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Let $G$ be a group and let $V$ be a complex vector space which is a representation of $G$. Let's write the (left) action of $g\in G$ on $v\in V$ as $gv$.

The dual vector space of $V$ is the set of linear maps from $V$ to $\mathbb{C}$, and is written as $V^*$. I'll use the notation $(\phi,v)$ to mean $\phi(v)$ if $\phi\in V^*$ and $v\in V$.

The dual representation of $V$ is defined to be the vector space $V^*$ with the (right) $G$-action $(\phi g,v):=(\phi,gv)$ for $\phi\in V^*$, $v\in V$ and $g\in G$.

A representation of $G$ is irreducible if the only $G$-invariant subspaces are the zero subspace or the whole vector space.

I can show that if $V^*$ is an irreducible representation, then so is $V$. Indeed, if $W$ is a $G$-invariant subspace of $V$ then its annihilator $W^\perp=\{\phi\in V^*\colon w\in W\implies (\phi,w)=0\}$ is a $G$-invariant subspace of $V^*$, so either $W^\perp=V^*$, which clearly implies $W=0$, or $W^\perp=0$, which (by fiddling around with a Hamel basis) yields $W=V$.

Does the converse hold? That is, if $V$ is irreducible, must $V^*$ also be irreducible?

If $\dim V<\infty$ then $V^{**}$ is equivalent to $V$, so the answer is yes in this case. But I don't know what happens if $\dim V=\infty$.

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If $G$ acts from the left on $V$, doesn't it act from the right on $V^*$? – Rasmus Jul 6 '11 at 18:26
I made the title more informative. Feel free to roll-back if you don't like it. – Rasmus Jul 6 '11 at 18:29
@Rasmus: That depends on your conventions: you can use the antihomomorphism $g \mapsto g^{-1}$ to turn the right action into a left one (and vice versa). – t.b. Jul 6 '11 at 18:33
@Rasmus: yes, it does seem to act from the right! I'll change the notation... but the question still seems to make sense, and be interesting. – irrep Jul 6 '11 at 18:34
up vote 23 down vote accepted

No. Let $G = S_{\infty}$ denote the group of permutations of $\mathbb{N}$ which fix all but finitely many elements. $G$ is countable, so any irreducible representation has at most countable dimension. $G$ has a countable-dimensional irreducible representation $V$ given by the subspace of $\bigoplus_{i=1}^{\infty} \mathbb{C}$ of sequences adding to $0$, and $V^{\ast}$ is of uncountable dimension, so cannot be irreducible. (Explicitly, there is an obvious bilinear pairing on $V$ giving $V^{\ast}$ a proper invariant subspace isomorphic to $V$.)

Edit: Here's a cute non-constructive argument. Suppose that $G$ has an infinite-dimensional irreducible representation $V$. If $V^{\ast}$ is reducible, we're done. Otherwise, $V$ is a proper invariant subspace of $V^{\ast \ast}$ (at least given the axiom of choice), so $V^{\ast}$ is an irreducible representation whose dual is not irreducible.

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Nice counter-example! I'm going to remember this one. – David Speyer Jul 6 '11 at 19:36
@David: I think the second argument I just came up with is cuter, though. – Qiaochu Yuan Jul 6 '11 at 19:43
This is indeed really nice. Maybe the most obvious example of a group with an infinite-dimensional irreducible representation is obtained by taking $G = \operatorname{GL}(V)$ for any infinite-dimensional vector space $V$? – Pete L. Clark Jul 6 '11 at 21:40
@paul: can you define topological and algebraic irreducibility for me? Using the definitions I have in my head, it seems that algebraic irreducibility is a priori a stronger, not a weaker, condition... – Qiaochu Yuan Jul 6 '11 at 21:55
Answering my own comment above: $V$ is irreducible because it is spanned by the set $G(1,-1,0,0,\dots)$, and if $v\in V$ with $v\ne0$, say $v=(v_1,\dots,v_n,0,0\dots)$ then by applying an element of $G$ you can arrange that $v_1\ne v_n$. Then if $g$ is the $n$-cycle $g=(1,2,\dots,n)$ and $h=(1,2)g^{-1}$, I think that $(v_1-v_n)^{-1}(e+g+g^2+\dots+g^{n-2}+h)v = (1,-1,0,0,\dots)$, hence the smallest invariant subspace containing $v$ is $V$. – irrep Jul 7 '11 at 12:20

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