paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9912153 | Recall from CITE that the suspension theorem in morphic cohomology implies that morphic cohomology can be represented by morphisms into spaces of zero cycles in projective spaces. Since zero cycles are given by points in symmetric products this can be interpreted in the following way. Let MATH be the infinite symmetric... |
math/9912153 | We first prove that the total NAME class MATH is injective. So suppose that for some MATH, we have that MATH . So each NAME class MATH for MATH. Now recall from REF that in the algebra of operations between MATH and MATH, that the that the NAME classes and NAME character are related by a formula of the form MATH where ... |
math/9912153 | The space of rational maps in the morphism space MATH stabilizes if and only if the group completion of its space of rational maps is the two fold loop space, MATH . But by definition, the left hand side is equal to MATH . But by Rowland's theorem CITE or by the more general ``projective bundle theorem" proved in CITE ... |
math/9912153 | Consider the NAME character defined on the MATH . Now the morphic cohomology of a point is equal to the usual cohomology of a point, MATH, so this group is non zero if and only if MATH. So the NAME character gives an isomorphism MATH . Let MATH be the NAME character of the NAME class, MATH. Since the NAME character is ... |
math/9912153 | Consider the commutative diagram involving the total NAME character MATH . By REF , if MATH is NAME periodic, then the top horizontal map MATH is an isomorphism. But by REF we know that the two vertical maps in this diagram are isomorphisms. Thus if MATH is NAME periodic, then the bottom horizontal map in this diagram ... |
math/9912157 | We begin with REF : The map MATH has the RLP with respect to the map MATH if and only if each map MATH factors through MATH, that is, if and only if each MATH-element of MATH is in the image of MATH. Now REF : The map MATH has the RLP with respect to the map MATH if and only if for each MATH-element MATH of MATH whose ... |
math/9912157 | In this proof we use the terms weak equivalence, (acyclic) fibration, (acyclic) cofibration, and cofibrant as defined in REF . We check the hypotheses of the Recognition Lemma from the previous section. Since MATH is complete and cocomplete, so is MATH; limits and colimits are taken degreewise. The class of weak equiva... |
math/9912157 | We construct a factorization MATH of the diagonal map MATH in the following way. Let MATH be the chain complex which has MATH in degree MATH. We describe the differential by saying that it sends a generalized element MATH in MATH to MATH. Let MATH be the map which sends MATH to MATH and let MATH be the map which sends ... |
math/9912157 | The group MATH may be calculated by choosing a cofibrant replacement MATH for MATH and computing the homotopy classes of maps from MATH to MATH. (Recall that all objects are fibrant, so there is no need to take a fibrant replacement for MATH.) A MATH-projective MATH-exact resolution MATH of MATH serves as a cofibrant r... |
math/9912157 | That MATH is triangulated follows from NAME 's result, since we have shown that MATH is a self-equivalence. The identification of the triangles uses the definition of fibre sequences from CITE and the construction of the path object from the proof of REF . There is also a dual proof using cofibre sequences and cylinder... |
math/9912158 | See CITE and CITE. |
math/9912158 | See CITE . |
math/9912158 | See CITE. |
math/9912158 | See CITE . |
math/9912158 | Suppose MATH, MATH have the same image under REF. We choose representatives MATH, MATH which have closed MATH-orbit. Let us define MATH REF by MATH . Choose complementary subspaces MATH of MATH in MATH. We choose a MATH-parameter subgroup MATH as follows: MATH on MATH and MATH on MATH. Then the limit MATH exists and it... |
math/9912158 | See CITE or CITE. |
math/9912158 | See CITE, CITE. |
math/9912158 | See CITE. (REF therein is a misprint of REF .) |
math/9912158 | CASE: See CITE for the first assertion. During the proof of CITE, we have shown the second assertion, using CITE = REF. CASE: Consider the alternating sum of dimensions of the complex MATH. It is equal to the alternating sum of dimensions of cohomology groups. It is nonnegative, if MATH by REF . On the other hand, it i... |
math/9912158 | Consider MATH satisfying MATH for any MATH, MATH. If we set MATH then MATH is MATH-invariant and contained in MATH by REF. Thus we have MATH by the stability condition. |
math/9912158 | We prove MATH by the induction on MATH. The assertion is trivial when MATH. Take a point in MATH and its representative MATH. As in REF, there exists MATH such that MATH . We decompose MATH into eigenspaces of MATH: MATH . We also decompose MATH into eigenspaces of MATH as MATH. Then REF holds where MATH is replaced by... |
