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

Here is a topic in the vein of Describe a topic in one sentence and Fundamental examples : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model.

A classical example is Galois theory for solving polynomial equations.

Any examples for homological algebra ? For Fourier analysis ? For category theory ?

share|improve this question
3  
I think that this should be community wiki, for reasons similar to those given by others in the comments at Fundamental Examples. –  Jonas Meyer Dec 13 '09 at 11:06
4  
@Harry: I think your position concerning honesty is rather extreme. –  S. Carnahan Dec 13 '09 at 16:21
2  
I've converted this question to wiki, but please make such questions community wiki in the future. See the FAQ and the discussion on meta.MO (tea.mathoverflow.net/discussion/6) for more on when a question should be community wiki. –  Anton Geraschenko Dec 13 '09 at 17:13

27 Answers 27

up vote 32 down vote accepted

[In front of a blackboard, in an office at Real College]

Skeptic: And why should I care about holomorphic functions?

Holomorphic enthusiast:$\;$ Can you compute $\quad$ $\sum_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}$ ? Here $a$ is one of your cherished real numbers, but not an integer.

Skeptic: Well, hm...

Holomorphic enthusiast, nonchalantly: Oh, you just get

$$\sum_{n={-\infty}}^{\infty} \frac{1}{(a+n)^2}=\pi^2 cosec^2 \pi a $$

It's easy using residues.

Skeptic: Well, maybe I should have a look at these "residues".

Holomorphic enthusiast (generously): Let me lend you this introduction to Complex Analysis by Remmert, this one by Lang and this oldie by Titchmarsh. As Hadamard said: "Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe".You can look for a translation at Mathoverflow. They have a nice list of mathematical quotations, following a question there.

Skeptic: Mathoverflow ??

Holomorphic enthusiast (looking a bit depressed) : I think we should have a nice long walk together now. [Exeunt]

share|improve this answer
    
There's an error in either the top display or the bottom, as one of these has a minus and one of them has a plus. It doesn't change the answer, since we're integrating over the integers, but I figured you should know. –  Harry Gindi Dec 13 '09 at 16:01
    
@ Harry: I don't know what you mean.The denominator is the square of a+n in both cases.There are no "minus" signs. –  Georges Elencwajg Dec 13 '09 at 17:35
    
oh, sorry, I have my jsmath font rendering at 100% instead of 130%, and the plus sign is invisible. –  Harry Gindi Dec 13 '09 at 17:44
    
I can't reproduce the invisible plus sign. Even when I set jsMath to render math at 50%, both plus signs are perfectly visible. –  Anton Geraschenko Dec 14 '09 at 0:13
1  
Consider $g(x) := cos(ax)$ for $a \in \mathbb{R} \setminus \mathbb{Z}$. Calculate the Fourier coefficients: $a_0 = \frac{2}{a \pi} \sin(a\pi)$, $a_k = \frac{2 a (-1)^k}{(k^2 - a^2) \pi} \sin(a\pi)$ for $k \geq 1$. Simplify the Fourier series for $g$ and plug in $x := \pi$ to obtain $\pi \cot(a \pi) = \sum_{k=-\infty}^{\infty} \frac{1}{a - k}$. Taking the derivative term-wise on both sides gives us the required identity for $\frac{\pi^2}{\sin(\pi x)^2}$. –  santker heboln Aug 3 '10 at 2:52

Let us call "division algebra over $\mathbb R$" a finite-dimensional vector space $A$ equipped with a bilinear map $A \times A \to A: (a,b) \mapsto a \bullet b$ , such that $a\bullet b=0$ implies $a=0$ or $b=0$. ( Associativity is not required).

Examples : the reals, the complexes, the real quaternions and the octonions of Graves-Cayley.

Any such division algebra must necessarily have dimension 1,2,4 or 8 (as in the examples). This was proved indepently in 1958 by Kervaire and Milnor using Bott's periodicity theorem, a fantastic result in algebraic topolgy which had just been proved.

