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math/0107085 | Since MATH is uniformly MATH close to MATH, its inverse is a contraction of MATH and has a unique fixed point, which after a change of coordinates we may assume is MATH. Since MATH is uniformly close to MATH it is fixed point free and after a further change of co-ordinates we may assume MATH . Consider the space MATH of MATH maps MATH such that MATH and MATH . Any MATH satisfies MATH for all MATH . Moreover, such a MATH is completely determined by its values on the interval MATH, since MATH . Indeed any MATH map MATH is the restriction of some element of MATH. Clearly, a sequence in MATH will converge if and only if it converges on MATH . Hence MATH can be considered a closed subset of the complete metric space of continuous functions on MATH with the MATH sup norm. Consider the map MATH on MATH given by MATH. Then MATH . It follows MATH . Moreover in the MATH sup norm this map is easily seen to be a contraction. It follows that there is a unique fixed point MATH for MATH in MATH. The map MATH is continuous and satisfies MATH for all MATH. The first of these equations implies MATH has image which is unbounded above and below, so MATH is surjective. This equation also implies that there is a uniform bound on the size of MATH for MATH . Indeed if MATH is an integer and MATH for MATH then MATH or MATH implies MATH . So the sets MATH have diameters with an upper bound independent of MATH . In particular, MATH is proper. We can now see that MATH is injective. Since MATH implies MATH which in turn implies MATH and MATH is is a uniform expansion, it follows that if MATH fails to be injective then the sets MATH do not have a diameter with an upper bound independent of MATH . This contradiction implies MATH must be injective. The fact that MATH is proper implies MATH is continuous. Hence MATH is a topological conjugacy from the standard affine action of MATH on MATH to the action of the group generated by MATH and MATH . |
math/0107086 | Without loss of generality, we may assume that MATH. Let MATH be a MATH-invariant open neighborhood of the orbit MATH. Our task is to construct an open neighborhood MATH of MATH such that MATH for all MATH. By the Slice Theorem (see, for example, CITE), there exists a MATH-invariant open neighborhood MATH of MATH and a MATH-equivariant projection map MATH that makes MATH a disc bundle. Let MATH denote the fiber MATH. Properties LS REF, LS REF, and the NAME Lemma guarantee the existence of positive number MATH such that the connected component MATH of MATH is an open disk embedded in MATH and contained in MATH. Let MATH denote the connected component of MATH containing the identity. Since MATH is MATH-invariant, MATH is contained in MATH. Using the associated bundle structure of MATH (see CITE), one can show that MATH is open and is the connected component of MATH containing MATH. Let MATH, and let MATH be an element of MATH. Since MATH is an element of MATH, LS REF implies that MATH is in MATH for all MATH, so that MATH is in the connected component of MATH containing MATH, and, hence, MATH. This implies that MATH is in MATH for all MATH. Since MATH, this proves that MATH is MATH-stable. |
math/0107086 | We first prove the case when MATH is positive semi-definite on MATH. The condition EM REF is necessary for the following construction. By the Slice Theorem, there exists a MATH-invariant open neighborhood MATH of MATH and a MATH-equivariant projection map MATH that makes MATH a vector bundle. Let MATH denote the natural isomorphism. Define MATH as MATH. The condition EM REF guarantees that MATH is well defined. This is called NAME 's velocity map CITE. We incorporate it in the definition of the MATH-invariant function used in the proof of the equivariant NAME stability test. Let MATH where MATH is a real number sufficiently large so that MATH satisfies LS REF (see below), and MATH and MATH are defined by MATH . The MATH-equivariance of MATH and MATH guarantees that MATH is MATH-invariant. The MATH-invariance of MATH follows from the equivariance of MATH and MATH-invariance of MATH and MATH. Hence, MATH is MATH-invariant. We now prove that MATH satisfies conditions LS REF, LS REF, and LS REF, for sufficiently large values of MATH. Condition LS REF. That MATH is a critical point of MATH is obvious. Hence, it suffices to prove that MATH is a critical point of MATH. Let MATH denote the slice MATH. Then MATH . Since MATH is MATH-invariant, this implies that we only need to show that MATH is a critical point of MATH. Since MATH (where MATH is the natural isomorphism) we have MATH for MATH in MATH, so that MATH . Therefore, by EM REF, MATH is a critical point of MATH. This proves that MATH is a critical point of MATH. Condition LS REF. From the decomposition REF, we see that the condition LS REF is satisfied if and only if MATH is positive-definite on MATH. Let MATH and MATH denote the restrictions of MATH and MATH to MATH, respectively. We claim that MATH is positive semi-definite with kernel MATH, and that MATH is positive-definite on MATH. Then MATH is positive-definite for sufficiently large values of MATH, so that MATH satisfies LS REF. The claim is proved as follows. Since MATH is positive semi-definite with kernel MATH, we conclude that the quadratic form MATH is positive semi-definite with kernel MATH. Since MATH contains MATH, the decomposition REF implies that MATH . Therefore, by EM REF, MATH is positive-definite on MATH. Combining this with REF, we conclude that MATH is positive-definite on MATH. Condition LS REF. Let MATH denote the flow of MATH. Since MATH, we have MATH, and MATH . Since MATH is in MATH for any MATH, we have MATH, and MATH . The inequality follows from REF and MATH-invariance of MATH. Hence, LS REF is satisfied by the continuity of MATH, MATH, and MATH. If MATH is negative semi-definite on MATH with kernel MATH, then we redefine MATH by MATH . The proof proceeds as in the positive semi-definite case. |
math/0107092 | Given MATH, we construct MATH and its desingularization MATH explicitly. Let MATH be the unit circle bundle of MATH. Set MATH in MATH and MATH with MATH. Then MATH is a principal MATH-bundle over MATH. In general, the unit circle bundle MATH is an orbifold rather than a manifold. MATH is a quotient of MATH by the action of MATH defined by MATH. It follows that the isotropy group of a point in the quotient of MATH is MATH for all points MATH. When the isotropy group is trivial REF the quotient is smooth. In the case that MATH, MATH is a usual line bundle around that loop, but the MATH-manifold still has a nontrivial orbifold structure. Set MATH. Let MATH be the meridian loop of MATH before the quotient is taken. Denote the class it represents in the quotient MATH by MATH. The homeomorphism MATH determines a section MATH which is specified up to homology by the relation: MATH where MATH is the meridian of MATH in MATH, MATH is a fiber of MATH, and MATH. The local invariants MATH specify MATH up to orientation-preserving equivariant homeomorphism. The bundle MATH can be extended to the unit circle bundle of MATH by equivariantly attaching MATH with a bundle isomorphism MATH. Bundle isomorphisms covering the identity are classified up to vertical equivariant isotopy by homotopy classes of maps in MATH. However we can change MATH by a bundle automorphism classified by MATH; these maps change MATH by a multiple of the fiber. Therefore the resulting bundle MATH can be completely specified by the map MATH for some MATH. Thus we determine the principal MATH bundle of MATH by specifying that MATH. In summary: CASE: The unit circle bundle of MATH is obtained by gluing the quotient MATH using maps MATH . Note that this bundle depends only on the section MATH as well. CASE: The unit circle bundle of MATH is obtained by gluing in the quotient MATH into MATH using maps MATH . CASE: The unit circle bundle of the desingularization MATH is obtained by gluing in MATH using maps MATH . Next we show that two orbifold line bundles MATH and MATH with the same isotropy representations and equivalent desingularizations MATH are equivalent as orbifold line bundles. Construct two principal MATH-bundles MATH and MATH from MATH to form unit circle bundles MATH and MATH. The construction depends on choices of the class MATH coming from sections MATH for MATH. Let MATH be the obstruction to extending these sections over MATH. Let MATH be the primary difference of MATH and MATH. A diagram chase MATH shows that MATH. Thus there is an element MATH such that MATH, and MATH. Therefore MATH implying that MATH is homotopic to MATH through a homotopy in MATH. Since the construction of the unit circle bundle of the orbifold line bundle in REF depended only on these sections, MATH and MATH are equivalent. |
math/0107092 | If we can split MATH into a REF-dimensional real line bundle and a complex orbifold line bundle MATH, then MATH is just the mod REF reduction of MATH for some orbifold line bundle MATH. Hence MATH lifts. We need to find a nowhere zero section of MATH. Note that each MATH comes with a MATH-invariant oriented nonzero vector field that is tangent to MATH at each point in MATH. This vector field induces a nonzero section MATH. Remove an extra MATH from the interior of MATH and put a similar nonzero section on the boundary. The obstruction to extending the section into the interior of MATH is an element of MATH . Using the homology relation MATH the obstruction can be removed by changing the framing on the boundary of MATH. Thus MATH admits a nowhere zero vector field. |
math/0107092 | Let MATH and MATH be two MATH structures which are lifts of the frame bundle. Away from the MATH's, the difference of two MATH structures is a complex line bundle as in the smooth case. Because MATH for all MATH, we can extend the complex line bundle over the desingularization MATH using techniques in REF . Thus we can investigate locally to show that any two lifts of isotropy representations into MATH differ by a representation into MATH. Note that this is not immediately obvious because there are many different representations of MATH into MATH. Let MATH be the unit vector field on MATH which is tangent to the circle MATH at each point. We use the fact that MATH is a rotation which leaves the nonzero vector field MATH invariant. Identify MATH with the unit quaternions. The map MATH is given by MATH for all MATH and is the double cover of MATH. Thus MATH can be thought of as the unit quaternions modulo the equivalence MATH. Without loss of generality, we may assume that the invariant vector field MATH is generated by MATH at each point in MATH. It is easy to see that elements of MATH which rotate the second two components while leaving MATH invariant are of the form MATH. Hence MATH where MATH is a MATH-root of unity in MATH and MATH. The MATH representation MATH is given by MATH for all MATH. Here we have used the fact MATH. Using this identification, MATH projects to MATH by the adjoint map as well, MATH for all MATH. Thus the representation MATH lifts to MATH: MATH given by MATH (or equivalently MATH) for some representation MATH. The representation MATH is given by MATH for some MATH. Hence the difference of two MATH-structures MATH locally is a representation MATH. Globally, MATH differs by a complex line bundle over MATH and local isotropy representations into MATH, that is, an element in MATH as described in REF . |
math/0107092 | Extend MATH to an orthonormal coframe MATH on a patch of MATH so that MATH, and MATH are horizontal lifts of an orthonormal coframe MATH on MATH. Let MATH be the dual vector fields with respect to the metric MATH. The difference MATH-form MATH can be thought of as an element in MATH via the vector space isomorphism MATH defined by MATH . The action of MATH on the bundle MATH is modeled on MATH. Thus we can NAME multiply the MATH component of MATH to get MATH where MATH is a linear map defined by MATH for a basis element MATH. This map can be conveniently reformulated as MATH where MATH is contraction with the vector field which is MATH-dual to MATH. Let MATH be the functions defined by MATH . We can use REF and the first NAME Structure equation MATH to calculate the connection matrix for MATH . For example, we can write MATH as MATH to get the top row of the connection matrix MATH . The MATH's in the second, third, and forth row are pulled-back from the connection MATH-form for the NAME connection on MATH. The connection matrix for MATH is MATH . Using the isomorphism MATH, the difference MATH can be written as MATH . A straight forward calculation gives MATH. |
math/0107092 | We work with the full NAME operator first. By using the definition of MATH from REF , we see from MATH that MATH is MATH self-adjoint. The NAME operator decomposes into a sum of two self-adjoint operators: MATH where MATH. Squaring and noting that MATH and MATH yields MATH where MATH denotes the anticommutator. The last term simplifies using NAME relations and REF : MATH . One can use the connection matrix REF to calculate that MATH for MATH. Therefore the curvature reduces to MATH and we can see that MATH . By the definition of MATH, the action of MATH for MATH commutes with NAME multiplication. Therefore MATH is a scalar endomorphism, so MATH . Restricting attention to MATH gives the formula. |
math/0107092 | Let MATH be a local coframe where MATH are pulled back from the base. First we take the adjoints of both terms on the left hand side of REF . Applying the adjoint of MATH in the first term of REF gives MATH . We used REF in the first line, and REF in the second. We also used the definition of MATH from REF . Similarly, we take the adjoint of MATH in the second term of REF to find MATH . Next we show that the sum of the right hand sides of REF is equal to MATH . First, note that for MATH, MATH . This holds for MATH since MATH is the curvature of a principal orbifold circle bundle so is pulled back from MATH; it holds for the remaining MATH since MATH are pulled back from MATH. Hence, MATH . Combining REF , and REF and projecting onto the self-dual REF-forms we get: MATH . Using REF , we can reduce further MATH . The last equality is the same calculation as in REF . |