math/9912158 | The proof is essentially contained in CITE. See also REF for a similar result. |
math/9912158 | We have a surjective homomorphism MATH of codimension MATH over MATH. This gives a morphism from MATH to the fiber product of NAME bundles. By a straightforward modification of the arguments in CITE, one can show that it is an isomorphism. The detail is left to the reader. The assumption MATH is used to distinguish MAT... |
math/9912158 | We prove the assertion by induction on MATH, MATH. (The result is trivial when MATH.) We first make a reduction to the case when MATH . Fix a MATH and consider MATH . Since MATH is not a root of unity, we have MATH for MATH. Hence the above MATH is well-defined. Suppose MATH. By REF and the choice of MATH, we have MATH... |
math/9912158 | We first show that MATH is surjective. Choose a closed subvariety MATH of MATH so that MATH is a trivial MATH-bundle over MATH. There is a diagram MATH with exact rows by REF. By a diagram chase it suffices to prove the surjectivity for the restrictions of MATH to MATH and to MATH. By repeating the process on MATH, it ... |
math/9912158 | By CITE the moment map MATH is a NAME function, and critical manifolds are the fixed point MATH. Let MATH, MATH, be the components of MATH. By CITE, stable and unstable manifolds for the gradient flow of MATH coincide with MATH-attracting sets of NAME decomposition CITE: MATH . These are invariant under the MATH-action... |
math/9912158 | Let MATH denote the projection to the MATH-th factor (MATH). Let MATH be the diagonal embedding MATH. Then we have MATH. Hence MATH . If we substitute REF into the above, we get MATH . In particular, MATH is spanned by MATH's. If MATH for some MATH, then MATH. Hence we have MATH. The above equality REF implies MATH. Th... |
math/9912158 | Since MATH is a union of connected components (possibly single component) of the fixed point set of the MATH-action on a nonsingular variety MATH, MATH is nonsingular. Suppose that MATH is a fixed point of the MATH-action. It means that MATH lies in the closed orbit MATH. But MATH converges to MATH as MATH. Hence the c... |
math/9912158 | Let MATH be the structure sheaf of the diagonal considered as a sheaf on MATH. By the above argument, the NAME complex of MATH gives a resolution of MATH: MATH where MATH. Thus we have the following equality in the NAME group MATH . Since MATH is injective and MATH is surjective, we have MATH . Each factor of the right... |
math/9912158 | We apply REF. By CITE, the metric on MATH defined in REF is complete. By the construction, it is invariant under MATH, where MATH is the maximal compact subgroup of MATH. (Note that the hyper-Kähler structure is not invariant under the MATH-action, but the metric is invariant.) The moment map for the MATH-actions is gi... |
math/9912158 | In order to relate MATH relative to MATH, MATH, MATH and MATH relative to MATH, MATH, MATH, we replace MATH by MATH factor by factor. CASE: First we want to replace MATH by MATH. We consider the following fiber square: MATH where MATH is the projection. We have MATH where we have used the base change CITE in the second... |
math/9912158 | As in the proof of REF, we replace MATH by MATH factor by factor. CASE: First we replace MATH by MATH. Consider the following fiber square: MATH where MATH is the projection. By base change CITE and REF, we have MATH where MATH and MATH are projections. CASE: Consider the fiber square MATH where MATH is the projection.... |
math/9912158 | See CITE. |
math/9912158 | The proof below is modelled on CITE. We give the proof for MATH. Then the same result for MATH follows by MATH, MATH. We consider the complex REF for MATH and MATH: MATH where we put suffixes MATH, MATH to distinguish endomorphisms. We have sections MATH and MATH of MATH and MATH respectively. Identifying these vector ... |
math/9912158 | Fix a subspace MATH with MATH. Let MATH be the parabolic subgroup of MATH consisting elements which preserve MATH. We also fix a complementary subspace MATH. Thus we have MATH. We will check the assertion for MATH. The assertion for MATH follows if we exchange MATH and MATH. We consider MATH . It is a linear subspace o... |
math/9912158 | The latter statement follows from the first statement and REF . Thus it is enough to check the first statement. And the first statement follows from the transversality of intersections REF in MATH. Let MATH be the MATH-part of the moment map MATH. It induces a map MATH for MATH. Let us denote it by MATH. Thus we have M... |
