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