math/0107092 | Recall that for a generic choice of metric and perturbation MATH the moduli space satisfies MATH for all solutions MATH to REF-dimensional NAME equations (see subsection REF for more details). For this good pair the moduli space MATH is a smooth manifold with no reducible solutions. Since we can choose a perturbation generically such that the projection of MATH onto the harmonic REF-forms is not a multiple of the harmonic representative of MATH for all solutions in MATH, we have that MATH on MATH as well, hence MATH does not contain reducible solutions either. By Theorem B, MATH is diffeomorphic to a smooth manifold without reducible solutions. We have in effect shown that MATH is a good pair and that this moduli space can be used to calculate the SW invariant. Choose a specific MATH structure MATH on MATH such that MATH is pulled back and MATH. There exists a MATH structure MATH on MATH such that MATH by Theorem B, and MATH . From this the formula follows. |
math/0107094 | For each prime MATH, let MATH, the congruence subgroup of level MATH (here MATH denotes the identity matrix of MATH). Then MATH is a residually finite MATH-group. Set MATH. Then MATH is a residually finite MATH-group for all MATH. Furthermore MATH has finite index in MATH, so by CITE we see that MATH is not right orderable. This completes the proof of REF . |
math/0107094 | Observe that MATH is torsion free. Let MATH be a normal solvable subgroup of finite index in MATH with minimal derived length. We shall prove the result by induction on the derived length of MATH, the result being obvious when this is zero because then MATH and thus MATH will be finite. We now assume that the derived length of MATH is at least one. Let MATH denote the penultimate term of the derived series of MATH. Then MATH is a normal abelian subgroup of MATH and MATH has strictly smaller derived length than that of MATH. Let MATH be a maximal normal abelian subgroup of MATH containing MATH. We shall let MATH denote the pro-MATH completion of a group MATH. Then the exact sequence MATH yields an exact sequence MATH . Let MATH denote the image of MATH in MATH. Since MATH is abelian, we see that MATH is abelian and we deduce that MATH is abelian. Also we may view MATH as a subgroup of MATH because MATH is a residually finite MATH-group. Therefore MATH is an abelian normal subgroup of MATH containing MATH and we conclude that MATH. But MATH can also be described as the kernel of the natural map MATH. Therefore MATH is isomorphic to a subgroup of MATH and we deduce that MATH is a residually finite MATH-group. Similarly MATH is a residually finite MATH-group. Since the derived length of MATH is strictly less than the derived length of MATH, induction shows that MATH has a series MATH with MATH and MATH torsion free abelian for all MATH. By setting MATH, we obtain the required series for the first part of REF . The assertion that MATH is right orderable now follows from CITE and the fact that torsion free abelian groups are right orderable. |
math/0107095 | Let MATH be the tiles of type MATH in MATH whose lowest squares lie just above a given diagonal MATH. It is obvious from the definition of MATH that the ordering of these tiles with respect to each other is independent of MATH. Similarly, let MATH be the tiles of type MATH that lie just below the diagonal MATH. Now, let MATH be the totally ordered set of all the boundary squares and all of the MATH tiles that are incident to MATH. Since we know that the ordering restricted to the set of tiles of form MATH (or the set of tiles of form MATH, or the set of boundary squares) is independent of MATH, it follows from REF that the total ordering MATH is independent of MATH. Since any two tiles of the same type or any tile and any square that are comparable by MATH must be incident to some common diagonal MATH, REF covers all possible cases in which the MATH relation occurs between tiles of the same type or between tiles and boundary squares. |
math/0107095 | We must show that, given a tiling MATH, there exists no sequence MATH where each MATH is either a tile or a square in MATH. Since there can clearly be no cycle containing one or two elements, we may assume MATH. Now, suppose a diagonal incident to both MATH and MATH is also incident to some MATH with MATH. If MATH, then MATH is a shorter cycle. If MATH, then MATH (by transitivity, since MATH, MATH, and MATH are on the same diagonal and hence totally ordered), and MATH is a shorter cycle. Thus, if MATH is a cycle of minimal length in MATH, then any infinite diagonal that is incident to both MATH and MATH can be incident to no other tile or square. Denote by MATH the infinite diagonal between square levels MATH and MATH and let MATH and MATH be the lowest and highest values of MATH for which MATH is incident to both MATH and MATH. All tiles and squares other than MATH and MATH must lie entirely below MATH or entirely above MATH. Since no tile or square below MATH can be comparable with any tile or square above MATH, the sequence MATH must lie entirely above MATH or entirely below MATH. Suppose without loss of generality that the former is the case and that the lowest square in MATH is at least as low as the lowest square in MATH. Then none of the elements MATH below MATH is comparable with MATH, and this is a contradiction. |
math/0107095 | Obviously, the first statement implies the second. Now, suppose MATH. Since these orientations are acyclic, there must exist at least one leftmost tile, that is, there exists a MATH for which there exists no MATH with MATH or MATH. It follows that MATH and MATH must each contain the leftmost squares in MATH in the rows of level MATH; thus, their positions are determined and they are equal. Similarly, we can now choose a MATH that is leftmost among the remaining set of tiles, and its position is also completely determined. Repeating this argument, we conclude that MATH for all MATH and MATH for which these values are defined. |
math/0107095 | Suppose MATH is a leftmost tile vertex in MATH. As in the previous proof, we will choose the leftmost squares of each of the levels MATH and declare them to be our first tile MATH. That there exists at least one square in MATH on each of these levels follows from the consistency conditions of MATH. However, we must verify that this set of squares is actually a ribbon tile. Since it contains one square on each of levels MATH, it is sufficient to show that it is connected. Suppose otherwise; then we can assume without loss of generality that there exists some MATH such that the leftmost square MATH in the row of level MATH is left of the square MATH in the row of level MATH but that these two squares are not adjacent. Now, let MATH REF be the square of level MATH that is right of (left of) and adjacent to MATH. Both MATH and MATH must be squares in MATH, since they are adjacent to MATH and they are left of MATH, the leftmost square of their level in MATH. Now, by the consistency conditions, there is exactly one tile MATH containing a square on level MATH such that MATH; every other tile with a square on level MATH must be right of MATH. If MATH, then MATH, violating acyclicity. Hence MATH. However, since MATH is left of every square of level MATH, one may deduce from the consistency conditions that every tile vertex MATH of MATH that crosses level MATH must satisfy MATH. Thus MATH, a contradiction, and it follows that the squares chosen do in fact form a tile. Now we form a new region MATH by removing the squares in this tile from MATH. We form a new graph MATH by removing MATH from the vertex set of MATH - along with all boundary squares of MATH that are not boundary squares of MATH - and adding all the boundary squares of MATH that were not boundary squares of MATH. (These are necessarily members of our removed tile.) We define the edges of MATH and their orientations in a new orientation MATH in the most natural possible way. Two vertices of MATH are adjacent whenever the corresponding edges in MATH were adjacent, and MATH is the orientation induced by MATH on such edges. Furthermore, if MATH is one of the new boundary squares and is in a row MATH, we declare it to be adjacent in MATH to every tile MATH that crosses rows MATH, MATH, or MATH (with MATH) and every boundary square MATH of those levels (with MATH if and only if MATH is directly or indirectly left of MATH and MATH when the opposite is true). Finally, if MATH was a vertex of MATH with MATH, we relabel it MATH in MATH. It follows easily from the definition that the MATH thus defined - together with the partial orientation on forced edges induced by MATH - is a ribbon tile graph consistent with MATH. If we can further prove that MATH is necessarily acyclic, then REF will follow by induction. Suppose MATH has a cycle; then this cycle must include a square MATH of MATH that was a member of the tile MATH - otherwise the cycle would also be a cycle in MATH. Also, although the cycle may contain multiple squares from the removed tile, it must contain at least one element that is not a square of MATH, since the squares do not themselves contain any cycles. Suppose the cycle contains a sequence MATH of former squares of MATH but that the elements MATH immediately preceding this sequence and MATH immediately following it are not squares of MATH. Then we must have had MATH in MATH, and MATH too must have had a cycle. |
math/0107095 | First of all, we can determine whether a local replacement move involving tiles MATH and MATH can occur by applying the theory we have already developed to MATH, the region tiled by just these two tiles. Now, if MATH admits two distinct tilings, then MATH must admit two distinct orientations that extend the orientation on the forced edges. In particular, MATH must contain at least one free edge, and that can only be the edge MATH. For this edge to be free, we must have MATH. Without loss of generality, suppose MATH. Notice that if MATH, then MATH contains two squares of level MATH. Inspection shows that if tile MATH contains the leftmost square of one of these same-level pairs, it must contain the leftmost square of every such pair. (See REF .) Moreover, if a tile contains one square on a row higher than MATH (lower than MATH), it must contain every square with this property. Thus, a local replacement move must consist of swapping the squares in each same-level pair. Such a swap produces a new pair of ribbon tiles if and only if the single square on row MATH (respectively, MATH) is adjacent to both of the squares in row MATH (respectively, MATH). Inspection shows that whenever there is no square MATH with either MATH or MATH, then a local move can in fact occur. Now, if MATH is the tiling obtained by applying such a move to MATH, and MATH then inspection shows that we must have MATH; it is also easy check that if MATH is any tile or square outside of MATH, then MATH if and only if MATH. Similarly, MATH if and only if MATH. It follows that if MATH and MATH differ by a local replacement move, MATH and MATH do in fact differ by the reversal of exactly one edge: the edge MATH. For the converse, note that if the orientation of an edge MATH of the orientation MATH can be reversed without creating a cycle, we know there must be no vertex MATH of MATH with either MATH or MATH; we have already shown that this is a sufficient condition for there to exist a local move involving the tiles MATH and MATH, and so MATH must be the orientation obtained by reversing the edge. |
math/0107097 | This follows by induction on MATH: it follows from the definition of the MATH in the case where MATH. In the general case, it follows by applying the associativity identities and the identities involving the unit map, MATH. Consider the diagram MATH . The associativity condition implies that we can shuffle copies of MATH to the immediate left of the rightmost term, and shuffle the MATH on the right to get a factor on the left of MATH and one on the right of MATH (this factor of MATH exists because MATH) and we can evaluate MATH on each of these before evaluating MATH on their tensor product. The conclusion follows from the fact that each copy of MATH that appears in the result has been composed with the unit map MATH so the left factor is MATH and the right factor is MATH so the entire expression becomes MATH which is what we wanted to prove. |
math/0107097 | This is a straightforward consequence of REF and the associativity condition in REF . |
math/0107097 | We prove the claim when MATH and use the functoriality of MATH with respect to morphisms of MATH to conclude it in the general case. In this case MATH and MATH. We claim that the diagram MATH commutes, where MATH, MATH, and MATH are as defined in REF so that the lower row and right column are the same as REF . Clearly, the upper sub-triangle of this diagram commutes since MATH. On the other hand, the lower sub-triangle also clearly commutes by the definition of MATH and the fact that MATH is a MATH-coalgebra. It follows that the entire diagram commutes. But this implies that MATH (in the notation of REF ) satisfies the condition that MATH so that MATH is MATH-closed - see REF . It follows that MATH since MATH is maximal with respect to this property (see REF ). This implies both of the statements of this lemma. |
math/0107097 | Let MATH if MATH is unital and MATH otherwise. It is very easy to see that REF commutes with MATH as defined in REF . Suppose that MATH is another coalgebra morphism that makes REF commute. We claim that MATH must coincide with MATH. The component MATH isomorphically maps MATH to the direct summand MATH . For MATH to be a coalgebra morphism, we must have (at least) MATH for all MATH. This requirement, however, forces MATH. REF and the argument above verify all of the conditions in REF . |
math/0107097 | Consider the endomorphism MATH . The operad identities imply that the diagram MATH commutes since MATH is a unit of the operad and MATH must preserve the coproduct structure (acting, effectively, as the identity map). It follows that MATH and that MATH. The hypotheses imply that MATH and we claim that MATH. Otherwise, suppose that MATH. Then MATH so MATH, which is a contradiction. The conclusion follows. |
math/0107097 | REF of the sub-coalgebra MATH is stated in an invariant way, so that any coalgebra morphism must preserve it. |
