Compact form of $\mathfrak{su}(2)$ Apologies for the completely elementary things, but I am stuck. Please help.
Consider $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$ and let $X_+,X_-,H$ be as in Theorem 6.6 of [1],
$$
[X_+,X_-]=H,\quad[X_\pm,H]=\pm X_\pm.
$$
Let
$$
J_0=\imath H,\quad J_1=X_+-X_-,\quad J_2=\imath(X_++X_-).
$$
According to Theorem 6.11 of [1], $\mathfrak{u}_0=\mathbb{R}\{J_0,J_1,J_2\}$ is a compact form of $\mathfrak{g}$. But a direct computation shows
$$
[J_0,J_1]=-J_2,\quad[J_1,J_2]=2J_0,\quad[J_2,J_0]=-J_1.
$$
Are these not the commutation relations of the non-compact form $\mathfrak{sl}(2,\mathbb{R})$?
Thanks.
[1] A. Knapp. Lie groups beyond an introduction. 2nd edition, Birkhaeuser 2002.

Yes, you have an error in your original commutation relations. On the top of page 98, you have the equations
$$
[H_\alpha, E_\alpha] = 2E_\alpha, \quad [H_\alpha, E_{-\alpha}] = 2E_{-\alpha}, \quad [E_\alpha, E_{-\alpha}] = H_\alpha.
$$
Then on the bottom of page 295, $X_\alpha = a_\alpha E_\alpha$, with $a_\alpha a_{-\alpha} = 1$. This implies the commutation relations
$$
[H_\alpha, X_\alpha] = 2X_\alpha, \quad [H_\alpha, X_{-\alpha}] = 2X_{-\alpha}, \quad [X_\alpha, X_{-\alpha}] = H_\alpha.
$$
To map this to the usual $\mathfrak{su}(2)$ algebra, you can just take (for example)
$$
J_1 = \frac{1}{2}(X_\alpha-X_{-\alpha}), \quad J_2 = \frac{i}{2}(X_\alpha+X_{-\alpha}), \quad J_3 = \frac{i}{2}H_\alpha.
$$
These then satisfy $[J_i, J_j] = \epsilon_{ijk}J_k$.
