearly independent complex unit tangent vectors MATH, MATH, MATH to MATH at MATH that depend continuously on MATH. As a consequence of REF and the chain rule, we have : MATH . Previously, we already noted that MATH. Thus, the known numbers MATH are solution of the following system of MATH equations : MATH . The determinant MATH of the matrix to the left also exists for MATH, and it depends continuously on MATH. It is nowhere zero, and MATH is compact, hence its norm is bounded from below. The vectors MATH, MATH are linearly independent, as any complex line passing through MATH intersects MATH transversally. Hence we can use NAME 's rule to express MATH in terms of MATH, MATH, MATH, MATH and the integrals MATH. Each of those terms can be estimated from above with MATH, hence MATH. Since there is only a finite number of MATH's, we have that MATH. |
math/0106235 | It is easy to see that the function MATH is properly defined : let MATH, MATH uniformly on MATH, MATH respectively. Then MATH. MATH, hence MATH. Thus MATH. Since the sequence of polynomials MATH converges uniformly on MATH, their limit MATH is in MATH. We also have that MATH for MATH. As we can repeat this argument for every point MATH with corresponding neighborhood MATH, the proof is complete. |
math/0106235 | This a fairly standard application of the dominated convergence theorem of NAME. In detail : let MATH, let MATH. Then MATH for all MATH. Define MATH and MATH in the following way : MATH . Then MATH . For fixed MATH and MATH, MATH converges to MATH. Furthermore, for all MATH and MATH one has that MATH . For MATH there is a similar estimate on MATH. The function MATH is integrable on MATH. Applying NAME 's theorem yields that MATH thus MATH. |
math/0106235 | Choose a compact set MATH that contains MATH, as in the proof of REF . Choose a compact set MATH such that MATH. Let MATH, MATH, MATH be a sequence of polynomials (all vanishing at REF) that converges uniformly to MATH on MATH. Then MATH uniformly on MATH, hence, because of the chainrule, MATH . In the proof of REF we saw that for a polynomial MATH, the solution MATH, MATH, MATH of the system MATH is indeed MATH ,MATH, MATH. Taking the limit on both sides of the system of REF yields that MATH. |
math/0106235 | Let MATH be the determinant of the matrix to the left in MATH. We again use NAME 's rule to express MATH in terms of MATH, MATH, MATH, MATH and the integrals MATH. These are all continuous functions of MATH on MATH. Therefore MATH is in MATH, and repeating this argument for all MATH yields that MATH. Hence MATH. |
math/0106235 | For a MATH (or MATH), such that MATH, we have that MATH and MATH (or MATH). |
math/0106235 | From the conditions it follows immediately that MATH is connected. Suppose MATH is not MATH-convex. Then it is not weakly linearly convex either, meaning there is a point MATH such that every complex hyperplane MATH through MATH intersects MATH. We take for MATH the complex tangent space to MATH at MATH. It contains a complex line that is tangential to MATH at MATH and intersects MATH. This contradicts our assumption that such a line intersects MATH transversally. |
math/0106236 | Clearly, if MATH, then MATH, hence MATH. Let MATH be some edge edge with MATH (MATH is the vertex fixed by MATH). Choose MATH such that MATH, and choose MATH such that REF . With these definitions, we have MATH . We have MATH, which implies MATH. Hence, for MATH, there is no loss in assuming that MATH (otherwise, we could simply modify our choice of MATH). Finally, we have MATH, which implies MATH, which in turn implies MATH for some MATH. |
math/0106236 | We first construct a subtree MATH with MATH edges that projects to MATH. Choose MATH such that MATH. Let MATH, and define MATH if MATH, and MATH if MATH. Let MATH be the union of the edges MATH (REF, note that the union MATH of MATH projects to a spanning tree of MATH). Now, choose MATH such that MATH (we choose MATH). With these definitions, we have MATH and our claim concerning the images of vertex stabilizers follows immediately. We still need to compute the images of the generators MATH. We have MATH, which implies MATH. If MATH, we have MATH and MATH, so there is no loss in assuming that MATH. If MATH, then MATH and MATH, which implies MATH, so we can let MATH. If MATH and MATH, then MATH and MATH, so we have MATH, hence MATH. This implies MATH. Finally, there exists some MATH such that MATH, and an argument similar to the previous one shows that MATH. |
