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 \(\lambda\).
Find the decomposition of \[ A^{(2,1^2)}\times A^{(2,2)}\uparrow_{S_4\times S_4}^{S_8} \] as a sum of irreducible representations of \(S_8\).
Let \(T\) denote the trivial representation on the Young subgroup \(S_3\times S_2\times S_1\) of \(S_7\) and \(\mathrm{Alt}\) denote the alternating representation on the Young subgroup \(S_3\times S_2\times S_1\) of \(S_7\). Find the decomposition of \[ T\uparrow_{S_3\times S_2\times S_1}^{S_7}\quad\text{and}\quad\mathrm{Alt}\uparrow_{S_3\times S_2\times S_1}^{S_7} \] as a sum of irreducible representations of \(S_7\).
Find the decomposition of the Kronecker product \(A^{(3,2)}\otimes A^{(3,2)}\) as a sum of irreducible representations of \(S_5\).
In the source, part (b) is internally inconsistent: the displayed formulas carry superscript \(S_6\), while the prose twice describes \(S_3\times S_2\times S_1\) as a Young subgroup of \(S_7\) and asks for decompositions “of \(S_7\)”. The prose reading is taken here: under the exam’s own preamble convention a Young subgroup \(S_{\lambda_1}\times\cdots\times S_{\lambda_k}\) sits in \(S_n\) for \(\lambda\) a partition of \(n\), so the subgroup of \(S_7\) is the one for the padded partition \((3,2,1,1)\), and the induction target is \(S_7\).
Solution
We use the Frobenius characteristic map. Under this map, induction from a Young subgroup corresponds to multiplication of Schur functions, while tensoring by the sign representation sends \(s_\lambda\) to \(s_{\lambda'}\).
1. For part (a),\[ \operatorname{ch}\left(\left(A^{(2,1,1)}\boxtimes A^{(2,2)}\right)\uparrow_{S_4\times S_4}^{S_8}\right) =s_{(2,1,1)}s_{(2,2)}. \]Moreover\[ s_{(2,2)}=h_2^2-h_3h_1. \]
The first assertion is the induction-product property of the Frobenius characteristic. The second is the Jacobi–Trudi determinant \[ s_{(2,2)}= \det\begin{pmatrix}h_2&h_3\\ h_1&h_2\end{pmatrix} =h_2^2-h_3h_1. \]
2. Repeated application of the Pieri rule gives\[ \begin{aligned} s_{(2,1,1)}h_2^2 ={}&s_{(2,2,2,1,1)}+2s_{(3,2,1,1,1)}+2s_{(3,2,2,1)}+2s_{(3,3,1,1)}+s_{(3,3,2)}\\ &+s_{(4,1,1,1,1)}+4s_{(4,2,1,1)}+s_{(4,2,2)}+2s_{(4,3,1)}\\ &+2s_{(5,1,1,1)}+2s_{(5,2,1)}+s_{(6,1,1)}, \end{aligned} \]and\[ \begin{aligned} s_{(2,1,1)}h_3h_1 ={}&s_{(3,2,1,1,1)}+s_{(3,2,2,1)}+s_{(3,3,1,1)}+s_{(4,1,1,1,1)}\\ &+3s_{(4,2,1,1)}+s_{(4,2,2)}+s_{(4,3,1)}\\ &+2s_{(5,1,1,1)}+2s_{(5,2,1)}+s_{(6,1,1)}. \end{aligned} \]
The Pieri rule says that \(s_\lambda h_r\) is the sum of \(s_\mu\) over partitions \(\mu\) for which \(\mu/\lambda\) is a horizontal \(r\)-strip, each with coefficient \(1\). First, \[ s_{(2,1,1)}h_2 =s_{(2,2,1,1)}+s_{(3,1,1,1)}+s_{(3,2,1)}+s_{(4,1,1)}, \] while \[ s_{(2,1,1)}h_3 =s_{(3,2,1,1)}+s_{(4,1,1,1)}+s_{(4,2,1)}+s_{(5,1,1)}. \] Applying Pieri once more, with \(h_2\) in the first line and \(h_1\) in the second, and collecting equal partitions gives exactly the two displayed expansions.
3. Therefore\[ \begin{aligned} \left(A^{(2,1,1)}\boxtimes A^{(2,2)}\right)\uparrow_{S_4\times S_4}^{S_8} \cong{}&A^{(2,2,2,1,1)}\oplus A^{(3,2,1,1,1)}\oplus A^{(3,2,2,1)}\\ &\oplus A^{(3,3,1,1)}\oplus A^{(3,3,2)}\oplus A^{(4,2,1,1)}\oplus A^{(4,3,1)}. \end{aligned} \]
1. All terms not displayed cancel, and each surviving Schur function has coefficient \(1\). The Frobenius characteristic is injective on the complex representation ring of \(S_8\), so the Schur expansion gives the stated irreducible decomposition.
