# Do two dimensional representations with the same adjoint representation differ by a character?

Let $K$ be a field of characteristic not equal to $2$. Let $\text{ad} : \text{GL}_2(K) \to \text{GL}_3(K)$ be the adjoint representation, obtained by $\text{GL}_2(K)$ acting on $2 \times 2$ matrices with trace $0$ by conjugation. Suppose $\rho_1, \rho_2 : G \to \text{GL}_2(K)$ are representations of a group $G$ such that $\text{ad}\rho_1 \cong \text{ad}\rho_2$ over $K$. Is there a character $\eta : G \to K^\times$ such that $\rho_1 \cong \rho_2 \otimes \eta$ over $K$? That is, if $x \in \text{GL}_3(K)$ such that $\text{ad}\rho_1 = x(\text{ad}\rho_2) x^{-1}$, can one produce $y \in \text{GL}_2(K)$ and $\eta$ as above such that $\rho_1 = y \eta (\otimes \rho_2) y^{-1}$?

I am most interested in the case when $K$ is a finite field. In that case, I believe one can treat the case when the projective image of $\rho_i$ is isomorphic to $\text{PSL}_2(K)$ or $\text{PGL}_2(K)$ fairly easily by using that those groups have automorphism group $\text{P}\Gamma\text{L}_2(K)$. But I'd like a proof that does not break into cases based on the subgroups of $\text{PGL}_2(K)$.

• This is an interesting question! It seems ripe for attack by character computations, but I don't see how to do it immediately. One can still ask it in characteristic $2$ by considering the adjoint representation $\mathrm{GL}_2(K) \to \mathrm{GL}(\mathfrak{pgl}_2)(K)$. Apr 13, 2018 at 13:46
• I suspect that it is false for silly reasons if you do not require the representations $\rho_1$ and $\rho_2$ to be irreducible; do you want to require that? Apr 13, 2018 at 13:49
• Certainly the question may be asked in characteristic $2$ as well. I left that out since it is not so relevant to me at the moment, and since I believe I have a proof that is specific to characteristic $2$. Apr 13, 2018 at 14:08
• I think the answer is easy when both $\rho_i$ are reducible. If $\rho = \chi_1 \oplus \chi_2$ then $\text{ad}\rho = \chi_1\chi_2^{-1} \oplus 1 \oplus \chi_1^{-1}\chi_2$. Using this, I think it's easy to see that $\rho_1$ and $\rho_2$ are twists of each other under the hypothesis that $\text{ad}\rho_1 = \text{ad}\rho_2$. Apr 13, 2018 at 14:11
• Good point. I also just noticed that I'm thinking of complex representations, but that these are (in the finite-field case) actually modular representations. Apr 13, 2018 at 15:26

## 2 Answers

$\DeclareMathOperator\ad{ad}$This is true. The way I am viewing it, one has to make two cases, the representation $\ad V$ (and hence $\ad W$) is irreducible, or both are dihedral.

Part I: Assume that $\ad V$ and $\ad W$ are irreducible ($\rho_1 : G \rightarrow GL(V)$ and $\rho_2:G \rightarrow GL(W)$. Since $V$, $W$ are two dimensional, $V^*=V\otimes \bigwedge^2V^*$, $W^*=W\otimes \bigwedge^2 W^*$. Then $1\oplus \ad V= V\otimes V^*=V\otimes V\otimes \bigwedge^2V*=1\oplus S^2V\otimes \bigwedge^2V^*$. By assumption, this is isomorphic to $W\otimes W^*$, hence we get $S^2V\otimes \bigwedge^2V^*=S^2W\otimes \bigwedge ^2 W^*$, i.e. $S^2V\otimes \bigwedge ^2W= S^2W\otimes \bigwedge ^2V$.

Claim: $V\otimes W=E\oplus F$ is not possible if $E$, $F$ are two-dimensional representations. For, if such $E$, $F$ exist, then, taking second exterior on both sides, we get

$$S^2V\otimes \bigwedge^2 W\oplus S^2W\otimes \bigwedge^2V=\bigwedge^2E\oplus \bigwedge^2 F \oplus E\otimes F.$$ Now use the preceding para to conclude that $S^2V\otimes \bigwedge^2W$ contains $\bigwedge^2E$, contradicting irreducibility: $\ad V=S^2V\otimes \bigwedge^2V^*\supset \bigwedge^2W^*\otimes \bigwedge^2 E \otimes \bigwedge^2 V^*$.

