Let $k$ be a fixed positive integer and $\Sigma_k$ the $k$th symmetric group. By letting $\Sigma_k$ permuting an orthonormal basis of a $k$dimensional Euclidean space, there is a "regular representation" $$ r_k: \Sigma_k\longrightarrow O(k) $$ which induces a map between classifying spaces $$ \rho_k: B\Sigma_k\longrightarrow BO(k). $$ Let $X$ be a finite CWcomplex and a map $$ f: X\longrightarrow B\Sigma_k. $$ For any positive integer $n$, we produce a map by taking selfproduct of $f$ for $n$times $$ \prod_n f: X\longrightarrow \prod_n B\Sigma_k $$ and have a composition $$ g_n: X\overset{\prod_n f}{\longrightarrow}\prod_n B\Sigma_k\overset{\prod_n \rho_k}{\longrightarrow}\prod_n BO(k)\longrightarrow BO(nk)=G_{nk}(\mathbb{R}^\infty). $$ where the last map is induced by the inclusion $$ \prod_n O(k)\longrightarrow O(nk). $$ Question: I want to prove that for any $k$, any finite CWcomplex $X$ and any map $f$, there exists a positive integer $n$ such that $g_n$ is nullhomotopic. Is it true? How to prove?

2$\begingroup$ Are you the same person as haohaizi (mathoverflow.net/users/65800/haohaizi) and, if so, why do you keep creating new accounts? This is, by my count, your third one. $\endgroup$ – Qiaochu Yuan Nov 14 '15 at 17:02
Yes, this is true. Your map $B\Sigma_k\rightarrow BO(k)$ gives rise to a map $B\Sigma_k\rightarrow BO$, i.e. an element in the reduced Ktheory group $\tilde{ko}^0(B\Sigma_k)$.
Now that whole group is probably not torsion, but the AtiyahHirzebruch spectral sequence tells you that, if $K$ is a finitedimensional $CW$ complex with torsion homology groups, $\tilde{ko}^0(K)$ is torsion.
So if you take $K$ to be a suitably modified skeleton of $B\Sigma_k$ (see below), it follows that there is $n$ such that the map $$K\rightarrow B\Sigma_k\rightarrow BO(k)\rightarrow BO(nk)$$ is null homotopic, where the last map is the times $n$ map.
But if $X$ is a finitedimensional $CW$complex, the map $X\rightarrow B\Sigma_k$ will factor over $K$ if we choose $K$ to be a sufficiently highdimensional "skeleton".
I was imprecise above about $K$ being a "suitably modified skeleton", let me clarify this:
If $Y$ is a CWcomplex with torsion homology groups, then, for each $n$, there is an $n+1$dimensional CW complex $K$ with torsion homology groups and a map $K\rightarrow Y$ that induces isomorphisms on homology up to degree $n1$.
To build that $K$, simply take $\tilde{K}$ to be the $n$skeleton of $Y$. Now the $n$th homology of that is not necessarily torsion, but since rationalized homotopy and rational homology agree, we can choose maps from $S^n$ to $\tilde{K}$ such that they are nullhomotopic in $Y$, and their images in $H_n(\tilde{K})$ generate it rationally. But then form $K$ by attaching cells along those maps. $K$ comes with a map to $Y$, and has torsion homology.

2$\begingroup$ An equivalent argument goes like this: The Chern character from $\tilde{ko}^0(K)$ to the product of the rational cohomology groups $H^{4j}(K)$ is rationally an isomorphism if $K$ is a finite complex. Therefore in order for a vector bundle on $K$ to be such that some multiple of it is trivial it is sufficient if the rational Pontryagin classes vanish. Of course they vanish if the bundle is pulled back from a bundle on $B\Sigma_k$ (or any space having trivial rational cohomology). $\endgroup$ – Tom Goodwillie Nov 14 '15 at 14:06

1$\begingroup$ By the way, for finite $G$ the group $\tilde{ko}^0(BG)$ is indeed not a torsion group; in fact it is torsionfree, isn't it? According to the AtiyahSegal completion theorem it is a completion of the real representation ring of $G$. $\endgroup$ – Tom Goodwillie Nov 14 '15 at 14:09

1$\begingroup$ Sorry, I'm confused, maybe you find my mistake. If you take the inclusion $\mathbb{R} P^2\rightarrow B\Sigma_2$, then the associated bundle is $E = \gamma_{\mathbb{R} P^2} \oplus \mathbb{R}$ and $w(kE) = (1+w_1)^k = 1 + kw_1 + (k1)w_1^2$ which is never zero. $\endgroup$ – user_11437 Nov 15 '15 at 12:55

2$\begingroup$ It is $1+kw_1+\frac{k(k1)}{2}w_1^2$, which is $1$ for $k=4n$ or $k=4n+1$. $\endgroup$ – Achim Krause Nov 15 '15 at 15:57

$\begingroup$ @TomGoodwillie Thanks Prof. Goodwillie. In your comment, I do not understand the first two steps: "The Chern character from $\tilde{KO}^0(K)$ to the product of the rational cohomology groups $H^4j(K)$ is rationally an isomorphism if $K$ is a finite complex. Therefore in order for a vector bundle on $K$ to be such that some multiple of it is trivial it is sufficient if the rational Pontryagin classes vanish." Could you explain? Thanks! $\endgroup$ – Quan Nov 16 '15 at 5:39