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Note: All Young diagrams are to use the English notation scheme.

Suppose I have two tableaux $T$ and $T'$ on the same Young diagram (we insert the numbers $1, 2, \ldots, n$ in two different ways in the diagram). We say that $T > T'$ if the first entry (while reading left to right, top to bottom) in which $T$ and $T'$ differs is such that the entry of $T$ is larger than that entry of $T'$.

Is it true that if $T > T'$, then there exist $a \neq b$, $\{a, b\} \subset \{1, 2, \ldots, n\}$ such that $\{a, b\}$ appears in the same row of $T$ and $\{a, b\}$ appears in the same column of $T'$?

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If you are going to say «left to right, top to bottom» you need explain how you draw the diagrams, for there are two different conventions (the English and the French conventions...) which will result in two different orders. –  Mariano Suárez-Alvarez Feb 23 '12 at 2:21
    
Woah. Did not realize that there were English/French conventions. Thanks, edited. –  678293 Feb 23 '12 at 2:24
    
I suspect you mean standard tableaux? If not, you might as well write $T\ne T'$, since the order wouldn't matter; also the statement would be evidently false, since you could have different tableaux on a diagram with only one row or only one column. –  joriki Feb 23 '12 at 2:33

1 Answer 1

Suppose $1,2,3,\dots,17$ appear in the same places in $T$ and $T'$, but $17+1=18$ is in location $X$ in $T$ and location $Y$ in $T'$, with $X\ne Y$. Convince yourself that $X$ and $Y$ can't be in the same row, and can't be in the same column. Assume without loss of generality that $X$ is above and to the right of $Y$ (I don't know if this is English or French; my entries increase left to right, and top to bottom). Consider the location at the intersection of the row containing $X$ and the column containing $Y$. The number in that location is in the same row as 18 in $T$ and in the same column as 18 in $T'$.

Of course, 17 is a variable.

EDIT: Let's look at an example. Suppose your two tableaux both look like $$\matrix{1&2&5&6&12&13&a\cr3&7&8&10&17&b&\cr4&14&15&c&&&\cr9&16&d&&&&\cr11&e&&&&&\cr f&&&&&&\cr}$$ That is, they agree up to 17. In each, 18 must go to one of the six locations $a,b,c,d,e,f$. Suppose it goes to $b$ in $T$ and to $d$ in $T'$. The row containing $b$ and the column containg $d$ meet at $8$, so $8$ and $18$ are in the same row in $T$ and in the same column in $T'$. Similarly for any other choice of two of the 6 possible locations for $18$.

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Are you sure this works for non-prime values of $17$?! –  Mariano Suárez-Alvarez Feb 24 '12 at 7:03
    
I'm sure you're kidding, but just for the record the divisibility status of 17 was not used anywhere in the proof. –  Gerry Myerson Feb 24 '12 at 11:31
    
This is the English style. –  Brian M. Scott Feb 25 '12 at 5:28

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