230
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There are certain things in mathematics that have caused me a pleasant surprise -- when some part of mathematics is brought to bear in a fundamental way on another, where the connection between the two is unexpected. The first example that comes to my mind is the proof by Furstenberg and Katznelson of Szemeredi's theorem on the existence of arbitrarily long arithmetic progressions in a set of integers which has positive upper Banach density, but using ergodic theory. Of course in the years since then, this idea has now become enshrined and may no longer be viewed as surprising, but it certainly was when it was first devised.

Another unexpected connection was when Kolmogorov used Shannon's notion of probabilistic entropy as an important invariant in dynamical systems.

So, what other surprising connections are there out there?

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  • $\begingroup$ It should be mentioned that the connection you refer to is due to Furstenberg (ams.org/mathscinet-getitem?mr=498471). Later Furstenberg and Katznelson together used this connection to derive other combinatorial results, including a multidimensional extension of Szemeredi's theorem and a density version of the Hales-Jewett's theorem. $\endgroup$ Commented Aug 23, 2015 at 17:46
  • $\begingroup$ This question is off topic. Please consult the "don't ask" part of the help page, which instructs us all to "avoid asking subjective questions where every answer is equally valid, like 'What’s your favorite ______?'" mathoverflow.net/help/dont-ask $\endgroup$ Commented Jan 15 at 2:54

89 Answers 89

190
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As well known as the connection is, I am constantly amazed by the power of analytical geometry (developed by Descartes and Fermat) to make connections between geometrical ideas and algebraic ideas. It seems remarkable to me that so much geometrical information (as for example in the case of the conic sections) can be represented so succinctly (via quadratic equations in two variables). The geometry suggests things to think about in algebra and the algebra suggests things to think about in geometry. It is just amazing!!

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    $\begingroup$ This is the observation that should have occurred to everyone first! (It didn't to me either.) It is so familiar we forget how amazing it is. $\endgroup$ Commented Feb 8, 2010 at 3:02
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    $\begingroup$ I totally agree. To put another exclamation point on this idea, Algebra and Geometry co-existed for around a thousand years before this observation was made mainstream by Descartes and Fermat. I wonder what other yet-unseen mathematics we'll weave, in a thousand years, into middle school education. $\endgroup$ Commented Sep 29, 2013 at 22:34
163
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Monstrous Moonshine.

I mean why should the Fourier series of the $j$-invariant have coefficients related to the dimensions of the representations of the largest sporadic simple group? And why should the proof of this fact drag in mathematics from String Theory?

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111
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I'll recycle one I mentioned in a thread last week, connecting an elementary problem about polynomials to the classification of finite simple groups:

Definition: A polynomial $f(x) \in \mathbb{C}[x]$ is indecomposable if whenever $f(x) = g(h(x))$ for polynomials $g$, $h$, one of $g$ or $h$ is linear.

Theorem. Let $f, g$, be nonconstant indecomposable polynomials over $\mathbb C$. Suppose that $f(x)-g(y)$ factors in $\mathbb{C}[x,y]$. Then either $g(x) = f(ax+b)$ for some $a,b \in \mathbb{C}$, or $$\operatorname{deg} f = \operatorname{deg} g = 7, 11, 13, 15, 21, \text{ or } 31,$$ and each of these possibilities does occur.

The proof uses the classification of the finite simple groups [!!!] and is due to Fried ["Exposition on an arithmetic-group theoretic connection via Riemann's existence theorem", 1980, in the proceedings of the 1979 Santa Cruz conference on finite groups], following a the reduction of the problem to a group/Galois-theoretic statement by Cassels [1970]. [W. Feit, "Some consequences of the classification of finite simple groups," 1980.]

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My favorite surprise, which is perhaps the record-holder for the longest time it took for the two ideas to be brought together, is the connection between regular n-gons and Fermat primes. The Greeks knew how to construct regular n-gons by ruler and compass for n=3,4,5,6. Fermat introduced numbers of the form $2^{2^m}+1$ around 1640 in the mistaken belief they were prime for all m. Then in 1796 Gauss discovered how to construct the regular 17-gon, and a few years later showed that the n in a constructible n-gon is the product of some power of 2 by distinct Fermat primes.

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99
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From an essay of Arnol'd:

Jacobi noted, as mathematics' most fascinating property, that in it one and the same function controls both the presentations of a whole number as a sum of four squares and the real movement of a pendulum.

(Source: On teaching mathematics, V. Arnol’d, 1997, trans. A. Goryunov.)

