Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:49:28Z http://mathoverflow.net/feeds/question/14634 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14634/why-should-algebraic-objects-have-naturally-associated-topological-spaces-forme Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) Kevin Lin 2010-02-08T11:42:00Z 2011-08-10T22:41:15Z <p>In <a href="http://mathoverflow.net/questions/14314/algebraic-topologies-like-the-zariski-topology" rel="nofollow">this question</a>, Harry Gindi states:</p> <blockquote> <p>The fact that a commutative ring has a natural topological space associated with it is a really interesting coincidence.</p> </blockquote> <p>Moreover, in the answers, Pete L. Clark gives a list of other "really interesting coincidences" of algebraic objects having naturally associated topological spaces.</p> <p>Is there a deeper explanation of the occurrence of these "really interesting coincidences"? It seems to suggest that the standard definition of "topological space" (collection of subsets, unions, intersections, blah blah), which somehow always seemed kind of a weird and artificial definition to me, has some kind of deeper significance or explanation, since it pops up everywhere...</p> <p>The (former) title of this question is meant to be provocative ;-)</p> <p><hr /></p> <p>See also:</p> <p><a href="http://mathoverflow.net/questions/10102/what-are-interesting-families-of-subsets-of-a-given-set" rel="nofollow">What are interesting families of subsets of a given set?</a> </p> <p><a href="http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/" rel="nofollow">How can I really motivate the Zariski topology on a scheme?</a> --- particularly Allen Knutson's <a href="http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/14354#14354" rel="nofollow">answer</a></p> <p><hr /></p> <p><strong>Edit 1:</strong> I should clarify a bit. Let me be more explicit: Is there a unified explanation (mathematical ... or perhaps not) for why various <em>algebraic</em> (where "algebraic" is loosely defined) objects should have naturally associated topological spaces? Pete in the comments notes that he does not like the use of the word "coincidence" here --- but if these things are not coincidences, then what's the explanation?</p> <p>Of course I do understand the intuitive idea behind the definition of "topological space", and how it abstracts for example the notions of "neighborhood" and "near" and "far". It is not surprising that the formalism of topological spaces is useful and ubiquitous in situations involving things like R^n, subsets of R^n, manifolds, metric spaces, simplicial complexes, CW complexes, etc. </p> <p>However, when you start with algebraic objects and then get topological spaces out of them --- I find that surprising somehow because <em>a priori</em> there is not necessarily anything "geometric" or "topological" or "shape-y" or "neighborhood-y" going on.</p> <p><hr /></p> <p><strong>Edit 2:</strong> Somebody has voted to close, saying this is "not a real question". I apologize for my imprecision and vagueness, but I still think this is a real question, for which real (mathematical) answers can conceivably exist.</p> <p>For example, I'm hoping that maybe there is a theorem along the lines of something like: </p> <blockquote> <p>Given an algebraic object A satisfying blah, define Spec(A) to be the set of blah-blahs of A such that blah-blah-blah. There is a natural topology on Spec(A), defined by [something]. When A is a commutative ring, this agrees with the Zariski topology on the prime spectrum. When A is a commutative C^* algebra, this agrees with the [is there a name?] topology on the Gelfand spectrum. When A is a Boolean algebra... When A is a commutative Banach ring... etc.</p> </blockquote> <p>Of course, such a theorem, if such a theorem exists at all, would also need a definition of 'algebraic object'.</p> http://mathoverflow.net/questions/14634/why-should-algebraic-objects-have-naturally-associated-topological-spaces-forme/14635#14635 Answer by Harry Gindi for Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) Harry Gindi 2010-02-08T11:47:38Z 2010-02-09T00:00:22Z <p>Well, this isn't a full answer, but I think it's worth posting.</p> <p>We can identify a Hausdorff space with its poset of open sets because every convergent (thanks JDH) ultrafilter converges uniquely to a point, so in fact, all of the "set data" of the space is contained in the poset of opens itself. We can define a lattice structure on this poset that takes the place of our algebra defined by intersections and unions. This is why the category of Hausdorff spaces is actually a category monadic over sets. Hausdorff spaces are actually totally described by their algebras, which seems pretty cool to me. </p> <p>Some speculation on the question:</p> <p>We have a natural poset structure on subobjects of algebraic objects, which at least gets us partway to having a topology. </p> <p>Further, we can classify