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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:

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

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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.

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  • $\begingroup$ Yes, I have edited the mistakes in the question. $\endgroup$ Commented May 25, 2021 at 4:51

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