Blackbox Theorems - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-21T18:22:49Z http://mathoverflow.net/feeds/question/99506 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99506/blackbox-theorems Blackbox Theorems Benjamin Steinberg 2012-06-13T20:58:36Z 2012-09-15T16:00:15Z <p>By a <i>blackbox theorem</I> I mean a theorem that is often applied but whose proof is understood in detail by relatively few of those who use it. A prototypical example is the Classification of Finite Simple Groups (assuming the proof is complete). I think very few people really know the nuts and bolts of the proof but it is widely applied in many areas of mathematics. I would prefer not to include as a blackbox theorem exotic counterexamples because they are not usually applied in the same sense as the Classification of Finite Simple Groups. </p> <p>I am curious to compile a list of such blackbox theorems with the usual CW rules of one example per answer. </p> <p>Obviously this is not connected to my research directly so I can understand if this gets closed. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99508#99508 Answer by Dirk for Blackbox Theorems Dirk 2012-06-13T21:30:53Z 2012-06-13T21:30:53Z <p>Does <a href="http://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="nofollow">FLT</a> count?</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99509#99509 Answer by Jan Weidner for Blackbox Theorems Jan Weidner 2012-06-13T21:31:25Z 2012-09-15T16:00:15Z <p>The existence of resolution of singularities in characteristic zero is certainly used by many more people than those who know the details of its proofs, especially the original one. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99512#99512 Answer by D K for Blackbox Theorems D K 2012-06-13T21:34:43Z 2012-06-13T21:34:43Z <p>Determinacy of Borel Games seems like a good example of this.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99513#99513 Answer by Jan Weidner for Blackbox Theorems Jan Weidner 2012-06-13T21:36:02Z 2012-06-13T21:36:02Z <p>The decomposition theorem for perverse sheaves is used in many areas of mathematics, for example representation theory, while the details of the weights machinery involved in its proofs are notoriously hard.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99514#99514 Answer by Karl Schwede for Blackbox Theorems Karl Schwede 2012-06-13T21:36:28Z 2012-06-14T14:15:10Z <p>I think the main statements of the MMP (Minimal Model Program ) in algebraic geometry qualify for this. </p> <p>It will even become more of a black box in the future, as people will understand better how to apply it.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99515#99515 Answer by Santiago for Blackbox Theorems Santiago 2012-06-13T21:43:05Z 2012-06-13T21:43:05Z <p>Fixed point theorems (such as Brouwer and Kakutani's) are very frequently invoked, specially in Econ. I am not sure how many people are familiar with the proofs. There are many nice proofs available, by the way.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99516#99516 Answer by Karl Schwede for Blackbox Theorems Karl Schwede 2012-06-13T21:43:14Z 2012-06-13T21:43:14Z <p>I think also many people treat certain tools in homological algebra this way. For example various facts about spectral sequences and how to use them.</p> <p>In the spectral sequences example, I feel like many people once learned the background, and then forgot it (perhaps could reconstruct if forced). But regardless, they still know how to apply the machines in the problems relevant to them.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99517#99517 Answer by Ralph for Blackbox Theorems Ralph 2012-06-13T21:45:18Z 2012-06-13T21:45:18Z <p>Most mathematicans know that the axiom of choice is independent from ZF axioms, but I guess most non-set-theorists don't know details of the proof. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99520#99520 Answer by Zsbán Ambrus for Blackbox Theorems Zsbán Ambrus 2012-06-13T22:08:47Z 2012-06-15T16:24:07Z <blockquote> <p><strong>Jordan's curve theorem</strong> is used as a blackbox. </p> </blockquote> <p>This topology theorem states that a looped continuous path in the plane partitions the points of the plane, such that any continuous path going from a point in one partition to a point in the other intersects the loop. </p> <p>There seem to be a lot of theorems in calculus of which I don't fully understand the proof, though some of this shows my ignorance of calculus. Jordan's theorem seem to be an extreme example though. Let me list some other examples.