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 [math]\displaystyle{ V^{\otimes n} }[/math] obtained from the action of [math]\displaystyle{ S_n }[/math] on [math]\displaystyle{ V^{\otimes n} }[/math] 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 [math]\displaystyle{ S_n }[/math] given by permuting the boxes of [math]\displaystyle{ \lambda }[/math]. Define two permutation subgroups [math]\displaystyle{ P_\lambda }[/math] and [math]\displaystyle{ Q_\lambda }[/math] of S_n as follows:

[math]\displaystyle{ P_\lambda=\{ g\in S_n : g \text{ preserves each row of } \lambda \} }[/math]

and

[math]\displaystyle{ Q_\lambda=\{ g\in S_n : g \text{ preserves each column of } \lambda \}. }[/math]

Corresponding to these two subgroups, define two vectors in the group algebra [math]\displaystyle{ \mathbb{C}S_n }[/math] as

[math]\displaystyle{ a_\lambda=\sum_{g\in P_\lambda} e_g }[/math]

and

[math]\displaystyle{ b_\lambda=\sum_{g\in Q_\lambda} \sgn(g) e_g }[/math]

where [math]\displaystyle{ e_g }[/math] is the unit vector corresponding to g, and [math]\displaystyle{ \sgn(g) }[/math] is the sign of the permutation. The product

[math]\displaystyle{ c_\lambda := a_\lambda b_\lambda = \sum_{g\in P_\lambda,h\in Q_\lambda} \sgn(h) e_{gh} }[/math]

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 [math]\displaystyle{ V^{\otimes n}=V \otimes V \otimes \cdots \otimes V }[/math] (n times). Let S_n act on this tensor product space by permuting the indices. One then has a natural group algebra representation [math]\displaystyle{ \C S_n \to \operatorname{End} (V^{\otimes n}) }[/math] on [math]\displaystyle{ V^{\otimes n} }[/math] (i.e. [math]\displaystyle{ V^{\otimes n} }[/math] is a right [math]\displaystyle{ \C S_n }[/math] module).

Given a partition λ of n, so that [math]\displaystyle{ n=\lambda_1+\lambda_2+ \cdots +\lambda_j }[/math], then the image of [math]\displaystyle{ a_\lambda }[/math] is

[math]\displaystyle{ \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. }[/math]

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

[math]\displaystyle{ a_\lambda = e_{\text{id}} + e_{(1,2)} + e_{(3,4)} + e_{(1,2)(3,4)}. }[/math]

For any product vector [math]\displaystyle{ v_{1,2,3,4}:=v_1 \otimes v_2 \otimes v_3 \otimes v_4 }[/math] of [math]\displaystyle{ V^{\otimes 4} }[/math] we then have

[math]\displaystyle{ 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). }[/math]

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

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

[math]\displaystyle{ e_{\text{id}} + e_{(1,2)} = \mathbb{1} + S }[/math]

maps into [math]\displaystyle{ \operatorname{Sym}^2 V }[/math], more precisely

[math]\displaystyle{ \frac{1}{2}(\mathbb{1} + S) }[/math]

is the projector onto [math]\displaystyle{ \operatorname{Sym}^2 V }[/math]. Then

[math]\displaystyle{ \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) }[/math]

which is the projector onto [math]\displaystyle{ \operatorname{Sym}^2 V\otimes \operatorname{Sym}^2 V }[/math].

The image of [math]\displaystyle{ b_\lambda }[/math] is

[math]\displaystyle{ \operatorname{Im}(b_\lambda) \cong \bigwedge^{\mu_1} V \otimes \bigwedge^{\mu_2} V \otimes \cdots \otimes \bigwedge^{\mu_k} V }[/math]

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

The image [math]\displaystyle{ \C S_nc_\lambda }[/math] of [math]\displaystyle{ c_\lambda = a_\lambda \cdot b_\lambda }[/math] in [math]\displaystyle{ \C S_n }[/math] is an irreducible representation of S_n, called a Specht module. We write

[math]\displaystyle{ \operatorname{Im}(c_\lambda) = V_\lambda }[/math]

for the irreducible representation.

Some scalar multiple of [math]\displaystyle{ c_\lambda }[/math] is idempotent,^[1] that is [math]\displaystyle{ c^2_\lambda = \alpha_\lambda c_\lambda }[/math] for some rational number [math]\displaystyle{ \alpha_\lambda\in\Q. }[/math] Specifically, one finds [math]\displaystyle{ \alpha_\lambda=n! / \dim V_\lambda }[/math]. In particular, this implies that representations of the symmetric group can be defined over the rational numbers; that is, over the rational group algebra [math]\displaystyle{ \Q S_n }[/math].

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

[math]\displaystyle{ c_{(2,1)} = e_{123}+e_{213}-e_{321}-e_{312}. }[/math]

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

See also

• Representation theory of the symmetric group

Notes

References

• William Fulton. Young Tableaux, with Applications to Representation Theory and Geometry. Cambridge University Press, 1997.
• Lecture 4 of Fulton, William; Harris, Joe (1991) (in en-gb). Representation theory. A first course. Graduate Texts in Mathematics, Readings in Mathematics. 129. New York: Springer-Verlag. doi:10.1007/978-1-4612-0979-9. ISBN 978-0-387-97495-8. OCLC 246650103. https://link.springer.com/10.1007/978-1-4612-0979-9. 
• Bruce E. Sagan. The Symmetric Group. Springer, 2001.

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