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Let $A:=\mathbb{C}S_n$ be the symmetric group aglebra. Let $T$ be a standard Young tableaux of shape $\lambda$. Denote $R(T)$ and $C(T)$ as row and column stabilizers of $T$. For a set $S \subseteq S_n$ we write $S^+ := \sum_{x \in S} x \in \mathbb{C}S_n$ and $S^- := \sum_{x \in S} sgn(x) x \in \mathbb{C}S_n$. Define Young symmetrizer by $$ c_{T} = R(T)^+ C(T)^-. $$ Young symmetrizer $c_T$ generates minimal left ideal $Ac_T \cong V_\lambda$, i.e. irreducible representation of $S_n$. Denote by $P(x) := (n!)^{-1}\sum_{g \in S_n} g x g^{-1}$ the averaging by conjugation, which is a projection to the center of group algebra $P: A \to Z(A)$. It is known, that $$ P(c_T) = \frac{1}{\dim V_\lambda} \chi_\lambda $$

Question. What is the image of projector $P$ on $x_T := R(T)^+ C(T)^+$ or $y_T = R(T)^- C(T)^-$? Is it true that it would be the scaled version of character of the representation $Ax_T$? If so, what are the multiplicities of that representation?

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    $\begingroup$ Let $n = 3$ and $\lambda = \left(2,1\right)$. Let $T$ be any tableau of shape $\lambda$ with entries $1,2,3$. (It does not matter which tableau we choose, since $P\left(c_T\right)$ depends only on $\lambda$.) Then, $P\left(x_T\right) = 6 + 4t + 3c$, where $t$ is the sum of all three transpositions, and $c$ is the sum of both $3$-cycles. Meanwhile, ... $\endgroup$
    – darij grinberg
    Commented Jul 13 at 23:35
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    $\begingroup$ ... the representation $Ax_T$ has dimension $3$. Thus, if its character was a scalar multiple of $P\left(x_T\right)$, then it would have to be $\dfrac12 P\left(x_T\right) = 3 + 2t + \dfrac32 c$. But this is impossible, since the character of a representation must have integer entries. $\endgroup$
    – darij grinberg
    Commented Jul 13 at 23:37
  • $\begingroup$ @darijgrinberg thank you for the reply. I see, the character should have a dimension coefficient at the identity permutation. Probably, this property only holds for idempotents. $\endgroup$
    – Alimzhan
    Commented Jul 15 at 11:53
  • $\begingroup$ Yeah, looks like this. And if $T$ is a skew (rather than straight) standard tableau (e.g., of shape $(3,2)/(1)$), then $P(c_T)$ is not a scalar multiple of a character either (since the dimension of the Specht module usually does not divide $n!$). $\endgroup$
    – darij grinberg
    Commented Jul 15 at 15:36

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