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    $\begingroup$ See also: "My Lunch with Arnol'd" which helps put this surprise in perspective by showing it through the eyes of an amateur mathematician... gomboc.eu/99.pdf $\endgroup$ Commented May 19, 2011 at 20:08
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    $\begingroup$ Also related :), there's a very nice article (in french) by Alain Chenciner where he starts discussing the pendulum and ends up connecting to Riemann surfaces. Even if you know very little french the figures are enjoyable and suggestive. As a bonus, there's a short rant, on the state of physics in the math curricula in France which probably has some universal qualities. $\endgroup$ Commented Jul 5, 2017 at 17:12
  • $\begingroup$ @PabloLessa sadly that link is broken! is there another place where one might be able to find them? $\endgroup$
    – Samantha Y
    Commented Nov 27, 2018 at 22:11
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    $\begingroup$ @SamanthaY perso.imcce.fr/alain-chenciner/Pendule_a_Gazette.2001.pdf $\endgroup$ Commented Dec 27, 2018 at 19:18
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    $\begingroup$ @Bumblebee Here's a very bad one: cmat.edu.uy/~lessa/terrible-translations/chenciner.txt $\endgroup$ Commented Sep 1, 2020 at 22:17
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Quillen's result in Elementary proofs of some results of cobordism theory using Steenrod operations that the ring of cobordism classes of (stably) complex manifolds is isomorphic to Lazard's ring (i.e. the universal ring classifying formal group laws). This seems so mysterious to me. Why should cobordism classes of complex manifolds have anything to do with the algebraic geometry of formal group laws? Nevertheless this has been one of the most important observations for modern homotopy theory. It is the driving force behind Chromatic Stable Homotopy which tries to build a dictionary between the algebraic geometry of FGLs and structures present in the stable homotopy category. It is shocking how successful this has been.

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82
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Here is one of my favorites. If you consider a singular node of an algebraic curve locally it looks like the curves $xy=0$ in $\mathbb{C}^2$, or $x^2+y^2=0$. This consists of two smooth arcs intersecting to each other transversally (reducible in particular).

Now, one step further, if we consider a cusp which is analytically equivalent to the origin in the curve $y^2+x^3=0$ in $\mathbb{C}^2$, it is locally irreducible. However, here comes the interesting point, if we intersect the singularity with a small ball $$[(x,y)\in \mathbb C^2:\ |x|^2+|y|^2=\epsilon]\cong S^3$$ what we've got is that such an intersection is $$(ae^{2i\theta},a^{3/2}e^{3i\theta})\subset S^1\times S^1\subset S^3$$ which is contained in a torus winding two times in one direction in the torus and three times in the other direction, in other words, we have an trefoil knot. alt text

Now in the case of surfaces, all these facts give rise to an amazing relation between topology and algebraic geometry. The underlaying space topological space in $\mathbb C^4$ of $$x^2+y^2+z^2+w^3=0$$ is a manifold!! (note it is singular at the origin in the context of AG!). As far as I know, if one intersects a small ball with the singularity, as I did above, one gets a topological sphere whose differential structure is NOT the standard one. Even more, considering in $\mathbb C^5$ the following hypersurface $$x^2+y^2+z^2+w^3+t^{6k-1}=0$$ and carrying out the intersection with a small sphere around the origin, for $k=1,2,\ldots 28$ one may get all the 28 possible exotic differential structures on the 7-sphere that Milnor found.

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    $\begingroup$ Wow,that IS pretty whack,Csar.This example alone is a testament to the power of modern topology and geometry and the incredible connections it has uncovered. $\endgroup$ Commented Jul 15, 2010 at 19:55
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    $\begingroup$ There's an interesting related result -- let $SS_n(X)$ denote the space of subsets of a topological space $X$ where the subsets have cardinality between $1$ and $n$. As a space, you can consider it to be $X^n / \Sigma_n$. Then $SS_3(S^1)$ is the 3-sphere. By design, $SS_n(S^1)$ has a fixed-point free $SO_2$-action, so it's a Seifert-fibred space. A non-singular orbit of this $SO_2$-action on $SS_3(S^1)$ is the trefoil knot. $\endgroup$ Commented Aug 16, 2010 at 0:00
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    $\begingroup$ Your claim about taking the link of the singular point of the hypersurface $x^2 + y^2 + z^2 + w^3 = 0$, and getting an exotic five-sphere, sounds wrong to me. Is there a reference for this? $\endgroup$
    – Sam Nead
    Commented Mar 13, 2011 at 22:08
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    $\begingroup$ @RyanBudney I'm confused by the claim that $SS_3 (S^1)$ is the 3-sphere. If $SS_3 (S^1)$ is really the same as the symmetric product $(S^1)^3/\Sigma_3$, then there are maps $S^1\to (S^1)^3/\Sigma_3 \to S^1$ defined by $z \mapsto [z, 1, 1]$ and $[x, y, z]\mapsto xyz$ which compose to the identity. So $(S^1)^3/\Sigma_3$ isn't simply connected. $\endgroup$
    – Dan Ramras
    Commented Sep 1, 2019 at 1:51
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    $\begingroup$ @DanRamras, apologies, I somehow conflated the two spaces. Delete my 2nd sentence -- those two aren't quite the same. $\endgroup$ Commented Sep 1, 2019 at 7:06
66
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The beautiful analogy between number fields and function fields (and in general, algebra and geometry) that one learns about in arithmetic algebraic geometry.

