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Let $\mathfrak{g}_1$ and $\mathfrak{g}_2$ be semisimple Lie algebras over $\mathbb{C} $. Let $V_1$ and $V_2$ be simple modules over $\mathfrak{g}_1$ and $\mathfrak{g}_2$, respectively. They can be infinite dimensional.

$V_1 \otimes V_2$ is a $\mathfrak{g}_1 \oplus \mathfrak{g}_2$-module with the action given by $$ (x_1, x_2)(v_1 \otimes v_2)=(x_1 v_1)\otimes v_2 + v_1 \otimes (x_2 v_2). $$

Is it true that $V_1 \otimes V_2$ is a simple $\mathfrak{g}_1 \oplus \mathfrak{g}_2$-module? If yes, how do I prove it?

H R
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    Where exactly are you getting stuck? This seems to be a pretty straightforward problem to solve. – David Hill Sep 10 '19 at 19:18
  • @DavidHill if M is a submodule and has a vector like $v_1 \otimes v_2$, then it's okay. But if it has only sums? – H R Sep 11 '19 at 21:25

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This is not such a trivial problem. The solution I know uses the Jacobson density theorem: Let $M$ be a simple $R$-module, $D = \mathrm{End}_R(M)$. If $\varphi \in \mathrm{End}_D(M)$, $x_1,\ldots, x_n \in M$, then there exists $r \in R$ such that $rx_i = \varphi(x_i)$ for all $i$. I will show how this reduces to the case of principal tensors.

Let $\mathcal U_1$ and $\mathcal U_2$ be the enveloping algebras of $\mathfrak g_1$ and $\mathfrak g_2$. By Dixmier's Lemma, even if $V_1$ and $V_2$ are infinite-dimensional, $$ \mathrm{End}_{U_i}(V_i) = \mathbb C$$ for $i = 1,2.$ (This is a generalization of the usual finite-dimensional Schur lemma to modules of countable dimension over $\mathbb C$). Now suppose that $$v = \sum_{i=1}^n v^1_i \otimes v^2_i \in V_1 \otimes V_2$$ is nonzero. We want to show $\mathcal U(\mathfrak g_1 \oplus \mathfrak g_2) v = V_1 \otimes V_2$. Without loss of generality, we may choose $\{v^1_i\}$ to be linearly independent over $\mathbb C$ and $v^2_i \neq 0$ for all $i$. By Jacobson density, there exists $u \in \mathcal U_1$ such that $$ uv^1_i = \begin{cases} v^1_1 & i= 1 \\ 0 & i > 1 .\end{cases}$$ Hence for $u\otimes 1 \in \mathcal U_1 \otimes \mathcal U_2 =\mathcal U(\mathfrak g_1 \oplus \mathfrak g_2)$, $$(u \otimes 1)v = v^1_1 \otimes v^2_1,$$ which reduces to the case of when $v$ is a principal tensor.

Joshua Mundinger
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  • By Dixmier's Lemma, all homomorphism of representations $V_1 \rightarrow V_1$ are scalars operators, i.e., if $\varphi \in End_{U_1} (V_1)$, then there existis $\lambda \in \mathbb{C} $ such that $\varphi(v^1_i)= \lambda v_i^1$. So I don't know how you can find $\varphi \in End_{U_1} (V_1)$ such that $\varphi(v_i^1) = \delta_{1 i} v_1^1$. – H R Sep 16 '19 at 15:06
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    Jacobson density is for approximating elements in double shriek, not single shriek. Single shriek is $\mathbb C$, so double shriek is the $\mathbb C$-linear endomorphisms. – Joshua Mundinger Sep 16 '19 at 15:08
  • @JoshuaMundinger, can your proof be extended to any lie algebras, i.e. not just semi-simple lie algebras as was asked in the question ? – alpha Jul 01 '23 at 22:02
  • @alpha yes, in fact the proof given doesn't use anything about Lie algebras, only that we have two algebras $\mathcal U_1$ and $\mathcal U_2$ of countable dimension over $\mathbb C$. – Joshua Mundinger Jul 02 '23 at 07:12