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$ \newcommand{\g}{\mathfrak{g}} \newcommand{\h}{\mathfrak{h}} \newcommand{\k}{\mathfrak{k}} $

Let $\g$ and $\h$ be two semisimple Lie algebras, and let $\k := \g \oplus \h$. Show that a $\k$-module is simple iff it is the tensor product of simple $\g$-module and $\h$-module.

We may assume that all Lie algebras/vector spaces involved are finite-dimensional, and the involved field is both algebraically closed and of characteristic $0$.

We first note that if $V$ is a $\g$-module and $W$ is a $\h$-module, then $\k$ acts on $V \otimes W$ by the following: $$ q = g \oplus h \implies q \cdot (v \otimes w) := (g \cdot v) \otimes w + v \otimes (h \cdot w) $$ It's easy to check that this action indeed makes $V \otimes W$ a $\k$-module. Unfortunately, I do not have much progress on this from here. For $\impliedby$, initial idea is to assume $V \otimes W$ is not simple, and let $U \subseteq V \otimes W$ be a non-trivial proper submodule. Then, perhaps we can impose some kind of projection of $U$ onto $V$ and $W$ respectively, and contradict their simplicity by showing that this induces a non-trivial proper submodule in $V$ or $W$, contradicting their simplicity. However, as far as I know, there isn't such a notion for tensor product.

For $\implies$, my idea is also similar, but it's clear that $V$ and $W$ are not submodules of $V \otimes W$, so I'm not sure how to proceed with the construction of the $V$ and $W$ given an arbitrary $\k$-module either.

Any help is appreciated.

Note: The following questions are related.

  1. Tensor product of irreducible representations of semisimple Lie algebras: The solution uses Jacobson Density Theorem, which is probably overpowered in the case of finite-dimensional Lie algebras/vector spaces. I am looking for a proof without using this theorem.
  2. Tensor product of irreducible representations: This also seems related, but the question assumes that the field involved is $\Bbb{C}$. Furthermore, the two (unaccepted) answers utilise Clebsch-Gordon formula, which I'm not familiar with.

EDIT: Following Torsten Schoeneberg's hint, I was able to almost construct a proof of the $\implies$ direction. We consider $U$, a $\k$-submodule, as a $\g$-module via the action: $$ g \cdot u := \underbrace{(g \oplus 0)}_{\in \g \oplus \h} \cdot u $$ Then since $\g$ is semisimple, $U$ is semisimple as a $\g$-module, so we can write $U = \bigoplus_{i=1}^n X_i$ for simple $X_i$. Similarly, we can write $U = \bigoplus_{i=1}^m Y_i$ as with $Y_i$ being simple $\h$-modules. If all of $X_i$ are isomorphic (and similarly all of $Y_i$), then we can construct an explicit isomorphism from $X_1 \otimes Y_1 \to U$, which completes the proof. However, I have not been able to prove the claim thus far.