math/0107097 | Since the kernel of the the structure map of MATH vanishes MATH so that MATH is mapped isomorphically by the projection MATH. It is first necessary to show that MATH can serve as the cogeneration map, that is, that REF commutes. This conclusion follows from the commutativity of the diagram MATH where MATH is the canonical classifying map of MATH. The upper (curved) triangle commutes by the definition of MATH, the lower left triangle by the fact that MATH splits the classifying map. The lower right square commutes by functoriality of MATH. We must also show that MATH is pointed irreducible. The sub-coalgebra generated by MATH is group-like. Claim: If MATH is an arbitrary element, its coproduct in MATH for MATH sufficiently large, contains factors of MATH. This follows from the fact that MATH must have terms MATH for MATH - see REF with MATH. It follows that every sub-coalgebra of MATH must contain MATH so that MATH is the unique sub-coalgebra of MATH generated by a group-like element. This implies that MATH is pointed irreducible. The statement about any pointed irreducible coalgebra mapping to MATH follows from REF . |
math/0107097 | Let MATH and MATH be free MATH-modules. Elements of MATH can be described as MATH, where MATH, MATH are basis elements. So the list in this case has two elements and the set of lists contains a single element: MATH . Elements of MATH are functions from MATH to MATH - that is, a set of ordered pairs MATH where MATH is a basis element, MATH (not necessarily a basis element) and every basis element of MATH occurs as the left member of some ordered pair. The general statement follows from the recursive definition of MATH in REF and the definition of in-order traversal in REF . Now we prove REF : Let MATH be given by MATH as in REF . We consider the effect of the transformations defined in REF through REF on this element. CASE: transformations insert or remove terms equal to MATH into each list in the set. CASE: transformations have no effect on the lists (they only affect the predicates used to define the module whose elements the lists represent). CASE: transformations permute portions of each list in MATH. In the DG case, whenever an element MATH is permuted past an element MATH, the list is multiplied by MATH. Note that, in no case is the data in the lists altered. Furthermore, we claim that the equality of the trees resulting from performing MATH and MATH on MATH implies that: CASE: the permutations of the lists from the type-REF transformations must be compatible CASE: the copies of MATH inserted or removed by the type-REF transformations must be in compatible locations on the tree. Consequently, the lists that result from performing MATH and MATH on the lists of MATH must be the same and MATH . The isomorphism of final expression trees also implies that the predicates that apply to corresponding element of these lists are also the same. Since this is true for an arbitrary MATH we conclude that MATH . In the DG case, we note that type-REF transformations may introduce a change of sign. Nevertheless, the fact that the elements in the lists are in the same order implies that they have been permuted in the same way - and therefore have the same sign-factor. |
math/0107097 | Let MATH be given by MATH . We get MATH where MATH are permutations. In each of these lists, we replace MATH by a set of lists MATH representing the value of MATH and apply MATH and MATH, respectively - possibly permuting the resulting longer lists. As in REF , the result is two copies of the same set of lists. This is because both sets of operations result in the expression tree MATH, implying that the permutations must be compatible. As in REF , the key fact is that the data in the lists is not changed (except for being permuted). |
math/0107098 | The right integral MATH is itself an element of MATH REF . Since MATH, the result of the left action of MATH on MATH is again in MATH . NAME, suppose MATH . Then MATH which has to be equal to MATH for all MATH, and this holds if and only if MATH. |
math/0107098 | A central element MATH acts on a simple MATH-module as multiplication by a scalar, which is nonzero whenever MATH is not nilpotent. Therefore, as a MATH- module, MATH coincides with the maximal semisimple submodule of MATH, which is annihilated by the radical of MATH. By restriction, MATH is an isomorphism of MATH-modules, and REF provides an isomorphism of MATH-module structures on MATH and MATH, the latter given by multiplication. In particular, MATH coincides with the annihilator of the radical (that is, the socle) of MATH. |
math/0107098 | Let MATH, which means that MATH for any MATH. Then in general, MATH CITE. Write MATH for some coefficients MATH, and MATH . Then MATH, for any MATH. Solving the system, we get MATH, where MATH are the minors of MATH, and MATH. By assumption this implies MATH, therefore MATH and MATH. Clearly all MATH, and hence MATH is generated by MATH, and MATH is nondegenerate on MATH. |
math/0107098 | NAME for a finite dimensional NAME algebra means that MATH is trivial. Since MATH is a homomorphic image of MATH and MATH by REF , we get an isomorphism. |
math/0107098 | The finite dimensional simple modules of type MATH (that is, all MATH act on them by MATH) over MATH are parametrized by the dominant integral weights MATH. The algebra of characters, spanned by MATH, is isomorphic to MATH. By NAME 's tensor product theorem CITE, MATH where MATH and MATH for all MATH. By restriction, simple MATH-modules of type MATH are modules over MATH. The subquotient MATH of MATH acts trivially on the simple modules MATH, whose weights are all in MATH CITE. Therefore, the algebra of characters MATH of MATH-modules can be obtained from that of MATH by setting the characters of simple modules with MATH- multiple highest weights equal to their dimensions. This leads to the formula above. |
math/0107098 | Since MATH is assumed to be simply laced, we will always identify the root lattice MATH with MATH, the MATH-span of the coroots MATH. Denote by MATH the real vector space spanned by the simple roots, and let MATH be its dual space with the basis MATH such that MATH for any MATH. Here MATH is the restriction of the Killing form in MATH. Then MATH, the MATH-span of MATH, is a lattice in MATH dual to MATH. Let MATH denote the maximal torus of the simply connected group MATH with MATH. Then the elements of MATH can be considered as characters on MATH by setting MATH, for any MATH. Recall that MATH. Then we are in the setting of REF (in our case MATH in the notations of CITE), and we have MATH where the map MATH is given by the evaluation at MATH. The notation MATH stands for the set of representatives of the MATH orbits in MATH, which has a natural MATH action. The subgroup MATH is generated by the reflections MATH. We can rewrite the result parametrizing the blocks by the MATH-orbits in the restricted root lattice. Indeed, the set MATH contains all MATH where MATH. Therefore, it is parametrized by MATH, which under duality corresponds to MATH, with the orbits of the MATH-action in MATH translated to the orbits of the MATH-action in MATH. Thus the maximal ideals of MATH can be parametrized by MATH, the representatives of MATH-orbits in the restricted root lattice. The stabilizer subgroup MATH is generated by those MATH such that MATH, or MATH for MATH. In the dual picture this corresponds to those MATH, such that MATH for MATH. These are exactly the reflections MATH that stabilize the element MATH with respect to the MATH-action. Denote by MATH the subgroup generated by such reflections for a fixed MATH. Then we get MATH . However, if we want to consider the subalgebra MATH, which is isomorphic to MATH, then it will be more natural to parametrize its blocks by weights in MATH. This agrees with the decomposition of MATH as a module over itself, where the block corresponding to an element MATH contains only the composition factors of highest weights in the MATH-orbit of MATH. The set MATH under the MATH-action has the same orbit structure as MATH under the MATH-action. Indeed, with the conditions on MATH formulated in REF , for any MATH there exists a unique MATH such that MATH, MATH. This provides a one-to-one correspondence between the orbits of the MATH actions in MATH and MATH. Clearly the MATH-action on MATH has the same orbit structure. Therefore we can translate the parametrization by MATH to MATH. In this interpretation, the subgroup MATH is generated by the reflections in MATH (not necessarily simple) which stabilize the weight MATH with respect to the MATH-action in MATH, and we obtain the statement of the theorem. |
math/0107098 | The algebra of functionals MATH is mapped homomorphically onto MATH by the isomorphism MATH. A functional in MATH is not a trace, and hence is zero when evaluated on any element MATH. Any such functional is mapped to a nilpotent central element by MATH. Therefore, since MATH is a bijection, all idempotents in MATH have their pre-images in MATH, or equivalently all central idempotents belong to the subalgebra MATH. As a module over itself by left multiplication, MATH decomposes into a direct sum of two-sided ideals MATH, corresponding to the blocks of the category of its finite dimensional modules. This induces a decomposition of the center MATH. By REF , MATH with each MATH being a local algebra with the idempotent MATH. Since all idempotents are in MATH, the element MATH is the idempotent in MATH, and MATH. In particular, each MATH is a local algebra, MATH. |
math/0107098 | REF is equivalent to the analogous statement in CITE: Set MATH, MATH, then MATH. Let MATH. For a central element MATH write MATH . Here we have used the fact that MATH is a left integral, the property of left-invariant functionals MATH and the identities MATH CITE. REF follows from REF . For MATH write MATH where we have used MATH. For MATH, MATH. Consider the algebra isomorphism from MATH to MATH, given by the action of MATH. Then by CITE for any MATH, MATH. Therefore we have MATH . Now let MATH, MATH. Then MATH as required. |
math/0107098 | Consider the block decomposition of the center MATH. The block MATH consists of elements which act nontrivially only on indecomposable modules with highest weights in the MATH-orbit of MATH. Since this weight is stabilized by the MATH action, and the NAME module is simple, projective and injective in the category of finite dimensional modules over MATH CITE, the block MATH is one-dimensional. By the structure of MATH REF and MATH REF , we have MATH. Let MATH. Then MATH. Write the sequence of mappings MATH . The first arrow is the definition of MATH, and the last equality follows by REF , since MATH. Then we have MATH by the definition of MATH, and hence MATH. Also, MATH. Now since MATH, we have MATH. The ideal MATH annihilates the radical of MATH by REF , and therefore its intersection with MATH belongs to the annihilator of the radical of MATH. We obtain MATH. The map MATH is an algebra isomorphism between MATH and MATH, and therefore MATH. By the definition of MATH, MATH, and hence this element annihilates the radical of MATH, or equivalently, MATH annihilates the radical of MATH. Here we used the isomorphism of algebras MATH, given by the action of MATH REF . |
math/0107098 | Recall that MATH is a direct sum of local NAME algebras. Its socle (annihilator of the radical) is spanned by the socles of each of the MATH blocks. Only one of these blocks is semisimple, the one corresponding to the orbit of the NAME weight MATH. The socle of this block is an idempotent, while the socles of all other blocks are nilpotent of second degree. Therefore, the square of the socle of MATH is one-dimensional. Since MATH by REF , MATH and it is nonzero as a character of a direct sum of modules. Therefore, MATH spans MATH. |
math/0107098 | of the Theorem. Obviously MATH. We will show that MATH. For each MATH, denote by MATH an element of the one- dimensional socle of the block MATH. We want to show that each MATH belongs to MATH, or equivalently that MATH is in MATH. We will use the lemmas above, recalling that MATH, the isomorphism given by the action of MATH which maps characters of MATH-modules to MATH-characters. As before let MATH. Then the element MATH spans the square of MATH, and therefore, by REF the isomorphism MATH maps MATH to the subspace spanned by MATH. By REF , MATH is mapped by MATH to an element MATH for some coefficients MATH. Then up to nonzero scalar multiples one can write the following sequence of mappings: MATH . Here set MATH. The last equality is a consequence of REF since MATH and its square are invariant under the antipode. Since the square of MATH is one-dimensional, MATH coincides up to a scalar multiple with the product MATH for any MATH. Therefore by weight considerations, MATH decomposes into a sum containing elements in all blocks of the category, and hence all MATH in MATH are nonzero. Act on both sides of the last equality by an idempotent of one block MATH. Then we get MATH, the projection of the MATH-character of MATH to the block MATH, which is spanned by the MATH-characters of simple modules with highest weights in the MATH orbit of MATH. This means that each MATH belongs to MATH. This completes the proof of the Theorem. |
math/0107098 | The restriction of MATH to MATH is a projective indecomposable module and the projective cover of MATH CITE. By the definition of the block MATH, its elements act as endomorphisms of a projective module with composition factors in the corresponding MATH-orbit of MATH, while elements of the other blocks MATH act on it by zero. Let MATH. By REF , MATH. For any MATH, it is possible to find MATH such that MATH contains MATH with a nonzero coefficient in its decomposition with respect to the basis of simple characters. Therefore, any character of a simple module appears in some combination in the linear span of MATH. This means that MATH acts nontrivially in each projective cover of a simple module with the highest weight in the MATH-orbit of MATH, mapping it to its simple socle. Since MATH is NAME, for any MATH there exists MATH such that MATH. Then the kernel of the mapping of MATH to MATH is trivial. Now compare the dimensions. First note that MATH, where MATH is the stabilizer of MATH with respect to the MATH action. This follows from the structure of filtrations of projective modules over MATH. Each projective module has a filtration by NAME modules MATH, which are the universal highest weight modules for MATH CITE. Then we have the reciprocity relation for the projective cover of a simple module MATH REF. The NAME filtration for a projective module in the MATH alcove has the following form: MATH . The last formula of course can be deduced from the general result on the characters of tilting modules in CITE (since all projective modules over MATH are tilting, CITE). It also follows by direct computation, using the fact that any projective module appears as a direct summand of a tensor product MATH for some finite dimensional module MATH CITE. We need to check that there are no extra endomorphisms of the restriction of MATH over MATH. Suppose there are additional composition factors in MATH isomorphic to its maximal semisimple quotient MATH. Then by the tensor product decomposition for simple modules CITE, a simple MATH-module MATH has to appear in the filtration of MATH for some MATH. By the strong linkage principle (REF , CITE), this means that the weight MATH should be less than or equal to at least one of the weights MATH of the NAME composition factors in MATH, and belong to the same MATH-orbit. For MATH, we should have then MATH which is false for any MATH and any dominant MATH. Therefore, there are no composition factors of the form MATH in any of the MATH. The statement for other weights MATH such that MATH is regular, is derived using the translation principle CITE which states that the structure of the filtration of a NAME module is the same for all highest weights in the same alcove. The proof for the lowest walls of MATH is obtained similarly, replacing MATH by a suitable fundamental weight MATH in the argument above. This means that MATH for any MATH. Therefore, we get MATH and the two algebras are isomorphic. |
