Lecture 17

Problem 89. Decompose (π,ℳ(3,2))(\pi,{\mathcal{M}}^{(3,2)}) into irreducible representations of S5S_{5}.

Solution: We first note that by (17.7), ℳ(3,2){\mathcal{M}}^{(3,2)} has dimension 1010 with basis

\ytableausetup⁢t⁢a⁢b⁢l⁢o⁢i⁢d⁢s,s⁢m⁢a⁢l⁢l⁢t⁢a⁢b⁢l⁢e⁢a⁢u⁢x⁢\ytableaushort⁢123,45,\ytableaushort⁢124,35,\ytableaushort⁢125,34,\ytableaushort⁢134,25,\ytableaushort⁢135,24,\ytableaushort⁢145,23,\ytableaushort⁢234,15,\ytableaushort⁢235,14,\ytableaushort⁢245,13,\ytableaushort⁢345,12.\ytableausetup{tabloids,smalltableaux}\ytableaushort{123,45},\quad% \ytableaushort{124,35},\quad\ytableaushort{125,34},\quad\ytableaushort{134,25}% ,\quad\ytableaushort{135,24},\quad\ytableaushort{145,23},\quad\ytableaushort{2% 34,15},\quad\ytableaushort{235,14},\quad\ytableaushort{245,13},\quad% \ytableaushort{345,12}.

We then compute how many elements of the basis a representative of each conjugacy class fixes to find the character values of χπ\chi_{\pi}. These are:

e(12)(12)⁢(34)(123)(123)⁢(45)(1234)(12345)χπ10411100\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ \hline\cr\chi_{\pi}&10&4&1&1&1&0&0\end{array}

We already know that the trivial representation is a subrepresentation of π\pi, the sum of all tabloids is invariant under the action of S5S_{5}, so we can consider the character of the resulting quotient representation. By Problem 66(b) this is

e(12)(12)⁢(34)(123)(123)⁢(45)(1234)(12345)χπ−triv93000−1−1\begin{array}[h]{c|rrrrrrr}&e&(12)&(12)(34)&(123)&(123)(45)&(1234)&(12345)\\ \hline\cr\chi_{\pi}-\mathrm{triv}&9&3&0&0&0&-1&-1\end{array}

Now it is just a case of inspecting the character table of S5S_{5}, see Section 12.2, and determining which sum of irreducible characters gives the above values. Thus

π=triv⊕π′⊕ρ,\pi=\mathrm{triv}\oplus\pi^{\prime}\oplus\rho,

where π′\pi^{\prime} is the representation labelled π\pi in the character table in Section 12.2. We note that this is unique, up to reordering, by Theorem 5.12.

Problem 90. Show that (π,𝒮(n))(\pi,{\mathcal{S}}^{(n)}) is the trivial representation of SnS_{n} and that (π,𝒮(1,…,1))(\pi,{\mathcal{S}}^{(1,\ldots,1)}) is the sign representation.

Solution: There is only one tabloid of shape (n)(n), \ytableausetup⁢t⁢a⁢b⁢l⁢o⁢i⁢d⁢s,s⁢m⁢a⁢l⁢l⁢t⁢a⁢b⁢l⁢e⁢a⁢u⁢x⁢\ytableaushort⁢1⁢…⁢n\ytableausetup{tabloids,smalltableaux}\ytableaushort{1\ldots n}, so (π,ℳ(n))(\pi,{\mathcal{M}}^{(n)}) is irreducible thus (π,ℳ(n))=(π,𝒮(n))(\pi,{\mathcal{M}}^{(n)})=(\pi,{\mathcal{S}}^{(n)}). Each element of SnS_{n} acts as the identity on \ytableausetup⁢t⁢a⁢b⁢l⁢o⁢i⁢d⁢s,s⁢m⁢a⁢l⁢l⁢t⁢a⁢b⁢l⁢e⁢a⁢u⁢x⁢\ytableaushort⁢1⁢…⁢n\ytableausetup{tabloids,smalltableaux}\ytableaushort{1\ldots n} so this is the trivial representation.

Now note that for any tableau tt of shape (1,…,1)(1,\ldots,1) we have C⁢(t)=SnC(t)=S_{n}. Let tt be the standard tableau (11 to nn in order going top to bottom), then

𝐞t=∑σ∈Snsgn⁢(σ)⁢σ⋅\ytableausetup⁢n⁢o⁢t⁢a⁢b⁢l⁢o⁢i⁢d⁢s,s⁢m⁢a⁢l⁢l⁢t⁢a⁢b⁢l⁢e⁢a⁢u⁢x⁢\ytableaushort⁢1,⋅,⋅,⋅,n.{\bf e}_{t}=\sum_{\sigma\in S_{n}}\mathrm{sgn}(\sigma)\sigma\cdot% \ytableausetup{notabloids,smalltableaux}\ytableaushort{1,\cdot,\cdot,\cdot,n}.

