9
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @Aaron Exactly- thanks. I'm afraid I lacked the technical knowledge to define this precisely. $\endgroup$ – Devlin Mallory Aug 19 '13 at 14:21
7
$\begingroup$

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 $Aa_Tb_T \cong Ab_Ta_T$.

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.