Young Symmetrizer - Construction

Construction

Let V be any vector space over the complex numbers. Consider then the tensor product vector space (n times). Let Sn act on this tensor product space by permuting the indices. One then has a natural group algebra representation on .

Given a partition λ of n, so that, then the image of is

\text{Im}(a_\lambda) := a_\lambda V^{\otimes n} \cong \text{Sym}^{\lambda_1}\; V \otimes \text{Sym}^{\lambda_2}\; V \otimes \cdots \otimes \text{Sym}^{\lambda_j}\; V.

For instance, if, and, with the canonical Young tableau . Then the corresponding is given by . Let an element in be given by . Then

The latter clearly span .

The image of is

\text{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, and are the symmetric and alternating tensor product spaces.

The image of in is an irreducible representation of Sn, called a Specht module. We write

for the irreducible representation.

Some scalar multiple of is idempotent, that is for some rational number . Specifically, one finds . In particular, this implies that representations of the symmetric group can be defined over the rational numbers; that is, over the rational group algebra .

Consider, for example, S3 and the partition (2,1). Then one has

If V is a complex vector space, then the images of on spaces provides essentially all the finite-dimensional irreducible representations of GL(V).

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