# Why is the exterior power $\bigwedge^kV$ an irreducible representation of $GL(V)$?

$$\newcommand{\GL}{\operatorname{GL}}$$

Let $$V$$ be a real $$n$$-dimensional vector space. For $$1 we have a natural representation of $$\GL(V)$$ via the $$k$$ exterior power:

$$\rho:\GL(V) \to \GL(\bigwedge^kV)$$, given by $$\rho(A)=\bigwedge^k A$$. I am trying to show $$\rho$$ is an irreducible representation. Let $$0\neq W \le \bigwedge^kV$$ be a subrepresentation. If we can show $$W$$ contains a non-zero decomposable element, we are done.

Indeed, suppose $$W \subsetneq \bigwedge^kV$$. Then, there exist a decomposable element $$\sigma=v_1 \wedge \dots \wedge v_k \neq 0$$, such that $$\sigma \notin W$$. We assumed $$W$$ contains a non-zero decomposable element $$\sigma'=u_1 \wedge \dots \wedge u_k \neq 0$$. Define a map $$A \in \GL(V)$$ by extending $$u_i \to v_i$$. Then

$$\rho(A) (\sigma')=\bigwedge^k A(u_1 \wedge \dots \wedge u_k)=\sigma \notin W,$$

while $$\sigma' \in W$$, con

So, the question reduces to the following: Why does every non-zero subrepresentation contain a non-zero decomposable element?

I asked an even more naive question here-whether or not every subspace of dimension greater than $$1$$ contains a non-zero decomposable element?

• A nice conceptual way to work with this is to first decompose the representation as a module for the group of diagonal matrices (into so-called weight spaces). Then note what happens to the weight of a vector in one of these subspaces when one acts by suitable upper triangular unipotent matrices. – Tobias Kildetoft Jan 3 at 9:09
• Thanks. Unfortunately, I really know barely nothing about the machinery of representation theory. Can you please elaborate on this or give me a reference? (I don't know what a weight of a vector is, and naive googling only found something in the context of representations of Lie algebras, not Lie groups). – Asaf Shachar Jan 3 at 9:17
• The definition is essentially the same. The representation decomposes as a sum of $1$-dimensional subspaces, and a vector in such a subspace will be acted on via a scalar. This scalar depends on the element acting, giving a linear character of the subgroup of diagonal matrices, and this character is what is called the weight of the vector. It may be a bit much to get into if none of this is familiar, but I would still advice you to try writing it up explicitly for $k=1$ when $\dim(V) = 2$ to get a feel for what happens. – Tobias Kildetoft Jan 3 at 9:43

Pick a basis $$e_1, \dots e_n$$ of $$V$$ so that we can identify $$GL(V)$$ with $$GL_n(F)$$ (we'll start out working with an arbitrary base field $$F$$ and then restrict $$F$$ later). Write $$T$$ for the subgroup of $$GL_n(F)$$ consisting of diagonal matrices. An element of $$T$$ consists of some diagonal elements $$(t_1, \dots t_n)$$ and acts on $$\Lambda^k(V)$$ by sending $$e_i$$ to $$t_i e_i$$, then extending multiplicatively.
What this means is that each pure tensor $$e_{i_1} \wedge e_{i_2} \wedge \dots \wedge e_{i_k} \in \Lambda^k(V)$$ is a simultaneous eigenvector for every element of $$T$$; said another way, it spans a $$1$$-dimensional (hence simple) subrepresentation of $$\Lambda^k(V)$$, considered as a representation of $$T$$. (These are the "weight spaces" of this representation.) Since $$\Lambda^k(V)$$ is the direct sum of these $$1$$-dimensional subspaces, it follows that $$\Lambda^k(V)$$ is semisimple as a representation of $$T$$.
The significance of semisimplicity is that any $$GL(V)$$-subrepresentation of $$\Lambda^k(V)$$ is also a $$T$$-subrepresentation, and subrepresentations of semisimple representations are semisimple; they must also have the same simple components, in the same or smaller multiplicities. Moreover, if $$F$$ is any field except $$\mathbb{F}_2$$ (over $$\mathbb{F}_2$$, unfortunately, $$T$$ is the trivial group), the different $$1$$-dimensional representations above are all nonisomorphic. The conclusion from here is that any $$GL(V)$$-subrepresentation of $$\Lambda^k(V)$$ must be a direct sum of weight spaces.
But now we're done (again, for any field $$F$$ except $$\mathbb{F}_2$$), for example because $$GL(V)$$ acts transitively on these weight spaces.
• Thanks. Can you please elaborate on your use of the fact that the different $1$-dimensional representations are all nonisomorphic? How do you conclude from that that any $GL(V)$-subrepresentation is a direct sum of some of the $1$-dimensional weight spaces? – Asaf Shachar Jan 8 at 11:38