Note that for any other tableaux t′t^{\prime} of the same shape we have 𝐞t′=±𝐞t{\bf e}_{t^{\prime}}=\pm{\bf e}_{t}, where sign is positive if t′t^{\prime} differs from tt by an even permutation and negative if they differ by an odd permutation. Thus (π,𝒮(n))(\pi,{\mathcal{S}}^{(n)}) is 11-dimensional. Pick 𝐞t{\bf e}_{t} to be our basis element, where tt is the standard tableau, and we have that even permutations fix 𝐞t{\bf e}_{t} and odd ones send it to −𝐞t-{\bf e}_{t}. Thus we have the sign representation.

Problem 91. Show that (π,ℳ(n−1,1))(\pi,{\mathcal{M}}^{(n-1,1)}) is isomorphic to the standard permutation representation of SnS_{n} on ℂn{\mathbb{C}}^{n} and that (π,𝒮(n−1,1))(\pi,{\mathcal{S}}^{(n-1,1)}) is isomorphic to the usual irreducible n−1n-1-dimensional subrepresentation W0W_{0} of ℂn{\mathbb{C}}^{n}.

Solution: Let ϕ:ℂn→𝐘𝐓𝐃(n−1,1)\phi:{\mathbb{C}}^{n}\rightarrow\mathbf{YTD}^{(n-1,1)} be given by the linear extension of

\ytableausetup⁢t⁢a⁢b⁢l⁢o⁢i⁢d⁢s,s⁢m⁢a⁢l⁢l⁢t⁢a⁢b⁢l⁢e⁢a⁢u⁢x⁢𝐞i⟼\ytableaushort⁢⋯⁢⋯⁢⋯⁢⋯,i,\ytableausetup{tabloids,smalltableaux}{\bf e}_{i}\longmapsto\ytableaushort{% \cdots\cdots\cdots\cdots,i},

for each 1≤i≤n1\leq i\leq n. Then as it sends one basis to another it is a bijection and thus ϕ\phi is an isomorphism of vector spaces. Next we note that elements of the form (j⁢k)(jk) generate SnS_{n} so we just need to consider the action of these on each basis. We now consider the two cases, ii is permuted or it is not. If j≠i≠kj\neq i\neq k then

ϕ⁢((j⁢k)⋅𝐞i)=ϕ⁢(𝐞i)=\ytableaushort⁢⋯⁢⋯⁢⋯⁢⋯,i=(j⁢k)⋅\ytableaushort⁢⋯⁢⋯⁢⋯⁢⋯,i=(j⁢k)⋅ϕ⁢(𝐞i).\phi((jk)\cdot{\bf e}_{i})=\phi({\bf e}_{i})=\ytableaushort{\cdots\cdots\cdots% \cdots,i}=(jk)\cdot\ytableaushort{\cdots\cdots\cdots\cdots,i}=(jk)\cdot\phi({% \bf e}_{i}).

Suppose now, without loss of generality that i=ki=k, then

ϕ⁢((i⁢j)⋅𝐞i)=ϕ⁢(𝐞j)=\ytableaushort⁢⋯⁢⋯⁢⋯⁢⋯,j=(i⁢j)⋅\ytableaushort⁢⋯⁢⋯⁢⋯⁢⋯,i=(i⁢j)⋅ϕ⁢(𝐞i).\phi((ij)\cdot{\bf e}_{i})=\phi({\bf e}_{j})=\ytableaushort{\cdots\cdots\cdots% \cdots,j}=(ij)\cdot\ytableaushort{\cdots\cdots\cdots\cdots,i}=(ij)\cdot\phi({% \bf e}_{i}).

Thus ϕ\phi is an isomorphism and (π,ℳ(n−1,1))(\pi,{\mathcal{M}}^{(n-1,1)}) and (π,ℂn)(\pi,{\mathbb{C}}^{n}) are isomorphic.

