Signs and tensor products

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 Tor1A(M, M) = I/I2. 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., Mp, q = Mq, 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 d1 + (-1)pd2 (again a convention). Thus when we move an element from Mp, q to Mq, 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 Mp, q —> Mq, p.

In order to use this for our example of Tor1A(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

LSymn(K) = K⊗ nLA[Sn] A

In this situation the above shows that H-n(LSymn(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.]

8 thoughts on “Signs and tensor products

    • Yes, in Chapter I, Section 4 using bisimplicial methods and local inductive limits he defines it on D. I think this is different from the above, as Bhargav pointed out.

      As usual I claim no originality. The construction in the post above can be found by searching the web also. I was just trying to understand it and in fact, I had misunderstood!

      • Maybe it’s worth pointing out that, in characteristic 0, these constructions *are* the same (which is explained in Quillen’s notes on the cotangent complex, I think).

  1. A simple question to which the answer is well known but I don’t know where to find it since I am a beginner: if $B, B’$ are $C$-algebras is $L_{(B \otimes_C B’)/C} = L_{B/C} \oplus L_{B’/C})$? Or does the direct sum of the cotangent complexes merely fit into some exact triangle of various cotangent complexes? A reference?

    • First of all, I think you mean L_{B/C} \otimes_C B’ \oplus B \otimes_C L_{B’/C}).

      Then the answer is yes if B and B’ are tor independent over C.

      If you mean derived tensor product everywhere (so now B \otimes_C B’ is not just a ring) then yes in general provided you have a suitable framework to work in (eg simplicial rings in this case would be OK).

      • thanks, that was the conclusion I reached too, I just wasn’t sure (and yes, you interpreted my question correctly despite my typos/inaccuracies).

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