[proofplan] We prove the identity by pairing both sides with an arbitrary Schur function $s_\rho$. The left-hand side becomes $\langle s_\mu s_\nu, s_\lambda s_\rho\rangle$ by the defining adjointness of $s_\lambda^\perp$. The right-hand side is evaluated by expanding the Hall-dual coproduct of $s_\rho$ and then applying adjointness separately to the two skewing operators. Both pairings reduce to the same coefficient of $s_\mu \otimes s_\nu$ in $\Delta(s_\lambda s_\rho)$, so orthonormality of the Schur basis identifies the two symmetric functions. [/proofplan]

[step:Declare the Hall-dual coproduct and its Schur expansion]Let $\Delta: \Lambda \to \Lambda \otimes \Lambda$ denote the coproduct adjoint to multiplication with respect to the Hall [inner product](/page/Inner%20Product), meaning that for all $a,b,h \in \Lambda$, \begin{align*} \langle ab,h\rangle = \langle a \otimes b,\Delta h\rangle, \end{align*} where the [tensor product](/page/Tensor%20Product) inner product is determined on Schur basis tensors by \begin{align*} \langle s_\alpha \otimes s_\beta, s_\gamma \otimes s_\delta\rangle = \langle s_\alpha,s_\gamma\rangle \langle s_\beta,s_\delta\rangle. \end{align*} Since the Schur basis is orthonormal and \begin{align*} s_\alpha s_\beta=\sum_\kappa c_{\alpha\beta}^{\kappa}s_\kappa, \end{align*} the coproduct satisfies, for every partition $\kappa$, \begin{align*} \Delta s_\kappa = \sum_{\alpha,\beta} c_{\alpha\beta}^{\kappa}s_\alpha \otimes s_\beta. \end{align*}[/step]

[guided]The product formula is easiest to prove by using the structure dual to multiplication. Define $\Delta: \Lambda \to \Lambda \otimes \Lambda$ to be the [linear map](/page/Linear%20Map) adjoint to multiplication under the Hall inner product: \begin{align*} \langle ab,h\rangle = \langle a \otimes b,\Delta h\rangle \end{align*} for all $a,b,h \in \Lambda$. The inner product on $\Lambda \otimes \Lambda$ is the product inner product, so on Schur basis tensors it is \begin{align*} \langle s_\alpha \otimes s_\beta, s_\gamma \otimes s_\delta\rangle = \langle s_\alpha,s_\gamma\rangle \langle s_\beta,s_\delta\rangle. \end{align*} We now compute $\Delta s_\kappa$ for a partition $\kappa$. Write \begin{align*} \Delta s_\kappa = \sum_{\alpha,\beta} d_{\alpha\beta}^{\kappa}s_\alpha \otimes s_\beta \end{align*} for uniquely determined integers $d_{\alpha\beta}^{\kappa}$, since the tensors $s_\alpha \otimes s_\beta$ form an [orthonormal basis](/page/Orthonormal%20Basis) of $\Lambda \otimes \Lambda$. Pairing with $s_\alpha \otimes s_\beta$ gives \begin{align*} d_{\alpha\beta}^{\kappa} &= \langle s_\alpha \otimes s_\beta,\Delta s_\kappa\rangle \\ &= \langle s_\alpha s_\beta,s_\kappa\rangle. \end{align*} By the definition of Littlewood-Richardson coefficients, \begin{align*} s_\alpha s_\beta = \sum_\tau c_{\alpha\beta}^{\tau}s_\tau. \end{align*} Using Schur orthonormality, this gives \begin{align*} \langle s_\alpha s_\beta,s_\kappa\rangle = c_{\alpha\beta}^{\kappa}. \end{align*} Hence \begin{align*} \Delta s_\kappa = \sum_{\alpha,\beta} c_{\alpha\beta}^{\kappa}s_\alpha \otimes s_\beta. \end{align*}[/guided]

[step:Compute the pairing of the left-hand side with a Schur test function] Fix an arbitrary partition $\rho$. By the defining adjointness of $s_\lambda^\perp$, \begin{align*} \left\langle s_\lambda^\perp(s_\mu s_\nu),s_\rho\right\rangle = \left\langle s_\mu s_\nu,s_\lambda s_\rho\right\rangle. \end{align*} Using the coproduct adjointness of multiplication, this becomes \begin{align*} \left\langle s_\mu s_\nu,s_\lambda s_\rho\right\rangle = \left\langle s_\mu \otimes s_\nu,\Delta(s_\lambda s_\rho)\right\rangle. \end{align*} [/step]

