I am slightly confused over the semantics of the definition of complete reducibility of a representation of a Lie algebra. Here's what I understand it to mean:

First, A representation $\phi:g \rightarrow gl(V)$ of a Lie algebra $g$ is irreducible if the vector space $V$ has no invariant subspaces.

Where we say a subspace $W$ of $V$ is invariant if it is invariant under $\phi(x)$ for all $x$ in $g$.

Now, it is said that a representation is completely reducible if it decomposes into a direct sum of irreducible representations (Wikipedia).

I struggle to understand what the last sentence means. What I think it means is that $V$ decomposes into a direct sum of vector spaces $V_i$ which are all irreducible (IE have no invariant subspaces).

This makes sense to me, but I also am not quite sure that I understand why it's not trivial. Should it not be the case that for any representation we will always be able to decompose $V$ into a direct sum of subspaces which are irreducible? I mean, the answer is obviously not (unless $g$ is semisimple), but I'm struggling with an intuitive understanding of that. Any help is appreciated.

  • $\begingroup$ Surely you meant indecomposable. Yes, that is correct. But it may not mean irreducible. $\endgroup$
    – orangeskid
    Oct 22, 2017 at 0:01

1 Answer 1


I'll give you an example of a representation of a non-semisimple Lie algebra that is not completely reducible.

Suppose $V = \mathbb C^2$ and suppose $\mathfrak g$ is the one-dimensional Lie algebra $$ \mathfrak g = \left\{ \begin{bmatrix} a & a \\ 0 & a \end{bmatrix} :a \in \mathbb C \right\}.$$ Let $\phi : \mathfrak g \to gl(V)$ be the natural representation defined by matrix multiplication, $$ \phi(g)(v) = gv,$$

Then $W = \{ (c, 0) : c \in \mathbb C \}$ is a proper non-trivial invariant subspace of $V$, because $$\phi(g)(w) \in W$$ for every $g \in \mathfrak g $ and $w \in W$.

However, it is impossible to find a complementary invariant subspace $W'$ such that $$V = W \oplus W'.$$ Therefore, $V$ is not completely reducible.

This example illustrates that the key obstacle to complete irreducibility is not the ability to find an invariant subspace, but the ability to find a complementary subspace that is also an invariant subspace.


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