# A basic question about Young symmetrizers

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 $$\alpha=\beta=Id$$?

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 $$T=\begin{matrix} 1 & 3 & 5 \\ 2 & 4 & \\ \end{matrix} \qquad, \qquad T'=\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & \\ \end{matrix}$$
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 $$\alpha=(24)(35)$$, $$\beta=(34)$$.
• Thank you. Great answer! In fact, the two questions are equivalent, i.e., $Y(T)Y(T')\neq 0$ iff $\exists \alpha\in C(T),\exists \beta\in R(T), T'=\alpha\beta T$. Aug 21, 2020 at 21:14