Some specific examples:

  • The idea that a Galois group and a fundamental group (one algebro-number theoretic, the other geometric/topological) are two instances of the same thing.

  • The use of the term ramification in both number theory and geometry. Describing $\mathbb{Z}$ as simply connected because $\mathbb{Q}$ has no unramified extensions.

  • The appearance of integral closure in both algebraic geometry and algebraic number theory. The integral closure, in the former case, actually corresponds to a distinct geometric idea: non-singularity.

  • The idea of considering a prime number to be a point; then viewing localization at that prime, -adic completion at that prime, and the residue field of that prime as if they were the corresponding geometric objects. In particular, using the term "local" in number theory, as if we were talking about geometry! This idea is built into scheme theory.

There are many more examples.

Galois Groups and Fundamental Groups by Szamuely looks deeply into the relationships between Galois groups and fundamental groups and eventually develops a theory which covers both.

An Invitation to Arithmetic Geometry by Lorenzini explores the beautiful relation between algebraic curves and algebraic number theory.

This post ("Mazur's knotty dictionary" on the neverendingbooks.org blog) explores the analogy between prime numbers and knots.

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59
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How about something simple: $e^{i\pi}=-1$.

Like when you first hear that, what the hell does the ratio of circumference to diameter of circles have to do with the square root of negative one and the base of the natural exponent?

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    $\begingroup$ ...or between exponential and trigonometric functions generally. $\endgroup$ Commented Apr 23, 2011 at 1:05
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    $\begingroup$ In my opinion, this becomes a lot less mysterious as soon as you think of the exponential and trigonometric functions as eigenfunctions of the differentiation operator (respectively, its square), which is really the reason they're both so important. The basic properties and interrelationships of these functions - including the above identity - are natural consequences of this formulation. $\endgroup$ Commented Jul 29, 2011 at 3:06
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    $\begingroup$ Well yes, but that's the way it is with all these surprising results, isn't it? They all indicate a connection that no one had suspected, but is undoubtedly important. Once that connection is chewed over enough and becomes something you learn as a matter of course, then the original surprising result becomes "understandable", or sometimes even "trivial". But it certainly wasn't originally, and often isn't even to people first encountering these things today. $\endgroup$ Commented Sep 30, 2013 at 0:54
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    $\begingroup$ Well, last time I had to teach that and tried to remember how it is proved, I realized that in fact that is the definition of $\pi$ (as soon as one parametrizes the unit circle by $e^{it}$ and sees that this curve has constant speed one, that is). $\endgroup$ Commented Apr 12, 2015 at 11:29
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    $\begingroup$ @CarlOffner Feynman in his book "Surely you're joking..." recalls his amusement while he and other physics students watched two maths students discuss: the first asks the second why a certain result follows from the assumptions. "It's trivial! It's trivial!" replies the second, and begins a complicated explanation. The first one is trying hard to follow. But after 15 minutes, finally he admits "You were right, it's trivial". Physicists then would joke that mathematicians only prove trivial theorems. $\endgroup$
    – Del
    Commented Aug 31, 2019 at 8:23
53
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The ubiquity of Littlewood-Richardson coefficients. Given three partitions $\lambda, \mu, \nu$ each with at most $n$ parts, there is a combinatorial definition for a number $c^\nu_{\lambda, \mu}$ which is nonzero if and only if any of the following statements are true:

  • There exist Hermitian matrices $A, B, C$ whose eigenvalues are $\lambda, \mu, \nu$, respectively and $A + B = C$ (one can also replace Hermitian by real symmetric)
  • The irreducible representation of ${\bf GL}_n({\bf C})$ with highest weight $\nu$ is a subrepresentation of the tensor product of those irreducible representations with highest weights $\lambda$ and $\mu$.
  • Indexing the Schubert cells of the Grassmannian ${\bf Gr}(d,{\bf C}^m)$ (where $d \ge n$ and $m-d$ is at least as big as any part of $\lambda, \mu, \nu$) by $\sigma_\lambda$ appropriately, the cycle $\sigma_\nu$ appears in the intersection product $\sigma_\lambda \sigma_\mu$.
  • There exists finite Abelian $p$-groups $A,B,C$ and a short exact sequence $0 \to A \to B \to C \to 0$ such that $B \cong \bigoplus_i {\bf Z}/p^{\nu_i}$, $A\cong \bigoplus_i {\bf Z}/p^{\lambda_i}$, and $C\cong \bigoplus_i {\bf Z}/p^{\mu_i}$.

