How to compute homology of symmetric products of complexes?

Question

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.

Answer 1

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.

Answer 2

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.