For \(n\ge 1\), let \[ p_n=\frac{1}{n!}\sum_{\sigma\in S_n}\sigma, \] and let \(d=\operatorname{diag}(x_1,\ldots,x_N)\), where \(V=\mathbb C^N\). The matrix \(d\) acts on each factor of \(V^{\otimes n}\), and \(S_n\) acts on \(V^{\otimes n}\) by permuting tensor factors.
Calculate \[ \operatorname{Tr}_{V^{\otimes n}}(p_n d^{\otimes n}). \]
The value of \[ \operatorname{Tr}_{V^{\otimes 10}} \bigl((p_4\otimes p_3\otimes p_3)d^{\otimes10}\bigr) \] can be written as a linear combination of Schur functions. Calculate the coefficient of \(s_{[6,3,1]}\).
Calculate the multiplicity of the simple \(S_{10}\)-module \(S^{[3,3,3,1]}\) in \(V^{\otimes10}\) for \(\dim V=N=5\) and for \(N=3\).
Solution
1. The trace in part (a) is\[ h_n(x_1,\ldots,x_N)=s_{(n)}(x_1,\ldots,x_N). \]
The operator \(p_n\) is the averaging idempotent for the \(S_n\)-action, so its image is \[ \operatorname{Sym}^n(V)\subseteq V^{\otimes n}. \] The operator \(d^{\otimes n}\) commutes with the permutation action and hence with \(p_n\). Therefore \[ \operatorname{Tr}_{V^{\otimes n}}(p_nd^{\otimes n}) =\operatorname{Tr}_{\operatorname{Sym}^n(V)}(d). \] A basis of \(\operatorname{Sym}^n(V)\) is indexed by weak compositions \[ a_1+\cdots+a_N=n, \qquad a_i\ge0, \] and the corresponding basis vector has \(d\)-eigenvalue \[ x_1^{a_1}\cdots x_N^{a_N}. \] Summing these eigenvalues gives the complete homogeneous symmetric function \(h_n=s_{(n)}\).
2. The trace in part (b) is the symmetric function\[ s_{(4)}s_{(3)}s_{(3)}. \]
The operator \(p_4\otimes p_3\otimes p_3\) projects \[ V^{\otimes10}=V^{\otimes4}\otimes V^{\otimes3}\otimes V^{\otimes3} \] onto \[ \operatorname{Sym}^4(V)\otimes\operatorname{Sym}^3(V)\otimes\operatorname{Sym}^3(V). \] By 1, the characters of the three factors are \(s_{(4)},s_{(3)},s_{(3)}\), so the character of their tensor product is their product.
3. The coefficient of \(s_{(6,3,1)}\) in \(s_{(4)}s_{(3)}s_{(3)}\) is \(3\).
By the Pieri rule, \[ s_{(4)}s_{(3)} =s_{(7)}+s_{(6,1)}+s_{(5,2)}+s_{(4,3)}. \] Multiplication by the final \(s_{(3)}\) adds a horizontal \(3\)-strip. For the target partition \[ \lambda=(6,3,1), \] the partitions in the preceding sum for which \(\lambda/\mu\) is a horizontal \(3\)-strip are exactly \[ \mu=(6,1),\qquad(5,2),\qquad(4,3). \] Each occurs with Pieri coefficient \(1\). Hence \[ [s_{(6,3,1)}]\,s_{(4)}s_{(3)}s_{(3)}=3. \]
4. The multiplicity of \(S^{(3,3,3,1)}\) in \(V^{\otimes10}\) is\[ 175\quad\text{when }N=5, \qquad 0\quad\text{when }N=3. \]
Schur-Weyl duality gives \[ V^{\otimes10} \cong \bigoplus_{\substack{\lambda\vdash10\\\ell(\lambda)\le N}} \mathbf S_\lambda(V)\otimes S^\lambda \] as a \(GL(V)\times S_{10}\)-module. Thus the multiplicity of \(S^\lambda\) is \[ \dim\mathbf S_\lambda(\mathbb C^N)=s_\lambda(1^N). \] For \(N=3\), the partition \((3,3,3,1)\) has four nonzero parts, so the Schur functor vanishes and the multiplicity is \(0\).
For \(N=5\), the hook-content formula gives \[ s_{(3,3,3,1)}(1^5) =\prod_{(i,j)\in(3,3,3,1)}\frac{5+j-i}{h_{ij}}. \] The hook lengths, row by row, are \[ (6,4,3),\qquad(5,3,2),\qquad(4,2,1),\qquad(1), \] and the corresponding numerators \(5+j-i\) are \[ (5,6,7),\qquad(4,5,6),\qquad(3,4,5),\qquad(2). \] Therefore \[ \frac{5\cdot6\cdot7\cdot4\cdot5\cdot6\cdot3\cdot4\cdot5\cdot2} {6\cdot4\cdot3\cdot5\cdot3\cdot2\cdot4\cdot2\cdot1\cdot1} =175. \]