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 $D\neq \mathbb{F}$?

Thank you in advance!

  • 3
    $\begingroup$ 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. $\endgroup$
    – Callum
    Jul 19, 2023 at 15:59
  • $\begingroup$ 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. $\endgroup$ Jul 19, 2023 at 16:04

1 Answer 1


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).


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