# Can the defining rep of $E_7$ split over a finite subgroup while the adjoint remains simple?

Does the (simply connected compact) Lie group $$E_7$$ contain a finite subgroup $$G \subset E_7$$ such that the $$56$$-dimensional irrep of $$E_7$$ splits over $$G$$ as $$28 \oplus \overline{28}$$, but the $$133$$-dimensional adjoint representation remains simple when restricted to $$G$$? By "$$28 \oplus \overline{28}$$" I mean of course a $$28$$-dimensional complex irrep plus its dual.

Such a subgroup is definitely Lie primitive (meaning it doesn't fit inside a proper Lie subgroup $$L \subset E_7$$), since an imprimitive subgroup would split $$133 = \mathfrak{l} \oplus (\dots)$$. According to Griess and Ryba - Finite simple groups which projectively embed in an exceptional Lie group are classified!, of the (quasi)simple subgroups of $$E_7$$, only three are primitive, and none of them work. ($$133$$ splits over $$SL_2(29)$$ and $$SL_2(37)$$, and $$56$$ remains simple over $$PSU_3(8)$$.) But I don't know about nonsimple subgroups.

This feels like a good homework problem, but in fact came up in my research: such a subgroup would allow me to build a superconformal field theory with some nice properties.

• I love the title of that paper. I know Bob as a very calm person, so can only imagine how much the result must have excited him for him to put that exclamation mark. Jun 22, 2019 at 2:01
• I don't think this is homework problem. I haven't worked out all details, and I will probably write up some sort of answer. I think the difficult case might be if the $28$-dimensional irreducible representation is induced from $1$-dimensional representation of a subgroup of index $28$. Jun 22, 2019 at 17:43
• @GeoffRobinson I take it that you think that there is no such $G$ (which also seems most likely to me). I look forward to your write up. Jun 23, 2019 at 16:17
• I am not certain at present that there is no such $G$, though it would be my inclination to believe that. I am currently trying to prove that such a $G$ would have a quasisimple subnormal subgroup acting irreducibly on the $28$-dimensional module and on the $133$-dimensional module, which you have essentially already excluded. If the $28$-dimensional rep of $G$ is not monomial, I think I can do it. If it is monomial, there seems to be more to be done. Jun 23, 2019 at 16:46

## 1 Answer

I know this is an old question, but I can answer it in the negative. First, I have mostly completed a list of the Lie primitive subgroups of $$E_7(k)$$ for all $$k$$, including $$k=\mathbb C$$, and there is no such example.

But even before that, a paper of Liebeck and Seitz, entitled 'Subgroups of exceptional algebraic groups which are irreducible on an adjoint or minimal module', appearing in J. Group Theory, shows that only simple subgroups act irreducibly on the adjoint.

In fact, for $$p=5$$ there are examples, namely the sporadic groups $$M_{22}$$, $$HS$$ and $$Ru$$. But there is none over $$\mathbb{C}$$.

• I've noticed that subgroups of a Lie group with trivial center acting irreducibly on the adjoint always seem to be either simple, as you describe in this answer (or almost simple, as in the case of $Aut(PSU(3,3))$ in $G_2$) or else the normalizer of an elementary abelian $p$ group, or the commutator subgroup of such a normalizer. Is some result like this known? Mar 10, 2023 at 13:18
• All subgroups of any simple algebraic group are either contained in a normalizer of an elementary abelian $p$-group, contained in some algebraic subgroup (and hence cannot act irreducibly on the adjoint) or is almost simple, or is one particular subgroup of $E_8$ (which is also not irreducible). This is due to Borovik and Liebeck-Seitz. Mar 10, 2023 at 13:55
• Ah I see this result for Ad irreducible subgroups and even more generally Lie primitive finite subgroups is exactly Theorem from the beginning of section 6 of the paper Griess and Ryba - Finite simple groups which projectively embed in an exceptional Lie group are classified! linked to by the OP Mar 15, 2023 at 16:56