Is there a known explicit description of the abelian $2$-group $\mathsf{Ho}(\mathbb{S})\overset{\mathrm{def}}{=}\mathsf{Ho}(QS^0)\cong\Pi_{\leq1}(QS^0)$?
Homotopy Category of the Sphere Spectrum in Algebraic Topology
at.algebraic-topologyhigher-algebrahomotopy-theoryinfinity-categories
Related Solutions
Yes, for the same reason. Let me sketch a proof.
1- $QS^0\otimes X$ is group-complete. Indeed, its $\pi_0$ is $\mathbb Z\otimes \pi_0(X)$, and that's a group for the usual reasons. Another way to prove it is to prove that the shear map for $X\otimes Y$ is (the shear map of $X)\otimes Y$, which can be seen by noting that $\otimes$ commutes with coproducts and hence finite products in each variable.
2- There is a natural transformation $X\to QS^0\otimes X$ given by tensoring $\mathbb F\to QS^0$ by $X$, and this induces a natural transformation $X^{gp}\to QS^0\otimes X$.
3- Both sides commute with colimits (a colimit of grouplike $E_\infty$-spaces is grouplike so I don't have to worry about whether I'm talking about colimits in monoids or grouplike monoids), therefore to check that this map is an equivalence, it suffices to do so for $X= \mathbb F$, and for that one it is a tautology.
Another way to phrase this is to use the following sequence of natural equivalences (and using point 1- for the last one):
$X^{gp} = QS^0\otimes_{QS^0} X^{gp} = (QS^0\otimes_\mathbb F X)^{gp}= QS^0\otimes X$
The second natural equivalence comes from the fact that group completion is symmetric monoidal, and $(QS^0)^{gp}\simeq QS^0$.
One of the default examples of ordinary graded commutative rings is the polynomial ring $\mathbf Z[t]$. Let us first examine the analogue of that, and then see where else that leads!
1. $S$-grading on $S\{t\}$
For the sake of clarity, allow me to denote the underlying infinite loop space (equivalently: grouplike $\mathbb E_\infty$-space) of the sphere spectrum $S$ by $\Omega^\infty(S)$. Recall that, by the Barrat-Priddy-Quillen Theorem, the ininite loop space $\Omega^\infty(S)$ can be described as the group completion of the groupoid of finite sets and isomorphisms $\mathcal F\mathrm{in}^\simeq$. Consider the constant functor $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$ with value $S$ - since the latter is the monoidal unit, this is a symmetric monoidal functor. Left Kan extension along the canonical map $\mathcal F\mathrm{in}^\simeq \to (\mathcal F\mathrm{in}^\simeq)^\mathrm{gp}\simeq \Omega^\infty(S)$ produces a symmetric monoidal functor $\Omega^\infty(S)\to \mathrm{Sp}$, and as such exhibits an $S$-graded $\mathbb E_\infty$-ring.
But which one? To figure out which one, note that the passage to the "underlying $\mathbb E_\infty$-ring" of an $S$-graded $\mathbb E_\infty$-ring is given by passage to the colimit. Thus the underlying $\mathbb E_\infty$-ring, which we have just adorned with an $S$-grading, is $$ \varinjlim_{\Omega^\infty(S)}\mathrm{LKan}^{\Omega^\infty(S)}_{\mathcal{F}\mathrm{in}^\simeq}(S)\simeq \varinjlim_{\mathcal F\mathrm{in}^\simeq}S, $$ where we used that the left Kan extension of a functor does not change its colimit. Now we can use the explicit description of the groupoid of finite sets $\mathcal F\mathrm{in}^\simeq \simeq \coprod_{n\ge 0}\mathrm B\Sigma_n$ to obtain $$ \varinjlim_{\mathcal F\mathrm{in}^\simeq} S\simeq \bigoplus_{n\ge 0}S_{h\Sigma_n}. $$ We may recognize this as the free $\mathbb E_\infty$-ring on a single generator $S\{t\}$, corresponding in terms of spectral algebraic geometry to the smooth affine line $\mathbf A^1$ (as opposed to the flat affine line $\mathbf A^1_\flat = \mathrm{Spec}(S[t])$ for the polynomial $\mathbb E_\infty$-ring $S[t]\simeq \bigoplus_{n\ge 0}S$).
