Why are finiteness conditions important (and how to recognize them)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T01:47:57Z http://mathoverflow.net/feeds/question/63715 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63715/why-are-finiteness-conditions-important-and-how-to-recognize-them Why are finiteness conditions important (and how to recognize them)? Yuhao Huang 2011-05-02T18:09:49Z 2012-04-03T17:42:24Z <p>I think everybody here has met lots of finiteness conditions, like those requiring a vector space to be finite dimensional, an abelian group to be finitely generated, a ring to be Noetherian, a manifold to be compact, a sheaf to be coherent, and a complex to be bounded. And there are lots of good theorems once you assume some finiteness condition. (e.g. the Serre duality and Hodge decomposition for compact Kähler manifolds.) And removing those finiteness conditions seems to be non-trivial and interesting. (This applies to e.g. the two theorems mentioned above.)</p> <p>So my question is, </p> <blockquote> <p>why are finiteness conditions so important? </p> </blockquote> <p>This question baffled me for a long time. I remember before I learnt compactness, when doing proofs in calculus, I felt I needed some finite covering of the closed unit interval, but I somehow thought I should avoid using that in my proof. Even after I learnt compactness for a while, the only thing I felt it gives me, was some "combinatorial advantage" ---- I mean, I didn't understand the necessity of assuming compactness in many theorems in elementary analysis, although I was sure I used it in the proofs and I can make some tricky counterexample if compactness wasn't assumed. [I don't feel I got a better understanding on that even now, I can only say I got used to it, i.e. assuming compactness then good things happen.]</p> <p>The remarks on compactness also apply to me when I first learn the condition of a ring being Noetherian. Somehow the condition looks unnatural to me at the beginning, although after getting used to it I felt examples of non-noetherian rings are crazy.</p> <p>And one more thing, I think one thing that Hartshorne/EGA make (early-level) readers confused is that they spent lots of time proving finiteness conditions, like proper pushforward of a coherent sheaf is coherent, or the cohomology of a coherent sheaf on a proper scheme over A is a coherent A-module. One can only appreciate them if he/she is sophisticated enough. (If you are about to prove these theorem in your algebraic geometry class, how do you motivate them and describe why people care about them?)</p> <p>===============</p> <p>A related question, maybe I should ask this in a separated thread, is, how do we recognize good finiteness conditions? Some are "easy", like compactness, and finite generation. But some are tricky, like the condition of a triangulated category being compactly generated. By recognizing good finiteness conditions one might hope to prove some good theorem, but how do we know whether the conditions are too restrictive or not? (I guess this requires hard work, but is there any convincing sign of a good condition before one dives into the details?) Anybody here knows the history of compactness (for topological spaces) and coherence (for sheave of modules)? [Judging by the name, coherent sheaves may come before quasi-coherent ones.]</p> <blockquote> <p>Please re-tag it.</p> </blockquote> http://mathoverflow.net/questions/63715/why-are-finiteness-conditions-important-and-how-to-recognize-them/63721#63721 Answer by Pete L. Clark for Why are finiteness conditions important (and how to recognize them)? Pete L. Clark 2011-05-02T18:54:13Z 2011-05-03T02:34:54Z <p>The fact that various finiteness conditions lead to good theorems which are manifestly false in their absence seems like a good explanation of why they are important. (In fact, I am having trouble thinking of a wholly different kind of explanation for why anything in pure mathematics is important.)</p> <p>I think you are on to something to the extent that we need to give nonexamples and counterexamples along with our theorems in order to give students even a fighting chance at appreciating them. In the realm of commutative algebra this was something that was notoriously underappreciated until relatively recently: I recall well Rota writing about the "hygienic theorems" [Rota, <em>Indiscrete Thoughts</em>, pp. 215-216] in algebra, e.g. things like "Every regular domain is normal". As he wrote, we have no chance of grasping results like this unless we see examples -- preferably several -- of domains which are not regular, not normal, and normal but not regular. In this particular example this is easily done, but unfortunately many of the core counterexamples in the subject have a reputation of being too difficult to show beginners. At this point I feel the need to quote directly from p. 136 of Reid's <em>Undergraduate Commutative Algebra</em>:</p> <blockquote> <p>The catch-phrase "counterexamples due to Akizuki, Nagata, Zariski, etc. are too difficult to treat here" when discussing questions such as Krull dimension and chain conditions for prime ideals, and finiteness of normalisation is a time-honoured tradition in commutative algebra textbooks (comparable to the use of fascist letters $\mathfrak{P}$ and $\mathfrak{m}$ etc., for prime and maximal ideals). This does little to stimulate enthusiasm for the subject, and only discourages the reader in an already obscure literature; I discuss here three counterexamples (taken, with some simplifications, from the famous "unreadable" appendix to [Nagata]) to show some of the ideas involved.</p> </blockquote> <p>This is very well said (well, except that I honestly don't know what's wrong with $\mathfrak{m}$...): most of the standard texts in commutative algebra leave unanswered the natural questions an alert reader will have: is this hypothesis necessary? is the converse of this result true? What happens if we <em>don't</em> assume that $M$ is a finitely generated module over a Noetherian domain? and so forth.