Boutot's theorem says that if $X$ is a variety over a field of characteristic 0 with rational singularities, and if $G$ is a reductive group acting on $X$, then the quotient $X/G$ has rational singularities as well.


Is it known whether an analog of this result is true for log terminal singularities?


Namely suppose $X$ is a variety over a field of characteristic 0 with log terminal singularities and an action of a reductive group $G$. Then must $X/G$ also have log terminal singularities?



Does the quotient of a variety with log terminal singularities also have log terminal singularities?
Now I can give you a definite answer. In general, the quotient of a klt singularity by a reductive group is not klt, because for instance, the canonical divisor of the quotient may not be $\mathbb{Q}$-Cartier.


However, one can define a broader notion: klt type. A singularity $(X;x)$ is said to be of klt type if there exists a boundary $B$ through $x$ for which $(X,B;x)$ is klt.
Then, one gets the following theorem:


**Theorem:** Let $X$ be an affine variety with klt type singularities over an algebraically closed field of characteristic zero and $G$ be a reductive group acting on $X$. Then $X/\!/G$ is of klt type again.


The previous is Theorem 1 in <https://arxiv.org/abs/2111.02812>.
I should also mention that the klt type property is an etale property, i.e., if you can check it in an etale cover of $X$ then it holds for $X$. This is Proposition 4.1. in <https://arxiv.org/abs/2111.02812>.
Furthermore, the klt type condition


