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Let $G$ be a complex semisimple Lie group and let $\rho: G \longrightarrow \mathrm{SL}(n,\mathbb{C})$ be a faithful irreducible representation of $G$ with $n \geq 3$. Suppose that $G$ contains a copy of $\mathrm{SL}(2,\mathbb{C})$ such that

$\bullet$ its action preserves a decomposition $\mathbb{C}^n = \mathbb{C}^2 \oplus \mathbb{C}^{n-2}$

$\bullet$ its action is the natural action of $\mathrm{SL}(2,\mathbb{C})$ on $\mathbb{C}^2$ and it acts trivially on $\mathbb{C}^{n-2}$.

My question is: must $\rho(G)$ be equal to the whole $\mathrm{SL}(n,\mathbb{C})$? or do you know any counterexamples?

Obviously, $\mathrm{SL}(n,\mathbb{C})$ contains $\mathrm{O}(n, \mathbb{C})$ the group of automorphisms of a non degenerate quadratic form and $\mathrm{Sp}(n, \mathbb{C})$ the symplectic group if $n$ is even. So for my question not to be trivial, I make the extra assumption that $\rho(G)$ is not contained in any conjugate of these two groups.

I understand that diging deep enough in the theory of representations should provide an answer to my question, but being no expert in Lie theory I was wondering if I was missing any simple argument or counter-example. Thanks for your attention.

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    $\begingroup$ Maybe I am missing something here, but if $\rho(G)$ were the whole $SL(n,\mathbb{C})$, then $\rho(G)$ would not preserve the decomposition $\mathbb{C}^n = \mathbb{C}^2 \oplus \mathbb{C}^{n-2}$, right? $\endgroup$
    – Matthias Klupsch
    Commented Jun 11, 2016 at 16:12
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    $\begingroup$ @MatthiasKlupsch Apparently, the subspace invariance assumption concerns the image of the $SL_2$ subgroup. $\endgroup$
    – Andrei Smolensky
    Commented Jun 11, 2016 at 16:18
  • $\begingroup$ I misunderstood your comment sorry. It is the copy of $\mathrm{SL}(2, \mathbb{C})$ who must preserve the decomposition, not the whole $\rho(G)$. $\endgroup$
    – Selim G
    Commented Jun 11, 2016 at 16:19
  • $\begingroup$ Aha, I see, that makes more sense then. $\endgroup$
    – Matthias Klupsch
    Commented Jun 11, 2016 at 16:19
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    $\begingroup$ Up to conjugacy, your $SL_2$ is the standard $SL_2 \leq SL_n$. So the question can be rephrased "Let $SL_2 \leq G \leq SL_n$, $G$ semisimple and acting irreducibly. Must the representation be self-dual?" $\endgroup$
    – Allen Knutson
    Commented Jun 11, 2016 at 18:13

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The only groups $G$ of the type you are looking for are $SL(n)$, $SO(n)$ or $Sp(n)$.

This is proved in a paper by Beukers and Heckman (see Proposition (6.4) in that paper). See the math review

http://www.ams.org/mathscinet-getitem?mr=974906

for a reference.

Some explanation: Beukers and Heckman talk about a "reflection" normalising the irreducible subgroup. In their setting, a (complex) reflection is an element $g\in GL(n)$ such that $g-1$ has kernel of dimension $n-1$. In that case, they prove that a connected irreducible algebraic subgroup of $SL(n)$ which is normalised by a complex reflection is either $SL(n)$ or $SO(n)$ or $Sp(n)$. Now the standard unipotent element $u$ in your SL(2) contained in $G$, is indeed a complex reflection and so your group $G$ is normalised by the "reflection" $u$.

Incidentally, in your case the situation $SO(n)$ cannot arise since this unipotent reflection cannot lie in $SO(n)$.

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