Please scroll down to the bold section if you are too bored to read the whole details.

Aiming to explain the mathematical structure of Kahler manifolds, Freedman and Van Proeyen, in their book Supergravity said the following on p. 262-263:

One can show that a real manifold $M_{2n}$ equipped with an almost complex structure is a complex manifold (of dimension $n$) if and only if the Nijenhuis tensor vanishes.

The almost complex structure is then called the complex structure of $M_{2n}$. There us then a covering of $M_2n$ by coordinate charts $U_I$ with complex coordinates $z^{\alpha}$, $\bar{z}^{\bar{\alpha}}$. In the coordinate basis $\partial/\partial{z}^{\alpha}$, $\partial/\partial{\bar{z}}^{\bar{\alpha}}$, the almost complex structure takes the form J=$\pmatrix{i\delta_{\alpha}^{\beta}&0\\0&-i\delta_{\bar{\alpha}}^{\bar{\beta}}}$.

They add

We now suppose that $M_{2n}$ has a torsion-free (i.e. symmetric) connection $\Gamma_{ij}^k$ and that $J_{i}^{j}$ is covariantly constant $$\nabla_kJ_i^j=\partial_kJ_i^j-\Gamma^l_{ki}J_l^j+\Gamma_{kl}^jJ_i^l=0. \hspace{0.6cm}(1)$$ We now assume that $M_{2n}$ has a Riemanian metric $g_{ij}(\phi)$. This metric is hermitian if it is invariant under the action of the almost complex structure, namely $JgJ^T=g$.

Then, they say

Finally, we demand that the affine connection in (1) is the Levi-Civita connection. We then have two covariantly constant tensors: $$\nabla_kJ_i^j=0$$ $$\nabla g_{ij}=0$$

My question is concerned with the last quote, why did the author assume that the affine connection is the Levi-civita connection and why was it concluded from that that $\nabla_kJ_i^j=0$ and $\nabla g_{ij}=0$?

  • $\begingroup$ In $(1)$ it was already assumed that $\nabla_k J_i^j = 0$. Whereas $\nabla g_{ij} = 0$ is the condition that the connection is a metric connection; in particular, it is true of the Levi-Civita connection. $\endgroup$ – Michael Albanese Feb 13 at 3:30

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