[proofplan]
We encode a standard tableau by the saturated chain of Young diagrams obtained from its subtableaux with entries at most $k$. The essential external input is Fomin's growth-diagram theorem for Schützenberger operators: for straight standard tableaux, the chain model identifies promotion, inverse promotion, and evacuation with three boundary operators $P$, $P^{-1}$, and $E$ satisfying the dihedral relation $E\circ P=P^{-1}\circ E$. After verifying that the hypotheses of that theorem match the present straight-shape jeu de taquin convention, we transfer the boundary identity back through the chain bijection. Finally, evacuation is an involution, so the intertwining identity is equivalent to the displayed conjugation identity.
[/proofplan]
[step:Encode a standard tableau by its chain of shapes]
Fix $T\in \operatorname{SYT}(\lambda)$. For each $k\in \{0,1,\dots,n\}$, define $\lambda_k(T)$ to be the Young diagram consisting of the boxes of $T$ whose entries lie in $\{1,\dots,k\}$. Thus $\lambda_0(T)=\varnothing$, $\lambda_n(T)=\lambda$, and $\lambda_k(T)$ is obtained from $\lambda_{k-1}(T)$ by adding exactly one box for every $1\le k\le n$.
Define
\begin{align*}
\Gamma: \operatorname{SYT}(\lambda) &\to \mathcal C_\lambda
\end{align*}
by sending $T$ to the saturated chain
\begin{align*}
\Gamma(T)=(\lambda_0(T),\lambda_1(T),\dots,\lambda_n(T)),
\end{align*}
where $\mathcal C_\lambda$ denotes the set of saturated chains of Young diagrams from $\varnothing$ to $\lambda$ of length $n$. The map $\Gamma$ is bijective: given a chain $(\mu_0,\mu_1,\dots,\mu_n)$, place the label $k$ in the unique box of $\mu_k\setminus \mu_{k-1}$ for each $1\le k\le n$. Since each $\mu_k$ is a Young diagram, the resulting filling is increasing along rows and columns, hence is a standard Young tableau.
[guided]
The point of passing from tableaux to chains is that promotion and evacuation become geometric operations on boundary data. For a tableau $T$, define $\lambda_k(T)$ to be the shape occupied by entries at most $k$. Since $T$ is standard, the entries $1,\dots,k$ form a Young diagram: if a box is present, every box weakly above and weakly to its left has a smaller entry and is therefore also present. Hence
\begin{align*}
\varnothing=\lambda_0(T)\subset \lambda_1(T)\subset \cdots \subset \lambda_n(T)=\lambda
\end{align*}
is a saturated chain, meaning that one box is added at each step.
Conversely, suppose we are given a saturated chain
\begin{align*}
\varnothing=\mu_0\subset \mu_1\subset \cdots \subset \mu_n=\lambda.
\end{align*}
For each $k\in\{1,\dots,n\}$, the skew difference $\mu_k\setminus \mu_{k-1}$ consists of one box. Put the entry $k$ in that box. Because every $\mu_k$ is a Young diagram, whenever a box is added, all boxes above it and to its left have already appeared earlier in the chain. Therefore their labels are smaller than $k$. This proves that the filling is strictly increasing along rows and columns. Thus the chain construction and the tableau construction are inverse to each other, so $\Gamma$ is a bijection.
[/guided]
[/step]
[step:Invoke Fomin's growth-diagram theorem for Schützenberger operators]
We use the following precise external result, Fomin's growth-diagram theorem for Schützenberger promotion and evacuation on straight standard tableaux. Let $\lambda$ be a Young diagram with $n$ boxes, and let $\mathcal C_\lambda$ be the set of saturated chains from $\varnothing$ to $\lambda$. There exist maps
\begin{align*}
P: \mathcal C_\lambda \to \mathcal C_\lambda
\end{align*}
a map
\begin{align*}
P^{-1}: \mathcal C_\lambda \to \mathcal C_\lambda
\end{align*}
and a map
\begin{align*}
E: \mathcal C_\lambda \to \mathcal C_\lambda
\end{align*}
with the following properties. First, under the chain bijection $\Gamma$, $P$ represents Schützenberger promotion, $P^{-1}$ represents inverse promotion, and $E$ represents Schützenberger evacuation:
\begin{align*}
\Gamma(\operatorname{pr}(T))=P(\Gamma(T)).
\end{align*}
\begin{align*}
\Gamma(\operatorname{pr}^{-1}(T))=P^{-1}(\Gamma(T)).
\end{align*}
\begin{align*}
\Gamma(\operatorname{evac}(T))=E(\Gamma(T)).
\end{align*}
Second, these boundary operators satisfy the dihedral growth-diagram relation
\begin{align*}
E\circ P=P^{-1}\circ E
\end{align*}
as maps $\mathcal C_\lambda\to \mathcal C_\lambda$.
The hypotheses of this theorem match the present setting: $T$ is a straight standard Young tableau of fixed shape $\lambda$ with entries $\{1,\dots,n\}$, promotion is the jeu de taquin operation specified in the statement, and evacuation is Schützenberger evacuation on the same set $\operatorname{SYT}(\lambda)$. The theorem's operators $P$, $P^{-1}$, and $E$ are therefore not being defined by an informal unrectified shift of positions; they are the growth-diagram boundary operators whose local rules include the necessary rectification and complementary-label conventions. Hence the displayed relation already includes the nontrivial compatibility of rectification with reversal.