math/0107098 | We know that the subalgebra MATH contains all idempotents of MATH REF . We will estimate from above the dimension of the radical of the center, computing the number of inequivalent nilpotent edomorphisms of all projective modules. Fix MATH. If MATH, then it is fixed by the MATH action and the corresponding block contains the single projective module MATH which is simple and has no nilpotent endomorphisms. Suppose that MATH is regular, then its MATH orbit contains two weights. Denote the corresponding indecomposable projective modules by MATH and MATH. In the case MATH, all indecomposable projective modules have highest weights in MATH. Therefore, they all satisfy the condition of REF . Then since MATH for any regular MATH by REF , we have MATH, and each algebra of endomorphisms contains a nilpotent element and an identity. Therefore, the center of MATH contains at most two nilpotent elements. On the other hand, the dimension of the regular block MATH of MATH is equal to MATH, which means that all nilpotent central elements are contained in MATH, and hence MATH. |
math/0107100 | Consider the quotient map MATH. By REF , the map MATH is étale of degree equal to MATH, hence MATH is smooth and we have MATH and MATH. The formulas for MATH, MATH follow easily from this remark. In addition, MATH is a minimal surface of general type iff MATH is nef and big, iff MATH is nef and big, iff MATH and MATH have genus MATH. The irregularity of MATH is equal to the dimension of the MATH-invariant subspace of MATH. Since MATH acts separately on MATH and MATH, one has MATH. We set, as usual, MATH. The second projection MATH induces a pencil MATH whose general fibre is isomorphic to MATH, hence MATH is rational and MATH is a double plane. |
math/0107100 | By REF, there are the following possibilities for MATH: CASE: MATH, MATH has a pencil MATH of hyperelliptic curves of genus REF with REF double fibres and MATH restricts to the hyperelliptic involution on the general fibre MATH of MATH; CASE: MATH, MATH has a pencil MATH of hyperelliptic curves of genus REF with REF double fibres and MATH restricts to the hyperelliptic involution on the general fibre MATH of MATH. Consider REF first. Let MATH be the double cover branched on REF points corresponding to the double fibres of MATH. MATH is a smooth curve of genus REF. By taking base change and normalization, one gets a diagram: MATH where MATH is an étale double cover and MATH is a pencil with general fibre isomorphic to MATH. One has MATH, hence by CITE the fibration MATH is smooth and isotrivial. The above diagram shows that MATH is also isotrivial, that REF double fibres have smooth support and they are the only singular fibres of MATH. So, in the terminology of CITE, MATH is a quasi-bundle and by REF (compare also CITE) there exist a curve MATH and a finite group MATH that acts faithfully on MATH and MATH in such a way that the diagonal action on MATH is free and MATH is isomorphic to MATH. The pencil MATH is induced by the second projection MATH and the involution MATH is induced by MATH, where MATH is the hyperelliptic involution of MATH. Since MATH is regular, we have MATH (compare proof of REF ). The formula for MATH follows from REF . If MATH is as in REF , then we let MATH be a MATH-cover branched over REF points corresponding to the double fibres of MATH (the existence of such a cover can be easily shown by using REF). MATH is again a smooth curve of genus REF and one argues exactly as before. |
math/0107100 | By REF MATH is not birational iff it has degree REF, hence MATH is not birational iff it is composed with an involution MATH of MATH. By the results of CITE this happens iff MATH has MATH isolated fixed points. Finally, by REF if MATH has degree REF then MATH is a rational surface, namely the involution associated with MATH is rational. The statement now follows from REF . |
math/0107100 | It is easy to check that if MATH is homogeneous of degree MATH and it vanishes on an orbit of order MATH, then it is of the required form. Conversely, the zero locus of a polynomial MATH is invariant for MATH, hence it is a union of orbits. The points of MATH that have nontrivial stabilizer are: MATH, MATH and the points MATH, MATH. A polynomial MATH vanishes at MATH iff MATH and it vanishes at MATH iff MATH. A straightforward computation shows that if MATH vanishes at one of these points then it has a multiple root there. This remark completes the proof. |
math/0107100 | Assume by contradiction that MATH has two distinct rational involutions MATH and MATH. By REF , to each involution MATH there corresponds an isomorphism of MATH with a quotient MATH, where MATH is hyperelliptic and MATH is induced by MATH, MATH being the hyperelliptic involution of MATH. The projections of MATH onto the two factors induce pencils MATH, MATH, MATH. A base point free pencil of MATH is determined uniquely by the span of the cohomology class of one of its fibres, which is an isotropic subspace of MATH. Since MATH, MATH has exactly two free pencils and thus we have MATH. Since the involutions MATH and MATH are different, we have MATH, MATH, or, equivalently, MATH and MATH. Looking at the pair MATH for the types listed in REF one sees that this cannot happen. This proves REF. REF follows from REF and from REF by the same argument, since the pair MATH is different for each type of double planes. Now REF follows from REF . |
math/0107100 | The proof is long and it is divided in several steps. By REF there exist a curve MATH, an hyperelliptic curve MATH of genus REF or REF and a group MATH that acts faithfully on MATH and MATH such that MATH and MATH is induced by the involution MATH of MATH, where MATH is the hyperelliptic involution of MATH. We denote by MATH (respectively, MATH) the subset of elements MATH that do not act freely on MATH (respectively, MATH). The following conditions are satisfied: CASE: MATH is empty; CASE: MATH and MATH are a union of conjugacy classes of MATH; CASE: both MATH and MATH generate MATH (in particular both MATH and MATH are nonempty); CASE: the elements of MATH have order MATH. REF is equivalent to the fact that the diagonal action of MATH on MATH is free (compare REF). REF follows from the definition of MATH and MATH. REF follows from the fact that by REF the curves MATH and MATH are rational. In fact, if MATH is the subgroup of MATH generated by MATH, then MATH is a connected étale cover, hence MATH. The same argument shows that MATH is generated by MATH. To prove REF, we recall that for a point MATH the multiplicity of the fibre of MATH over MATH is equal to the order of ramification at MATH of the quotient map MATH, which in turn is equal to the order of the stabilizer of MATH in MATH. Since all the multiple fibres of MATH are double (compare proof of REF ) MATH consists of elements of order REF. MATH is not one of the following groups: MATH . By REF, MATH is generated by elements of order MATH. So if MATH is cyclic, then it is equal to MATH, but then MATH does not have two disjoint sets of generators, contradicting REF. The same argument rules out MATH. So assume that MATH is a dihedral group MATH, MATH. Looking at the conjugacy classes of the elements of order REF of MATH, one sees that REF imply that MATH contains all the reflections. Then MATH does not generate MATH, a contradiction to REF. As in REF, we denote by MATH the finite subgroup of automorphisms of MATH induced by MATH. The group MATH is one of the following: MATH . This follows by REF and by the classification of the finite subgroups of MATH that we have recalled at the beginning of REF (MATH is excluded since it is not generated by elements of order MATH, contradicting REF). We denote by MATH, MATH the images of MATH, MATH in MATH. We notice that by REF MATH and MATH are sets of generators of MATH and are stable under conjugacy. The elements of MATH have order MATH. If MATH, MATH, then MATH acts freely on the branch locus MATH of MATH. By REF , if MATH then MATH is not isomorphic to MATH and thus the exact sequence REF becomes: MATH . Arguing as in the proof of REF , one sees that MATH contains all the reflections of MATH. By REF, this is the same as saying that every reflection can be lifted to an element of order REF of MATH that acts freely on MATH. On the other hand, if MATH is fixed by an element MATH, then the inverse image MATH of MATH is fixed by both the elements MATH that lift MATH. Hence the reflections of MATH act freely on MATH. Assume now that MATH is fixed by a rotation MATH of MATH. We may assume that MATH has order MATH. Let MATH be the lifts of MATH. The stabilizer MATH of MATH in MATH is cyclic, since MATH acts faithfully on the smooth curve MATH, and it is generated by MATH, MATH and MATH. Hence, up to exchanging MATH and MATH, we may assume that MATH generates MATH and, in particular, MATH. Let MATH be a reflection and let MATH be an element of order REF that lifts MATH. Since MATH, we either have MATH or MATH. In the former case, MATH is isomorphic to MATH, contradicting REF . The latter case can occur only for MATH even, since for MATH odd MATH does not generate MATH. In this case the elements MATH and MATH have order REF for MATH odd and thus they are not contained in MATH, contradicting the fact that MATH contains all the reflections. If MATH, then MATH is even. By REF , if MATH then MATH and MATH are not isomorphic and the exact sequence REF gives: MATH . Assume by contradiction that MATH is odd. Then a rotation MATH of order MATH can be lifted to MATH of order MATH. Arguing as in the proof of REF one shows that MATH is dihedral, contradicting REF . As we have already remarked, the curve MATH is defined by an equation of the form MATH in the weighted projective plane MATH, where MATH and MATH is a homogeneous polynomial of degree MATH. Up to isomorphism, there are the following possibilities for MATH, MATH and MATH: CASE: MATH, MATH, MATH, MATH; CASE: MATH, MATH, MATH, MATH, MATH; CASE: MATH, MATH, MATH, MATH; CASE: MATH, MATH, MATH, MATH; CASE: MATH, MATH, MATH where MATH (the zeroes of MATH are the points of the orbit of order REF of MATH); CASE: MATH, MATH, MATH is a polynomial of degree REF whose zeroes are the elements of the orbit of order REF of MATH. In all cases, the exact sequence REF is split. We recall that the zero set MATH of MATH is the branch locus of the hyperelliptic double cover MATH and it is invariant under the action of MATH. If MATH, then we have necessarily REF , since the smallest orbit of MATH on MATH has REF elements and there is only one such orbit. If MATH, then a similar argument shows that we have REF . By REF , the only remaining possibility is that MATH is a dihedral group MATH. If this is the case, then REF imply that the only possibilities for MATH, and MATH are REF - iv). The fact that MATH is of the form stated above follows from REF , it has been proven while describing type Ib for REF and it can be proven in the same way for REF . The fact that the exact sequence REF is split has already been checked in REF while describing the various types of double planes except for REF , where one can use exactly the same argument. If MATH, then MATH is of type Ia. Up to a suitable choice of homogeneous coordinates, we may assume that the action of MATH on MATH is the one described for surfaces of type Ia. By REF , the groups MATH and MATH are not isomorphic, hence REF gives the exact sequence: MATH which is split by REF . The curve MATH is as in REF or REF and, with respect to the canonical splitting MATH, each of the nonzero elements of MATH has REF fixed points on MATH. Thus MATH contains MATH and three elements MATH that map to the three nontrivial elements of MATH. By REF , MATH contains the remaining nonzero elements MATH of MATH, which are a set of generators. We consider the quotient map MATH. By REF, the image via MATH of the fixed set of MATH has cardinality divisible by MATH for MATH. Then the NAME formula implies MATH. Hence REF is excluded and we have REF . Now, using again REF, it is easy to check that we have a double plane of type Ia. If MATH, MATH, then MATH is of type Ib. Here we have either REF or REF . We will show that REF corresponds to type Ib and that REF does not occur. We consider REF first. By REF the groups MATH and MATH are not isomorphic, hence REF gives the exact sequence: MATH which is split by REF . The curve MATH is the same as in type Ib and MATH. Using the canonical splitting MATH one sees that MATH contains MATH and MATH, MATH (compare the description of type Ib in REF. It follows by REF that MATH. Let MATH be the projection onto the quotient and let MATH (respectively, MATH) be the number of branch points of MATH that are images of the fixed points of MATH (respectively, of the elements MATH). So MATH and the NAME formula gives MATH. Set MATH and consider the quotient map MATH: the NAME formula gives MATH and thus MATH. Let MATH be the subgroup generated by MATH and MATH and set MATH. The subgroup MATH is isomorphic to MATH and the NAME formula gives: MATH. Since MATH, we have: MATH, MATH, MATH and MATH. The group MATH is isomorphic to MATH and it acts on MATH in such a way that only the reflections have fixed points. It follows that the action on MATH