# Restrictions on coordinate basis of $T_pM$ required for a manifold to admit a Hermitian metric?

I asked this question about relating the Riemannian metric on a manifold $$M$$ to the Hermitian metric that arises when $$M$$ is thought of as a complex manifold (i.e. with integrable complex structure).

One point that keeps puzzling me is about the nature of the (real) coordinate basis on $$T_pM$$. A lot of discussion about this seems to treat this basis as somehow diagonalizing the metric $$g$$ (though one exception is this answer to a similar question).

I.e. if we have $$\frac{\partial}{\partial z^i} = \frac{\partial}{\partial x^i} - i \frac{\partial}{\partial y^i}$$ and $$\frac{\partial}{\partial \bar{z}^i} = \frac{\partial}{\partial x^i} + i \frac{\partial}{\partial y^i}$$, then the real Riemannian metric $$g$$ associated to a Hermitian metric on $$M$$ seems to be of the form $$g = f(x,y) dx^i \otimes dx^j + h(x,y) dy^i \otimes dy^j$$, without mixing $$dx^i$$ and $$dy^j$$.

There's some mention of this in Huybrechts Lemma 1.2.7 (admittedly not on the tangent space of a manifold, but on a real vector space).

Is this a necessary result of the metric being compatible with an almost complex structure $$J$$, which acts via $$J(\frac{\partial}{\partial x^i}) = \frac{\partial}{\partial y^i}$$, $$J(\frac{\partial}{\partial y^i}) = - \frac{\partial}{\partial x^i}$$? Is this not actually the case?

Yes, essentially the story goes like this: you have a real vector space $$V$$ with an endomorphism $$J:V\to V$$ so that $$J^2=-I$$. Then by linear algebra this linear transformation has eigenvalue $$i,-i$$ and you can find a basis $$\{x_1,...,x_n,y_1,...,y_n\}$$ so that $$J(x_k)=y_k$$ and $$J(y_k)=-x_k$$. If you extend this linearly to an endomorphism of the complexified vector space $$J:V_{\mathbb C}\to V_{\mathbb C}$$. Then we can see that $$z_k=x_k-iy_k$$ and $$\bar z_k=x_k+iy_k$$ naturally are eigenvectors of $$J$$ corresponding to eigenvalues $$i$$ and $$-i$$ respectively. The story for manifold is really just a family version of this linear algebraic result, where we are able to locally find coordinates so that the tangent vectors $$\frac{\partial}{\partial x_k},\frac{\partial}{\partial y_k}$$ at each point satisfies the above relation, and hence we have eigenvectors like before on the complexified tangent bundles.
Edit: I am not one hundred percent sure, but I think it is true that $$g$$ takes the special form. It should follows from that $$g$$ is given by the real part of the Hermitian metric, and the Hermitian metric takes a special form, e.g. as shown in the link in your post. You can then just expand the expression $$g_{i\bar j}dz_j\otimes d\bar z_k$$ etc to see its real part. Similarly, if you look at the Kaehler form/Hermitian form, it is a real $$(1,1)$$-form as well. The computation is something along the line of $$i(dz_j\wedge d\bar z_k+dz_k\wedge d\bar z_j)=2(dx_j\wedge dy_k+dy_j\wedge dx_k)$$, this is also due to this local formula for the metric.
For your question about whether $$x_i,y_i$$ are orthonormal with respect to the Hermitian metric. You can always pick $$x_i,y_i$$ so that it is true for the vector space case, but for manifold, you cannot pick local coordinates so that $$\frac{\partial}{\partial x_j},\frac{\partial}{\partial y_j}$$ form an orthonormal basis at every point. We don't even have such things for usual Riemannian manifold (i.e. we can't pick coordinates so that $$g(\frac{\partial}{\partial x_j},\frac{\partial }{\partial x_k})(p)=\delta_{jk}(p)$$ for all $$p$$ in some neighborhood.) Also let me mention that in the Kaehler case, there is something like a complex normal coordinates.
• Thanks, but I guess I'm asking about the inner product $\langle . , .\rangle$ on $V$ that appears to be extended to $\langle . , . \rangle_{\mathbb{C}}$, and whether associating $\langle . , . \rangle_{\mathbb{C}}|_{V^{1,0}}$ with a Hermitian product $( . , . )$ means that the above $x_i$, $y_i$ have to be orthogonal/orthonormal wrt $\langle . , . \rangle$, and if this extends to the case of tangent vectors. Basically I am asking if being compatible with complex structure and hermitian metric forces a Riemannian metric to take a particular form. Nov 1, 2021 at 14:51