math/0106236 | As usual, we begin by lifting MATH to MATH. Let MATH be the (unique) vertex with MATH, and let MATH be some edge with MATH. Let MATH. Let MATH and MATH for MATH. Then the union MATH of MATH is the desired lift of MATH. There exists some MATH such that MATH. Choose MATH such that MATH for MATH. Using computations analogous to those in previous sections, we can immediately read off that MATH if MATH. Moreover, we have MATH, which implies that MATH . Using the identity MATH, we immediately see that MATH for some MATH, hence MATH . |
math/0106236 | As in the previous cases, we first construct a lift of the graph MATH. Pick some vertex MATH and some edge MATH such that MATH. Let MATH. Since MATH, there exists some edge MATH such that MATH and MATH is in the MATH-orbit of MATH. Let MATH. There is no loss in assuming that REF . There exists some MATH such that MATH. For MATH, we now recursively define MATH by letting MATH if MATH, MATH otherwise. There exists some MATH such that MATH. Moreover, there is no loss in assuming that MATH, that is, MATH. (here MATH denotes the integral part of MATH) We now choose MATH for MATH such that MATH. In particular, we can let MATH. The elements MATH and the vertex groups generate MATH. With these definitions, a careful analysis of the action of the MATH-s on MATH yields the above description of MATH. The computations are essentially the same as before. |
math/0106236 | CASE: Suppose MATH is not hyperbolic. Since an automorphisms is hyperbolic if and only if it is atoroidal by REF, there exists some word MATH such that MATH is conjugate to MATH for some MATH. There is no loss in assuming that MATH, where MATH is even, MATH and MATH. Clearly, both MATH and MATH are cyclically reduced, and there exists some even number MATH such that MATH (indices modulo MATH). But this implies that MATH, hence neither MATH nor MATH are atoroidal, so they are not hyperbolic, which contradicts our hypothesis. CASE: Suppose that MATH is not hyperbolic. Then, by REF, there exists some word MATH such that MATH is conjugate to MATH for some MATH. We can write MATH where either MATH, or MATH for some MATH, or MATH, or MATH. Moreover, we may assume that MATH is cyclically reduced and that MATH is minimal. In particular, there is no cancellation between successive subwords MATH, MATH (indices modulo MATH). Then for any MATH, there is no cancellation between MATH and MATH, and MATH is cyclically reduced. Moreover, the image of MATH is of the same form as MATH, for example, a subword of the form MATH is mapped to a subword of the form MATH. This implies that MATH maps each subword MATH to a conjugate of itself. Since MATH is hyperbolic, this rules out subwords MATH as well as subwords of the form MATH. The choice of MATH also rules out subwords of the form MATH and MATH, which implies MATH, a contradiction. |
math/0106238 | Given MATH-module isomorphisms MATH, then MATH is a MATH-automorphism of the MATH-module MATH and so MATH by REF . If MATH respect Hermitian metrics, then MATH and so MATH. |
math/0106238 | If MATH lie in the fibers over MATH of the projection REF, respectively, then MATH is an isometric automorphism of MATH while MATH is a Hermitian automorphism of MATH, yielding a commutative diagram MATH . Thus, just as in REF, these automorphisms obey MATH . Because MATH has determinant one, it preserves the volume form. Thus identity REF implies that MATH preserves the splitting MATH and hence, we may write MATH . Identity REF then implies that MATH on MATH, for MATH, and since the isomorphism MATH of complex lines is independent of the choice of unit-norm MATH, we must have MATH . Recall that MATH. Thus, MATH . Therefore, MATH is a principal MATH bundle, as claimed. |
math/0106238 | Suppose MATH is a smooth curve in MATH with initial point MATH. Let MATH, let MATH, and let MATH be a spin connection on MATH. Because MATH is a NAME module derivation of MATH, we have (see REF ) MATH where MATH also denotes the NAME connection on MATH. Thus, if MATH and MATH are parallel along MATH with respect to MATH, then MATH must be parallel along MATH with respect to MATH. Let MATH denote parallel translation, with respect to the NAME connection on MATH or spin connection on MATH, along MATH from MATH to a point MATH. Therefore, MATH and MATH. The sections of MATH over MATH given by MATH agree when MATH and are parallel along the curve MATH from MATH. Hence, parallel translation along curves in MATH with respect to spin connections commutes with NAME multiplication and thus yields NAME isomorphisms of the fibers of MATH. |
