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Suppose $A:V\to W$ is a linear transformation between vector spaces. Define the dual map $A^*$ to be a linear transformation $A^*:W^*\to V^*$ between dual spaces $W^*$ and $V^*$ defined by $(A^*\phi)(v)=\phi(Av)$ for all functionals $\phi\in W^*$ (i.e. $\phi:W\to K$ where $K$ is the scalar field). Observe $A^*\phi$ takes $v\in V$ as input and outputs ...

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The tensor product of (the underlying vector spaces of) two Lie algebras does not have a natural Lie algebra structure. You have a $\mathfrak{g} \oplus \mathfrak{g}'$-action. It's defined by extending $$(X \oplus X')(v \otimes v') = Xv \otimes v' + v \otimes X' v'$$ linearly, where $X, X' \in \mathfrak{g}, \mathfrak{g}'$ and $v, v' \in V, V'$. In ...

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For a more general point of view: a (left) module over a Lie algebra $\mathfrak{g}$ is the same thing as a (left) module over the universal enveloping algebra $\mathbb{U}\mathfrak{g}$, an associative algebra defined as the quotient of the free associative algebra on $\mathfrak{g}$ quotiented out by an ideal: \mathbb{U}\mathfrak{g} = T(\mathfrak{g}) / ...

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This is the correct definition. See, for example, page 7 of Isaac Goldbring's notes here. One thing to note is that this is just the usual $L^2$ space on $G$ with respect to $G$'s Haar measure $\mu$, normalized so that $\mu({1}) = 1$ (although the particular normalization doesn't matter). Since $G$ is discrete, $\langle f, g \rangle = \int f\bar{g} d\mu = ... 1 This is typically not true for$n>2$. For instance, suppose$R=k$is a field,$M=k^2$, and the$M_i$are a bunch of distinct$1$-dimensional subspaces. Then the$M_i$satisfy your hypotheses, but$M/\bigcap M_i=M$is$2$-dimensional while$\bigoplus M/M_i$is$n$-dimensional. 1 Here's another hint: Prove that$\mathrm{End}(V) \simeq V^\ast \otimes V$. Also prove that$(V \otimes V)^\ast \simeq V^\ast \otimes V^\ast$. 1 The analogue for the translation action of$\mathbb{R}^2$on$L^2(\mathbb{R})$is the subspace of compactly supported functions. More generally you can pick just about any function and consider the subspace spanned by its translates. 1 You're right. Many books give definitions that incorrectly specialize to the empty set and related objects; see too simple to be simple on the nLab for a discussion. My preferred phrasing of the correct definition is that a simple module$V$is a module that has exactly two submodules (which must therefore be$0$and$V$), so the zero module is never ... 1 If$V$and$W$are$\mathfrak{g}$-modules, then the usual formula you have applied defines the "diagonal"$\mathfrak{g}$-action on the tensor product$V\otimes W$, i.e., it gives the tensor product of two Lie algebra representations - see Definition$1.12$here. I think the word "diagonal" comes from the group case, where the action is really$g(v_1\otimes ...

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The main observation to make is that if $\rho: \mathfrak{g} \to \mathfrak{h}$ is a homomorphism of lie algebras then $\rho([\mathfrak{g},\mathfrak{g}]) \subset [\mathfrak{h},\mathfrak{h}]$. In your case $\mathfrak{g}=L$ is semisimple, so $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$, and $\mathfrak{h} = \mathbb{C}$, so $[\mathfrak{h},\mathfrak{h}]=0$. Hence ...

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I finally got it thanks to Nephry's comment. The text "if $X = \mathbb C$" refers to if $X$ is the trivial representation corresponding to the trivial homomorphism $G \to \operatorname{GL}(X)$. With that, here's the proof: Since $P$ commutes with all elements of $g$, it follows that $P : X \to X$ is intertwining. By Schur's Lemma, $P$ is either an ...

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