2. $S$-grading on symmetric algebras
The previous example can be easily generalized by observing that $\mathcal F\mathrm{in}^\simeq$ is the free $\mathbb E_\infty$-space ( = symmetric monoidal $\infty$-groupoid) on a single generator. That means that a symmetric monoidal functor $\mathcal F\mathrm{in}^\simeq \to \mathrm{Sp}$ (which always factors through the maximal subgroupoid $\mathrm{Sp}^\simeq\subseteq\mathrm{Sp}$) is equivalent to the data of a spectrum $M\in \mathrm{Sp}$ (the image of the singleton set). The functor is given by sending a finite set $I$ to the smash product $M^{\otimes I}$, and is evidently symmetric monoidal. The same Kan extension game as before now gives rise to an $S$-graded $\mathbb E_\infty$-ring spectrum, this time with underlying $\mathbb E_\infty$-ring $$ \mathrm{Sym}^*(M)\simeq \bigoplus_{n\ge 0}M^{\otimes n}_{h\Sigma_n}, $$ the free $\mathbb E_\infty$-ring generated by $M$. We recover the prior situation by setting $M=S$. For $M = S^{\oplus n}$, we obtain an $S$-grading on $S\{t_1, \ldots, t_n\}$, corresponding to the spectral-algebro-geometric smooth affine $n$-space $\mathbf A^n$.
3. $S$-grading on $S\{t^{\pm 1}\}$
Another way to generalize the example of the $S$-grading on $S\{t\}$ is to take directly the constant functor $\Omega^\infty(S)\to \mathrm{Sp}$ with value $S$, instead of starting with a constant functor on $\mathcal F\mathrm{in}^\simeq$ and Kan-extending it. This produces a perfectly good $S$-graded $\mathbb E_\infty$-ring, which let us denote $S\{t^{\pm 1}\}$.
From the group-completion relationship between $\Omega^\infty(S)$ and $\mathcal F\mathrm{in}^\simeq$, it may be deduced that $S\{t^{\pm 1}\}$ and $S\{t\}$ are related in terms of $\mathbb E_\infty$-ring localization as $S\{t^{\pm 1}\}\simeq S\{t\}[t^{-1}]$, justifying our notation. Here we are localizing $S\{t\}$ at the element $t\in \mathbf Z[t] = \pi_0(S\{t\})$. In terms of spectral algebraic geometry, this is encoding the spectral scheme $\mathrm{GL}_1$, the smooth punctured line.
4. Remark on non-negative grading
In algebraic geometry, we often prefer to think about non-negatively graded commutative rings than graded commutative rings. Just as the latter are equivalent to lax symmetric monoidal functor $\mathbf Z\to\mathrm{Ab}$, so are the former equivalent to lax symmetric monoidal functors $\mathbf Z_{\ge 0}\to \mathrm{Ab}$.
By analogy, the "non-negatively $S$-graded $\mathbb E_\infty$-rings" are lax symmetric monoidal functors $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$. Indeed, just as $\mathbf Z_{\ge 0}$ is the free commutative monoid on one generator, so is $\mathcal F\mathrm{in}^\simeq$ the free $\mathbb E_\infty$-space on one generator. That is the reason why we were encountering such functors above (and Kan extending them along group completion, as we would to view as $\mathbf Z_{\ge 0}$-grading as a special case of a $\mathbf Z$-grading).
5. Some actual "non-tautological" examples though
So far, a not-completely-unreasonable complaint might be that all the examples of $S$-graded $\mathbb E_\infty$-rings were sort of tautological.
For a very non-tautological example, see the main result of this paper of Hadrian Heine. It shows that there exists an $S$-graded $\mathbb E_\infty$-ring spectrum such that its $S$-graded modules are equivalent to the $\infty$-category of cellular motivic spectra. In fact, much more is proved: this situation is very common, and under some not-too-harsh compact-dualizable-generation assumptions, a symmetric monoidal stable $\infty$-category will be equivalent to $S$-graded modules over some $S$-graded $\mathbb E_\infty$-ring. So you may just as well view this result as a wellspring of potentially interesting examples of $S$-gradings "occurring in nature"! :)
Best Answer
This is the groupoid given by the 1-truncation $\tau_{\leq 1}(QS^0)$. This groupoid has $\mathbb Z$-many objects (since $\pi_0^s = \mathbb Z$), and each one has automorphism group $C_2$ (since $\pi_1^s = C_2$). The tensor product on objects is given by addition in $\mathbb Z$, and on morphisms by addition in $C_2$. One way to see this is to consider the universal functor $\Sigma \to QS^0$ given by the Barratt-Priddy-Quillen theorem (i.e. the fact that $K(\Sigma) = QS^0$; here $\Sigma$ is the groupoid of finite sets with the disjoint union monoidal structure), and to postcompose with the truncation functor $QS^0 \to \tau_{\leq 1} (QS^0)$; the fact that this functor is symmetric monoidal yields this description of the category. This perspective is discussed a bit more here.
From the description I've given, I suppose it follows that $\tau_{\leq 1} (QS^0)$ splits
symmetricmonoidally as $\tau_{\leq 1} (QS^0) = \mathbb Z \times BC_2$, (where $\mathbb Z$ is a discrete symmetric monoidal groupoid and $BC_2$ is a 1-object symmetric monoidal groupoid), which is maybe a little surprising. This is not to say that $\tau_{\leq 1} \mathbb S$ splits...