# Tangent space on complex manifolds

1. Nov 4, 2013

### mnb96

Hello,

I understand the concepts of real differentiable manifold, tangent space, atlas, charts and all that stuff. Now I would like to know how those concepts generalize in the case of a complex manifold.

First of all, what does a coordinate chart for a complex manifold look like? Is it a function $\xi : U\subseteq \mathbb{C}^m \rightarrow \mathbb{C}^n$ where U is a open subset of ℂm ?

Secondly, how do we obtain a tangent space? What are the tangent vectors? In case of real manifolds we obtain them through directional derivatives, but for a complex manifold how do we define a directional derivative?

Thanks.

2. Nov 4, 2013

### lavinia

The transition functions between coordinate charts of a complex manifold are required by definition to be holomorphic. This is a stringent condition. Tangent spaces are defined in the usual way. In the case of a complex manifold the tangent spaces are also complex vector spaces.

The tangent spaces of a manifold may still be complex vector spaces even when the manifold is not a complex manifold. This is called an almost complex structure on the manifold.

http://en.wikipedia.org/wiki/Almost_complex_manifold#Examples

More generally, the fibers of an even dimensional real vector bundle may be complex vector spaces. Any real vector bundle may be "complexified" by tensoring the fibers with the complex numbers. The complexified tangent bundle of a manifold (even or odd dimensional) is used to define the Pontryagin classes of the manifold.

3. Nov 4, 2013

### mnb96

But what is the "usual way"?
I used to think that for a real manifold, I could obtain the tangent space from the set of directional derivatives of a chart evaluated at a point. What is the definition of "directional derivative" for complex functions?

4. Nov 4, 2013

### fzero

In order to formally define derivatives and the like, it is usually convenient to use the almost complex structure to pass from complex coordinates to real coordinates. Holomorphicity then translates to the Cauchy-Riemann equations and differential calculus is straightforward. Translation back to expressions in complex coordinates involves using the chain rule, which in term ends up being a linear transformation involving the appropriate form of the almost complex structure.

Edit: After using the above translation, one can establish rules for taking derivatives with respect to complex variables. These turn out to be equivalent to the Wirtinger derivatives.

5. Nov 4, 2013

### lavinia

A holomorphic map is a smooth map. So one constructs the tangent bundle as usual.

6. Nov 5, 2013

### mnb96

Ok...I will try to make the simplest example that comes to my mind.
We have an atlas formed by only one chart $$\xi :U\subset \mathbb{C} \rightarrow M\subseteq \mathbb{C}$$

defined as:

$$\xi(z)=z^2 \quad\quad; \, \mathcal{Im}(z)>0$$

First question: is $\xi(z)$ a 1-dimensional complex manifold?

Second question: is $\xi'(z)=2z$ a basis vector for the "tangent space" of M at $p=\xi(z)$ ?

7. Nov 6, 2013

### WWGD

I guess you mean a submanifold of $\mathbb C$.I think first of all, it is necessary for $\chi(z)$ to be a Real 2-manifold (and you can check, e.g., using the regular-value theorem, using the Jacobian). And then the Real 2-manifold has to admit a complex structure. There are some theorems on the existence of complex structures for even-dimensional Real-manifolds; I think one was named Neuland-Niremberg, but I'm not 100%. There are other obvious conditions, like the Real 2-manifold must be orientable, since every complex manifold is orientable.

8. Nov 7, 2013

### mnb96

I am not sure what you mean by "checking that it is a real 2-manifold" but I assume you mean that I should check whether or not the following $M \subset \mathbb{R}^2$ is a manifold:

$$x(u,v) = u^2-v^2$$
$$y(u,v) = 2uv$$

where I used $\xi(z)=z^2 = (u+iv)^2 = (u^2-v^2) + 2iuv$.
If we assume the domain of $\xi$ is the open subset $\Omega = \left\{ (u,v) | \quad v>0, \; u\neq 0\right\}$, then M should be a manifold, yes (probably a submanifold of ℝ2).

