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By a blackbox theorem 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.

I am curious to compile a list of such blackbox theorems with the usual CW rules of one example per answer.

Obviously this is not connected to my research directly so I can understand if this gets closed.

Benjamin Steinberg
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    I have heard the resolution of singularities being mentioned as an example of this (although there seem to be simpler proofs around now). – darij grinberg Jun 13 '12 at 21:28
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    The simpler proofs are still at least 20 pages of fairly technical mathematics though. – Karl Schwede Jun 13 '12 at 21:38
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    Domain of use is important here. Many theorems are invoked by physicists who have no idea of the actual proofs. – Steve Huntsman Jun 13 '12 at 22:05
  • Quite related: http://mathoverflow.net/questions/32409/examples-of-folk-theorems – Martin Brandenburg Jun 14 '12 at 09:58
  • While we're at algebra, would the theorem that all finite skew fields are fields count? Or is that a curiosity that has no actual applications? – Zsbán Ambrus Jun 14 '12 at 22:02
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    @Zsbán: That theorem has nice consequences, e.g., in finite geometry. But treating it as a blackbox is just laziness, since the proof is just a couple of pages of basic graduate algebra. – Felipe Voloch Jun 15 '12 at 00:19
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    It's interesting to compare this question with other MO questions about personally verifying results that you cite: http://mathoverflow.net/questions/23758/published-results-when-to-take-them-for-granted and http://mathoverflow.net/questions/98821/how-often-do-people-read-the-work-that-they-cite for example. – Timothy Chow Jun 15 '12 at 15:39
  • If would set quantative measure: " widely applied means one knows more than N papers which use theorem" within N = 25 or 50 or whatever... would it change answers? It seems alll hard theorems are in the list:) are they really so widely applied? PS any way great question PS PS is classification of simple groups applied some where? – Alexander Chervov Jun 16 '12 at 12:57
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    The classification is used by people working on permutation groups and graph theory all the time. Also they are used in profinite group theory. – Benjamin Steinberg Jun 16 '12 at 19:21
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    In my mind I was hoping for things used in at least 100 papers and understood in all technical detail by fewer than 5% of people in the general area to which the theorem belongs. But it need not be this rigid. – Benjamin Steinberg Jun 17 '12 at 03:08
  • 100 papers ... so many ... are there really so many theorems which are black box and used so wide ? If you would require to provide a paper where the theorem has been proved, then by number of citations from e.g. scholar.google one can make answer more veriafiable more quantative :) :) – Alexander Chervov Jun 17 '12 at 16:58
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    I think older well known theorems are often not cited from the source and also are not well counted by citation counts. I am pretty sure the classification of simple groups is not always cited in a way that checking bibliographical references will catch. – Benjamin Steinberg Jun 17 '12 at 22:46

59 Answers59

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The graph minor theorem and the graph structure theorem 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.

Gjergji Zaimi
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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.

Charles Staats
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Jan Weidner
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Deligne's Theorem, found at Wikipedia under the heading of Weil conjectures, 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.

Gerry Myerson
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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.

Kestutis Cesnavicius
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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 S3 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 (e.g. Reshetikhin-Turaev invariants).

All known proofs of Kirby's Theorem are a nightmare (see this MO question). 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.

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.

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 Lurie's 2009 notes.

