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Suppose given a variety $X$ over an algebraically closed field $K$, $\mathrm{char}K = 0$, equipped with an action of a reductive group $G$. Suppose also that $X$ admits a good quotient $p\colon X\to Y:= X//G$. How can we describe the algebra of differential forms on $Y$? A naive guess is that at least when $X$ and $Y$ are both non-singular the natural map

$$ \Omega^\bullet_Y \to (p_*\Omega^\bullet_X)^G $$

is an isomorphism. Is this guess true? A similar question for tangent sheaves: if we denote by $\mathcal G$ the subsheaf of $\mathcal TX$ generated by the image of the Lie algebra of $G$, is it true that

$$ \mathcal TY \simeq p_*(\mathcal TX/\mathcal G) $$

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    $\begingroup$ Consider $X=G$ with $G$ acting by translation, then $p\colon X\to Y=\mathrm{pt}$ is a good quotient. In this case, the map which you wanted to be an isomorphism is $ 0 \to \mathfrak{g}^*$ (the zero space mapping to the dual of the Lie algebra of $G$). $\endgroup$
    – Piotr Achinger
    Commented Oct 15, 2019 at 9:18

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The statement I made in the question is obviously false as Piotr pointed. However, the following holds when both $X$ and $Y$ are non-singular:

$$ \Omega^\bullet Y \simeq (p_*(Ann(\mathcal G)\subset \Omega^\bullet X))^G $$

This result is carefully proved in

Brion, M., Differential forms on quotients by reductive group actions, Proceedings of the American Mathematical Society, 126 (9) (1998), 2535 -- 2539

and its generalizations to the singular case are discussed.

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