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You need two conditions for a field to be a splitting field for a specific irreducible representation (in characteristic zero to begin with): It must contain the character values of the representation …

[This was intended to be comment to Ben's reply but I exceeded the allowable limit for comments.] Actually it doesn't work over any ring. Just take any ring $R$ for which $GL(V)(R) \to PGL(V)(R)$ is …

Yes, it is true in general. I found it as Thm 3.1.2 of Brion, Michel(F-GREN-IF); Kumar, Shrawan(1-NC) Frobenius splitting methods in geometry and representation theory. Progress in Mathematics, 231. B …

The positive results seem at most to be tied to finite dimensional representations in positive characteristic: Let $\frak H$ be the Heisenberg algebra with basis $x,y,$ where $c$ is central and $[y,x …

As there seem to be some differing opinions in the comments as to whether all irreducible representations are finite-dimensional let me give the argument I had in mind. A module over the path algebra …

The problem of classifying irreducible $sl_2(\mathbb C)$-representations is essentially untractable as it contains a wild subproblem. Indeed, the action of the Casimir element $C$ on any irreducible r …

I am not sure to which extent your questions are really related to positive characteristic. The obvious difference between positive characteristic and characteristic zero related to the question is o …

This is very far from a solution but just some ideas that could be possibly be used as a start. First a simple observation on an Artin-Schreier type extension $X\rightarrow Y$, where $X:=\mathrm{Spec …

An irreducible representation is real or quaternionic precisely when its character is real-valued. By the Peter-Weyl theorem all characters are real-valued precisely when every element in the group is …

We have that $\det T_k$ is a fixed (depending on $n=\dim V$ and $k$ only) power of $\det T$. To see this, as well as getting the power, one can for instance note that $\mathrm{SL}(V)$ is the commutato …

It's actually easier to go the other way around. Finite dimensional modules over an algebra $A$ fulfils the Krull-Remak-Schmidt theorem of being isomorphic to a direct sum of indecomposable modules wi …

I think that an example of non-equivalent permutation sets is given by $G=(\mathbb Z/p\mathbb Z)^n$ for $n>2$ (and $p$ a prime). Then the automorphism group is $\mathrm{GL}_n(\mathbb Z/p\mathbb Z)$, t …

If you by "cone" mean exactly that $A$ should be isomorpic to $\mathrm{gr}_{\mathfrak m}A$ it seems that the following is counterexample: Let $G=\mathbb G_m$, $A=k[x,y,z]/(x^2+y^3+z^5)$ with $tx=t^{15 …

I think that the relation is that if $\psi\colon A \to TM$ is the action of the algebroid, then we have that $df(a)=\psi(a)(f)$ for $f$ in degree zero of the CE complex and $a\in\Gamma(A)$ and $d\omeg …

It seems clear to me that blocks should have something to do with the decomposition of the category as a direct product of subcategories. A decomposition into a product of two factors corresponds exac …