Timeline for What is the most general "two in one row for A & in one column for B" theorem?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Feb 8, 2012 at 18:56 | vote | accept | darij grinberg | ||
| Jul 18, 2010 at 21:41 | comment | added | darij grinberg | So, we can permute the entries within each column of $B$ in such a way that the element that appears in row $1$ of $A$ ends up in row $1$ of $B$, the element that appears in row $2$ of $A$ ends up in row $2$ of $B$, ..., that appears in row $i$ of $A$ ends up in row $i$ of $B$. In other words, every element in the $j$-th row of (my modified tableau) $B$ lies in the $j$-th row of $A$, for every $j$. But this modified tableau $B$ can also be obtained from the tableau $A$ by permuting the entries within each row. So (a) is proven. | |
| Jul 18, 2010 at 21:31 | comment | added | darij grinberg | Okay, it doesn't help with proving (c). But let me remark that your proof also yields (a): Again suppose that any two entries in the same row in $A$ are in different columns in $B$. Then, your argument shows that each column of $B$ has exactly one element in row $1$ of $A$, exactly one element in row $2$ of $A$, ..., exactly one element in row $i$ of $A$, where $i$ is the length of this column (because otherwise, your proof of $\lambda_1+...+\lambda_i\geq \mu_1+...+\mu_i$ could be strengthened to a proof of the strict inequality $\lambda_1+...+\lambda_i > \mu_1+...+\mu_i$ which is absurd). | |
| Jul 18, 2010 at 21:14 | comment | added | darij grinberg | Thanks for the nice proof. I'm going to see how much of it survives for different-sized tableaux. | |
| Jul 18, 2010 at 21:14 | comment | added | darij grinberg | Since you require the tableaux to have the same number of cells, this "dominance" is indeed the same as "majorization". I am more used to the latter notion since I took interest at elementary inequalities long before I had an idea what a representation is. | |
| Jul 13, 2010 at 12:52 | comment | added | Matt Fayers | I've edited in response to your comments; I hope this makes sense. I've heard 'majorization' as a synonym for dominance, but in representation theory everyone says dominance. | |
| Jul 13, 2010 at 12:50 | history | edited | Matt Fayers | CC BY-SA 2.5 |
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| Jul 13, 2010 at 12:33 | history | edited | Matt Fayers | CC BY-SA 2.5 |
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| Jul 13, 2010 at 6:55 | comment | added | darij grinberg | Wait, do you require $\lambda_1+...+\lambda_n=\mu_1+...+\mu_n$? So is this the majorization order? | |
| Jul 13, 2010 at 6:54 | comment | added | darij grinberg | Thank you, I didn't know of the dominance order. (It seems, however, that the usual trick for proving (a), (b), (c) won't work here...) | |
| Jul 12, 2010 at 21:06 | history | answered | Matt Fayers | CC BY-SA 2.5 |