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Suppose $G$ is an affine group scheme over an algebraically closed field $k$, and $V$ is a finite dimensional representation of $G$. Let $K$ be an extension of $k$, and assume that $V_K$ is reducible …

The functor of injective homomorphisms from $H$ to $G$ is represented by a scheme of finite type over $\mathbb Z$ (a locally closed subscheme of the product $G^H$). If this has points over $\overline{ …

This is not true, in general. For example, take $G = H = \mathbb Z$. Let $G \times H$ act on $\mathbb C^3$ in such a way that a generator of $G$ carries $e_1$ to $e_2$, and $e_2$ and $e_3$ to $0$, whi …

Sure. If you have an algebraic group $G$ defined over $\mathbb Q$ acting on a $\mathbb Q$-algebra $A$, the $\mathbb Q$-algebra of invariants $A^G$ is the equalizer of two usual homomorphisms of algebr …

The split Grothendieck group for vector bundles on a complete variety appears in Nori's PhD thesis on the fundamental group scheme; this was published in the Proceedings of the Indian Academy of Scien …

If by $V/G$ you mean the space of orbits, this is not true. Consider $\mathbb C^*$ acting on the affine space $\mathbb A^1$ by multiplication; the space of orbits has two points, but the only variety …

The embedding of the unitary group $U_n$ into $GL_n(\mathbb C)$ is a homotopy equivalence; this is easily seen to imply that $H^*_{U_n}(X)$ is isomorphic to $H^*_{GL_n}(X)$. So the result for compact …

[Edit]: my previous counterexample was irredeemably wrong; hopefully this one works. Suppose that there exists an étale $G$-equivariant map $V' \times^{P}G \to V$; then there exists an invariant neig …

I don't think this is literally true. For example, suppose that $G$ is a finite cyclic group of order 2 generated by $s$, suppose that $L$ is an non-trivial 2-torsion invertible $A$-module over a Dede …

All this is described very nicely in Serre's book on representations of finite groups.