math/9912158 | The latter statement follows from the former one together with the projection REF . Thus it is enough to prove the former statement. By definition, MATH consists of MATH such that there exists MATH satisfying REF. We fix representatives MATH, MATH. Then the above MATH is uniquely determined. Recall that we have chosen ... |
math/9912158 | As is explained in CITE, the result follows from REF. The factor MATH is introduced to make the differential in the NAME complex equivariant. |
math/9912158 | CASE: Generalizing REF, we have the following formula for MATH: MATH where the summation runs over the set of ordered MATH-tuples MATH such that MATH, MATH for MATH. Choose MATH so that MATH . Consider the following term appeared in the above formula: MATH . By CITE it is equal to MATH where MATH is the NAME polynomial... |
math/9912158 | It is enough to check that MATH, MATH, MATH and the coefficients of MATH are mapped to MATH. For MATH and the coefficients of MATH, the assertion is clear from the definition. For MATH and MATH, we can use a reduction to rank MATH case as in REF. Namely, it is enough to show the assertion when the graph is of type MATH... |
math/9912158 | As in REF, we may assume that the graph is of type MATH. Now REF together with the fact that NAME polynomials form a basis of symmetric polynomials implies the assertion. |
math/9912158 | The following proof is an adaptation of proof of CITE, which was inspired by CITE in turn. We need the following notation: MATH . We prove MATH by induction on the dimension vector MATH. When MATH, the result is trivial since MATH. Consider MATH and suppose that MATH . Take MATH. We want to show MATH. We may assume tha... |
math/9912158 | The assertion is proved exactly as CITE. Note that the regularity assumption of MATH is not used here. |
math/9912158 | CASE: If we restrict the complex MATH to MATH, MATH is surjective and MATH is injective by REF. Thus MATH is represented by a genuine MATH-module, and MATH is a polynomial in MATH. The degree is equal to MATH by the definition of MATH. CASE: The first equation is the consequence of MATH, which follows from REF. The rem... |
math/9912158 | It is enough to show that there exists a simple l-integrable l-highest module with given NAME polynomials MATH. We can construct it as the quotient of the standard module MATH by the unique maximal proper submodule. (The uniqueness can be proved as in the case of NAME modules.) Here the parameter MATH is chosen so that... |
math/9912158 | REF follows from REF . We show REF . Note that MATH is mapped to MATH under MATH. And MATH is mapped to the fundamental class MATH under MATH. Combining with the projection REF , we find that the operator MATH is mapped to MATH under the homomorphism REF. Thus as an operator on MATH, it is equal to MATH where MATH is t... |
math/9912158 | By REF, MATH is a geneneralized eigenspace for MATH for a homomorphism MATH. Thus it is enough to study the eigenvalue. We consider MATH, MATH as MATH-modules via MATH as before. Let MATH, MATH be weight space as in REF. By the definition of MATH and MATH, we have MATH . By REF we have MATH where MATH, MATH are defined... |
math/9912158 | We have the following equality in MATH: MATH where MATH is (the restriction of) the natural line bundle over MATH. The assertion follows immediately. |
math/9912158 | Recall that we have a distinguished vector (we denote it by MATH) in the standard module MATH REF . It has the properties listed in REF. In particular, it is the eigenvector for MATH, and the eigenvalues are given in terms of MATH therein. In the present setting, MATH is equal to MATH. Let MATH . We have MATH. We want ... |
math/9912158 | Since MATH is equal to MATH in the NAME group, the assertion follows from the discussion above. |
math/9912158 | We use the transversal slice in REF. The idea to use transversal slices is taken from CITE. Choose and fix a point MATH. Suppose that MATH is contained in a stratum MATH for some MATH. We first show If MATH denote the constant local system on MATH, the corresponding vector space MATH is nonzero. If we restrict MATH to ... |
math/9912158 | See CITE and CITE. |
math/9912158 | By the decomposition theorem for a semi-small map CITE, the left hand side of REF decomposes as MATH where MATH is a component of MATH and MATH is the intersection complex associated with an irreducible local system MATH on MATH. Moreover, by CITE, we have MATH where MATH runs over the set of irreducible local systems ... |
math/9912167 | Checking that the diagonals are NAME and conelike is complicated but routine; the more significant issue is self-transversality. We describe the minimal strata containing a configuration MATH. Let MATH be the connected components of the inverse images in MATH of points in MATH, not including components which consist of... |
