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?
[Math] the difference between real and complex manifolds
complex-analysisdifferential-geometryreal-analysis
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The reason we have charts on a topological manifold is so that if we pick a chart $U \to \mathbb{R}^n$ that is a continuous function, we can say that the topological space $U$ is homeomorphic to the topological space $\mathbb{R}^n$, so any questions about the topology of $U$ can be moved to questions about the topology of Euclidean space, which we understand very well.
For differentiable manifolds, we want charts that make $U$ diffeomorphic to $\mathbb{R}^n$, so that any questions about the differentiable structure on $U$ can be moved to questions about the differentiable structure of Euclidean space, which we understand well.
For smooth manifolds, we want $U$ to be isomorphic to $\mathbb{R}^n$ in the appropriate sense as well.
This is why we don't consider all charts on a differentiable manifold or a smooth manifold.
Of course, for this to make sense, we need a way to be able to define what it means for a map to be diffeomorphic. We know how to define diffeomorphisms of subspaces of Euclidean space. The transition map idea is a device to let us take that definition and use it to define what a diffeomorphism is from a subspace of our manifold. (and happily, that turns out to be all we need to make the definition work)
It may be a fun exercise to show the analogous fact for topological manifolds. Suppose you didn't even bother defining a topology on the set $M$: you just had an atlas of charts that are just ordinary bijective functions. Try using the atlas to define a topology on $M$. You'll find that you need the transition maps to be continuous, and that's all you need.
(EDIT: you also need the domain and image of the transition maps to be open)
At a first glance, it sounds like smoothness (having all derivatives) implies holomorphy (having one derivative). However, there are two different meanings of derivative being used here. A complex function $f:\mathbb C\to \mathbb C$ is smooth if, when considered as a function $f:\mathbb R^2\to \mathbb R^2$, both component functions of the output have all higher order partial derivatives. On the other hand, a complex function $f$ is called holomorphic if it is complex differentiable, meaning its first partial derivatives exist and satisfy the Cauchy-Riemman equations.
To see these are not equivalent, note that $z\mapsto \overline{z}$ is smooth (it is a linear map $\mathbb R^2\to \mathbb R^2$), but not holomorphic.
It is true that every holomorphic function is smooth.
Best Answer
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 $X$.