# Why do we require that a complex manifold has the structure of a real manifold?

I am taking a course in complex manifolds, heavily influenced by Huybrechts' book "Complex Geometry", and in it we define a complex manifold $X$ to be a smooth, real manifold $M$ together with an equivalence class of holomorphic charts. As far as I can see, there is no compatibility condition on the two atlases.

My question is: Why do we do this? Is this definition useful?

Since holomorphic functions are smooth when considered as real functions, any complex atlas should define a real atlas, so assuming that the manifold already has a real atlas seems superfluous. I suppose this means that as a topological space the manifold is Hausdorff and second-countable, but to me it would seem nicer to just define a complex manifold as such a topological space with a holomorphic atlas. This is the approach taken in a Riemann surfaces course I have attended previously, and just after the definition in his book, Huybrechts mentions that we could use this approach to define complex manifolds. It seems then that the definition as given must have been, at least to him, somehow preferable.

Moreover, I know that in the real case there are topological spaces that have several different smooth structures ($S^7$ is such an example) so if we don't assume that the complex and real atlases are compatible, I'm not sure how we could use real atlas to do anything useful since there is not in general a single choice of real atlas for each complex manifold. If we do assume compatibility, then what's the point of requiring it in the definition instead of just mentioning it as a consequence?

• I agree that it's superfluous. Maybe he just wanted to emphasize that there's a forgetful functor to smooth real manifolds. – Qiaochu Yuan Jan 28 '15 at 9:46
• You can also have almost-complex manifolds. – Orest Bucicovschi Jan 28 '15 at 12:51