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.



Flips on standard Young tableaux and descent sets
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.