To the best of my knowledge there is still no purely algebraic proof of this theorem on possible dimensions of real division algebras, although the statement is completely algebraic and elementary.

share|improve this answer

If you're a combinatorialist and you want to know the asymptotics of a sequence $a_n$ with a nice generating function $A(z) = \sum_{n \ge 0} a_n z^n$, the very first thing you should do is find out if $A$ is meromorphic, since then one can analyze the asymptotics of $a_n$ using its poles. Even if $A$ isn't meromorphic, if one has sufficiently good information about its singularities then there are transfer theorems that translate information about the behavior of $A$ near its poles to the behavior of $a_n$ for large $n$. In other words, combinatorialists (and by extension computer scientists) should learn complex analysis.

For example, let $E_n$ be the number of alternating permutations on $n$ letters. Then $E(z) = \sum_{n \ge 0} E_n z^n = \sec z + \tan z$ is meromorphic with poles $z = \frac{\pi}{2} + 2k \pi, k \in \mathbb{Z}$. The dominant singularity is at $z = \frac{\pi}{2}$ and one now knows without doing any other computations that $E_n \sim n! \left( \frac{2}{\pi} \right)^n$. Even better one can write down an exact series converging to $E_n$ with one term for each pole. The corresponding expansion of the Bernoulli numbers $B_n$ gives the classical evaluation of the zeta function at even integers.

share|improve this answer
    
Another example for this (using Tauberian Theorems) is www2.math.uni-paderborn.de/fileadmin/Mathematik/AG-Klueners/…. –  Timo Keller Jan 16 '10 at 13:20

One can use the machinery of the fundamental group and of covering spaces to easily prove that any subgroup of a free group must be free.

share|improve this answer

There are a number of algebraic theorems which are easier to prove using topology. The best known is probably the fundamental theorem of algebra, but there are others. For example, an $n\times n$ matrix with positive entries always has an eigenvector all of whose enetries are nonnegative. (The matrix defines a continuous function from the standard simplex to itself which always has a fixed point.) I learned about this from Elmer Rees.

share|improve this answer
2  
Can the fundamental theorem of algebra be proven not using topology/trascendent methods? Mabe it is not made easier by topology but *possible :) –  Mariano Suárez-Alvarez Dec 13 '09 at 16:35
    
There are analytic and algebraic proofs of the FTA, but not as simple as the topological one. Perhaps someone here knows the history better and can put the three proofs in chronological order? –  José Figueroa-O'Farrill Dec 13 '09 at 16:38
1  
All the "algebraic" proofs I know of depend on things like the fact that odd-degree real polynomials have a real root (that is, on the mean value theorem). –  Mariano Suárez-Alvarez Dec 13 '09 at 17:16
    
You'd need to use some special fact about the real numbers though, since it is essentially a special property of $\mathbb{C}=\mathbb{R}+i\mathbb{R}. I think the impressive thing is that all that is needed is just the (really quite weak) MVT plus some algebra (elementary Sylow theorem and Galois theory stuff). –  Thomas Bloom Dec 13 '09 at 17:20
    
There were "algebraic" proofs that predated Gauss, but they're really technical (they aim to be constructive!) and have wide gaps. I think they can be patched up, but not without some pretty heavy machinery (at least Galois theory). Are the topological and analytic proofs that different? The complex-analytic proof I know is basically an application of Cauchy's formula, which is a corollary of Stokes' theorem... Unfortunately I'm not that familiar with the topological proof, but Gauss' proof is more or less topological in nature. Fully analytical proofs I think require Cauchy. –  Harrison Brown Dec 13 '09 at 21:47

Problem: You need to multiply large numbers, with $10^9$ digits (or take products of power series). Your computer doesn't have the ability to do $10^{18}$ calculations.

Solution: Recognize multiplication of a power series as a convolution. Take a discrete Fourier transform of the digit sequences, multiply, and apply the inverse Fourier transform. Then perform the carries. This should take under $10^{12}$ calculations. The Fast Fourier Transform takes about $n \log n$ calculations.

The point is not that this is a fast algorithm or a clever trick. It's that you start out with a basic question about integers you can explain to someone comfortable with grade school math, and you end up dealing with complex or at least real numbers, characters of $\mathbb Z/n$, and properties of convolutions.

share|improve this answer

Problem: Suppose you care about the real world and objects you can hold in your hands. Show that any flexible polyhedron maintains a constant volume while it is flexed. This was known as the Bellows Conjecture.