Community
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Daniel Moskovich
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    Freedman's theorem "Casson handles are handles" is also used as a black box by many people. Once this is known, standard arguments from higher dimensions can be pushed down to 4 dimensions to prove h-cobordism and Poincare. Hopefully this will be rectified next summer when an extended workshop will go over the proof (I think at Bonn). – Ian Agol Jun 14 '12 at 17:20
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    I confess that I have no looked at the proof of the Kirby calculus theorem either recently. But I personally think that the difficulty of the Thom transversality theorem and Cerf theory are overplayed. Is the Reidemeister move theorem for smooth knots a difficult theorem? It's the same sort of thing. Yes, there are a lot of details if you want to be very rigorous, but the lemmas all have natural statements. For instance, you can prove Thom transversality in the setting of a finite-dimensional vector space of polynomial functions, using algebraic geometry. – Greg Kuperberg Jun 19 '12 at 06:11
  • @Greg Kuperberg: I'm uneasy about my Frechet manifold intuition, so e.g. playing with Thom-Boardman stratification or multijet transversality looks hard to me right now (presentation of the MCG is also difficult). Maybe one day I'll feel more comfortable with it; maybe all topologists should (is there a nice intro text?). Anyway, these aren't standard things which "consumers" of Kirby's Theorem know much about. Reidemeister's Theorem's proof is at least combinatorial (a smooth proof uses weaker tools). What's the algebraic geometry proof of Thom transversality? I'm really curious about that! – Daniel Moskovich Jun 19 '12 at 07:36
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I think the Uniformization theorem is an example of blackbox theorem : any simply connected Riemann surface is conformally equivalent to either the open unit disk, the complex plane or the Riemann sphere.

Malik Younsi
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    The standard proof of the Uniformization theorem with the Green's function, while rather involved, shouldn't really surpass the ability of most who come across it. There also exists a short and elegant proof that uses certain rather more advanced tools: the Mayer-Vietoris sequence and the celebrated Newlander-Nirenberg theorem. But NN for surfaces is just the existence of isothermal coordinates, which is much simpler to prove. This proof can be found in Demailly's "Complex Analytic and Differential Geometry" (available at http://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf). – HeWhoHungers Jun 16 '12 at 01:24
  • The only proof I've seen is the one with the Green's function. Thank you for the reference! – Malik Younsi Jun 18 '12 at 16:25
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    @HeWhoHungers: I agree that the proof in Demailly's book is marvellous and elegant, but it is neither easy or short in any sense. I talked about exactly this proof in a lecture course on Teichmüller theory some years ago, addressing an audience of very bright graduate students. I needed 3 or 4 hours to communicate the proof and I remember it to be a tour de force, both for me and the audience. Even if you take the advanced tools for granted (as I did), the details (many of which are thrown under the carpet in the book) are very, very subtle. – Johannes Ebert Jun 26 '12 at 17:40
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    The ration #{people who quote the theorem on a daily basis} / #{people who know the details of the proof offhand} is very high, so it is a perfect example of a blackbox theorem. – Johannes Ebert Jun 26 '12 at 17:47
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Often Serre's GAGA (translation between algebraic and analytic land) is treated as a black box.

Martin Brandenburg
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Faltings' Theorem, 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.

Gerry Myerson
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Existence and uniqueness of invariant Haar measure on a locally compact topological group.

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.

Abhishek Parab
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    This is pretty easy to avoid in practice. eg, Haar measure on a manifold is easily constructed using invariant differential forms. Similarly, differential forms lift measure from $\mathbb Q_p$ to $p$-adic groups. Adeles are a little trickier (eg, naive choices on $G_m(A)$ yield the zero measure). – Ben Wieland Jun 14 '12 at 01:13
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    I think that probably most people in harmonic analysis more or less know how it works (as compared to the classification of finite simple groups). – Benjamin Steinberg Jun 14 '12 at 02:11
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    Actually, the proof is quite widely available. – Felix Goldberg Jun 15 '12 at 01:28
  • Again, in mild dissent from the previous comments, I have seen a fair number of papers where $L^1(G)$ is name-checked by people who are not au fait with the proofs of existence and uniqueness for Haar, apart from some hasty revision for qualifying exams or similar. – Yemon Choi Jun 15 '12 at 04:08
  • That said, I agree that this example doesn't really fit Benjamin Steinberg's original requirement. – Yemon Choi Jun 15 '12 at 05:46
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    What is not terribly well known (or exposited in very many books) is the constructive proof of existence and uniqueness of Haar measure that does not use the axiom of choice. While I imagine the vast majority of people who make use of Haar measure either don't care about the axiom of choice or have nicer constructions as Ben Wieland suggests, it is at very least an interesting curiosity that the axiom of choice is not needed at all, since the usual proof one sees relies so crucially on it. – Evan Jenkins Jun 15 '12 at 17:44
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    @EvanJenkins: do you have a reference for the non-AC proof? When studying Haar measure construction I found a lot of texts doing only the compact case, and one text with an AC proof of the locally compact case. – Emilio Pisanty Jun 28 '12 at 11:56
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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.

