# Invariant subspaces of isotypic representation

I am trying to give a general expression for an invariant subspace of a completely reducible representation. I think that this may be known, but I have not been able to find a reference for this, so I have tried to write two proofs regarding the case that our representation is isotypic. I would be grateful if you could check if they are correct.

For context, assume that $$\mathfrak{g}$$ is a semisimple Lie algebra over $$\mathbb{F}=\mathbb{R}$$ or $$\mathbb{C}$$. Then, by Weyl's theorem, any finite-dimensional representation of $$\mathfrak{g}$$ is completely reducible. Thus, let $$W$$ be one such representation. We may write $$W=V_{1}^{\oplus k_{1}}\oplus\dots\oplus V_{n}^{\oplus k_{n}}$$, where the $$V_{i}$$ are irreducible. In particular, the submodules $$S_{V}(V_{i})=V_{i}^{\oplus k_{i}}$$ are the isotypic submodules of $$W$$.

Now, consider an invariant subspace $$E\subseteq W$$. Since the isotypic decomposition of $$W$$ is functorial, it follows that $$E=\bigoplus_{i=1}^{n}(V_{i}^{\oplus k_{i}}\cap E)$$, and clearly every $$V_{i}^{\oplus k_{i}}\cap E$$ is an invariant subspace of $$S_{V}(V_{i})$$. This means that we only need to understand the structure of the invariant subspaces of isotypic modules.

First question: determining (irreducible) invariant subspaces

Assume $$V$$ is an irreducible $$\mathfrak{g}$$-module, and that $$W=V^{\oplus n}$$ is an isotypic module for $$V$$. Take a nonzero proper invariant subspace $$E\subseteq W$$, and for the moment suppose that $$E$$ is irreducible. The projection maps $$\pi_{i}\colon E\to V$$ are equivariant, so they are either isomorphisms or the zero map by Schur's Lemma. In particular, let $$J$$ be the set of indices $$j$$ such that $$\pi_{j}\colon E\to V$$ is nonzero. Fix $$j_{0}\in J$$, so that for every $$j\in J\setminus \{j_{0}\}$$ the map $$\varphi_{j}=\pi_{j}\circ (\pi_{j_{0}})^{-1}$$ is an isomorphism of $$V$$ into itself (if $$J=\{j_{0}\}$$ then simply $$E=V$$ is the $$j_{0}$$-th copy of $$V$$ inside $$W$$). Then, any element $$x\in E$$ takes the form $$x=\sum_{j\in J}\pi_{j}(x)=\sum_{j\in J}\pi_{j}\pi_{j_{0}}^{-1}\pi_{j_{0}}(x)=\pi_{j_{o}}(x)+\sum_{j\in J\setminus \{j_{0}\}}\varphi_{j}(\pi_{j_{0}}(x)),$$(here I'm abusing the notation, what I mean is that $$\pi_{j_{0}}(x)$$ lives in the $$j_{0}$$-th copy of $$V$$ inside $$W$$, and so on) and since $$\pi_{j_{0}}$$ is an isomorphism, we may write $$E=\{ v+\sum_{j\in J\setminus \{ j_{0} \}}\varphi_{j}(v) \mid v\in V \}$$. We can also set $$\varphi_{i}=0$$ for $$i\notin J$$ so that $$E=\{ v+\sum_{i\neq j_{0}}\varphi_{i}(v) \mid v\in V \}$$(again, abusing notation). By setting $$\varphi_{j_{0}}$$ to be the identity map, we conclude:

A subspace $$E\subseteq W$$ is invariant and irreducible if and only if there exist endomorphisms $$\varphi_{1} , \dots , \varphi_{n} \in \operatorname{End}_{\mathfrak{g}} (V)$$ (one of which is nonzero) such that $$E=\{(\varphi_{1}(v),\dots,\varphi_{n}(v)) \mid v\in V\}.$$In general, a subspace $$E\subseteq W$$ is invariant if and only if it is a sum of subspaces of the previous form.

Is my proof correct?

Second question: giving a compact form for the invariant subspaces

I was wondering if there is a more compact expression for a general invariant subspace of an isotypic module. So far I have the following:

Assume $$\operatorname{End}_{\mathfrak{g}}(V)=\mathbb{F}$$. For example, this is always the case when $$\mathbb{F}=\mathbb{C}$$. Then, we know that $$W$$ is isomorphic to $$V\otimes \mathbb{F}^{n}$$, where $$\mathfrak{g}$$ acts trivially on $$\mathbb{F}^{n}$$, and since all endomorphisms are scalars, an irreducible subspace of $$V\otimes \mathbb{F}^{n}$$ takes the form $$E=\{(a_{1} v,\dots,a_{n} v) \mid v\in V\}=V\otimes \langle (a_{1},\dots,a_{n})\rangle$$ for some scalars $$a_{1},\dots,a_{n}\in\mathbb{F}$$. In particular, invariant subspaces are of the form $$V\otimes U$$, where $$U\subseteq \mathbb{F}^{n}$$ is any subspace.