Solution: With a little commutative algebra, you can prove that 12*volume is an algebraic integer in $\mathbb Q$ adjoin the lengths of the sides. Any continuous function from $\mathbb R$ to a countable set is constant. In fact, the volume is a root of a single polynomial.

share|improve this answer

For me personally, the first time I "got" Fourier analysis was when I understood how it could be used to prove Roth's theorem on arithmetic progressions (that any dense set of integers contains an arithmetic progression of length 3). It can be proved in several ways, but when you see this way you immediately realize that the technique is likely to be useful for many other problems. And in fact, Roth's theorem can also be used to justify Szemerédi's regularity lemma (which is not quite the same as justifying a whole theory, but it is a very useful technique) in a similar way.

share|improve this answer
1  
I thought that it was the other way round, that Szemerédi's regularity lemma was one of the tools used in the original proof of Szemerédi's theorem for progressions of arbitrary length ? –  Ewan Delanoy Jan 16 '10 at 14:54
3  
Well, yes, but nowadays we realise that the regularity lemma and the Fourier-analytic method are closely related, so that the fact that they are both useful for tasks such as finding arithmetic progressions. In particular, one can prove the regularity lemma by a spectral decomposition of the adjacency matrix (a generalisation of the Fourier decomposition of a function). There is also a similarity of technique between the basic idea of Roth's argument (density increment argument) and the basic idea behind the regularity lemma (energy increment argument). –  Terry Tao Jan 18 '10 at 5:07

Suppose you are interested in random walks on an extremely structured graph such as a hypercube graph or a cycle graph. If your graph happens to be the Cayley graph of an abelian group $G$, as in both of the above examples, then it is easy to describe the behavior of random walks on it because the eigenvectors of the adjacency matrix are precisely the characters of $G$ and the eigenvalues depend in a simple way on the characters; in other words, you should learn about the discrete Fourier transform.

Edit: Some elaboration. Let $G$ be a finite abelian group with $|G| = n$. A character of $G$ is a homomorphism $G \to \mathbb{C}$, and it is a basic fact of character theory that the characters form a basis of the space of functions $G \to \mathbb{C}$; this is the discrete Fourier transform. Now let $\mathbf{A}(G)$ be the adjacency matrix of a Cayley graph of $G$ using generators $\{ s_1, ... s_k \}$. The group $G$ acts on the space of functions $G \to \mathbb{C}$ by sending a function $f : G \to \mathbb{C}$ to $f(gx)$. Call this representation $\rho$; then (and this is the important connnecting observation) one may regard $\mathbf{A}(G)$ as the linear operator $\displaystyle \sum_{i=1}^{k} \rho(s_i)$.

Proposition: Let $\chi_j : G \to \mathbb{C}$ be a character of $G$. Then $\chi_j$ is an eigenvector of $\mathbf{A}(G)$ with eigenvalue $\displaystyle \sum_{i=1}^{k} \chi_j(s_i)$, and these are all the eigenvectors.

Proof. Just observe that $\rho(s_i) \chi_j(g) = \chi_j(s_i g) = \chi_j(s_i) \chi_j(g)$. The fact that these exhaust the set of eigenvectors follows from the basic fact cited above.

For example, the cycle graph $C_n$ is the Cayley graph of the cyclic group $\mathbb{Z}/n\mathbb{Z}$ with generators $\{ 1, -1 \}$, so its eigenvectors are just the rows of the discrete Fourier transform matrix on $\mathbb{Z}/n\mathbb{Z}$ and its eigenvalues are $e^{ \frac{2\pi i k}{n} } + e^{- \frac{2\pi ik}{n} } = 2 \cos \frac{2\pi k}{n}$. (Note that I have implicitly identified the space of functions $G \to \mathbb{C}$ with the free vector space on the elements of $G$ in the usual way.)

share|improve this answer
1  
The connection with the "Fourier discrete transform" is not quite clear to me. Could you elaborate ? –  Ewan Delanoy Dec 13 '09 at 20:35
    
I've added some elaboration. I hope it helps. –  Qiaochu Yuan Dec 13 '09 at 21:19

Well for a long time there was no proof of the Burnside theorem avoiding Representation theory. Now there are methods to proof it without Representation theory, but they are still a lot harder then the original representation theoretic one.

http://en.wikipedia.org/wiki/Burnside_theorem

share|improve this answer

Mathematical logic can be motivated by other areas of math in at least two different ways:

[1] It allows you to formulate (and prove) results about the unsolvability of certain problems. These are obviously essential, since they tell you that you shouldn't spend too much time trying to solve those problems, which are often very natural problems.