math/0106238 | The embedding MATH is MATH-equivariant if the circle acts on MATH by the action REF and on MATH by the composition of multiplication with weight two on the fibers of MATH and the circle action REF. The splicing map is MATH-equivariant if the circle acts by scalar multiplication on the sections both before and after cutting off the sections. Hence, the splicing map MATH is MATH-equivariant if the circle acts on MATH by the action REF and by the action REF in the lemma statement on the domain, namely: for MATH and MATH, by MATH . Because MATH acts diagonally in the definition of MATH, the action REF on the domain is equal to the circle action given by scalar multiplication with weight two on the fibers of MATH and by the action of the constant MATH in MATH with weight negative one on the factor MATH. This last action is equal to the action REF on the domain in the statement of the lemma by REF comparing the action of the constant MATH subgroup of MATH on MATH with the standard action: MATH . This completes the proof. |
math/0106238 | Because MATH is good, the union of strata MATH is disjoint from the union of strata MATH (see remarks in REF). Hence, for sufficiently small parameters MATH and MATH in the definition of MATH, the link MATH is disjoint from these strata. The geometric representatives MATH do not intersect the lower levels MATH of MATH except at points in MATH or at points in MATH by CITE. Therefore, the intersection REF is contained in the top stratum, MATH. The geometric representatives are transverse to MATH. Hence, the intersection MATH is a smooth submanifold of MATH. For generic values of MATH and MATH, the preceding intersection will be transverse to the edge and thus, by dimension-counting, disjoint from the edge. |
math/0106238 | The standard orientation for MATH is defined, through the diffeomorphism MATH by applying the convention REF to the submanifold MATH of MATH, while the orientation MATH is defined by applying it to MATH. Thus, to compare the two orientations of MATH, it suffices to compare the orientations of MATH and MATH which induce these orientations of MATH. An orientation for MATH is given by an orientation for the index bundle of the deformation operator MATH defined in CITE, whose kernel gives the tangent spaces of MATH. Suppose that the point MATH at which we do the orientation comparisons is obtained by gluing a framed MATH connection MATH onto the background pair MATH at a point MATH: MATH . Note that we abuse notation here and omit explicit mention of the frame for MATH. Then the excision argument CITE yields the isomorphism MATH . There are isomorphisms (see CITE), MATH where in the last isomorphism we have used the identification MATH and the identification of a ball in MATH with a ball around MATH in MATH via the gluing map. REF implies that the orientation MATH of MATH is given through the isomorphisms REF by the complex orientations of MATH, MATH, and MATH, the orientation of MATH determined by the homology orientation MATH CITE, the standard orientation of MATH defined in CITE, and the orientation for MATH given by that of MATH. We now describe the orientation of MATH inducing the standard orientation of MATH. First, the orientation of MATH is induced by one for MATH by identifying the normal bundle of MATH in MATH with the obstruction bundle MATH and using the complex orientation of this obstruction bundle. With MATH as in REF, then by REF, and REF there is an isomorphism of obstruction bundle fibers, MATH . Hence, the complex orientation of the obstruction bundle matches the complex orientation of the factors MATH and MATH in REF. We now compare the tangent space MATH with the remaining factors on the right-hand-sides of the isomorphisms REF. The moduli space MATH can also be written as the disk bundle appearing on the left-hand-side of the following diagram: MATH . We further suppose, without loss of generality, that MATH corresponds, via the gluing map MATH, to a point MATH in the base of the bundle on the left-hand-side of REF , so MATH . The fiber of the projection MATH in REF is MATH, while the same diagram identifies the fiber of the projection MATH with the fiber MATH of the projection MATH. Thus, MATH . If we compare the isomorphisms REF with REF, we see that the standard orientation of MATH is induced, through the convention REF, by the orientation of MATH given by the complex orientation of the normal bundle of MATH and the orientation of MATH defined through the isomorphism REF by the complex orientation of MATH, the orientation of MATH determined by the homology orientation MATH, the standard orientation of MATH, and the orientation of MATH. This orientation matches the orientation MATH as described in the paragraph following REF. This completes the proof. |