maps out of the space by looking at kernels of maps. There is also a canonical action for normal subgroups and ideals (product of normal subgroups and sum of ideals). These have the nice property that they are closed under this operation. They are also closed under intersections. This gives us a complete modular lattice on kernels. The interesting case about rings is that we have a third operation, the product of ideals. The interesting thing about the product of ideals is that it is only defined for finite products. Then we have a somewhat natural structure to start working in (at least for rings). </p> <p>I believe what Qiaochu was talking about with the Galois connections is that for any algebraic structure, you can associate this poset structure on subobjects, and even better, sharpen the characerization by looking at kernels (at least in the case of groups and rings). However, my point was that the additional operation of multiplication, which is restricted to finite products (I guess I would say it has finite arity, but that's not exactly right), gives us an operation on the poset that looks like "finite unions" or "finite intersections". </p> http://mathoverflow.net/questions/14634/why-should-algebraic-objects-have-naturally-associated-topological-spaces-forme/14654#14654 Answer by Tim Porter for Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) Tim Porter 2010-02-08T15:21:21Z 2010-02-08T19:49:36Z <p>I will take the question at face value, but not in the sense of justifying the definition.</p> <p>A topological space is a <b>convenient</b> way of encoding, or perhaps better, organising, <i>certain types of information</i>. (Vague but true! I will give some instances. the data is sometimes `spatial' but more often than not, is not.)</p> <p>Perhaps we should not think of spaces as 'god given' merely 'convenient', and there are variants that are more appropriate in various contexts. </p> <p>A related question, coming from an old Shape Theorist (myself) is : <i>when someone starts a theorem with 'Given a space \$X\$...', how is the space 'given'?</i> As an algebraic topologist I sometimes need to use CW-complexes, but face the inconvenience that if I could give the CW structure precisely I could probably write down an algebraic model for its homotopy type precisely, and vice versa, so a good model is exactly the same as the one I started with. I hoped for more insight into what the space 'was' from my modelling. <b>Giving</b> the space is the end of the process, not the beginning. Strange. A space is a pseudo-visual way of thinking about 'data', which encodes important features, or at leastsome features that we can analyse, partially.</p> <p>If someone gives me a compact subspace of \$\mathbb{R}^n\$, perhaps using some equations and inequalities, can I work out algebraic invariants of its homotopy type, rather than just its weak homotopy type? The answer will usually be no. Yet important properties of \$C^*\$ algebras on such a space, can sometimes be related to algebraic topological invariants of the homotopy type.</p> <p>Spaces can arise as ways of encoding actual data as in topological data analysis, where there is a 'cloud' of data points and the practitioner is supposed to say something about the underlying space from which the data comes. There are finitely many data points, but no open sets given, they are for the data analyst to 'divine'.</p> <p>Not all spatial data is conveniently modelled by spaces as such and directed spaces of various types have been proposed as models for changing data. Models for space-time are like this, but also models for concurrent systems.</p> <p>Looking at finite topological spaces is again useful for encoding finite data (and I have rarely seen infinite amounts of data). For instance, relations between finite sets of data can be and are modelled in this way. Finite spaces give all homotopy types realisable by finite simplicial complexes. Finite spaces can be given precisely (provided they are not too big!) How do invariants of finite spaces appear in their structure? (Note the problem of infinite intersections does not arise here!!!)</p> <p>At the other extreme, do we need points? Are locales not cleaner beasties and they can arise in lots of algebraic situations, again <i>encoding</i> algebraic information. Is a locale a space?</p> <p>I repeat topological spaces are convenient, and in the examples you cite from algebraic geometry they happen to fit for good algebraic reasons. In other contexts they don't. Any Grothendieck topos looks like sheaves on a space, but the space involved will not usually be at all `nice' in the algebraic topological sense, so we use the topos and pretend it is a space, more or less.