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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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Dear BenjaLim, Do you know highest weight theory for $\mathfrak sl_2$? If so, have you tried computing the weights of your $8$-dimensional representation? Regards, –  Matt E Sep 14 '12 at 3:09
    
@MattE I have had a look at the releveant section in Fulton and Harris. Now I am trying to see what would span $V_0$ (as in Michael's answer below). Such a vector to span it would have to be a simultaneous eigenvector for $\textrm{ad}_{H_1},\textrm{ad}_{X_1},\textrm{ad}_{Y_1}$. Suppose such a vector is $(a,b,c,d)$. Then let $\textrm{ad}_{H_1}$ act on it. We see that necessarily $a = b = 0$. Letting $\textrm{ad}_{Y_1}$ act on it we see that $d =0$. Finally letting $\textrm{ad}_{X_1}$ act on it shows that $c =0$ and I get no such vector. What's happening? –  fpqc Sep 14 '12 at 8:50
    
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 $V_0 = \mathbb{C} \cdot H$. –  Michael Joyce Sep 14 '12 at 12:34
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1 Answer

up vote 3 down vote accepted

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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Yes in my matrix for $\textrm{ad}_{H_1}$ the entry $a_{88}$ should be $-1$ and not $1$. –  fpqc Sep 14 '12 at 7:22
    
I have read the relevant chapter in Fulton and Harris. Now we can concentrate on the four dimension subrepresentation spanned by $\{H_1,H_2,X_1,Y_1\}$. Now by looking at the eigenvalues of $H_1$ that appear in this block, I know that it must decompose as a direct sum of a 3-dimensional representation and a $1$ - dimensional representation. Now for my 3 -dimensional representation I get it is spanned by $2X_1,-2Y_1$ and $-H_1$. However, I don't seem to be able to see what will span $V_0$. –  fpqc Sep 14 '12 at 8:35
    
@BenjaLim: $V_0$ will be spanned by a weight vector with weight $0$, so an element of $\mathfrak{h} \subset \mathfrak{sl}_3$, where $\mathfrak{h}$ is the subalgebra of diagonal matrices. So you can directly compute which linear combination of $H_1$ and $H_2$ is closed under $\text{ad}_{H_1}, \text{ad}_{X_1}, \text{ad}_{Y_1}$. Or you can take any $H \in \mathfrak{h}$ and use, say, $X_1$ to project it into the weight $2$ space. Then add an appropriate multiple of $H_1$ so that it projects to the zero vector in the weight $2$ space. This will be your generator for $V_0$. –  Michael Joyce Sep 14 '12 at 11:53
    
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.... –  fpqc Sep 14 '12 at 11:58
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