Subject: Re: homology of Hex Date: Mon, 20 Nov 2000 23:51:20 -0500 (EST) From: Tom Goodwillie Let B be the blue territory and let B_0 be the blue solid torus. Let R be the red territory and let R_0 be the red solid torus. Let b\in H_{k-1}(B) generate the image of H_{k-1}(B_0). Let r\in H_{n-k-1}(R) generate the image of H_{n-k-1}(R_0). Duality implies that the sequence 0 -> H_k(B) -> H_k(B,B_0) -> \Z -> H_{k-1}(B) -> H_{k-1}(B,B_0) -> 0 is isomorphic to the sequence 0 -> H^{n-k-1}(R,R_0) -> H^{n-k-1}(R) -> \Z -> H^{n-k}(R,R_0) -> H^{n-k}(R) -> 0 Using this, and assuming b sits in H_{k-1}B in a certain way, you can figure out how r sits in H_{n-k-1}(R). For example, as we know, b has infinite order iff r has finite order. b is primitive (i.e. generates an infinite cyclic summand) iff r=0. b is nx, x primitive, n>0, iff r generates a cyclic summand of order n. b=nx+y, x primitive, y generating a cyclic summand of order m iff r has order n and belongs to a cyclic summand of order mn. But there are other cases; an element of finite order in a finitely generated abelian group need not belong to any cyclic summand. (Smallest example: Z/8 x Z/2 has an element of order 4 that is not divisible by 2.) I claim that the way b sits in its group determines (and is determined by) the way r sits in its group, where "the way g sits in (the f. g. ab. group) G" means the class of (G,g) w.r.t. the equivalence relation on pairs (G,g) generated by isomorphism and (GxH,(g,0)) ~ (G,g). A key point is that (because every chain complex of free abelian groups splits as a direct sum of two-term complexes, each of which is a free resolution of a single homology group) it is OK to pretend that H_{k-1}(B) is the only homology group of B. Now the rule is this: Given G with distinguished element g, let G be the cokernel of the map R->F of free ab groups and let g be represented by \Z -> F. Apply Hom( ,\Z) to get F^* -> R^* and F^* -> \Z^*. The answer is the cokernel of F^* -> R^* x \Z^*, with distinguished element represented by the obvious \Z = \Z^* -> R^* x \Z^*. Tom Goodwillie P.S. To answer the question, I do not think there are any other constraints on those numbers. > > So my question is: Which values for the quadruple > (n_blue,d_blue,n_red,d_red) are possible? Alexander duality over Q tells > us that exactly one of n_blue and n_red is 0. Change of coefficients > tells us that player X loses over Z/p if and only if p divides n_X but > not d_X, and Alexander duality tells us that this happens for exactly > one of the two players. I have the feeling that this is not a complete > characterization, but I ran out of simple ideas. > -- > /\ Greg Kuperberg (UC Davis) > / \ > \ / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ > \/ * All the math that's fit to e-print * > >