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.

Karl Schwede
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Jordan's curve theorem is used as a blackbox.

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.

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.

  • the existance and basic properties of the Lebesgue measure and infinite product measures
  • the fact that a Wiener process is almost surely everywhere continuous (mentioned below as a separate answer by weakstar)
  • 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)
  • the spectral theorem about linear maps on a possibly infinite-dimensional Hilbert-space
  • the proof that a convex function (from reals to reals) is always continuous everywhere and has a left and right derivative everywhere (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.)
  • 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.)
  • 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.)
Zsbán Ambrus
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    This example is not really what I want because all basic algebraic topology books do it. – Benjamin Steinberg Jun 13 '12 at 22:59
  • Uh… just because books have the proof doesn't mean we take the time to read and understand it. – Zsbán Ambrus Jun 14 '12 at 09:30
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    Most of those are genuinely laborious proofs, but the one about convex functions can be done in a few lines. A convex function clearly has at most two intervals of monotonicity, which implies that the left and right limits at each point exist. If they aren't the same for some point then we can find a chord between two points of the graph close to the discontinuity which passes below the graph (on the left if the jump is downwards, or to the right if it is upwards) contradicting convexity. Differentiability is obtained by showing that (f(x+r)−f(x))/r is monotone in r. – Ian Morris Jun 14 '12 at 13:12
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    Perhaps you could replace your fifth example with Rademacher's theorem: every Lipschitz function from an open subset of $\mathbb{R}^n$ to $\mathbb{R}^m$ is differentiable almost everywhere. This is a more serious result which people use all the time, and I'm not sure everyone really knows the proof (though maybe I should speak for myself). – Paul Siegel Jun 14 '12 at 23:39
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Embedding theorems for abelian categories (Freyd, Mitchell, Lubkin, ...) seem to qualify.

Fred Rohrer
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How is the proof of the Poincare' Conjecture (in all dimensions) not yet anywhere on this list?

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).

tweetie-bird
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Nagata embedding is another black box - its statement is very simple and useful, but its proof is hard.

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.

S. Carnahan
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My immediate thought upon seeing the question, and, I believe, one of the biggest examples of this phenomenon, is:

Class Field Theory

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.

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.

See, for example, this quote from the Syllabus to Brian Conrad's course on class field theory:

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.

David Corwin
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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.

Dori Bejleri
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  • Isn't Grothendieck's proof in FGA quite well-known? But I have to admit that some details are still unclear to me, and yes, the existence is often used as a black box. – Martin Brandenburg Jun 15 '12 at 06:51
  • I'm sure it is well known, but at least every reference I've seen on Hilbert schemes for students basically says don't bother learning the proof unless you're interested and just take existence as a black box. – Dori Bejleri Jun 17 '12 at 23:56
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When learning algebraic geometry and in particular the notion of smooth varieties, you will probably stumble upon the following Theorems:

  • Regular local rings are factorial.
  • Localizations of regular local rings are regular, too.
  • A local ring is regular iff its residue field has finite projective dimension (Serre).

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.

Martin Brandenburg
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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.

Ricky
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I think the main statements of the MMP (Minimal Model Program ) in algebraic geometry qualify for this.

It will even become more of a black box in the future, as people will understand better how to apply it.

Karl Schwede
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    Can you be a little more specific? do you refer to minimal models in algebraic topology? – Gil Kalai Jun 14 '12 at 11:40
  • No, sorry. I mean the minimal model program in algebraic geometry. For example, finite generation of the canonical ring, forming relative minimal models, etc. – Karl Schwede Jun 14 '12 at 11:58
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The well-definedness of the connected sum of two manifolds. After all, we choose two arbitrary balls for gluing; why should the result be independent? The proof depends on the nontrivial Disc theorem.