And probably many more things.

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Ehud Hrushovski's proof, using model theory, of the geometric Mordell-Lang conjecture in algebraic geometry.

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45
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Connection between the typical number of isolated nonzero solutions ($N$) of a system of equations $$f_1=f_2=\cdots=f_n=0,$$ where each $f_k$ is a polynomial in $n$ complex variables, and the mixed volume ($V$) of the Newton polytopes of $f_k$: $$N=(n!)\cdot V.$$

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  • $\begingroup$ Oh, this is surprising! $\endgroup$
    – Unknown
    Commented May 27, 2010 at 7:13
43
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Here is a copypaste of something I've already mentioned in another question.

The fastest known solution of the word problem in braid groups originated from research on large cardinal axioms; the proof is independent of the existence of large cardinals, although the first version of the proof did use them. See Dehornoy, From large cardinals to braids via distributive algebra, Journal of knot theory and ramifications, vol. 4 (1995), no. 1, 33–79 (MR).

To me this is an absolute mystery! Large cardinals are usually considered an esoteric subject situated on the border of the observable universe. So why should they have any relevance to braids, a very down to earth part of mathematics? Let alone give an algorithm for distinguishing braids, and what's more, the fastest algorithm known.

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I think the disparity between the world-views in low-dimensional topology versus high-dimensional topology are surprising. Even after you learn the reasons why, IMO they should still be surprising. Examples:

1) Teichmuller space exists, yet hyperbolic manifolds in dimension $3$ and larger are rigid. There are many interesting connections here such as the link between conformal geometry, complex analysis and hyperbolic geometry in dimension 2.

2) Exotic smooth structures on $\mathbb R^4$ but not on $\mathbb R^n$ for $n\neq 4$.

3) Why the Poincare conjecture/hypothesis is "hard" in dimensions $3$ and $4$ yet relatively "easy" in other dimensions.

4) Geometry being particularly relevant to $2$ and $3$-dimensional manifolds yet less so in higher dimensions.

I could go on. Some of these are connections, some I suppose are disconnections. But a connection is only a surprise if you have reason to think otherwise. :)

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    $\begingroup$ @Ryan: is the fact that geometry is less useful in high dimensions an empirical observation, or is there more mathematical content to this? $\endgroup$
    – Jim Conant
    Commented Mar 13, 2011 at 17:44
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    $\begingroup$ @Jim: From Hillman's "Four-manifolds, Geometries and Knots", up to homeomorphism there are only 11 geometric 4-dimensional manifolds with finite fundamental group. In dimension 4 a finite-volume hyperbolic manifold's volume is a function of its Euler characteristic. I see those as having a fair bit of content. Sorry for being slow to reply. $\endgroup$ Commented Sep 1, 2011 at 6:40
  • $\begingroup$ >>3) Because dimension 4 has infinite differential structure? $\endgroup$ Commented Jan 5, 2017 at 13:09
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The work of Nabutovsky and Weinberger applying computability theory (a.k.a. recursion theory) to differential geometry. For example one of their results is that if you consider the space of Riemannian metrics on a smooth compact manifold $M$ of dimension at least 5 and sectional curvature $K\le 1$, then there are infinitely many extremal metrics. This is a purely geometric statement, but the only known proof uses concepts from computability theory. Moreover the results from computability theory that are used in their work are very deep; prior to their work, some skeptics regarded this area of computability theory as being overly specialized and having no hope of being connected to other areas of mathematics. See the exposition of Robert Soare (available on his website) for more information.

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The pair correlation function between Riemann zeta function zeros is the same as the pair correlation function between eigenvalues of random Hermitian matrices.

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    $\begingroup$ A caveat, this is not a theorem. Montgomery showed this for test functions whose Fourier transform had restricted support (in fact, support [-2,2] iirc.) Montgomery conjectured the same holds for more general test functions. Odlyzko's computations provided spectacular numerical evidence. And Katz-Sarnak proved an analogous statement for function fields. $\endgroup$
    – Stopple
    Commented Mar 13, 2011 at 20:02
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    $\begingroup$ Yes, it is indeed a fascinating relation, but one should add it assumes RH $\endgroup$ Commented Oct 24, 2015 at 10:32
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McKay's observation that the special fiber in the desingularization of du Val singularities is a bunch of $\mathbb P^1$s linked according to the Dynkin diagram corresponding to the group of the singularity.

