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.

Now, let’s consider derived tensor product. There is a canonical map M ⊗^{L} N —> M ⊗^{L} N which gives an involution of M ⊗^{L} M when M = N. For example, if M = A/I, then we get an involution Tor_{1}^{A}(M, M) = I/I^{2}. 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.

To figure out what the sign is, suppose we have a double complex M^{*, *} which is symmetric, i.e., M^{p, q} = M^{q, p} switching the two differentials. (My convention: the two differentials of a double complex commute.) OK, so now we want this flipping map M^{*, *} —> M^{*, *} to induce a map of associated total complexes

Tot(M

^{*, *}) ——> Tot(M^{*, *})

but in the construction of Tot there are signs. Namely, emanating from the (p,q) spot is the differential d_{1} + (-1)^{p}d_{2} (again a convention). Thus when we move an element from M^{p, q} to M^{q, p} without signs, this isn’t compatible with the differential d on Tot. What works is to throw in a sign (-1)^{pq} for the map M^{p, q} —> M^{q, p}.

In order to use this for our example of Tor_{1}^{A}(A/I, A/I) assume for the moment that I is flat. Then the double complex

I ⊗ A —> A ⊗ A

| |

I ⊗ I —> A ⊗ I

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.

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} 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.

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^{*} 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^{*}. (Carefully take the total complex and use group generated by flipping *adjacent indices* and the sign I used above for those.) Call this complex of A[S_n]-modules K^{⊗ n}. Then set

LSym

^{n}(K) = K^{⊗ n}⊗^{L}_{A[Sn]}A

In this situation the above shows that H^{-n}(LSym^{n}(M[1])) = ∧^{n}(M).

[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.]