I recently learned that $GL_2(\mathbb{F}_3)$ can be embedded into $GL_2(\mathbb{C})$; specifically,

$$ \left(\begin{array}{cc} -1 & -1 \\ -1 & 0 \end{array} \right) \mapsto \left(\begin{array}{cc} -1 & -1 \\ -1 & 0 \end{array} \right) $$


$$ \left(\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right) \mapsto \left(\begin{array}{cc} 1 & -1 \\ -\sqrt{-2} & -1 + \sqrt{-2} \end{array} \right). $$

I think same goes for $\mathbb{F}_5$.

Can this be done for other larger primes?

(Motivation: I am interested in making some mod-$l$ representations complex. )


First, let me expand slightly on the last paragraph of Geoff Robinson's answer. Your question is equivalent to the question whether $G=\text{GL}_2(\mathbb{F}_p)$ has a faithful 2-dimensional complex representation. Now, such a representation cannot be the sum of two one-dimensional ones, since they would all factor through $G'$, so you are looking for an irreducible representation.

The irreducible representations of $\text{GL}_2(\mathbb{F}_p)$ are actually fairly easy to write down. If you induce all possible one-dimensional representations from $B=\begin{pmatrix}* & *\\0 & *\end{pmatrix}$, then the inductions will either be irreducible (hence $p+1$-dimensional) or they will be direct products of a one-dimensional and another irreducible. This way you get the characters lifted from $G/G'$ and some $p$-dimensional irreducible ones. All the remaining irreducible representations of $G$ have dimension $p-1$ (they can also be written down explicitly, but slightly less easily). In summary, the dimensions of the irreducible representations of $G$ are 1, $p-1$, and $p$.

Now, to your motivation: usually when people try to "make mod $l$ representations complex", what they mean is lift them to characteristic 0. In other words, if $\overline{\rho}:\Gamma\longrightarrow \text{GL}_n(\mathbb{F}_l)$ is a representation, find a representation $$ \rho:\Gamma\longrightarrow \text{GL}_n(\mathbb{Z}_l)\hookrightarrow \text{GL}_n(\mathbb{Q}_l) $$ that reduces to $\overline{\rho}$ modulo $l$. When this can be done is a fairly delicate question, in general. After that, you can fix some embedding of $\mathbb{Q}_l$ into $\mathbb{C}$ if you want, to make $\rho$ into a complex representation.

  • $\begingroup$ I think there are typos in the displayed equation in the last paragraph. Youwant $n$ there, not $2$, I suppose. $\endgroup$ – Geoff Robinson Aug 26 '11 at 12:01
  • $\begingroup$ @Geoff: certainly, thank you! $\endgroup$ – Alex B. Aug 26 '11 at 12:25
  • $\begingroup$ Thank you Alex B. for this great answer. I really appreciate your clarification of what lifting representations means. $\endgroup$ – Giuseppe Aug 26 '11 at 13:46
  • $\begingroup$ @Giuseppe You are very welcome. If you google for "lift representation characteristic zero", you will find lots of literature. $\endgroup$ – Alex B. Aug 26 '11 at 14:43

No, unfortunately, $3$ is the largest prime for which you can do this, although ${\rm SL}(2,\mathbb{F}_5)$ embeds in ${\rm GL}(2,\mathbb{C}).$ There are many ways to see this, but one result which puts strong lower bounds on dimensions of complex representations is a result of Feit and Thompson (bulding on work of Brauer), around 1960 or so: If $G$ is a finite subgroup of ${\rm GL}(n,\mathbb{C}),$ then $G$ has a normal Sylow $p$-subgroup for each prime $p > 2n+1.$

It is also possible to see directly from its character table, that ${\rm GL}(2,\mathbb{F}_{p})$ has no non-linear irreducible character of degree less than $p-1.$

  • $\begingroup$ Thank you for your swift response. I wish I could accept both of these great answers, but I've gone for Alex B.'s because I found his comments on my 'motivation' very illuminating. I hope you don't mind. $\endgroup$ – Giuseppe Aug 26 '11 at 13:44
  • $\begingroup$ Of course I don't mind $\endgroup$ – Geoff Robinson Aug 26 '11 at 18:17

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