About Young symmetrizer $c_{\lambda}$

Question

I'm reading the Fulton and Harris's book "Representation Theory". I want to ask about the proof of lemma 4.25.

Let $c_{\lambda}$ be the young symmetrizer, and let $V_{\lambda} = {\mathbb C}S_d$. According to the proof, if $W \subset V_{\lambda}$ is a subrepresentation, $c_{\lambda}W$ is either ${\mathbb C}c_{\lambda}$ or $\{ 0 \}$.

But, I can't prove this. In particular, I can't see ${\mathbb C}c_{\lambda} \subset c_{\lambda}W$ when I suppose $W \neq \{ 0 \}$.

Could anyone help me?

Answer 1

I found this proof really difficult to follow myself, but on the third reading (with a bit of googling) I managed to get the hang of things. I'll explain in a bit more detail here, using the shorthand $A := \mathbb{C}\mathcal{S}_d$ and $c := c_\lambda$.

Let $W$ be an invariant subspace of $V = Ac$. According to Lemma 4.23 (ii), for every $x$ in $A$, $cxc$ is a scalar multiple of $c$. It follows that $cAc$ is the one dimensional subspace of $A$ consisting of scalar multiples of $c$, i.e. $cAc = \mathbb{C} c$ for short.

Since $W \leq Ac$ (i.e $W$ is a subspace of $Ac$), $cW \leq cAc$. In other words, $cW$ is a subspace of the one-dimensional subspace $cAc = \mathbb{C}c$. Since a subspace of a one-dimensional vector space is either zero or the whole thing, this entails $cW = \{0\}$ or $cW = \mathbb{C}c$.

Suppose $cW = \mathbb{C}c$. Then $Ac = A\mathbb{C}c = A cW \subset W$, where the final inclusion follows from the invariance of $W$. It follows in this case that $W = Ac =: V$.

Suppose now $cW = \{0\}$. We are going to show this implies $W = \{0\}$. Now since $W \leq Ac$, $WW \leq AcW$. But if $cW = \mathbb{C}c$ then $WW \leq (Ac)W = A(cW) = 0$. But $W^2 = 0$ implies $W = 0$, thanks to the following lemma:

Lemma: If $W$ is an invariant subspace of a general group ring $\mathbb{C}[G]$, and $W^2 = 0$, then $W = 0$. Proof: Let $U$ be a complementary invariant subspace, so that $\mathbb{C}[G] = > U+W$ is a direct sum of representations. Then we can write $1 = u + w$ for some $u \in U, w \in W$. In particular, $w = wu + w^2$. Since $W^2 > = 0$, $w^2 = 0$, and hence $w = wu$. It follows from the invariance of $U$ that $w \in U$. In particular, $1 = u + w$ is also contained in $U$, and by invariance, $U$ is all of $\mathbb{C}[G]$. Hence $W = 0$.

We have shown that if $W$ is an invariant subspace of $V = Ac$, then $W$ is either $0$ or all of $Ac$. Hence $Ac$ is irreducible.