must be the one described for type Ib. In addition, we have a commutative diagram: MATH . The branch points of MATH are the images of the points of MATH fixed by MATH and the branch points of MATH are the images of the points fixed by the elements of type MATH. In particular, MATH and MATH have no common branch point and thus their fibre product is smooth and connected. It follows that MATH is isomorphic to the fibre product of MATH and MATH and that this is type Ib. Now we consider REF . By REF the groups MATH and MATH are not isomorphic, hence REF gives the exact sequence: MATH which is split by REF . Arguing as in the description of type Ib, one sees that with respect to the canonical decomposition MATH the elements of order MATH that act freely on MATH are the following: MATH, MATH and MATH. Since these elements do not generate MATH, this case does not occur by REF . In REF is either of type Ic or of type Id. In this case, the curve MATH and its automorphism group have been analyzed in detail in the description of surfaces of type Ic in REF. In particular, it has been shown that, with respect to the canonical splitting MATH, the elements of order MATH that act freely on MATH are those of the form MATH with MATH of order MATH. Then, by REF , we have two possibilities: CASE: MATH and MATH consists of all the elements MATH with MATH of order MATH. CASE: MATH, where MATH denotes the sign of MATH, and MATH consists of all the transpositions. We consider REF first. One has MATH and by the NAME formula the branch locus of the quotient map MATH consists of MATH points. Thus each transposition of MATH fixes MATH points. Let MATH be a subgroup of MATH isomorphic to MATH. By the NAME formula, the curve MATH is rational. In addition, the induced map MATH is a degree MATH morphism with MATH simple branch points. This shows that REF corresponds to type Ic Now assume we are in REF and denote by MATH the number of branch points that are the images of the fixed points of the elements MATH with MATH a double cycle and by MATH the number of branch points that are the images of the fixed points of the elements MATH with MATH a transposition. So MATH has MATH fixed points if MATH is a double cycle and it has MATH fixed points if MATH is a transposition. We have MATH and the NAME formula gives again MATH. Denote by MATH the subgroup of MATH generated by the elements MATH with MATH a double cycle. MATH is isomorphic to MATH and the NAME formula applied to the quotient map MATH gives MATH, which implies: MATH, MATH, MATH. Denote by MATH (respectively MATH) the branch points corresponding to the elements MATH with MATH a transposition (respectively a double cycle). Denote by MATH the subgroup of MATH generated by the elements MATH, where MATH is a transposition. MATH is isomorphic to MATH and the quotient curve MATH is rational by the NAME formula. The quotient map MATH is a degree MATH cover branched over MATH and MATH. We set MATH. The quotient map MATH is an étale double cover and thus MATH has genus MATH. Summing up, we have a commutative diagram: MATH . Thus the curve MATH is obtained from MATH and MATH by base change and normalization. To show that this is type Id, we consider the action of MATH on MATH. The quotient map MATH is branched over MATH. If MATH is a subgroup of MATH isomorphic to MATH and we write MATH, then MATH has genus MATH and the induced map MATH is a degree MATH cover with one simple ramification point above MATH and MATH simple ramification points over MATH and MATH, whose NAME closure is MATH, as required. In REF is of type II As explained in the description of type II, there is a unique splitting MATH and the elements of order MATH that act freely on MATH are the elements MATH, where MATH is a double cycle of MATH. The subgroup generated by these is MATH, thus we have MATH and MATH. Arguing as in the previous steps, it is easy to check that this is type II. |
math/0107100 | Denote by MATH the blow-up of MATH at the isolated fixed points of MATH, by MATH the involution of MATH induced by MATH, and by MATH the quotient surface MATH. MATH is the minimal resolution of MATH. We denote by MATH the geometrically ruled surface MATH, MATH, by MATH the class of a section of MATH with self-intersection MATH and by MATH the class of a ruling. Arguing as in the proof of REF, one can show that MATH is obtained from a surface MATH by blowing up MATH pairs of points MATH such that: CASE: MATH lie on different rulings of MATH if MATH; CASE: for every MATH the point MATH is infinitely near to MATH in the direction of MATH; CASE: the pull back of the fibration MATH is the hyperelliptic fibration of genus REF MATH. The singular fibres of MATH consist of a MATH-curve counted twice (the inverse image of MATH) and of two disjoint MATH-curves, contained in the branch locus of MATH. So each singular fibre of MATH contains a pair of disjoint MATH-curves, one of which is contracted by the map MATH while the other one is mapped to a ruling of MATH. If we denote by MATH the branch locus of MATH, then MATH, where MATH are distinct rulings of MATH and MATH is effective. In the proof of REF it is shown that MATH is linearly equivalent to MATH. For MATH has a MATH point at MATH with the same tangent as MATH and MATH are the only singularities of MATH. Notice that the strict transform of MATH on MATH is smooth by construction. The curve MATH is the image of the strict transform in MATH of the divisorial part of the fixed locus of MATH on MATH, hence by REF it has two irreducible components if MATH is of type Ia and it is irreducible otherwise compare the description of the different types of double planes in REF. The map MATH is not determined uniquely: two such maps MATH and MATH are related by elementary transformations centered at some the points MATH (an elementary transformation of MATH centered at MATH of consists in blowing up MATH and then blowing down the strict transform of the ruling through MATH). If MATH, then we can perform an elementary transformation at, say, MATH and get MATH. The fact that the points MATH are the only singularities of MATH and that they have ``vertical" tangent implies that no section of MATH is contained in MATH. Thus we have MATH, namely MATH and at least MATH of the MATH are not in MATH. Thus, if MATH, one can perform an elementary transformation at such a point MATH, replace MATH by MATH and finally arrange that MATH. If MATH, then the required plane model can be obtained by composing MATH with the birational morphism MATH that contracts the exceptional curve of MATH and considering the NAME factorization MATH. We denote by MATH the image of the exceptional curve of MATH, by MATH the image of MATH, by MATH the image of MATH and by MATH the image of MATH, which is a curve of degree REF. Since MATH, MATH is a point of multiplicity REF of MATH and at most two of the MATH are infinitely near to MATH. The branch locus of MATH is the image of MATH and it is easy to check that it has the stated properties. In particular, if MATH is of type Ia, then the two components of MATH, being numerically equivalent to one another, are both mapped to curves of degree REF. It is well known that the singularities of a double cover of a smooth surface can be solved by repeatedly blowing up the base of the cover at the singularities of the branch curve and taking base change and normalization, and there are formulas for the numerical invariants of such a resolution (compare for instance REF. Applying this method to MATH one sees that a resolution of MATH has nonzero geometric genus iff there exists a curve MATH of degree REF with multiplicity MATH at MATH and with a double point at MATH such that MATH is tangent to MATH at MATH for MATH. The intersection multiplicity of such a MATH with a line MATH is MATH, hence MATH must be equal to MATH, where MATH is a conic containing MATH. This proves REF. A standard computation using again the formulas for double covers of CITE shows that if MATH satisfies properties REF - iv) then the double cover MATH is plane model of a minimal double plane of general type MATH with MATH and MATH (beware, the resolution of MATH as a double cover is not minimal!). In addition, the pencil of lines through MATH pulls back on MATH to a hyperelliptic pencil of genus REF, hence MATH is of type I by REF . |
math/0107100 | The proof is very similar to the proof of REF , hence it is left to the reader. |
math/0107100 | Follows from REF , since by REF the set of surfaces with non birational bicanonical map is MATH. |
math/0107100 | REF is the content of REF. The remaining properties will be a consequence of the Lemmas that follow. |
math/0107100 | By REF , MATH is the set of surfaces MATH admitting an involution with REF isolated fixed points and MATH is the set of surfaces admitting an involution with REF isolated fixed points. We remark that by NAME 's formula REF a minimal surface MATH with MATH and MATH contains no MATH-curve, hence it coincides with its canonical model. Therefore, given a smooth family MATH of minimal surfaces of general type with MATH and MATH, where MATH is the unit disk, we have to show that if the class of MATH is in MATH (respectively, MATH) for each MATH, then also MATH (respectively, MATH). Since by REF the surface MATH admits exactly one rational involution MATH, there is a birational map MATH that restricts to MATH on MATH for MATH. By REF MATH is actually biregular and we denote by MATH the restriction of MATH to MATH. Thus we have to show that the number of isolated fixed points of MATH is the same for all MATH. Clearly one has MATH. Let MATH be a fixed point of MATH. By NAME 's Lemma and by the fact that MATH preserves the fibres of MATH, there exist local analytic coordinates near MATH such that MATH is given by MATH and MATH acts by MATH, where MATH or MATH. Hence the fixed locus of MATH is the disjoint union of two smooth closed sets MATH and MATH of dimensions respectively REF and the restriction of MATH to MATH and MATH is a smooth map. So the MATH-dimensional part of the fixed locus of MATH on MATH is equal to MATH for every MATH. By the above remarks the map MATH is proper and étale, hence the cardinality of MATH is constant. |
math/0107100 | The map MATH is flat and finite. Given a MATH-linearized line bundle MATH on MATH, MATH is locally free of rank equal to MATH and MATH acts on MATH via the regular representation. One can define a trace map MATH by MATH. Clearly MATH splits the inclusion MATH and thus MATH is a line bundle. If we consider MATH then there is a natural inclusion MATH which is an isomorphism on MATH. So sections of MATH are rational sections of MATH whose divisor of poles is supported on MATH and they are characterized by the fact that they pull back to regular sections of MATH. The statement follows from this remark by an easy local computation. |
math/0107100 | For a variety MATH we denote by MATH the functor of deformations of MATH; if there is a group MATH that acts on MATH we denote by MATH the deformations of MATH that preserve the MATH-action. It is well known that the tangent space to MATH is MATH and the obstruction space is MATH. Consider a point MATH of MATH. By REF , MATH is a quotient MATH, where MATH, MATH and MATH are as explained in REF. As usual we denote by MATH the subgroup of MATH generated by MATH and by the hyperelliptic involution MATH of MATH. Consider the map of functors MATH defined by taking fibre products and dividing by the ``diagonal" MATH-action. Clearly, given an object MATH, MATH is a family of surfaces whose smooth fibres have a double plane structure of the same type as MATH. Since MATH and MATH have dimension REF, MATH and MATH are unobstructed. The tangent space to MATH is MATH. Since the quotient map MATH is étale, MATH is naturally isomorphic to MATH, which is in turn isomorphic to MATH by the NAME formula and by the fact that MATH acts on MATH and on MATH separately. The map on tangent spaces induced by MATH is the inclusion MATH. The claim follows if we show that this map is an isomorphism, that is, that MATH. For types Ia, Ib, Id one has MATH and there is nothing to prove. If MATH is of type Ic or II we show that MATH. By NAME duality, the dual of MATH is MATH and thus it is equivalent to show MATH. In turn this follows by applying REF to the map MATH, which has three branch points for both types (see REF). The above computations also show that the NAME family of MATH is smooth, hence the moduli space, which is locally a finite quotient of the NAME family, is normal. |
math/0107100 | By the proof of REF , if MATH is in MATH then the dimension of MATH (and of MATH) at the point corresponding to MATH is equal to the dimension of MATH. This dimension can be computed as in the proof of REF , using NAME duality and REF . |
math/0107100 | The proof of REF shows that, in order to prove that MATH (respectively, MATH) is irreducible it is enough to show the existence of two families of smooth curves MATH and MATH with a MATH-action such that: CASE: MATH maps each fibre of MATH and MATH to itself and the action on the fibre is the one required for type Ia (respectively, Ib, Ic, Id); CASE: every MATH, respectively, MATH, with a MATH-action of the right type is isomorphic to a fibre of MATH, respectively, MATH. The family MATH can be easily constructed using the equations given in REF. We sketch briefly a construction of MATH for the various types. The reader can easily supply the details. For each type we use the notation introduced in REF. CASE: the curve MATH is determined by the three pairs of points MATH, MATH. The only condition is that the six points are distinct; CASE: the curve MATH depends on the choice of the elliptic curve MATH, of the point MATH of order MATH and of a pair of points of MATH. The only condition is that the two points are not branch points of the quotient map MATH; CASE: the curve MATH is determined by the choice of a homogeneous polynomial of degree REF with generic branching; CASE: given the curve MATH, the map MATH is determined by the two double fibres, MATH and MATH. Clearly MATH is MATH-torsion and it is nonzero, since otherwise all the fibres of MATH over a branch point would be double. So MATH is determined by the choice of MATH, MATH, MATH and of a reduced divisor MATH. The condition that MATH and MATH are the only double fibres is clearly open, hence also in this case we have an irreducible parameter space for MATH. |
