I'm trying to understand the space of Verma modules for the $\mathfrak{sl}(2,\mathbb C)$ lie algebra and specifically which modules possess hermitian forms. To fix conventions, let's take the algebra to be spanned by $H,E,F$ with commutation relations \begin{align} [ H , E ] = 2 E, \qquad [ H , F ] = - 2 F , \qquad [ E , F ] = H , \end{align} and the quadratic casimir to be \begin{align} C_2 = H^2 + 2 E F + 2 F E . \end{align} On a Verma $M_\lambda$ generated by highest weight vector $v_\lambda$ \begin{align} E v_\lambda = 0, \qquad H v_\lambda = \lambda v_\lambda \end{align} $C_2$ will act as multiplication by a complex number \begin{align} C_2 = \lambda ( \lambda + 2) . \end{align} My question is if $M_\lambda$ possesses a hermitian form for all $\lambda$, or if they only possess one when $\lambda ( \lambda + 2)$ is real. It seems to me that the latter is the case, but I'm having a hard time finding this issue discussed elsewhere to confirm this. I'm a physicist, not a mathematician, so I would like to check with others with more expertise to be sure.
The reason I think this is the following. If I'm understanding correctly, hermitian forms on $M_\lambda$ are in one-to-one correspondence with a choice of an anti-linear involution $\omega$ of the lie algebra (a choice of "hermitian adjoint"). However, any anti-linear involution is conjugate to either \begin{align} \omega_1 ( H ) = - H , \qquad \omega_1 (E) = - F, \qquad \omega_1 (F) = - E, \end{align} or \begin{align} \omega_2 ( H ) = - H , \qquad \omega_2 (E) = F, \qquad \omega_2 (F) = E. \end{align} The first one gives $\mathfrak{sl}(2, \mathbb R)$ as the real subspace of $\mathfrak{sl}(2, \mathbb C)$ and the second gives $\mathfrak{su}(2)$. With either choice, $C_2$ is hermitian and so $\lambda( \lambda+2)$ must be real. It seems then that representations with complex $C_2$ do not possess a hermitian form.