Hermitian manifold counterexample

Question

I'm trying to come to come to grips with the notion of a hermitian manifold. Although I know some examples of hermitian manifolds, I am more interested in counterexamples: naturally occurring Riemannian manifolds which are not hermitian.

Could anyone check that my (counter)example below is indeed correct? I fear that I am misunderstanding something.

Relevant definitions: A linear map $J: V \to V,$ for $V$ a vector space, is said to be an almost complex structure if $J^2 =-Id.$

A Riemannian manifold $M$ is said to be hermitian if the metric $g$ is compatible with an almost complex structure $J$ on the manifold's tangent space. This means that $\langle X, Y \rangle = \langle JX, JY \rangle,$ for all $X,Y \in TM.$

On a complex manifold, there is a natural almost complex structure on $TM$ induced by the holomorphic charts. This is simply the map $\partial_x \mapsto \partial_y$ and $\partial_y \mapsto -\partial_x.$

Main Question: Let us now consider a simple 2D real surface embedded in $\mathbb{R}^3:$ the (upper half) cylinder parametrized by $(x,y) \mapsto (x, y, \sqrt{1-y^2}).$ The induced Riemannian metric can be written in the basis $(\partial_x, \partial_y)$ as

\begin{pmatrix} 1 & 0 \\ 0 & 1/(1-y^2) \end{pmatrix}

Now, the above parametrization can also be thought of as providing maps from $\mathbb{C}\cap \{|\text{Re}(z)|<1\}$ to the upper half cylinder, with the transition maps simply the identity and hence trivially holomorphic.

The induced complex structure, in matrix form, is \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}

We easily compute that $J^*GJ \neq G$ (where $G$ is the matrix of the metric $g$). Hence the Riemannian metric is NOT compatible with the induced almost complex structure.

Hence, the cylinder described is not a hermitian manifold.

Supplemental note (feel free to ignore): Following the above arguments, I get that a necessary and sufficient condition for a real surface in $\mathbb{R}^3$ which has holomorphic transition functions to be a hermitian manifold is for the metric to be of the form \begin{pmatrix} a & b \\ -b & a \end{pmatrix} Is this right?

Answer 1

There seem to be two different questions here, and it's not clear from your phrasing which one you meant to be asking. (Your "main question" doesn't actually include a question!)

First question: Suppose $M$ is a Riemannian manifold endowed with a given almost-complex structure $J$. Which Riemannian metrics are Hermitian with respect to this given $J$?

The answer to this is just a computation: $g$ is Hermitian with respect to $J$ if and only if $\left < X,Y \right> = \left< JX, JY \right>$ for all tangent vectors $X$ and $Y$. In your cylinder example, you've given $M$ the induced Riemannian metric (which is perfectly natural) and you've given it an almost complex structure induced by a particular parametrization (not so natural). As you noted, this metric is not Hermitian with respect to this almost complex structure.

Second question: However, a more interesting question (and the one I suspect you might have intended) is, given a particular metric $g$, is there some almost complex structure with respect to which $g$ is Hermitian. In the case of surfaces, this question has a very satisfying answer.

First note that an almost complex structure $J$ automatically determines an orientation -- For any nonzero tangent vector $X$, we can declare the ordered basis $(X, JX)$ to be positively oriented. It's a standard exercise to show that the resulting orientation is independent of the choice of $X$.

It turns out that if $M$ is an oriented $2$-manifold and $g$ is a Riemannian metric on $M$, then there is always an almost complex structure on $M$ with respect to which $g$ is Hermitian.

To prove this, simply define $JX$ to be the unique vector $Y$ such that $|Y|=|X|$, $\left <X,Y\right >=0$, and $(X,Y)$ is a positively oriented basis; in other words, $JX$ is obtained from $X$ by rotating $90^\circ$ "counterclockwise." Another standard computation shows that $J^2 = -\text{Id}$ and $\left <X,Y\right > = \left <JX,JY\right >$ for all $X,Y$.

Thus, the answer to the second question is "If $M$ is an orientable surface, then every Riemannian metric on $M$ is Hermitian with respect to some almost complex structure. If $M$ is not orientable, then no Riemannian metric is."

In your cylinder example, if we use instead the almost complex structure determined by the parametrization $(x,y)\mapsto (x, \sin y, \cos y)$, then the induced metric is Hermitian.

In higher dimensions, the situation is more subtle, because there are topological obstructions to the existence of any almost complex structure at all (beyond the obvious one of orientability). And in higher dimensions, your definition of a "Hermitian manifold" is actually more properly called an almost Hermitian manifold. To be a Hermitian manifold, you also have to require that the almost complex structure be integrable. (For surfaces this is not an issue, because every almost complex structue is integrable in that case.)

One might wonder whether, given a Riemannian metric $g$ and some almost complex structure $J$, there exists another almost complex structure $J'$ with respect to which $g$ is Hermitian. Off the top of my head, I don't know the answer to that one.

Answer 2

For the Riemannian metric on a complex $1$-dimensional (real $2$-dimensional) manifold to come from a hermitian metric, it is necessary and sufficient that it be conformally equivalent to the flat metric on $\Bbb R^2$, i.e., of the form $\lambda(x,y)(dx\otimes dx+dy\otimes dy)$ for some positive function $\lambda$. That is, the matrix you have written above needs to be a scalar multiple of the identity. (If the hermitian metric is given by $h(z)|dz|^2$, then $h=\lambda$.)

In your supplemental note, I'm not sure what you're thinking. The metric needs to be a positive definite symmetric matrix; you're giving the matrix for multiplication by a complex number $a-ib$.