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Let $T_{\lambda}$ be the set of standard young tableaux (SYT) of shape $\lambda_1\geq \lambda_2\cdots\geq \lambda_n$. Now consider pushing a row $\mu$ with $\mu\geq \lambda_1$ onto $Y$ to give shape $\mu\geq \lambda_1\geq\cdots \lambda_n$. Call the resulting set of SYT $T_{\mu\cup \lambda}$. I'm curious whether there's any natural correspondence between $T_{\lambda}$ and $T_{\mu\cup\lambda}$. Specifically, if we count the cardinalities $|T_\lambda|=f_{\lambda}$ and $|T_{\mu\cup \lambda}|=f_\lambda$ via the hook-length formula, then there's an easy relation of the form:

$$f_{\mu\cup \lambda}=f_{\lambda}\frac{(|\lambda|+\mu)!}{|\lambda|!}\frac{1}{\prod_{j=1}^\mu h(1,j)},$$

where $h(1,j)$ are the hooks of the first row of shape $\mu\cup\lambda$.

Question 1: Is there a direct way to go from uniformly random $T_{\lambda}$ to $T_{\lambda\cup\mu}$? In other words, given an SYT $t\in T_{\lambda}$, is it possible to construct a uniformly random $q\in T_{\mu\cup \lambda}$? OR, is it possible to go from uniformly random $q$ to uniformly random $t$ by killing off row $\mu$ and then applying a flattening procedure to the rest of the entries?

Question 2: The formula for $f_{\mu\cup \lambda}$ factors into a piece $f_{\lambda}$ and $\frac{(|\lambda|+\mu)!}{|\lambda|!}\frac{1}{\prod_{j=1}^\mu h(1,j)}$. Thus it looks like there's a bijection between $T_{\mu\cup\lambda}$ and the pair $(T_{\lambda},H)$ where $H$ is some space of multiplicities with $|H|=\frac{(|\lambda|+\mu)!}{|\lambda|!}\frac{1}{\prod_{j=1}^\mu h(1,j)}$.

My thoughts on both these questions are limited to thinking that perhaps there's a way to apply (modify?) the Novelli-Pak-Stoyanovskii shuffling algorithm to go from $q$ to $t$. Specifically, the algorithm presents a bijection for the identity:

$$|\lambda|!=f_{\lambda}\prod_{(i,j)\in\lambda}h(i,j)$$

via $S_{|\lambda|} \rightarrow (T_{\lambda},H)$, where $S_{|\lambda|}$ is the space of permutations of length $|\lambda|$ and $H$ is a complementary space of certain multiplicities involved during the algorithm.

So maybe the right thing to look at is:

$$|\lambda|!\cdot f_{\mu\cup\lambda}=(|\lambda+\mu)!\cdot f_{\lambda}\cdot \frac{1}{h(1,j)}?$$

To use the algorithm for a shape $\lambda$, one picks a uniformly random permutation of size $|\lambda|$ and then applies a kind of bubble-sorting algorithm to the result. It's clear to me that if one looks at $\lambda\cup\mu$ instead, then the only difference is when this bubble sorting is working with entries on the first row $\mu$.

Unfortunately, I can't quite see how to make sense of this because somehow one needs to keep track of what's going on in row $\mu$. Is there a better way?

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