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What do the derived functors of the symmetric powers look like? I understand that this is related to the homology of the symmetric groups, but I don't know a reference for that.

Namely, I'm interested in the homotopy groups of the free simplicial commutative ring on a simplicial set. Let $X_\bullet$ be a simplicial set; I'd like to know the homotopy groups of $\mathbb{Z}[X_\bullet]$. This is the symmetric algebra on the free simplicial abelian group $\mathbb{Z}X_\bullet$, which is weakly equivalent to a product of Eilenberg-MacLane simplicial abelian groups corresponding to the homology of $X_\bullet$ (and is cofibrant). In particular, we have a weak equivalence of simplicial commutative rings $$\mathbb{Z}[X_\bullet] \simeq \bigotimes \mathbb{L} \mathrm{Sym}^\bullet K( H_n(X_\bullet, n)),$$ which brings up the question of what $\mathbb{L} \mathrm{Sym}^\bullet$ looks like. Tyler Lawson points out in answering this question that the answer is somewhat complicated and describes it in low degrees.

Is a complete answer known?

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I think some computations of higher homotopy groups of spheres can be rephrased as the derived functors of symmetric powers: arxiv.org/abs/1103.4580v1 –  John Wiltshire-Gordon May 15 '12 at 19:15
    
I believe that "I'm interested in the homotopy groups of the free simplicial commutative ring on Let $X_\bullet$ be a simplicial set" should probably be "I'm interested in the homotopy groups of the free simplicial commutative ring on A SIMPLICIAL SET. Let $X_\bullet$ be a simplicial set". Is this correct? –  Theo Johnson-Freyd May 15 '12 at 23:03
    
Thanks for the correction and the reference to the paper. –  Akhil Mathew May 16 '12 at 0:12
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The symmetric groups have complicated cohomology taken individually, but taken all together the homology possesses extra structure (Dyer-Lashof operations) that makes it simple to describe. A nice reference, if I remember correctly, is Bisson and Joyal's "Q-rings and the homology of the symmetric groups." (preprint here: hopf.math.purdue.edu/Bisson-Joyal/Luminy.pdf) –  Tyler Lawson May 16 '12 at 5:23
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arxiv.org/abs/0911.0638 may be of interest –  m_t May 16 '12 at 7:23
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1 Answer

up vote 6 down vote accepted

The homology of all of the symmetric groups together is well understood, as Tyler says. Taking mod $p$ coefficients, that is the special case when $X = S^0$ of the calculation of $H_*(CX)$ as a functor of $H_*(X)$, where $C$ is the monad on based spaces associated to any $E_{\infty}$ operad of spaces. The calculation in this form is given as Theorem 4.1, page 40, of [Cohen, Lada, May. The homology of iterated loop spaces, SLN Vol 533. 1976] which is available on my web page. The functor is not all that complicated, but you do have to understand the Dyer-Lashof operations, which are very much like Steenrod operations and can be seen with those as special cases of a general construction of Steenrod operations [A general algebraic approach to Steenrod operations. In SLN Vol. 168. 1970] also on my web page. The paper of Bisson and Joyal cited by Tyler gives a reformulation of this functor in the case $p=2$. If you want the integral homology, that is a mess to write down in closed form, but the mod $p$ Bockstein spectral sequence of $CX$ is entirely determined by that of $X$, as explained in Theorem 4.13 op cit above, so that integral information is also available. It is worth emphasizing that viewing the homology of symmetric groups as a special case of $H_*(CX)$ substantially simplifies both the calculation and understanding the answer.

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Very interesting; thanks. –  Akhil Mathew May 20 '12 at 3:24
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