</p> <p>By a coincidence I have just finished -- that is, within the last half hour -- teaching a first graduate course on commutative algebra. I tried to spend a lot of time on examples, and I was not afraid to make "technical" digressions about what happens when $M$ is not a finitely generated....Especially I spent an extra long amount of time on module-theoretic questions, which made me feel closer to the heart of the subject. It is easy to motivate the need for modules to be finitely generated: there is a structure theorem for finitely generated modules over a PID but there is no structure theorem for infinitely generated abelian groups. The example of $\mathbb{Q}_p$ as a $\mathbb{Z}_p$-module shows that even over a DVR infinitely generated modules can have a complicated structure. Then, when I got to Noetherian rings I motivated them in part by showing that the Noetherian condition was <strong>equivalent</strong> to many seemingly innocuous and desirable properties, like every submodule of a finitely generated module being finitely generated. At the same time I discussed plenty of examples of non-Noetherian rings, including rings which are very nice "except that they are non-Noetherian" like the ring of all algebraic integers. So I think I gave my students at least an opportunity to feel their way around finiteness conditions in the subject.</p> <p>Let me add that there are some recent texts which do a much better job at this. Most of all I can enthusiastically recommend T.Y. Lam's <em>Lectures on Modules and Rings</em>. As with all of his books, his skill at balancing theory and examples is superior and makes for very pleasant, stimulating reading. </p> <p>It goes much the same for compactness in elementary analysis, but it seems easier to me to supply the necessary counterexamples: every time you encounter a theorem which holds on a compact interval $[a,b]$, ask yourself whether it holds on noncompact intervals (and, if applicable, compact non-intervals!). In all the instances I can think of now, such counterexamples are well known and relatively easy to supply. </p> http://mathoverflow.net/questions/63715/why-are-finiteness-conditions-important-and-how-to-recognize-them/63725#63725 Answer by Charles Matthews for Why are finiteness conditions important (and how to recognize them)? Charles Matthews 2011-05-02T19:06:07Z 2011-05-02T19:06:07Z <p>Two types of points, I think. (1) Counting dimensions is usually a lot more interesting for finite-dimensional vector spaces than for the rest. (2) Where finiteness conditions can be removed, as often they can, a "principle of diminishing returns" may operate. But in this case it may be something of a matter of taste. Doing everything for separable metric spaces, for example.</p> http://mathoverflow.net/questions/63715/why-are-finiteness-conditions-important-and-how-to-recognize-them/63727#63727 Answer by Inna for Why are finiteness conditions important (and how to recognize them)? Inna 2011-05-02T19:12:42Z 2011-05-02T19:12:42Z <p>Generally I think that in fact you're looking at "small" vs. "big", rather than finite vs. infinite. In each field there is a different notion of what small means: in category theory, for example, we tend to distinguish categories with a set of objects rather than a class. For manifolds we are often looking at only finite dimensional anyways, so we care about compactness vs. non-compactness.</p> <p>But the big thing is that small things behave very differently from big things, quite often because we can count/classify the small things but we can't the big things. So we try to keep ourselves to the cases that we know and rule out the things we don't, because otherwise we get very lost.</p> <p>And as a last note I want to point to the <a href="http://en.wikipedia.org/wiki/Eilenberg%25E2%2580%2593Mazur_swindle" rel="nofollow">Eilenberg-Mazur swindle</a>, which shows that infinite sums generally are not associative. Very weird things happen with infinity: $\infty+1=\infty=2\infty$, for example. This means that in order to deal with infinite (big) things we need different tools than when dealing with finite (small) things.</p> http://mathoverflow.net/questions/63715/why-are-finiteness-conditions-important-and-how-to-recognize-them/63736#63736 Answer by Martin Brandenburg for Why are finiteness conditions important (and how to recognize them)? Martin Brandenburg 2011-05-02T20:27:38Z 2011-05-02T20:34:22Z <p>I don't know if I really have enough mathematical background to give a profound answer, but there are some things I would like to mention here since I have also pondered a lot about questions when and <em>why</em> finiteness conditions are so important.</p> <p>Basically the answer to the question "why are finiteness conditions so important?" is very, very simple: Because they make it possible to <em>do</em> mathematics. A mathematical theory which tries to overcome natural finiteness conditions tends to be isolated and narrow. In contrast to that, when you impose good finiteness conditions, the theory becomes rich, very beautiful (which is, of course, subjective). Also "which finiteness conditions" has a very simple answer: Exactly the ones which you need to do the mathematics you want to develope or, at least, imagine. There is no general recipe to produce a good finiteness condition, except that it should fit best to your situation. I've choosen the word "exactly" in order to exclude too restrictive resp. strong finiteness conditions here. On the other hand, we don't always have to look for the most general finitness conditions, unless for some application, we really need more general ones.