In general, we only know that $$D=\operatorname{End}_{\mathfrak{g}}(V)$$ is a finite-dimensional division algebra over $$\mathbb{F}$$, and it may be the case that $$W$$ does not have the same dimension as a representation of the form $$V\otimes D^{n}$$ as above. However, we can consider the map $$\psi\colon V\otimes D^{n}\to W$$ given by $$v\otimes (g_{1},\dots,g_{n})\mapsto \sum_{i=1}^{n}g_{i}(v)$$, and the argument from before would imply that invariant subspaces of $$W$$ are now of the form $$\psi(V\otimes U)$$, where $$U$$ is an $$\mathbb{F}$$-vector subspace of $$D^{n}$$.

Is my argument correct? Is there a neater way to write down a general invariant subspace when $$D\neq \mathbb{F}$$?

• You almost show but don't quite (unless I've misunderstood) that an invariant $E \subset V^{\oplus n}$ must be of the form $V^{\oplus k}$ for some $k \leq n$ and this is indeed the case. Of course it doesn't have to align with the original decomposition as the natural "diagonal" example $\{(v,v)| v \in V\} \subset V \oplus V$ would show but otherwise we are done. Then of course such a subspace can also be written $E \cong V^{\oplus k} \cong V \otimes \mathbb{F}^k$ where $\mathbb{F}^k$ is understood as a trivial representation. Jul 19, 2023 at 15:59
• Indeed, a general invariant subspace $E\subseteq V^{\oplus n}$ is isomorphic to $V^{\oplus k}$. I don't say it explicitly, but I did prove that when $E$ is irreducible, then it is isomorphic to $V$, from which the other isomorphism follows. I am interested in understanding how $E$ can be embedded into $V^{\oplus n}$, so just knowing the isomorphism class of $E$ is not enough for what I want. Jul 19, 2023 at 16:04
I am not sure whether this is what you are looking for, but for $$\mathbb F=\mathbb C$$ things can be understood very well in terms of highest weight vectors. Take the decomposition $$\mathfrak g=\mathfrak n_-\oplus\mathfrak h\oplus\mathfrak n_+$$, where $$\mathfrak h$$ is a Cartan subalgebra and $$\mathfrak n_+$$ ($$\mathfrak n_-$$) is the sum of all positive (negative) root spaces. Then for any representation $$W$$ you call a vector $$w\in W$$ a highest weight vector of weight $$\lambda\in\mathfrak h^*$$ if and only if $$\mathfrak n_+\cdot w=0$$ and $$H\cdot w=\lambda(H)w$$ for any $$H\in\mathfrak h$$. Then it is easy to see that any finite dimensional representation contains at least one highest weight vector and for an irreducible representation $$V$$, the space of highest weight vectors is one-dimensional. The corresponding weight $$\lambda$$ then determines $$V$$ up to isomorphism (and there is an irreducible $$V$$ for a given $$\lambda$$ if and only if $$\lambda$$ is dominant and algebraically integral). It then follows that for an isotypical representation $$W\cong V^{\oplus n}$$ the space of highest weight vectors is $$n$$-dimensional (and all vectors in there have the same weight $$\lambda$$) and the $$\mathfrak g$$-invariant subspaces of $$W$$ are in bijective correspondence with linear subspaces in the space of highest weight vectors. (Similarly, the space of $$\mathfrak g$$-equivariant maps between two isotypical representations is isomorphic to the space of all linear maps between the spaces of highest weight vectors.) For a general representation $$W$$ of $$\mathfrak g$$ you can then define a decomposition into isotypical components. The isotypical component of weight $$\lambda$$ is most easily defined as the $$\mathfrak g$$-reprsentation generated by the subspace of highest weight vectors of weight $$\lambda$$. As above, invariant subpaces of $$W$$ are then in bijective correspondece with families of linear subspaces in the spaces of highest weight vectors of some fixed wight.
Over $$\mathbb F=\mathbb R$$ you can use similar arguments based on complexifications, but things certainly get a bit more complicated there (and I am not sure whehter there is a universal explicit answer).