# Blackbox Theorems [closed]

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.

• 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
• 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
• @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 0:19
• 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
• 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 3:08

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.

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.

• There are many papers by H. Hauser whose message is "You can understand Hironaka's proof!". There is even a game-theoretic interpretation. Very recommended. – Martin Brandenburg Jun 14 '12 at 8:15
• There are also now books (one by Kollar and another by Cutkosky) that aim to present the proof at a graduate student level. I think they don't prove the most general/detailed statements from Hironaka's original paper, though. – Dan Ramras Jun 15 '12 at 0:34
• Update: the new proofs of resolution by Abramovichâ€“Temkinâ€“WÅ‚odarczyk and McQuillanâ€“Marzo using stacky weighted blowups are again supposed to be much easier to understand. – R. van Dobben de Bruyn May 13 at 23:01

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.

• You can add to that all the étale cohomology machinery. – Felipe Voloch Jun 14 '12 at 0:13
• I had guessed this might be a blackbox, but being outside the area I wasn't sure. – Benjamin Steinberg Jun 14 '12 at 2:13

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.

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.

• 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
• 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 6:11
• @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 7:36

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.

• 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 www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf). – HeWhoHungers Jun 16 '12 at 1:24
• 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
• @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
• 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

Often Serre's GAGA (translation between algebraic and analytic land) is treated as a black box.

• Are you sure? At least in the original version by Serre is highly accessible, easy and enjoyable to read through. – Filippo Alberto Edoardo Jun 27 '12 at 0:17
• @Filippo: Indeed Serre's paper is very nice. – Martin Brandenburg Jun 27 '12 at 6:56

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.

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.

• 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 1:13
• 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 2:11
• Actually, the proof is quite widely available. – Felix Goldberg Jun 15 '12 at 1:28
• 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
• @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

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.

• Always about homological algebra, I think that the whole "derived functors" story is used by many people but proofs are rarely read. – Filippo Alberto Edoardo Jun 14 '12 at 2:27
• That's very true as well. – Karl Schwede Jun 14 '12 at 2:30
• Definitely spectral sequences come to you in black boxes first ... and perhaps they stay black boxes ;). – Martin Brandenburg Jun 14 '12 at 9:57

Embedding theorems for abelian categories (Freyd, Mitchell, Lubkin, ...) seem to qualify.

• Plus the Gabriel-Quillen-Laumon theorem, embedding an exact category into an abelian one. – Matthias Künzer Jun 14 '12 at 15:58

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.)
• This example is not really what I want because all basic algebraic topology books do it. – Benjamin Steinberg Jun 13 '12 at 22:59
• Uhâ€¦ just because books have the proof doesn't mean we take the time to read and understand it. – Zsbán Ambrus Jun 14 '12 at 9:30
• 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)âˆ’f(x))/r is monotone in r. – Ian Morris Jun 14 '12 at 13:12
• 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

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.

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).

• I think the proof of the Geometrization conjecture is a better answer since it is more applicable. – Benjamin Steinberg Jun 13 '12 at 23:24
• Oh, ok. Do you want to declare the two answers merged for purpose of this voting people are doing? (since this is community wiki anyway) – tweetie-bird Jun 14 '12 at 0:39
• Sure. Why don't you just edit your answer. – Benjamin Steinberg Jun 14 '12 at 2:10

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.

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.

• 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 6:51
• 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

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.

• Also, the Serre conjecture a.k.a. Quillen-Suslin theorem. – darij grinberg Jun 14 '12 at 10:18

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 Ã‰tale 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.

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.

• 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

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.

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.

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.

• Is this theorem actually applied frequently inside or outside of set theory? – Jan Weidner Jun 13 '12 at 21:59
• 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
• @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
• @Ralph: you may say the same about any axiom in any theory. – Michal R. Przybylek Jun 13 '12 at 23:41
• 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 0:24

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.

• for two dimensional compact manifolds (the case most non topologists see) proving that the connected sum is independent of choices is easy – Daniel Barter Jun 15 '12 at 2:29
• @Daniel: how is it easier than in the high-dimensional case? Just curious.... – Johannes Ebert Jun 26 '12 at 17:23
• @JohannesEbert By inspection. – Loki Clock Jan 15 '15 at 21:32

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

• 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

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
• 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: mathoverflow.net/questions/70248/… – Daniel Moskovich Jun 15 '12 at 0:24
• 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

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

• A perfect example! – Joël Jun 28 '12 at 12:52

The existence of Brownian Motion.

• 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 9:49
• @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
• 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

The Feitâ€“Thompson Theorem stating that every finite group of odd order is solvable.

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.

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 :)