Suppose that \(\lambda=(\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_k)\) is a partition of \(n\). Then \(A^\lambda\) denotes the irreducible representation of the symmetric group \(S_n\) such that the Frobenius image of \(\chi^{A^\lambda}=\chi^\lambda\) is the Schur function \(S_\lambda(x_1,\ldots,x_N)\) where \(N>n\), and \(S_{\lambda_1}\times\cdots\times S_{\lambda_k}\) denotes the Young subgroup of \(S_n\) corresponding to \(\lambda\).
Use the Murnaghnam–Nakayama rule to compute the character table of \(S_4\).
Express \(\chi^{A^{(2,2)}}\downarrow_{S_2\times S_2}^{S_4}\) as a sum of irreducible characters of \(S_2\times S_2\). (Hint: First write out the character table for \(S_2\times S_2\).)
Solution
Let the conjugacy classes of \(S_4\) be indexed by cycle type in the order \[ (1^4),\quad (2,1^2),\quad (2^2),\quad (3,1),\quad (4). \] Their sizes are respectively \[ 1,6,3,8,6. \]
1. By the Murnaghan–Nakayama rule, the irreducible character table of \(S_4\) is\[ \begin{array}{c|rrrrr} \lambda &(1^4)&(2,1^2)&(2^2)&(3,1)&(4)\\ \hline (4) &1& 1& 1& 1& 1\\ (3,1) &3& 1&-1& 0&-1\\ (2,2) &2& 0& 2&-1& 0\\ (2,1,1) &3&-1&-1& 0& 1\\ (1^4) &1&-1& 1& 1&-1 \end{array}. \]
We use the Murnaghan–Nakayama rule in the form \[ \chi^\lambda_{(r,\mu)} = \sum_{\nu}\,(-1)^{\operatorname{ht}(\lambda/\nu)-1}\chi^\nu_\mu, \] where the sum is over partitions \(\nu\) obtained from \(\lambda\) by removing a rim hook of size \(r\).
For the identity class, the values are the dimensions \(f^\lambda\), computed from the hook-length formula: \[ f^{(4)}=1,\quad f^{(3,1)}=3,\quad f^{(2,2)}=2,\quad f^{(2,1,1)}=3,\quad f^{(1^4)}=1. \]
For the one-row partition \((4)\), every removable rim hook has height \(1\), so every character value is \(1\). For the one-column partition \((1^4)\), a rim hook of length \(r\) has height \(r\), hence contributes \((-1)^{r-1}\); therefore this row is \[ 1,-1,1,1,-1, \] the sign character.
For \(\lambda=(3,1)\): - at cycle type \((2,1^2)\), the unique removable \(2\)-rim hook is the horizontal strip consisting of the two rightmost boxes of the first row. It has height \(1\) and leaves \((1,1)\), so the value is \(f^{(1,1)}=1\); - at \((2^2)\), after that removal the remaining partition is \((1,1)\), whose value on a transposition is \(-1\), so the value is \(-1\); - at \((3,1)\), there is no removable \(3\)-rim hook leaving a Young diagram, so the value is \(0\); - at \((4)\), the whole diagram is a rim hook of height \(2\), giving \(-1\). Thus the row is \(3,1,-1,0,-1\).
For \(\lambda=(2,2)\): - at \((2,1^2)\), the two removable \(2\)-rim hooks have opposite signs and leave one-box-count contributions of equal size, so the value is \(0\); - at \((2^2)\), the two successive removals contribute \(1+1=2\); - at \((3,1)\), the unique admissible \(3\)-rim hook has height \(2\), so the value is \(-1\); - there is no rim hook of length \(4\) because the \(2\times2\) diagram contains a \(2\times2\) block, so the value is \(0\). Hence the row is \(2,0,2,-1,0\).
Finally, \((2,1,1)\) is conjugate to \((3,1)\), and conjugating a partition tensors the corresponding irreducible representation with the sign representation. Therefore its row is obtained by multiplying the \((3,1)\) row by the sign row, giving \[ 3,-1,-1,0,1. \] This yields the displayed table.
2. Let\[ H=S_2\times S_2 =\{1,(12),(34),(12)(34)\}. \]The restricted character of \(A^{(2,2)}\) on these four elements is\[ \chi^{(2,2)}\!\downarrow_H=(2,0,0,2). \]
The identity has cycle type \((1^4)\), each of \((12)\) and \((34)\) has cycle type \((2,1^2)\), and \((12)(34)\) has cycle type \((2^2)\). Reading the \((2,2)\) row of 1 gives the values \[ 2,0,0,2. \]
3. The four irreducible characters of \(H\cong C_2\times C_2\) are\[ \begin{array}{c|rrrr} &1&(12)&(34)&(12)(34)\\ \hline \mathbf 1\boxtimes\mathbf 1&1&1&1&1\\ \mathbf 1\boxtimes\operatorname{sgn}&1&1&-1&-1\\ \operatorname{sgn}\boxtimes\mathbf 1&1&-1&1&-1\\ \operatorname{sgn}\boxtimes\operatorname{sgn}&1&-1&-1&1 \end{array}. \]
Each factor \(S_2\) has exactly two irreducible characters, the trivial and sign characters. Since \(H\) is a direct product, its irreducible characters are their external tensor products, producing the four rows displayed.
4. Therefore\[ \chi^{A^{(2,2)}}\!\downarrow_{S_2\times S_2}^{S_4} = (\mathbf 1\boxtimes\mathbf 1) + (\operatorname{sgn}\boxtimes\operatorname{sgn}). \]
For an irreducible character \(\theta\) of \(H\), its multiplicity in the restriction is \[ \langle (2,0,0,2),\theta\rangle_H = \frac14\sum_{h\in H}(2,0,0,2)(h)\overline{\theta(h)}. \] For \(\mathbf 1\boxtimes\mathbf 1\) this equals \[ \frac14(2+2)=1, \] and for \(\operatorname{sgn}\boxtimes\operatorname{sgn}\) it equals \[ \frac14(2+2)=1. \] For each of the two mixed characters it equals \[ \frac14(2-2)=0. \] Hence the restriction is exactly the sum displayed above. This proves part (b).