</p> <ul> <li>the existance and basic properties of the Lebesgue measure and infinite product measures</li> <li>the fact that a Wiener process is almost surely everywhere continuous (mentioned below as a separate answer by weakstar)</li> <li>the fact that the roots of a complex polynomial (or the eigenvalues of a complex matrix) are continuous in the coefficients (though I should learn the proof for this because the more precise statements on how well conditioned the roots are on the coefficients is useful)</li> <li>the spectral theorem about linear maps on a possibly infinite-dimensional Hilbert-space</li> <li><s>the proof that a convex function (from reals to reals) is always continuous everywhere and has a left and right derivative everywhere</s> (Update: okay, remove this last one because Ian Morris gave a simple proof below. I seemed to remember it was more difficult than that. Thanks, Ian.)</li> <li>Rademacher's theorem: every Lipschitz function from an open subset of $\mathbb{R}^m$ to $\mathbb{R}^n$ is differentiable almost everywhere. (Added on Paul Siegel's suggestion. For some reason I haven't heared of this theorem before, but it sure sounds useful.) </li> <li>Lebesgue's criterium which claims that a bounded function from reals to reals is Riemann-integrable iff it's continuous almost everywhere. (The proof is elementary and doesn't require any ideas, but it's laborous.)</li> </ul> http://mathoverflow.net/questions/99506/blackbox-theorems/99521#99521 Answer by Noah S for Blackbox Theorems Noah S 2012-06-13T22:08:50Z 2012-06-14T00:26:03Z <p>Saharon Shelah has a series of results he actually calls "black boxes," and uses accordingly (see his paper, "Black Boxes," <a href="http://arxiv.org/abs/0812.0656" rel="nofollow">http://arxiv.org/abs/0812.0656</a>); my understanding is that these are Diamond-like theorems that are provable in ZFC. </p> <p>(Diamond, for clarification, is a sort of guessing principle: it asserts that there exists a single sequence $(A_\alpha)_{\alpha\in\omega_1}$ such that $A_\alpha\subseteq\alpha$ such that, for any $A\subseteq \omega_1$, the set $$\lbrace \alpha: A_\alpha=A\cap\alpha\rbrace$$ is "large" (specifically, stationary - intersects every closed unbounded subset of $\omega_1$). This principle is not provable in ZFC; it follows from $V=L$ and implies $CH$, but both of these implications are strict. My understanding, which is quite limited, is that Diamond is used in constructions of $\omega_1$-sized structures where one needs to "guess correctly" stationarily often, and that Shelah developed the black boxes to perform many of these same constructions in ZFC alone.)</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99523#99523 Answer by Zsbán Ambrus for Blackbox Theorems Zsbán Ambrus 2012-06-13T22:21:19Z 2012-06-14T10:09:43Z <blockquote> <p><strong>Recognizing hamiltonian graphs is NP-complete.</strong> </p> </blockquote> <p>(A hamiltonian graph is a graph that has a cycle passing through every node.) Everyone likes to use this theorem for proving other NP-completeness proofs, but few people would know an actual proof. Even the simplest proof is somewhat messy. The theorem that 3-colorable graphs are NP-complete is similar. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99526#99526 Answer by tweetie-bird for Blackbox Theorems tweetie-bird 2012-06-13T22:42:59Z 2012-06-14T02:23:39Z <p>How is the proof of the Poincare' Conjecture (in all dimensions) not yet anywhere on this list? </p> <p>Edit: in light of the comments below, this answer is now being upgraded to the proof of the Geometrization Conjecture (which implies the Poincare' Conjecture, among other things).</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99528#99528 Answer by Gerry Myerson for Blackbox Theorems Gerry Myerson 2012-06-13T23:00:40Z 2012-06-13T23:00:40Z <p><a href="http://en.wikipedia.org/wiki/Faltings%27_theorem" rel="nofollow">Faltings' Theorem</a>, to the effect that a curve of genus greater than 1 over the rationals has only finitely many rational points, is often invoked, I suspect often by people who haven't gone through a proof in detail. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99529#99529 Answer by Benjamin Steinberg for Blackbox Theorems Benjamin Steinberg 2012-06-13T23:01:00Z 2012-06-14T00:32:31Z <p>I think the solution to Hilberts 5th problem is an example. For a while Gromov's polynomial growth theorem was an example because the proof invoked Hilberts 5th. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99530#99530 Answer by Asaf Karagila for Blackbox Theorems Asaf Karagila 2012-06-13T23:03:55Z 2012-06-13T23:03:55Z <p>Most mathematicians can recite the construction of a Vitali set and state that the axiom of choice is needed. Very few of them would know to describe the proof that the axiom of choice is <em>really</em> needed, i.e. Solovay's model (or even the Feferman-Levy in which every set is Borel).