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Consider $T$ to be a standard Young tableau of shape $\lambda$ (in English notation). The descent set of $T$, $Des(T)$, is defined to the set of all positive integers $i$ such that $i+1$ lies strictly south (and weakly west) of $i$ in $T$.

A flip on $T$ is defined to be a move where we exchange the positions of $i$ and $i+1$ in $T$ if $i\in Des(T)$, provided that $i$ and $i+1$ are not the same column. Let $T'$ be another standard Young tableau obtained by doing a sequence of flips starting from $T$.

$\textbf{Question}$: Is it true that $Des(T)\neq Des(T')$?

It seems to be true from the examples I worked out, and seemed to be sort of result that would be present in literature. If this is indeed the case, I'd appreciate any reference that states this.

Thanks.

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  • $\begingroup$ The tag 'reference-request' might be appropriate for this question. $\endgroup$ – Ricardo Andrade Aug 1 '13 at 3:01
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Here is a counterexample copied out of a Sage session:

sage: T = StandardTableau([[1,2,3,7,9],[4,5,8],[6,11],[10]]); T.pp()
  1  2  3  7  9
  4  5  8
  6 11
 10
sage: T.standard_descents()                                         
[3, 5, 7, 9]
sage: T = StandardTableau(T.bender_knuth_involution(9)); T.pp()     
  1  2  3  7 10
  4  5  8
  6 11
  9
sage: T.standard_descents()                                    
[3, 5, 7, 8, 10]
sage: T = StandardTableau(T.bender_knuth_involution(10)); T.pp()    
  1  2  3  7 11
  4  5  8
  6 10
  9
sage: T.standard_descents()
[3, 5, 7, 8]
sage: T = StandardTableau(T.bender_knuth_involution(8)); T.pp()     
  1  2  3  7 11
  4  5  9
  6 10
  8
sage: T.standard_descents()                                    
[3, 5, 7, 9]

Some comments on the syntax:

T.pp() is short for "pretty-print of T"; this prints the tableau T as a table rather than as a list of lists.

T.standard_descents() gives the descent set of T. Why is it not called T.descents() ? Because T.descents() computes something different (some kind of measure for the deviation of T from semistandardness -- hence, completely useless for standard tableaux). It is an artefact of history that it is the latter method, not the former, which has the shorter name.

Calling the Bender-Knuth involution method is a bit of an overkill here since all we are actually doing is switching two entries; an alternative would be to use T.symmetric_group_action_on_values(p) for a permutation p. Either way one has to explicitly cast the result into the StandardTableau class in order to apply the standard_descents() method, because T.bender_knuth_involution(i) returns a SemistandardTableau, while T.symmetric_group_action_on_values(p) returns just a Tableau.

If you find a descent statistic that is actually strictly semiinvariant under these flips, I'm curious to know.

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  • $\begingroup$ Oh, I forgot: this requires a sage-5.10 at least and patch #7983. $\endgroup$ – darij grinberg Aug 1 '13 at 14:20
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To the contrary, it can be shown that any two SYT of the same shape are related by a sequence of flips. An explicit negative answer is given by $$ \begin{array}{ll}1246 & 1246\\ 35 & 37\\ 7 & 5 \end{array}. $$

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  • $\begingroup$ Thanks for the response. I might be wrong but I don't think I can go from the tableau on the left to the one on the right (or the other way around) by doing flips of the type I mentioned. $\endgroup$ – user61318 Aug 1 '13 at 6:56
  • $\begingroup$ If the shape of the tableau is (4,2,1) as above, then the tableau whose column reading word is 3215467 can not be taken to any other tableau, as all descents are the sorts that can not be flipped. Hence, I don't think that any two SYTs are related by a sequence of flips. $\endgroup$ – user61318 Aug 1 '13 at 6:57
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    $\begingroup$ The flips Stanley is talking about are supposed to be invertible, while the ones of the OP are not. $\endgroup$ – darij grinberg Aug 1 '13 at 13:48

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