TITLE: Stable homotopy groups of $QX$
QUESTION [5 upvotes]: 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.

REPLY [10 votes]: 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 $\Bbb Z[t^{\pm 1}]$.