## August 17, 2023

### 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.

Posted at August 17, 2023 8:55 AM UTC

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### 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 $\overline k$.

Posted by: Nick Olson-Harris on August 17, 2023 7:24 PM | Permalink | Reply to this

### Re: Representation Theory Question

For any two reps $V,W$ over $k$, $\mathop{Hom}_{\mathop{Rep}_{\bar k}}(V \otimes_k \bar{k}, W \otimes_k \bar{k}) \simeq \mathop{Hom}_{\mathop{Rep}_k}(V, W) \otimes_k {\bar k}$, where $\mathop{Rep}_k$ denotes the category of representations of $\mathop{GL}_n(k)$ over $k$. In particular, $\dim \mathop{Hom}$ doesn’t depend on the field.

Now, consider $\langle V, W \rangle := \dim \mathop{Hom}(V, W)$ as a scalar product on the Grothendieck group of representations. It is positive definite (with irreducibles forming an orthogonal basis - which depends on $k$, of course, but we just need the positivity). Therefore, just by basic properties of Euclidean scalar products, if $\langle V, W \rangle = \langle V, V \rangle = \langle W, W \rangle$ then $V \simeq W$, independently of the field.

Posted by: Alexander Shamov on August 17, 2023 11:33 PM | Permalink | Reply to this

### Re: Representation Theory Question

Everything I wrote in the last paragraph depends heavily on semisimplicity, but I think the claim that if $V \simeq W$ over $bar{k}$ then $V \simeq W$ over $k$ as well holds in much greater generality. Here’s an argument that works whenever $\mathop{char} k = 0$ and $\mathop{Hom}(V,W)$ is finite dimensional.

If $V \simeq W$ over $\bar{k}$ then the isomorphisms form a Zariski open dense subvariety in $\mathop{Hom}(V,W)$. On the other hand, the set of $k$-linear homomorphisms is Zariski dense in $\bar{k}$-linear ones, so it intersects the set of isomorphisms nontrivially.

Posted by: Alexander Shamov on August 18, 2023 12:50 AM | Permalink | Reply to this

### Re: Representation Theory Question

As long as $k$ is a perfect field we have $V\cong W$ if and only if $V\otimes_k\bar k\cong W\otimes_k\bar k$.

For, as Alexander stated, we have $\mathrm{Hom}_{G\otimes\bar k}(V\otimes\bar k,W\otimes\bar k) \cong \mathrm{Hom}_G(V,W\otimes\bar k) \cong \mathrm{Hom}_G(V,W)\otimes\bar k,$ where the latter isomorphism can be seen by taking a $k$-basis for $\bar k$. In particular, $\dim_k\mathrm{Hom}_G(V,W) = \dim_{\bar k}\mathrm{Hom}_{G\otimes\bar k}(V\otimes\bar k,W\otimes\bar k).$

Next, regarding $\mathrm{Hom}_G(V,W)$ as a subspace of $\mathrm{End}_G(V\oplus W)$, which is a finite dimensional algebra, we can intersect with the Jacobson radical and define $\mathrm{rad}_G(V,W)$. Then $V$ and $W$ share no isomorphic direct summands if and only if $\mathrm{rad}_G(V,W)=\mathrm{Hom}_G(V,W)$.

We are reduced to showing that the Jacobson radical behaves well with respect to base change: $J\mathrm{End}_{G\otimes\bar k}(V\otimes\bar k)=J\mathrm{End}_G(V)\otimes\bar k.$ Now $\mathrm{End}_G(V)$ is a finite dimensional algebra, so $J\mathrm{End}_G(V)$ is nilpotent and $\mathrm{End}_G(V)/J\mathrm{End}_G(V)$ is semisimple, so a direct product of matrices over division algebras.