</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99531#99531 Answer by Gerry Myerson for Blackbox Theorems Gerry Myerson 2012-06-13T23:05:27Z 2012-06-13T23:05:27Z <p>Deligne's Theorem, found at Wikipedia under the heading of <a href="http://en.wikipedia.org/wiki/Weil_conjectures" rel="nofollow">Weil conjectures</a>, which is the Riemann Hypothesis for zeta-functions of algebraic varieties over finite fields, is often applied to estimate exponential sums in Number Theory, I suspect often by people (like me) who haven't gone through a proof in detail.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99533#99533 Answer by Abhishek Parab for Blackbox Theorems Abhishek Parab 2012-06-13T23:25:21Z 2012-06-13T23:25:21Z <blockquote> <p>Existence and uniqueness of invariant Haar measure on a locally compact topological group.</p> </blockquote> <p>It is used in harmonic analysis and number theory. It is not so difficult a result to state but a proof is not so commonly seen in books. The measure allows one to define an integral on the group and do analysis. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99537#99537 Answer by Anthony Quas for Blackbox Theorems Anthony Quas 2012-06-13T23:51:02Z 2012-06-13T23:51:02Z <p>I've got to put in 2c for ergodic theory: the Multiplicative Ergodic Theorem is widely quoted, but locating a complete proof is hard.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99538#99538 Answer by Gjergji Zaimi for Blackbox Theorems Gjergji Zaimi 2012-06-13T23:54:44Z 2012-06-13T23:54:44Z <p>The <a href="http://en.wikipedia.org/wiki/Robertson%E2%80%93Seymour_theorem" rel="nofollow">graph minor theorem</a> and the <a href="http://en.wikipedia.org/wiki/Graph_structure_theorem" rel="nofollow">graph structure theorem</a> are two results which are invoked quite often in combinatorics/graph theory. Much like the classification of finite simple groups they are excellent ways of sweeping hundreds of pages of technical proofs under just a few sentences.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99541#99541 Answer by weakstar for Blackbox Theorems weakstar 2012-06-14T01:12:50Z 2012-06-14T01:12:50Z <p>The existence of Brownian Motion.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99542#99542 Answer by Daniel Moskovich for Blackbox Theorems Daniel Moskovich 2012-06-14T01:26:13Z 2012-06-14T01:26:13Z <p>Low dimensional topology is unfortunately full of such theorems. Maybe the archetypal example is the Kirby Theorem, which states that surgery on two framed links in S<sup>3</sup> give diffeomorphic 3-manifolds if and only if the links are related by a specific set of combinatorial moves. The result is used routinely, in order to prove that invariants of framed links descend to topological invariants of the manifold (<i>e.g.</i> Reshetikhin-Turaev invariants).</p> <p>All known proofs of Kirby's Theorem are a nightmare (see <a href="http://mathoverflow.net/questions/16848/proofs-of-kirbys-theorem" rel="nofollow">this MO question</a>). You need to use some heavy tool (Cerf's Theorem/ explicit presentation of Mapping Class Groups) in order to show that some expansion of the space of Morse functions (a Frechet space) is path connected. This is outside the toolbox of most topologists.</p> <p>I would be surprised if there were 20 people in the world who have read through and understood the details of the proof of Kirby's Theorem. Yet it's routinely used.</p> <p>There are more mild examples too. The proof that PL 3-manifolds can be smoothed, and that the resulting smooth structure is unique up to isotopy (the exact statement is in Kirby-Seibenmann), is used routinely as though it were obvious, but it is actually quite a hard theorem which is not covered in any of the standard textbooks (Thurston's "3-Manifolds" being an exception). See <a href="http://www.math.harvard.edu/~lurie/937.html" rel="nofollow">Lurie's 2009 notes</a>.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99543#99543 Answer by S. Carnahan for Blackbox Theorems S. Carnahan 2012-06-14T01:26:22Z 2012-06-14T01:26:22Z <p><a href="http://arxiv.org/abs/0706.1907" rel="nofollow">Nagata embedding</a> is another black box - its statement is very simple and useful, but its proof is hard.</p> <p>By combining Nagata embedding with Hironaka's resolution of singularities (mentioned in another answer), you get "any smooth variety over a characteristic zero field admits an open immersion into a proper smooth variety", which is concise enough that people often use it without citing the authors' hard work.