math/0107101 | Take the first variation of MATH: MATH from REF . But the variation is within a fixed cohomology class so MATH. Thus MATH and the variation vanishes for all MATH if and only if MATH . |
math/0107101 | If MATH are the weights of the MATH-dimensional vector representation of MATH, the weights of the spin representations MATH are MATH where there is an even number of minus signs for MATH and an odd number for MATH. If MATH are the positive roots of MATH then substituting MATH we have from REF for MATH and MATH the same weights MATH as the adjoint representation. |
math/0107101 | The covariant derivative of the MATH-form MATH at any point can be written MATH where MATH and acts on MATH by identifying MATH with the NAME algebra of MATH. Since MATH is fixed by MATH we may as well assume that MATH where MATH is the orthogonal complement in MATH to the NAME algebra of MATH. Since MATH is not covariant constant, MATH will not vanish. However, the harmonicity condition on MATH will force many of its components (as representations of MATH) to vanish. We shall index representations by their highest weight - the adjoint representation MATH of MATH has highest weight MATH. If we decompose the tensor product MATH into irreducible representations as in REF of NAME 's book CITE, we find that MATH is the direct sum of two MATH-dimensional irreducibles MATH with highest weight MATH and MATH respectively. These are interchanged under a change of orientation, just like self-dual and anti-self-dual forms in four dimensions. Similarly MATH breaks up into irreducibles with highest weights MATH and each with multiplicity one. NAME informed me of the following result: If MATH, the components of MATH with highest weight MATH, MATH, MATH and MATH all vanish. Since we are considering exterior powers of the NAME algebra it is convenient to think of these as spaces of left-invariant forms on the group. Then, for example, MATH is the space of coclosed invariant MATH-forms. The MATH-form MATH is built out of the structure constants of the NAME algebra MATH and as a consequence of this, the action of MATH on MATH can be rewritten as MATH if we consider MATH as a MATH-form on the group. Thus, if MATH then MATH and MATH . Now if MATH denotes a root vector for the root MATH, the vector MATH lies in MATH. Let MATH be an element of the NAME subalgebra, then because MATH etc. MATH . Since we can find MATH for which MATH this shows that the irreducible representation MATH maps non-trivially into MATH under MATH. That representation is also contained in MATH, interchanging the roles of MATH and MATH. However, since we can choose MATH such that MATH and MATH and vice-versa, MATH appears twice in MATH, and so if MATH, then both of these components in MATH vanish. We now work similarly with MATH using the vector MATH . We deduce that if MATH, then both of the MATH components in MATH vanish. Now consider MATH. The inclusion is unique and is given by MATH for an orthonormal basis MATH of the NAME algebra MATH. In this case MATH and MATH is an invariant map from the irreducible MATH to the irreducible MATH and thus must vanish. Hence MATH is the NAME MATH which is a non-zero scalar multiple of MATH. It follows that if MATH, the component MATH (and similarly MATH) in MATH vanishes. This completes the proof of the lemma. To return to the theorem, consider MATH. The right hand side, since MATH, is a vector bundle associated to the representation MATH. The skew part we calculated to have highest weights MATH and MATH. The calculation gives at the same time the symmetric part to have weights MATH, MATH and MATH. But MATH is the image under an invariant map of MATH which from the lemma only has components with highest weights MATH and MATH. Since these do not occur in MATH we deduce that MATH. Similarly, since MATH, we see that MATH. |
math/0107101 | From REF , the first variation of MATH at MATH is MATH and the first variation of the quadratic form MATH is MATH . Thus, introducing a Lagrangian multiplier, the constrained critical point is given by MATH and from CITE, this is equivalent to the structure of a manifold with weak holonomy MATH (sometimes called a nearly parallel MATH structure). |
math/0107101 | From REF , the first variation of MATH is MATH . The first variation of MATH is MATH where MATH and MATH. Using NAME 's theorem and a NAME multiplier MATH we find the constrained critical point to be given by the equations MATH and MATH, that is, MATH . The compatibility REF actually follow from these equations. From REF we have MATH since MATH is closed. Moreover, from REF MATH using REF and the fact that MATH, which, as we have seen, follows from MATH. Thus MATH as in REF . REF give a metric of weak holonomy MATH, sometimes called a nearly NAME metric. To see this, consider the MATH-form on MATH . This is stable, and from REF closed. Moreover, by comparing with the normal forms above, we see that the MATH metric it defines is MATH where MATH is the MATH metric. From REF we have MATH and then from REF , MATH. It follows that the cone metric above is a Riemannian metric of holonomy MATH. However, from CITE, this implies that MATH is nearly NAME. |
math/0107101 | The gradient flow is the solution of the equation MATH where MATH for any vector field MATH. In our case we have MATH described by a MATH-form MATH so using the inner product above with MATH, MATH since MATH. But this equation holds for all MATH-forms MATH and thus yields the gradient flow equation MATH . But MATH, so we obtain MATH and from CITE we see that the MATH-form MATH defines a metric with holonomy MATH. In this metric MATH. Conversely, if a MATH manifold is foliated by equidistant hypersurfaces, defining the function MATH to be the distance to a fixed hypersurface MATH, we can write the defining MATH-form in the form MATH and the statement that this is closed is equivalent to the gradient flow REF . |
math/0107101 | As before, we have the derivative of the Hamiltonian MATH given by MATH and the Hamiltonian vector field MATH is defined by MATH so in our case we have MATH and this gives the equations: MATH . We shall see that if the compatibility REF between MATH and MATH hold for MATH, then they hold for all subsequent time. First consider the condition MATH. This can be viewed as the vanishing of the moment map for the natural action of MATH on the symplectic manifold MATH. To see this, for a vector field MATH on MATH, consider the function MATH . We have MATH . But the first term is MATH since MATH. Thus MATH . Now since MATH and MATH are closed, MATH so REF can be written MATH using the definition of the symplectic form. We deduce that MATH is the moment map for MATH, evaluated on MATH. Since MATH is non-degenerate, from REF MATH vanishes for all MATH if and only if MATH. Now since the functional MATH is diffeomorphism invariant, MATH . NAME commutes with all the functions MATH. Hence the Hamiltonian flow of MATH is tangential to the zero set of all the functions MATH, that is, the space of pairs MATH such that MATH. Thus if MATH holds for MATH, it holds for all time. Note that it follows then that MATH for all time too. Next consider the second compatibility REF . The form MATH is defined by the derivative of MATH, MATH, so the derivative of MATH can be expressed via MATH. The volume MATH is homogeneous of degree MATH, so its derivative is homogeneous of degree one and hence MATH . We have MATH . So from REF MATH from REF . But MATH, and hence MATH . Thus MATH and so if MATH at MATH then it holds for all MATH. The evolution equation thus preserves the MATH geometry on MATH, and we define the MATH-form MATH . From REF , MATH. From REF , using the MATH metric defined by MATH, we have MATH and from REF , MATH. Thus MATH defines a metric of holonomy MATH on MATH for some interval. |
math/0107103 | Assume that MATH is a stable MATH-Higgs bundle. This means that it represents a smooth point of the moduli space (an extra argument, similar to the one given by NAME in CITE, is required to deal with the case of non-smooth points of the moduli space). It is known from CITE that MATH represents a critical point of MATH on MATH if and only if it is a variation of NAME structure. In our case this means that it is a MATH-Higgs bundle of the form MATH where MATH maps MATH to MATH and each MATH is contained in either MATH or MATH. Define MATH . It follows from the results of CITE that the eigenvalues of the Hessian of MATH at MATH are all even integers and that its MATH-eigenspace is isomorphic to the first hypercohomology of the complex MATH . The key ingredient in the proof is then the following vanishing criterion: the first hypercohomology MATH vanishes if and only if MATH is an isomorphism. This is proved using the fact that MATH is a semi-stable NAME bundle. Using this criterion it is then a matter of elementary linear algebra to show that MATH. It follows that MATH is not a local minimum of MATH whenever MATH, thus concluding the proof. |
math/0107103 | The proof of this Theorem applies the strategy used by Thaddeus CITE in his proof of the NAME formula. First one obtains a relatively simple description of the moduli space MATH for an extreme value of the parameter MATH. Next one studies the variation of the moduli spaces MATH as MATH varies, in order to obtain information about MATH for the value of MATH in which one is interested. We note that it is sufficient to consider the case MATH, since the case MATH can be dealt with via duality of triples. Via a careful analysis involving the stability condition for triples we obtain a bound on the values of MATH: for MATH the moduli space MATH is empty, unless MATH . For MATH there is no upper bound on MATH, however, the moduli spaces MATH stabilizes for MATH sufficiently large. Thus it makes sense to consider the ``large MATH moduli space", MATH, in both cases. Analysis of MATH shows that it is non-empty and irreducible - for MATH this is the main result of CITE. There is a finite number of so-called critical values of the parameter MATH, these are values for which strict MATH-semi-stability is possible. The MATH-stability condition remains the same between critical values. Thus we need to study how MATH varies as MATH crosses a critical value and, in particular, show that it remains non-empty and irreducible. The locus where MATH changes consists of triples which are strictly MATH-semi-stable for the critical value of MATH, and what we need to show is that this locus has strictly positive codimension. The category of triples is an Abelian category and we study strictly semi-stable triples via their NAME filtration by stable triples. This involves developing the theory of extensions of triples; in particular we show that the set of extensions of a triple MATH by a triple MATH is isomorphic to the first hypercohomology of the complex MATH . One can show that the codimension of the locus where MATH changes as MATH crosses a critical value is strictly positive if this first hypercohomology group is non-vanishing for any extension. An important fact which we prove in the course of these arguments is that for MATH any MATH-stable triple is a smooth point of the moduli space. Finally the non-vanishing of the above first hypercohomology is proved using a vanishing criterion which is reminiscent of the one described in the proof of REF above. |
math/0107104 | We follow the definition of a non-unital MATH-ring given above. Assume that MATH for some elements MATH, MATH and MATH in MATH. Then MATH in MATH. Since MATH is a MATH-ring, we get an element MATH in MATH such that MATH. Note now that the element MATH is invertible in MATH, whence MATH . It is then easy to see that MATH. Therefore MATH is a MATH-ring. |
math/0107104 | CASE: Take quasi-inverses MATH in MATH and MATH in MATH for MATH and MATH respectively. Then, by using the relations REF we easily check that the element MATH is a quasi-inverse for MATH. Indeed, we get that MATH . Thus MATH and MATH . Since MATH, MATH, MATH and MATH, we conclude that MATH. CASE: Assume that MATH for some MATH and MATH in MATH. Left and right multiplication by MATH, coupled with the fact that MATH yields MATH which we can rewrite as MATH. Since MATH, there exists an element MATH in MATH such that MATH. Similarly (using the hypothesis that MATH), we find an element MATH in MATH such that MATH. Recalling that MATH and MATH, we have MATH . Now MATH and by REF we have MATH, so MATH and therefore MATH. Similarly (using this time that MATH), MATH. Using these relations, we find that MATH . Set MATH. Since MATH, we have MATH, and so MATH (by REF). Therefore the element MATH is invertible (with inverse MATH) and it follows from the relations REF that MATH. Consequently MATH . Since MATH and MATH, it follows from REF that MATH, whence also MATH and this shows that MATH, as desired. |
math/0107104 | Let MATH be an element in MATH with quasi-inverse MATH, so that MATH, MATH and if MATH, MATH, we have MATH. Now take MATH in MATH and assume that there is an equation: MATH for some MATH and MATH in MATH. By using that MATH, we can rewrite this as MATH . Right and left multiplication by MATH yields MATH. Observe that MATH. Since MATH is a MATH-ideal of MATH (by REF ), there exists therefore an element MATH in MATH (using CITE if necessary) such that the element MATH (and actually MATH). By computation (and using again that MATH) MATH where MATH is an invertible element in MATH and MATH. Note that MATH. Rearranging the previous equality, we get MATH . Let MATH, and note that MATH . Let MATH be a quasi-inverse for MATH in MATH (and actually MATH). The defect idempotents for MATH and MATH are MATH and MATH, both belonging to MATH. Note that in fact MATH and MATH are defect idempotents of the quasi-invertible element MATH (in MATH), with quasi-inverse MATH. Now consider the idempotent MATH, which is equivalent to MATH, hence also centrally orthogonal to MATH. Similarly, we see that it is a defect idempotent for the quasi-invertible element MATH (with quasi-inverse MATH). Compute that MATH where MATH, as before, is an invertible element. Observe that MATH and MATH since MATH. Therefore, MATH . Now, MATH where MATH is an invertible element. We also have that MATH and that MATH. Thus, MATH . Set MATH and MATH, and note that the element MATH matches with our notation in REF . Since MATH and MATH belong to MATH, our hypothesis implies that either MATH (respectively MATH) or else MATH (respectively MATH) is a MATH-corner, where MATH, MATH, MATH and MATH are as in REF . It follows from REF that MATH and so MATH and therefore MATH. Since MATH there is an element MATH in MATH such that MATH, and so MATH . This shows that MATH, as desired. |
