### Representation Theory Question

#### Posted by John Baez

I’m working with Todd Trimble and Joe Moeller on categories and representation theory, and I’ve run into this question:

Suppose $k$ is a field of characteristic zero. Then any algebraic representation of $\mathrm{GL}(n,k)$ restricts to give an algebraic representation of the subgroup $D \subset \mathrm{GL}(n,k)$ consisting of invertible diagonal matrices. If two algebraic representations of $\mathrm{GL}(n,k)$ restrict to give equivalent representations of $D$, do they have to be equivalent as representations of $\mathrm{GL}(n,k)$?

I think the answer is yes, and maybe I can even string together a proof. But for programmatic reasons I’m seeking a proof that avoids the theory of Young diagrams and the theory of roots and weights. I want to only use easy general stuff. I think I see such a proof for $k = \mathbb{C}$. But it doesn’t seem to generalize. Let me explain.

To show that two algebraic representations of $\mathrm{GL}(n,\mathbb{C})$ must be equivalent if they become equivalent when restricted to the subgroup of diagonal matrices, we can use some standard facts:

an algebraic representation of $\mathrm{GL}(n,\mathbb{C})$ is determined by its restriction to the maximal compact subgroup $\mathrm{U}(n)$.

finite-dimensional continuous representations of compact Lie groups are determined up to isomorphism by their characters.

the character of a finite-dimensional continuous representation of $\mathrm{U}(n)$ is determined by its restriction to the diagonal matrices, since characters are continuous and constant on conjugacy classes, and diagonalizable matrices are dense in $\mathrm{U}(n)$.

I would like a similarly elementary proof for any field of characteristic zero — if one exists! But for this we would need to eliminate the analysis. I can do some of this, but I get stuck at this point: the diagonalizable matrices are not Zariski dense in $\mathrm{GL}(n,k)$ when $k$ is not algebraically closed. So what, if anything, is the easy conceptual reason for why representations that become isomorphic when restricted to the diagonal subgroup $D \subset \mathrm{GL}(n,k)$ have to be isomorphic to begin with?

Of course if this is actually *false* that would explain my puzzlement.

But I think the problem is that while I’ve seen stuff about how the theory of roots and weights generalizes from reductive Lie groups to reductive algebraic groups over other fields, I haven’t read through all the proofs. So I don’t know how you deal with that fact that when $k$ is not algebraically closed, not all semisimple matrices are diagonalizable. I’ve seen Milne’s discussion of ‘tori’ over arbitrary fields here:

- James Milne,
*Algebraic Groups*, Chapter 14: Tori; groups of multiplicative type; linearly reductive groups.

but I don’t see how to use it to help solve my problem.

## Re: Representation Theory Question

Any algebraic rep of $\mathrm{GL}(n, k)$ on a $k$-vector space $V$ extends uniquely to one of $\mathrm{GL}(n, \overline k)$ on $\overline k \otimes_k V$, where $\overline k$ is the algebraic closure. This is determined up to isomorphism by its restriction to the diagonal matrices $D(n, \overline k)$…but I think $D(n, k)$ is Zariski dense in $D(n, \overline k)$ so it should be determined by the restriction of the original representation to diagonal matrices. And obviously, isomorphic representations of $\mathrm{GL}(n, \overline k)$ remain isomorphic when restricted to the subgroup $\mathrm{GL}(n, k)$. So the question seems to be equivalent to asking whether two non-isomorphic algebraic reps of $\mathrm{GL}(n, k)$ can become isomorphic upon tensoring with $\overline k$.