Xander Henderson
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Clement Yung
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  • No need to be familiar with Clebsch-Gordan. Have a look at wikipedia. – Dietrich Burde Sep 15 '20 at 16:57
  • @DietrichBurde sorry, I don't see how Wikipedia answers my question. – Clement Yung Sep 16 '20 at 05:19
  • I am sorry, I just wanted so say that Clebsch-Gordan is explained there. The arguments can be used not only for $\Bbb C$ but for any algebraically closed field of characteristic zero. – Dietrich Burde Sep 16 '20 at 08:36
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    I would think that if you start with a $\mathfrak g$-module $V$ and an $\mathfrak h$-module $W$, and then consider that $V\otimes W$ via restriction e.g. just as $\mathfrak g$-module, it's isomorphic to $V^{\dim W}$. That motivates the following idea for $\implies$: Given a simple $\mathfrak k$-module $U$, view it via restriction as $\mathfrak g$-module. It's still semisimple; show that all its simple components are isomorphic. That simple component is your candidate for $V$. And then likewise to get $W$. Finally, construct some map between $U$ and $V \otimes W$ and show it's an iso. – Torsten Schoeneberg Sep 16 '20 at 20:09
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    For $\impliedby$, I think the method in your first link is sort of the way to go. For finite-dimensional $V$, you could avoid explicit use of Jacobson density and Dixmier's or even Schur's Lemma and instead show that the image of the representation $\mathfrak g \rightarrow End_{\mathbb C}(V)$ generates the entire endomorphism ring as associative algebra. That would give you that step where one reduces to principal tensors. – Torsten Schoeneberg Sep 16 '20 at 20:20
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    Finally, though, probably the best way to view this theory is via universal enveloping algebras; and somehow all this seems to be encapsulated by $$U(\mathfrak g \oplus \mathfrak h) \simeq U(\mathfrak g) \otimes U(\mathfrak h).$$ – Torsten Schoeneberg Sep 16 '20 at 20:23
  • @TorstenSchoeneberg Thank you, your comments were helpful. For $\implies$, following your hint I was able to construct the proof (and the isomorphism at the end), except that I can't show all the simple components of $U$ (as a $\mathfrak{g}$-module) are isomorphic. Is it possible for you to provide further hints? – Clement Yung Sep 17 '20 at 03:12
  • @TorstenSchoeneberg also, I apologise but I made a serious typo on the action of $q$ on $v \otimes w$. Please see the revised post. – Clement Yung Sep 17 '20 at 04:38
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    I thought that $\Rightarrow$ was simple, but the equivalent result for finite group representations uses their characters, and someone more expert than me told me (independently of what was said here) that surjectivity of the map, $U(\mathfrak g)\to\operatorname{End}\Bbb C(V)$ is the heart of the argument, which _is the Jacobson density theorem applied to this relatively simple (finite dimensional) case. So essentially I am just confirming that Torsten Schoeneberg is right. – Marc van Leeuwen Sep 17 '20 at 16:52
  • To show that all simple components are isomorphic, in other situations one uses some extra group action which turns out to permute those components. The closest thing to that we have here would be the action of $0\oplus \mathfrak h$ (or the assoc. algebra it generates, which again is $End_{\mathbb C}(W)$, or then essentially its unit group $GL(W)$). That action is transitive on the second factor $W$, but commutes with the $\mathfrak g \oplus 0$-action on the whole thing, and that should somehow give what we want. – Torsten Schoeneberg Sep 17 '20 at 17:03
  • So we might end up using @MarcvanLeeuwen's "heart of the argument" for both directions! -- And another thing that just struck me is: In the case $\mathfrak g = \mathfrak h$, what we have here is a simple $\mathfrak g \oplus \mathfrak g$-module $V\otimes W$ whose restriction to $\mathfrak g \oplus 0$ we decompose as $V^{\dim W}$ and whose restriction to $0 \oplus \mathfrak g$ we decompose as $W^{\dim V}$; whereas what Clebsch-Gordan theory does is figuring out how its restriction to the diagonal $(x,x) \in \mathfrak g \oplus \mathfrak g$ decomposes, and that's highly non-trivial! – Torsten Schoeneberg Sep 17 '20 at 17:11
  • I see I said $\Rightarrow$ where I meant $\Leftarrow$. I know about implications, but I can never tell my left hand from my right ;-( Apologies for the possible confusion. – Marc van Leeuwen Sep 18 '20 at 13:35

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To show the reverse implication, consider the associative algebra $A\subseteq End(V)$ generated by the image $\rho: \mathfrak{g} \rightarrow End(V)$. Since $\mathfrak{g}$ is semi-simple, the image $\rho(\mathfrak{g})$ is also semi-simple. In particular, one can find the Casimir operator on $\rho(\mathfrak{g})$, and $A$ contains identity. Note that $V$ is also a simple $A$-module. Lastly, we can use some result from representation theory of associative algebra to show that (check Etingof Introduction to Representation Theory Corollary 3.2.1) to see that $A = End(V)$ and if $U\subseteq V \otimes W$, we can find some $a \in End(V)$ and $b \in End(W)$ such that $b(a(u)) \in U$ is a pure tensor, and finding appropriate $a\in End(V)$ and $b \in End(W)$ one can obtain all basis vectors of $V\otimes W$.

To prove the forward implication, first regard $U$ as a $\mathfrak{g}$ module. Note that $U = \bigoplus_{i=1}^n V_i $ for some simple $V_i$ by Weyl. Choose any one of them and say $V= V_1$. Now consider $\hom_{\mathfrak g}(V,U) \otimes V$ as a $\mathfrak h \oplus \mathfrak g$ module, where $h \in \mathfrak h$ acts on $\phi\in \hom_{\mathfrak g}(V,U)$ by $h \phi$ (and $\mathfrak g$ acts trivially). Then consider the map $f_V: \hom_{\mathfrak g}(V,U) \otimes V \rightarrow U$ given by $$f_V(\phi \otimes v) = \phi(v).$$ One can show that this map is a $\mathfrak{g \oplus h}$ module isomorphism by showing that this linear map is injective and $f_V$ commutes with the action of $\mathfrak{g \oplus h}$ hence it must be surjective. We conclude that it must be the case that $\hom_{\mathfrak g}(V,U)$ is a simple $\mathfrak{h}$ module, for otherwise, $\hom_{\mathfrak{g}}(V,U) \otimes V$ splits into direct sum of $\mathfrak{g} \oplus \mathfrak{h}$ modules.

Clement Yung
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monikernemo
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