What is Gelfand-Tsetlin basis  for an irreducible representation of sl(n)? Let us  consider, for example,  the standard irreducible $\mathfrak{sl}_3$-module 
$\Gamma_{1,1}$ with  the  highest weight $(1,1),$ $\dim \Gamma_{1,1}=8.$ 
The set of all weight of $\dim \Gamma_{1,1}$  is  $(0,0),(1,1),(2,-1),(1,-2),(-2,1),(-1,2),(-1,-1).$
Question. What is the Gelfand-Tsetlin basis for $\Gamma_{1,1}$?
As  I understood  from literature (Zhelobenko )  there is a combinatorial structure $\Lambda$  depended of $(1,1)$   such that a  basis  of $\dim \Gamma_{1,1}$   can be  labeled via  the $\Lambda$  but I cant do  it. Anybody can help?
 A: Let $V$ denote the standard representation of $\mathrm{sl}_3$. Then your $\Gamma_{1,1}$ is $V\otimes V^*/k$. To obtain a GT-basis we have to choose a flag of subspaces $0 \subset V_1 \subset V_2 \subset V_3 = V$ ($\dim V_i = i$) and restrict to $\mathrm{sl}(V_i) \subset \mathrm{sl}(V)$. First, let us take $U = V_2$. Then $V = U \oplus k$ (as $\mathrm{sl}(U)$-module), hence $\Gamma = \mathrm{sl}(U) + U + U^* + k$.
Now if $e_1,e_2,e_3$ is a basis of $V$ such that $V_i = \langle e_j\rangle_{j \le i}$ and $e^i$ is the dual basis of $V^*$ then 
$$
\mathrm{sl}(U) = \langle e_1\otimes e^2, e_2\otimes e^1, e_1\otimes e^1 - e_2\otimes e^2\rangle.
$$ 
Further,
$$
U = \langle e_1\otimes e^3, e_2\otimes e^3\rangle,\
U^* = \langle e_3\otimes e^1, e_3\otimes e^2\rangle,\
k = \langle e_1\otimes e^1 + e_2\otimes e^2 - 2e_3\otimes e^3\rangle.
$$
So, collecting all these vectors you get the GT-basis.
A: 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 right) are $1$s (resp $-1$), and all other 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).
