Young symmetrizer
# Young symmetrizer

In mathematics, a Young symmetrizer is an element of the group algebra of the symmetric group, constructed in such a way that, for the homomorphism from the group algebra to the endomorphisms of a vector space $V^{\otimes n}$ obtained from the action of $S_{n}$ on $V^{\otimes n}$ by permutation of indices, the image of the endomorphism determined by that element corresponds to an irreducible representation of the symmetric group over the complex numbers. A similar construction works over any field, and the resulting representations are called Specht modules. The Young symmetrizer is named after British mathematician Alfred Young.

## Definition

Given a finite symmetric group $S_n$ and specific Young tableau λ corresponding to a numbered partition of n, and consider the action of $S_{n}$ given by permuting the boxes of $\lambda$. Define two permutation subgroups $P_{\lambda }$ and $Q_{\lambda }$ of $S_n$ as follows:

    $P_{\lambda }=\{g\in S_{n}:g{\text{ preserves each row of }}\lambda \}$

and

    $Q_{\lambda }=\{g\in S_{n}:g{\text{ preserves each column of }}\lambda \}.$

Corresponding to these two subgroups, define two vectors in the group algebra $\mathbb {C} S_{n}$ as

    $a_{\lambda }=\sum _{g\in P_{\lambda }}e_{g}$

and

    $b_{\lambda }=\sum _{g\in Q_{\lambda }}\operatorname {sgn}(g)e_{g}$

where $e_{g}$ is the unit vector corresponding to g, and $\operatorname {sgn}(g)$ is the sign of the permutation. The product

    $c_{\lambda }:=a_{\lambda }b_{\lambda }=\sum _{g\in P_{\lambda },h\in Q_{\lambda }}\operatorname {sgn}(h)e_{gh}$

is the Young symmetrizer corresponding to the Young tableau λ. Each Young symmetrizer corresponds to an irreducible representation of the symmetric group, and every irreducible representation can be obtained from a corresponding Young symmetrizer. (If we replace the complex numbers by more general fields the corresponding representations will not be irreducible in general.)

## Construction

Let V be any vector space over the complex numbers. Consider then the tensor product vector space $V^{\otimes n}=V\otimes V\otimes \cdots \otimes V$ (n times). Let $S_n$ act on this tensor product space by permuting the indices. One then has a natural group algebra representation $\mathbb {C} S_{n}\to \operatorname {End} (V^{\otimes n})$ on $V^{\otimes n}$ (i.e. $V^{\otimes n}$ is a right $\mathbb {C} S_{n}$ module).

Given a partition λ of n, so that $n=\lambda _{1}+\lambda _{2}+\cdots +\lambda _{j}$, then the image of $a_{\lambda }$ is

    $\operatorname {Im} (a_{\lambda }):=V^{\otimes n}a_{\lambda }\cong \operatorname {Sym} ^{\lambda _{1}}V\otimes \operatorname {Sym} ^{\lambda _{2}}V\otimes \cdots \otimes \operatorname {Sym} ^{\lambda _{j}}V.$

For instance, if $n=4$, and $\lambda =(2,2)$, with the canonical Young tableau $\{\{1,2\},\{3,4\}\}$. Then the corresponding $a_{\lambda }$ is given by

    $a_{\lambda }=e_{\text{id}}+e_{(1,2)}+e_{(3,4)}+e_{(1,2)(3,4)}.$

For any product vector $v_{1,2,3,4}:=v_{1}\otimes v_{2}\otimes v_{3}\otimes v_{4}$ of $V^{\otimes 4}$ we then have

    $v_{1,2,3,4}a_{\lambda }=v_{1,2,3,4}+v_{2,1,3,4}+v_{1,2,4,3}+v_{2,1,4,3}=(v_{1}\otimes v_{2}+v_{2}\otimes v_{1})\otimes (v_{3}\otimes v_{4}+v_{4}\otimes v_{3}).$

Thus the set of all $a_{\lambda }v_{1,2,3,4}$ clearly spans $\operatorname {Sym} ^{2}V\otimes \operatorname {Sym} ^{2}V$ and since the $v_{1,2,3,4}$ span $V^{\otimes 4}$ we obtain $V^{\otimes 4}a_{\lambda }=\operatorname {Sym} ^{2}V\otimes \operatorname {Sym} ^{2}V$, where we wrote informally $V^{\otimes 4}a_{\lambda }\equiv \operatorname {Im} (a_{\lambda })$.

Notice also how this construction can be reduced to the construction for $n=2$. Let $\mathbb {1} \in \operatorname {End} (V^{\otimes 2})$ be the identity operator and $S\in \operatorname {End} (V^{\otimes 2})$ the swap operator defined by $S(v\otimes w)=w\otimes v$, thus $\mathbb {1} =e_{\text{id}}$ and $S=e_{(1,2)}$. We have that

    $e_{\text{id}}+e_{(1,2)}=\mathbb {1} +S$

maps into $\operatorname {Sym} ^{2}V$, more precisely

    ${\frac {1}{2}}(\mathbb {1} +S)$

is the projector onto $\operatorname {Sym} ^{2}V$. Then

    ${\frac {1}{4}}a_{\lambda }={\frac {1}{4}}(e_{\text{id}}+e_{(1,2)}+e_{(3,4)}+e_{(1,2)(3,4)})={\frac {1}{4}}(\mathbb {1} \otimes \mathbb {1} +S\otimes \mathbb {1} +\mathbb {1} \otimes S+S\otimes S)={\frac {1}{2}}(\mathbb {1} +S)\otimes {\frac {1}{2}}(\mathbb {1} +S)$

which is the projector onto $\operatorname {Sym} ^{2}V\otimes \operatorname {Sym} ^{2}V$.

The image of $b_{\lambda }$ is

    $\operatorname {Im} (b_{\lambda })\cong \bigwedge ^{\mu _{1}}V\otimes \bigwedge ^{\mu _{2}}V\otimes \cdots \otimes \bigwedge ^{\mu _{k}}V$

where μ is the conjugate partition to λ. Here, $\operatorname {Sym} ^{i}V$ and $\bigwedge ^{j}V$ are the symmetric and alternating tensor product spaces.

The image $\mathbb {C} S_{n}c_{\lambda }$ of $c_{\lambda }=a_{\lambda }\cdot b_{\lambda }$ in $\mathbb {C} S_{n}$ is an irreducible representation of $S_n,$ called a Specht module. We write

    $\operatorname {Im} (c_{\lambda })=V_{\lambda }$

for the irreducible representation.

Some scalar multiple of $c_{\lambda }$ is idempotent, that is $c_{\lambda }^{2}=\alpha _{\lambda }c_{\lambda }$ for some rational number $\alpha _{\lambda }\in \mathbb {Q} .$ Specifically, one finds $\alpha _{\lambda }=n!/\dim V_{\lambda }$. In particular, this implies that representations of the symmetric group can be defined over the rational numbers; that is, over the rational group algebra $\mathbb {Q} S_{n}$.

Consider, for example, $S_3$ and the partition (2,1). Then one has

    $c_{(2,1)}=e_{123}+e_{213}-e_{321}-e_{312}.$

If V is a complex vector space, then the images of $c_{\lambda }$ on spaces $V^{\otimes d}$ provides essentially all the finite-dimensional irreducible representations of GL(V).