I think that one usually uses this result as a black box.

Martin Brandenburg
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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.

Jan Weidner
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Perhaps Atiyah-Singer's index theorem can qualify. Another candidate is Gromov's h-principle.

John D. Cook
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Gil Kalai
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    The situation with the h-principle is certainly complicated by the fact that there are multiple h-principles with different proofs. – Igor Khavkine Jun 18 '12 at 20:47
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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].

I think

the Riesz representation theorem (about the dual of $C_c(X)$)

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 wonderful notes by Terry Tao.

wildildildlife
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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.

Ralph
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    Is this theorem actually applied frequently inside or outside of set theory? – Jan Weidner Jun 13 '12 at 21:59
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    Would you use AC if it contradicts ZF ? The magnitude of the independence theorem is that we use it implicitely whenever we apply AC, since it tells us that AC doesn't lead to logical contradictions that weren't already present in ZF. – Ralph Jun 13 '12 at 22:20
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    @Ralph, I disagree. First, the independence of AC from ZF is not the same as the consistency of ZFC relative to ZF (indeed, the latter is very easy). Second, I'm not certain that even this is used frequently outside of set theory; when non-logicians use the axiom of choice, they are not tacitly assuming that it is consistent with ZF, they are tacitly assuming that it is part of some consistent set theory - for example, how many non-logicians know the ZF axioms off the top of their head? I think in practice the set theory that is actually used is generally some high-but-finite-order arithmetic. – Noah Schweber Jun 13 '12 at 22:56
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    @Ralph: you may say the same about any axiom in any theory. – Michal R. Przybylek Jun 13 '12 at 23:41
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    In fact, building off of Michal, perhaps the consistency of the axioms of powerset, replacement, and separation would be better, since these are implicitly used whenever comprehension (forming the set of all $x$ such that $P(x)$) is used, and full comprehension actually is contradictory! But I still don't feel that these are good examples. Roughly speaking, either you're Platonist - in which case mere consistency of AC isn't sufficient to justify using it - or one is interested in proving theorems from axioms, in which case "ZFC proves X" is valuable even if ZFC isn't known to be consistent. – Noah Schweber Jun 14 '12 at 00:24
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    (cont'd) So I don't think that the relative consistency of ZFC is actually being invoked whenever Choice is used. – Noah Schweber Jun 14 '12 at 00:25
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    This is often invoked informally in my field in the following manner: this proof uses ultrafilters, but it doesn't really use "choice" because any theorem of arithmetic/combinatorics that has a proof with choice has a proof without choice. – Kevin O'Bryant Jun 30 '12 at 16:29
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I think differential topology has dozens of these results. Here are some examples that immediately come to mind:

  • 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 $N$
  • 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 $f \leq b$
  • Every continuous isomorphism of smooth vector bundles is homotopic to a smooth isomorphism (and other such "continuous equivalence = smooth equivalence" results)

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.

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:

  • De Rham's theorem: the De Rham cohomology groups of a manifold are isomorphic to the singular cohomology groups with real coefficients
  • The Hodge theorem: every De Rham cohomology class on a Riemannian manifold has a harmonic representative
  • Whitehead's result that every smooth manifold has a unique PL structure
Paul Siegel
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    Definitely, definitely true. Even more so for "the second part" of "the fundamental theorem of Morse Theory", that passing a critical point attaches a handle: http://mathoverflow.net/questions/70248/searching-for-an-unabridged-proof-of-the-basic-theorem-of-morse-theory – Daniel Moskovich Jun 15 '12 at 00:24
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    This doesn't quite seem to be in the spirit of what the original questioner was going for. This is more like, "Mathematical facts that people think that they understand but are actually a bit trickier than they think," which is interesting, but perhaps deserves its own thread. – Dan Lee Jun 15 '12 at 20:36
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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.

Marty Isaacs
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Deligne's construction of Galois representations corresponding to modular forms.

