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Endomorphisms Of The Cohomology Ring Of Generalized Grassmannians

Posted on:2009-04-16Degree:MasterType:Thesis
Country:ChinaCandidate:F LiFull Text:PDF
GTID:2120360242480071Subject:Basic mathematics
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In this paper, for generalized Grassmannians G/H[10], according Gr(o|¨)bner basis theory, we classify all endomorphisms of the integral cohomology ring of G/H. The Lefschetz numbers of these endomorphisms are computed.The first chapter is introduction. Firstly, we introduce the key idea of Algebraic Topology is to turn the abstract geometric problems into computable algebraic problems. Specially, we say the problem of determining End(H*(X)) is a step toward a homotopy classification of all self-maps of X. Finally, some results on the computation of End(H*(X)) are list.In the second chapter, we introduce generalized Grassmannians first. In this paper, we pay attention toF4/C3·S1; F4/B3·S1; E6/A6·S1; E6/D5·S1; E7/E6·S1; E7/D6·S1; E8/E7·S1; E7/A7·S1.whose integral cohomology rings are given by Duan.H.B and Zhao.X.Z[10][11]H*(G/H) = Z[y1,…,yn]/ < r1,…,rm > .where, G/H is one of generalized Grassmannians above, deg(yi) = 2i, ri is a homogeneous polynomial in degree 2i.Let f be an endomorphisms of H*(G/H). Since y1 is the only generator in dimension 2, f sends y1 to a multiple of itself. The main result of this paper is as follows:Theorem 1. Let f be an endomorphisms of H*(G/H), f(yi) = ky1, k∈Z.(1) If G/H≠E6/A6·S1 then f(yi) = kiyi, i = 1,…n;(2) If G/H≠E6/A6·S1, then f(yi) = kiyi, i= 1,3,4,6 orThe third chapter is the proof of Theorem 1. The base of the proof is Gr(o|¨)bner basis theory.Let R = F[x1,…,xn] be the polynomial ring on the field F. Fix a monomial order. Every nonzero ideal I of R has a Gr(o|¨)bner basis. From a generator system {g1,…,gt} of I we can get a Gr(o|¨)bner basis of I by Buchberger algorithm.The main idea about Gr(o|¨)bner basis we use in this paper is that for all f in R , we can determine f is in I or not. Specifically, by division algorithm[12], use the Gr(o|¨)bner basis of I to divide f. If the remainder r = 0, then f∈I. If the remainder r≠0, then f (?) I.Let f be an endomorphism of H*(G/H). Since H*(G/H, Q) = H*(G/H,Z)(?) Q, we know f is an endomorphism of H*(G/H,Q) too. So we compute End(H*(G/H,Q)) first, then come back to End(H*(G/H,Z)).Suppose H*(G/H, Q) = Q[y1,…, yn]/ < g1,…,gm >. Let Q[y1,…,yn]i be the sub-module of homogeneous polynomials in degree 2i. The set of monomials form a basis of Q-module Q[y1,…,yn]i.Since y1 is the only generator in dimension 2, f sends y1 to a multiple of itself. Let f(y1) = ky1,k∈Q. Every f(yi)∈Q[y1,…,yn]i can be expanded uniquely as:According f is a ring homomorphism and (2), we can expand f(gi) as a homogeneous polynomial in degree 2i. It is easy to see:f(gi)∈1,…,gm>, i = 1,…,m.(3)By (Maple), we can get a Grobner basis G of the ideal < g1,…,gm>. Divide f(gi) by G according division algorithm. Let the remainder be hi. hi is a homogeneous polynomial in degree 2i. (3) is true(?) hi=0. Let the coefficients of the monomials in hi be zero. We get polynomial equations about cαi, k. Finally, use the "solve" and "gsolve" in Maple to solve these equations. So we get f.The fourth chapter is the application of the main result. Let f be a self-map of G/H. f* is the endomorphism of H*(G/H,Q) induced by f.When f*(yi) = kiyi, k∈Q, the Lefschetz number iswhereβq is the qth Betti number. Pk(G/H) is obtained from Pt(G/H) (the poincare polynomial of G/H) by substituting the t2 by k.Considering E6/A6·S1, we compute the base b(i) of H2i(E6/A6·S1) by dimH2i(E6/A6·S1) first. According b(i) , L(f) is easy to get. Finally, by Lefschetz Fixed Point Theorem we have:Theorem 2. Let f be a self-map of G/H. f* is the endomorphism of H*(G/H,Q) induced by f.(1)If G/H = E6/D5·S1, then f has a fixed point.(2)If G/H≠E6/D5·S1, f*(y1)≠-y1,then f has a fixed point.
Keywords/Search Tags:Endomorphisms
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