math/0106238 | The result follows from REF and the identity MATH given in CITE. |
math/0106238 | From REF, the cohomology class MATH is the first NAME class of MATH . By REF , the gluing map MATH is circle-equivariant when the circle acts on MATH by the action REF and on MATH by the action in REF , but with multiplicity two. Thus, MATH and the conclusion follows. |
math/0106238 | Let MATH be the MATH structure REF obtained by splicing, over a neighborhood of MATH, the MATH structure MATH over MATH with the MATH structure MATH over MATH, where MATH. We thus obtain an associated MATH bundle MATH as in REF and a bundle isomorphism, MATH . Hence, if MATH is the pre-image of MATH under the obvious projection, we have a bundle map MATH where MATH is the pre-splicing map REF. The map MATH is gauge equivariant with respect to the action of MATH on the domain and the action of MATH on the range. The group MATH of gauge transformations over MATH (see REF) acts trivially on the restriction of MATH to the complement of the splicing point MATH. From REF we have MATH (where MATH is the inclusion in REF) and so the bundle map REF descends to a bundle map on gauge-group quotients, MATH . By REF , the bundle map REF is circle-equivariant with respect to the circle action on the domain induced by the action REF on the factor MATH in MATH together with the trivial action on MATH and the circle action on MATH induced by REF. The bundle map REF thus descends to a bundle map on circle quotients, MATH . The argument in the proof of REF which shows that the two circle actions on MATH (described in its hypothesis) are equal then implies that the circle action on MATH induced by the circle action REF on the factor MATH in MATH and trivial actions on MATH and on MATH is equal to the twice the circle action described following REF. The multiplicity of this action does not affect the quotient, so the bundle given as the domain of the map REF is isomorphic to MATH and hence the restrictions of MATH and MATH to MATH are isomorphic. Finally, because the gluing map MATH and the splicing map MATH are MATH-equivariantly homotopic, there is a bundle isomorphism MATH. This completes the proof of the lemma. |
math/0106238 | The restriction of the difference, MATH to the subspace MATH defined in REF vanishes by that lemma. Therefore, by considering the exact sequence of the pair, MATH we see that the difference REF lies in the image of the homomorphism MATH appearing in the exact sequence of the pair REF. Because the map MATH is transverse to MATH, the image of the homomorphism REF (see CITE and CITE) is generated by MATH. The difference REF is therefore a multiple of MATH. One can calculate this multiple by evaluating the difference REF on a chain which intersects MATH transversely at a single point. Such computations are carried out in the proof of CITE and in CITE, giving REF . |
math/0106238 | The argument yielding REF implies that if MATH acts diagonally on MATH with weight negative two on MATH, then we have an isomorphism of line bundles over MATH: MATH where MATH is the projection. By the preceding isomorphism and REF of MATH, we have an isomorphism of line bundles over MATH: MATH . Consequently, from this isomorphism we see that, as bundles over MATH, MATH . Now, consider the circle quotient of the bundle on the right-hand side of the preceding isomorphism, MATH where the circle acts diagonally on the factors MATH and MATH. The definition of MATH as the line bundle associated to the circle bundle REF and the isomorphism REF then yield the bundle isomorphisms over MATH: MATH . One can check that the map defined below is an isomorphism of line bundles over MATH, MATH given for MATH, MATH, MATH, and MATH by MATH . Therefore, the isomorphisms REF give the desired isomorphism REF. |
math/0106238 | If MATH, then MATH by CITE. Thus, MATH . The assertions now follow from REF and standard computations (see, for example, CITE). |
math/0106238 | The MATH quotient of the bundle MATH by the action described at the end of REF can be written as MATH with the factor of negative one appearing because the action is diagonal, as explained in REF . Note that the circle action described in REF is twice the action in the definition of MATH (see REF ), so the weight two action on MATH described at the end of REF becomes a weight one action here. REF then follows immediately from REF and the definition of MATH. |
