# Homology exponents for $QX$

We say that a space $X$ has a homology $p$-exponent if some power of $p$ annihilates the $p$-torsion in $H_\ast(X;\mathbb{Z})$.

I am interested in the homology exponents of the free infinite loop space $QX=\Omega^\infty\Sigma^\infty X$. Apparently, by examining the Bockstein spectral sequence, it should be possible to determine the integral homology of $QX$ as a functor of the integral homology of $X$ (see May's Homology operations on infinite loop spaces, Remark 2.6). Before I get into the gory details of this calculation, however, I would like to ask if the following is known.

If $X$ is a space with a homology $2$-exponent, must $QX$ have a homology $2$-exponent also?

By May's Remark 2.6, the answer should be no. If $X$ is a Moore space with $\tilde H_*(X; Z) = Z/p$ concentrated in an odd degree $2q-1$, and $\tilde H_*(X; Z/p) = Z/p\{x, y\}$ with $\beta_1(y) = x$, then in $H_*(Q(X); Z/p)$ the $r$-th Bockstein $\beta_r$ is defined and nonzero on $y^{p^{r-1}}$ for each $r\ge1$, so $H_*(Q(X); Z)$ contains a copy of $Z/p^r$ in degree $2p^{r-1}q-1$. Hence there is no uniform power of $p$ that annihilates all the $p$-torsion.

I think that the answer is no, but possibly for dumb reasons. There is a model for $QX$ (when $X$ is connected):

$$QX \simeq \coprod_n E\Sigma_n \times_{\Sigma_n} X^{\times n} / \sim$$

where $\Sigma_n$ is the symmetric group, and so $E\Sigma_n$ is taken to be the ordered configuration space of $n$ points in $\mathbb{R}^\infty$. We think of this as the space of points in $\mathbb{R}^\infty$ labelled by points of $X$. The relation $\sim$ drops points when their label is the basepoint.

Snaith has shown that this space stably splits along the filtration by the number $n$. When $X = S^k$ (with $k>0$), the split summands are identifiable as the Thom space $B\Sigma_n^{k\gamma_n}$, where $\gamma_n = \mathbb{R}^n$ is the permutation representation of $\Sigma_n$, and $k\gamma_n$ is its $k$-fold sum. Its homology is then a shift by $nk$ of the homology of $\Sigma_n$ with coefficients in the $k$th tensor power of the sign representation:

$$\tilde{H}_{*+nk}(B\Sigma_n^{k\gamma_n}) = H_*(\Sigma_n, sign^{\otimes k})$$

since the action of $\Sigma_n$ on the top dimensional homology of $\gamma_n \cup \{ \infty\}$ is by the sign representation. In particular, if $k$ is even, this is the trivial representation.

So take $X = S^2$; it has a homology exponent $0$, since $1 = p^0$ annihilates all of the $p$-torsion in $H_*(X)$. However, the homology of $QX$ contains $H_*(S_n, \mathbb{Z})$ for all $n$. There is an argument that I learned in Remark 1.11 of this nice paper by Pakianathan which shows that the homology $p$-exponent of the Sylow $p$ subgroup $S(p^n) \leq \Sigma_{p^n}$ is $p^n$. Now take an element in $H_*(S(p^n), \mathbb{Z})$ realizing that exponent, and transfer it up to an element $x_n \in H_*(\Sigma_{p^n})$ to get a family of elements $\{x_1, x_2, \dots\}$ of unbounded homology $p$-exponent in $H_*(QS^2)$.

Now, if you restrict your attention to connected spaces $X$ with entirely torsion homology, I think that if $X$ has a $p$-exponent, then so does $QX$. The reason is that we can again use the Snaith splitting to get

$$H_*(QX) = \bigoplus_n H_*(E\Sigma_{n+} \wedge_{\Sigma_n} (X^{\wedge n}))$$

and you can compute $H_*(E\Sigma_{n+} \wedge_{\Sigma_n} (X^{\wedge n}))$ as the target of a spectral sequence whose $E_2$ term is $H_*(\Sigma_{n}, \tilde{H}_*(X^{\wedge n}))$. If the homology $p$-exponent of $X$ is $m$, then $p^m$ kills the module $\tilde{H}_*(X^{\wedge n})$ (using the Künneth theorem), and hence this group homology. There might be some extension problems in the SS that I'm ignoring here, but if they can be resolved, then $p^m$ must kill $H_*(QX)$.

• In light of John's answer, I'm guessing that I can't ignore the extension problems. – Craig Westerland May 8 '14 at 14:37