# How to compute homology of symmetric products of complexes?

First, I would appreciate references on the notion of derived symmetric powers of perfect modules over various kinds of derived commutative algebras (say cdgas in characteristic zero, simplicial commutative rings, or $E_{\infty}$-rings).

Here is a very special case of the kind of computation one would like to do:

Let $R$ be a (discrete) commutative $k$-algebra over a field $k$ of characteristic zero, and let $M^{*}=M^{i} \rightarrow M^{i+1}$ be a two-term complex of free or projective $R$-modules. There is a notion of symmetric power, $Sym^{p}(M^{*})$, which people say is just the quotient of ${M^{*}}^{\otimes p}$ by the action of the symmetric group $\Sigma_{p}$. I take this to mean that we should take the quotient of the subcomplex generated by the images of $Id -\sigma$ for all $\sigma \in \Sigma_{p}$. (Is that right?) (Here, tensor products are taken over $R$, not $k$.)

QUESTION 1: How is the homology of $Sym^{p}M^{*}$ related to the homology of $M^{*}$?

QUESTION 2: The symmetric product of classical modules commutes with base change. Is that true for these derived versions?

I'm assuming characteristic zero, since otherwise I gather homology of the symmetric group causes interesting complications. Comments on such complications are also welcome.

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Let's start by looking at any action of a group $G$ on a chain complex $C$. You can, as you say, kill the image of $1-g$ for each $g\in G$ to get the complex of coinvariants, call it $C_G$: the largest quotient complex on which $G$ acts trivially. In general this functor from $G$-complexes to complexes does not preserve weak equivalences (quasi-isomorphisms), but if $G$ is finite and everything is linear over a field of char zero (or any ring in which $|G|$ is invertible) then not only is it homotopy-invariant in this sense but $H(C_G)=H(C)_G$. The key is averaging over $G$ to split things up. –  Tom Goodwillie Mar 9 '12 at 13:56
OK. This is helpful. A similar thing for simplicial modules? Can the coinvariants be made to preserve weak equivalences after somehow resolving, like tensoring with chains on $EG$ over $kG$? –  dhagbert Mar 9 '12 at 20:49
@Dhagbert: Yes, tensoring with chains on EG makes coinvariants "derived" and invariant under weak equivalence. –  Tyler Lawson Mar 10 '12 at 15:19
Good. Could you point out a place in the literature where this is discussed? –  dhagbert Mar 11 '12 at 16:38
@Dhagbert: Well, because $C_*(EG)$ is levelwise free over $\mathbb{Z}[G]$, this "derived coinvariants" is just the derived tensor product of $\mathbb{Z}$; any reference on derived tensor products or "hypertor" should show that it preserves weak equivalences. So far as the reference to McCleary below, I was more thinking of it as a reference for derived tensor product of modules over a DGA in order to take powers. (Topologists do work a lot with symmetric powers, but it is usually the "homotopy" symmetric power. There are a lot of sticky issues related to it which you may not want to know.) –  Tyler Lawson Mar 13 '12 at 17:03

If $n!$ is invertible in $H_0(R)$, then the homology of the symmetric power is isomorphic to the $\Sigma_n$-invariants in the homology of $M^{\otimes n}$, and the latter is the derived tensor product of $M^*$ with itself $n$ times (so you generally need a projective resolution, or the Kunneth spectral sequence, to compute it). This is because the "$\Sigma_n$-coinvariants" functor that you described is exact, and so commutes with taking homology.

The homology of this does, indeed, commute with base change.

If $n!$ is not invertible in $R$, then things get much more complicated. For example, you have to decide whether you want the derived symmetric power (where you don't take orbits, but instead derived-tensor over $R[\Sigma_n]$ with the trivial module $R$) or you want an actual symmetric power (which you would form for simplicial modules over a constant ring by making sure that it is levelwise projective and then taking symmetric power levelwise).