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99544#99544 Answer by Malik Younsi for Blackbox Theorems Malik Younsi 2012-06-14T01:36:13Z 2012-06-14T01:36:13Z <p>I think the <a href="http://en.wikipedia.org/wiki/Uniformization_theorem" rel="nofollow"><strong>Uniformization theorem</strong></a> is an example of <em>blackbox theorem</em> : any simply connected Riemann surface is conformally equivalent to either the open unit disk, the complex plane or the Riemann sphere.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99551#99551 Answer by Kestutis Cesnavicius for Blackbox Theorems Kestutis Cesnavicius 2012-06-14T05:02:14Z 2012-06-14T05:02:14Z <p>The existence of Neron models. This gets used all the time when one talks of abelian varieties, but familiarity with the proof is almost never needed.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99557#99557 Answer by Fred Rohrer for Blackbox Theorems Fred Rohrer 2012-06-14T06:20:07Z 2012-06-14T06:20:07Z <p>Embedding theorems for abelian categories (Freyd, Mitchell, Lubkin, ...) seem to qualify.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99560#99560 Answer by Michael Greinecker for Blackbox Theorems Michael Greinecker 2012-06-14T07:02:27Z 2012-06-14T07:02:27Z <p>The Borel isomorphism theorem says that any two Polish (complete and separable metrizable) spaces endowed with their Borel $\sigma$-algebra are isomorphic as measurable spaces if and only if they have the same cardiality and this cardinality is either countable or the cardinality of the continuum.</p> <p>The result is extremely useful and widely applied in probability theory. It allows one to prove many results for general Polish spaces by proving them for the real line or the unit interval. The proof is actually not that hard, but somewhat messy and gives little useable insight for those not working in descriptive set theory.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99566#99566 Answer by Martin Brandenburg for Blackbox Theorems Martin Brandenburg 2012-06-14T08:29:59Z 2012-06-14T08:29:59Z <p>When learning algebraic geometry and in particular the notion of smooth varieties, you will probably stumble upon the following Theorems:</p> <ul> <li>Regular local rings are factorial.</li> <li>Localizations of regular local rings are regular, too.</li> <li>A local ring is regular iff its residue field has finite projective dimension (Serre).</li> </ul> <p>Many texts on algebraic geometry take this as a black box, quoting standard sources of commutative algebra. The reason seems to be that you don't have to understand the methods of the proof (e.g. Koszul homology) in order to apply these results.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99568#99568 Answer by Martin Brandenburg for Blackbox Theorems Martin Brandenburg 2012-06-14T08:37:04Z 2012-06-14T08:37:04Z <p>Doesn't Zorn's Lemma count? Of course in ZF this is not a Theorem (rather it is undecidable), but in ZF+AC it is a real Theorem which is often mentioned without proof, especially in classes outside of mathematical logic. For example, in commutative algebra it is quoted in order to get enough maximal ideals in rings, etc.</p> <p>Of course it is not hard to understand the proof of AC => Zorn, but many students take this on faith. I don't know if this also applies to mathematicians.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99573#99573 Answer by Jeffrey Kuan for Blackbox Theorems Jeffrey Kuan 2012-06-14T09:00:06Z 2012-06-14T09:00:06Z <p>One of the seminal results in random matrix theory is that in the edge scaling limit, the largest eigenvalue of random Hermitian matrices is the Tracy-Widom distribution. The original proof by Tracy and Widom is full of so many unintuitive technical details that most people who cite it don't understand it. (Or so I've been told).</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99579#99579 Answer by Denis Serre for Blackbox Theorems Denis Serre 2012-06-14T09:46:49Z 2012-06-14T09:46:49Z <p>The Open Mapping Theorem (known as Banach-Schauder Theorem) is used daily by zillions of analysts. But its proof is far from trivial and is often overlooked by users. It is not just a straightforward consequence of Baire's Theorem.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99584#99584 Answer by Martin Brandenburg for Blackbox Theorems Martin Brandenburg 2012-06-14T10:07:10Z 2012-06-14T10:07:10Z <p>The well-definedness of the <a href="http://en.wikipedia.org/wiki/Connected_sum" rel="nofollow">connected sum</a> of two manifolds. After all, we choose two arbitrary balls for gluing; why should the result be independent? The proof depends on the nontrivial <a href="http://en.wikipedia.org/wiki/Disc_theorem" rel="nofollow">Disc theorem</a>.</p> <p>I think that one usually uses this result as a black box.