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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 $T'$?

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  • $\begingroup$ Why is this map an injection? In your example, swapping 6 with 7 gives the same SYT after flattening. $\endgroup$ – Siddharth Venkatesh Dec 2 '14 at 9:26
  • $\begingroup$ @siddharth Venkatesh: meant to say surjection, pardon! $\endgroup$ – Alex R. Dec 2 '14 at 15:34
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For any partition (n) or (k,1^n-k) alias (k,1,1,..,1) with n-k '1's, the resulting tableaux are always () alias empty tableau or ((1),(2), ...,(n-k)) alias a vertical one-column tableau. Bur 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.

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