For example, as a group theorist, you might often want to know whether two particular groups given by generators and relations are isomorphic. It would be nice to have some set of tools that allowed you to solve the problem mechanically, but no such tools exist (Novikov's Theorem).

Or, as a number theorist, you might wish for a set of tools allowing you to decide effectively whether a given polynomial equation has integer solutions. This is ruled out by the Davis-Putnam-Robinson-Matiyasevich Theorem.

[2] It can give you easier proofs of theorems in seemingly unrelated subfields. I hope somebody can provide/confirm examples here...for example, I think Gödel's Compactness Theorem gives some mileage in algebraic geometry (Nullstellensatz?), and nonstandard analysis can simplify a number of proofs (Tychonoff's Theorem?). (Although nonstandard analysis isn't exactly mathematical logic, the fact that proofs of standard results using nonstandard analysis can be trusted is a theorem of logic.)

share|improve this answer
    
Lowenheim--Skolem gives a very quick proof of Ax's Theorem that every injective polynomial map from $\mathbb{C}^n$ to itself is surjective. –  HJRW Dec 13 '09 at 23:24

The spectral theory of commutative Banach algebras led to an elegant proof, due to Gelfand, of the following (previously difficult) theorem of Wiener: If $f$ is a nowhere vanishing complex valued function on the unit circle whose Fourier coefficients are absolutely summable, then the Fourier coefficients of $1/f$ are also absolutely summable.

share|improve this answer

One problem that one can solve with Fourier analysis very easily is the isoperimetrical inequality and the corresponding characterization of the circle. Of course, this can also be done in many. many other ways, but using Fourier series it becomes particularly simple.

share|improve this answer

One set of problems to which homological algebra applies surprisingly well is those posed by topology, as evinced by algebraic topology :P

This is of course quite anhistorical...

share|improve this answer

Set theory provides by far the easiest proof of the existence of transcendental numbers -- just show that the algebraic numbers and the integers can be put into 1-1 correspondence, but the real numbers and the integers can not. Liouville's proof isn't too hard, but it's nowhere near as elegant.

share|improve this answer
8  
Right -- and not only that, if you follow the proof through, you get an explicit example of a transcendental number. That's because (1) you can enumerate the algebraic numbers explicitly, and (2) the usual proof of the uncountability of the reals gives you, for any list of reals, an explicit real not on the list. But I don't know how many times I've read that the set theory proof doesn't give you an explicit transcendental number. –  Tom Leinster Dec 24 '09 at 0:59

This example is more closely related to a question of mine, but I'll give it here anyway.

A graded poset $P$ is Sperner if no antichain is larger than the largest level $P_i$ of $P$. This property is named after Sperner, who proved that the Boolean posets $B_n$ of subsets of $\{ 1, 2, ... n \}$ are Sperner. the Boolean posets $B_n$ are also rank-symmetric because they satisfy $(B_n)_i = (B_n)_{n-i}$, i.e. ${n \choose i} = {n \choose n-i}$ and unimodal because the sequence $(B_n)_i$ is at first increasing and then decreasing.

Let $G$ be a group acting on $\{ 1, 2, ... n \}$. Then $G$ acts on $B_n$ in the obvious way by order- and grade-preserving automorphisms. Define the quotient poset $B_n/G$ in the obvious way, which inherits the grading on $B_n$. It's not hard to see that $B_n/G$ is also rank-symmetric.

Theorem: $B_n/G$ is unimodal and Sperner.

The only known proofs of this theorem are algebraic (the one I know uses linear algebra), and even for special cases a purely combinatorial proof is not known. For example, in the case that $n = ab$ is a product of two positive integers and $G = S_b \wr S_a$ we recover the poset $L(a, b)$ of Young diagrams that fit into an $a \times b$ box; then the above theorem implies that the coefficients of the q-binomial coefficient ${a + b \choose b}_q$ are unimodal. This was first proven by Sylvester in 1878, and a combinatorial proof was not found until 1989. A combinatorial proof of the Sperner property is still not known. This example is all the more remarkable because the statement that $L(m, n)$ is Sperner requires almost no mathematics to understand.

