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For a finite group $G$ with a representation $\pi:G\to GL_n(\mathbb C)$ one can define the adjoint representation $Ad$ as the non-trivial summand in $End(\pi)$, i.e. $End(\pi)=\pi\otimes \pi^{\vee}=1\oplus Ad(\pi)$.

For example, if $\pi$ is self-dual then $\pi\otimes \pi=Sym^2(\pi)\oplus \bigwedge^2(\pi)$, and for $n>2$ it is possible to show that $Ad(\pi)$ is reducible.

Are there any known examples when $Ad(\pi)$ is irreducible ?

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  • $\begingroup$ There are examples where $\pi$ is self-dual and $n=2$. $\endgroup$
    – Tom Goodwillie
    Commented Apr 10, 2013 at 3:12
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    $\begingroup$ If $\pi$ is irreducible, then it is not difficult to show that $M^n(\mathbb C)$ is an irreducible $G\times G$-module via $(g,h)A = \pi(g)A\pi(h)^{-1}$. Now the $G$-module structure you are considering is the restriction of this action to the diagonal $\Delta(G)$. So basically, this can be thought of as a Frobenius reciprocity type of question. $\endgroup$
    – Benjamin Steinberg
    Commented Apr 10, 2013 at 15:21

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Yes there are lots of examples. Here are a few.

The 3-dimensional representations of ${\rm PSL}_2(7)$.

The 3-dimensional representations of the triple cover $3.A_6$ of $A_6$. (The adjoint in that case is an irreducible 8-dimensional representation of $A_6$.)

The 4-dimensional representations of the double cover $2.A_7$ of $A_7$. (The adjoint is an irreducible 15-dimensional representation of $A_7$.)

The 4-dimensional representations of the double cover $2.{\rm PSU}_4(2)$ of ${\rm PSU}_4(2)$.

The 5-dimensional representations of ${\rm PSU}_4(2)$.

The ATLAS of Finite Groups is a ggood place to look for examples. I expect there are some infinite families as well.

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Derek Holt has given a bunch of examples for $n>2$. In the case $n=2$, $Ad \pi$ is irreducible if and only if it does not contain a character $\chi$ (since it has dimension 3), that is if and only if the two condition holds: $Hom_G(\pi,\pi)$ has dimension $1$, and $Hom_G(\pi,\pi(\chi))$ has dimension $0$ for every non-trivial character $\chi$. Now the first condition is automatic by Schur's lemma (since $\pi$ is assumed irreducible) and the second one is equivalent to $ \pi \not \simeq \pi(\chi)$ for every non-trivial character $\chi$. Now if for instance you take a group which has no non-trivial character $\chi$ (that is a group whose abelianization is trivial, a.k.a a prefect group), and an irreducible representation $\pi$ of dimension $2$ of that group, you get an example as you asked for, that is such that $Ad \pi$ is irreducible. For example if consider a subgroup $G$ of $Gl_2(\mathbb C)$ is isomorphic to $A_5$, and $\pi$ the canonical representation, then you get an example. (Edit: actually, "2.A_5", see the comments below)

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  • $\begingroup$ I think you want $G=2.A_5 = {\rm SL}_2(5)$ rather than $A_5$. $\endgroup$
    – Derek Holt
    Commented Apr 10, 2013 at 13:43
  • $\begingroup$ @Derek: why? Isn't $A_5$ a subgroup of $\Gl_2(\C)$? $\endgroup$
    – Joël
    Commented Apr 10, 2013 at 13:45
  • $\begingroup$ No. It is a subgroup of ${\rm GL}_3({\mathbb R})$, which comes from the adjoint of the 2-dimensional representation of $2.A_5$. $\endgroup$
    – Derek Holt
    Commented Apr 10, 2013 at 14:34

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