</p> http://mathoverflow.net/questions/14634/why-should-algebraic-objects-have-naturally-associated-topological-spaces-forme/14712#14712 Answer by François G. Dorais for Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) François G. Dorais 2010-02-08T23:47:38Z 2010-02-10T01:49:12Z <p>I think a reasonable partial explanation comes from universal algebra. The lattice Con(A) of <a href="http://en.wikipedia.org/wiki/Congruence_relation" rel="nofollow">congruences</a> of an algebra is always a <a href="http://en.wikipedia.org/wiki/Algebraic_lattice" rel="nofollow">complete algebraic lattice</a>. Therefore, it is meet continuous in the sense that <code>\$\bigvee_i a \wedge b_i = a \wedge \bigvee_i b_i\$</code> whenever the \$b_i\$ form a <em>directed</em> family of congruences. When Con(A) happens to be finitely distributive, then one can drop the 'directed' requirement. In this case, Con(A) becomes a <a href="http://en.wikipedia.org/wiki/Pointless_topology" rel="nofollow">frame</a> and it can thus be viewed as an abstract topological space (i.e. a locale). In fact, since Con(A) is algebraic the corresponding locale is always spatial and it always corresponds to a concrete spectral space.</p> <p>In the case of a commutative ring A, the lattice Con(A) is isomorphic with the lattice Id(A) of ideals of A. The lattice Id(A) is not always distributive. (Though it is when A is a Prüfer domain and hence when A is a Dedekind domain, for example.) To remedy this, one looks at the radical ideals of A, which are always better behaved, to define the Zariski spectrum. In my humble opinion, the existence of radicals makes commutative rings very special among algebras.</p> http://mathoverflow.net/questions/14634/why-should-algebraic-objects-have-naturally-associated-topological-spaces-forme/14718#14718 Answer by Tom Leinster for Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) Tom Leinster 2010-02-09T00:21:54Z 2010-02-09T06:31:35Z <p>This is an excellent question, I think.</p> <p>Topological spaces could be crudely --- and I mean crudely --- divided into two kinds:</p> <ul> <li><p><strong>The geometric</strong>. These are the spaces that come up all over geometry and algebraic topology &mdash; manifolds, CW complexes, configuration spaces, CW complexes, etc. These are almost invariably Hausdorff, though there are plenty of compact Hausdorff spaces that are often thought of as "pathological", such as the Cantor set. I don't much like terms such as "nice space" and "pathological", because although they might be intended harmlessly, they sound to me a bit dismissive towards the second kind of space.</p></li> <li><p><strong>The spectra</strong>. I'm using this term loosely (and <em>not</em> in the sense of homotopy theory), but I mean things like the spectrum of a ring, the spectrum of a Boolean algebra (= a compact Hausdorff totally disconnected space), the maximal spectrum of a C*-algebra, and then things such as Julia sets, dynamical attractors, and solutions to iterated function systems, all of which have a spectrummy feel to me. </p></li> </ul> <p>The spectra seem a bit unloved. No one denies the importance of, say, Spec of a commutative ring, but still, I reckon that most mathematicians subconsciously regard geometric spaces as the primary kind, and sometimes the spectra simply get swept away as "pathological". </p> <p>__</p> <p><strong>Edit</strong>: Don't read this without also reading Ilya Grigoriev's comment below!</p> http://mathoverflow.net/questions/14634/why-should-algebraic-objects-have-naturally-associated-topological-spaces-forme/14728#14728 Answer by Sridhar Ramesh for Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) Sridhar Ramesh 2010-02-09T02:12:44Z 2010-02-09T02:12:44Z <p>From the perspective of locale theory, a topological space is nothing more than a model of the theory of finitary conjunctions and infinitary disjunctions (corresponding, perhaps, to the intuition that semidecidable propositions are closed under precisely these operations, and the idea that "open" is somehow just a geometric casting of "semidecidable"); that is, a topological space is little more than a lattice with finite meets and infinitary joins, the former distributing over the latter (i.e., a frame). It is perhaps not all that surprising that many algebraic structures should give rise to (complete) lattices satisfying this distributive property, is it? Well, that's a subjective judgement. Perhaps it isn't the best answer. But it's what I'm going with for now. [That is, it's not so surprising that many algebraic objects should give rise to frames; the more surprising thing is the realization that frames could be understood (contravariantly) in geometric terms in the first place.]</p> http://mathoverflow.net/questions/14634/why-should-algebraic-objects-have-naturally-associated-topological-spaces-forme/14759#14759 Answer by Andrew Stacey for Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) Andrew Stacey 2010-02-09T11:07:05Z 2010-02-09T11:07:05Z <p>(I have deleted my original answer to this question because the question was changed in such a way that it made my answer irrelevant. I still think that my basic point was valid, and so am posting a new answer to make that point again. As my original post gained quite a few votes, I judge it not ethical to completely reword my answer but keep those votes. I am also making this answer "community wiki" not because I think anyone else should edit it but to remove it from the reputation/vote game.)