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    $\begingroup$ This seems to be part of the more general phenomenon of the ADE classification. Worth a separate answer (which points back to this one)? $\endgroup$ Commented Jan 13, 2015 at 15:25
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This is probably not the most serious of applications, but I found the equivalence (in game theory) of the determinacy of Nash's board game Hex with the Brouwer Fixed Point theorem to be a surprising, if somewhat lighthearted, connection.

You can read David Gale's paper.

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  • $\begingroup$ I thought of another (in my opinion, simpler) proof of the Brouwer Fixed Point theorem, but I don't know where I could write it. It goes through the standard equivalence to the nonexistence of a retract from $I^2$ to its boundary. The main idea is the following fact: Suppose that $A,B\subset I^2$ are disjoint closed sets, where $A\cap\partial I^2=(0,1)$ and $B\cap\partial I^2=(1,0)$. Then $(0,0)$ and $(1,1)$ are in the same connected component of $I^2\setminus(A\cup B)$. This can be proven by drawing a (sheared) Hex board on the square with small enough hexagon size. $\endgroup$ Commented May 30, 2017 at 1:46
  • $\begingroup$ You then suppose a retract $f:I^2\to\partial I^2$ exists. The inverse images of $(0,1)$ and $(1,0)$ fit the hypotheses of the above fact, and so $(0,0)$ and $(1,1)$ must be in the same connected component of $f^{-1}\big(\partial I^2\setminus\{(0,1),(1,0)\}\big)$. Applying the retract to this, one obtains that $(0,0)$ and $(1,1)$ are in the same connected component of $\partial I^2\setminus\{(0,1),(1,0)\}$, which is clearly false; contradiction. $\endgroup$ Commented May 30, 2017 at 1:51
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I was recently amazed at a quick connection between two facts I've known since high school. The Euler characteristic of a sphere, thought of as #vertices + #faces - #edges on a polyhedron, buckyball, etc., is 2; I re-deduced this from the fact that the derivative of $f(x)=1/x$ is $f'(x)=-1/x^2$.

The steps of the proof are as follows: construct the Riemann sphere using two complex charts, both C, with the holomorphic transition map $f(z)=1/z$ on each neighborhood minus its origin. Now we want to look at the Chern class of the cotangent bundle, which in standard orientation is the negative of the Euler class of the tangent bundle, i.e. the sphere. Well, assuming complex analysis, look at $df=\frac{-1}{z^2}dz$ to see the effect of the transition map on the cotangent bundles: as a ``holomorphic'' 1-form, that has a double pole at one point and no zeros. Thus we know that a section of the cotangent bundle of the sphere has divisor degree $-2$. So $\chi(S^2)=2$ and I now cannot separate this fact from $f'(x)=-1/x^2$ in my mind. It seem somehow more mysterious, ridiculous, and delightful that this connection is so short.

(Everyone I've mentioned this to prefers their own proof and perhaps it's better to do this slightly more directly to get a self-intersection 2 for a section of the tangent bundle, i.e. vector fields vanish twice, which gives the Euler class in $H^2(S^2)$ as a multiple of the orientation class.)

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My favorite connection in mathematics (and an interesting application to physics) is a simple corollary from Hodge's decomposition theorem, which states:

On a (compact and smooth) riemannian manifold $M$ with its Hodge-deRham-Laplace operator $\Delta,$ the space of $p$-forms $\Omega^p$ can be written as the orthogonal sum (relative to the $L^2$ product) $$\Omega^p = \Delta \Omega^p \oplus \cal H^p = d \Omega^{p-1} \oplus \delta \Omega^{p+1} \oplus \cal H^p,$$ where $\cal H^p$ are the harmonic $p$-forms, and $\delta$ is the adjoint of the exterior derivative $d$ (i.e. $\delta = \text{(some sign)} \star d\star$ and $\star$ is the Hodge star operator). (The theorem follows from the fact, that $\Delta$ is a self-adjoint, elliptic differential operator of second order, and so it is Fredholm with index $0$.)