But so far we have been treating this case as a real manifold.
What about the second question on tangent spaces?

9. Nov 7, 2013

### lavinia

An analytic function may be thought of as smooth function that takes values in the Euclidean space and which has further structure - for instance its Jacobian is complex linear. But whatever additional structure it may have it is still just a smooth function.

10. Nov 7, 2013

### mnb96

uh... I confess that I feel a bit confused now.
Perhaps I should have not started this thread in the first place, as it seems that my deceivingly simple questions keep on triggering answers containing concepts I am not familiar with (e.g. "tensoring of fibers", "Pontryagin classes of the manifold", "complex/almost complex structure", "regular-value theorem", "Neuland-Niremberg theorem"...).

It is probably wiser for me to thank all the users who have participated to this thread, admit my ignorance and then give up on this discussion. Maybe I am not ready for these topics and I should simply try to find a (gentle) introductory text.

It was quite interesting to notice that the apparently simple question "is the function f(z)=z2 a chart of a manifold, and what are the tangent spaces of such manifold?" is trivial to answer when $f:\mathbb{R}\rightarrow \mathbb{R}$, but seems very difficult to answer when $f:\mathbb{C}\rightarrow \mathbb{C}$.

Last edited: Nov 7, 2013
11. Nov 7, 2013

### fzero

It's not easy to answer that question without knowing what space we're talking about. If we want a topological space $X$ to be a complex manifold, we need to know what $X$ is. A topology on $X$ will tell us what the open sets $\{U_a\}$ in $X$ are. Charts $\phi_a$ map these open sets to the open polydisc in $\mathbb{C}^n$. The manifold is complex if the transition functions $\phi_a\circ \phi_b^{-1}$ on the intersection of two open sets $U_a\cap U_b$ are holomorphic.

Viewed in light of the complete definitions, $f(z)=z^2$ might be a chart or it might not. The answer depends on the open sets. It is certainly more natural to define the open sets and then look for charts. A question one can ask is whether two different choices of charts is equivalent in some way or whether they lead to different structures on $X$, but this is more difficult than just showing that some manifold is complex. However, we can say that $f(z)$ maps the open unit disk in $\mathbb{C}$ to itself, so any proper chart can be composed with $f(z)$ to give another acceptable chart.

I think part of your confusion is that you're looking at charts as some sort of functions. They are, but the more important role they have is as local coordinates on the manifold. Maybe it would be useful to discuss this in terms of a definite example like $\mathbb{CP}^1 = S^2$. Take a look at the standard discussion for that (there's many lecture notes available online that discuss it) and maybe we can come back to it here and discuss some other choice of charts other than the standard ones and see if we can make sense of the differential structure. It's not something I've thought about myself before.

12. Nov 7, 2013

### WWGD

Sorry, mnb96, I thought you were asking wether the graph of $f(z)=z^2$ was a submanifold of $\mathbb C$. Then later I though you may have been asking wether the fibers ; regular fibers of the map were submanifolds of $\mathbb C$ (a standard way of obtaining submanifolds is as the fibers of smooth functions at non-critical points; a non-critical point x of a function $f$ is a point x where $Df_x$ has maximal rank). My point was that every complex n-manifold is a Real 2n-manifold, but not necessarily viceversa. Maybe my post was confused. If you can expand, maybe we can clarify--and I can clarify a few things for myself. But $$f(z)=z^2$$ cannot be a chart for the whole of $\mathbb C$ because it is not invertible; you will have to restrict it to a subset where it is 1-1.

BTW, there is a Result in Lee's Smooth that the graph of a smooth function f: $\mathbb R^p\rightarrow \mathbb R^q$ is a submanifold of $\mathbb R^p \times \mathbb R^q$.

BTW: the standard topology in $\mathbb C$ ( or any subset/subspace of it) that I know of is the one generated by the usual metric $d((x_1+iy_1),(x_2+iy_2))=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Last edited: Nov 7, 2013