[step:Compute the pairing of the right-hand side with the same Schur test function] For the right-hand side, bilinearity of the Hall inner product gives \begin{align*} &\left\langle \sum_{\alpha,\beta} c_{\alpha\beta}^{\lambda} \bigl(s_\alpha^\perp s_\mu\bigr) \bigl(s_\beta^\perp s_\nu\bigr), s_\rho \right\rangle \\ &\qquad = \sum_{\alpha,\beta} c_{\alpha\beta}^{\lambda} \left\langle \bigl(s_\alpha^\perp s_\mu\bigr) \bigl(s_\beta^\perp s_\nu\bigr), s_\rho \right\rangle. \end{align*} Applying coproduct adjointness to each product inside the sum gives \begin{align*} \left\langle \bigl(s_\alpha^\perp s_\mu\bigr) \bigl(s_\beta^\perp s_\nu\bigr), s_\rho \right\rangle = \left\langle s_\alpha^\perp s_\mu \otimes s_\beta^\perp s_\nu, \Delta s_\rho \right\rangle. \end{align*} Using \begin{align*} \Delta s_\rho = \sum_{\gamma,\delta} c_{\gamma\delta}^{\rho}s_\gamma \otimes s_\delta, \end{align*} we obtain \begin{align*} &\left\langle \sum_{\alpha,\beta} c_{\alpha\beta}^{\lambda} \bigl(s_\alpha^\perp s_\mu\bigr) \bigl(s_\beta^\perp s_\nu\bigr), s_\rho \right\rangle \\ &\qquad = \sum_{\alpha,\beta,\gamma,\delta} c_{\alpha\beta}^{\lambda}c_{\gamma\delta}^{\rho} \left\langle s_\alpha^\perp s_\mu,s_\gamma\right\rangle \left\langle s_\beta^\perp s_\nu,s_\delta\right\rangle. \end{align*} By adjointness of $s_\alpha^\perp$ and $s_\beta^\perp$, \begin{align*} \left\langle s_\alpha^\perp s_\mu,s_\gamma\right\rangle = \left\langle s_\mu,s_\alpha s_\gamma\right\rangle, \qquad \left\langle s_\beta^\perp s_\nu,s_\delta\right\rangle = \left\langle s_\nu,s_\beta s_\delta\right\rangle. \end{align*} Therefore \begin{align*} &\left\langle \sum_{\alpha,\beta} c_{\alpha\beta}^{\lambda} \bigl(s_\alpha^\perp s_\mu\bigr) \bigl(s_\beta^\perp s_\nu\bigr), s_\rho \right\rangle \\ &\qquad = \sum_{\alpha,\beta,\gamma,\delta} c_{\alpha\beta}^{\lambda}c_{\gamma\delta}^{\rho} \left\langle s_\mu,s_\alpha s_\gamma\right\rangle \left\langle s_\nu,s_\beta s_\delta\right\rangle. \end{align*} [/step]

[step:Identify the right-hand pairing as the same coproduct coefficient] Using Schur orthonormality and the [Littlewood-Richardson rule](/theorems/5183), \begin{align*} \left\langle s_\mu,s_\alpha s_\gamma\right\rangle = c_{\alpha\gamma}^{\mu}, \qquad \left\langle s_\nu,s_\beta s_\delta\right\rangle = c_{\beta\delta}^{\nu}. \end{align*} Thus the right-hand pairing is \begin{align*} \sum_{\alpha,\beta,\gamma,\delta} c_{\alpha\beta}^{\lambda}c_{\gamma\delta}^{\rho} c_{\alpha\gamma}^{\mu}c_{\beta\delta}^{\nu}. \end{align*} On the other hand, since $\Delta$ is adjoint to multiplication, it is multiplicative with respect to the multiplication on $\Lambda \otimes \Lambda$. Hence \begin{align*} \Delta(s_\lambda s_\rho) = \Delta s_\lambda \, \Delta s_\rho. \end{align*} Expanding both factors gives \begin{align*} \Delta(s_\lambda s_\rho) &= \left(\sum_{\alpha,\beta} c_{\alpha\beta}^{\lambda}s_\alpha \otimes s_\beta\right) \left(\sum_{\gamma,\delta} c_{\gamma\delta}^{\rho}s_\gamma \otimes s_\delta\right) \\ &= \sum_{\alpha,\beta,\gamma,\delta} c_{\alpha\beta}^{\lambda}c_{\gamma\delta}^{\rho} (s_\alpha s_\gamma) \otimes (s_\beta s_\delta). \end{align*} Pairing with $s_\mu \otimes s_\nu$ yields \begin{align*} \left\langle s_\mu \otimes s_\nu,\Delta(s_\lambda s_\rho)\right\rangle = \sum_{\alpha,\beta,\gamma,\delta} c_{\alpha\beta}^{\lambda}c_{\gamma\delta}^{\rho} c_{\alpha\gamma}^{\mu}c_{\beta\delta}^{\nu}. \end{align*} This is exactly the value of the right-hand pairing computed above. [/step]

[step:Conclude equality from Schur orthonormality] Combining the computations, for every partition $\rho$ we have \begin{align*} \left\langle s_\lambda^\perp(s_\mu s_\nu),s_\rho\right\rangle = \left\langle \sum_{\alpha,\beta} c_{\alpha\beta}^{\lambda} \bigl(s_\alpha^\perp s_\mu\bigr) \bigl(s_\beta^\perp s_\nu\bigr), s_\rho \right\rangle. \end{align*} The Schur functions form an orthonormal basis of $\Lambda$, so a symmetric function is determined by all of its Hall inner products with Schur functions. Therefore \begin{align*} s_\lambda^\perp(s_\mu s_\nu) = \sum_{\alpha,\beta} c_{\alpha\beta}^{\lambda} \bigl(s_\alpha^\perp s_\mu\bigr) \bigl(s_\beta^\perp s_\nu\bigr). \end{align*} This proves the skewing product formula. [/step]