</p> <p>I think that the basic answer to this question is that there are connections between algebraic and topological things <strong>because we look for them</strong>. And we look for them because we have, in the past, found them useful. Something I continually (and I mean "continually", just ask one) tell my students is that mathematicians are fundamentally lazy. If we have a good theorem, we don't just use it for what it was first proved for, we look for other ways to use it, ways to extend it, ways to push it further than it was ever intended to be pushed. So if, as a topologist, I see the algebraists doing wonderful things with classifying and studying rings, then I'll do my best to make a ring out of my topological space so that I can steal (sorry, "use") their ideas and save myself a lot of bother. Thus: cohomology theory and the whole area of homotopy theory.</p> <p>That the reverse is true is no surprise. Again, mathematicians are lazy so if we see a bridge with lots of useful stuff going in one direction, we ignore the "one way" signs and go the other way.</p> <p>You could then ask "Why do the bridges exist at all?". Well, they don't always exist. Sometimes we can construct them and sometimes not. It feels a bit like you are looking at a bridge and say "Wow! Who would ever have thought of putting a bridge there?!" but ignoring all the stumps and collapsed half-bridges that litter the riverbank. Of course, one can ask about a <em>specific</em> bridge and ask why that one didn't collapse, but the question feels much more general than that.</p> <p>So, in conclusion, that bridges exist is, I feel, more down to the downright mulishness of mathematicians determined to build a bridge wherever they can, regardless of how many collapses and "Pont d'Avignon"s they create in the process.</p> <p><hr /></p> <p>The above, clearly, works for any two areas of mathematics. Thinking particular of topological spaces, then I think that the questioner is missing the point of "near" and "far" a little when he says:</p> <blockquote> <p>However, when you start with algebraic objects and then get topological spaces out of them --- I find that surprising somehow because a priori there is not necessarily anything "geometric" or "topological" or "shape-y" or "neighborhood-y" going on.</p> </blockquote> <p>(I should point out that the "near" and "far" bit added in the question is in response to my original answer.)</p> <p>Consider this scenario:</p> <ol> <li>I have something I don't know anything about.</li> <li>Can I find something out about something <strong>like it</strong> but simpler?</li> <li>Yes! Great! But how do I <strong>measure</strong> which things are better <strong>approximations</strong> of my unknown thing than others?</li> </ol> <p>Isn't that just what is going on in studying these algebraic objects? The language is fundamentally topological so there's no surprise at all that topological spaces result.</p> <p>So as soon as an area of mathematics becomes interesting in that there are things that you can't figure out easily and simply then the question of finding enough approximations comes in and thus topology.</p> <p>In conclusion of this second part, "interesting = topological" so making the (bizarre) assumption that algebra is interesting, it must thus be topological.</p> http://mathoverflow.net/questions/14634/why-should-algebraic-objects-have-naturally-associated-topological-spaces-forme/72632#72632 Answer by Yasha for Why should algebraic objects have naturally associated topological spaces? (Formerly: What is a topological space?) Yasha 2011-08-10T22:41:15Z 2011-08-10T22:41:15Z <p>I know this is very late, I just happened to run into this question. </p> <p>I think Von Neumann might answer this question like this (as he once has): "One does not understand anything in mathematics one simply gets used to it." Then he may add, we have a natural functor from the category of topological spaces to the category of rings, it is an interesting natural phenomenon that this functor is strictly invertible if we restrict to suitable subcategories of the category of top spaces, and of the category of rings. This is Gelfand-Naimark.</p> <p>One does not "understand" phenomena anymore then one understands why the universe exists.</p> <p>By the way a topological space is also naturally an algebraic object, we can take its partially ordered set of open subsets. This often completely determines the space: <a href="http://math.nie.edu.sg/dszhao/Research%20papers/conference%20proceeding/posetmodels.pdf" rel="nofollow">http://math.nie.edu.sg/dszhao/Research%20papers/conference%20proceeding/posetmodels.pdf</a> (Although I didn't really read this.)</p>