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Consider the group algebra of the symmetric group $ \mathbb{C} S_k$. Given some Young tableau $T$ of shape $\lambda$, let $a_{\lambda,T}$ and $b_{\lambda,T}$ be the row symmetrizer and column antisymmetrizer of the tableau respectively.

It is known that the Young symmetrizer $c_{\lambda,T} = a_{\lambda,T} b_{\lambda,T}$ is proportional to an idempotent. That is, $c_{\lambda,T}^2 = m_\lambda c_{\lambda,T}$ with $m_\lambda \in \mathbb{R}$.

Using character theory, one can show that the element $$\omega_\lambda = \sum_{\pi \in S_k} \pi c_{\lambda,T} \pi^{-1}$$ is proportional to a centrally primitive idempotent (see e.g. Proposition 2 in the notes by Graham Gill, representation theory of the symmetric group: basic elements). It therefore projects onto the isotypic component associated to $\lambda$.

Is there a more straightforward way (i.e. one that doesn't use character theory) to show that $\omega_\lambda$ is proportional to an idempotent, that is to show that $\omega_\lambda^2 = n_\lambda \omega_\lambda$ with $n_\lambda \in \mathbb{R}$?

edit: I feel one should be able to take advantage of the averaging operation / Reynolds operator $\alpha \mapsto \sum_{g \in G} g \alpha g^{-1}$. I don't quite know how to however.

edit2: Apologies, I have opened a question on mathoverflow just a moment before the first comment got in after a couple of days.

Felix Huber
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Probably the easiest way to see this is using the isomorphism \begin{align}\mathbb{C}S_n&\cong\bigoplus_i M_{n_i}(\mathbb{C}),&(1)\end{align} which is proved using Maschke's Theorem and the Wedderburn-Artin Theorem (I assume you know this). Throughout, I'll write $c_\lambda=c_{\lambda,T}$.

Starting with a fact about the right-hand side.

Lemma: Let $A=M_n(\mathbb{C})\oplus M_n(\mathbb{C})$. If $I$ is a left ideal of $M_n(\mathbb{C})$ and $I\oplus 0\cong 0\oplus I$ as a left $A$-module, then $I=(0)$.

It follows that

Corollary: If $I$ is any nonzero ideal of $A=\bigoplus_i M_{n_i}(\mathbb{C})$ with support in $M_{n_i}(\mathbb{C})$ for some $i$, then $I$ is not is isomorphic (as a left $A$-module) to any left ideal $J$ with support in $M_{n_j}(\mathbb{C})$, $j\neq i$.

Now, turning to the right-hand side of the isomorphism, recall the minimal left ideal $\mathcal{I}_\lambda=\mathbb{C}S_n c_\lambda$.

Proposition: If $w\in S_n$, then the map $$ \rho_w:\mathcal{I}_\lambda\to \mathcal{I}_\lambda w, \;\;\;\rho_w(x)=xw $$ is an isomorphism of left $\mathbb{C}S_n$-modules with inverse $\rho_{w^{-1}}$.

Now, since $\mathcal{I}_\lambda$ is a minimal left ideal, its image on the right-hand side of the isomorphism (1) has support in a single column of $M_{n_i}(\mathbb{C})$ for some $i$. For $w\in S_n$, $\mathcal{I}_\lambda w\cong \mathcal{I}_\lambda$. Since $\mathcal{I}_\lambda w$ is a minimal left ideal, its image under (1) has support in a single column of $M_{n_j}(\mathbb{C})$ for some $j$. Therefore, by the corollary, the image of $\mathcal{I}_\lambda w$ under isomorphism (1) has support in a single column of $M_{n_i}(\mathbb{C})$ for the same $i$ (it is worth noting at this point a standard linear algebra fact: right multiplication by an invertible matrix implements a combination of elementary column operations). In particular,

Proposition: The image of $\sum_{\pi\in S_n} \mathcal{I}_\lambda \pi^{-1}$ under isomorphism (1) has support in $M_{n_i}(\mathbb{C})$ for some $i$.

Next, observe that $w_\lambda=\sum_{\pi}\pi c_\lambda \pi^{-1}\in \sum_{\pi\in S_n} \mathcal{I}_\lambda \pi^{-1}$. Therefore, the image of $w_\lambda$ under (1) has support in a single $M_{n_i}(\mathbb{C})$. This element is clearly central since, for $\sigma\in S_n$, \begin{align} \sigma w_\lambda&=\sum_\pi \sigma\pi c_\lambda \pi^{-1}\\ &=\sum_\pi \sigma\pi c_\lambda \pi^{-1}\sigma^{-1}\sigma\\ &=\sum_\pi (\sigma\pi) c_\lambda (\sigma\pi)^{-1}\sigma\\ &=w_\lambda \sigma. \end{align}

Since the center of $\bigoplus_i M_{n_i}(\mathbb{C})$ consists of matrices of the form $\oplus_i d_iI_{n_i}$, it follows that the image of $w_\lambda$ under (1) is of the form $0\oplus dI_{n_i}\oplus 0$. In particular, $$w_\lambda^2=d^2w_\lambda.$$

To show that $d\neq 0$, it is enough to show that $w_\lambda\neq 0$. But, this follows by expressing $c_\lambda=a_\lambda b_\lambda$ as a linear combination of elements of $S_n$. Indeed, if $\sigma$ is in the row stabilizer of $T$ (so $\sigma$ appears in $a_\lambda$), then $\sigma^{-1}$ is not in the column stabilizer of $T$ (so $\sigma^{-1}$ does not appear in $b_\lambda$). Therefore, the coefficient of the identity element in $c_\lambda$ is $1$. Since no non-identity element of $S_n$ is conjugate to the identity, it follows that the coefficient of the identity in $w_\lambda$ is $n!$, showing that $w_\lambda\neq0$.

David Hill
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    Artin--Wedderburn is very close to character theory, which OP wanted to avoid. For instance, the isotypic projector formula can be proved from it in a few lines. – Joshua P. Swanson Sep 10 '19 at 06:08
  • Thanks for the detailed explanation. So if I understand correctly, the bottom line is that when considering a semisimple group ring, any element that is central is necessarily also proportional to some central idempotent / the identity matrix on an isotypic component. Further, the map $\alpha \mapsto \sum_{g\in G} g \alpha g^{-1}$ will always yield a central element by construction, so this can be seen as a default construction. Considering the Young symmetrizer, it is then clear that $\sum_{\pi\in S_k} \pi c_\lambda \pi^{-1}$ projects onto the isotypic component / 2-sided ideal. – Felix Huber Sep 10 '19 at 12:24
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    That's not quite right. A central element of the group ring maps to a direct sum of scalar matrices, which are not necessarily proportional to an idempotent (e.g. $2I_2\oplus 3I_2\in M_2\oplus M_2$). The key fact here is that $w_\lambda$ has nonzero entries concentrated in a single block. – David Hill Sep 10 '19 at 15:45
  • Thanks, I missed that aspect. So it is fair to say that, as long as $\alpha$ has support in one isotypical component only, the Reynolds operation $\sum_{g \in G} g\alpha g^{-1}$ will yield a scalar of the identity of that component. For the symmetric group it is clear that $c_\lambda$ fulfills that requirement. I think that's indeed a nice way to see that $\omega_\lambda$ is proportional to an idempotent. Thanks a lot! – Felix Huber Sep 10 '19 at 16:56
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    you bet. Don't forget the last point: we had to check that $w_\lambda\neq 0$! – David Hill Sep 10 '19 at 19:11