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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?

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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 $J : V \to V$.

Definition 1. An inner product $g$ on $V$ is said to be compatible with the linear complex structure $J$ if $J$ is orthogonal with respect to $g$ (i.e. $g(Jx,Jy) = g(x,y)$ for all $x,y \in V)$.

Note that this is what you call a Hermitian inner product but I prefer to avoid this terminology for the sake of clarity. Instead:

Definition 2. A Hermitian inner product on $(V,J)$ is a real bilinear map $h : V \times V \to \mathbb{C}$ which is sesquilinear in the sense that $h(Jx, y) = -h(x, Jy) = ih(x,y)$ for all $x,y \in V$ and Hermitian-symmetric in the sense that $h(y,x) = \overline{h(x,y)}$ for all $x, y \in V$.

Now, the point is that these two notions are essentially the same:

Proposition. If $h$ is a Hermitian inner product on $(V,J)$, then $g := \operatorname{Re}(h)$ is a compatible inner product on $(V,J)$. Conversely, is $g$ is a compatible inner product on $(V,J)$, then there is a unique Hermitian inner product $h$ on $(V,J)$ such that $g = \operatorname{Re}(h)$.

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 $J$).

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  • $\begingroup$ Thanks for your answer. However, it is the extension of $g$ to the complexified tangent space that specifically interests me. (I'm reading a paper that makes use of it.) $\endgroup$ – Nate Eldredge Feb 4 '14 at 0:50
  • $\begingroup$ Okay, but what do you wish to know about it? $\endgroup$ – Seub Feb 4 '14 at 2:23
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    $\begingroup$ Am I correct that we usually consider the bilinear, rather than the sesquilinear, extension (in the sense I describe in my question)? And why is this the "correct" choice? $\endgroup$ – Nate Eldredge Feb 4 '14 at 3:23
  • $\begingroup$ I don't think it really matters, since they are so easily related. The complex bilinear extension is slightly easier to define. In any case I want to stress again that this story ($g$ gives rise to $g^\mathbb{C}$) is unrelated to $J$, in other words it doesn't matter that you are on a complex manifold. $\endgroup$ – Seub Feb 4 '14 at 18:58
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I guess the main point of passing to complexifications is to get rid of conjugate linearity properties and have erverything complex (multi)-linear. The main issue that is missing in your description of the whole situation is the decomposition of the complesified tangent bundle into a holomorphic and an antiholomorphic part. (In fact, one would not even need to do this is the setting of manifolds, it is a pure linear-algebra-gadget, but I'll stay in the manifold setting.)

You have correctly observed that $T_pM\otimes\mathbb C$ carries two distinct almost complex structures, namely the one coming from $J_p$ and the one coming from multiplication by $i$ in the second factor. But this gives rise to a decomposition $T_pM\otimes\mathbb C=T^{1,0}_pM\oplus T^{0,1}_pM$, with the summands being the complex subspaces on which the two structures coincide respectively are negatives of each other (and both spaces have complex dimension $n$). from the bases built up from $\tfrac{\partial}{\partial x_i}$ and $\tfrac{\partial}{\partial y_j}$ you mention, one can build up a complex basis $\{\tfrac{\partial}{\partial z_i}\}$ for $T^{1,0}_pM$ (the "holomorphic tangent space") and a complex basis $\{\tfrac{\partial}{\partial\bar z_i}\}$ for $T^{0,1}_pM$ (the "anti-holomorphic tangent space"). Now the fact that a metric on $g$ on $T_pM$ is Hermitian is equivalent to the fact that its complex bilinear extension to the complexified tangent space vanishes on $T^{1,0}_pM\times T^{1,0}_pM$ and on $T^{0,1}_pM\times T^{0,1}_pM$, so it actually defines a paring $T^{1,0}_pM\times T^{0,1}_pM\to\mathbb C$.

The whole technique is more familiar in the setting of differential forms. Here one first looks at real linear maps $T_pM\to\mathbb C$, which can be canonically identified with complex linear maps $T_pM\otimes\mathbb C\to\mathbb C$, and such a map is complex linear (with respect to $J_p$) if and only if it restricts to zero on $T^{0,1}$ and conjugate linear if and only if it restricts to zero on $T^{0,1}$. This can be viewed as a decomposition of the complexified cotangent space and the induced decomposition of exterior powers gives the decomposition of differential forms on an almost complex manifold into $(p,q)$-types.

So as I said initially, the whole issue is to arrive at a setting in which one only deals with complex multilinear maps, but has the possibility to encode linearity and conjugate linearity properties of underlying objects or natural analogs of such properties as vanishing on certain subspaces.

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  • $\begingroup$ I gave this answer the bounty since I think it was closer to an answer than the other response. However, I have gained a (I think) deeper understanding which I will write up an post in the next few days. $\endgroup$ – Steven Gubkin Dec 31 '14 at 21:29
  • $\begingroup$ @StevenGubkin: I'll be happy to add to the answer if you can give me more details on what you are interested in. $\endgroup$ – Andreas Cap Jan 1 '15 at 13:05

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