Clearly $J\mathrm{End}_G(V)\otimes\bar k$ is nilpotent, so is contained in $J\mathrm{End}_{G\otimes\bar k}(V\otimes\bar k)$, and since base change behaves well with respect to direct products and forming matrix algebras, it is enough to show that if $A$ is a finite dimensional division algebra over $k$, then $A\otimes\bar k$ is semisimple. Let $K\leq A$ be a splitting field. Then $A\otimes\bar k=A\otimes_K(K\otimes_k\bar k)$, and this is semisimple if and only if $K\otimes\bar k$ is reduced.

Thus if $k$ is perfect, then each $K/k$ is separable, so each $K\otimes\bar k$ is reduced, and the result follows.

Posted by: andrew hubery on August 18, 2023 9:28 AM | Permalink | Reply to this

### Re: Representation Theory Question

Nick wrote:

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 $D(n, \overline k)$.

What’s the slickest proof of the last fact: when $K$ is algebraically closed and of characteristic zero, an algebraic rep of $\mathrm{GL}(n,K)$ is determined up to isomomorphism by its restriction to the diagonal matrices $D(n,K)$? I sketched a fairly slick proof above for $K = \mathbb{C}$, but that uses some features of the complex numbers. I can also prove it using the classification of irreducible representations of $\mathrm{GL}(n,K)$ in terms of Young diagrams, showing that their characters are linearly independent by a kind of brute-force combinatorial calculation, but that’s exactly what I’m trying to avoid here. I can probably also prove it using some mathematical logic, by noting that any algebraically closed field of characteristic zero is elementarily equivalent to $\mathbb{C}$, but that somehow seems a bit odd.

Posted by: John Baez on August 18, 2023 6:16 PM | Permalink | Reply to this

Every representation of GL(n,K) is semi-simple, so is fully determined by its trace map. The semi-simple elements of GL(n,K) form a Zariski dense set, so this trace map is fully determined by its restriction to them. Every semi-simple element of GL(n,K) is conjugated to a diagonal element, and as trace maps are constant up to conjugation, the trace map on the diagonal matrices determines the representation.

Posted by: lieven le bruyn on August 18, 2023 9:37 PM | Permalink | Reply to this

As to “Every representation of GL(n,K) is semi-simple”: GL(n,K) is a reductive group (trivial unipotent radical), and in characteristic zero this is equivalent to all finite dimensional reps being direct sums of irreducible reps (Milne, Linear Algebraic Groups, Thm. 22.138). See also ‘Reductive group’ on Wikipedia

Posted by: lieven le bruyn on August 19, 2023 7:20 PM | Permalink | Reply to this

Thanks! Milne’s Thm. 22.138 comes after a lot of material on roots, Borel subgroups etc., which is the sort of material I’d like to see if I can avoid right now — but luckily his proof of this theorem seems not to rely on that material. I’ll have to check, though.

Following the logic backward, it seems to go like this. Remember, all I care about now is $G = GL(N,k)$ where $k$ has characteristic zero and is also algebraically closed (if that helps).

Theorem 22.138: if a connected algebraic group $G$ over a field of characteristic zero is reductive, all its finite-dimensional representations are semisimple.

Proof: show $G = Z \cdot G'$ where the center $Z$ is a ‘group of multiplicative type’ (whatever that means) and the derived group $G'$ is semisimple. All this should be easy for $GL(n,K)$. Separately show that all representations of $Z$ and $G'$ are semisimple, and combine these facts. For $Z$ we use:

Theorem 14.50: A commutative algebraic group is linearly reductive if it is of multiplicative type.

Proof: “We saw in Theorem 14.22 that $Rep(H)$ is semisimple if the algebraic group $H$ is of multiplicative type.”

Here Milne is not assuming the field is algebraically closed; he proves Theorem 14.22 using Galois descent to reduce to the case where $H$ is split by $k$, which I guess means that $H$ takes some very tractable standard form where it’s easy to work out $Rep(H)$. Again all this should be easier for our particular example where $H$ is the center of $GL(n,k)$; when $H$ is split by $k$ it should then be the group of diagonal invertible $n \times n$ matrices.