From this it is now easy to proof, that every not trivial deRham cohomology class $[\omega] \in H^p$ has a unique harmonic representative $\gamma \in \cal H^p$ with $[\omega] = [\gamma]$. Please note the equivalence $$\Delta \gamma = 0 \Leftrightarrow d \gamma = 0 \wedge \delta \gamma = 0.$$

Besides that this statement implies easy proofs for Poincaré duality and what not, it motivates an interesting viewpoint on electro-dynamics:

Please be aware, that from now on we consider the Lorentzian manifold $M = \mathbb{R}^4$ equipped with the Minkowski metric (so $M$ is neither compact nor riemannian!). We are going to interpret $\mathbb{R}^4 = \mathbb{R} \times \mathbb{R}^3$ as a foliation of spacelike slices and the first coordinate as a time function $t$. So every point $(t,p)$ is a position $p$ in space $\mathbb{R}^3$ at the time $t \in \mathbb{R}$. Consider the lifeline $L \simeq \mathbb{R}$ of an electron in spacetime. Because the electron occupies a position which can't be occupied by anything else, we can remove $L$ from the spacetime $M$.

Though the theorem of Hodge does not hold for lorentzian manifolds in general, it holds for $M \setminus L \simeq \mathbb{R}^4 \setminus \mathbb{R}$. The only non vanishing cohomology space is $H^2$ with dimension $1$ (this statement has nothing to do with the metric on this space, it's pure topology - we just cut out the lifeline of the electron!). And there is an harmonic generator $F \in \Omega^2$ of $H^2$, that solves $$\Delta F = 0 \Leftrightarrow dF = 0 \wedge \delta F = 0.$$ But we can write every $2$-form $F$ as a unique decomposition $$F = E + B \wedge dt.$$ If we interpret $E$ as the classical electric field and $B$ as the magnetic field, than $d F = 0$ is equivalent to the first two Maxwell equations and $\delta F = 0$ to the last two.

So cutting out the lifeline of an electron gives you automagically the electro-magnetic field of the electron as a generator of the non-vanishing cohomology class.

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    $\begingroup$ Wow, this was very interesting! $\endgroup$ Commented Aug 29, 2011 at 20:30
32
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Complex multiplication of elliptic curves and the explicit construction of the maximal abelian extension of a quadratic imaginary number field.

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  • $\begingroup$ Could someone give a brief explanation of what's surprising? A total ignoramus like me just sees "something in arithmetic geometry/number theory connected to something in arithmetic geometry/number theory". (Maybe even that's wrong.) $\endgroup$ Commented Feb 8, 2010 at 4:43
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    $\begingroup$ If you evaluate (appropriately normalized) elliptic functions at points lying in quadratic imaginary fields, the values you obtain are algebraic numbers, lying in abelian extensions of said quadratic imaginary fields; and all such extensions can be obtained in this way. (Compare with: $e^{2\pi i z}$ evaluated at rational numbers gives algebraic numbers, which generate abelian extension of ${\mathbb Q}$, and all abelian extension of ${\mathbb Q}$ are obtained in this way. $\endgroup$
    – Emerton
    Commented Feb 8, 2010 at 5:03
  • $\begingroup$ Thanks, Emerton. But I still don't get a sense of why this is a "surprising connection". I'm not disputing that it is - I'd just like to learn, at least, where the element of surprise lies. $\endgroup$ Commented Feb 8, 2010 at 6:15
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    $\begingroup$ Another way of saying it is that (with some slight fiddling) coordinates of points of finite order on an elliptic curve with complex multiplication give abelian extensions of the appropriate quadratic imaginary field. This was Kronecker's Jugendtraum (dream of his youth). Only in few cases is this explicit description of abelian extensions possible. Why do I think it is surprising? Compare what Emerton said above, $\exp(2 \pi iz)$ generating abelian extensions of $\mathbb{Q}$. Tell this to someone and ask them to guess how you'd generalise! It really is surprising that it is possible at all. $\endgroup$ Commented Feb 8, 2010 at 8:37
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    $\begingroup$ Elliptic curves (over $\mathbb{C}$) have their origins in studying elliptic integrals. As analytic objects they are $\mathbb{C}$ modulo a lattice. It's not immediately obvious to me that this is an algebraic object, and that the Weierstrass $\mathcal{P}$ function, which is an infinite sum, should compute anything number theoretic. So perhaps the connection is between analysis and algebra/arithmetic from this point of view. $\endgroup$
    – Zavosh
    Commented Feb 8, 2010 at 21:16
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My personal favorite is Multiple Zeta Values $$ \zeta(s_1,\ldots,s_d) = \sum_{n_1>\ldots>n_d} \frac{1}{n_1^{s_1}\ldots n_d^{s_d}} $$ They appears in relation with

  • Quantum groups (they are coefficient of Drinfeld's KZ associator)
  • Deformation quantization (Kontsevich's formula for the affine space)
  • Feynmann diagrams (a large class of diagrams evaluate to MZV's)
  • Kashiwara-Vergne conjecture (representation theory of Lie groups)
  • Modular forms (Zagier noticed that the space of relations in depth 2 is canonically isomorphic to the space of cusp forms on $SL_2$ through their period polynomials)
  • Moduli spaces of curves of genus 0 $\mathcal{M}_{0,n}$

the list goes on and on... the reason for all this lies in the theory of mixed Tate motives over $\mathbb{Z}$.