Claim: $V\otimes W$ is reducible. For $V\otimes W\otimes V^*\otimes W^*= (1\oplus \ad V)\otimes (1\oplus \ad W)\supset 1\oplus \ad V\otimes \ad W$ contains a two-dimensional space of invariants: clearly $\ad W=\ad W^*$, and $\ad W^*=\ad V^*$ by assumption.

The above two claims imply that $V\otimes W$ contains a one-dimensional invariant subspace $E$. Therefore, $W=V^*\otimes E=V\otimes \bigwedge^2V^*\otimes E$, and thus $W$ is got from $V$ by tensoring with a character.

Part II : When $\ad V=\ad W$ is not irreducible, $V$, $W$ are dihedral (I have a long, but easy proof) and the equality of $\ad V$ and $\ad W$ implies the equivalence of $V$ and $W$ by arguments similar to the preceding (it is long, but again easy).

The first few sections of Murty and Prasad - Tate cycles on a product of two Hilbert modular surfaces (MSN) contain most of what I have written (and also answer your question).

• Are you assuming that $K$ is finite? Or is it still true that if $V$ is an (absolutely?) irreducible two-dimensional representation over any field $K$ (not of characteristic $2$?), then either ${\rm ad} V$ is irreducible or else $V$ is dihedral? (And I guess $V$ is dihedral if it is is the induction of a character on an index-$2$ subgroup of $G$...) Apr 17, 2018 at 7:32
• @sibilant: $K$ is an algebraically closed field of char not two (otherwise, the direct sums are problematic).. Then yes, for an irreducible two dimensional $V$, either $ad V$ is irreducible or $V$ is dihedral. My earlier proof was long but I think now I have a simpler explanation: this is equivalent to $S^2V$ having a one dimensional invariant subspace; that is there is a quadratic form Q (necessarily non-degenerate) which is almost invariant hence we get a rep of G into GO(Q)=GO(2) which is dihedral. Apr 17, 2018 at 8:35
• Thanks for your answer! This looks like exactly the kind of argument I was looking for. I'm a bit overwhelmed at the moment, so it may take me a week or so for me to think through it carefully. I'm going to leave the post open until I do that in case I have follow-up questions, but hopefully by the end of next week I'll be comfortable accepting this answer. Thanks for your understanding. Apr 17, 2018 at 12:02
• @Dudelsack: That is Schur's lemma; If $E$ is irreducible, the space of invariants in $E\otimes E^*$ is one dimensional. The point is $E\otimes E^*=Hom (E,E)$, and hence the invariants are just $G$ equivariant linear transformations from $E$ into itself. By Scur's lemma, this space consists of scalars. Apr 20, 2018 at 13:17
• If $S^2V$ has a one dimensional invariant subspace, then so does $S^2V^*$; but the latter are exactly symmetric bilinear forms (equal quadratic forms). Thus there is a quadratic form $Q$ which is semi-invariant (under the action of the group $G$, the quadratic form goes into a multiple of itself). But over the algebraic closure all non-degenerate quadratic forms are equivalent, hence you may take $Q$ to be the quadratic form $Q(x,y)=xy$. The group which preserves this is precisely the dihedral group (the diagonals together with the "Weyl group element , the permutation matrix of size two). Apr 27, 2018 at 14:53

Not an answer, but a possible start to a character computation @LSpice, too long for a series of comments.

