I have a question regarding the equivalence of two definitions of the young symmetrizer. First, some notation: let $\lambda$ be a partition of $n$. Given a $\lambda$-tableau $T$ (that is, a tableau of shape $\lambda$ filled with the entries $1,2,\ldots,n$), we define the row stabilizer of $T$ by $R(T)=S_{r_1} \times \dots \times S_{r_l}$ where $r_1, \dots, r_l$ are the rows of $T$, and analogously, we define the column stabilizer of $T$ as $C(T)=S_{c_1} \times \dots \times S_{c_k}$, where $c_1, \dots, c_k$ are the columns of $T$. (We regard both $R(T)$ and $C(T)$ as subgroups of the symmetric group $S_n$.) Let $a_T=\sum_{\sigma \in R(T)} \sigma $ and $b_T=\sum_{\tau \in C(T)} \mbox{sgn}(\tau) \tau $ be two elements of the group algebra $\mathbb C\left[S_n\right]$.

Now, for my question: some references I've seen define $c_T = a_T b_T$ as the Young symmetrizer. The important thing for my purposes here is that this symmetrizer corresponds to an irreducible representation of $S_n$ indexed by $\lambda$. However, I've seen other sources define the Young symmetrizer $b_T a_T$. The latter definition is more useful in something I'm working on, so I wanted to verify that this is correct. Does the second definition still correspond in the same way to the same irreducible representation of $S_n$? All of my intuition tells me it should, and I've tried out several examples with a computer algebra system, but I just want to be sure.

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    $\begingroup$ To be precise, you are asking if $\mathbb{C}[S_n]a_Tb_T\cong \mathbb{C}[S_n]b_Ta_T$ as left $\mathbb{C}[S_n]$-modules? $\endgroup$ – Aaron Aug 19 '13 at 14:15
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    $\begingroup$ @Aaron Exactly- thanks. I'm afraid I lacked the technical knowledge to define this precisely. $\endgroup$ – Devlin Mallory Aug 19 '13 at 14:21

Let $A = \Bbb{C}[S_n]$, we wish to prove that $Aa_Tb_T \cong Ab_Ta_T$ as left $A$ - modules. To do this we need the crucial fact that $a_Tb_Ta_Tb_T = n_T (a_Tb_T)$ for some constant $n_T \neq 0$. This comes upon knowing that $Aa_Tb_T$ is an irreducible representation of $S_n$ and using Schur's Lemma. Now define maps

$$f : Aa_Tb_T \stackrel{\cdot \frac{a_T}{\sqrt{n_T}}}{\longrightarrow} Ab_Ta_T \hspace{5mm} \text{and}\hspace{5mm} g: Ab_Ta_T \stackrel{\cdot \frac{b_T}{\sqrt{n_T}}}{\longrightarrow} Aa_Tb_T$$

which are simply just right multiplication by $a_T/\sqrt{n_T}$ and $b_T/\sqrt{n_T}$ respectively. Note that division by $n_T$ is ok because we are just dividing by a number. Then for any $x \in Aa_Tb_T$, write $x = y a_Tb_T$ for some $y \in A$. Then $$g(f(x)) = \frac{ya_Tb_Ta_Tb_T}{n_T} = \frac{n_T \cdot (ya_Tb_T)}{n_T} = x$$ and similarly for any $z \in Ab_Ta_T$, we find $f(g(z)) = z$. It follows that $f$ and $g$ define mutual left $A$ - module inverses and so $Aa_Tb_T \cong Ab_Ta_T$.

  • $\begingroup$ Thanks, this is very helpful and clear. To ask about one point: the reason that $a_T b_T$ is quasi-idempotent follows from Schur's Lemma because left-multiplication by $a_T b_T$ is an $S_n$-isomorphism and hence either zero or the identity, up to a scalar. Is this about correct? $\endgroup$ – Devlin Mallory Aug 19 '13 at 14:26
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    $\begingroup$ Yes that is absolutely spot on. When I learnt about this stuff in the past I was confused too. Fulton and Harris used one definition, and Fulton's Young Tableaux used another! $\endgroup$ – user38268 Aug 19 '13 at 14:28

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