Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given an even-dimensional (smooth) manifold, what is the difference between its (real) smooth structure and its complex structure? I realize that in the real case, the overlap functions of charts need only be smooth, while in the complex case they need to be holomorphic. However, this doesn't provide a satisfactory answer- it begs the question of why holomorphicity is a stronger condition than smoothness in the first case. The answer to which is just that "in the real case, limits can only go from either side, whereas in the complex case they can be taken from all sorts of directions"- but this doesn't not seem very rigorous. Any thoughts on what is really at work here?

share|improve this question
4  
Another difference is that the complex structure gives the manifold a canonical orientation whereas there are lots of real even-dimensional manifolds that aren't orientable at all. –  Matt Nov 30 '11 at 6:20
    
VERY good question that it's hard to get a straight answer to.Believe me,I've tried to get one and failed. –  Mathemagician1234 Nov 30 '11 at 6:37

4 Answers 4

Holomorphic functions are much more "rigid" than smooth functions. For example, let $U \subset \mathbb{C}^n$ be a a connected open set and let $V_1,V_2 \subset U$ be open subsets whose closures are disjoint (think disjoint small balls). If $f : U \rightarrow \mathbb{R}$ is just a smooth function, then the values of $f$ on $V_1$ and $V_2$ have no relationship whatsoever; indeed, if $f_i : V_i \rightarrow \mathbb{R}$ is smooth for $i=1,2$, then you can find a smooth function on $U$ which restricts to $f_i$ on $V_i$. However, if $f : U \rightarrow \mathbb{C}$ is holomorphic, then there is a very strong relationship between the values of $f$ on $V_1$ and $V_2$. In fact, just knowing what $f$ does on $V_1$ determines what it does everywhere on $U$, no matter how small $V_1$ is!

For manifolds, this rigidity manifests itself in the fact that complex manifolds have "extra structure" that a naked smooth manifold does not have. As an important example of this, consider a $1$-dimensional complex manifold $X$ (thus $X$ is a 2d-dimensional real manifold, i.e. a topological surface). A naked smooth topological surface has no "geometry"; in particular, you can't measure lengths of tangent vectors on it or angles between tangent vectors. On a $1$-dimensional complex manifold $X$, there is still no notion of "length". However, amazingly there is a notion of "angle" between tangent vectors! In the end, this comes down to the fact that bijective holomorphic maps between open subsets of $\mathbb{C}$ actually are conformal, i.e. they preserve angles between tangent vectors. This implies that the usual conformal (=angle measurement) structure on $\mathbb{C}$ induces a conformal structure on $X$.

share|improve this answer

There's a sort of "standard" answer to this question, involving the notion of an almost complex manifold.

A complex manifold has the property that each tangent space is endowed with the structure of a complex vector space. Roughly speaking, the difference between a real vector space of dimension $2n$ and a complex vector space of dimension $n$ is that in the complex vector space, you know what it means to multiply by $i$. More formally, you can put a complex structure on any real vector space by choosing any linear transformation $J$ such that $J^2 = -1$, with $J$ serving as multiplication by $i$.

Note that this $J$ adds a lot of structure to a real vector space. For example, each vector $v$ now has an associated $2$-dimensional subspace, namely the span of $v$ and $Jv$ (i.e. the set of complex multiples of $v$), and it's possible to measure angles between vectors on this subspace. Also, we can now classify linear transformations of the vector space into two types -- those that commute with $J$ (i.e. complex linear transformations) and those that don't.

A complex manifold of dimension $n$ can be thought of as a real manifold of dimension $2n$ together with a smoothly varying choice of $J$ for each tangent space. Indeed, a real manifold endowed with such a smooth choice of $J$'s is called an almost complex manifold.

The word "almost" is pertinent because not every smooth choice of $J$'s actually yields a complex manifold. You have to choose a set of $J$'s that are "integrable" in the sense that a certain tensor defined in terms of $J$ vanishes. My impression is that the geometric difference between almost complex manifolds and complex manifolds is not well-understood.

share|improve this answer
    
Saying that the geometric difference is "is not well-understood" is misleading. First, it is easy to tell whether or not a manifold supports an almost complex structure (it's just checking if a bundle has a nonvanishing section, which is just homotopy theory). Second, while it is true that we don't know if some famous simple manifolds support a complex structure, we know an enormous amount about the topological restrictions on which manifolds can have one (eg through Hodge theory). As an easy example of this, their odd Betti numbers have to be even. It's a huge subject with more to –  Adam Smith Nov 30 '11 at 16:43
    
do, but there is an enormous literature and we do know an awful lot. –  Adam Smith Nov 30 '11 at 16:44
    
That's a reasonable point. It's not my subject, and when I said that "it's not well-understood", I meant roughly the same thing as "it's an area of active research", though I suppose the latter statement slightly different and a bit more fair. Also, it's quite possible that I'm overestimating how much still needs to be done in the subject because of the question about $S^6$. –  Jim Belk Nov 30 '11 at 22:20
    
Oh, there's quite a bit left to be done, and it's an active area of research. Your comment just made it sound like there wasn't much known, while in fact there's a lot that is known. –  Adam Smith Dec 1 '11 at 1:57

Given any continuous function $f : M \to \mathbb R$ it can be approximated by a $C^\infty$-smooth function. But continuous functions $f : M \to \mathbb C$ can't be approximated by complex differentiable functions, since those are constant (provided $M$ is compact and connected).

That's the kind of thing that's different with complex manifolds. Complex differentiability is a huge restriction on functions, while real differentiability (at least from the perspective of topology) is not a major restriction.

Regarding existence of smooth structures, see:

http://mathoverflow.net/questions/1973/is-there-a-complex-structure-on-the-6-sphere

and

http://mathoverflow.net/questions/11664/complex-structure-on-sn

for examples of how existence of complex structures is a far more subtle question than that of smooth structures for "everyday" manifolds.

share|improve this answer

«why holomorphicity is a stronger condition than smoothness?»

Becase there are fewer holomorphic functions than smooth ones.

If we were to consider atlases whose transition functions were polynomial, the restriction would be stronger. If we were to consider only affine linear transition functions..., well, you get the picture :D

share|improve this answer
    
Might as well go all the way to talking about manifolds where the only transition maps are identity maps. –  Ryan Budney Nov 30 '11 at 7:09
    
@RyanBudney, well, that one is just too rigid. But even linear gives interesting things :) –  Mariano Suárez-Alvarez Nov 30 '11 at 17:28
1  
You get some interesting things this way -- things like the Riemann surface of $\log z$. –  Ryan Budney Nov 30 '11 at 17:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.