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?


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Nov 1, 2021 at 14:51
  • $\begingroup$ @nonreligious I misunderstood your question. I just edited my answers. $\endgroup$
    – lEm
    Nov 1, 2021 at 17:04

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