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99585#99585 Answer by Zsbán Ambrus for Blackbox Theorems Zsbán Ambrus 2012-06-14T10:09:00Z 2012-06-14T10:09:00Z <blockquote> <p>The proof for <strong>Hilbert's tenth problem</strong>, that is, that there is no algorithm to solve general Diophantiane equations.</p> </blockquote> <p>Benjamin Steinberg has mentioned this above in a comment. I believe the proof is complicated.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99586#99586 Answer by Martin Brandenburg for Blackbox Theorems Martin Brandenburg 2012-06-14T10:18:35Z 2012-06-14T10:18:35Z <p>Often Serre's <a href="http://en.wikipedia.org/wiki/GAGA#Serre.27s_GAGA" rel="nofollow">GAGA</a> (translation between algebraic and analytic land) is treated as a black box.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99589#99589 Answer by Chandrasekhar for Blackbox Theorems Chandrasekhar 2012-06-14T10:32:35Z 2012-06-14T10:32:35Z <p><a href="http://en.wikipedia.org/wiki/Fundamental_lemma_%28Langlands_program%29" rel="nofollow">Fundamental lemma (Langlands program)</a> which Ngô Bảo Châu proved and got the Fields medal in 2010.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99592#99592 Answer by Mikhail Bondarko for Blackbox Theorems Mikhail Bondarko 2012-06-14T10:49:11Z 2012-06-14T10:49:11Z <p>There are a lot of complicated 'motivic' statements. I would say that the Milnor Conjecture (and its generalization, the Bloch-Kato conjecture) are easy to understand and apply; yet its proof is very hard.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99594#99594 Answer by Ricky for Blackbox Theorems Ricky 2012-06-14T11:17:46Z 2012-06-14T11:17:46Z <p>Faltings' almost purity theorem. The proof given, for the smooth case, in $p$-adic Hodge theory has some problems, and the proof of the general case in the Asterisque paper Almost Étale Extensions is completely unreadable (at least to me) and also contains some mistakes. We now finally have a very good proof (by Peter Scholze), but the almost purity theorem has been used as a black box for years.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99615#99615 Answer by darij grinberg for Blackbox Theorems darij grinberg 2012-06-14T15:05:14Z 2012-06-14T15:05:14Z <p>For a long time, the <a href="http://en.wikipedia.org/wiki/Littlewood-Richardson_rule" rel="nofollow">Littlewood-Richardson rule</a> has been a black box. See <a href="http://www-math.univ-poitiers.fr/~maavl/pdf/lrr.pdf" rel="nofollow">van Leeuwen's wonderful article</a> for its history (and a rather involved, even if enlightening proof). This really changed with Stembridge's 2-pages long slick (although far from straightforward!) <a href="http://www.emis.de/journals/EJC/Volume_9/PDF/v9i1n5.pdf" rel="nofollow">proof (2002)</a> and Gasharov's <a href="http://www.math.cornell.edu/~vesko/papers/lrrule.ps" rel="nofollow">3-pages long proof (1998)</a>. (I have read Stembridge and can vouch for its good exposition; it's not short by virtue of being unreadable, but short by virtue of being short. I have not yet read Gasharov, and I am in the middle of van Leeuwen.)</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99617#99617 Answer by darij grinberg for Blackbox Theorems darij grinberg 2012-06-14T15:16:31Z 2012-06-14T22:01:16Z <p>Two results I have seen used without proof in undergraduate lectures were:</p> <ul> <li><p><a href="http://en.wikipedia.org/wiki/Tychonoff%27s_theorem" rel="nofollow">Tychonoff's theorem</a> in I don't remember what courses (but can be many).</p></li> <li><p>IIRC <a href="http://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_extension_theorem" rel="nofollow">Carathéodory's extension theorem</a> in probability theory (but might also have been a different fact from measure theory).</p></li> </ul> <p>In both cases the proofs are not long, but were deemed not useful enough to be taught. I am wondering whether this is special to the courses I had or generally common.</p> <p>Also, various courses on graph theory use some versions of the Jordan curve theorem; even the ones not requiring analysis (speaking of piecewise linear paths) are usually not proven. And several analysis courses don't prove the basic properties of real numbers, instead treating them as axioms.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99654#99654 Answer by Gerry Myerson for Blackbox Theorems Gerry Myerson 2012-06-14T23:08:09Z 2012-06-14T23:08:09Z <p>Many people apply the theorem $1+1=2$, but how many understand in detail the proof given by Russell and Whitehead in <a href="http://en.wikipedia.org/wiki/Principia_Mathematica" rel="nofollow">Principia Mathematica</a>? Well, I suppose there are other proofs available....