TITLE: A basic question about Young symmetrizers
QUESTION [5 upvotes]: 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 $\alpha=\beta=Id$?

REPLY [8 votes]: 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 $\alpha=(24)(35)$, $\beta=(34)$.