[[Representation theory of finite symmetric groups]]
# Young operator
Let $\Theta^p_{\lambda}$ be a [[Young tableau]] with $n$ boxes,
$H_{\lambda}^p$ be the subgroup of row permutations,
and $V^p_{\lambda}$ be the subgroup of column permutations.
The **row symmetrizer** $\mathfrak{s}_{\lambda}^p \in \mathbb{C}[S_{n}]$ is given by
$$
\begin{align*}
\mathfrak{s}_{\lambda}^p = \sum_{h \in H_{\lambda}^p} \delta_{h}
\end{align*}
$$
and the **column antisymmetrizer** $\mathfrak{a}_{\lambda}^p \in \mathbb{C}[S_{n}]$ by
$$
\begin{align*}
\mathfrak{a}_{\lambda}^p = \sum_{v \in V_{\lambda}^p} \sgn(v) \delta_{v}
\end{align*}
$$
then the **Young operator** is given by #m/def/rep/sym
$$
\begin{align*}
\mathfrak{e}_{\lambda}^p = \mathfrak{s}_{\lambda}^p \mathfrak{a}_{\lambda}^p
= \sum_{v \in V_{\lambda}^p} \sum_{h \in H_{\lambda}^p} \sgn(v) \delta_{hv}
\end{align*}
$$
> [!note]- Notation
> The partition subscripts will typically be written out as a sum for the purpose of these notes, e.g. $\mathfrak{e}_{3+2+2}^{(12)} = \delta_{(12)} * \mathfrak{e}_{3+2+2} * \delta_{(12)}$.
> In the literature, it is common to use an entire [[Young diagram]] as a subscript.
A practical way to do pen-and-paper calculations is with a [[Birdtrack notation]].
If single-box symmetrizers and antisymmetrizers are drawn, each line passes through exactly one symmetrizer and exactly one antisymmetrizer.
Each (anti)symmetrizer corresponds to a different row (column), with the number of lines passing through given by the number of boxes therein.
## Properties
1. $H^p_{\lambda} = pH_{\lambda}p^{-1}$ and $V^p_{\lambda} = pV_{\lambda}p^{-1}$ are subgroups of $S_{n}$ with $H_{\lambda}^p \cap V_{\lambda}^p = \{ e \}$.
Thus $\mathfrak{e}_{\lambda}^p = \delta_{p} * \mathfrak{e}_{\lambda} * \delta_{p^{-1}}$.
2. $\mathfrak{s}_{\lambda}^p$ and $\mathfrak{a}_{\lambda}^p$ are total [[Symmetrizer and antisymmetrizer elements]] for the subgroups $H_{\lambda}^p$ and $V_{\lambda}^p$.
3. $\mathfrak{s}^p_{\lambda}$ and $\mathfrak{a}^p_{\lambda}$ are essentially idempotent but in general not primitive.
4. The young operators $\mathfrak{e}_{\lambda}^p$ are **essentially idempotent** and primitive.
5. The irreps generated by $\mathfrak{e}_{\lambda}^p$ and $\mathfrak{e}_{\mu}^q$ are equivalent iff $\lambda = \mu$, regardless of $p$ and $q$.
Thus, the young operators for standard tableaux generate [[Ideal of the complex group ring|minimal left ideals]] for every non-equivalent irrep. #m/thm/rep/sym
> [!check]- Proof
> The proof is most intuitive with birdtrack arguments.
