Simultaneous triangularisation of an exterior power of a set of matrices

Question

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, representation theory, and 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. I previously posted this question on Stack Exchange, but it sank like a stone:

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 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 its Zariski closure or with the group which $X$ generates. Doing first the latter and then the former is equivalent to replacing $X$ 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 $G$. In my understanding this implies that $G$ 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 have to be a proper nonzero linear subspace of $\mathbb{R}^d$ which turns out to be preserved by $G$. So it seems to me that 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 $k \leq \frac{d}{2}$.

Answer 1

Here's a simple counterexample to Question 1: Let $d=4$ and $k=2$. Let $X\subset\mathrm{GL}_4(\mathbb{R})$ consist of a single element $J$ where $J^2=-I$. Then $J$ is not conjugate to any upper triangular matrix (over $\mathbb{R}$) since it does not have real eigenvalues. Meanwhile, since $$ \bigl(\Lambda^2(J)\bigr)^2 = \Lambda^2(J^2) = \Lambda^2(-I) = I, $$ the eigenvalues of $\Lambda^2(J):\Lambda^2(\mathbb{R}^4)\to\Lambda^2(\mathbb{R}^4)$ are $\pm 1$, and $\Lambda^2(\mathbb{R}^4)$ has a basis of $\Lambda^2(J)$-eigenvectors, so $\Lambda^2(J)$ is not just 'triangularizable', it's diagonalizable.

On the other hand, if $X$ is a connected Lie subgroup of $\mathrm{GL}_d(\mathbb{R})$ such that $$ \Lambda^k(X) = \{ \Lambda^k(x)\ |\ x\in X\ \} $$ is conjugate to an upper triangular subgroup of $\mathrm{GL}_m(\mathbb{R})$ where $m = {d\choose k}$ and $1< k < d$, then $X$ must, itself, be conjugate to an upper triangular subgroup of $\mathrm{GL}_d(\mathbb{R})$. This essentially follows from the Lie-Engel Theorem, since $X$ has to be solvable (because $\Lambda^k(X)$ is and it is either isomorphic to $X$ or, if $k$ is even and $-I$ lies in $X$, then $X$ is a double cover of $\Lambda^k(X)$), and the reality of its roots in the $\Lambda^k$-representation forces the reality of its roots in the $\Lambda^1$-representation.

As far as Question 2 goes, it's not at all clear what the best estimate would be, but it's clearly better than the trivial bound. To get any meaningful bound, you'd need to know something about the reductive group $G$. For example, if $G$ is finite, then there are only a finite number of irreducible (faithful) representations of $G$ and so there's only a finite list of things to consider. in that case, 'asymptotic bounds' won't make much sense. If $G$ is, say, a simple Lie group, such as $\mathrm{SL}(2,\mathbb{R})$, then the number of irreducible components of $\Lambda^k(V)$ could only grow polynomially (maybe even only quadratically, but I'm not sure about that) with the dimension of $V$, so the binomial coefficients estimate is pretty far off. Probably, for higher rank groups, there is an even lower estimate that can be got via multiplicity formulae, but I don't know whether it's qualitatively better than crude estimates based on primary $sl(2)$-subgroups.