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

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

and

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

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

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

and

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

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

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

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 ${\displaystyle V^{\otimes n}=V\otimes V\otimes \cdots \otimes V}$ (n times). Let S_n act on this tensor product space by permuting the indices. One then has a natural group algebra representation ${\displaystyle \mathbb {C} S_{n}\to \operatorname {End} (V^{\otimes n})}$ on ${\displaystyle V^{\otimes n}}$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The image of ${\displaystyle b_{\lambda }}$ is

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

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

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

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

for the irreducible representation.

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

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

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

If V is a complex vector space, then the images of ${\displaystyle c_{\lambda }}$ on spaces ${\displaystyle V^{\otimes d}}$ 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). Representation theory. A first course. Graduate Texts in Mathematics, Readings in Mathematics. Vol. 129. New York: Springer-Verlag. doi:10.1007/978-1-4612-0979-9. ISBN 978-0-387-97495-8. MR 1153249. OCLC 246650103.
• Bruce E. Sagan. The Symmetric Group. Springer, 2001.