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Let $\mathfrak{g}$ be the complex Lie algebra $\mathfrak{sl}_3(\Bbb{C})$. Consider adjoint representation $\textrm{ad} : \mathfrak{sl}_3(\Bbb{C}) \to \textrm{gl}(\mathfrak{g})$. $\mathfrak{g}$ has the usual complex basis

$\begin{eqnarray*} H_1 &=& \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right), H_2 &=& \left(\begin{array}{ccc} 0& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right),\\ X_1 &=& \left(\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right), X_2 &=& \left(\begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right), X_3 &=& \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right),\\ Y_1&=& \left(\begin{array}{ccc} 0& 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right), Y_2 &=& \left(\begin{array}{ccc} 0& 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right), Y_3 &=& \left(\begin{array}{ccc} 0& 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right)\end{eqnarray*}.$

Now by restricting $\textrm{ad}$ to just the vectors $H_1,X_1$ and $Y_1$ I can get an 8 dimensional representation of $\mathfrak{sl}_2(\Bbb{C})$. Suppose I wish to decompose this representation into irreducibles. Now I have checked that $\textrm{span}\{X_2,X_3\}$ and $\textrm{span}\{Y_2,Y_3\}$ are irreducible 2 - dimensional subrepresentations. Now there are still the vectors $H_1,H_2,X_1,Y_1$ whose span I have tried to check is irreducible. If we write down

$\textrm{ad}_{H_1}, \textrm{ad}_{X_1}, \textrm{ad}_{Y_1}, $

in the basis $H_1,H_2,X_1,Y_1,X_2,X_3,Y_2,Y_3$ (in this order) we get

$\begin{eqnarray*} \textrm{ad}_{H_1} &=& \left(\begin{array}{cccc|cc|cc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0& 0 & 2 & 0 \\ 0& 0& 0 & -2 \\ \hline &&&& -1 & 0 \\ &&&& 0 & 1 \\ \hline &&&&&&& 1 & 0 \\&&&&&&& 0 & -1 \end{array}\right) \textrm{ad}_{X_1} &=& \left(\begin{array}{cccc|cc|cc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ -2& 1 & 0 & 0 \\ 0& 0& 0 & -2 \\ \hline &&&& 0 & 0 \\ &&&& 1 & 0 \\ \hline &&&&&&& 0 & -1 \\&&&&&&& 0 & 0 \end{array}\right) \\ \textrm{ad}_{Y_1} &=& \left(\begin{array}{cccc|cc|cc} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 2& -1& 0 & 0 \\ \hline &&&& 0 & 1 \\ &&&& 0 & 0 \\ \hline &&&&&&& 0 & 0 \\&&&&&&& 1 & 0 \end{array}\right). \\ \end{eqnarray*}$

From looking at the first $4 \times 4$ block in each matrix it seems that the span $\{H_1,H_2,X_1,Y_1\}$ is irreducible. How do I prove this formally in an elegant way without bashing?

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    We know that the eigenvector $H$ must be in $\mathfrak{h}$, so let $H = \text{diag}(a,b,c)$, $a + b + c = 0$. Automatically, $\text{ad}_{H_1}(H) = 0$ and I get that $\text{ad}_{X_1}(H) = \text{ad}_{Y_1}(H) = 0$ gives the condition $a = b$. Thus, you can take $H$ to be $\text{diag}(1,1,-2)$ and $V_0 = \mathbb{C} \cdot H$.2012-09-14

1 Answers 1

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There are some typos in your formulas.

The way to decompose your representation is to first look for the eigenvalues of the Cartan subalgebra. In the case of $\mathfrak{sl}_2$, the Cartan subalgebra is one dimensional, spanned by $H_1$. It looks like you have eigenvalues of 2, 1, 1, 0, 0, -1, -1, and -2 (accounting for typographical error as posted). This tells you that your representation decomposes as $ V_2 \oplus V_1 \oplus V_1 \oplus V_0, $ where $V_k$ denotes the $k+1$-dimensional irreducible representation with highest weight $k$. (That is, $V_k = \text{Sym}^k V$ where $V$ is the standard representation of $\mathfrak{sl}_2$.)

You've found the two copies of $V_1$ already. To identify $V_2$, first find a highest weight vector $v$ of weight $2$ (i.e. an eigenvector of $\text{ad}_{H_1}$ with eigenvalue $2$) -- you've already done this in your calculation -- and then successively apply $Y_1$ to it: $Y_1(v)$ will have weight $0$ and $Y_1^2(v)$ will have weight $-2$. Then $v, Y_1(v), Y_1^2(v)$ will be a basis of $V_2$.

I would recommend reading Chapters 11-13 in Fulton and Harris' text, as they go into these types of calculations in great depth.

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    Thanks, I'll try to do that. Can you have a look at my comment in reply to MattE's comment above? Somehow I get$0$out of this calculation....2012-09-14