</p> <p>For example there is nothing wrong with Hartshorne's book in the definition of coherent sheaves when we restrict ourselves to noetherian schemes - everything works out nicely. But if we jump, some day, to non-noetherian schemes, then we have to reconsider the notions of "coherent", "of finite type", "of finite presentation", etc. In the affine case, this also motivates the definition of noetherian rings: I also agree that the definition involving increasing chaings of ideals might not be the most natural one, but what about the equivalent one which Pete has mentioned: Every submodule of a module of finite type is, again of finite type. Actually exactly this property is ofted needed and maybe it has motivated the definition of noetherian rings. No obscure chains. Besides, from a more modern perspective, it is a relative condition, which talks about objects "over" the ring.</p> <p>Also quite useful in practice (for example when surjectivity comes from an abstract argument and the injectivity just does not work out): A surjective endomorphism of a noetherian ring is an automorphism. Well I expect that you can list thousands of nice properties here. Remark that this property illstruates that often the finiteness condition is used to conclude something which says nothing at all about finitness. Another example: If $X$ is a <em>compact</em> topological space, then for every topological space $Y$ the map $X \times Y \to Y$ is a closed map. But of course this fits well since in the proof we want to use a <em>finite</em> intersection of open subsets etc., and the definition of a topology with this restricted intersection property is again based on basic examples out of which this notion was developed. So this fits together very well. Also, the property above characterizes compact topological spaces and probably has motivated the corresponding notion of proper schemes in algebraic geometry.</p> <p>Some general remarks about finitness: One of the most natural object of our mathematical universe is the set of natural numbers $\mathbb{N}$ (I hope no one already here objects and wants to generalize everything to regular cardinals), and the most basic proof involving natural numbers is induction (by the definiton of $\mathbb{N}$ as the smallest inductive set). In order to use induction in more sophisticated situations, we have to give our mathematical objects a measure in $\mathbb{N}$, for example dimension, length, depth, height, etc.. One of the most beautiful and basic examples for this is Grothendieck's vanishing result in sheaf cohomology for <em>finite</em> dimensional topological spaces. So basically you induct on the complexity of the topological space, which you cannot do for arbitrary topological spaces.</p> <p>Finally I have to admit that my oppinion on finiteness conditions has changed in the last months. For years, I wanted to generalize every notion, theorem or even theory in order to avoid all the occuring finiteness conditions. See <a href="http://mathoverflow.net/questions/11767/infinite-tensor-products" rel="nofollow">this MO question</a> for a very clear example: What about infinite tensor products of vector spaces? We can write them down and prove some basic stuff, but in the end there is nothing interesting which we can do with them and there are no useful connections or applications. So let's just forget about them! :-) The same goes for schemes which are not quasi-separated (<a href="http://mathoverflow.net/questions/37964/extending-local-modules-to-global-modules" rel="nofollow">link</a>, <a href="http://mathoverflow.net/questions/41789/what-is-the-center-of-qcohx" rel="nofollow">link</a>).</p> http://mathoverflow.net/questions/63715/why-are-finiteness-conditions-important-and-how-to-recognize-them/93019#93019 Answer by Ostap Chervak for Why are finiteness conditions important (and how to recognize them)? Ostap Chervak 2012-04-03T17:42:24Z 2012-04-03T17:42:24Z <p>You can think of finiteness conditions from other point of view. Suppose you have some category (say, <strong>Set</strong> or <strong>Top</strong>) which have both "nice" objects and "pathological" objects. Then, it is natural to ask if there is a subcategory ("smaller" one) in which every set is "nice". That way, if your only goal is to study a single "nice" space you can study it's properties in subcategory and conclude something about larger one.</p> <p>To study "smaller" category you need it to have some "nice" properties, like being cartesian-closed or something. That way <strong>Comp</strong> is "nice" subcategory of <strong>Top</strong> where you can use a lot of toopological constructions.</p> <p>Now suppose, you want to study Von-Neumann universe of all sets. The only other, "smaller" von-neumann universe you can build is universe of heredetarily finite sets. That way $H_\omega$ is a "nice" sub-universe of $V$ where you can use almost all constructions fro mset theory (the only axiom which isn't true in $H_\omega$ is axiom of infinity).</p> <p>But original question was stated not in the form "Why "niceness" properties are important" but "why finiteness conditions are important". Given that an answer to first question is much more understandable, we can say that finiteness conditions are important because all known "niceness" conditions are finiteness in nature.</p> <p>So , for example following "niceness" condition in <strong>Top</strong> which doesn't look like finiteness condition is in fact equivalent to compactness:</p> <p>$X$ is "nice" iff for every topological space Y projection $X\times Y\to Y$ is closed.</p> <p>This situation ("niceness" conditions are hiddenly "finiteness") because almost all categories we study are set-like, so given a "nice" sub-universe in set-like category we can construct an induced "nice" sub-universe in Von-Neumann universe $V$ and the only nontrivial von-neumann universe is precisely $H_\omega$.</p> <p>So, finiteness conditions are important because every regularity property in set-like category arises from a refularity property in universe of sets $V$, and the "best" regularity property in $V$ is a condition of being hereditarily finite so any good regularity property is esentially a finiteness condition.</p> <p>Also note, that most of named finiteness conditions are actually "heredetarily finiteness" ones, they are usualy inherited by sum sub-objects.</p>