0

Suppose $L=L_1\oplus L_2$, where $L_1,L_2$ are finite-dimensional complex semisimple Lie algebras. Suppose $V$ is a finite-dimensional irreducible representation of $L$ (through $\rho:L\to \mathfrak{gl}(V)$), then I need to show either $L_1$ or $L_2$ act on $V$ trivially.

My proof follows like this: we first compute, for $X_1\in L_1,X_2\in L_2$, that $$ \rho(X_2)\rho(X_1)=\rho(X_1)\rho(X_2)-[\rho(X_1),\rho(X_2)]=\rho(X_1)\rho(X_2)-\rho[X_1,X_2]=\rho(X_1)\rho(X_2). $$ The last step follows from the structural theorem of semisimple Lie algebras, which gives $[X_1,X_2]=0$. But how to reason from here?

I am also curious whether the step where I used semisimplicity is correct. My argument comes from the decomposition of semisimple Lie algebras into direct sum of simple ideals. Is this how semisimplicity in the assumption is used?

EDIT: I just revisited the definition of a direct sum of Lie algebras, and it seems that the last equality directly follows from definition, but not semisimplicity. How does one apply the semisimplicity condition then?

EDIT on Sept. 26 It seems that one could prove $V$ is irreducible as a $L_1-$module. To prove this, we use Weyl's theorem and Schur's lemma. The previous one uses the complex semisimple property to induce that $V$ is completely irreducible as a $L_1-$module. The latter one and the commutative property of $\rho(X_1),\rho(X_2)$ shows that $\rho(X_2)$ acts on the irreducible submodules as scalars. This shows $V$ must be irreducible as a $L_1$ module. Is this argument correct? And how should I continue?

barbatos233
  • 157
  • @MarianoSuárez-Álvarez Yes. The computation shows $\rho(L_1)V$ is a subrepresentation and by irreducibility it is either $0$ or $V$. But how should I proceed in the case $\rho(L_1)V=V$? – barbatos233 Sep 22 '23 at 16:46
  • 2
    This isn't true. Neither summand needs to act trivially. The simplest example being $\mathfrak{sl}_2\oplus\mathfrak{sl}_2$ which acts on $\mathbb{C}^2 \otimes\mathbb{C}^2 \cong \mathbb{C}^4 $ as the natural representation of $\mathfrak{so}_4$ on $\mathbb{C}^4 $ – Callum Sep 22 '23 at 19:55
  • 1
    I think the result that you are looking for is instead that if $V$ is irreducible for $L$ and for $L_1$ then $L_2$ acts trivially – Callum Sep 23 '23 at 09:29
  • Cf. https://math.stackexchange.com/q/3826725/96384 and https://math.stackexchange.com/q/3350849/96384 as follow-up to Callum's comment. – Torsten Schoeneberg Sep 23 '23 at 22:55
  • @Callum I believe I can reduce the original problem to this case. Please kindly refer to my update, and if it is correct, may I ask how should I proceed? – barbatos233 Sep 26 '23 at 11:53
  • 1
    You cannot prove $V$ is automatically irreducible as an $L_1$ module because it doesn't have to be true. In the example I mentioned above $\mathbb{C}^2 \otimes\mathbb{C}^2$ is irreducible as an $L$-module but as a $L_1$-module it is the union of invariant subspaces of the form $\mathbb{C}^2 \otimes\langle v\rangle$ so is definitely not irreducible. – Callum Sep 26 '23 at 12:04

0 Answers0