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31
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The Jones polynomial of knot theory and Feynman path integrals.

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    $\begingroup$ I'm not sure why this is surprising. It was originally defined via subfactors, but the path integral formalism followed very closely behind. Also, I'm not sure that Feynman path integrals count as mathematics... $\endgroup$ Commented Feb 8, 2010 at 5:07
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    $\begingroup$ Well, you are right. I think "surprising" is a subjective property, perhaps an experience of facing one's own ignorance. I don't know much about subfactors and first saw this polynomial in the context of knot invariants, divorced from its origins. For this reason, the physics connection seemed like a big surprise. It appears that you are an expert and so it is not surprising to me that you are not surprised. $\endgroup$
    – Zavosh
    Commented Feb 8, 2010 at 15:50
  • $\begingroup$ Well then couldn't you say that it's the connection to knot theory that's the surprise? $\endgroup$ Commented Aug 15, 2010 at 14:00
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    $\begingroup$ I always found it ironic that knot theory began with Lord Kelvins model of atoms as knots in ether (loosely speaking). After a 360 degree rotation (or make that 720 degree :-) we're at string theory now. $\endgroup$ Commented Jul 25, 2011 at 12:00
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It is possible to compute the Betti numbers of a smooth complex variety $X(\mathbb{C})$ by computing the cardinality of $X(\mathbb{F}_{p^n})$ for a prime $p$ with good reduction and a finite number of positive integers $n$; in other words, by brute force.

The above claim is wrong, so I'll phrase it the other way around. The Betti numbers of a smooth complex variety control the behavior of the number of points on $X(\mathbb{F}_{p^n})$; for example, for a smooth projective curve of genus $g$ we have $|\text{Card}(X(\mathbb{F}_q))| - q - 1| \le 2g \sqrt{q}$.

Generally I find the relationship between the arithmetic and topological properties of varieties surprising, although maybe this is a temporary kind of surprise that arithmetic geometers are used to. Another example: if $X$ is a curve, then whether the curvature of $X(\mathbb{C})$ is positive, zero, or negative determines whether $X(\mathbb{Q})$ is rationally parameterizable, a finitely generated group, or finite (unless it's empty).

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  • $\begingroup$ I'm a bit confused: why does one know the zeta function of a variety by just knowing finitely many values of $\#X({\bf F}_{p^n})$? $\endgroup$
    – Steven Sam
    Commented Feb 8, 2010 at 4:52
  • $\begingroup$ Hmm. You're right; I'd need a bound on the sum of the Betti numbers to conclude that. My mistake. $\endgroup$ Commented Feb 8, 2010 at 5:01
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    $\begingroup$ Because if you enough such values, you can solve for the eigenvalues of Frobenius on etale cohomology, and hence compute the zeta-function. But the number of values you need will depend on the number of eigenavlues, hence on the dimension you are trying to compute (if I understand things correctly). $\endgroup$
    – Emerton
    Commented Feb 8, 2010 at 5:05
  • $\begingroup$ Now that I think about it, it's not obvious to me that a less naive algorithm wouldn't be able to recover the entire zeta function from finitely many values of #X(F_{p^n}) knowing only the dimension of the variety. Knowing the possible absolute values of the eigenvalues might help out enough. $\endgroup$ Commented Feb 8, 2010 at 5:43
  • $\begingroup$ Can't one just exhibit curves of very high genus in $\mathbb A_2 (\mathbb F_{p^n})$ that pass through any given set of points? Choose the $\mathbb F_{p^n}$ points of some low-genus curve, you then need to go to $F_{p^m}$, for $m$ not a divisor of $n$, to tell the difference. $\endgroup$
    – Will Sawin
    Commented Feb 2, 2012 at 0:53
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Special values of the Riemann zeta function and class numbers of cyclotomic fields.

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Ulam's problem on determining the length of the longest increasing subsequence of a random permutation. The solution and the full description of the answer brought together ideas from integrable systems, combinatorics, representation theory, probability (appearing in the form of polynuclear growth model for instance), and random matrix theory.

