13
$\begingroup$

Why do algebraic geometers still use the term "quasi-compact" when they almost never deal with Hausdorff spaces? They certainly use "local" rather than "quasi-local" (local = quasi-local + noetherian), so is there any reason other than historical contingency?

Do algebraic geometers who do work in other fields still follow this convention when they write other papers? If they do, do they write at the beginning of the paper something along the lines of "by compact, we mean quasi-compact and Hausdorff"?

Improve this question
$\endgroup$

5 Answers 5

Reset to default
18
$\begingroup$

The condition of quasi-compactness in the Zariski topology bears little resemblance to the condition of compactness in the classical analytic topology: e.g. any variety over a field is quasi-compact in the Zariski topology, but a complex variety is compact in the analytic topology iff it is complete, or better, proper over $\operatorname{Spec} \mathbb{C}$.

I think many algebraic geometers think to themselves that a variety is "compact" if it is proper over the spectrum of a field. I have heard this terminology used and occasionally it shows up in (somewhat informal) writing.

So a perhaps more accurate brief answer is that in algebraic geometry the distinction between quasi-compact and quasi-compact Hausdorff is very important, whereas in other branches of geometry non-Hausdorff spaces turn up more rarely.

Anyway, many mathematicians have been happy with the quasi-compact / compact distinction for about 50 years, so I don't think this usage is going away anytime soon.

To address the last question: when writing for a general mathematical audience, it is a good idea to give an unobtrusive heads up as to your stance on the quasi-compact / compact convention. (The same probably goes for other non-universal conventions in mathematics.) If I were speaking about profinite groups, I would say something like:

"A profinite group is a topological group which can be expressed as an inverse limit of finite discrete groups. Equivalently, a topological group is profinite if it is compact (Hausdorff!) and totally disconnected."

This should let people know what side I'm on, and thus be able to understand me. When writing for students, I might take pains to be more explicit, using a "By compact I mean..." construction as you have indicated above.

Improve this answer
$\endgroup$
7
  • 3
    $\begingroup$ It really is a standard convention in mathematics. For instance, one of the standard references for the subject of its title is Ryszard Engelking's General Topology (latest edition published 1989). In Engelking's text, compact implies Hausdorff. And so forth. To ardently campaign for one position or the other is, in my view, just advertising which texts you haven't read. When writing about compact spaces, it is important to make clear what "compact" means. Either usage, without any explanation, will be confusing to some people. $\endgroup$
    – Pete L. Clark
    Commented Mar 3, 2010 at 17:48
  • 3
    $\begingroup$ @Pete: I found your remark surprising. Some googling led to a page with this on it: Bourbaki and Engelking [Bou66, Eng89] use the term "quasi-compact" instead, suggesting that all other topology textbooks use "compact" and "compact Hausdorff". $\endgroup$
    – Gerald Edgar
    Commented Mar 3, 2010 at 18:10
  • 3
    $\begingroup$ @Gerald: if we're talking serious standard references, yes, I believe this is correct. But the list of serious standard references for GT is very short: I would include also Willard and Kelley. I would say Munkres' book is an excellent undergraduate text. Note by the way that the latest edition of Engelking's book has 706 citations on MathSciNet -- it is indubitably a standard reference. (Kelley: 112. Willard: 139 total from the last two editions. Munkres: 156.) $\endgroup$
    – Pete L. Clark
    Commented Mar 3, 2010 at 18:43
  • 1
    $\begingroup$ It would be nice if "analytically compact" and "algebraically compact" became standard. $\endgroup$
    – JBorger
    Commented Mar 3, 2010 at 21:27
  • 1
    $\begingroup$ By the way, I attended an excellent algebraic geometry seminar talk today: it was about various "compactifications" of various moduli spaces. I resisted the temptation to point out that they were already quasi-compact. (P.S.: yes, I am not that far from being an algebraic geometer, if that was not already clear. Lots of mathematicians in adjacent fields would say the same.) $\endgroup$
    – Pete L. Clark
    Commented Mar 4, 2010 at 1:09
7
$\begingroup$

Personally, I was never a big fan of the Bourbaki convention of including the Hausdorff axiom into compactness. I would be happy if we dropped the "quasi", but perhaps it would lead to too much confusion at this point in time. On the other hand, schemes were once called preschemes, so algebraic geometers are capable of change.

(This topic is a bit a diversion I think, and I'm not seriously suggesting that the terminology should be changed in this instance. In general, however, I think it is OK to occasionally break with tradition and modify terminology when it's genuinely ungainly. Of course mathematics, like any human endeavor, is full of choices which in hindsight may seem a little odd and perhaps inconsistent. Most of us can live with that. As a complex geometer, I am happy to use "Riemann surface" and "elliptic curve" in the same sentence.)

Improve this answer
$\endgroup$
0
5
$\begingroup$

I think it comes from Bourbaki. In french, "compact" is what you call "quasi-compact and Hausdorff". Calling a non-Hausdorff space compact would make no sense hence "quasi-compact".

I imagine it is the same thing for "local = quasi-local + noetherian". For me, a commutative ring is local if it has only one maximal ideal. There is no noetherian hypothesis in the definition.

Improve this answer
$\endgroup$
1
  • $\begingroup$ @YBL: I fixed a typo. $\endgroup$
    – Harry Gindi
    Commented Mar 3, 2010 at 17:22
4
$\begingroup$

It is a useful technique in algebraic geometry to work over the complex numbers with the analytic topology...

Improve this answer
$\endgroup$
6
  • 3
    $\begingroup$ But that does not justify using a definition of compact which is different from everyone else, surely? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 3, 2010 at 15:26
  • $\begingroup$ Now I have augmented my answer to clarify that I meant to also use the usual topology of the complex field. $\endgroup$
    – BCnrd
    Commented Mar 3, 2010 at 15:47
  • 1
    $\begingroup$ I guess you are commenting on the "they almost never deal with Hausdorff spaces" rather than answering his question? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 3, 2010 at 15:49
  • 2
    $\begingroup$ The question contains a flawed premise -- algebraic geometers certainly do care about Hausdorff spaces, because they care about complex manifolds, Kahler metrics, Hodge theory, etc. Pointing out the flaw in the premise is an important part of the answer. Another (implicit, here, but explicit in other recent comments) premise is that algebraic geometers are the only ones who use quasi-compact and compact in this way. That's false: I do this, and I am a number theorist. $\endgroup$
    – Pete L. Clark
    Commented Mar 3, 2010 at 16:01
  • 5
    $\begingroup$ I'm doing both. As a contrast, algebraic geometers use the word "immersion" to mean (essentially) what differential geometers called "embedding", and in diff geom. "immersion" has a much broader meaning (for foliations, in Lie correspondence, etc.) which tends to not be relevant in A.G., hence no confusion (except for diff. geometers who need to read stuff in alg. geom.!). $\endgroup$
    – BCnrd
    Commented Mar 3, 2010 at 16:01
2
$\begingroup$

This is far from standard, but in my mind, when I'm doing algebraic geometry, "quasicompact" means "compact in the Zariski topology," and "compact" means "compact in the analytic topology" (or is not used at all if I'm not working over a topological field). Thus, in principle, having explained that the terms were being used this way, one could write statements like

"$\mathbb{A}^n_{\mathbb{C}}$ is quasicompact but not compact"
and
"Projective varieties are both compact and quasicompact."

But I would be very hesitant to do so, since I've never seen the terminology used this way so explicitly.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .