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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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  • $\begingroup$ See arXiv:math/0211289v2 [math.RT]. $\endgroup$ Jun 23, 2011 at 19:25
  • $\begingroup$ @Claudio I have read the preprint $\endgroup$
    – Melania
    Jun 23, 2011 at 20:08

2 Answers 2

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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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  • $\begingroup$ Thank you. How to reduce $gl_n$ module to $sl_n$ module? $\endgroup$
    – Melania
    Jun 24, 2011 at 19:47
  • $\begingroup$ 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. $\endgroup$ Jun 25, 2011 at 5:11
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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.

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