math/9912167 | Let MATH be the union of all faces of MATH (both finite and infinite) and let MATH be its image in MATH. Consider the cohomology exact sequence of the triple MATH: MATH . On the other hand, MATH since MATH cuts MATH into the pieces of MATH, and on each of these we have a unique top cohomology class, the fundamental cla... |
math/9912167 | The existence of MATH originates with the geometry of the configuration space MATH. This is a manifold with corners whose interior is MATH, the space of pairs of distinct points in MATH. If MATH, it is clearly homotopy equivalent to MATH. In the general case it has the same homology by a NAME argument. Each of the glui... |
math/9912167 | We can measure MATH by pairing it with REF-cycles in MATH. There are several kinds of these, but the only kind that can have non-zero pairing is represented by a torus MATH, where MATH and MATH are two disjoint knots in MATH. In this case MATH measures their linking number in MATH: MATH . Likewise MATH measures their l... |
math/9912167 | Say that a vertex of MATH provides a dollar to the component MATH if it lies in the knot MATH in MATH, and that it provides REF cents if it lies in the NAME surface MATH. By REF , each component MATH, of which there are MATH, needs a dollar in order for MATH to depend on MATH at the configuration MATH. Each vertex, of ... |
math/9912168 | Suppose MATH is a section of MATH with MATH. Such a section is the same as a MATH-equivariant function MATH on the universal cover MATH of MATH. Let MATH. Then MATH drops down to a closed real-valued one-form on MATH and MATH where MATH is the lift of MATH to MATH, a path from MATH to MATH. On the other hand, suppose M... |
math/9912168 | Suppose MATH is a section of MATH and an eigenspinor of MATH with eigenvalue MATH. Then MATH so MATH is an eigenspinor of MATH, also with eigenvalue MATH. If MATH is an eigenspinor of MATH then MATH is an eigenspinor of MATH with the same eigenvalue. |
math/9912168 | As before let MATH be a normalized section of MATH and let MATH. Let MATH where MATH and denote by MATH the vectorfield MATH on MATH. The spin structure MATH on MATH induces a spin structure on MATH, the associated spinor bundle is equal to MATH. To shorten the notation we will write this as MATH. Let MATH be a smooth ... |
math/9912168 | Since MATH we have from REF MATH . The integrand MATH only contains terms of degrees MATH, MATH so the integral over MATH which has dimension MATH vanishes, and the corollary follows. |
math/9912168 | From REF we have MATH . REF tells us that MATH and using REF we see that MATH . Since MATH has dimension MATH and MATH only contains terms of degrees MATH, MATH, the integral vanishes and we are left with MATH which proves the theorem. |
math/9912171 | Denote an outermost sub-disk of some MATH by MATH and suppose it is cut off by an arc MATH on MATH. By the "Facts" proved in REF pp REF, any such outermost arc MATH must have both end points on a single disk MATH which belongs to some n-float of genus MATH. Furthermore MATH must separate the boundary components of MATH... |
math/9912171 | Since the MATH part of an outer-most disk must be contained in a float of genus greater than one we must have a MATH on the float to create the genus. The MATH-handle is the tunnel disjoint from the decomposing annulus MATH. |
math/9912171 | No one of the two knots has a NAME splitting where the meridian is a primitive element since MATH. A primitive meridian would mean that the NAME splittings of the knots will induce a NAME splitting of the connected sum which would make the tunnel number additive or less. Now drill a tunnel in MATH with end points on op... |
math/9912171 | We can assume that the decomposing annulus MATH intersects the NAME splitting MATH as follows: It intersects MATH in two vertical annuli and MATH in one meridional annulus. (This is a consequence of the fact that MATH is induced by the respective NAME splittings). Choose two essential disks MATH and MATH for MATH on bo... |
math/9912171 | Assume in contradiction that MATH is a strongly irreducible NAME splitting for MATH. Let MATH be the NAME surface and let MATH be the decomposing annulus for the connected sum minimizing the intersection with MATH. We can assume (see REF ) that after an isotopy of the annulus MATH is a collection of essential curves on... |
math/9912171 | Assume first that the annuli MATH are simultaneously primitive in MATH. The proof will be by induction on MATH. For MATH we can glue MATH to MATH along MATH and MATH to obtain a manifold MATH. Since the annuli MATH and MATH are incompressible we have that MATH. The generator of the MATH is a primitive element in the fr... |
