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What are the most overpowered theorems in mathematics?

By "overpowered," I mean theorems that allow disproportionately strong conclusions to be drawn from minimal / relatively simple assumptions. I'm looking for the biggest guns a research mathematician can wield.

This is different from "proof nukes" (that is, applying advanced results to solve much simpler problems). It is also not the same as requesting very important theorems, since while those are very beautiful and high level and motivate the creation of entire new disciplines of mathematics, they aren't always commonly used to prove other things (e.g. FLT), and if they are they tend to have more elaborate conditions that are more proportional to the conclusions (e.g. classification of finite simple groups).

Answers should contain the name and statement of the theorem (if it is named), the discipline(s) it comes from, and a brief discussion of why the theorem is so good. I'll start with an example.

The Feit-Thompson Theorem. All finite groups of odd order are solvable.


Solvability is an incredibly strong condition in that it immediately eliminates all the bizarre, chaotic things that can happen in finite nonabelian simple groups. Solvable groups have terminating central and derived series, a composition series made of cyclic groups of prime order, a full set of Hall subgroups, and so on. The fact that we can read off all that structure simply by looking at the parity of a group's order is amazing.

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closed as primarily opinion-based by Najib Idrissi, user26857, user91500, user147263, yoknapatawpha Oct 25 '15 at 14:45

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I heard Feit-Thompson is getting nerfed in the next version. $\endgroup$ – Federico Poloni Nov 7 '13 at 9:13
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    $\begingroup$ disproportionately strong conclusions, minimal / relatively simple assumptions, Fermat's little theorem? $\endgroup$ – Paul S. Nov 7 '13 at 19:03
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    $\begingroup$ @FedericoPoloni: nerfed? next version? can you explain what that means? I conjecture it relates to video game patches. $\endgroup$ – Nikolaj-K Nov 8 '13 at 9:20
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    $\begingroup$ @NickKidman It's a joke, you guessed well. "Overpowered" is (also) common computer game jargon for some powers/features in a game that are much stronger than the others and should be reduced in power ("nerfed") to preserve the balance of the game. Feel free to remove the comment if you think it's not funny or it shouldn't be allowed here because it's gratuitously off-topic. $\endgroup$ – Federico Poloni Nov 8 '13 at 9:32
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    $\begingroup$ @FedericoPoloni: I will sue you in federal court. $\endgroup$ – Nikolaj-K Nov 8 '13 at 9:40

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there are no theorems whose conclusions contain more that their fully-spelled-out premises.

all purported examples of such just use, e.g., definitions that allow a short syntactic statement of the theorem while shoving all the actual work of it under the rug.

well, unless math is inconsistent at any rate - in which case we have bigger problems.

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  • $\begingroup$ How would you define "fully-spelled-out-premises"? I think it's never possible. For example, suppose I write 10+4=x. How do you know that I'm not thinking of a clock where 10 o'clock + 4 o'clock = 2 o'clock? Notation (including the English language) always requires interpretation and is never self-explanatory. This sentence may look like English but perhaps it's actually another language or just random nonsense? $\endgroup$ – Andrius Kulikauskas Sep 4 '18 at 11:08
  • $\begingroup$ My Ph.D. thesis, Symmetric Functions of the Eigenvalues of a Matrix, explored a very curious situation where substitution (of eigenvalues) into a symmetric function actually gave it more meaning combinatorially. For example, consider the product x1*x2*x3 and plug in the eigenvalues of a 3x3 matrix. The determinant of a matrix Aij is the product of its eigenvalues. And that determinant carries information about 3! permutations. The product became richer from substitution. And if we set the off-diagonal elements to zero, then the determinant is just the product of the diagonal elements. $\endgroup$ – Andrius Kulikauskas Sep 4 '18 at 11:15
  • $\begingroup$ I suppose my thought is that premises can't and don't come "fully spelled out" but rather it seems that theorems are what do the spelling out. $\endgroup$ – Andrius Kulikauskas Sep 4 '18 at 11:16
  • $\begingroup$ That just isn't how logic works, alas. The concept of "good inference" has built into it "get out less-or-equal to what you put in". $\endgroup$ – sherifffruitfly Sep 12 '18 at 14:02
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The Frobenius Theorem in differential topology. The workhorse behind this theorem is the so-called flowbox theorem (or "rectification theorem for vector fields"), which is arguably the real hero here.

There are several deep applications. Here are just a few:

  • Riemannian geometry: Riemannian manifolds with zero curvature are locally isometric to $\mathbb{R}^n$.
  • Elementary differential geometry: Cartan's Theorem on maps into Lie groups, which in turn yields (among other things) the fundamental theorem on space curves, as well as the fundamental theorem on surfaces.
  • Lie theory: If $\mathfrak{h} \subset \mathfrak{g}$ is a Lie subalgebra, where $\mathfrak{g} = \text{Lie}(G)$, then there is a Lie subgroup $H \leq G$ with $\mathfrak{h} = \text{Lie}(H)$.
  • Overdetermined systems of first-order PDE.
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The Maximum Principle in PDE theory.

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