Edit: I should probably provide a reference. This material is from some notes on algebraic combinatorics by Stanley.

share|improve this answer
    
"The only proof I know uses linear algebra": I haven't thought this through all the way, but I would think that $B_n/G$ would inherit a weighting of its elements from the LYM inequality, from which the fact that $B_n/G$ is Sperner would follow from some easy counting argument. Does this not work? Or is it what you were referring to? –  Harrison Brown Dec 13 '09 at 23:14
    
Ah, wait, I think I see why that doesn't work. Never mind, then. –  Harrison Brown Dec 13 '09 at 23:17

Also, differential Galois theory for differential equations.

share|improve this answer

The Jones polynomial: a knot invariant that originally came from subfactor theory.

share|improve this answer
1  
Can you elaborate, please? Thanks! –  Jose Brox Dec 13 '09 at 11:57
    
To be honest, I'm not able to do a good job of explaining this example without more time than I currently have to study up. I was hoping that community wiki justified my light post; I'd be happy (but by no means expect) to have someone more knowledgeable elaborate. Otherwise I may do so late in the week. –  Jonas Meyer Dec 14 '09 at 3:15
    
mathoverflow.net/questions/544/why-are-subfactors-interesting/…, last paragraph, contains some elaboration. –  Jonas Meyer Feb 10 '10 at 15:37

The Poincaré conjecture is an example of a purely topological statement which apparently cannot be proved only by topological means.

share|improve this answer
3  
Many topologists were arguing the reverse -- the Poincare conjecture is naturally an analytic statement. It's a misleading coincidence (due to Moise) that you can think of it as a purely topological or combinatorial statement. Classification of 2-manifolds (uniformization) was first proven using analysis. The high-dimensional Poincare conjecture (Smale) used basics of ODEs, if anything it was remarkable for how little analysis his proof used. –  Ryan Budney Dec 13 '09 at 21:39
    
Yet certainly there were many topologists trying to prove the Poincaré conjecture by purely topological means. –  José Figueroa-O'Farrill Jan 20 '10 at 11:09

Many geometric problems cannot be solved without hard analysis. Perhaps the best known example is the Calabi Conjecture proved by Yau.

share|improve this answer

Sometimes you want to understand a group $G$, but the only thing you know is that there is an extension $1 \to A \to G \to E \to 1$. If everything is abelian, $G$ corresponds to an element in $Ext^1(E,A)$. If at least $A$ is abelian, then $E$ acts on $N$ by conjugation and $G$ corresponds to an element in $H^2(E,A)$. Thus the classification of groups naturally leads to cohomology groups, which have a rich theory.

There is also a topological motivation: Which spheres act freely on finite groups?

share|improve this answer

As for category theory, I don't think that there is a motivating example which has not already a category theoretic flavor. The leading theme is to unify and then generalize constructions resp. arguments, which come up in all areas of mathematics. Historically, natural transformations were introduced for the foundations of homology theory of topological spaces. But to start with an easy example, you may observe that for abelian groups $A,B,C$ there is a canonical isomorphism $(A \oplus B) \oplus C \cong A \oplus (B \oplus C)$, which reminds you of other associativity results such as $(X \cup Y) \cup Z \cong X \cup (Y \cup Z)$ for sets (here $\cup$ means disjoint union). Within category theory, you can see what's the real content of this: direct sum and disjoint unions are examples of coproducts, and coproducts are always associative. Even more striking, Yoneda's Lemma, which lies at the heart of foundations of category theory, tells you that the case of sets already settles the general case!

But category theory is more than just a language, it also provides general constructions: Assume you want to approximate a theory with another theory. This may be formalized by finding an adjunction between two categories. Freyds/Special Adjoint Functor Theorem tell you when this is possible. Although in some situations you can't write down the adjunction, the only thing you need is to know that it exists. For example, what is the categorical coproduct of an infinite family of compact hausdorff spaces? Can you write it down without using Stone-Cech?

There are also somewhat global motivations: Some Theories behave like other theories, and thus you may develope a theory for a large class of categories at once: monodial categories, topological categories, algebraic categories, locally presentable categories, etc. Of course, the same is true for other notions of category theory (functors, natural transformations, types of morphisms, etc.).