[guided]
The central issue is that promotion is not merely a raw cyclic shift of the entries of the chain. A cyclic shift usually produces boundary data that must be completed by the local growth-diagram rules, and that completion is the rectification step. Therefore the proof uses Fomin's growth-diagram theorem in its full form, rather than trying to replace it by the elementary identity for shifts of a finite cyclically ordered set.
The theorem applies to exactly the objects in this proof. We have a fixed Young diagram $\lambda$ with $n$ boxes and a straight standard Young tableau $T\in\operatorname{SYT}(\lambda)$. The map $\Gamma$ sends $T$ to its saturated chain of shapes in $\mathcal C_\lambda$. Fomin's theorem provides three maps on this same chain set,
\begin{align*}
P: \mathcal C_\lambda \to \mathcal C_\lambda,
\end{align*}
\begin{align*}
P^{-1}: \mathcal C_\lambda \to \mathcal C_\lambda,
\end{align*}
and
\begin{align*}
E: \mathcal C_\lambda \to \mathcal C_\lambda.
\end{align*}
These maps are defined by the growth-diagram local rules. Those local rules are the part of the theorem that handles rectification and the complementary order in evacuation.
The theorem then gives the three intertwining identities
\begin{align*}
\Gamma(\operatorname{pr}(T))=P(\Gamma(T)),
\end{align*}
\begin{align*}
\Gamma(\operatorname{pr}^{-1}(T))=P^{-1}(\Gamma(T)),
\end{align*}
and
\begin{align*}
\Gamma(\operatorname{evac}(T))=E(\Gamma(T)).
\end{align*}
It also gives the boundary relation
\begin{align*}
E\circ P=P^{-1}\circ E.
\end{align*}
This is the precise dihedral statement needed here. It says not only that reversal changes the direction of a formal shift, but also that the growth-diagram rectification procedures and evacuation's complementary-label convention are compatible with that reversal. Thus the nontrivial growth-diagram content is supplied by the external theorem, and the remaining argument is only the transfer of this identity through the bijection $\Gamma$.
[/guided]
[/step]
[step:Transfer the boundary identity back to tableaux]
Let $T\in \operatorname{SYT}(\lambda)$. Using the intertwining identities from the growth-diagram model and the boundary relation just proved, we compute
\begin{align*}
\Gamma((\operatorname{evac}\circ \operatorname{pr})(T))=(E\circ P)(\Gamma(T)).
\end{align*}
By the boundary relation,
\begin{align*}
(E\circ P)(\Gamma(T))=(P^{-1}\circ E)(\Gamma(T)).
\end{align*}
Using the growth-diagram identifications again,
\begin{align*}
(P^{-1}\circ E)(\Gamma(T))=\Gamma((\operatorname{pr}^{-1}\circ \operatorname{evac})(T)).
\end{align*}
Since $\Gamma$ is injective, this gives
\begin{align*}
(\operatorname{evac}\circ \operatorname{pr})(T)=(\operatorname{pr}^{-1}\circ \operatorname{evac})(T).
\end{align*}
Because $T$ was arbitrary, the maps satisfy
\begin{align*}
\operatorname{evac}\circ \operatorname{pr}=\operatorname{pr}^{-1}\circ \operatorname{evac}.
\end{align*}
[/step]
[step:Use involutivity of evacuation to obtain the equivalent conjugation formula]
By the [Schützenberger evacuation theorem](/theorems/8461) [citetheorem:8461], evacuation is an involution on standard Young tableaux:
\begin{align*}
\operatorname{evac}\circ \operatorname{evac}=\operatorname{id}_{\operatorname{SYT}(\lambda)}.
\end{align*}
Composing the identity
\begin{align*}
\operatorname{evac}\circ \operatorname{pr}=\operatorname{pr}^{-1}\circ \operatorname{evac}
\end{align*}
on the right by $\operatorname{evac}$ gives
\begin{align*}
\operatorname{evac}\circ \operatorname{pr}\circ \operatorname{evac}=\operatorname{pr}^{-1}\circ \operatorname{evac}\circ \operatorname{evac}.
\end{align*}
Using $\operatorname{evac}^2=\operatorname{id}_{\operatorname{SYT}(\lambda)}$, the right-hand side becomes $\operatorname{pr}^{-1}$. Hence
\begin{align*}
\operatorname{evac}\circ \operatorname{pr}\circ \operatorname{evac}=\operatorname{pr}^{-1}.
\end{align*}
Conversely, composing this conjugation identity on the right by $\operatorname{evac}$ and again using $\operatorname{evac}^2=\operatorname{id}_{\operatorname{SYT}(\lambda)}$ recovers
\begin{align*}
\operatorname{evac}\circ \operatorname{pr}=\operatorname{pr}^{-1}\circ \operatorname{evac}.
\end{align*}
Thus the two displayed formulations are equivalent, and the theorem follows.
[/step]