math/9912171 | Let MATH be the NAME splitting of MATH determined by the minimal tunnel system which intersects the decomposing annulus MATH in a single point. We can therefore assume that MATH. The once punctured annulus MATH has two boundary components coming from the vertical annuli MATH and denoted by MATH respectively and one bou... |
math/9912171 | Since MATH has a genus two NAME splitting (as MATH is a tunnel number one knot) and is irreducible, the NAME splitting is strongly irreducible. Otherwise we could compress the NAME surface to both sides and obtain an essential MATH-sphere in contradiction. Similarly any NAME splitting of minimal genus three of a hyperb... |
math/9912173 | The independence of the definition with respect to the ordering of the classical crossings and the invariance of MATH under NAME moves of type II and III follows exactly as in CITE, the only difference being an exchange of the variable MATH for MATH. The behaviour under NAME moves of type I is depicted in REF . Since t... |
math/9912173 | REF follows as in CITE. REF is an immediate consequence of the definition of the matrix MATH. MATH . |
math/9912173 | The formula can easily be checked by verifying the corresponding relation for every state MATH in the state sum model used in CITE to define the "partition function" MATH. MATH . |
math/9912173 | REF follow from the fact that summing up the columns (rows) of the determinant belonging to MATH REF gives the trivial column (row) vector. REF is an immediate consequence of setting MATH in REF (or in the definition). MATH . |
math/9912173 | Assume the above skein relation has a non-trivial solution. Consider the skein triples MATH where MATH is a classical knot diagram with one crossing and MATH where MATH is a standard diagram, with arbitrary orientations of the components, of the NAME link (the latter triple corresponding to an arbitrary of the diagram'... |
math/9912174 | To set up notation, let MATH intersect MATH in points MATH, MATH, MATH, and MATH, where MATH is joined to MATH and MATH is joined to MATH via oriented arcs MATH and MATH in MATH. Similarly, MATH is joined to MATH and MATH is joined to MATH-via arcs MATH and MATH in MATH. Fix a pair of arcs on MATH, MATH and MATH, joini... |
math/9912174 | The idea of the proof is simple and follows the approach of CITE or CITE. It is actually implicit in CITE. Basically, if MATH denotes a REF - manifold used in the computation of MATH then a REF - manifold that can be used to compute MATH is built from MATH by adding copies of a REF - manifold MATH along the lifts of MA... |
math/9912174 | The idea of the proof is related to that of the previous theorem. In the definition of MATH one considers the infinite cyclic cover of MATH. The infinite cyclic cover of MATH is built from that of MATH by removing MATH copies of MATH and replacing it with MATH - copies of the infinite cyclic cover of MATH. Hence it is ... |
math/9912174 | First note that a subgroup MATH of MATH, a torsion group with nonsingular linking pairing, is a metabolizer if and only if the linking form vanishes on MATH and the order of MATH is the square root of the order of MATH. (From the exact sequence MATH we have that MATH. If the linking form vanishes on MATH, then MATH.) I... |
math/9912174 | If MATH is cg - slice, let MATH be the appropriate metabolizer. Similarly, let MATH be the metabolizer that exists because MATH is cg - slice. Then the previous theorem provides a metabolizer MATH. For any MATH there is a MATH such that MATH. Hence, MATH. Hence, MATH is cg - slice as desired. |
math/9912174 | The quotient of MATH under the MATH - action of MATH is a MATH - fold branched cover of the quotient of MATH under the action of MATH. This quotient is a REF - sphere and the branch set has quotient consisting of REF points. The MATH - fold branched cover of a MATH over REF points is again a REF - sphere, so the quotie... |
math/9912174 | The cover of MATH decomposes as MATH, but note that the cyclic action of the deck transformation restricted to MATH is the inverse of the action that arises when we consider the cover of MATH, since the character MATH has been replaced by MATH. That the MATH action is well defined on the union implies that MATH. Consid... |
math/9912174 | The action of MATH and MATH generate a dihedral action on the rational vector space MATH. It follows from representation theory that any such rational representation of the dihedral group splits as the direct sum of three types of representations CITE. The first, denoted simply as MATH, is the trivial action on MATH. T... |
math/9912174 | As expected, the proof begins by decomposing the manifolds used to compute the NAME - NAME invariants. Hence: CASE: Decomposing MATH . We begin by describing how to construct a knot MATH as the union of MATH and a trivial twisted tangle, MATH, (see REF , illustrating a trivial twisted tangle) so that the restricted cha... |
