Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $h_1\subset gl(V_1)$ and $h_2\subset gl(V_2)$ be two irreducible representations of reductive real Lie algebras.

When the representation of $h_1\oplus h_2$ in $V_1\otimes V_2$ is not irreducible?

I guess that the both representations $h_1\subset gl(V_1)$ and $h_2\subset gl(V_2)$ must be complex, i.e. on $V_i$ exists a complex structure commuting with $h_i$. But I can prove only that at least one of these representation is complex (otherwise the complisifications $h_1^\mathbb{C}\subset gl(V_1^\mathbb{C})$ and $h_2^\mathbb{C}\subset gl(V_2^\mathbb{C})$ are irreducible, and consequently $h_1^\mathbb{C}\oplus h_2^\mathbb{C}$ is irreducible in $V_1^\mathbb{C}\otimes_\mathbb{C} V_2^\mathbb{C}$, which gives a contradiction).

share|cite|improve this question
I always thought that it always irrep over any field for any groups, am I wrong ? (at least for Lie groups, for some exoctic groups it is not true as I heard from Olshaskii, but these are some exotics - infinite-dims groups or something...) –  Alexander Chervov Jun 20 '12 at 17:49
Consider, e.g., the representations $u(n)\subset gl(2n,R)$ and $u(m)\subset gl(2m,R)$. The tensor product $R^{2n}\otimes_R R^{2m}=C^{n}\otimes_R C^{m}$ contains invariant subspace $C^n\otimes_C C^m$. –  Anton Galaev Jun 21 '12 at 8:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.