math/0106238 | Because MATH retracts onto MATH, as one can see from REF of MATH, it suffices to compute the NAME class of the restriction of MATH to MATH. It is easier to compute the NAME class of the bundle MATH. The observation (following from REF) that MATH . REF , and the description of MATH in REF imply that MATH . By the description of the circle action on the bundle MATH prior to its REF , the preceding isomorphism is circle-equivariant if the circle acts on the factor MATH by the action REF and with weight two on the fiber MATH and trivially on the remaining factors; the weight of the circle action on the fiber MATH is two because MATH appears in the tensor-product square on the left-hand side. The action of MATH on the vector bundle MATH is given, for MATH and MATH, by MATH where the homomorphisms MATH and MATH are defined in REF, respectively. To see this, observe that CASE: The action of MATH on MATH covers the action of MATH on MATH, CASE: The central MATH in MATH acts on MATH by scalar multiplication with weight negative one. REF implies that the action of MATH on the component MATH of MATH is given by the projection MATH and the action of MATH on MATH. REF implies that the central MATH acts on MATH by scalar multiplication on the fibers with weight negative two, just as in the definition of the homomorphism MATH before REF . Any group action satisfying the above two properties will differ from the action REF by a representation of MATH on MATH. However, by CITE there are no such non-trivial representations. Hence, these two properties characterize the above group action. REF yield the bundle isomorphisms, MATH and so MATH . The preceding isomorphism is MATH-equivariant, where MATH acts on MATH as in the description of the instanton obstruction bundle REF, if MATH acts trivially on MATH and on MATH by the standard action on MATH. (The action is trivial on MATH because elements of MATH act on the fiber of MATH by elements of MATH which are in the kernel of the homomorphism MATH.) Therefore, MATH . The bundle isomorphism REF is circle-equivariant and so, from REF of MATH, we see that the isomorphism REF descends to an isomorphism MATH where the circle acts with weight two on the fibers of MATH and the action REF on the factor MATH in MATH. By the equivalence of the circle actions on MATH given by REF and because the circle action of REF is twice the action in REF , the circle quotient on the right-hand side of REF is equivalent to one where the circle acts diagonally by scalar multiplication on the fibers of MATH and by the action of REF on MATH. The desired REF for the NAME class of MATH then follows from the isomorphism REF, the preceding description of the circle quotient on the right-hand side of REF, and REF . |
math/0106238 | Because MATH is good, there are no zero-section pairs in MATH. Thus, if the parameters MATH and MATH in the definition of MATH are sufficiently small, then the neighborhood MATH is disjoint from MATH. Hence, for any point MATH, the section MATH is not identically zero and MATH is not a reducible connection. By CITE or CITE, the restriction of MATH to any open subspace of MATH cannot be reducible. |
math/0106238 | REF and the dimension-counting argument used in the proof of CITE imply that the intersection MATH is disjoint from the lower level, MATH. The first assertion then follows from the proof of REF . The image MATH is the intersection of MATH with the union of the lower levels, MATH. Therefore, the second assertion follows from the first and the fact that MATH preserves strata on MATH. |
math/0106238 | Both sides of REF are linear in MATH, so we may assume without loss of generality that MATH for MATH. First, we note that the intersection numbers on both sides of REF do not change if we decrease the parameter MATH defining MATH (to another positive generic value) because of the obvious cobordism defined by this change of parameter. Thus, in the proof we may assume that the parameter MATH is as small as desired. There is a cobordism MATH with boundaries given by MATH . Such a cobordism exists because MATH and MATH are defined by pullbacks (by the appropriate restriction maps) of zero loci of sections of the same line bundle, for MATH, or by the degeneracy locus of sections of the same vector bundle for MATH. If MATH, where MATH and MATH, we replace MATH with MATH in the intersection REF as follows. By perturbing the cobordism MATH, we can assume that MATH is transverse to MATH . The dimension-counting arguments in the proof of REF show that the closure MATH in MATH of the cobordism MATH will not intersect MATH . Then, for a sufficiently small parameter MATH, the space MATH will not intersect MATH . Therefore the intersection MATH is contained in the transverse intersection of MATH with the space REF. Hence, this last intersection is a family of smooth, compact, oriented one-dimensional submanifolds with one boundary given by the set of points in the intersection on the left-hand side of REF and the other boundary given by the set of points in the intersection MATH . We now repeat this process with each MATH in the product MATH until MATH has been replaced by MATH. This completes the proof. |
