# Unitary irreducible representation of $\mathfrak{sl}\left( 3, \mathbb{C} \right)$ and the inner product of some weight vectors

I came across this question in the book "Lie Groups, Lie Algebra and Representations" by Brian C Hall.

Let $$\sigma: \mathfrak{sl} \left( 3, \mathbb{C} \right) \rightarrow \mathfrak{gl} \left( V \right)$$ be an irreducible representation with highest weight $$\mu = \left( m_1, m_2 \right)$$. Let $$V$$ be an inner product space such that the representation $$\sigma$$ is unitary, i.e., for all $$Z \in \mathfrak{su} \left( 3 \right)$$, we have $$\sigma \left( Z \right)^* = - \sigma \left( Z \right)$$. Let $$v_0 \in V$$ be a unit weight vector corresponding to the highest weight $$\mu$$. Define $$u_1 = \sigma \left( Y_1 \right) \sigma \left( Y_2 \right) v_0$$ and $$u_2 = \sigma \left( Y_2 \right) \sigma \left( Y_1 \right) v_0$$. Then, we have the following: $$\langle u_1, u_1 \rangle = m_2 \left( m_1 + 1 \right), \ \langle u_2, u_2 \rangle = m_1 \left( m_2 + 1 \right), \ \langle u_1, u_2 \rangle = m_1m_2.$$

The hint is given to prove that $$\sigma \left( X_i \right)^* = \sigma \left( Y_i \right)$$. I do quite understand how to prove this. If we want to use the commutation relations, then $$\sigma \left( H_1 \right)$$ or $$\sigma \left( H_2 \right)$$ will come into the expression. Any help with this is appreciated!

For reference, the symbols $$\alpha_1 = \left( 2, -1 \right)$$ and $$\alpha_2 = \left( -1, 2 \right)$$ are the positive simple roots of $$\mathfrak{sl}\left( 3, \mathbb{C} \right)$$.

For $$\mathfrak{sl}\left( 3, \mathbb{C} \right)$$, we have use the following basis

$$H_1 = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{matrix} \right], \ H_2 = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right],$$ $$X_1 = \left[ \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right], \ X_2 = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right], \ X_3 = \left[ \begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right],$$ $$Y_1 = \left[ \begin{matrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right], \ Y_2 = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right], \ Y_3 = \left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix} \right].$$

Also, we have the following commutation relations:

$$[H_1, X_1] = 2X_1, [H_1, Y_1]=-2Y_1, [X_1, Y_1] = H_1, [H_2, X_2] = 2X_2, [H_2, Y_2] = -2Y_2, [X_2, Y_2] = H_2, [H_1, H_2] = 0,$$ $$[H_1, X_2] = -X_2, [H_1, X_3] = X_3, [H_1, Y_2] = Y_2, [H_1, Y_3] = -Y_3, [H_2, X_1] = -X_1, [H_2, X_3] = X_3, [H_2, Y_1] = Y_1,$$ $$[H_2, Y_3] = -Y_3, [X_1, X_2] = X_3, [X_1, Y_2] = 0, [X_1, Y_3] = -Y_2, [X_2, X_3] =0, [X_2, Y_1] = 0, [X_2, Y_3] = Y_1,$$ $$[X_3, Y_1] = -X_2, [X_3, Y_2] = X_1, [X_3, Y_3] = H_1 + H_2, [Y_1, Y_2] = -Y_3, [Y_1, Y_3] = 0, [Y_2, Y_3] = 0.$$

You have a misstatement in the question and in the hint. The question should read "$$\sigma(Z)^*=-\sigma(Z)$$ for all $$Z\in\mathfrak{su}(3)$$" (not its complexification $$\mathfrak{sl}(3,\mathbb{C})$$). Also the hint asks you to show $$\sigma(X_i)^* = \sigma(Y_i)$$. So what you need to do is express $$X_i$$ and $$Y_i$$ as complex linear combinations of elements of $$\mathfrak{su}(3)$$. For example, $$X_1 = \frac{1}{2}(A_1-iB_1), \qquad Y_1 = -\frac{1}{2}(A_1+iB_1),$$ where $$A_1 = \begin{pmatrix}0&1&0\\-1&0&0\\0&0&0 \end{pmatrix}, \qquad B_1 = \begin{pmatrix}0 & i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},$$ are both elements of $$\mathfrak{su}(3)$$. Then $$\sigma(X_1)^* = \frac{1}{2}[\sigma(A_1)-i\sigma(B_1)]^* = \frac{1}{2}[-\sigma(A_1)-i\sigma(B_1)] = \sigma(Y_1).$$ Once you've established this, you can write $$\langle u_1,u_1\rangle = \langle \sigma(Y_1)\sigma(Y_2) v_0, \sigma(Y_1)\sigma(Y_2)v_0\rangle = \langle v_0, \sigma(X_2)\sigma(X_1)\sigma(Y_1)\sigma(Y_2)v_0\rangle,$$ and then use the commutation relations, and the fact that $$\sigma(X_i)v_0=0$$, to establish that $$\langle u_1,u_1\rangle = m_2(m_1+1)$$, and similarly with the other identities.