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I'm having trouble finding some elementary results on the following. Let $Y$ be a standard Young Tableau of shape $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_n)$ with $N:=\sum_{i=1}^n\lambda_i$.

My question is, for a given element $k$, what can be said about the location of $k$ in terms of bounds on its $(i,j)$ coordinates in the tableau? For example, 1 must be at $(1,1)$, 2 can be at $(2,1)$ or $(1,2)$, 3 can be at $(2,1),(3,1),(1,2),(1,3)$ and so on. What I'm looking for is some crude statement of the form: $k$ lies in the region bounded by intervals $[(i_1,j_1),(i_1,j_1')],\ldots,[(i_n,j_n),(i_n,j_n')]$ where the intervals are defined between the two bracketed coordinates. References would be highly appreciated!

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There exists a standard tableau of given shape with $k$ in the $(i,j)$ position if and only if $i+j\le k+1$. It is not clear to me what a strip is defined to be. –  anon Jun 4 '13 at 22:31
    
@anon: could you clarify your condition? If $k=2$ it says that 2 can be at $(1,1)$ which is impossible because 1 must be there. I've renamed strips to intervals. To clarify I am trying to isolate the intervals on each horizontal slice that are allowable for $k$ to be in –  Alex R. Jun 4 '13 at 23:14
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Ah, "standard" here meaning not only that columns and rows are strictly increasing, but also that every number $1$ to $n$ appears precisely once. –  anon Jun 4 '13 at 23:25
    
Is the shape $\lambda$ fixed? So the bounds will depend on the geometry of $\lambda$? Or are you asking, for all SYTs of all shapes, what is the set of $(i,j)$ that occur as the coordinates of $k$? –  S123 Jun 5 '13 at 0:47
    
@Steve: the shape is fixed. For simplicity you can assume that it's triangular (if this is more tractable) –  Alex R. Jun 5 '13 at 6:39
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1 Answer

up vote 3 down vote accepted

The lower bound is $k\ge ij$, as we need to put the numbers $1,\ldots, k-1$ "below" $k$. The upper bound depends on the tableau. We cannot place $k$ at $(i,j)$ if we cannot place $k-1$ elements in the columns less than $i$ and the rows less than $j$. This translates to $$k \leq \sum_{k=1}^{i-1}\lambda_k + \sum_{l=1}^{j-1}\lambda'_l - (i-1)(j-1)$$ where $\lambda'_1,\lambda'_2,\ldots$ describe the conjugate tableau. (I forget the standard notation for this.) We need to subtract $(i-1)(j-1)$ to avoid double counting.

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Thanks! This looks perfect. Do you happen to know of a reference that talks about this and possibly similar results? –  Alex R. Jun 6 '13 at 20:06
    
Unfortuantely, no. My knowledge of Young tableau comes mostly from volume 2 of Enumerative Combinatorics, but I don't remember if this is addressed. –  deinst Jun 6 '13 at 20:28
    
The upper bound is more easily given as $k\leq N-|\mu|$ where $\mu$ is the diagram obtained by chopping off the first $i-1$ row and the first $j-1$ columns from the diagram of$~\lambda$ (which makes $(i,j)$ the corner square). Both lower and upper bounds are easily seen to be sharp. –  Marc van Leeuwen Jun 7 '13 at 12:00
    
I interpreted, and edited the answer, as giving bounds on $k$ for a given square. If instead $k$ is given then the same conditions are easily interpreted as restricting to candidate squares $(i,j)$ to some region. –  Marc van Leeuwen Jun 7 '13 at 12:04
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