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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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    $\begingroup$ 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, $\endgroup$ – Matt E Sep 14 '12 at 3:09
  • $\begingroup$ @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? $\endgroup$ – user38268 Sep 14 '12 at 8:50
  • $\begingroup$ 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 $V_0 = \mathbb{C} \cdot H$. $\endgroup$ – Michael Joyce Sep 14 '12 at 12:34
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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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  • $\begingroup$ Yes in my matrix for $\textrm{ad}_{H_1}$ the entry $a_{88}$ should be $-1$ and not $1$. $\endgroup$ – user38268 Sep 14 '12 at 7:22
  • $\begingroup$ 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 $V_0$. $\endgroup$ – user38268 Sep 14 '12 at 8:35
  • $\begingroup$ @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 $V_0$. $\endgroup$ – Michael Joyce Sep 14 '12 at 11:53
  • $\begingroup$ 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.... $\endgroup$ – user38268 Sep 14 '12 at 11:58

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