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Let $X$ be a finite CW complex, $n \ge 2$, and $\Sigma_n$ be the permutation group on $n$ symbols. Let $X^{(n)}=X^n/\Sigma_n$ be the quotient of the natural action of $\Sigma_n$ on $X^n$. We call $X^{(n)}$ the $n$-th symmetric power of $X$.

It is well-known that $\pi_1(X^{(n)})=\widetilde{H_1}(X;\mathbb{Z})$ for any $X$. Furthermore, if $X$ is simply connected, then it is also known that $\pi_i(X^{(n)})=\widetilde{H_i}(X;\mathbb{Z})$ for $1\le i \le 2n$, see Theorem 5.9 here. The "simply connectedness" hypothesis here cannot be dropped (just see Remark 5.10 on the next page for an example).

I want to understand what can be said about higher homotopy groups of $X^{(n)}$ when $X$ is not simply connected. In particular, are there some techniques to estimate $\pi_2(X^{(n)})$ when $X$ is path-connected but not simply connected? I have been trying to show that $\pi_2(X^{(n)})\ne 0$ for all $n \ge 2$ and finite CW complexes $X$ but I have not suceeded yet! Any help will be appreciated.

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  • $\begingroup$ I think this is an artifact of the fact that if $X$ is $k$-connective then the pair $(X^{(n)}, X^{(n-1)}) $ should have a certain explicit connectivity that grows with $k$ and $n$, and so $\pi_d(X^{(n)}) $ will stabilize (to $H_d(X)$) for large $n$, depending on $k, d$. I would expect that for $k=1$ and $d=2$ it stabilizes with $n\geq 3$, so the only subtle case should be $n=2$. $\endgroup$
    – Achim Krause
    Commented Aug 30 at 14:05
  • $\begingroup$ @AchimKrause thanks, I understand the highly connected cases. I am interested in the case $k=0$, i.e., when $X$ is NOT simply connected. $\endgroup$
    – Jeremy
    Commented Aug 30 at 14:18
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    $\begingroup$ By $k$-connective i mean $k-1$-connected, so $k=1$ is your case. Just wanted to point out that $n=2$ should be the only subtle case. $\endgroup$
    – Achim Krause
    Commented Aug 30 at 15:04

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