Jeff H
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The Feit–Thompson Theorem stating that every finite group of odd order is solvable.

Salvatore Siciliano
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Cech and Sheaf (derived-functor) cohomologies are isomorphic on a paracompact space $X$ with the sheaf being, for example, $\underline{\mathbb{C}}^*_M$, the sheaf of $\mathbb{C}^*$-valued functions on $X$.

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 :)

cheyne
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The existence of Brownian Motion.

weakstar
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  • I think it's not so much the existence that's hard to prove but that it being continuous everywhere has probability 1. – Zsbán Ambrus Jun 14 '12 at 09:49
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    @Zsbán: Continuity of BM is part of the standard definition, so proving that is the same as proving that it exists. However, the proof that BM is almost-surely nowhere differentiable is probably less well known. – George Lowther Jun 14 '12 at 22:03
  • Okay, you're right. I accept that you have to prove existence and continuity together. – Zsbán Ambrus Jun 15 '12 at 16:12
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    The proof of continuity usually follows from the "Kolmogorov Criterion": If there exists strictly positive constants $\varepsilon$, $p$ and $C$ such that $$\mathbb{E}|X_t - X_s|^p \leq C|t-s|^{1+\varepsilon}$$ then almost surely $X$ has a modification which has $\alpha$-Hölder continuous paths for any $\alpha \in (0,\frac{\varepsilon}{p})$ – Felipe Olmos Jun 15 '12 at 23:54
  • The proof of continuity usually follows from the "Kolmogorov Criterion"

    ... which itself is a little technical, but nevertheless fairly intuitive chaining argument.

     – Alexander Shamov Jun 16 '12 at 21:44
  • Every book I've ever seen that proves the existence of BM goes on to prove it is nowhere differentiable. – Nate Eldredge Jun 17 '12 at 03:32
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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.

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.

Martin Brandenburg
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  • I admit that I take it on faith. – Johannes Ebert Jun 14 '12 at 09:51
  • Right, but besides Zorn's lemma, some tricky applications of it also count, such as proving that any vector space (over any field) has a basis, or that any Banach space (real or complex) has an orthonormal basis. – Zsbán Ambrus Jun 14 '12 at 09:53
  • @Zsban: What do you mean by "tricky"? Existence of (orthonormal) bases is a trivial consequence of Zorn's lemma and I guess that everyone knows how that works. – Martin Brandenburg Jun 14 '12 at 10:10
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    Okay, but in ZF+Zorn's lemma it's a tautology! I don't think in practice most people use specifically the fact that AC implies Zorn's lemma but just the fact that it's generally considered okay to prove results that depend on Zorn's lemma. – Qiaochu Yuan Jun 14 '12 at 10:30
  • @Qiaochu: I agree with that. – Martin Brandenburg Jun 14 '12 at 11:20
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For a long time, the Littlewood-Richardson rule has been a black box. See van Leeuwen's wonderful article 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!) proof (2002) and Gasharov's 3-pages long proof (1998). (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.)

darij grinberg
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In theoretical computer science, possibly the best example is the PCP Theorem: 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).

Scott Aaronson
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  • Is Dinur's proof not really well-understood by practitioners in this area? – Ian Agol Jun 18 '12 at 23:39
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    Her proof is certainly well-understood at a high level, and it's also certainly better-understood than the algebraic proof. But by the standard of how many people could fully reconstruct her proof within (say) two months, if locked in a room by a mad scientist with only pencils and blank paper ... well, the requisite experiment hasn't been done, and I shouldn't speak for everyone else, but I wouldn't want to be put to the test. :-) – Scott Aaronson Jun 19 '12 at 06:15
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Recognizing hamiltonian graphs is NP-complete.

(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.

Zsbán Ambrus
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    Probably the unsolvability of Hilbert's 10th goes here as well. – Benjamin Steinberg Jun 14 '12 at 00:30
  • It's not very hard to get from colourability to 3-SAT - I don't claim that there is an obvious gadget construction, but if you play around for a bit you can find one of the (many) options easily enough. And the proof that 3-SAT is NP-complete is quite nice and conceptual. – user36212 Mar 28 '17 at 13:31
7

C*-algebra theory has a number of good examples of this.

