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There is a well-known construction of minimal idempotents in the group algebra of the symmetric group $\mathbb C[S_n]$ using row symmetrizers and column antisymmetrizers. But these idempotents are not *-idempotents with respect to the *-structure on $\mathbb C[S_n]$ which sends $g$ to $g^{-1}$. (In other words, they are not orthogonal projections with respect to the usual inner product.)

Is there in the literature somewhere a construction of minimal *-idempotents for $\mathbb C[S_n]$? My goal is to do some explicit calculations, so simple and explicit constructions are preferred.

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There is a well-known construction of the primitive idempotents for the symmetric group over the rationals that is due to Murphy (and possibly Jucys?). It can be found in his paper "A new construction of Young's seminormal representation of the symmetric group", J. Algebra, 69 (1981), 287-297.

To describe this for $1\le k\le n$ let $L_k=(1,k)+\dots+(k,k-1)$ be the $k$th Jucys-Murphy element. Notice that $L_k^*=L_k$ for all $k$. You can check that $L_kL_m=L_mL_k$, for $1\le k,m\le n$. Now for each standard tableau $t$ define $$ E_t = \prod_{k=1}^n\prod_{c\ne c_k(t)}\frac{L_k-c}{c_k(t)-c}, $$ where $c_k(t)$ is the content of $k$ in $t$ --- that is, $c_k(t)=c-r$ if $k$ appears in row $r$ and column $c$ of $t$. In the second product $c$ runs over all possible possible contents in all standard tableaux of size $n$.

Murphy shows that $\{E_t\mid t \text{ standard}\}$ is a complete set of pairwise orthogonal idempotents in $\mathbb{Q}S_n$. In fact, this is almost immediate if you just consider how these elements act on the seminormal representations of the symmetric group. For a more modern account see Okounkov and Vershik's paper "A new approach to representation theory of symmetric groups", Selecta Math. (N.S.), 2 (1996), 581–605.

By construction, $E_t^*=E_t$ so these idempotents have the property that you want.

For calculations these idempotents can be constructed recursively. They become a mess, however, if you expand them in the permutation basis of the group algebra, so you should try and avoid this if possible.

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  • $\begingroup$ Thanks, that's helpful. Do you know whether there is an analogous construction in the Hecke algebra deformation of the group algebra? $\endgroup$ May 22, 2014 at 16:34
  • $\begingroup$ Yes, everything works for the Hecke algebras of types A and B and more generally of type $G(r,1,n)$. In these cases it is probably first due to Hoefsmit, Murphy, Ariki-Koike. I am biased, but see for example arxiv:1304.0906. $\endgroup$
    – Andrew
    May 22, 2014 at 23:40
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I believe that if $\rho: S_n \to GL_d(\mathbb{C})$ is a unitary representation with respect to $\langle \, , \, \rangle$, and $v \in \mathbb{C}^d$, then $$e_v = \frac{d}{n!} \sum_{g \in S_n} \langle \rho(g)(v), v \rangle g$$ will be a minimal idempotent -- basically projection to $v$. Since $g^* = g^{-1}$, then $e_v^* = e_v$.

To check it is an idempotent, one can do a calculation similar to Serre (page 14 of ``Linear Representations of Finite Groups", the usual summing over the group and applying Schur's Lemma). Also, if $P$ is some unitary (change of basis) matrix of the group algebra, you can check, extending $\rho$ linearly that $\frac{d}{n!} \sum_{g \in S_n} \langle \rho(Pg) v, v \rangle Pg$ is the same idempotent. (If $P$ is not unitary, just replace the second $P$ with $(P^{-1})^T$.)


I don't have enough ``reputation" to comment for real, but to comment on Andrew's answer: regarding the Hecke algebra, on page 14 of 1304.0906, they set $*$ to be the anti-automorphism such that $T_i^* = T_i$. But as you have $\langle T_i v, v \rangle = \langle v, T_i^{-1}v \rangle$, I believe you have $T_i^* = T_i^{-1}$. I do not think it is immediate (or even true?) that the formula for an idempotent using JM elements will be a $*$-idempotent for your $*$. However, this idempotent still should project to one of the seminormal basis elements with given spectrum.

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  • $\begingroup$ The Hecke algebra has a unique anti-isomorphism $*$ such that $T_i\mapsto T_i$, for $1\le i<n$. Then $T_w^*=T_{w^{-1}}$ and, consequently, $L_k^*=L_k$, for $w\in S_n$ and $1\le k\le n$. As the Jucys-Murphy elements are $*$-invariant so are the idempotents given above. The map $T_i\mapsto T_i^{-1}$ does not extend to an anti-automorphism in general. The inner products in 1304.0906 have the property that $\langle uh,v\rangle=\langle u,vh^\star\rangle$. (This paper uses right modules.) $\endgroup$
    – Andrew
    May 28, 2014 at 0:06

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