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Let $G$ be a finite group acting linearly on affine space $\mathbb{A}^n_k$ over $k = \mathbb{C}$. Since the action is linear, it can be extended to an action of $\operatorname{GL}_n(\mathbb{C})$, which is connected, and therefore the induced action on cohomology $H_c^k(\mathbb{A}^n_k;\mathbb{Z})$ must be trivial. In particular, this shows that the Hodge–Deligne polynomials of $\mathbb{A}_k^n$ and the GIT quotient $\mathbb{A}_k^n \;//\; G$ both equal $(uv)^n \in \mathbb{Z}[u, v]$.

Now, a more general invariant than the Hodge–Deligne polynomial is the class of a variety in the Grothendieck ring of varieties $\operatorname{K}(\operatorname{Var}_k)$. My question is: is it true that

$$ [\mathbb{A}^n_k \;//\; G] \overset{?}{=} [\mathbb{A}^n_k] $$

in $\operatorname{K}(\operatorname{Var}_k)$ (with still $k = \mathbb{C}$)? I can show this equality for $G$ abelian, but have not yet found a counterexample for $G$ non-abelian. Furthermore, I have seen examples (Saltman) of a finite group $G$ acting on $\mathbb{A}_k^n$ such that the quotient is not a rational variety, but I'm not sure if that yields a counterexample here.

Edit: apparently (see Jason Starr's comment) the answer is no in general. What about if we restrict to the symmetric group $G = S_n$?

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    $\begingroup$ Saltman's examples (and similar examples by Bogomolov, Ojanguren, et al.) are not even stably rational. Now apply Larsen-Lunts. $\endgroup$
    – Jason Starr
    Commented Jun 30, 2022 at 13:57
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    $\begingroup$ About your Edit: do you mean for the standard action of $S_n$ on $\Bbb{A}^n$? Then the quotient is isomorphic to $\Bbb{A}^n$... $\endgroup$
    – abx
    Commented Jun 30, 2022 at 15:41
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    $\begingroup$ @JasonStarr: Larsen-Lunts is for smooth projective varieties, and it's not immediately applicable here, as the quotient is not smooth. That is, if $[U] = \mathbf{L}^n$ for a quasi-projective variety $U$ we may not immediately deduce that $U$ is stably rational. $\endgroup$
    – Evgeny Shinder
    Commented Jun 30, 2022 at 18:19
  • $\begingroup$ @EvgenyShinder I believe there is a "Kirwan desingularization" of $\mathbb{A}^n$ (or perhaps something simpler) that makes the action "pseudoreflective" while only changing the class in the Grothendieck group by an integer-coefficient polynomial in $\mathbb{L}$. Now we can apply Larsen-Lunts to the group quotient (which is smooth by Chevalley-Shephard-Todd). $\endgroup$
    – Jason Starr
    Commented Jun 30, 2022 at 20:06
  • $\begingroup$ @abx No, I mean any linear action $\endgroup$
    – jessetvogel
    Commented Jul 1, 2022 at 7:24

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