  • Voiculescu's theorem: an ample representation of a C*-algebra essentially absorbs any nondegenerate representation
  • 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 ALL THE TIME, e.g. for excision or the existence of Kasparov products.
  • Stinespring's theorem: any completely positive map into $B(H)$ dilates to a representation

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.

Paul Siegel
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    Nice answers. An addendum/afterthought: it probably isn't used very often by practising operator algebraists, but the equivalence of nuclearity and amenability for C*-algebras gets invoked a lot of the time by people in Banach algebras, and I rather doubt many of them have actually worked through all the details. – Yemon Choi Jun 15 '12 at 00:26
6

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 really needed, i.e. Solovay's model (or even the Feferman-Levy in which every set is Borel).

Asaf Karagila
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  • But how much is the necessity of AC actually used? Within set theory it is used, but I would imagine set theorists could describe Solovay's model. Is the necessity of AC used outside of set theory? – Noah Schweber Jun 13 '12 at 23:17
  • By using the consistency of these exotic models, you can conclude that all sets are measurable, which sounds convenient. I'm not sure I've ever seen anyone do this, but Thurston does something more extreme in his notes: he uses the consistency of constructive math, in which all functions are continuous, to conclude that a particular function is continuous. Not only do I not know this black box, I don't know how to check that the rest of the argument is actually constructive. I'd be (less) nervous about checking that an argument doesn't use the axiom of choice. – Ben Wieland Jun 14 '12 at 01:47
  • Noah S, I cannot say about actual research but I can testify about teaching (from the studential side) that I saw this argument (and similar AC uses) in several courses. While not "active research" I do think that teaching is part of the work of most mathematicians. – Asaf Karagila Jun 14 '12 at 05:40
  • Fair, but is the specific result (that AC is needed to construct a Vitali set) important, or is it the general fact that constructing "pathological" objects often cannot be done within ZF that the teacher actually cares about? (And I agree with you wholeheartedly re: teaching as part of mathematics.) – Noah Schweber Jun 14 '12 at 09:43
  • Noah, the importance of non-measurable sets is - in my view - the same importance of Russell's paradox. These things tell us that we cannot always talk about "everything" and we must limit ourselves to smaller and better-behaved objects. – Asaf Karagila Jun 14 '12 at 10:21
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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.

Mikhail Bondarko
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6

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 linear parabolic PDE.

Deane Yang
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  • I understand where you're coming from, because I doubt anyone has published the details of this argument for, say, Ricci flow or mean curvature flow. BUT I expect that a lot of people who work in the field really do understand the argument, which is to reduce nonlinear existence to linear existence using approximation. Terry Tao actually sketched this in his lecture notes http://terrytao.wordpress.com/2008/03/28/285g-lecture-1-ricci-flow/ See Remark 2. As for linear parabolic existence, I'm not sure what you mean. What about Evans, for example? – Dan Lee Jun 15 '12 at 20:30
  • Dan, the reduction of nonlinear to linear is a reasonably straightforward argument. And maybe solving a linear parabolic system on a complete Riemannian manifold is, too. I haven't tried that hard but it seems like the infinite speed of propagation makes it trickier to reduce the proof to an open set in $R^n$. Or is that not so? – Deane Yang Jun 15 '12 at 21:51
  • I didn't scan the word "complete" properly and consequently misunderstood you, so I apologize for that. Frequently, one can solve a parabolic equation on a complete Riemannian manifold by taking the limit of solutions on a compact exhaustion with Dirichlet boundary conditions. For example, this works for the linear heat equation and also for Ricci-deTurck flow (with bounded curvature). Also, in the linear self-adjoint case (with time-independent coefficients), I think that general nonsense let's you construct a heat kernel via exponentiation of operators. – Dan Lee Jun 18 '12 at 18:48
  • Btw, the original purpose of Shi's Ricci flow estimates was to prove that this limit makes sense for Ricci-deTurck flow (with bounded curvature). I agree that Shi's short-time existence result is a black box that is not so widely known, but I should point out that many who work in Ricci flow don't need to quote this result, e.g. I don't believe it is needed to prove geometrization, uniformization (compact case), the differential sphere theorem, or most of Hamilton's big results. – Dan Lee Jun 18 '12 at 18:49
5

Determinacy of Borel Games seems like a good example of this.

