If M and N are modules over a ring A there is a canonical map M ⊗ N —> N ⊗ M by flipping tensors. If M = N this map is an involution but not the identity. For example, if V is a vector space of dimension n then flipping tensors gives an involution of V ⊗ V whose eigenvalues are 1 and -1 with multiplicity n(n + 1)\/2 and n(n – 1)\/2.<\/p>\n
Now, let’s consider derived tensor product. There is a canonical map M ⊗L<\/b><\/sup> N —> M ⊗L<\/b><\/sup> N which gives an involution of M ⊗L<\/b><\/sup> M when M = N. For example, if M = A\/I, then we get an involution Tor1<\/sub>A<\/sup>(M, M) = I\/I2<\/sup>. In this case, it seems clear that this map is either 1 or -1. My guess would be it is -1… Let’s see if I am right.<\/p>\n To figure out what the sign is, suppose we have a double complex M*, *<\/sup> which is symmetric, i.e., Mp, q<\/sup> = Mq, p<\/sup> switching the two differentials. (My convention: the two differentials of a double complex commute.) OK, so now we want this flipping map M*, *<\/sup> —> M*, *<\/sup> to induce a map of associated total complexes<\/p>\n Tot(M*, *<\/sup>) ——> Tot(M*, *<\/sup>)<\/p><\/blockquote>\n but in the construction of Tot there are signs. Namely, emanating from the (p,q) spot is the differential d1<\/sub> + (-1)p<\/sup>d2<\/sub> (again a convention). Thus when we move an element from Mp, q<\/sup> to Mq, p<\/sup> without signs, this isn’t compatible with the differential d on Tot. What works is to throw in a sign (-1)pq<\/sup> for the map Mp, q<\/sup> —> Mq, p<\/sup>.<\/p>\n In order to use this for our example of Tor1<\/sub>A<\/sup>(A\/I, A\/I) assume for the moment that I is flat. Then the double complex<\/p>\n \nI ⊗ A —> A ⊗ A computes the tor group. Note that in degrees (-1, 0) and (0, -1) we have I and that a cocycle is of the form f ⊕ -f with f ∈ I. Thus flipping this gives -f ⊕ f, i.e., the opposite. So it seems my hunch was correct.<\/p>\n Ok, but now what if K = M[1] in D(A) for some flat A-module M and we consider the action of flipping on H^{-2}(K ⊗L<\/sup> K) = M ⊗ M. It is clear from the above that the action of flipping is by -1 times the usual flipping map of M ⊗ M. Thus the S_2-coinvariants on this gives the second exterior square of M over A.<\/p>\n And now I’ve finally gotten to the point I wanted to make in this blog post. Let’s use the above to define derived symmetric powers of K in D(A). Choose a K-flat complex K*<\/sup> representing K and use the above to get an action of S_n on the total complex associated to the n-fold tensor product of K*<\/sup>. (Carefully take the total complex and use group generated by flipping adjacent indices<\/em> and the sign I used above for those.) Call this complex of A[S_n]-modules K⊗ n<\/sup>. Then set<\/p>\n \nLSymn<\/sup>(K) = K⊗ n<\/sup> ⊗L<\/sup>A[Sn<\/sub>]<\/sub> A\n<\/p><\/blockquote>\n In this situation the above shows that H-n<\/sup>(LSymn<\/sup>(M[1])) = ∧n<\/sup>(M).<\/p>\n [Edit: Bhargav points out that this isn’t the derived symmetric power you get in the symplicial world. For example, if K = A[0], then we get S_n group homology. Whereas if you think if A as a constant simplicial module, then Sym_n(A) = A.]<\/p>\n","protected":false},"excerpt":{"rendered":" If M and N are modules over a ring A there is a canonical map M ⊗ N —> N ⊗ M by flipping tensors. If M = N this map is an involution but not the identity. For example, … Continue reading
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\nI ⊗ I —> A ⊗ I\n<\/p><\/blockquote>\n