Let $K$ be a field of characteristic $0$ (and maybe algebraically closed...) and $G$ be a linear algebraic group, that is a group $G$ which is an affine variety where the multiplication and inversion are morphisms of affine varieties. By $K[G]$ we denote the coordinate ring of $G$.
The Reynolds operator of the group $G$ is defined as a map $R_G \colon K[G] \twoheadrightarrow K$, with the following propoerties:
- $R_G$ is a linear projection of $K[G]$ onto $K$
- $R_G$ is invariant under the action from the left, that is if we define $\sigma.f(\tau) := f(\sigma^{-1}\tau)$ for $\sigma,\tau \in G$ and $f \in K[G]$, we have $R_G(\sigma.f) = R_G(f)$ for all $\sigma \in G$ and $f \in K[G]$
If this operator exists, does it follow that we then also have invariance under the action from the right, which is defined by $\sigma\dot{\phantom{.}}f(\tau) := f(\tau\sigma)$ for $\sigma,\tau \in G$ and $f \in K[G]$? That is, is it true that we then have $R_G(\sigma\dot{\phantom{.}}f) = R_G(f)$ for all $\sigma \in G$ and $f \in K[G]$?
Note that $G$ is not assumed to be reductive or linearly reductive.