math/0107104 | The forward implication holds by combining CITE with CITE and CITE. Conversely, REF above implies that MATH, by REF . Thus, by REF imply that MATH is a MATH-ring. |
math/0107104 | Let MATH and MATH be idempotents in MATH, and assume that MATH and that MATH. We have to prove that MATH and MATH. To see this, assume first that we have an equation MATH, where MATH, MATH and MATH. Since MATH is an exchange ideal of MATH (see CITE), there is an idempotent MATH in MATH such that MATH. Write MATH and MATH for some elements MATH and MATH in MATH. Note that MATH. If MATH, then MATH and therefore MATH, with MATH as a quasi-inverse. Otherwise, MATH. Since MATH is essential and simple, MATH, and since MATH is purely infinite there exists an infinite idempotent MATH in MATH. It follows that MATH. Write MATH with MATH in MATH and MATH in MATH. Now consider the element MATH (MATH), and compute that MATH . Therefore MATH, and in fact MATH . This shows that MATH. The proof that MATH uses similar arguments. |
math/0107104 | The condition is necessary by CITE. To establish the converse, we show that if MATH is a regular element in MATH, then there exists a quasi-invertible element MATH such that MATH and then apply CITE. Let MATH be a regular element in MATH. Set MATH and MATH, and note that evidently MATH. If MATH, then MATH, and in this case MATH and MATH. Otherwise, there exists a quasi-invertible element MATH in MATH, by hypothesis. Now, CITE implies that MATH (with quasi-inverse MATH, where MATH and is a quasi-inverse for MATH). Set MATH and compute that MATH. |
math/0107104 | Evidently REF implies REF . The implication MATH follows from CITE and CITE. CASE: Let MATH be an element of MATH, and assume that MATH is regular. Thus we may write MATH for some MATH in MATH (and actually MATH). Now the argument in the previous proposition carries through. The argument that REF implies REF if MATH is an exchange ideal is very similar to the one used in the implication MATH, and we therefore omit the details. |
math/0107104 | Take idempotents MATH and MATH in MATH, and assume that MATH. Then, in particular, both MATH and MATH are non-zero, so MATH. Thus MATH and MATH for some MATH in MATH and MATH in MATH. Clearly this implies that MATH (with MATH as its quasi-inverse). Now the result follows from REF . |
math/0107104 | Assume first that MATH is a MATH-ring. Then MATH is a MATH-ring by CITE. Moreover, since MATH is simple and essential, MATH is prime and so we have MATH. Now, if MATH, we first lift MATH to a quasi-invertible MATH in MATH (by CITE), which is in fact left or right invertible, as we have just observed. Therefore MATH, being the image of MATH is also left or right invertible. Conversely, assume that MATH is a MATH-ring and that MATH. By REF , MATH is a MATH-ideal and by REF , for any pair of (defect) idempotents MATH and MATH with MATH or MATH in MATH, either MATH or else MATH is a MATH-corner. Hence, in order to apply REF we only have to show that one-sided invertible elements lift. For any MATH in MATH, denote by MATH its equivalence class in MATH. Assume that MATH. Since MATH is an exchange ideal of MATH, we may use CITE to find elements MATH and MATH in MATH such that MATH, MATH and MATH, MATH. Without loss of generality, we may assume that the idempotents MATH and MATH are both non-zero. Since MATH and MATH is purely infinite and simple, there is an infinite idempotent MATH such that MATH. Thus, if we write MATH and MATH for some elements MATH and MATH in MATH, we get that MATH is right invertible (with MATH as a right inverse), and is the required lift for MATH. |
math/0107104 | Let MATH denote the quotient morphism. By CITE we can find an element MATH in MATH such that MATH. Now choose a quasi-inverse MATH for MATH and an element MATH in MATH with MATH. By CITE the element MATH and MATH is a quasi-inverse for MATH. Evidently, MATH . But we also have MATH. Thus, both MATH REF and MATH are partial inverses to MATH (MATH), whence, by CITE MATH . It follows that MATH, MATH is the required lift of the pair MATH, MATH. |
math/0107104 | If MATH and MATH are centrally orthogonal idempotents in MATH, then MATH in MATH, since MATH is surjective. Since MATH is primely embedded in MATH this implies that MATH in MATH. Thus MATH so MATH. But then MATH . Similarly MATH, so MATH in MATH, proving that MATH is primely embedded in MATH. If MATH is quasi-primely embedded in MATH and MATH with quasi-inverse MATH, then MATH since MATH is surjective, and thus MATH by assumption. Moreover, as noted in REF, the element MATH is a quasi-inverse for MATH. It follows that if MATH and MATH then MATH so that MATH. However, MATH since MATH is an isomorphism and MATH in MATH. Consequently MATH in MATH, whence MATH, as desired. To show that MATH is a MATH-ring we shall verify REF for the extension MATH. By REF is a MATH-ring, and so is MATH, being (isomorphic to) an ideal in the MATH-ring MATH, compare CITE. To verify that quasi-invertible elements lift, consider MATH in MATH and choose a quasi-inverse MATH for MATH. Then MATH in MATH, whence MATH in MATH, since MATH is (quasi-) primely embedded in MATH. Consequently MATH with quasi-inverse MATH. Since MATH is a MATH-ring we can use REF to lift the pair MATH, MATH to a pair MATH, MATH in MATH which are quasi-inverses for each other. By construction MATH and MATH. Moreover, MATH in MATH (indeed in MATH), so MATH, as desired. Finally, consider two defect idempotents MATH and MATH in MATH and assume that MATH. Choose MATH and MATH in MATH with quasi-inverses MATH and MATH, respectively, such that MATH . To prove that MATH is a MATH-corner (if MATH and MATH are not centrally orthogonal) consider an equation MATH with MATH in MATH, MATH in MATH and MATH in MATH. Applying MATH we obtain the equation MATH and since MATH and MATH are defect idempotents in MATH, because MATH is (quasi-) primely embedded in MATH, we know from CITE that MATH is a MATH-corner in MATH. There is therefore an element MATH in MATH such that MATH . Now observe that MATH so that MATH for some element MATH in MATH. The same computation shows that MATH is an isomorphism between MATH and MATH. Consequently, the statement MATH is equivalent to the assertion that MATH . We have proved that MATH and a symmetric argument shows that also MATH, whence MATH is a MATH-corner in MATH, as desired. |
math/0107104 | Consider the pullback diagram MATH obtained as in the previous discussion from the extension MATH . Let MATH, and note that MATH. Thus the map MATH is injective and MATH is isomorphic to MATH. It follows that in the extension MATH is essential in MATH. Observe that MATH is a MATH-ring, since it is a quotient of MATH. If MATH, then we can lift MATH to a quasi-invertible element MATH in MATH, because MATH is a MATH-ring. Since moreover it is a prime ring, MATH must be either left or right invertible and hence the same holds for MATH. Therefore MATH is a MATH-ring, by REF . Since now MATH is a subdirect product of MATH and MATH, we conclude by REF that MATH is a MATH-ring. |
math/0107104 | By CITE, we have that MATH, where MATH for all MATH in MATH and MATH. Now, since we have an isomorphism MATH, we see that the only order-ideals in MATH are the trivial ones and MATH. Using for example, CITE, we conclude that the only non-trivial ideal of MATH is MATH. Moreover, we know that MATH; this, coupled with the fact that MATH is an exchange ring CITE allows us to conclude that MATH is a purely infinite simple ring, hence a MATH-ring by CITE. Finally, the conclusion follows by applying REF to the extension MATH . |
math/0107104 | CASE: We have to show that the monoid MATH satisfies the condition in REF . First, we identify this monoid with MATH, and MATH with MATH, where MATH is non-zero in MATH and MATH, where MATH is a countable unit for MATH, as in REF . Assume then that MATH, for some MATH. By REF , we can assume that MATH for MATH, MATH, that is, MATH and so there exist elements MATH, MATH in MATH such that MATH . Hence, by definition of the operation in MATH, we get that MATH . After restricting to the extreme boundary of MATH and using CITE (and also that the scale is finite), we see that MATH . Therefore MATH in the quotient monoid, so we take MATH and MATH, and MATH. CASE: Let MATH, which is non-zero in MATH, and continue to use the notation of REF . Assume that MATH and that they are different. As in CITE, there exist elements MATH in MATH (for MATH, MATH) such that MATH and MATH, and that MATH and MATH. Observe that MATH. Assume, to reach a contradiction, that MATH is a MATH-ring. Then REF provides us with two order-ideals MATH REF of MATH and elements MATH in MATH such that MATH and MATH. If both MATH and MATH are non-zero, then MATH and MATH properly contain MATH so they must contain MATH. But then MATH. We may assume without loss of generality that MATH, hence we have the equation MATH. We then find elements MATH and MATH in MATH such that MATH, that is, MATH. Evaluating at MATH we get the contradiction. |
math/0107104 | Let MATH be a non-zero element in MATH, and set MATH and MATH, where MATH is a countable unit for MATH. Assume that MATH is non-elementary. Then REF ensures that MATH. Assume first that MATH is a MATH-ring and that MATH has finite scale. Then, using the arguments in CITE, it follows that MATH is stably finite. Taking into account that MATH is moreover a prime ring, we get that MATH. By CITE, MATH has stable rank one, but this contradicts REF . The scale is therefore not finite and so there exists at least one element MATH in MATH such that MATH. Now REF implies that there can be only one infinite extremal state, since otherwise MATH would not be a MATH-ring, so neither would be MATH, in contradiction with our hypothesis. Thus we have to deal with the case where MATH . Let MATH, take any MATH in MATH such that MATH, and consider the equation MATH. Another application of REF provides us with elements MATH and MATH (one of which is zero, but not both) such that MATH. Now, since MATH, necessarily MATH and so MATH. This implies that MATH, a contradiction. Assume now that MATH is elementary. We have already observed that in this case MATH for a division ring MATH, and this is a MATH-ring. |
math/0107104 | Again, we let MATH in MATH and adopt the notation of REF . Assume that MATH is closed and that MATH, for some MATH, MATH in MATH and some MATH. We may clearly assume that MATH for any MATH. We of course get that MATH in MATH. We have to distinguish three cases: If MATH, then MATH in MATH, hence MATH and therefore MATH in MATH. We let in this case MATH and MATH. If MATH for all MATH, then without loss of generality we may assume that the element MATH. Let MATH be a natural number such that MATH. By the arguments in CITE, there exists a function MATH in MATH such that MATH and MATH. Now, it is easy to check that MATH whence MATH . In the quotient monoid we therefore have MATH. Moreover, MATH, and thus MATH. We let in this case MATH, MATH, MATH, MATH and MATH. Finally, assume that MATH and MATH. Let MATH be a natural number. Since MATH is closed, we can construct a function MATH in MATH such that MATH and MATH. Now, MATH, and then MATH. As before, we also have MATH, so again MATH. In this case, MATH, MATH, MATH, MATH and MATH. To prove the converse direction, assume that MATH is a MATH-ring and that MATH is not closed. Then MATH, so we can take an element MATH in MATH. By CITE MATH equals the closure of the convex hull MATH. Therefore we can write MATH, where MATH. On the other hand, MATH is the direct convex sum of MATH and MATH (see CITE), so that there exist a real number MATH in MATH and an element MATH in MATH such that MATH . By using CITE, construct a function MATH such that MATH and MATH, so MATH. Since MATH is a MATH-ring, there exist order-ideals MATH (MATH, MATH) such that MATH in MATH and elements MATH in MATH such that MATH. Upstairs, this means that MATH for some MATH and MATH in MATH. As in the proof of REF , we see that MATH or MATH. Now MATH because MATH, and we therefore conclude that MATH, whence MATH (and we may actually assume that MATH). The equation MATH is equivalent to MATH and restricting to the extreme boundary of MATH we see that MATH . This implies (using the lower semi-continuity of MATH) that MATH which is plainly impossible. |
math/0107108 | First note that MATH. If MATH, then MATH and hence MATH, for some MATH in MATH (the MATH-torsion subgroup of MATH), and MATH. Hence, the fiber of MATH is isomorphic to MATH, and the lemma follows. |
math/0107108 | Since the NAME map is surjective, it gives an inclusion MATH . Since the fibers of MATH are projective, any function on MATH is constant on these fibers, hence it factors through the NAME space, and hence we have an equality. |
math/0107108 | This lemma is stated in CITE with the extra assumption that MATH (because NAME is interested in representations of the fundamental group), but the proof actually works for NAME bundles of any degree. For convenience of the reader, we will give the necessary details. Since MATH is a fixed point, by CITE we know that it is of the form MATH with MATH sending MATH to MATH. Since MATH, we have MATH and MATH and MATH are nontrivial. Since MATH and MATH are coprime, MATH is stable. Then, since MATH is MATH-invariant, the stability condition implies MATH . The subbundle MATH is also MATH-invariant, and the stability condition implies MATH . Combining both inequalities, we obtain MATH, and NAME implies that MATH is nontrivial. Furthermore, if MATH is the MATH-subbundle of MATH (that is, the first term in the NAME filtration), then the slope of MATH is bigger or equal than the slope of MATH, so we still have MATH . Let MATH be a nonzero element of MATH that maps to a nonzero element in MATH. For each MATH, let MATH be the extension given by MATH . The NAME bundle MATH is defined taking MATH and the NAME field MATH is given by MATH . By construction, we see that MATH is nilpotent. We define MATH. The rest of the proof is identical to the proof given in CITE, so we only give a sketch. First one checks that MATH is isomorphic to MATH, and that these NAME bundles are stable, hence they are in the nilpotent cone. Furthermore, MATH . Finally NAME checks that MATH is not isomorphic to MATH (here is where we use the fact that MATH has nonzero image in MATH). |
