# A technical step working on an almost Hermitian manifold

I am working on a 6 dimensional almost Hermitian manifold $(M,g,J)$. The set $\{e_1,e_2,e_3\}$ gives a local orthonormal basis of $(1,0)$ vectors. The goal is to prove that $\nabla_e f + \nabla_f e \in T^{1,0}$ for any $e,f \in T^{1,0}$ implies $[e_i,e_j]^{0,1} = -\bar{\lambda}e_k$ - $(ijk)$ cyclic permutation of $(123)$, $\lambda \in \mathbb{C}$ a constant, and $\nabla$ Levi-Civita connection.

Checking the steps: since $g$ is of type $(1,1)$ by definition, we get

$$g([e_i,e_j]^{0,1},e_k) = g([e_i,e_j],e_k) = g(\nabla_{e_i}e_j-\nabla_{e_j}e_i, e_k)$$

and the latter is equivalent to $2g(\nabla_{e_i}e_j,e_k)$. This is because of the assumption $\nabla_{e_j}e_i = -\nabla_{e_i}e_j \mod T^{1,0}$ and because $g$ is of type $(1,1)$. Now I am supposed to be able to prove

\begin{equation} 2g(\nabla_{e_i} e_j, e_k) = -\bar{\lambda}\varepsilon_{ijk} \quad (*) \end{equation} for a constant $\lambda \in \mathbb{C}$, where $\varepsilon_{ijk}$ is the Levi-Civita symbol - which gives the sign of the permutation $(1,2,3) \mapsto (i,j,k)$.

I have proved that the expression $g(\nabla_{e_i}e_j,e_k)$ is totally skew-symmetric. My problem is to show the existence of a constant $\lambda$ such that $(*)$ holds. In the last equation I write $-\bar{\lambda}$ because it is useful to write this constant in that form for other equivalences that crop up later. I see no reason for which $g(\nabla_{e_i}e_j,e_k)$ should be a constant and not a more general function. I have no further hypotheses on the manifold and its structure.

Any help is appreciated. Thanks.