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It's more common to talk about the GT basis of $\mathfrak{gl}_n$-modules. In the case of the adjoint representation, the first row of the GT scheme is $(1,0,\ldots,0,-1)$ and each subsequent row satisfies the interlacing condition, which implies that the left (resp, right) edge consists of a string of $1$s (resp $-1$s), followed by a string of $0$s (possibly empty), except that the bottom entry can be $1$ (resp $-1$) if all entries on the left (resp rightedge consist of a string of ) are $-1$s 1$s (resp $-1$), and all other entriees entries are zeros. It follows that the whole scheme can be reconstructed from the positions of the lowest $1$ along the left edge of the triangle and the lowest $-1$ along the right edge. They may occupy any of the $n$ possible rows/positions each, except that both cannot occur in the $n$th row, corresponding to the impossibility of the bottom entry being simultaneously $1$ and $-1.$

In your special case $n=3$ the diagrams will look like this:

$$ \begin{array}{rrrrr} 1 & & 0 & & -1\\ & 1 & & 0 & \\ & & 1 & & \end{array} \quad $$

(this is the highest weight; there are 7 more).

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It's more common to talk about the GT basis of $\mathfrak{gl}_n$-modules. In the case of the adjoint representation, the first row of the GT scheme is $(1,0,\ldots,0,-1)$ and each subsequent row satisfies the interlacing condition, which implies that the left (resp, right) edge consists of a string of $1$s (resp $-1$s), followed by a string of $0$s (possibly empty), the right edge consist of a string of $-1$s and all other entriees are zeros. It follows that the whole scheme can be reconstructed from the positions of the lowest $1$ along the left edge of the triangle and the lowest $-1$ along the right edge.

In your special case $n=3$ the diagrams will look like this:

$$ \begin{array}{rrrrr} 1 & & 0 & & -1\\ & 1 & & 0 & \\ & & 1 & & \end{array} \quad $$

(this is the highest weight; there are 7 more).