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Consider an $n$-dimensional complex vector space $V$ with a chosen basis $e_1,\ldots,e_n$. This basis defines a Cartan decompostion of $GL(V)\cong GL_n$ and for an (integral dominant) highest weight $\lambda$ we may consider the irreducible representation $L_\lambda$. In $L_\lambda$ we have the Gelfand-Tsetlin basis for the chain $$GL(V)\supset GL(\mathrm{span}(e_1,\ldots,e_{n-1}))\supset\ldots\supset GL(\mathbb{C}e_1).$$

If we also assume that $\lambda$ is a partition (its coordinates are nonnegative integers), then we may consider the Schur module $\mathbb S^\lambda V$ as a factor of the tensor product $$\Lambda=(\wedge^1V)^{\otimes(\lambda_1-\lambda_2)}\otimes(\wedge^2V)^{\otimes(\lambda_2-\lambda_3)}\otimes\ldots\otimes(\wedge^nV)^{\otimes\lambda_n}.$$ Tableaux of shape $\lambda$ with pairwise distinct elements from $\{1,\ldots,n\}$ in each column enumerate the obvious basis in $\Lambda$. Images of the vectors corresponding to semistandard tableaux are known to comprise a basis in $\mathbb S^\lambda V$.

Now, the representations $L_\lambda$ and $\mathbb S^\lambda V$ of $GL(V)$ can be naturally identified up to a scalar multiple. Thus, we have two bases in the same space. For many years I kind of assumed these bases were the same simply via the standard weight preserving bijection between GT patterns and SSYTs. When I finally decided to take a closer look, however, they turned out to be distinct even for $GL_3$.

So my question is: is there some concrete connection between these bases? Some sort of duality, perhaps?

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    $\begingroup$ I'm pretty certain that w.r.t. a correctly (and I expect robustly) chosen order, the change-of-basis matrix is upper triangular with $1$s on the diagonal. But I can't point you to a proof of this. $\endgroup$ – Allen Knutson Apr 30 '17 at 11:23
  • $\begingroup$ @AllenKnutson Yes, you're right, one may read GT patterns from top to bottom and from left to right within each row and compare the resulting words lexicographically. Accordingly, for an SSYT one may write down a word consisting of the parts of the subdiagram comprised of the elements no greater than $n-1$, then the parts of the subdiagram comprised of the elements no greater than $n-2$ and so on. One then again compares the words lexicographically. The corresponding matrix will be unitriangular. I am, of course, secretly hoping for a stronger connection... $\endgroup$ – imakhlin May 1 '17 at 18:02
  • $\begingroup$ I wouldn't hold out much hope. The point being, the most natural (e.g. group-independent) things you can do produce Lusztig's canonical and semicanonical bases, and those are likewise related by such upper-triangularity. The only stronger connection I might imagine, for your two bases, is if one of the two change-of-basis matrices has natural-number entries (as in the case of canonical expanded in semicanonical). I'm pretty certain that neither of your bases is the canonical or semicanonical, by the way. $\endgroup$ – Allen Knutson May 2 '17 at 13:00
  • $\begingroup$ I think the usual correspondence of SSYTs an GT patterns identifies the SSYT basis as described in Fultons book on Young tableaux with the GT basis for the chain $\mathrm{GL}(V)\supset\mathrm{GL}(\mathrm{span}(e_1,\dots,e_{n-1}))\supset\cdots\supset \mathrm{GL}(\mathbb C e_1)$. The correspondence works by considering the sequence of Young diagrams you obtain by first removing boxes labelled $n$, then $n-1$ etc., so you should do the same in your chain of subgroups. $\endgroup$ – Christoph May 31 '17 at 6:30
  • $\begingroup$ @Christoph Here's the counterexample I mentioned. Consider $\mathfrak{gl}_3$ and the highest weight $\rho=(2,1,0)$. If the claim were true, then the subspace of the Schur module spanned by all semistandard tableaux with "3"-s in a given set of boxes would be $\mathfrak{gl}_2$-invariant. Now consider $f_{1,2}\in\mathfrak{gl}_2$ mapping $e_1$ to $e_2$ and $e_2$ to 0. The straightening algorithm provides: $\endgroup$ – imakhlin Jun 4 '17 at 15:07

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