Since I have studied analysis as well as algebra recently, I am familiar to work on integrablities, and such concepts when I look at topologies. Currently, I am studying algebraic geometry, and I want to ask If it is meaningful to work on such concepts on the Zariski topology. I have searched quite a lot, and I have heard that the Zariski topology is meaningful as being a "topology", rather than being a set which sufficies the axioms of being a topology. Is there a understandable example which views the Zariski topology as a "topology", and is it meaningful to work on convergency, integration, etc. of some algebraic geometrical objects which includes the Zariski topoloogy?

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    $\begingroup$ The Zariski topology isn't Hausdorff, so convergence doesn't behave reasonably. You need a measure to talk about integration. $\endgroup$ Apr 19, 2014 at 16:50
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    $\begingroup$ The Zariski topology isn't there to talk about convergence, but to talk about sheaf theory. This is seen as a natural generalization of sheaf theory as used in complex analysis, as are more exotic generalizations like the étale topology (which isn't a topology in the classical sense). $\endgroup$ Apr 19, 2014 at 17:09
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    $\begingroup$ See also meta: meta.mathoverflow.net/questions/1650/…. Maybe OP can clarify what is meant by: the Zariski topology is meaningful as being a "topology", rather than being a set which sufficies the axioms of being a topology. Is there a understandable example which views the Zariski topology as a "topology". What is the notion of topology here that is significantly different from a set equipped with a topology? $\endgroup$
    – Todd Trimble
    Apr 20, 2014 at 17:09
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    $\begingroup$ Possibly relevant question on math.SE $\endgroup$
    – Asaf Karagila
    Apr 21, 2014 at 7:49
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    $\begingroup$ Another one from MSE which seems (at least to me) to formulate the question more clearly: math.stackexchange.com/questions/53852/… $\endgroup$
    – Lucia
    Apr 21, 2014 at 14:18

2 Answers 2


I've decided to expand my comment into an answer.

The point is what do you think a topology is for. If you think that a topology is for talking about convergence of sequences, then no the Zariski topology doesn't do anything for you in that regard. However I contend that in geometry (even differential geometry) people don't use a topology because of convergence issues, but to speak of local properties. That is you may want to say e.g. that a function is 0 not only at a point, but also in a neighbourood of that point. Or you may want to say that you can define an object locally and then try to patch it up to a global object. The Zariski topology gives you exactly this.

This is what I mean, loosely speaking, when I say that the Zariski topology is there to do sheaf theory. One could argue that the whole purpose of sheaf theory is to bridge the gap between local and global properties. If your only experience so far comes through a differentiable setting (for instance multivariable calculus) you may not appreciate how hard these questions are in general. This is because in a differentiable setting you have partitions of unity, which roughly speaking, tell you that the sheaf theory is trivial. However in an algebraic (or complex analytic) setting you do not have that luxury anymore.

As a final note, working in algebraic geometry people noticed that they don't really need a topology on the underlying set to do sheaf theory (that is, to speak about local properties), but that there is a more abstract notion of a Grothendieck topology that does the trick. So when people speak about the "étale topology" or the "fppf topology" as a refinement of the Zariski topology, they don't mean a genuine topological space but one of these more abstract generalizations.

EDIT: as requested in the comment, let me put here a proof that the cohomology of $C^\infty$-modules is trivial. To be precise what I'll prove is the following: if $0\to \mathcal{F}'\to \mathcal{F}\to \mathcal{F}''\to 0$ is a short exact sequence of sheaves of $C^\infty$ modules then for every open set $U$ the sequence $$ 0\to \mathcal{F}'(U) \to \mathcal{F}(U)\to \mathcal{F}''(U)\to 0$$ is exact. In fact, by general results, the only thing needed to prove is the surjection of the last map. Take a section $s''\in \mathcal{F}''(U)$. Since the sequence of sheaves is exact we may find a cover $\{U_i\}_i$ of $U$ and sections $s_i\in \mathcal{F}(U_i)$ such that the image of $s_i$ in $\mathcal{F}''(U_i)$ is $s|_{U_i}$. Let now $\psi_i\in C^\infty(U)$ a partition of unity associated with the cover. I claim that the expression $$ s = \sum_i \psi_i s_i$$ makes sense as an element of $\mathcal{F}(U)$. In fact this expression makes sense locally around every point of $U$ (since locally only a finite number of $\psi_i$ are nonzero) and the sheaf property says that these patch to give us a global section. Then it is easy to check that the image of $s$ in $\mathcal{F}''(U)$ coincides with $s''$ locally around any point, and so by the sheaf property, must coincide globally.

  • $\begingroup$ I've seen many people refer to partitions of unity as a way of doing exactly what you describe, but I've never seen a "partitions of unity argument", as such. Can you outline an example of such an argument showing that sheaf theory on, say, real manifolds are trivial? $\endgroup$
    – Will Chen
    Apr 19, 2014 at 20:09
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    $\begingroup$ I've added an explanation in the answer. $\endgroup$ Apr 19, 2014 at 20:25
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    $\begingroup$ @oxeimon 'sheaf theory on real manifolds are trivial' - depends what sort of sheaves you are interested in. Sheaves of continuous or smooth real-valued functions, alright. Analytic functions, or perhaps more 'interesting' sheaves valued in other things (e.g. germs of functions to Lie groups), not so trivial. $\endgroup$
    – David Roberts
    Apr 28, 2014 at 0:54

As the comments have said, the Zariski topology is not hausdorff, and hence is not metrizable, which means you can't talk about convergence. For example, the Zariski topology of any algebraic curve is just the cofinite topology!

Though, the real thing is that the Zariski topology doesn't exist for doing analysis. Generally, classic algebro-geometric spaces come in two types - the characteristic 0 type (varieties over a field of characteristic 0), and the positive characteristic type (varieties over a characteristic p field). You can consider the analytic properties of any characteristic 0 space by looking at the corresponding analytification, as described by Serre's GAGA theorems (e.g. http://en.wikipedia.org/wiki/GAGA#GAGA). These are "analytic spaces", which are the equivalent of complex manifolds with singularities. Positive characteristic spaces (for example, varieties over a finite field), generally only have finitely many points, so it doesn't make much sense to do analysis on it.

However, I think you can still talk about integration by looking at functionals on the space of differential forms, which can be defined in any characteristic! (after all, integration is just a way of assigning a number to a differential form). I know this is one way of defining the Jacobian variety in the characteristic 0 case, and I suspect this also makes sense in characteristic $p$.

If you're interested in integration in characteristic $p$, perhaps you should read about coleman integrals, which aren't really characteristic $p$, but at least it's $p$-adic! See for example:

$p$-adic integrals and Cauchy's theorem

or just google "coleman integral" or "p-adic integral".

  • $\begingroup$ The notion of convergence makes sense in any topological space (metrizability is irrelevant), so "you can't talk about convergence" seems a bit extreme to me. Of course, I agree that the notion of convergence is rarely used when thinking about / working with the Zariski topology. $\endgroup$ Apr 21, 2014 at 3:39
  • $\begingroup$ obvious remark: what you certainly cannot do is speak of "the limit" of a sequence but just of "a limit". this kind of kills any ordinary calculus you might like to perform. $\endgroup$ Apr 21, 2014 at 22:08

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