>
> For `4.`, we expands the convolution as follows
> $$
> \begin{align*}
> \mathfrak{e}_{\lambda}^p \mathfrak{e}_{\lambda}^p
> &= \mathfrak{a}_{\lambda}^p \mathfrak{s}_{\lambda}^p \mathfrak{a}_{\lambda}^p \mathfrak{s}_{\lambda}^p
> = \sum_{q \in S_{n}} [\mathfrak{s}^p_{\lambda} \mathfrak{a}_{\lambda}^p](q)\, \mathfrak{a}_{\lambda}^p \delta_{q} \mathfrak{s}_{\lambda}^p
> \end{align*}
> $$
> for the term $q = e$ we get $\mathfrak{e}_{\lambda}^p$, and for all other terms either
>
> 1. $q$ produces a zero connection (i.e. two lines intersect the same symmetrizer and antisymmetrizer)
> 2. $q$ switches lines connected to the same symmetrizer, giving $\mathfrak{e}^p_{\lambda}$
> 3. $q$ switches two lines connected to the same antisymmetrizer, giving $\mathfrak{-e}_{\lambda}^p$
> 4. A combination of `2.` and `3.` gives at most a sign change
>
> hence $\mathfrak{e}^p_{\lambda}\mathfrak{e}^p_{\lambda} = \eta_{\lambda}\mathfrak{e}_{\lambda}^p$, but $\eta_{\lambda} \neq 0$ since $\mathfrak{e}_{\lambda}^p(e) \neq 0$ and thus $\Tr \Rho(\mathfrak{e}_{\lambda}^p)\neq 0$.
> Hence $\mathfrak{e}_{\lambda}^p$ is essentially idempotent.
> Moreover the above argument already demonstrates the [[Idempotent primitivity criterion]], hence $\frac{1}{\eta_{\lambda}} \mathfrak{e}_{\lambda}^p$ is primitive.
>
> For `5.` we will show that if $\lambda \neq \mu$ then $\mathfrak{e}_{\lambda} q \mathfrak{e}_{\mu} = 0$ for all $q \in \mathbb{C}[G]$,
> i.e. the [[Equivalence of irreps on left ideals criterion]].
> Consider $\mathfrak{e}_{\lambda} = \mathfrak{s}_{\lambda} \mathfrak{a}_{\lambda}$ and $\mathfrak{e}_{\mu} = \mathfrak{s}_{\mu}\mathfrak{e}_{\mu}$ in birdtracks, with the symmetrizers and antisymmetrizers ordered from top to bottom from longest to shortest.
> Then the first symmetrizer of $\mathfrak{s}_{\lambda}$ has $\lambda_{1}$ lines,
> each of which must enter a different one of $\mathfrak{a}_{\mu}$'s $\mu_{1}$ antisymmetrizers if $\mathfrak{s}_{\lambda} \delta_{p} \mathfrak{a}_{\mu} \neq 0$,
> which by the [[Pigeonhole principle]] is impossible if $\lambda_{1} > \mu_{1}$.
> If $\lambda_{1} = \mu_{1}$, only antisymmetrizers of $\mathfrak{a}_{\mu}$ with at least two lines are available, of which there are $\mu_{2}$.
> For a nonzero connection, each of the second symmetrizers $\lambda_{2}$ lines must connect to a different one of these,
> which is impossible if $\lambda_{2} > \mu_{2}$.
> Continuing this argument we see a nonzero connection of $\mathfrak{s}_{\lambda}$ to $\mathfrak{a}_{\mu}$ is impossible if $\lambda > \mu$.
> The same goes for connecting $\mathfrak{a}_{\lambda}$ to $\mathfrak{s}_{\mu}$ if $\mu > \lambda$.
> Thus by lineärity, if $\lambda \neq \mu$ then $\mathfrak{e}_{\lambda} q \mathfrak{e}_{\mu} = 0$ for all $q \in \mathbb{C}[G]$.
>
> Now since $\mathfrak{e}_{\lambda}^p = \delta_{p} \mathfrak{e}_{\lambda} \delta_{p^{-1}}$, it follows that $\mathfrak{e}_{\lambda}^p \delta_{p} \mathfrak{e}_{\lambda} = \delta_{p} \mathfrak{e}_{\lambda} \mathfrak{e}_{\lambda} = \delta_{p}\eta_{\lambda}\mathfrak{e}_{\lambda} \neq 0$,
> hence we have equivalence by the [[Equivalence of irreps on left ideals criterion]].
> <span class="QED"/>
#
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