</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99657#99657 Answer by Paul Siegel for Blackbox Theorems Paul Siegel 2012-06-15T00:04:50Z 2012-06-15T00:04:50Z <p>I think differential topology has dozens of these results. Here are some examples that immediately come to mind:</p> <ul> <li>The tubular neighborhood theorem: every submanifold $N$ of a manifold $M$ has an open neighborhood which is diffeomorphic to the total space of the normal bundle of $N$</li> <li>The fundamental theorem of Morse theory: if $f: M \to \mathbb{R}$ is a Morse function and $[a,b]$ is an interval which contains no critical values of $f$ then the set of all points where $f \leq a$ is a deformation retract of the set of all points where $f \leq b$</li> <li>Every continuous isomorphism of smooth vector bundles is homotopic to a smooth isomorphism (and other such "continuous equivalence = smooth equivalence" results)</li> </ul> <p>Probably most topologists know the basic ideas behind the proofs of these results, but I think many would be hard-pressed to actually write down a complete argument. I say this with confidence because I know of several textbooks by good authors that have proofs which are either wrong or sketchy on some details.</p> <p>There are also some results with standard proofs that are widely known, but I think considerably more people use the results than know the proofs:</p> <ul> <li>De Rham's theorem: the De Rham cohomology groups of a manifold are isomorphic to the singular cohomology groups with real coefficients</li> <li>The Hodge theorem: every De Rham cohomology class on a Riemannian manifold has a harmonic representative</li> <li>Whitehead's result that every smooth manifold has a unique PL structure</li> </ul> http://mathoverflow.net/questions/99506/blackbox-theorems/99659#99659 Answer by Paul Siegel for Blackbox Theorems Paul Siegel 2012-06-15T00:14:49Z 2012-06-15T00:14:49Z <p>C*-algebra theory has a number of good examples of this.</p> <ul> <li>Voiculescu's theorem: an ample representation of a C*-algebra essentially absorbs any nondegenerate representation</li> <li>Kasparov's technical theorem: if anybody really cares, I'll repeat the statement. The point is that anybody who works with bivariant K-theory uses this result <em>ALL THE TIME</em>, e.g. for excision or the existence of Kasparov products.</li> <li>Stinespring's theorem: any completely positive map into $B(H)$ dilates to a representation</li> </ul> <p>I have been using Voicalescu's theorem and KTT for a about a year or so longer than I knew the proofs. I probably still wouldn't know the proofs if it hadn't become necessary. Stinespring's theorem is probably better known among the people who use it because it's not so difficult, but it could be tempting to use it as a black box.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99660#99660 Answer by Jeff Strom for Blackbox Theorems Jeff Strom 2012-06-15T00:20:02Z 2012-06-15T00:20:02Z <p>How about Haynes Miller's theorem resolving the Sullivan conjecture?</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99669#99669 Answer by David Harris for Blackbox Theorems David Harris 2012-06-15T02:20:33Z 2012-06-15T02:20:33Z <p>The Lovasz Local Lemma gives a very simple criterion for when certain random events have positive probability. Almost all applications of the Lemma automatically have an algorithm to find such events. The details of these proofs (especially in their most general forms) can be messy, but the LLL criterion is so simple you can use it basically as a black box.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99672#99672 Answer by Dori Bejleri for Blackbox Theorems Dori Bejleri 2012-06-15T03:17:45Z 2012-06-15T03:17:45Z <p>The existence of Hilbert and Quot schemes. These are arguably the most important objects in moduli/deformation theory but the proof of their existence is almost never even presented in books on the topic, let alone needed or used. All the properties and applications follow formally so the existence is used as a black box.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99674#99674 Answer by Yemon Choi for Blackbox Theorems Yemon Choi 2012-06-15T04:12:38Z 2012-06-15T04:12:38Z <p>Slightly debatable, but my impression is that the fact that injectivity, Property (P), and hyperfiniteness define the same class of von Neumann algebras is used by many people without feeling the need to learn the proofs.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99677#99677 Answer by Daniel McLaury for Blackbox Theorems Daniel McLaury 2012-06-15T06:55:55Z 2012-06-15T06:55:55Z <p>Chevalley's theorem: any algebraic group is the extension of a linear algebraic group by an abelian variety.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99682#99682 Answer by Gil Kalai for Blackbox Theorems Gil Kalai 2012-06-15T08:19:25Z 2012-06-15T08:19:25Z <p>Perhaps Atiya-Singer's index theorem can qualify. Another candidate is Gromov's h-principle.