math/0106238 | For any smooth manifold MATH, let MATH denote the chain complex of smooth singular chains CITE and let MATH be the complex of smooth singular cochains. For any smooth submanifold MATH, we define the complex of smooth singular, relative cochains by MATH . We will write MATH and MATH for the homology of the complexes MATH and MATH, respectively. Thus, MATH is the smooth singular cohomology of MATH and MATH is the smooth singular, relative cohomology of MATH. By the NAME theorem (see the discussion in CITE), there is a functorial isomorphism MATH. By applying the Five Lemma to the long exact sequences of the pair MATH in singular and smooth singular cohomology, we obtain an isomorphism MATH. We first define a smooth singular cocycle, that is, a closed MATH, which represents the cohomology class MATH. For any smooth singular chain MATH of dimension equal to that of MATH, set MATH where MATH is any smooth singular chain in MATH which is homologous in MATH to MATH and transverse to MATH. By definition of MATH, if MATH is a smooth singular cycle in MATH which represents a homology class MATH, we then have MATH so MATH is a smooth singular cocycle representing the cohomology class MATH on MATH. (By the NAME isomorphism, it does not matter whether we consider MATH as a singular cohomology class or a smooth singular cohomology class.) Because the restriction of MATH to MATH vanishes, the cocycle MATH defines an element MATH of the smooth singular, relative cohomology MATH which maps to MATH under the homomorphism MATH. Let MATH be the element of relative singular cohomology given by the image of MATH under the isomorphism MATH and let MATH be any representative of MATH. Then the element MATH given by the image of MATH under the homomorphism MATH will satisfy the conclusion of the lemma. |
math/0106238 | Let MATH be the inclusion map of pairs. The equality MATH implies that MATH is the image of the relative fundamental class of MATH under the inclusion MATH. By REF, the intersection on the left-hand-side of REF is a finite collection of points in the interior of MATH (with the multiplicities of the geometric representatives). The obstruction section MATH vanishes transversely on MATH by REF and the assumption that the parameters MATH and MATH in the definition of the link are generic, so the relative fundamental class of the manifold-with-boundary MATH is given by MATH . The geometric representatives intersect MATH transversely so, by the definition of a geometric representative and the cocycle MATH, the intersection number on the left-hand-side of REF is given by evaluating MATH on the relative fundamental class REF, yielding REF. |
math/0106238 | The definition of MATH in REF and the existence of a retraction from the uncompactified gluing data space MATH to the boundary MATH imply that MATH deformation-retracts to MATH. In turn, MATH deformation-retracts to the space MATH defined in REF. These retractions commute with MATH, so it suffices to prove the lemma for the restriction of the projection MATH to MATH. This restriction of MATH can be written as the composition of the projections MATH . The projection MATH induces an injective homomorphism on cohomology, so we need only verify that MATH induces an injective homomorphism on cohomology. The map MATH is the product of the identity on MATH and the projection MATH . As will be discussed the proof of REF , the space MATH can be identified with the projectivization of a complex rank-two vector bundle MATH. By CITE, the projection map MATH induces an injective homomorphism on cohomology. This proves the lemma. |