Denis
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  • Is this really a good example? The proof, especially the inductive version, is not particularly difficult, and my understanding is that descriptive set theorists - i.e., the ones who would use it - do tend to know it. – Noah Schweber Jun 13 '12 at 22:01
  • In fact I was more thinking of computer scientists and game theorists, who often use this as a hammer to have determinacy, although I think few know the proof. But you are right that this would probably not qualify as a "blackbox theorem" for people of set theory. – Denis Jun 14 '12 at 12:43
  • I didn't know cs people used Borel determinacy. Are there any easy-to-understand examples of this? – Noah Schweber Jun 14 '12 at 23:16
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    For instance in automata theory, models of alternating automata on infinte structures (like words or trees) are often used. The accepting condition is typically Borel (see for instance http://en.wikipedia.org/wiki/Parity_game), so thanks to Martin's Theorem, we can define in a sound way the language recognized by such an automaton as the sets of inputs where Player I wins, and the complement is the set of inputs where Player II wins. – Denis Jun 16 '12 at 13:40
4

The Cohen-Structure theorem in commutative algebra (classifying complete local rings in some sense).

Karl Schwede
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4

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.

Santiago
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  • Is Brouwer's fixed point theorem not taught in nearly every introductory algebraic topology course as a rather simple application of homology? – HeWhoHungers Jun 19 '12 at 23:05
4

I've got to put in 2c for ergodic theory: the Multiplicative Ergodic Theorem is widely quoted, but locating a complete proof is hard.

Anthony Quas
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4

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.

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.

Michael Greinecker
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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.

David Harris
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4

Chevalley's theorem: any algebraic group is the extension of a linear algebraic group by an abelian variety.

Daniel McLaury
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4

The sharp Sobolev inequality of Aubin and Talenti 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.

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 beautiful paper by Cordero-Nazaret-Villani.

For those who do not want to learn the full details of the Brenier map, my collaborators and I have included our paper titled "Sharp Affine $L_p$ Sobolev Inequalities" 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.

Willie Wong
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Deane Yang
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4

What about Carleson's theorem 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.

Nate Eldredge
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    I have a feeling that it isn't used that much per se, but rather is there to set the limits/scope of various theorems, and to motivate generalizations. However, I'm not really in the right circles to make an adequate judgment – Yemon Choi Jun 17 '12 at 09:10
3

Saharon Shelah has a series of results he actually calls "black boxes," and uses accordingly (see his paper, "Black Boxes," http://arxiv.org/abs/0812.0656); my understanding is that these are Diamond-like theorems that are provable in ZFC.

(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.)

Noah Schweber
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  • I'm not sure whether Shelah's result about undecidability of MSO over reals is concerned by your post, but it's another good example. – Denis Jun 14 '12 at 12:45
  • Another result of Shelah which may fit in this list is the Main Gap Theorem. – Haim Jun 14 '12 at 15:44
  • Also the whole family of preservation theorems for countable support iterations of proper forcings can be used without understanding their proofs, and I suspect several people doing forcing (myself included) do use them extensively before feeling the need to actually go and check/understand the proofs. – David Fernandez-Breton May 16 '16 at 12:05
3

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.