math/0107108 | The map that sends a vector bundle MATH to the NAME bundle MATH defines an inclusion of MATH in the nilpotent cone. Since they have the same dimension, this gives one component of the nilpotent cone. It does not have a nontrivial MATH action, because if it had, then it would produce a nontrivial section of the tangent bundle of MATH, but this is known to have no nontrivial sections (CITE and CITE). In the rest of the components we have the MATH action given by MATH. To show that this action is nontrivial, we use the previous lemma: Let MATH be a fixed point (with MATH), and let MATH be the NAME bundle given by REF . Since the limit of the action moves MATH to MATH, they are in the same irreducible component of the nilpotent cone, and since both bundles are stable and nonisomorphic, they correspond to different points in the moduli space, and hence the MATH action is nontrivial in this component. |
math/0107108 | Using the isomorphisms REF and the projection formula, we have MATH . |
math/0107108 | Consider the following diagram, constructed using REF MATH where MATH denotes the relative tangent bundle for the projection MATH. Note that the diagram is well defined, since MATH is a section of MATH (compare REF ). The diagram is commutative because the zero scheme of the two morphisms between the line bundles MATH and MATH are the same, namely the ramification divisor, hence the maps differ by a scalar, but this scalar is absorbed in the isomorphism between MATH and MATH. Since the tangent line bundle MATH has negative degree, MATH. Therefore, the previous diagram gives MATH where MATH and MATH are the isomorphism of REF , the isomorphism MATH is given by MATH, MATH is defined by composition, and MATH is the NAME map. The bottom row is the first exact sequence in the statement of the proposition. Now we will restrict the NAME map MATH to MATH. To do this, we need an explicit description of the morphism MATH. Using the isomorphism MATH (compare proof of REF ), an element of this group is written as MATH with MATH and MATH. On the other hand, using the isomorphism MATH, an element of this group will be written as MATH with MATH and MATH. Since MATH comes from multiplication with MATH, a short calculation using MATH gives MATH . The subspace MATH is the zero locus of the trace map sending MATH to MATH. Then we have a commutative diagram MATH where MATH is projection to the first summand. Now, if MATH, then MATH, and using the explicit REF for MATH, we obtain MATH and hence the following diagram is commutative MATH where MATH is projection to the first summand followed by multiplication by MATH. The top row is the second exact sequence in the statement of the proposition. |
math/0107108 | As we have seen, the standard action has this property. Now let MATH be a point giving a smooth spectral curve MATH. The tangent vector at MATH defined by the standard action is contained in the kernel of the NAME map MATH, because the standard action does not change the isomorphism class of the spectral curve. Now we will show that the tangent vector defined by any action that lifts to MATH is also in the kernel of the NAME map. Denote MATH and MATH. In general, if MATH is a covering map between differentiable manifolds, a holomorphic structure on MATH induces a holomorphic structure on MATH. Hence, using the morphism MATH of REF , a deformation of MATH gives a deformation of MATH. The map MATH between the deformation spaces is the composition MATH where the inclusion is induced by the adjunction map MATH and the isomorphism is given by the projection formula and the isomorphism MATH. We have several varieties associated to a point MATH, namely the spectral curve MATH, the Jacobian MATH and the NAME variety MATH. Hence the MATH action produces deformations of all these objects MATH . We will study the relationship between MATH and MATH. Using the projection formula and the NAME spectral sequence for the projections MATH and MATH from MATH to MATH and MATH, we obtain that the space of deformations of MATH is MATH . The first and last terms correspond to deformations of MATH and MATH. To understand the second term, view MATH as a trivial fiber bundle. The deformations of this as a fiber bundle (that is, keeping the fiber and the base fixed) are parametrized by MATH that is, the second term. Analogously, the third term corresponds to deformations of MATH as a fiber bundle. The class MATH lies in the fourth summand MATH. Indeed, if we move MATH to a nearby point MATH, the fiber MATH is deformed to MATH, hence the components of MATH in the first three summands of REF have to be zero. Hence MATH lies in MATH, and is equal to MATH. MATH A deformation of a curve produces a deformation of its Jacobian, hence there is a natural map MATH and the infinitesimal version of the classical NAME theorem says that this map is injective. Clearly, in our situation MATH is the image of MATH under this map. Combining the injectivity of this map with the injectivity of MATH, we obtain that if MATH, then MATH. Since the action MATH lifts, the fiber MATH of the NAME map MATH at MATH is isomorphic to the fiber MATH at MATH for all MATH. This implies that MATH, hence MATH. This means that the tangent vector at MATH defined by the action MATH is in the kernel of the NAME map MATH, and since by REF this has dimension MATH, we obtain that the orbit MATH of MATH through MATH is included in the orbit of the standard action through MATH. In particular, the origin is a limiting point of the orbit MATH, that is, it is in the closure of the orbit, but not in the orbit (note that the origin is not in the orbit, because the fiber over the origin is not isomorphic to the fiber over MATH). The limiting points of an orbit are fixed points of the action. Then the origin is a fixed point of the action, and by hypothesis is the only fixed point. |
math/0107108 | Let MATH be an algebraic variety isomorphic to the moduli space MATH. By REF , choosing a set of generators of the ring of global functions on MATH we obtain an isomorphism MATH, and then the natural morphism MATH gives a morphism MATH . Note that MATH depends on the set of generators chosen: a different choice will give a morphism that differs by an automorphism (as an algebraic variety) of MATH. Up to isomorphism, this is the NAME map. More precisely, if MATH is an isomorphism, then there is an isomorphism (as algebraic varieties) MATH such that the following diagram commutes MATH . Let MATH be a MATH action with exactly one fixed point MATH, and such that it admits a lift to MATH. We know that such an action exists (using the standard MATH action on MATH and the isomorphism MATH), and by REF , MATH is the origin MATH of MATH. Indeed, if MATH were a different point, using MATH and MATH we would have an action contradicting REF . Then the fiber MATH over MATH is isomorphic to the nilpotent cone. This has several irreducible components, but by REF , only one of them (call it MATH) does not admit nontrivial MATH actions, and furthermore MATH is isomorphic to MATH, the moduli space of semistable vector bundles on MATH of rank MATH and fixed determinant MATH. Now, if MATH is isomorphic to MATH, then MATH is isomorphic to MATH, hence MATH is isomorphic to MATH, and by the NAME theorem for the moduli space of vector bundles, MATH is isomorphic to MATH. |
math/0107109 | Standard arguments from homological algebra together with the fact that complexes MATH do not have cohomology in dimensions MATH imply that it is sufficient to check that the multiplication maps MATH are associative and commutative in MATH. By REF it is sufficient to check that the multiplication maps REF are commutative and associative in MATH. The associativity condition clearly holds on the level of sheaves. To prove commutativity we should show that permutation of coordinates on MATH acts trivially on MATH. The action of the permutation group on MATH extends to an action of MATH. A transposition is MATH-homotopic to the automorphism given by MATH. It is therefore sufficient to check that this automorphism is the identity in MATH. Consider for simplicity of notations the case of one variable that is, the automorphism MATH of MATH defined by MATH. The sheaf MATH is isomorphic to the sheaf MATH which is weakly equivalent to MATH where the embedding MATH corresponds to the point MATH. Under this weak equivalence our automorphism becomes the automorphism of MATH defined by MATH. Denote this automorphism by MATH. One can easily see now that to prove that MATH is identity in MATH it is sufficient to construct a section MATH of MATH on MATH such that MATH . Let MATH be the coordinates on MATH and MATH the coordinates on MATH. Then the cycle of the closed subscheme in MATH given by the equation MATH defines a section of MATH on MATH which satisfies REF . |
math/0107109 | Let MATH be the functor from sheaves of pointed sets to sheaves of abelian groups which sends a sheaf of sets to the freely generated sheaf of abelian groups with the distinguished point set to be zero. Let further MATH be the normalized chain complex functor from simplicial abelian sheaves to the complexes of abelian sheaves. Then for any MATH one has MATH where MATH is the derived category of complexes of abelian sheaves in the NAME topology. The fact that REF is an isomorphism follows from the fact that MATH takes smash product to tensor product (modulo a quasi-isomorphism) and that MATH is quasi-isomorphic to MATH. Consider the suspension morphism MATH given by multiplication with the class MATH. Since REF is an isomorphism and we have a MATH-weak equivalence MATH, to show that REF is an isomorphism it is sufficient to show that MATH is an isomorphism. A standard argument allows one to reduce the problem to the case when MATH for a smooth scheme MATH over MATH. Open excision implies that MATH and we get a split short exact sequence MATH . Consider the morphism of sequences: MATH where MATH is the restriction of MATH to MATH, the first vertical arrow is MATH and the rest of vertical arrows are identities. The fact that MATH is an isomorphism follows now from REF below. |
math/0107109 | By CITE we have natural isomorphisms MATH where the target are NAME 's higher NAME groups and the proof immediately shows that we have similar isomorphisms for all groups of coefficients MATH . Consider the diagram: MATH where in the lower line the first morphism is given by covariant functoriality for the closed embedding MATH at the infinity and the second morphism is given by the contravariant functoriality for the open embedding MATH. One verifies using the explicit form of the isomorphism REF that both squares in REF commute. We conclude that the upper line is a short exact sequence since the lower one is a short exact sequence by NAME 's Localization Theorem CITE. |
math/0107109 | Since MATH and MATH, the restriction of a bistable operation to groups of degree MATH satisfies REF . On the other hand, for a family MATH we can construct MATH as follows. For MATH we have MATH taking MATH and taking MATH to be greater or equal to MATH we get: MATH where MATH. Using these isomorphisms, the operation MATH defines a map MATH . REF implies that this map does not depend on the choice of MATH and that the maps MATH all MATH satisfy REF . |
math/0107109 | It follows by adjunction argument from the fact that the map of the free abelian groups defined by MATH is isomorphic, in the derived category of sheaves of abelian groups, to the diagonal map MATH. |
math/0107109 | Follows from REF using the naturality of MATH with respect to the map defined by the codiagonal MATH. |
math/0107109 | Follows from REF . |
math/0107109 | To prove the proposition it is sufficient to construct for any MATH and any MATH in MATH a MATH-homotopy from the map MATH such that these homotopies are natural in MATH. Consider the map MATH which sends MATH to MATH. This map is flat over the complement to MATH in the target. Consider the pull-back MATH of MATH along the map MATH. It is a cycle on the target of MATH and the support of this cycle does not intersect MATH. Therefore, the flat pull-back MATH is defined. One verifies easily that this cycle is equidimensional over MATH of relative dimension zero and hence defines a map: MATH . The restriction of MATH to MATH is MATH and the restriction to MATH is the zero cycle. Therefore, MATH is a homotopy of the required form. It is clear that our construction is functorial in MATH. |
math/0107109 | Follows immediately from REF . |
math/0107109 | The fact that MATH for MATH follows immediately from REF . Consider the case MATH. We have an obvious map from MATH to MATH. On the other hand, an element of MATH considered as an operation defines a map MATH . Since MATH and MATH, REF implies that this map is a homomorphism. Therefore, it is sufficient to show that an element MATH of MATH which acts trivially on MATH is zero. By REF any MATH is determined by its action on objects of the form MATH where MATH is a closed subset equidimensional of relative dimension MATH over MATH. Let MATH be the closed subset of singular points of MATH. Since MATH is perfect, MATH is of codimension at least MATH in MATH. REF implies that the motivic cohomology of weight MATH of MATH map to the motivic cohomology of weight MATH of MATH isomorphically. It remains to show that MATH acts trivially on MATH . The normal bundle to MATH in MATH is trivial. Hence, by the homotopy purity theorem CITE, we have a weak equivalence MATH let MATH, MATH be the connected components of MATH. Then we have a map MATH and MATH is non-zero only for MATH where it is MATH, this map defines an isomorphism on MATH. We conclude that MATH acts by zero since by assumption it acts by zero on MATH. |
math/0107109 | A standard argument shows that it is sufficient to prove the proposition for MATH. In this case it follows immediately from out definition of the NAME class and the projective bundle theorem. |
math/0107109 | The fact that the restriction of MATH to the generic point is the tautological class follows from REF . The fact that MATH is determined by this condition follows from REF and the fact that if MATH is the embedding of the generic points of MATH, then MATH defines an isomorphism on MATH. |
math/0107109 | Let MATH be the image of MATH in MATH. The restriction of MATH to MATH is MATH for MATH that is, zero. Therefore, MATH. On the other hand MATH is defined by the condition that the restriction of MATH to MATH is zero. Since the restriction of MATH is the class of MATH we conclude that MATH. |
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