# Stable summands of $\Omega^kS^{n+k}$

The answer to this question might be known, but I don't know of any reference and will appreciate any references.

Let $n>0$. By Snaith splitting there is a stable splitting of $\Omega^kS^{n+k}$ into wedge of spaces $D_{k,r}S^n=F(\mathbb{R}^k,r)\ltimes_{\Sigma_r}(S^n)^{\wedge r}$ as

$$\Sigma^\infty\Omega^kS^{n+k}\simeq\bigvee_{r=1}^{+\infty}\Sigma^\infty D_{k,r}S^n$$

where $F(\mathbb{R}^k,r)$ is the configuaration space of $r$ distinct points in $\mathbb{R}^k$. It is known that $D_{k,2}S^n=\Sigma^nP_{n}^{n+k-1}$ where $P_n^m$ is the truncated real projective space $P^m/P^{n-1}$.

What I would like to know is about other pieces and specially $D_{k,2^s}S^n$ with $s>0$. Is it known that for any positive $k,s,n$ the space $D_{k,2^s}S^n$ is suspension of some $CW$-complex? If the answer is positive, do we know how many suspension may appear?

• The space $D_{k,r}S^n$ is always an $n$-fold suspension. This is so because there is a homeomorphism of spaces with $\Sigma_r$-action $(S^n)^{\wedge r}\cong S^n\wedge (S^{r-1})^{\wedge n}$. Here $\Sigma_r$ acts by permutation on the left, and on the right hand side it acts on $S^{r-1}$ via the reduced standard representation. It follows that $D_{k,r}S^n \simeq \Sigma^n F({\mathbb R}^k, r)_+ \wedge_{\Sigma_r} S^{(r-1)n}$. Another way to put it is: $D_{k,r}S^n$ is the Thom space of a vector bundle that contains a trivial $n$-dimensional sub-bundle. Therefore it is an $n$-fold suspension. – Gregory Arone Nov 27 '16 at 20:10
• I did not know/have not thought of this $\Sigma_r$-homeomorphism. Does this homeomorphism correspond to an isomorphism between two representations of $\Sigma_r$? and if so, then other isomorphisms correspond to other homeomorphisms? – user51223 Nov 28 '16 at 7:10
• The decomposition of the unreduced standard representation into the direct sum of the reduced standard representation and one-dimensional trivial representation. – user43326 Nov 28 '16 at 10:41
• @user43326 You mean the decomposition you mention, after we compose with $n$-fold diagonal, right? – user51223 Nov 28 '16 at 11:44
• It is just the decomposition of $R^r$. The point, I guess, is that this leads to the decomposition of $R^{nr}=(R^r)^{\oplus n}$ – user43326 Nov 28 '16 at 14:32

## 2 Answers

The full story is eluded to in the comments of Arone and Rognes. First of all, $D_{k,r}\Sigma^n X$ is always an $n$-fold suspension. More exciting is that, for all pairs $(k,r)$, there is a natural number $d = d(k,r)$, such that $D_{k,r} \Sigma^dX = \Sigma^{rd}D_{k,r}X$.

So what is $d$? It is the order of the canonical $r$-dimensional vector bundle over the configuration space $F(\mathbb R^k, r)/\Sigma_r$. This is a bundle of finite order because it has finite structure group and a finite dimensional base space. The actual numbers $d(k,r)$ were computed a long time ago: see [Cohen, Cohen, Kuhn, Neisendorfer, Bundles over configuration spaces, Pac. J. Math. 104 (1983), 47-54]. And yes, $d(2,r)=2$ for all $r$, was known even earlier.

• I really appreciate this. I presume $d$ is bounded below by $2$? This seems to have interesting implications when $r$ is a power of $2$ and $X$ is a sphere. I wish I could accepted all answers and comments at once for accepted answers at once! – user51223 Dec 16 '16 at 18:03
• If $p$ is an odd prime less than or equal to $r$, then $p^n$ is the exponent of $p$ in $d(2n+1,r)$. The exponent of 2 is similar, but involves the Adams phi function (already clear when $r$ is 2). – Nicholas Kuhn Dec 16 '16 at 22:43

When $k=2$ and $n$ is even, $$D_{k,r} S^n \simeq BB_{r+} \wedge S^{rn}$$ is the $rn$-fold unreduced suspension of the classifying space of the braid group $B_r$ on $r$ strings. I once cited Cohen-Mahowald-Milgram The stable decomposition for the double loop space of a sphere'' (1978) for this, and if I recall correctly they credited Arnold for the idea. Hence $rn$ is the maximal number of suspensions that can appear in this case.

Edit: For $k=2$ and $n=2$ the map $F(\mathbb{R}^2,r) \times_{\Sigma_r} \mathbb{R}^{2r} \to \mathbb{R}^{2r}$ taking $[z,\xi] = [z_1, \dots, z_r, \xi_1, \dots, \xi_r]$ to $$(\sum_i \xi_i, \sum_i z_i \xi_i, \dots, \sum_i z_i^{r-1} \xi_i)$$ trivializes the source as an $\mathbb{R}^{2r}$-bundle over $F(\mathbb{R}^2,r)/\Sigma_r$. Here the $z_i$ and $\xi_i$ in $\mathbb{R}^2$ are viewed as complex numbers, and the products in the displayed formula are formed in $\mathbb{C}$. A similar formula works for any even $n$, by taking the direct sum of $n/2$ copies of this $\mathbb{R}^{2r}$-bundle. Hence the Thom complex of this bundle, which is your $D_{2,r} S^n$, is the $rn$-th suspension of $(F(\mathbb{R}^2,r)/\Sigma_r)_+ \simeq BB_{r+}$. I recalled this in Proposition 9.10 of my paper "Topological logarithmic structures" GTM 16 (2009).

For $k=2$ and $n=1$, or more generally for $n$ odd, the stable splitting of $\Omega^2 S^3$ is closely connected to the story of Brown-Gitler spectra.

Regarding the case when $k$ is a higher power of $2$, you might first try $k=4$ and identify $\mathbb{R}^4$ with the quaternions. However, in that case the quaternionic Vandermonde matrix might well be singular. For instance, with $r=3$ and $z = (i,j,k)$ the matrix $\begin{pmatrix} 1 & 1 & 1 \\ i & j & k \\ -1 & -1 & -1 \end{pmatrix}$ annihilates $\xi = (1,0,1)$, so Arnold's idea does not extend to this case.

• Do we know anything for the case when $k$ is a higher power of $2$? – user51223 Dec 3 '16 at 6:12