math/9912174 | We have already seen that there is a correspondence between characters on MATH and MATH and that, by REF , MATH is invariant under this correspondence. It remains to check that metabolizers are preserved under the correspondence. To see this we need to be look a little more closely at the homology and the linking form.... |
math/9912174 | The NAME form for MATH is given by MATH . A symplectic basis for the form is given by the pair of elements MATH. In this basis the NAME form has a MATH in the upper left-hand entry, that is, the vector MATH is a metabolizing vector for the NAME form. In particular MATH is algebraically slice, and hence lies in the kern... |
math/9912174 | The argument begins with some basic computations that follow readily from the techniques of CITE and CITE. Details are presented in CITE. The knot MATH is built from MATH by removing neighborhoods of curves MATH and MATH linking the bands and replacing them with the complements of MATH and MATH. REF - fold branched cov... |
math/9912174 | First we note that MATH and MATH cobound an annulus MATH embedded in the complement in MATH of the slice disk for MATH just constructed. To see this, perform the band move joining MATH and MATH along the arc MATH. The resulting circle is unknotted and unlinked from the pair of circles formed during the slicing of MATH,... |
math/9912174 | The NAME polynomial of MATH is MATH. This can be quickly seen by noticing that changing one of the two rightmost crossings on MATH in REF changes MATH into the connected sum of two copies of REF - twisted double of the unknot, which has NAME polynomial MATH. The NAME skein relation shows that changing this crossing doe... |
math/9912174 | The curves MATH each lift to three curves in MATH, and the NAME sequence shows that replacing the six solid tori lying above these curves with knot complements, MATH or MATH, does not affect the cohomology. |
math/9912174 | Each irreducible factor of MATH is of the form MATH for some MATH and MATH, since the irreducibility of a polynomial MATH implies the irreducibility of MATH. Each of these factors, MATH, satisfies MATH. Hence, if MATH for some MATH, then the exponent of each irreducible factor of MATH must be even. It remains to observ... |
math/9912174 | Using REF one sees that since the character MATH takes the value MATH on the lift MATH of MATH, it takes the value MATH on the translate MATH and takes the value MATH on MATH. Similarly MATH takes the value MATH on MATH and hence takes the value MATH on MATH and MATH on MATH. The lemma now follows from REF . |
math/9912174 | Given MATH, the set MATH equals MATH if MATH, and equals MATH if MATH. Suppose MATH. Then by REF and the choice of MATH . Where MATH. By REF MATH is a norm. Using the hypotheses on MATH, REF , and the fact that MATH is odd, we see that, modulo norms, MATH equals a product MATH where MATH is the set of those MATH so tha... |
math/9912174 | Suppose that MATH has dimension MATH. Choose some basis for MATH. Each basis element is written MATH which we consider as the MATH-th row of a MATH matrix MATH. Elementary row operations, and interchanging of columns, which does not affect whether a row is odd (that is, has an odd number of non-zero entries) transforms... |
math/9912174 | Since MATH for MATH, it follows that MATH takes MATH eigenvectors to MATH eigenvectors and vice-versa. Hence MATH and so MATH since MATH. Similarly MATH. The off - diagonal entries are equal since the form is symmetric, and MATH must be a unit since the form is non-singular. |
math/9912174 | Consider the columns of MATH as a basis MATH of MATH. Let MATH denote the standard basis of MATH. Let MATH-Hom-MATH. The condition that every linear combination of the rows of MATH is even can be restated by saying that for each MATH, the sets MATH have the same cardinality modulo MATH. Another way to state this is as ... |
math/9912176 | This is an immediate consequence of the last corollary. |
math/9912176 | By induction we easily prove that MATH . NAME the possible orders of elements of MATH are MATH and MATH . In particular the order of the element MATH is MATH. Let MATH be the subgroup generated by MATH . Assume MATH admits an action of genus zero on some NAME surface MATH. Then there exists a short exact sequence MATH ... |
math/9912176 | We need only show that the group MATH presented in REF does not have genus zero. The order MATH of MATH modulo MATH will be MATH for some MATH, MATH . If MATH then MATH commutes with MATH and since MATH it follows that MATH has a subgroup isomorphic to MATH namely the subgroup MATH . But this contradicts REF . Thus sup... |
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