Math 202C: Lecture 13

*** Problems due May 3 at 23:59 ***

Let $H \leq G$ be finite groups. In Lecture 12, we proved that restriction of irreducible representations of $G$ to $H$ is multiplicity-free if and only if the centralizer $Z(\mathcal{C}(H),\mathcal{C}(G))$ is a commutative subalgebra of $\mathcal{C}(G).$ Our objective now is to use this criterion to establish that restriction of irreducible representations of $S_n$ to $S_{n-1}$ is multiplicity free.

For the moment we stay in the general context where the group $G$ and its subgroup $H$ are arbitrary; eventually we will set $G=S_n$ and $H=S_{n-1}.$ By definition, the centralizer $Z(\mathcal{C}(H),\mathcal{C}(G))$ is the set of all $|A\rangle \in \mathcal{C}(G)$ which commute with every $|B\rangle \in \mathcal{C}(H),$ and it is clear that this condition is equivalent to

$|A\rangle|h\rangle = |h\rangle|A\rangle, \quad h \in H,$

which is in turn equivalent to

$|A\rangle = |h\rangle|A\rangle|h\rangle^*, \quad h \in H,$

with $|h\rangle^*=|h^{-1}\rangle.$ To understand those elements $|A\rangle \in \mathcal{C}(H)$ which have the above property, consider an equivalence relation on $G$ defined by

$g_1 \sim g_2 \iff g_2=hg_1h^{-1} \text{ for some} h \in H.$

That is, equivalence classes under this relation are orbits of $H$ acting on $G$ by conjugation. Let $\Lambda$ be a set parameterizing the distinct $H$ -conjugacy classes in $G$ ,

$\{H-\text{conjugacy classes in }G\} = \{C_\alpha \colon \alpha \in \Lambda\}.$

In the case where $H=G,$ this is just the set of conjugacy classes in $G$ as per usual, making the following a generalization of something we already know.

Problem 13.1. Prove that the set $\{|C_\alpha\rangle \colon \alpha \in \Gamma\}$ is an orthogonal basis of $Z(\mathcal{C}(H),\mathcal{C}(G)).$

In the case $H=G,$ the centralizer $Z(\mathcal{C}(H),\mathcal{C}(G))$ is the center $Z(\mathcal{C}(G)),$ which is commutative by definition, and the above problem recovers the 202B fact that the center of the convolution algebra of a group has a basis consisting of indicator functions of conjugacy classes. At the other extreme, if $H=\{\iota\}$ is the subgroup consisting solely of the identity element, then $Z(\mathcal{C}(H),\mathcal{C}(G))=\mathcal{C}(G),$ which is commutative precisely when $G$ is abelian. In between these two extremes, it may be more difficult to determine whether or not $Z(\mathcal{C}(H),\mathcal{C}(G))$ is commutative, but there is a particular circumstance which will force this to be the case.

Problem 13.2. Prove that if the class basis $\{|C_\alpha\rangle \colon \alpha \in \Lambda\}$ consists of selfadjoint elements, then $Z(\mathcal{C}(H),\mathcal{C}(G))$ is a commutative algebra.

We now specialize to the case $G=S_n$ and $H=S_{n-1}.$ We would like to find a concrete indexing of the $S_{n-1}$ -conjugacy classes in $S_n$ . For permutations, the operation of conjugation has a simple combinatorial meaning: conjugating $g$ by $h$ relabels the elements in the cycles of $g$ according to $h$ . If $h$ is confined to $S_{n-1},$ then it cannot relabel $n$ , hence in order for two permutations $g_1$ and $g_2$ in $S_n$ to be $S_{n-1}$ -conjugates it is necessary that the cycle containing $n$ in each have the same length.

Problem 13.3. Prove that the above condition is also sufficient for $g_1$ and $g_2$ to be $S_{n-1}$ -conjugates.

We can now conclude that $S_{n-1}$ -conjugacy classes in $S_n$ are indexed by pairs $(r,\alpha)$ consisting of an integer $1 \leq r \leq n$ corresponding to the length of the cycle containing $n$ and a partition $\alpha \vdash n-r$ consisting to the cycle-type of the remainder of the permutation. In particular, we get the dimension formula

$\dim Z(\mathcal{C}(S_{n-1},\mathcal{C}(S_n))=\sum\limits_{r=1}^n p(n-r),$

where $p(k)$ is the number of partitions of $k$ . Furthermore, it is easy to see that if an $S_{n-1}$ -conjugacy class $C_{(r,\alpha)}$ contains a permutation $\pi$ , then it also contains $\pi^{-1}$ , since the inverse of a permutation not only has the same cycle type, but actually the same elements within each cycle. Thus, the centralizer $Z(\mathcal{C}(S_{n-1}),\mathcal{C}(S_n))$ is commutative by Problem 13.2.

The significance of this commutativity is that, by Lecture 12, it is equivalent to multiplicity-free branching for irreducible representations of the symmetric groups: the isotypic decomposition of any irreducible representation $V^\lambda$ of $S_n$ when viewed as a representation of $S_{n-1}$ has the form

$V^\lambda = \bigoplus_{\mu \vdash n-1} \mathrm{Mult}(V^\mu,V^\lambda)V^\mu, \quad \mathrm{Mult}(V^\mu,V^\lambda) \in \{0,1\}.$

In Lecture 14, we will use this fact to construct a special orthonormal basis in $V^\lambda$ which we will then exploit mercilessly.

Math 202C: Lecture 13

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