For the semisimple group $G'$ we use:

Proposition 22.137: Let $G'$ be a semisimple algebraic group over a field $k$ of characteristic zero. Every finite-dimensional representation of $G'$ is semisimple.

Here’s where the actual work lies, and while some of the work is done in the proof, it also uses lots of references to previous results — too many for me to list right now.

Posted by: John Baez on August 22, 2023 12:48 PM | Permalink | Reply to this

### Re: Representation Theory Question

For finite-dimensional irreducible representations of any group $G$ I think you can argue as follows: let $G \times G$ act on the ring of functions on $G$ in the obvious way. The matrix coefficients of an irrep $V$ form a $(G \times G)$-invariant subspace isomorphic to $V \otimes V^*$, which in particular is itself irreducible. Thus these subspaces must intersect trivially for distinct irreps, so the matrix coefficients (and in particular the characters) of distinct irreps are linearly independent. (Now that I think of it, I guess this also proves that the irreps can’t become isomorphic upon tensoring with $\overline k$, since the matrix coefficients don’t change.)

However, I don’t think I’ve ever learned how to prove that finite-dimensional algebraic representations of the general linear group decompose into irreducibles without resorting to the unitary group crutch and/or Schur–Weyl duality.

Posted by: Nick Olson-Harris on August 19, 2023 4:15 PM | Permalink | Reply to this

### Re: Representation Theory Question

Thanks to everyone here for the helpful ideas! Todd Trimble has come up with a nice way to reduce this problem:

If $k$ is a field of characteristic zero, show that two algebraic representations of $GL(n,k)$ that restrict to give equivalent representations of the subgroup $D(n,k)$ of diagonal matrices must be equivalent as representations of $GL(n,k)$.

to the case $k = \mathbb{C}$, where the approach I sketched in my post gets the job done.

We may write this up.

By the way, a category theorist would phrase this by saying the restriction functor

$F: Rep(GL(n,k)) \to Rep(D(n,k))$

is essentially injective. Essentially injective functors are much less important than essentially surjective functors, but they are still useful at times.

For example, if we let $K(C)$ be the Grothendieck group of an abelian category, there is this fact: if

$F : C \to D$

is an additive functor and $D$ is semisimple, then $F$ is essentially injective iff

$K(F) : K(C) \to K(D)$

is injective.

This is the case for the restriction for the functor

$F: Rep(GL(n,k)) \to Rep(D(n,k))$

So, one way to prove this functor is essentially injective is to show

$K(F): K(Rep(GL(n,k))) \to K(Rep(D(n,k)))$

is injective — that is, the induced map between representation rings is injective. This approach is quite manageable using facts about Young diagrams and symmetric polynomials, but it’s quite combinatorial, and Todd wanted a more conceptual approach.

Posted by: John Baez on August 26, 2023 10:47 AM | Permalink | Reply to this

### Re: Representation Theory Question

Todd Trimble has come up with a nice way to reduce this problem

Does this use some kind of ‘Zariski-density of a restriction to rationals’ type argument? Or at least some kind of use of the rationals? Otherwise (without further details) I’d be skeptical: how else would one ‘conceptually’ find any relation at all between $k$ and $\mathbb{C}$?

Posted by: anonymous on August 27, 2023 9:41 PM | Permalink | Reply to this

### Re: Representation Theory Question

Yes, the idea is to reduce the problem for an arbitrary field $k$ of characteristic zero down to the case of $\mathbb{Q}$, and then reduce that to the case of $\mathbb{C}$. You might say each step is ‘morally speaking’ a Zariski density type argument.

Theoretically there could be another style of argument where we reduce the problem for $k$ to the problem for $\overline{k}$, and then reduce that to the case of $\mathbb{C}$ using the fact that theory of algebraically closed fields of characteristic zero is complete, so all such fields are elementarily equivalent. But the details are not clear to me and this is not the way I want to proceed!

Posted by: John Baez on August 28, 2023 10:45 AM | Permalink | Reply to this

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