Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
6  
I think it could be interpreted as restricting the breadth of the list. I would suggest removing the tag to avoid this. –  François G. Dorais Feb 8 '10 at 0:30

64 Answers 64

In topology, the connection between the fixed point property and the topological dimension (the covering dimension)--the following two theorems are equivalent:

  • the cube $\ I^n\ $ has the fixed point property;
  • there exists a normal topological space $\ X\ $ such that $\ \dim X\ \ge\ n$.

for every $\ n=0\ 1\ ...$

The connection is my notion of the universal function (or universal morphism in general). The beginning of the story is:

THEOREM $\ \dim X \ge n\ \Leftrightarrow\ \exists\ universal\,\ f:X\rightarrow I^n\ $ (for every completely regular space $\ X$).

share|improve this answer

That the Jeffreys' prior in the pole-zero parameterization of a transfer function is the hyperbolic transfinite diameter of the support of the poles and zeros.

It's my favorite because I just discovered it last month. I like laughing at my own jokes.

share|improve this answer

Connection:

  1. The Langrange polynomial interpolation formula;
  2. The Chinese Remainder Theorem.

Formulations:

  1. Let $\ K\ $ be a field of characteristic $\ 0.\ $ Let $\ \phi:A\rightarrow K\ $ be an arbitrary function, where $\ A\ $ is a non-empty finite subset of $\ K.\ $ Then there exists an exactly polynomial $\ f:K\rightarrow K\ $ of degree $\ n < |A|,\ $ such that $\ \forall_{x\in A}\ f(x)=\phi(x)$.
  2. Let $\ A\ $ be a nonempty finite set of positive integers such that $\ \gcd(a\ b)=1\ $ for every two different $\ a\ b\in A.\ $ Let $\ \phi:A\rightarrow\mathbb Z\ $ be arbitrary. Then there exists $\ f\in\mathbb Z\ $ such that $\ \forall_{a\in A}\ f\equiv \phi(a) \mod a.\ $ The integer $\ f\ $ is unique in the following sense: $$\ \forall_{a\in A}\ g\equiv\phi(a)\!\!\!\mod a\quad \Rightarrow\quad g\equiv f\!\!\!\mod \prod A$$

Crucial special (basic) cases:

  1. There exists exactly one polynomial $\ f_b:K\rightarrow K\ $ of degree $\ n < |A|,\ $ such that $\ f_b(b)=1,\ $ and $\ \forall_{x\in A\setminus\{b\}}\ f_b(x)=0\,\ $ for every $\ b\in A\ $ (actually $\ \deg(f)=|A|-1$).
  2. There exista exactly one integer $\ f_b\!\!\mod\prod A\ $ such that $\ f_b\equiv 1\!\!\mod b,\ $ and $\ \forall_{a\in B\setminus\{b\}}\ f_b\equiv 0\!\!\mod a\ $ for every $\ b\in A$.

Once we get the basic elements $\ f_b,\ $ then $\ f\ $ is uniquely obtained as the respective linear combination of elements $\ f_b\ $ both in the Lagrange and in the Chinese cases.

Construction (of the basis elements):

  1. $\ \forall_{t\in K}\ f_b(t):=\frac{L_b(t)}{L_b(b)},\ $ where $\ L_b(t):=\prod_{a\in A\setminus\{b\}}\ (t-a)$
  2. $\ f_b\ := C_b\cdot d_b,\ $ where $\ C_b:=\prod_{a\in A\setminus\{b\}} a,\ $ and $\ d_b\cdot C_b\equiv 1\mod b$.

We see that $\ C_b\ $ corresponds to $\ L_b,\ $ and $\ d_b\ $ to $\ \frac 1{L_b(b)}$.

This connection, when generalized, unites the algebraic number theory and the theory of algebraic functions.

share|improve this answer

protected by François G. Dorais Sep 30 '13 at 0:52

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.