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  • $\begingroup$ Yes, and the later connection between longest increasing subsequences with the theory of quantum gravity! $\endgroup$ Commented Feb 8, 2010 at 4:16
  • $\begingroup$ Victor, that's a truly strange connection. Do you have a reference? $\endgroup$ Commented Feb 8, 2010 at 5:08
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    $\begingroup$ I got them from a paper of Odlyzko and Rains: dtc.umn.edu/~odlyzko/doc/incr.subseq.pdf (actually listening to Odlyzko speak on it) $\endgroup$ Commented Feb 8, 2010 at 5:19
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    $\begingroup$ @QiaochuYuan, here is the reference you asked about (which didn't exist when you asked the question, though I was already working on it then). $\endgroup$
    – Dan Romik
    Commented Sep 13, 2015 at 3:00
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Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe. (The shortest path between two truths in the real domain passes through the complex domain.) — Jacques Hadamard Paul Painlevé

It is often credited to Hadamard, but in fact Painlevé said this exact quote, and Hadamard in one of his books merely paraphrased it: "it has been written that..." See https://philosophy.stackexchange.com/a/61924

Hadamard clearly had his proof of the prime number theorem, along the approach of Riemann, in mind. The ubiquity of complex numbers may deserve a full answer in itself; but here we might highlight its use in number theory: Riemann zeta function and other L-functions, modular forms (which would be a whole answer in itself).

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    $\begingroup$ Why down vote? I thought Jacques Hadamard expressed in his quote that in his time the connection of prime numbers to the zeroes of the Riemann zeta function was surprising and much of a shortcut to proving the Prime Number Theorem. $\endgroup$
    – Unknown
    Commented Jun 2, 2010 at 1:05
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    $\begingroup$ Probably it would not have been downvoted had the post included the content of your comment. As it is, it is pretty non-specific (even though the quote and its context are fairly widely known). $\endgroup$ Commented Jan 13, 2015 at 15:00
  • $\begingroup$ The original quote is by Paul Painlevé. See homepage.math.uiowa.edu/~jorgen/hadamardquotesource.html $\endgroup$
    – puzzlet
    Commented Jun 11, 2018 at 9:34
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Here is one of my favorite, that I learned from A. G. Khovanskii: let $f$ be a univariate rational function with real coefficients. Then, you can think of $f$ as inducing a continuous self-map of $\mathbb{RP}^1 \cong S^1$, in particular, it has a topological degree, say $[f]$, and if $f$ happens to be a polynomial, it is obvious that $[f]=0$ if $\deg(f)$ is even, and that $[f]=\pm 1$ if $\deg(f)$ is odd (depending on the sign of the main coefficient).

If the decomposition of $f$ in continued fraction is $$ f=P_0+\cfrac{1}{P_1+\cfrac{1}{P_2+\ddots}}$$ Then one can prove easily that $[f]$ is the (finite) sum: $[f]=\sum_{i \geq 0} (-1)^i[P_i]$. (Khovanskii himself taught this to high-schoolers in Moscow.)

The interesting connection for me follows: for any real polynomial $P$, the topological degree of the fraction $P'/P$ is clearly the (negative of the) number of real roots of $P$. Thus, the computation formula above applied to $[P'/P]$ allows us to recover Sturm's theorem.

I don't know if it really qualifies as a new proof of the theorem, but it's definitely a different point of view on that proof.

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The fact that the circumference of a unit circle is used to normalize the bell curve. Elementary compared to the other examples, yes, but how shocking was it when you first learned it?

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    $\begingroup$ To me, this isn't really shocking. It's a natural consequence of the cute (and, yes, maybe even surprising) fact that the square of $\int e^{x^2}\;dx$ is equal to $\int e^{x^2 + y^2}\;dx\;dy$, the integral of a function whose level sets are circles. $\endgroup$
    – Vectornaut
    Commented Feb 1, 2012 at 21:45
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    $\begingroup$ @Vectornaut: your point is that this connection can be understood; but it still strikes me as initially surprising. $\endgroup$ Commented Apr 12, 2015 at 11:33
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    $\begingroup$ @BenoîtKloeckner, I see. I was never surprised because, if I recall correctly, I never knew the normalization factor before being shown how to find it. $\endgroup$
    – Vectornaut
    Commented Apr 12, 2015 at 21:29
  • $\begingroup$ I think James Stirling may have been the first to know this. $\endgroup$ Commented Dec 15, 2021 at 18:12
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Goppa’s construction of error-correcting codes from curves, leading to the Tsfasman–Vladut–Zink bound (the first improvement over the Gilbert–Varshamov bound; see Modular curves, Shimura curves, and Goppa codes, better than Varshamov–Gilbert bound). An error-correcting code may be regarded as a combinatorial structure, and I think that this is a surprising connection between algebraic geometry and combinatorics.

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