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99683#99683 Answer by Deane Yang for Blackbox Theorems Deane Yang 2012-06-15T08:31:41Z 2012-06-15T08:31:41Z <p>I've already posted this as a question on MathOverflow, but it appears that everyone working on the Ricci flow, as well as other geometric heat flows, takes it for granted the existence in short time of a solution to a nonlinear parabolic PDE on a vector bundle over a complete Riemannian manifold. I have not been able to find a complete proof for even a <em>linear</em> parabolic PDE.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99686#99686 Answer by Deane Yang for Blackbox Theorems Deane Yang 2012-06-15T08:57:26Z 2012-06-15T12:16:27Z <p>The sharp Sobolev inequality of <a href="http://www.ams.org/mathscinet-getitem?mr=448404" rel="nofollow">Aubin</a> and <a href="http://www.ams.org/mathscinet-getitem?mr=463908" rel="nofollow">Talenti</a> plays a critical role in many important theorems in geometric analysis, including the Yamabe problem. Using the co-area formula, it is easy to reduce the proof to proving the inequality for functions on $R^n$ that are a function of the distance to the origin only. This is a $1$-dimensional inequality. But the proofs of this 1-d inequality given by Aubin and Talenti are extremely hard to follow. At least one of them simply cites a paper by Bliss that uses techniques of calculus of variations that I find rather obscure. For this reason, I believe very few people who have used and cited the Aubin-Talenti inequality have ever understood its full proof.</p> <p>The situation, however, has improved. For those who know the details of the construction of the so-called Brenier map in optimal transportation, there is a full proof of the Aubin-Talenti inequality in a <a href="http://www.ams.org/mathscinet-getitem?mr=2032031" rel="nofollow">beautiful paper by Cordero-Nazaret-Villani</a>.</p> <p>For those who do not want to learn the full details of the Brenier map, my collaborators and I have included our paper titled <a href="http://www.ams.org/mathscinet-getitem?mr=1987375" rel="nofollow">"Sharp Affine $L_p$ Sobolev Inequalities"</a> the full details of the Cordero-Nazaret-Villani proof applied to the 1-dimensional Bliss inequality. In this case, the Brenier map can be constructed using only the fundamental theorem of calculus.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99699#99699 Answer by wildildildlife for Blackbox Theorems wildildildlife 2012-06-15T12:23:27Z 2012-06-15T12:23:27Z <p>Someone mentioned existence and uniqueness of Haar measure on a locally compact topological group. But if one uses the Riesz representation theorem and Tychonoff, the standard proof is not so long or hard, and may even be considered conceptual. For example a clear proof is in Bourbaki's Integration, and in Principles of Harmonic Analysis [by Deitmar and Echterhoff]. </p> <p>I think </p> <p>the <a href="http://en.wikipedia.org/wiki/Riesz_representation_theorem" rel="nofollow">Riesz representation theorem</a> (about the dual of $C_c(X)$)</p> <p>is more often used as a Blackbox theorem. Of course this is a main result in analysis, and many standard books (Rudin, Folland, Appendix of Conway's Functional analysis) have a proof, but they are all long and technical, and in my opinion very difficult to remember. See also Remark 4 in these <a href="http://terrytao.wordpress.com/2009/03/02/245b-notes-12-continuous-functions-on-locally-compact-hausdorff-spaces/" rel="nofollow">wonderful notes</a> by Terry Tao.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99764#99764 Answer by HeWhoHungers for Blackbox Theorems HeWhoHungers 2012-06-16T01:40:19Z 2012-06-16T01:40:19Z <p>In several complex variables it is often desired to be able to <strong>solve the $\bar\partial$ equation</strong>. The standard tool for that is <strong>Hörmander's $L^2$ method</strong> and though I suppose that most who use it have at some point read at least a sketch of the proof, most probably aren't familiar with the tedious details that go into the proof. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99805#99805 Answer by Salvatore Siciliano for Blackbox Theorems Salvatore Siciliano 2012-06-16T20:41:12Z 2012-06-16T20:41:12Z <p>The Feit–Thompson Theorem stating that every finite group of odd order is solvable.