But if one has not heard of category theory before, the first motiviation should be to think in categories (in the colloquial sense). For example the set, which underlies a group, really differs from the group. In almost every book and lecture, this is absorbed by abuse of notation. The existence of bases in vector spaces is no reason at all to restrict linear algebra to vector spaces of the form $K^{(B)}$. Similarily, vector bundles should not be defined as bundles which are locally isomorphic to some $\mathbb{R}^n \times X$, which most topologists still ignore! Rather, it is first of all a vector space object in the category of bundles over $X$.

Let's conclude with an example which both introduces functors and algebraic geometry: Assume you have a system of polynomial equations $f_1(x)=...=f_n(x)=0$ in $m$ variables defined over $\mathbb{Z}$ and you want to study the solutions in arbitrary rings at once, using a single mathematical object, e.g. having in mind some local-global results of algebraic number theory. So for every ring $R$, we put $F(R) = \{x \in R^m : f_1(x)=...=f_n(x)=0\}$. Observe that for every ring homomorphism $R \to S$ there is a set map $F(R) \to F(S)$ and that this is compatible with composition of homomorphisms. This exactly means that $F$ is a functor from the category of rings to the category of sets. Algebraic geometry studies functors which locally look like the functor above.

share|improve this answer

Proving the termination of Goodstein sequences (a problem in natural numbers) via arithmetic on infinite ordinals.

share|improve this answer

Complex analysis and the theory of spaces of analytic functions on open subsets of $\mathbb{C}$ are used to prove the Riemann Mapping Theorem, which has the nice topological consequence that there is a unique homeomorphism class of (nonempty) simply connected open subsets of $\mathbb{C}$.

share|improve this answer

Root systems and Dynkin diagrams for classification matters (see both Root system and ADE classification at Wikipedia).

share|improve this answer
    
This example doesn't strike me as "seemingly unrelated." I think something more in the spirit of the question is a question that doesn't require any knowledge of, say, Lie groups to state, but which is easiest to solve using the classification of (insert adjective here) Lie groups. –  Qiaochu Yuan Dec 13 '09 at 19:03

The Laplace transform to systematically solve homogeneous ODE's with constant coefficients by transforming them in polynomial equations (and then transforming the solution back).

Solving a differential equation by the Laplace transform

share|improve this answer

This is not an answer as much as a request for answers. Here are some topics for which I don't know an example of this sort, but would really like to see one (and nobody seems to have done them yet). Feel free to edit this post if good examples appear for some of these topics, or if you have a topic to add to the list. They are in no particular order,

  • Symplectic geometry (I'd really like to know...)

  • Algebraic geometry (there should be a ton of stuff here, it's so diverse)

  • Category theory

  • Homological algebra (is there an example simpler than some monstrous calculation?)

  • Group theory (so far, all that comes to my mind is that it applies to Galois theory, which in turn solves equations. But I think there should be a ton of simpler and nicer applications, shouldn't there?)

share|improve this answer
1  
Homological algebra is hallowed by its applications in algebraic topology (singular homology) and algebraic geometry (sheaf cohomology). Singular homology is, e.g., a fundamental tool in the classification of manifolds. Sheaf cohomology can be used to prove the Riemann-Roch theorem for algebraic curves which allows one, e.g., find coordinates for elliptic curves. –  Lennart Meier Dec 13 '09 at 21:14
    
"Algebraic geometry" means a lot of different things to different people. What part and what level of the theory would you want a motivating problem for? –  Qiaochu Yuan Dec 13 '09 at 22:06
    
Ilya, your comment on "group theory" suggests to me that you want all mathematics to be motivated by applications to number theory! Don't you think that, say, the symmetries of platonic solids are a subject of basic mathematical interest? –  HJRW Jan 18 '10 at 6:18
    
@Qiaochu: Well, the simpler the example, and the more of algebraic geometry needed to understand it, the better. –  Ilya Grigoriev Jan 19 '10 at 3:07
1  
@Henry. No, I definitely do not want all mathematics to be motivated by number theory. On the other hand, I don't know anything about symmetries of platonic solids that excites me too much. But if you have an example that is exciting, I'd be very interested to see it. More to the point, if it's simple to understand, interesting, and requires group theory to answer, by all means write it down in this thread! –  Ilya Grigoriev Jan 19 '10 at 3:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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