I'm working on some research problems relating to random matrix products, and this is taking me into areas of mathematics I've not previously studied: Lie groups, and especially real algebraic groups. In this area I encounter a lot of questions which I don't have the tools for, but which are probably easy for people who do have the correct tools. Here's one:
Question 1. Let $X \subset GL_d(\mathbb{R})$ be nonempty, let $1<k<d$ and let $X^{\wedge k}:=\{A^{\wedge k}\colon A \in X\}$. If $X^{\wedge k}$ is simultaneously triangularisable - that is, if there exists one basis for $\wedge^k\mathbb{R}^d$ with respect to which all of the elements of $X^{\wedge k}$ are simultaneously upper triangular - does it follow that $X$ itself is simultaneously triangularisable?
It seems to me that the question is unchanged if we replace $X$ with the smallest Zariski-closed subgroup of $GL_d(\mathbb{R})$ which contains $X$, so we may as well assume $X$ to be a real algebraic group $G \leq GL_d(\mathbb{R})$. This turns Question 1 into a question about the representation $\rho \colon G \to GL(\wedge^k\mathbb{R}^d)$ defined by $\rho(g):=g^{\wedge k}$ and its implications for the structure of $G$, which is perhaps a fairly simple question for people knowledgeable in representation theory (a topic I haven't studied).
The following question seems to me to be related and can perhaps be addressed with similar techniques:
Question 2. Let $G \leq GL_d(\mathbb{R})$ be a real algebraic group which acts irreducibly on $\mathbb{R}^d$ in the obvious manner, i.e. such that there does not exist a proper nonzero linear subspace of $\mathbb{R}^d$ which is preserved by every element of $G$. Then $G$ is reductive, because if $N$ were a nontrivial normal unipotent subgroup of $G$ then the set of vectors fixed by every element of $N$ would be a proper nonzero linear subspace of $\mathbb{R}^d$ which is preserved by $G$. In particular the representation $\rho \colon G \to GL(\wedge^k\mathbb{R}^d)$ defined by $\rho(g):=g^{\wedge k}$ must decompose into a direct sum of irreducible representations. Other than the trivial bound ${d \choose k}=\dim \wedge^k\mathbb{R}^d$, is there any reasonable, general bound $n=n(k,d)$ for the number of irreducible parts into which the representation splits?
A positive answer to question 1 would seem to imply the bound $n(k,d)={d \choose k}-1$ in question 2, but I would be surprised if that were the best bound possible.
It seems to me that by appealing to Hodge duality the answer to both questions must remain unchanged when $k$ is replaced with $d-k$, so we should without loss of generality be able to assume $k \leq \frac{d}{2}$.