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I have just begun learning algebraic geometry, and there are a lot more connections to complex analysis than I expected. For example:

  • $\mathbb{CP}^1$ is the Riemann sphere
  • Elliptic curves are parametrized by Weierstrass's elliptic functions
  • Riemann-Roch is a theorem in complex analysis and in algebraic geometry

This surprised me a lot at first, but it does make some more sense when one thinks about it -- analytic (i.e. holomorphic) functions are in a crude sense just "polynomials with arbitrarily large degree", and analysis over $\mathbb{C}$ probably should have some more algebraic properties over $\mathbb{R}$ because complex numbers were first investigated due to the fact that they are algebraically closed. After all, polynomials are holomorphic, rational functions are meromorphic, and all smooth algebraic curves are (to the best of my knowledge) Riemann surfaces.

This leads me to the question:

What most importantly differentiates complex analysis from the study of smooth algebraic varieties over $\mathbb{C}$?

To quote Wikipedia, "Algebraic varieties are locally defined as the common zero sets of polynomials and since polynomials over the complex numbers are holomorphic functions, algebraic varieties over C can be interpreted as analytic spaces. Similarly, regular morphisms between varieties are interpreted as holomorphic mappings between analytic spaces. Somewhat surprisingly, it is often possible to go the other way, to interpret analytic objects in an algebraic way."

So another way to phrase the question:

What are the most important examples of when it is impossible to interpret analytic objects in an algebraic way?

The only ones I can think of with my limited knowledge:

  1. It's not possible to "homogenize" an arbitrary analytic function, like it is with a polynomial, because the terms of an arbitrary analytic function have unbounded degree.
  2. Only smooth algebraic varieties correspond to objects of study of complex analysis.

Related questions: Connection between algebraic geometry and complex analysis?
Complex analysis book for Algebraic Geometers
Complex Analysis and Algebra

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Chill2Macht
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    What sort of answer would you be looking for? By GAGA if you stick to the class of projective varieties then there is, essentially, no difference. I find this question (while very well-meaning) not particularly easy to answer since it's not clear if you want specific examples or a metaphysical powwow on how they 'feel different' (and this is coming from someone that loves such powwows). For example, the disk/upper half-plane is not algebrizable? Infinite covers of algebraic things are often not algebrizable, etc. – Alex Youcis Sep 25 '16 at 08:25
  • @AlexYoucis You could probably right this as an answer -- I know very little about both fields, so I wouldn't dispute anything that someone wrote. For instance, while I saw GAGA on the WIkipedia article, I did not realize that it meant that there is essentially no difference, which to me is a surprising and powerful result. Also I am not quite sure what it means for a set to be algebrizable or not, in any case I did not know that infinite covers of algebraic things are often not algebrizable. – Chill2Macht Sep 25 '16 at 08:38
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    Just to comment, GAGA only applies to SOME things. So, given a variety $X/\mathbb{C}$ there is a natural way to topologize $X(\mathbb{C})$ (the obvious thing--on affine's give it the topology from thinking about it as a closed subset of $\mathbb{C}^n$) and in fact, a way to give $X(\mathbb{C})$ the structure of a 'complex analytic space'. To save you from that technical definition suffice it to say that if $X$ is smooth then $X(\mathbb{C})$ with this structure is a complex manifold. One usually denotes this complex manifold $X^\text{an}$. The result GAGA – Alex Youcis Sep 25 '16 at 08:49
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    of Serre says that there is a natural equivalence of categories between smooth projective varieties $X/\mathbb{C}$ and smooth projective complex manifolds (here this just means that it's a complex submanifold of $\mathbb{C}\mathbb{P}^n$ for some $n$) given by $X\mapsto X^\text{an}$. This is widly false if you remove projective though. For example, this says that if $X$ is projective then every meromorphic function on $X$ is 'algebraic' (i.e. locally a quotient of polynomials). This is obviously hugely false for something like $(\mathbb{A}^1_\mathbb{C})^\text{an}=\mathbb{C}$ which has lots – Alex Youcis Sep 25 '16 at 08:51
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    of global holomorphic functions which are not algebraic (e.g. $e^z$). My comment about covers essentially comes from the following beautiful result of Grothendieck et. al. If $M$ is an algebraic complex manifold (meaning that it is $X^\text{an}$ for some $X/\mathbb{C}$ smooth variety) then every finite covering space $N\to M$ is also algebraic, which gives an equivalence between finite covering spaces of $M$ and what are called finite etale covers of $X$. This creates an isomorphism $\pi_1^{\acute{e}\text{t}}(X,x)\cong \widehat{\pi_1^\text{top}(M,m)}$ (where here the hat means profinite – Alex Youcis Sep 25 '16 at 08:53
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    completion). That said, this is truly a phenomenon of finite covers. For example, all the finite covers of $\mathbb{C}^\times=(\mathbb{A}^1-\mathbb{C}-{0})^\text{an}$ are of the form $\mathbb{C}^\times\xrightarrow{z\mapsto z^n}\mathbb{C}^\times$. But, the universal cover is $\exp:\mathbb{C}\to\mathbb{C}^\times$ which is not algebraic. Finally, one might wonder whether it's easy to describe the image of the analytification functor--which complex manifolds are algebraic. This is very difficult (if one enlarges the domain category from smooth varieties to so-called smooth algebraic – Alex Youcis Sep 25 '16 at 08:53
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    spaces then the answer is easy--something is algebraic [in this sense] if and only if it's Moishezon: $\text{tr.deg}(\mathcal{M}(X)/\mathbb{C})=\dim(X)$). That said, one can show that ALL compact Riemann surfaces are algebraic (since the theory of positive bundles shows that they are all projective) which allows one to give an essentially complex answer for Riemann surfaces. Namely, a connected Riemann surface $M$ is algebraic if and only if there is a compactification of $M$, call it $\overline{M}$, which is a connected compact Riemann surface and for which $\overline{M}-M$ is a finite number – Alex Youcis Sep 25 '16 at 08:56
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    of points. so, for example, you can check that this is not the case for the disk (equivalently the complex upper half-plane). Thus, this is a concrete example of a NON-ALGEBRAIC complex manifold. I'll convert this to an answer if no one else answers--I just don't feel like making all of the above cogent ATM (feeling very sick!). Hope this helps. – Alex Youcis Sep 25 '16 at 08:57
  • @AlexYoucis This is great, I really appreciate it. No worries about cogency, a list of examples is still interesting and illuminating; also I have been sick this weekend too. – Chill2Macht Sep 25 '16 at 08:59
  • If you want to continue chatting about this, you can go here: http://chat.stackexchange.com/rooms/45862/discussion-of-gaga – Alex Youcis Sep 25 '16 at 09:07

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Algebraic geometry is about polynomials, while complex analysis is about all analytic functions of complex variables. But complex analysis does not contain algebraic geometry because algebraic geometers study polynomials also over other fields, other than complex numbers.

But this distinction is not strict, since some problems about polynomials are traditionally considered as complex analysis. There is even a term "analytic theory of polynomials".

And of course, complex analysis is widely used in geometry. For example, Riemann-Roch theorem that you mention, was originally proved by complex analysis (using Abelian integrals), and then a purely algebraic proof was found (by Dedekind and Weber).

Alexandre Eremenko
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