If $X$ is a space, we can form $QX=\varinjlim \Omega^n\Sigma^nX$ which is an infinite loop space with homotopy groups $\pi_i(QX)=\pi^{s}_i(X)$ the stable homotopy groups of $X.$ But these are the unstable homotopy groups of $QX.$

Q: Is there any similar expression for the stable homotopy groups of $QX$?

Just to get some feeling, I would already be very happy to know just what $\pi_0^s(QS^0)$ is. If I understand things correctly, it should be the same as $\pi_0^s(B\Sigma_\infty^+),$ but I still don't know how to compute it.

I guess $\pi_i^s(QX)$ seems like some sort of "secondary stable homotopy group" of $X$. But there is no reason to stop there. How about groups $\pi_i(Q^nX)$ which should be "$n$-ary stable homotopy groups" of $X$ - what sort of information do they carry?

Of course then we could start forming other very silly objects such as about some sort of stabilizations $\pi_{i+n}(Q^nX)$ as $n\to\infty$ (does that make sense and/or produce anything sensible?) and so on, but at that point I am burried so deep in things that I don't understand that it's a bit pointless.


The "Snaith splitting" gives the following spectrum level statement: for a pointed connected space $X$, there is a weak equivalence: $$ \Sigma^\infty_+ (\Omega^\infty \Sigma^\infty X) \simeq \bigvee_{n \geq 0} \Sigma^\infty (X^{\wedge n})_{h\Sigma_n} $$ There is also an unbased version: $$ \Sigma^\infty (\Omega^\infty \Sigma^\infty X) \simeq \bigvee_{n \geq 1} \Sigma^\infty (X^{\wedge n})_{h\Sigma_n} $$ This gives an isomorphism on the level of stable homotopy groups: $$ \pi_*^{s}((QX)_+) = \pi_*(\Sigma^\infty_+ \Omega^\infty \Sigma^\infty X) \cong \bigoplus_{n \geq 0} \pi_*^s((X^{\wedge n})_{h\Sigma_n}) $$

When the space $X$ is not connected, one needs to add a group-completion to get the correct statement. The net effect is that on $\pi_0$ we get an isomorphism between $\pi_0^s((QX)_+)$ and the group algebra on the free abelian group on $\pi_0(X)$: it is freely generated as a commutative ring by generators coming from $\pi_0(X)$ and their inverses. In the case of $QS^0$ this becomes the Laurent polynomial ring $\Bbb Z[t^{\pm 1}]$.

  • $\begingroup$ Awesome! Thank you Tyler, this is pretty much pretty much precisely what I wanted. $\endgroup$ – A Rock and a Hard Place Oct 23 '16 at 3:27
  • $\begingroup$ If you wouldn't mind an after-question - what are $\pi_i^s((QS^0)_+)$ for $i\ge 1$? I imagine those nasty group actions should start being relevant, but then you mention group-completion is needed and I don't know how that mingles with the actions of symmetric groups. $\endgroup$ – A Rock and a Hard Place Oct 23 '16 at 7:03
  • $\begingroup$ @user82454 Well, it's hard to give exact statements. The whole thing splits off a big component which is essentially a bunch of copies of the stable homotopy groups of spheres; $\pi_1$ is $\Bbb Z[t^{\pm 1}] \otimes (\Bbb Z/2^2)$; and you're getting a colimit of the stable homotopy groups of $B\Sigma_n$. $\endgroup$ – Tyler Lawson Oct 23 '16 at 17:10
  • $\begingroup$ Particular questions might be answerable but the whole answer is probably not. $\endgroup$ – Tyler Lawson Oct 23 '16 at 17:10
  • $\begingroup$ It's lucky than that I wasn't really looking for the whole answer, but just a bunch of particulars to try to get some feeling for it. :) Maybe it would be helpful if you could explain a bit more explicitly where to apply group-completion? But in either case thank you immensely for all your time and effort in explaining these probably very simple things to me. Greatly appreciated. $\endgroup$ – A Rock and a Hard Place Oct 23 '16 at 19:05

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