Fix $g$ in $G$, and let $a_i$ and $b_i$ be the eigenvalues of $\rho_i(g)$ for $i = 1, 2$. Then the eigenvalues of ${\rm ad} \rho_i(g)$ are $1$, $a_i b_i^{-1}$ and $b_i a_i^{-1}$. By assumption the sets $\{a_1 b_1^{-1}, b_1 a_1^{-1}\}$ and $\{a_2 b_2^{-1}, b_2 a_2^{-1}\}$ are the same. If $a_i b_i^{-1} \neq \pm 1$ then these two sets consist of distinct elements, and up to switching $a_2$ and $b_2$, we can identify that $a_1 b_1^{-1} = a_2 b_2^{-1}$. Therefore, if such a character $\eta$ were to exists, we would have to have $\eta(g) = a_1 a_2^{-1} = b_1 b_2^{-1}$. This is still true if $a_i b_i^{-1} = 1$: now $a_i = b_i$, so we do not have to worry about picking which of the two eigenvalues of $\rho_2(g)$ to call $a_2$, as they are the same. It's only if $a_i b_i^{-1} = -1$ (equivalently, ${\rm tr} \rho_i(g) = 0$) that we do not know which eigenvalue to call $a_2$. But since $b_2 = - a_2$, the quantity $\pm a_2$ is well-defined; in this case $\eta(g) = \pm a_1 a_2^{-1}$ is only determined up to sign.

So we have now defined a quasi-function $\eta: G\ "\!\to\!"\ K^\times$ with $\eta(g) = a_1 a_2^{-1}$ whenever ${\rm tr} \rho_i (g) \neq 0$ and $\eta(g)$ defined only up to sign otherwise. (I am just using "quasi-function" informally here.) Note that this quasi-function satisfies ${\rm tr} \rho_1 = \eta \cdot {\rm tr} \rho_2$, so that if one could prove that the quasi-function can be refined to a function that is a character $G \to K^\times$, then by Brauer-Nesbitt --- assuming we are free to assume $\rho_i$ are semisimple --- we would be done.

Moreover, note that $\eta^2: G \to K^\times$ is not only a perfectly well-defined function but also a character: indeed, $\eta^2 = {\rm det} \rho_1 \otimes {\rm \det \rho_2}^{-1}$.

To be continued, as I must dash, but possibly someone already has an idea for how to take it from here? Post-dash sequel:

[At this point, by the way, we have a proof for $K$ of characteristic $2$: there is no sign ambiguity, so that $\eta: G \to K^\times$ is a proper function; and since $\eta^2$ is a character, so is $\eta$. Finally since $\rho_1$ and $\rho_2 \otimes \eta$ have the same trace and determinant, and we are assuming $\rho_i$ semisimple, Brauer-Nesbitt applies to conclude that they are isomorphic representations.]

Some ideas about how to proceed from here, to see that $\eta$ is multiplicative where defined:

From the definition, it follows that $\eta(1) = 1$.

The identity ${\rm tr} \rho_i(g^{-1}) = \det \rho_i(g)^{-1} {\rm tr} \rho_i(g)$ combined with the trace and determinant equations gives us ${\rm tr} \rho_2(g) \eta(g^{-1}) = {\rm tr} \rho_2(g) \eta(g)^{-1}$, so that $\eta(g^{-1}) = \eta(g)^{-1}$ so long as ${\rm tr} \rho_i(g) \neq 0$.

Since $\det \rho_i(g) = \frac{1}{2}\big(({\rm tr} \rho_i(g))^2 - {\rm tr} \rho_i(g^2)\big)$ and $\det \rho_1(g) = \det \rho_2(g) \eta(g)$ and ${\rm tr} \rho_1(g) = {\rm tr} \rho_2(g) \eta(g)$, we can conclude that ${\rm tr} \rho_2(g^2) \eta(g)^2 = {\rm tr} \rho_2(g^2) \eta(g^2)$ for all $g \in G$. Therefore, if ${\rm tr} \rho_i(g^2) \neq 0$, then $\eta(g)^2 = \eta(g^2)$. (I don't know how useful this is.)

The general version of the two identities above is the pseudocharacter identity $${\rm tr} \rho_i(gh) + \det \rho_i(g) {\rm tr} \rho_i(g^{-1} h) = {\rm tr} \rho_i(g) {\rm tr} \rho_i(h),$$ which holds for any $g, h$ in $G$. Whence, for $g, h \in G$, $${\rm tr} \rho_2(gh) \big(\eta(gh) - \eta(g) \eta(h)\big) = \det \rho_2(g) {\rm tr} \rho_2(g^{-1} h) \big(\eta(g) \eta(h) - \eta(g)^2 \eta(g^{-1} h) \big).$$

Can anything useful come out of this?