By Theorem 5.12 these two representations must have isomorphic irreducible subrepresentations. (π,ℳ(n−1,1))(\pi,{\mathcal{M}}^{(n-1,1)}) contains a copy of the trivial representation, generated by the sum of all tabloids, and so does the permutation representation. The remaining irreducible subrepresentation of the permutation representation is (π,W0)(\pi,W_{0}) which therefore must be isomorphic to (π,𝒮(n−1,1))(\pi,{\mathcal{S}}^{(n-1,1)}), which by Theorem 17.11 is also irreducible.

Problem 92. Show that (π,ℳλ)(\pi,{\mathcal{M}}^{\lambda}) is a unitary representation of SnS_{n} with respect to the inner product

⟨∑[t]∈𝐘𝐓𝐃λz[t]⁢[t],∑[s]∈𝐘𝐓𝐃λw[s]⁢[s]⟩λ:=∑[t]∈𝐘𝐓𝐃λz[t]⁢w[t]¯.\left\langle\sum_{[t]\in\mathbf{YTD}^{\lambda}}z_{[t]}[t],\sum_{[s]\in\mathbf{% YTD}^{\lambda}}w_{[s]}[s]\right\rangle_{\lambda}:=\sum_{[t]\in\mathbf{YTD}^{% \lambda}}z_{[t]}\overline{w_{[t]}}.

Solution: We note that for any σ∈Sn\sigma\in S_{n} we have that σ⋅[t]=[σ⋅t]\sigma\cdot[t]=[\sigma\cdot t] is another tabloid. Thus

σ⋅(∑[t]∈𝐘𝐓𝐃λz[t]⁢[t])=∑[t]∈𝐘𝐓𝐃λz[t]⁢[σ⋅t]=∑[t]∈𝐘𝐓𝐃λz[t]⁢[σ⋅t]=∑[t]∈𝐘𝐓𝐃λz[σ−1⋅t]⁢[t].\sigma\cdot\left(\sum_{[t]\in\mathbf{YTD}^{\lambda}}z_{[t]}[t]\right)=\sum_{[t% ]\in\mathbf{YTD}^{\lambda}}z_{[t]}[\sigma\cdot t]=\sum_{[t]\in\mathbf{YTD}^{% \lambda}}z_{[t]}[\sigma\cdot t]=\sum_{[t]\in\mathbf{YTD}^{\lambda}}z_{[\sigma^% {-1}\cdot t]}[t].

Crucially, the coefficients are not changed, just permuted. We therefore have

⟨σ⋅(∑[t]∈𝐘𝐓𝐃λz[t]⁢[t]),σ⋅(∑[s]∈𝐘𝐓𝐃λw[s]⁢[s])⟩λ\displaystyle\left\langle\sigma\cdot\left(\sum_{[t]\in\mathbf{YTD}^{\lambda}}z% _{[t]}[t]\right),\sigma\cdot\left(\sum_{[s]\in\mathbf{YTD}^{\lambda}}w_{[s]}[s% ]\right)\right\rangle_{\lambda} =⟨∑[t]∈𝐘𝐓𝐃λz[σ−1⋅t]⁢[t],∑[s]∈𝐘𝐓𝐃λw[σ−1⋅s]⁢[s]⟩λ\displaystyle=\left\langle\sum_{[t]\in\mathbf{YTD}^{\lambda}}z_{[\sigma^{-1}% \cdot t]}[t],\sum_{[s]\in\mathbf{YTD}^{\lambda}}w_{[\sigma^{-1}\cdot s]}[s]% \right\rangle_{\lambda}
=∑[t]∈𝐘𝐓𝐃λz[σ−1⋅t]⁢w[σ−1⋅t]¯\displaystyle=\sum_{[t]\in\mathbf{YTD}^{\lambda}}z_{[\sigma^{-1}\cdot t]}% \overline{w_{[\sigma^{-1}\cdot t]}}
=∑[t]∈𝐘𝐓𝐃λz[t]⁢w[t]¯\displaystyle=\sum_{[t]\in\mathbf{YTD}^{\lambda}}z_{[t]}\overline{w_{[t]}}
=⟨∑[t]∈𝐘𝐓𝐃λz[t]⁢[t],∑[s]∈𝐘𝐓𝐃λw[s]⁢[s]⟩λ.\displaystyle=\left\langle\sum_{[t]\in\mathbf{YTD}^{\lambda}}z_{[t]}[t],\sum_{% [s]\in\mathbf{YTD}^{\lambda}}w_{[s]}[s]\right\rangle_{\lambda}.

As σ\sigma was arbitrary the inner product is invariant with respect to the action of SnS_{n} and thus (π,ℳλ)(\pi,{\mathcal{M}}^{\lambda}) is unitary.