For part (b), interpret the source as in the remark: the subgroup is the Young subgroup of type \((3,2,1,1)\) in \(S_7\); the extra \(S_1\) factor is invisible as an abstract direct factor.
4. The induced trivial representation has Frobenius characteristic\[ s_{(3)}s_{(2)}s_{(1)}^2. \]Repeated Pieri gives\[ \begin{aligned} s_{(3)}s_{(2)}s_{(1)}^2 ={}&s_{(3,2,1,1)}+s_{(3,2,2)}+2s_{(3,3,1)}+s_{(4,1,1,1)}\\ &+4s_{(4,2,1)}+3s_{(4,3)}+3s_{(5,1,1)}+4s_{(5,2)}+3s_{(6,1)}+s_{(7)}. \end{aligned} \]
First Pieri gives \[ s_{(3)}s_{(2)}=s_{(3,2)}+s_{(4,1)}+s_{(5)}. \] Multiplying by one copy of \(s_{(1)}=h_1\) gives \[ s_{(3,2,1)}+s_{(3,3)}+s_{(4,1,1)}+2s_{(4,2)}+2s_{(5,1)}+s_{(6)}. \] A second multiplication by \(h_1\) and another application of Pieri gives the displayed degree-\(7\) expansion.
5. Hence\[ \begin{aligned} T\uparrow_{S_3\times S_2\times S_1}^{S_7} \cong{}&A^{(3,2,1,1)}\oplus A^{(3,2,2)}\oplus2A^{(3,3,1)}\oplus A^{(4,1,1,1)}\\ &\oplus4A^{(4,2,1)}\oplus3A^{(4,3)}\oplus3A^{(5,1,1)}\\ &\oplus4A^{(5,2)}\oplus3A^{(6,1)}\oplus A^{(7)}. \end{aligned} \]
This is the representation decomposition corresponding to the Schur expansion in 4.
6. The alternating induction is obtained from 5 by conjugating every partition:\[ \begin{aligned} \mathrm{Alt}\uparrow_{S_3\times S_2\times S_1}^{S_7} \cong{}&A^{(1^7)}\oplus3A^{(2,1^5)}\oplus4A^{(2,2,1,1,1)}\oplus3A^{(2,2,2,1)}\\ &\oplus3A^{(3,1,1,1,1)}\oplus4A^{(3,2,1,1)}\oplus2A^{(3,2,2)}\\ &\oplus A^{(3,3,1)}\oplus A^{(4,1,1,1)}\oplus A^{(4,2,1)}. \end{aligned} \]
Let \(H=S_3\times S_2\times S_1\times S_1\le S_7\). The alternating representation in the problem is the restriction of the sign representation \(\operatorname{sgn}_{S_7}\) to \(H\). For any \(H\)-module \(V\) and \(S_7\)-module \(W\) there is a natural isomorphism \[ \operatorname{Ind}_H^{S_7}(V\otimes\operatorname{Res}_H W) \cong \operatorname{Ind}_H^{S_7}(V)\otimes W. \] Taking \(V\) trivial and \(W=\operatorname{sgn}_{S_7}\) shows that the alternating induction is the sign twist of the trivial induction. Since \[ A^\lambda\otimes\operatorname{sgn}\cong A^{\lambda'}, \] conjugating the partitions in 5 gives exactly the displayed decomposition.
For part (c), use the conjugacy classes of \(S_5\) in the order \[ (5),(4,1),(3,2),(3,1,1),(2,2,1),(2,1,1,1),(1^5), \] whose sizes are \[ 24,30,20,20,15,10,1. \]
7. The character of \(A^{(3,2)}\) on these classes is\[ \chi^{(3,2)}=(0,-1,1,-1,1,1,5). \]Consequently the character of \(A^{(3,2)}\otimes A^{(3,2)}\) is\[ (0,1,1,1,1,1,25). \]
The first row follows from the Murnaghan–Nakayama rule applied to the diagram \((3,2)\); at the identity the value is \(5\) by the hook-length formula. Characters multiply pointwise under tensor product, so squaring the seven entries gives the second row.
8. The pointwise square in 7 is the sum of the irreducible characters indexed by\[ (5),\ (4,1),\ (3,2),\ (3,1,1),\ (2,2,1),\ (2,1,1,1), \]each with multiplicity \(1\), and has zero inner product with the sign character \((1^5)\).
7. The remaining irreducible character is the sign character. The six displayed rows together with the sign character form all seven irreducible characters of \(S_5\); since the tensor-square character already equals the sum of the six displayed rows, uniqueness of irreducible-character expansion gives sign multiplicity \(0\).
9. Therefore\[ \boxed{ A^{(3,2)}\otimes A^{(3,2)} \cong A^{(5)}\oplus A^{(4,1)}\oplus A^{(3,2)}\oplus A^{(3,1,1)}\oplus A^{(2,2,1)}\oplus A^{(2,1,1,1)}.} \]
This is precisely the irreducible-character identity proved in 8.