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 Sn 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 Sn 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


 * $$b_\lambda=\sum_{g\in Q_\lambda} \sgn(g) e_g$$

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


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

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 Sn act on this tensor product space by permuting the indices. One then has a natural group algebra representation $$\C S_n \to \operatorname{End} (V^{\otimes n})$$ on $$V^{\otimes n}$$ (i.e. $$V^{\otimes n}$$ is a right $$\C S_n$$ 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 $$\C S_nc_\lambda$$ of $$c_\lambda = a_\lambda \cdot b_\lambda$$ in $$\C S_n$$ is an irreducible representation of Sn, 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 $$c^2_\lambda = \alpha_\lambda c_\lambda$$ for some rational number $$\alpha_\lambda\in\Q.$$ 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 $$\Q S_n$$.

Consider, for example, S3 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).