Let $X$ be a variety over $k = \mathbb{C}$ with an action of a finite group $G$. According to [this paper](https://arxiv.org/pdf/1711.07909.pdf) (Section 4), the induced action of $G$ on the cohomology of $X$ respects the mixed Hodge structure, allowing to define a $G$-equivariant $E$-polynomial


$$
e^G(X) = \sum\_{k, p, q} (-1)^k [H\_c^{k; p, q}(X; \mathbb{Q})] u^p v^q \in Z[u, v] \otimes\_\mathbb{Z} R(G) ,
$$


where $R(G)$ is the representation ring of $G$.


The coefficients of $e^G(X)$ can be computed from the $E$-polynomials of the quotients $e(X / H)$, where $H$ is a subgroup of $G$. For example, the group $G = S\_3$ has three irreducible representations: $T$ (trivial), $S$ (sign) and $D$ (standard), so we can write $e^{S\_3}(X) = a \otimes T + b \otimes S + c \otimes D$. Using that $e(X) = a + b + 2c$, $e(X / S\_3) = a$, $e(X / \langle \tau \rangle) = a + c$ and $e(X / \langle \rho \rangle) = a + b$, where $\tau = (1 \; 2)$ and $\rho = (1 \; 2 \; 3)$, we find that


$$
a = e(X / S\_3), \quad b = e(X) - 2 \cdot e(X / \langle \tau \rangle) + e(X / S\_3), \quad c = e(X / \langle \tau \rangle) - e(X / S\_3) .
$$


However, note that $S\_3$ has more subgroups than representations, so there is a non-trivial relation between the $E$-polynomials that always holds:


$$
e(X) - 2 \cdot e(X / \langle \tau \rangle) - e(X / \langle \rho \rangle) + 2 \cdot e(X / S\_3) = 0
$$


Now, a more refined invariant than the $E$-polynomial, is the class of a variety in the Grothendieck of varieties $\textrm{K}(\textbf{Var}\_k)$, that is, there is a ring morphism $\textrm{K}(\textbf{Var}\_k) \to \mathbb{Z}[u, v]$ sending $[X]$ to $e(X)$.


My question is: does the analogue relation hold in $\textrm{K}(\textbf{Var}\_k)$ as well? That is,


$$
[X] - 2 \cdot [X / \langle \tau \rangle] - [X / \langle \rho \rangle] + 2 \cdot [X / S\_3] = 0
$$


and similar relations for other finite groups $G$. Or, does anyone know a counterexample where this relation fails?



Representation-induced relations in the Grothendieck of varieties
No.


Let $X$ be a curve of genus $6g+1$ for some $g>0$ with a free action of $S\_3$.


None of $X, X/\langle \tau \rangle$, $X/ \langle \rho \rangle$, and $X/ S\_3$ will be stably birational to each other, since they have genus, respectively, $6g+1$, $3g+1$, $2g+1$, $g+1$ by Riemann-Hurwitz and thus all have different genuses.


By Larsen-Lunts, it follows that there is no $\mathbb Z$-linear relation between $X, X/\langle \tau \rangle$, $X/ \langle \rho \rangle$, and $X/ S\_3$.


So, in particular, the relation fails in this case.


