1
$\begingroup$

At the beginning I should warn everybody reading this post: I don't know much about algebraic geometry so specialists in this subject may see my question as ignorant.


As far I understood one on the main themes in algebraic geometry is to pursue as far as it is possible the duality between geometric objects and algebras: most basic result is the Hilbert Nullstellensatz but the theory goes much further-to the definition of general schemes due to Grothendieck. The notion of space has evolved through the history of mathematics but as far as some topological space was around, the minimal requirement (at least for me) was that the space should be Hausdorff. This is quite natural due to the following characterisation: each net has at most one limit. Moreover, when one is interested in compact or locally compact spaces, the assumpion of being Hausdorff automatically implies better behaviour (normality or complete regularity resp.).

Finally, there is the theory (which is close to my heart) of $C^*$-algebras: in this theory a fundamental result is the Gelfand-Najmark theorem which establishes the duality between compact Hausdorff spaces and commutative unital $C^*$-algebras. This is another algebra-geometry duality and allows one to think of the theory of general $C^*$-algebras as noncommutative topology: but there are plenty of situations when one has a "pathological" topological space (with some non Hausdorff topology) which is hard to deal with. Then one switches to the realm of algebras and tries to say something about this space using the associated algebra: in other words one doesn't stick to a geometric picture.

It seems that algebraic geometry goes the other way around and works very often with topological spaces which are non-Hausdorff. So my (rather vague) question is the following:

Question. What is the geometric meaning and the intuition behind non-Hausdorff spaces in the realm of algebraic geometry? How to interpret such non Hausdorff topologies in this algebra-geometric context?

Let me give one example, which may clarify about what sort of things I'm asking: when one forms a quotient space one glues some points of the space to the another and in such a way one obtains a new set of points. In particular one can take some subset $A \subset X$ which is not closed and collapse it to the one point: then $X/A$ would be non Hausdorff and the special point in the quotient will be $\pi(a)$ where $a \in A$ is arbitrary and $\pi$ denotes the natural projection. My intuition behind this example is the following: the point $\pi(a) \in X/A$ was obtained from the richer set of data which was the set $A$ and the fact that $A$ was not closed. A more dramatic example would be $X/G$ where for each $g \in G$ its orbit is dense in $X$: then my intuition behind this example is the fact that the points in $X/G$ have some extra internal structure. So the operation of taking quotients very often gives a non-Hausdorff topology.

$\endgroup$

closed as primarily opinion-based by Fernando Muro, Daniel Loughran, Dietrich Burde, abx, Stefan Kohl Jan 4 '15 at 21:29

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ "What is the meaning of" is not a well-defined question. Neither is "how to interpret". Can you try to rephrase your question to be more precise? $\endgroup$ – Yemon Choi Jan 4 '15 at 18:06
  • 6
    $\begingroup$ The analog of Hausdorffness in algebraic geometry is the notion of a separated scheme. Definition: A scheme $X$ is separated if the diagonal $X \to X \times X$ is a closed immersion. Now prove the following. Let $Y$ be a topological space. Then $Y$ is Hausdorff iff the diagonal $\{(y,y) : y\in Y\}$ is closed in the product topology on $Y \times Y$. $\endgroup$ – Ben Lim Jan 4 '15 at 18:07
  • $\begingroup$ I still don't understand what it is you seek in an "interpretation" of non-Hausdorff topological spaces arising in algebraic geometry. Are you merely asking for some conceptual explanation of why these spaces are non-Hausdorff? $\endgroup$ – Yemon Choi Jan 4 '15 at 20:47
  • 1
    $\begingroup$ Not only I'm aksing for such an explanation but I would also like to know why algebraic geometers care so much about spaces which are not Hausdorff. As I explained before: in noncommutative topology when some space is non Hausdorff one rather try to investigate it by associating some algebra to it and then investigate this algebra. This is the first point. Second is that due to the existence of convergent nets with more than one limit my natural reaction to non Hausdorff spaces is to dismiss them as being pathological. By the way: maybe I should add "soft-question" tag to this question? $\endgroup$ – truebaran Jan 4 '15 at 21:00
  • 6
    $\begingroup$ Well we don't care about the Zariski topology because we want to do classical analysis on it, we care about it because we can define sheaves on it. After that, a whole wide world opens up. $\endgroup$ – Donu Arapura Jan 4 '15 at 21:37
7
$\begingroup$

One of the things to think about in Algebraic Geometry is that the natural topology (Zariski topology) is the wrong topology, at least in part for the reasons you describe. There are other "topologies" that you can use (à la Grothendieck) that rectify this.

From my perspective, I prefer to think in terms of complex geometry. Strictly speaking, when we do algebraic geometry over $\mathbb{C}$, we still end up with the exact same problems you describe. However, what we do get that is quite nice is that the closed sets are much nicer; specifically, they are the complex subvarieties of our space $X$. To me, this is much nicer than the space being Hausdorff; I would rather my closed sets be nice than worrying about my open sets, in particular since there is often still a close analogue of Hausdorff that exists (separated; see the comments) in this case anyhow.

I mean, think just about $\mathbb{C}$. Since we are only supposed to be working with holomorphic objects, how would you describe, say, a line such as the real axis in that space? You need to allow some level of an-holomorphicity to do so, which is not as nice to work with. The only relevant sub-objects of $\mathbb{C}$, from a holomorphic standpoint, are the points.


With respect to that point, it's worth noting---and this was a sticking point for a long time for me---that unlike in the case of topology in general, the topology on the product of two varieties is not the product topology. This is why a space can be non-Hausdorff despite the diagonal in $X\times X$ being closed.

$\endgroup$
  • $\begingroup$ Thank you for your answer: I agree that the coincidence of two properties: being closed and being subvariety is nice and I believe that it is convenient. But you said that you care more about closed sets rather than open: however topology is about "being close without measuring distances" so the central notion in topology is the notion of (small) neighbourhoods and the convergence. However in algebraic geometry open sets are always very big: so the question is whether algebraic geometry really cares about topology or maybe topology provides only a convenient way to formulate results in AG? $\endgroup$ – truebaran Jan 4 '15 at 21:09
  • 5
    $\begingroup$ Indeed, AG doesn't really care about topology. E.g. one is hardly interested in the set of all Zariski-continuous maps from one scheme to another, such as complex conjugation on $\mathbb C$. One extreme point of view on this (Grothendieck's, expressed in the letter where he introduced dessins d'enfants) is that general topology was a mistake from the get-go, as evidenced by theorems about $T_1$ spaces and whatnot, when what Poincar\'e wanted to study was manifolds. $\endgroup$ – Allen Knutson Jan 4 '15 at 23:05

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