math/0106238 | To prove the homomorphism REF is injective, we first observe that it appears in the NAME sequence for the open cover MATH . The intersection of these two open sets is MATH. Hence, the homomorphism REF will be injective if the restriction map MATH is surjective. As noted in the proof of REF , there is a deformation retraction from MATH to MATH which commutes with MATH. Thus, proving the map REF is surjective is equivalent to proving that the following restriction map is surjective: MATH . The surjectivity of the map REF follows from a discussion of the following diagram in which the vertical maps come from the long exact sequences of the pairs: MATH . Because MATH is a smooth, codimension-MATH submanifold of MATH, with the NAME class of its normal bundle given by the pullback of the NAME class MATH of the normal bundle of MATH in MATH, we see that the relative cohomology MATH is generated by MATH. By REF , the following class is non-zero: MATH . Therefore, the generator MATH of the relative cohomology REF is not in the image of MATH, implying that MATH is the zero map. Hence, the map REF is surjective. As discussed previously, this implies that the map REF is injective. |
math/0106238 | By the definition of MATH in REF, the restrictions of the cohomology classes in REF to MATH coincide. The identity REF will then follow from REF if we can prove that the difference of the two cohomology classes is in the kernel of the restriction map given by the second component of the map REF: MATH . First, we describe the image of MATH under this map. Because there is an inclusion MATH the restriction of MATH to MATH vanishes. The expression for MATH in REF then implies that MATH . We now consider the restriction of MATH. Because the geometric representative MATH is pulled back from the quotient space MATH of irreducible MATH connections over MATH by the map MATH the cocycle MATH is pulled back from a cocycle MATH on MATH. By the construction CITE of the geometric representatives, the cocycle MATH represents the cohomology class MATH defined by the universal MATH bundle: MATH . We conclude that MATH . To compare the cohomology classes REF, we compare the bundles MATH and MATH. If MATH is the inclusion map, then there is an isomorphism MATH . (The existence of this isomorphism follows from the method used to obtain the corresponding isomorphism in REF .) The isomorphism REF then implies that MATH which, together with the identities REF, yields the desired result, REF. |
math/0106238 | Let MATH be any cocycle in the cohomology class MATH. By REF , there is a cochain MATH of degree MATH on MATH such that MATH . Because MATH, there is an open subspace MATH with MATH . Hence, the following intersection is empty: MATH . This implies that the map MATH which is surjective because the pairs appearing in this diagram are excisive couples; see CITE and CITE, is actually a map to the space of absolute cochains, MATH . Thus, we can write MATH, where MATH . Because MATH is supported on the complement of MATH in MATH, we see that MATH defines a cochain of degree MATH on MATH (by extending MATH by zero) and if we set MATH then MATH also represents the cohomology class MATH and REF yields MATH . This completes the proof. |
math/0106238 | Because MATH, there is an inclusion map of pairs: MATH . The sections MATH and MATH are homotopic through non-vanishing sections on MATH (by the homotopy MATH, for MATH) and so the relative NAME classes are equal, MATH as elements of MATH . For MATH sufficiently small, the intersection MATH is empty by REF and the continuity of the gluing map and obstruction section MATH with respect to NAME limits. Because MATH is supported on MATH while the relative NAME class in REF is supported on the complement of MATH, the cup-product in REF thus defines an element of MATH (compare the argument giving REF), so the pairing REF is well-defined. We can then write MATH using REF in the last line. This proves the desired identity REF. |
math/0106238 | By construction and REF, the cocycles MATH and MATH differ by cocycles of the form (up to a re-ordering which will be seen to be irrelevant) MATH where MATH. Note that we assume MATH or MATH, so we actually have a difference term in REF and not just MATH. Because the intersection REF is empty unless REF holds and because MATH and MATH have support in MATH and MATH respectively, the term REF vanishes unless REF holds. Consequently, MATH . REF implies that the following intersection is empty: MATH . Hence, because of the continuity of the gluing map and of the obstruction section MATH on MATH and by the same argument used to establish REF, the following intersection is also empty for MATH sufficiently small: MATH . By the reasoning which gave REF, the preceding equality implies that the cup-product MATH vanishes on MATH for MATH sufficiently small. If we assume that the neighborhoods MATH and MATH are contained in MATH for all MATH and MATH, the assumption that MATH or MATH implies that the difference term REF is supported in MATH and hence the cup-product of the difference term with the relative NAME class vanishes. Therefore identity REF holds. |
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