In the former case, you get the same kind of Kunneth-type spectral sequence, which takes the form $$H_r(\Sigma_n; H^s M^{\otimes n}) \Rightarrow H^{s-r} Sym^n(M^*).$$ Obviously involving the homology of the symmetric groups complicates things. This commutes with base change.

In the latter case, it is much more complicated and I'm not familiar enough to know effective computational procedures (e.g. if $M$ is free and concentrated in one degree, it computes homology of symmetric powers of wedges of spheres). It doesn't have a general definition for modules over a commutative dga or an $E_\infty$-algebra, but only over a simplicial commutative ring. In this case it only commutes with base change along maps of simplicial commutative rings.

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An instructive thing to look at even in the $n=2$ case is this: For a simplicial module $M$ you have the levelwise $M\otimes M$ with its $\Sigma_2$-action, hence also the associated dg object; but you also have the tensor square of the associated dg object of $M$ (a differential graded object obtained from a differential bigraded object). These two dg objects with group action are weakly equivalent, but not in the strongest sense: their complexes of invariants are not weakly equivalent. For example, if $M$ is trivial below level $n$ then $M\otimes M$ has no homotopy below $2n$ ... –  Tom Goodwillie Mar 9 '12 at 15:18
Thanks. This should be helpful. Could you suggest a reference for this? Particularly the Kunneth spectra sequence. Also, I'm not sure that you finished the sentence ending 'and the latter is the derived tensor product of M∗ with itself n times'. That doesn't seem quite right to me. –  dhagbert Mar 9 '12 at 20:46
... but its coinvariants can have $2$-torsion in $\pi_n$; whereas the other dg thing has nothing below $2n$ before (or after) taking coinvariants. –  Tom Goodwillie Mar 9 '12 at 22:09
In the first sentence, do you really mean "invariants" ? To my understanding the OP uses $Sym^n M := M^{\otimes n}_{\Sigma_n} = k \otimes_{\Sigma_n} M^{\otimes n}$. But that would be the co-invariants. –  Ralph Mar 15 '12 at 3:53
@Ralph: Yes, it's most naturally the co-invariants (but if n! is invertible the two are isomorphic). –  Tyler Lawson Mar 15 '12 at 13:06

Building on the results established by Tom and Tyler, it seems to me that the structure of $H_\ast(Sym^n M)$ can be further explored:

Since the characteristic of $k$ is assumed to be zero, $H_\ast(Sym^n M) = H_\ast(M^{\otimes n})_{\Sigma_n}$ by Tom's and Tyler's answer and by Künneth formula $H_\ast(M^{\otimes n}) = H_\ast(M)^{\otimes n}$. Hence as an intermediate result
$$H_\ast(Sym^n M) = H_\ast(M)^{\otimes n}_{\Sigma_n}$$

Consider $H_\ast(M)$ as a graded $k$-vector space and choose a basis $\lbrace x_j \mid j \in J\rbrace$ where $J$ is assumed to be totally ordered (denote the order by $\le$). Then the elements $x_{j_1} \otimes \cdots \otimes x_{j_n}=: \underline{x}_j$, $\;j = (j_1,...,j_n) \in J^n$ form a basis of $H_\ast(M)^{\otimes n}$ and the action of $\Sigma_n$ is given by $$\sigma^{-1} \cdot x_{j_1} \otimes \cdots \otimes x_{j_n} = x_{i_1} \otimes \cdots \otimes x_{i_n}$$ where $$(i_1,...,i_n) = (\sigma(j_1),...,\sigma(j_n))=: \sigma^{-1} \cdot (j_1,...,j_n).$$

The latter defines an action of $\Sigma_n$ on $J^n$ as well. We can subdivide $J^n$ into those tuples having exactly $l$ equal elements $(l= 1,...,n)$. This leads to the following set of representatives $E = J^n/\Sigma_n = \coprod_{l=1}^n\; E_l$ where

$$E_1 = \lbrace (j_1,...,j_n) \mid j_1 < ... < j_n\rbrace$$ $$E_l = \lbrace (j,...,j,j_1,...,j_{n-l}) \mid j_1 < ... < j_{n-l},\;\; j \neq j_i\; \forall i\rbrace\quad(2 \le l \le n)$$

