This is probably elementary for experts on the representation theory of the symmetric group, but I did not find the answers I need by a cursory look at the usual textbooks (they could be there, but I gave up trying to decipher conflicting notations and conventions).

Let $\lambda$ be an integer partition of $n$. A Young tableau $T$ is a bijective filling of the corresponding Young diagram with the numbers $1,2,\ldots,n$. For a permutation $\sigma$, let $\sigma T$ denote the tableau obtained by replacing each entry $i$ by $\sigma(i)$. Standard tableaux are the ones where entries increase in each row and column. For a Young tableau $T$, let $C(T)$ denote the group of permutations which preserve the columns of $T$, and let $R(T)$ the group of permutations which preserve the rows of $T$. In the group algebra $\mathbb{C}\mathfrak{S}_n$ of the symmetric group define, as usual, the elements $$ P(T)=\sum_{\sigma\in R(T)} \sigma $$ and $$ N(T)=\sum_{\sigma\in C(T)} {\rm sgn(\sigma)}\ \sigma\ . $$ Finally, the convention for Young symmetrizer that I will use is $$ Y(T)=P(T)N(T)\ . $$

Q1: Is it always true that for two different standard Young tableaux $T,T'$, of the same shape $\lambda$, we have $Y(T)Y(T')=0$?

Q2: Let $T$ be a standard Young tableau and let $\alpha\in C(T)$, $\beta\in R(T)$ be such that $\alpha\beta T$ is also standard. Does this necessarily require $\alpha=\beta=Id$?


1 Answer 1


For Q1 the answer in general is no. Young symmetrizers can be used to give a decomposition of $\mathbb C[S_n]$ into a direct sum of minimal left ideals but in general they are not pairwise orthogonal. One can actually characterize precisely when $Y(T)Y(T')\neq 0$ holds: (i) the underlying shape of $T$ and $T'$ needs to be the same (ii) every row of $T$ must intersect every column of $T'$ in at most one element.

So an explicit example where they fail to be orthogonal is $$T=\begin{matrix} 1 & 3 & 5 \\ 2 & 4 & \\ \end{matrix} \qquad, \qquad T'=\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & \\ \end{matrix}$$

See Orthogonal sets of Young symmetrizers by Stembridge for more details.

For Q2 the answer is no. It is possible for $\alpha\beta T$ to be a different standard Young tableaux. For example you can take $$T=\begin{matrix} 1 & 2 & \\ 3 & 4 & \\ 5 & & \\ \end{matrix}$$ and also $\alpha=(24)(35)$, $\beta=(34)$.

  • 4
    $\begingroup$ Thank you. Great answer! In fact, the two questions are equivalent, i.e., $Y(T)Y(T')\neq 0$ iff $\exists \alpha\in C(T),\exists \beta\in R(T), T'=\alpha\beta T$. $\endgroup$ Aug 21, 2020 at 21:14

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