Cauchy-Littlewood identities

We follow Chapter 1 of [{macdonald98}].


\[ \sum_\lambda s_\lambda(x) s_\lambda(y) = \prod{1 \over 1 - x_i y_j} =: \Pi(x, y) \qquad (4.3) \]

Unless otherwise specified, sums like this are over all partitions. This is true for symmetric polynomial case \(x = x_{1 : n}\) and \(y = y_{1 : \ell}\) as well as in the symmetric function case.

recall the definition

\[ z_\lambda = \prod_{i \ge 1} i^{m_i} m_i! \]

where \(m_i\) is the number of parts of \(\lambda\) equal to \(i\), i.e. \(m_i = \lambda'_i - \lambda'_{i + 1}\).

Other similar stuff:

\begin{align} \sum_\lambda z_\lambda^{-1} p_\lambda(x) p_\lambda(y) &= \Pi(x, y) \qquad(4.1) \\ \sum_\lambda h_\lambda(x) m_\lambda(y) = \sum_\lambda h_\lambda(y) m_\lambda(x) &= \Pi(x, y) \qquad(4.2) \end{align}

Claim. \(\omega_x \Pi(x, y) = \omega_y \Pi(x, y) = \prod(1 + x_i y_j) =: \Pi'(x, y)\).


\begin{align} \omega_x \Pi(x, y) = \omega_x \prod_j H(y_j, x) = \omega_x \prod_j (\sum_n y_j^n h_n(x)) = \prod_j (\sum_n y_j^n \omega_x h_n(x))\\ = \prod_j \sum_n y_j^n e_n(x) = \prod_j E(y_j, x) = RHS. \end{align}



Define scalar product on \(\Lambda\) by

\[ \langle h_\lambda, m_\mu \rangle = \delta_{\lambda \mu} \]


Theorem (4.6). For \(n \ge 0\), let \((u_\lambda)_{\lambda \vdash n}\) and \((v_\lambda)_{\lambda \vdash n}\) be \(\mathbb Q\) bases of \(\Lambda^n_{\mathcal Q}\). Then the following are equivalent:


\[ \langle p_\lambda, p_\mu \rangle = \delta_{\lambda \mu} z_{\lambda}, \qquad \langle s_\lambda, s_\mu \rangle = \delta_{\lambda \mu}. \]

(4.9) the bilinear form is symmetric and positive definite.

Since \(\omega(p_\lambda) = \pm p_\lambda\),

(4.10) The involution \(\omega\) is an isometry, i.e. \(\langle \omega u, \omega v \rangle = \langle u, v \rangle\).

Therefore by the Claim we have three dual Cauchy-Littlewood identities:

\begin{align} \sum_\lambda \epsilon_\lambda z_\lambda^{-1} p_\lambda(x) p_\lambda(y) &= \Pi'(x, y) \\ \sum_\lambda m_\lambda(x) e_\lambda(y) = \sum_\lambda m_\lambda(y) e_\lambda(x) &= \Pi'(x, y) \\ \sum_\lambda s_\lambda(x) s_{\lambda'}(y) &= \Pi'(x, y). \end{align}