[proofplan]
We determine the Schur expansion of $h_r^\perp s_\nu$ and $e_r^\perp s_\nu$ by computing their inner products with an arbitrary Schur function $s_\mu$. Adjointness converts these coefficients into the coefficients of $s_\nu$ in $h_r s_\mu$ and $e_r s_\mu$. The ordinary Pieri rules identify those coefficients as the indicators of horizontal and vertical strips, and Schur orthonormality then forces the desired expansions.
[/proofplan]
[step:Compute the Schur coefficients of $h_r^\perp s_\nu$ using adjointness]
Let $\mu$ be an arbitrary partition. Since $\{s_\lambda\}$ is an [orthonormal basis](/page/Orthonormal%20Basis) for $\Lambda$ with respect to $(\cdot,\cdot)_\Lambda$, the coefficient of $s_\mu$ in the Schur expansion of $h_r^\perp s_\nu$ is
\begin{align*}
(h_r^\perp s_\nu, s_\mu)_\Lambda.
\end{align*}
By the defining adjointness of $h_r^\perp$ to multiplication by $h_r$,
\begin{align*}
(h_r^\perp s_\nu, s_\mu)_\Lambda
=
(s_\nu, h_r s_\mu)_\Lambda.
\end{align*}
Using the Pieri rule for multiplication by $h_r$ (citing a result not yet in the wiki: Pieri Rule for Schur functions),
\begin{align*}
h_r s_\mu
=
\sum_{\alpha \,:\, \alpha/\mu \text{ is a horizontal } r\text{-strip}} s_\alpha.
\end{align*}
Taking the [inner product](/page/Inner%20Product) with $s_\nu$ and using Schur orthonormality gives
\begin{align*}
(s_\nu, h_r s_\mu)_\Lambda
=
\begin{cases}
1, & \text{if } \nu/\mu \text{ is a horizontal } r\text{-strip}, \\
0, & \text{otherwise}.
\end{cases}
\end{align*}
Therefore the coefficient of $s_\mu$ in $h_r^\perp s_\nu$ is $1$ precisely when $\mu \subset \nu$ and $\nu/\mu$ is a horizontal $r$-strip, and is $0$ otherwise.
[guided]
We want to prove an identity between symmetric functions. Since the Schur functions form an orthonormal basis, it is enough to compute the coefficient of each basis vector $s_\mu$. Let $\mu$ be an arbitrary partition. The coefficient of $s_\mu$ in $h_r^\perp s_\nu$ is obtained by pairing with $s_\mu$:
\begin{align*}
\text{coefficient of } s_\mu \text{ in } h_r^\perp s_\nu
=
(h_r^\perp s_\nu, s_\mu)_\Lambda.
\end{align*}
The point of introducing $h_r^\perp$ is that it is defined as an adjoint operator. Since $h_r^\perp$ is adjoint to multiplication by $h_r$, we may move $h_r^\perp$ off the first factor and replace it by multiplication by $h_r$ on the second factor:
\begin{align*}
(h_r^\perp s_\nu, s_\mu)_\Lambda
=
(s_\nu, h_r s_\mu)_\Lambda.
\end{align*}
Now the ordinary Pieri rule describes $h_r s_\mu$. It states that multiplication by $h_r$ adds a horizontal $r$-strip to the Young diagram of $\mu$:
\begin{align*}
h_r s_\mu
=
\sum_{\alpha \,:\, \alpha/\mu \text{ is a horizontal } r\text{-strip}} s_\alpha.
\end{align*}
Here $\alpha$ ranges over partitions containing $\mu$. Pairing both sides with $s_\nu$ selects exactly the coefficient of $s_\nu$, because the Schur basis is orthonormal:
\begin{align*}
(s_\nu, h_r s_\mu)_\Lambda
=
\sum_{\alpha \,:\, \alpha/\mu \text{ is a horizontal } r\text{-strip}} (s_\nu,s_\alpha)_\Lambda
=
\begin{cases}
1, & \text{if } \nu/\mu \text{ is a horizontal } r\text{-strip}, \\
0, & \text{otherwise}.
\end{cases}
\end{align*}
Thus the Schur coefficient of $s_\mu$ in $h_r^\perp s_\nu$ is exactly the indicator of the condition that $\nu/\mu$ is a horizontal $r$-strip.
[/guided]
[/step]
[step:Recover the horizontal-strip expansion from the computed coefficients]
The preceding step shows that the Schur coefficient of $s_\mu$ in $h_r^\perp s_\nu$ equals the Schur coefficient of $s_\mu$ in
\begin{align*}
\sum_{\lambda \subset \nu \,:\, \nu/\lambda \text{ is a horizontal } r\text{-strip}} s_\lambda.
\end{align*}
Since the Schur functions form a basis of $\Lambda$, equality of all Schur coefficients implies
\begin{align*}
h_r^\perp s_\nu
=
\sum_{\lambda \subset \nu \,:\, \nu/\lambda \text{ is a horizontal } r\text{-strip}} s_\lambda.
\end{align*}
[/step]
[step:Compute the Schur coefficients of $e_r^\perp s_\nu$ using adjointness]
Let $\mu$ be an arbitrary partition. The coefficient of $s_\mu$ in the Schur expansion of $e_r^\perp s_\nu$ is
\begin{align*}
(e_r^\perp s_\nu, s_\mu)_\Lambda.
\end{align*}
By the defining adjointness of $e_r^\perp$ to multiplication by $e_r$,
\begin{align*}
(e_r^\perp s_\nu, s_\mu)_\Lambda
=
(s_\nu, e_r s_\mu)_\Lambda.
\end{align*}
Using the Pieri rule for multiplication by $e_r$ (citing a result not yet in the wiki: Pieri Rule for Schur functions),
\begin{align*}
e_r s_\mu
=
\sum_{\alpha \,:\, \alpha/\mu \text{ is a vertical } r\text{-strip}} s_\alpha.
\end{align*}
Taking the inner product with $s_\nu$ and using Schur orthonormality gives
\begin{align*}
(s_\nu, e_r s_\mu)_\Lambda
=
\begin{cases}
1, & \text{if } \nu/\mu \text{ is a vertical } r\text{-strip}, \\
0, & \text{otherwise}.
\end{cases}
\end{align*}
Therefore the coefficient of $s_\mu$ in $e_r^\perp s_\nu$ is $1$ precisely when $\mu \subset \nu$ and $\nu/\mu$ is a vertical $r$-strip, and is $0$ otherwise.
[/step]
[step:Recover the vertical-strip expansion from the computed coefficients]
The preceding step shows that the Schur coefficient of $s_\mu$ in $e_r^\perp s_\nu$ equals the Schur coefficient of $s_\mu$ in
\begin{align*}
\sum_{\lambda \subset \nu \,:\, \nu/\lambda \text{ is a vertical } r\text{-strip}} s_\lambda.
\end{align*}
Since the Schur functions form a basis of $\Lambda$, equality of all Schur coefficients implies
\begin{align*}
e_r^\perp s_\nu
=
\sum_{\lambda \subset \nu \,:\, \nu/\lambda \text{ is a vertical } r\text{-strip}} s_\lambda.
\end{align*}
Together with the horizontal-strip identity, this proves both adjoint Pieri rules.
[/step]