Benjamin Steinberg
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    This is a good example (I learned Gromov's proof as a student, but not Montgomery-Zippin/Gleason), but I'm not sure how many applications it's had. Recently, though, Green and Tao have had to generalize Montgomery-Zippin for applications, but they've had to delve into the details of the proof, so maybe the situation will be rectified. – Ian Agol Jun 14 '12 at 17:27
  • There are some theorems in "abstract" non-commutative harmonic analysis, where people use approximation of connected groups by Lie groups, in order to then make use of structure theory. I think one example might be the result of Dixmier-as-written-up-by-Connes that the von Neumann algebra of a connected group is injective. Whether or not this counts as an "application" is I guess debatable – Yemon Choi Jun 15 '12 at 08:34
3

The proof for Hilbert's tenth problem, that is, that there is no algorithm to solve general Diophantiane equations.

Benjamin Steinberg has mentioned this above in a comment. I believe the proof is complicated.

Zsbán Ambrus
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    Does this really qualify? Benjamin writes: "I mean a theorem that is often applied". How can you apply a Theorem which says that you cannot solve some general equation? In specific situations, you might be able to do so (perhaps even via algebraic stacks http://mathoverflow.net/questions/96957). – Martin Brandenburg Jun 14 '12 at 10:13
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    @Martin: The MRDP theorem says much more, namely that every r.e. set is Diophantine, and it (and its formalized versions in fragments of arithmetic) does in fact have useful applications in logic. – Emil Jeřábek Jun 14 '12 at 11:01
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    In a similar vein, the word problem for groups? – Jonny Evans Jun 14 '12 at 11:48
  • @Emil: Thank you, I didn't know that. I was only refering to the content of the answer. – Martin Brandenburg Jun 14 '12 at 14:18
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    It is quite common in group theory to reduce algorithmic problems involving nilpotent groups to Hilberts 10th – Benjamin Steinberg Jun 14 '12 at 23:49
  • I think undecidability of the word problem is not a blackbox in group theory anymore. Most people know roughly how some proof works. Cohen's proof with 2-counter machines is fairly easy going. Also Rotman gives a nice presentation in his book. – Benjamin Steinberg Jun 14 '12 at 23:51
  • Undecidability in group theory gets applied elsewhere though. For instance, Gromov shows that there exist infinitely many contractible geodesics on a manifold with unsolvable word problem. There are also applications in the work of Nabutovsky-Weinberger (see Weinberger's "Computers, rigidity, moduli" book or their paper on "The fractal nature of Riem/Diff"). I imagine to the authors the undecidability is not a black box, but many of the readers (at least I) may not remember much of the proof, and don't need to for the applications. – Jonny Evans Jun 15 '12 at 07:09
3

Fundamental lemma (Langlands program) which Ngô Bảo Châu proved and got the Fields medal in 2010.

C.S.
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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.

Yemon Choi
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In several complex variables it is often desired to be able to solve the $\bar\partial$ equation. The standard tool for that is Hörmander's $L^2$ method 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.

HeWhoHungers
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2

Two results I have seen used without proof in undergraduate lectures were:

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.

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.

Zsbán Ambrus
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darij grinberg
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1

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.

Owen Sizemore
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1

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.

Denis Serre
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    I think you are overestimating the difficulty of its proof. While it is not trivial, it is something standard and covered without any problem in a functional analysis course. If the proof is overlooked, that's probably usually because of, well, oversight, not because it is too complicated to be understood by the ordinary analyst. – Michal Kotowski Jun 14 '12 at 16:49
  • Certainly this was a standard part of the graduate analysis sequence at Stanford. – Dan Ramras Jun 14 '12 at 23:47
  • Slightly contrary to the previous comments: although this does not quite fit the OP's original description as something whose proof is understood by relatively few who use it, it (or the closed graph theorem) get used all the time by algebraic functional analysts with no thought whatsoever for the two-step process used in its proof, or for its generalizations to other TVS. In that sense it is treated as a black box. – Yemon Choi Jun 15 '12 at 04:04
1

Many people apply the theorem $1+1=2$, but how many understand in detail the proof given by Russell and Whitehead in Principia Mathematica? Well, I suppose there are other proofs available....

Gerry Myerson
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1

How about Haynes Miller's theorem resolving the Sullivan conjecture?

Jeff Strom
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0

Does FLT count?

Dirk
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