Flattening Young Tableaux

Let $\lambda=(\lambda_1,\lambda_2,\cdots,\lambda_k)$ be a partition with $|\lambda|=n$ and $\lambda_1\geq \lambda_2\geq\cdots\geq \lambda_k$. For any Standard Young Tableaux (SYT) $T$ of shape $\lambda$, define the "flattened tableaux" by deleting the first row $\lambda_1$, and then relabeling all entries in the tableaux with respect to their relative order, thus giving a SYT $T'$ of shape $\lambda':=(\lambda_2,\cdots,\lambda_n)$ and $|\lambda'|=n-\lambda_1$. For example, if $\lambda=(4,3,2,1,1)$ with $|\lambda|=10$, then $\lambda'=(3,2,1,1)$. An as an example, take

$T=$ \begin{array}{cccc} 1 & 3 & 5 & 6\\ 2 & 4 & 7 & \ \\ 8 & 11 & \ & \ \\ 9 & \ & \ & \ \\ 10 & \ & \ &\ \end{array}

Then,

$T'=$ \begin{array}{cccc} 1 & 2 & 3 & \ \\ 4 & 7 & \ & \ \\ 5 & \ & \ & \ \\ 6 & \ & \ &\ \end{array}

Does this operation have a common name in literature? Has it been studied before? As a map, the flattening operation $\phi: SYT(\lambda)\rightarrow SYT(\lambda')$ is clearly a surjection (and not a bijection). On the other hand, are there specific $\lambda$ for which $\phi$ is uniform in $SYT(T')$? In other words, $|\{T: \ \phi(T)=T'\}|$ is the same for all $T'$?

• Why is this map an injection? In your example, swapping 6 with 7 gives the same SYT after flattening. Dec 2, 2014 at 9:26
• @siddharth Venkatesh: meant to say surjection, pardon! Dec 2, 2014 at 15:34

For any partition $$(n)$$ or $$(k,1^{n-k})$$ , the resulting tableaux are respectively empty or a vertical one-column tableau with entries $$1,2,...,n-k$$. But for any partition $$(k,2,1,1,..,1)$$ with $$n-k-2$$ '$$1$$'s, the beheaded tableau can be either
$$((1,2),(3),..,(n-k-1))$$ or $$((1,3),(2),..,(n-k-1))$$
depending on $$T$$. So, beyond these special partitions $$(n)$$ and $$(k,1^n-k)$$ there can be no such cases.
• I think he's answering the last question of OP with "no" by showing $T'$ with differing numbers of pre-images. In any case, "beheaded tableau" is great! Aug 31, 2021 at 0:13