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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?

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See arXiv:math/0211289v2 [math.RT]. –  Claudio Gorodski Jun 23 '11 at 19:25
    
@Claudio I have read the preprint –  Melania Jun 23 '11 at 20:08
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2 Answers

Let $V$ denote the standard representation of $sl_3$. 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 $sl(V_i) \subset sl(V)$. First, let us take $U = V_2$. Then $V = U \oplus k$ (as $sl(U)$-module), hence $\Gamma = sl(U) + U + U^* + k$. Now if $e_1,e_2,e_3$ is a basis of $V$ such that $V_i = <e_j>_{j \le i}$ and $e^i$ is the dual basis of $V^*$ then $$ sl(U) = < e_1\otimes e^2, e_2\otimes e^1, e_1\otimes e^1 - e_2\otimes e^2>. $$ Further, $$ U = <e_1\otimes e^3, e_2\otimes e^3>,\ U^* = <e_3\otimes e^1, e_3\otimes e^2>,\ k = < e_1\otimes e^1 + e_2\otimes e^2 - 2e_3\otimes e^3 >. $$ So, collecting all these vectors you get the GT-basis.

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Thank you for the exaple –  Melania Jun 24 '11 at 19:46
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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 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).

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Thank you. How to reduce $gl_n$ module to $sl_n$ module? –  Melania Jun 24 '11 at 19:47
    
The difference is that $\mathfrak{gl}_n,$ unlike $\mathfrak{sl}_n,$ has one-dimensional center (scalar matrices). Thus every simple $\mathfrak{gl}_n$-module remains simple under restriction to $\mathfrak{sl}_n$, but the same $\mathfrak{sl}_n$-module can be extended to a one-parameter family of $\mathfrak{gl}_n$-modules by making the center act as an arbitrary scalar. –  Victor Protsak Jun 25 '11 at 5:11
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