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99816#99816 Answer by Nate Eldredge for Blackbox Theorems Nate Eldredge 2012-06-17T03:41:11Z 2012-06-17T03:41:11Z <p>What about <a href="http://en.wikipedia.org/wiki/Carleson_theorem" rel="nofollow">Carleson's theorem</a> that Fourier series of $L^2$ functions converge almost everywhere? I don't read the right literature to see whether this is frequently invoked, but it seems like a useful tool to have. </p> http://mathoverflow.net/questions/99506/blackbox-theorems/99920#99920 Answer by Scott Aaronson for Blackbox Theorems Scott Aaronson 2012-06-18T17:10:21Z 2012-06-18T17:10:21Z <p>In theoretical computer science, possibly the best example is the <a href="http://en.wikipedia.org/wiki/PCP_theorem" rel="nofollow">PCP Theorem</a>: it's used all over the place, from cryptography to quantum computing, yet very few of us understand the details (especially for the strong, "modern" versions of it).</p> http://mathoverflow.net/questions/99506/blackbox-theorems/100479#100479 Answer by Marty Isaacs for Blackbox Theorems Marty Isaacs 2012-06-23T18:14:43Z 2012-06-23T18:14:43Z <p>Perhaps the existence of "Tarski Monsters" qualifies as a "blackbox" theorem. The theorem is that for sufficiently large prime numbers $p$, there exist infinite groups $G$ such that every proper nonidentity subgroup has finite order $p$. Such a Tarski monster is clearly 2-generated and has finite exponent, so it provides counterexamples to the Burnside problem. Also it is a simple group of prime exponent and it provides counterexamples for many other attempts to generalize properties of finite groups to groups in general.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/100702#100702 Answer by Cheyne for Blackbox Theorems Cheyne 2012-06-26T17:15:42Z 2012-06-26T17:15:42Z <blockquote> <p>Cech and Sheaf (derived-functor) cohomologies are isomorphic on a paracompact space $X$ with the sheaf being, for example, <code>$\underline{\mathbb{C}}^*_M$</code>, the sheaf of <code>$\mathbb{C}^*$</code>-valued functions on $X$.</p> </blockquote> <p>The proof uses partititions of unity along with hypercohomology and results from spectral sequences. If this ISN'T a black box theorem, I'd love a concise explanation :)</p> http://mathoverflow.net/questions/99506/blackbox-theorems/100713#100713 Answer by David Corwin for Blackbox Theorems David Corwin 2012-06-26T19:33:55Z 2012-06-26T19:33:55Z <p>My immediate thought upon seeing the question, and, I believe, one of the biggest examples of this phenomenon, is:</p> <p><strong>Class Field Theory</strong></p> <p>Almost anyone working in algebraic number theory uses the main results of class field theory regularly. However, even if many people have sat through a course going through the proofs of the theorems, very few people remember the proofs, and even fewer use them.</p> <p>I recall hearing advice from various mathematicians that the most important thing is to learn the statements of class field theory, but not the proofs.</p> <p>See, for example, this quote from the <a href="http://math.stanford.edu/~conrad/249BPage/syl.html" rel="nofollow">Syllabus to Brian Conrad's course on class field theory</a>:</p> <blockquote> <p>While it is somewhat instructive to know what goes into the proofs of the main theorems (e.g., to see what obstacles prevent the proofs from being entirely constructive), it cannot be said that the grungy details of these proofs are particularly relevant to using the theory in practice. Thus, in the first half of the course we will emphasize an understanding of the statements of the main results (in their many different forms) and will not place much emphasis on how the main theorems are proven; precise references will be given for those who wish to read the details of the proofs of the main theorems. Once we have spent some time digesting what class field theory tells us, we will study some applications of the theory, such as in the context of imaginary quadratic fields and abelian coverings of algebraic curves.</p> </blockquote> http://mathoverflow.net/questions/99506/blackbox-theorems/100830#100830 Answer by Owen Sizemore for Blackbox Theorems Owen Sizemore 2012-06-28T02:04:50Z 2012-06-28T02:04:50Z <p>The fact that every von Neumann algebra is the direct integral of factors. Every operator algebraist knows this, and could probably more or less explain the proof, but the details are tedious (and kind of useless in practice.) There are similar facts about decomposition of non-singular actions into ergodic ones and representation into irreducible representations.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/100839#100839 Answer by Jeff H for Blackbox Theorems Jeff H 2012-06-28T03:09:21Z 2012-06-28T03:09:21Z <p>Deligne's construction of Galois representations corresponding to modular forms.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/100840#100840 Answer by Karl Schwede for Blackbox Theorems Karl Schwede 2012-06-28T04:00:20Z 2012-06-28T04:00:20Z <p>The Cohen-Structure theorem in commutative algebra (classifying complete local rings in some sense).</p>