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 $n_\lambda \in \mathbb{R}$?

Alternatively, is there a (published) reference for the construction of $\omega_\lambda$ by such averaging operation?

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.

  • $\begingroup$ I think (to help out representation theorists like me who follow Harish-Chandra's dictum that "characters tell all") it might be helpful to specify what tools are allowable as more straightforward than character theory. $\endgroup$
    – LSpice
    Commented Sep 9, 2019 at 18:57
  • $\begingroup$ I was hoping for some trick involving Young tableaux and/or the Reynolds operator, as the derivation in Graham Gill's notes relies on Theorem 4 and Lemma 2 which aren't that straightforward either (to me, I'm a physicist.) Maybe I should adapt my question a bit to alternative ask for a (published) reference for that construction of $\omega_\lambda$. $\endgroup$ Commented Sep 10, 2019 at 10:30
  • $\begingroup$ Answered on math.stackexchange. (math.stackexchange.com/questions/3345754/…) $\endgroup$ Commented Sep 10, 2019 at 12:26


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