I've been trying to learn some complex geometry, and was getting
confused in thinking about Hermitian metrics. In this post, I've
written up my current understanding, in hopes that someone can look it
over and verify/clarify where I have gone
right/wrong. $\newcommand{\pp}[1]{\frac{\partial}{\partial #1}}$
I'll start with definitions, to fix notation and for practice.
Let $M$ be a complex manifold of dimension $n$, and let us fix a point
$p \in M$. Around $p$ we can find a holomorphic coordinate chart: an
open set $U \subset M$ and holomorphic coordinates $z_1, \dots, z_n :
M \to \mathbb{C}$. If $x_j, y_j$ are the real and imaginary parts of
$z_j$, then $x_1, y_1, \dots, x_n, y_n : U \to \mathbb{R}$ give us a
smooth coordinate chart. So $M$ is also a differentiable manifold of
real dimension $2n$, and thus we can define a tangent space $T_p M$ in
the usual way as a real vector space of dimension $2n$; a basis for
$T_p M$ is given by $\pp{x_1}, \pp{y_1}, \dots, \pp{x_n}, \pp{y_n}$.
Now $T_p M$ carries a natural almost complex structure; namely, the
linear map $J$ defined by $J \pp{x_j} = \pp{y_j}$ and $J \pp{y_j} =
-\pp{x_j}$. If the differentiable manifold $M$ carries a Riemannian
metric $g$, so that $g_p : T_p M \times T_p M \to \mathbb{R}$ is a
bilinear symmetric positive definite form, then we say this metric is
Hermitian if it respects the almost complex structure: $g_p(J v, J
w) = g_p(v,w)$ (and the same holds for every other point $p$). So
far, so good?
We may also consider the complexified tangent space $T_p M \otimes
\mathbb{C}$, which is a complex vector space of complex dimension
$2n$. As I understand it, we typically extend $g_p$ to $T_p M \otimes
\mathbb{C}$ by bilinearity; in particular, for $\alpha, \beta \in
\mathbb{C}$, we have $g_p(\alpha v, \beta w) = \alpha \beta
g_p(v,w)$. So $g_p$ is now a bilinear symmetric form on $T_p M
\otimes \mathbb{C}$.
In particular, $g_p : (T_p M \otimes \mathbb{C})^2 \to \mathbb{C}$ is no longer positive definite: indeed, $$g_p\left(\pp{x_i} – i \pp{y_j}, \pp{x_i} – i \pp{y_j}\right) = 0.$$
This confused me for a while, because the more usual notion of an inner
product on a complex vector space is a positive definite sesquilinear symmetric form:
i.e., $\langle \alpha v, \beta w \rangle = \alpha \bar{\beta} \langle
v,w \rangle$. Indeed, the word "Hermitian" is often used to describe
such a form. So I would naively ask: Why is it better to take $g_p$
to be bilinear on $T_p M \otimes \mathbb{C}$, rather than
sesquilinear? If we took the sesquilinear extension instead, would
we still get a reasonable theory, or would something bad happen?
I think part of the reason I was confused is that $T_p M \otimes
\mathbb{C}$, considered as a real vector space of dimension $4n$ (with
basis $\left\{ \pp{x_j}, i \pp{x_j}, \pp{y_j}, i \pp{y_j} \right\}$)
actually carries two distinct almost complex structures. First is the
obvious one: multiplication by $i$, sending $\pp{x_j}$ to $i
\pp{x_j}$ and so on. The other one is the complex linear extension of
$J$, which sends $\pp{x_j}$ to $\pp{y_j}$, sends $i \pp{x_j}$ to $i
\pp{y_j}$ and so on. And the point is that we take our extension of
$g_p$ to be Hermitian with respect to the latter, not the former.
Have I understood all this correctly?
Best Answer
Let me try to clarify the point that is bothering you. I think it's best to look at it in the simplest perspective: linear algebra, because it all boils down to linear algebra.
Here's our situation: we have a real vector space $V$ with a linear complex structure $J : V \to V$.
Note that this is what you call a Hermitian inner product but I prefer to avoid this terminology for the sake of clarity. Instead:
Now, the point is that these two notions are essentially the same:
As you can see, this has absolutely nothing to do with the complexification of $V$. It's true that $g$ extends to a symmetric complex bilinear map $g^\mathbb{C} : V^\mathbb{C} \to V^\mathbb{C}$ where $V^\mathbb{C} = V \otimes_{\mathbb{R}} \mathbb{C}$, but it's a different subject (and has nothing to do with the complex structure $J$).