The stabilizer of $j \in E_l$ is $\Sigma_l \le \Sigma_n$. Now by Corollary III.5.4 in Brown, Cohomology of Groups:

$$H_\ast(M)^{\otimes n} = \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; \operatorname{Ind}^{\Sigma_n}_{Stab(j)}k\cdot \underline{x}_j$$

$$\hspace{60pt} = \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k[\Sigma_n] \otimes_{k[\Sigma_l]}\;k\cdot \underline{x}_j$$

Hence by Brown, II (2.1):

$\qquad H_\ast(M)^{\otimes n}_{\Sigma_n} = k \otimes_{k[\Sigma_n]} H_\ast(M)^{\otimes n}$ $= \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k \otimes_{k[\Sigma_l]} \;k\cdot \underline{x}_j$

$\hspace{60pt}= \bigoplus_{l=0}^n \;\bigoplus_{j \in E_l}\; k\cdot \underline{x}_j$

So as final result we keep hold:

$\qquad\qquad H_\ast(Sym^n M) = \bigoplus_{j \in E}\; k\cdot \underline{x}_j\quad, \qquad E= J^n/\Sigma_n$

In particular, if $m := \dim_k H_\ast(M) < \infty$ then

$\dim_k H_\ast(Sym^n M) = |E| = \binom{m}{n} + \sum_{l=0}^{n-2} \;\binom{m-1}{l}\;m$

For example, if $n > m$ then $\dim_k H_\ast(Sym^n M) = 2^{m-1}\;m$.

Remark: The question requests that there is at most one $r$ such that $N_1 := H_r(M), N_2 := H_{r+1}(M) \neq 0$. Using a (more or less obvious) result of Dold: Homology of Symmetric Products and other Functors of Complexes (1958), (8.4), the structure of $H_\ast(M)^{\otimes n}_{\Sigma_n}$ can be refined as $$H_\ast(M)^{\otimes n}_{\Sigma_n} = \bigoplus_{p+q=n} (N_1)_{\Sigma_p} \otimes_k (N_2) _{\Sigma_q}$$ Then each $(N_i)_{\Sigma_p}$ can be separately analyzed as before.

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Thank you. At some points, you are using that the base ring is a field. (The question was for a base simplicial commutative ring/cdga $R$. That wasn't clear from the notation, so I've edited.) At what points would this cause complications in the above? It almost certainly won't be the case for me that $H_{*}(M)$ is free over $H_{0}(R)$). –  dhagbert Mar 15 '12 at 9:14
@Dhagbert: It causes complications in the identification of the homology of the tensor power, which propagates forward. Here's an example: Let $R$ be a polynomial algebra $\mathbb{Q}[x]$ and $M = R/x[1]$, the former is concentrated in degree zero and the latter in degree one. Then $H_* M \otimes_R M$ is $\mathbb{Q}$ in degrees two and three. The class in degree two is negated by the action of the symmetric group, but the class in degree three is fixed. Therefore, $H_* Sym^2(M) \cong R/x[3]$. –  Tyler Lawson Mar 15 '12 at 13:15
What can be said in case of tensor products over $R$ is that $H_\ast(M)^{\otimes_R,n}_{\Sigma_n}$ is an $R$-submodule of $H_\ast(M^{\otimes_R,n})_{\Sigma_n} = H_\ast(M^{\otimes_R, n}_{\Sigma_n})$ (apply $Tor^{R\Sigma_n}$ to the short exact sequence in the Künneth theorem). Moreover, $H_\ast(M)^{\otimes_R, n}_{\Sigma_n}$ is a quotient of $H_\ast(M)^{\otimes_k, n}_{\Sigma_n}$ (as $k$-vector space). I don't know if much more can be said in general. –  Ralph Mar 15 '12 at 14:20