Most Popular
1500 questions
48
votes
1 answer
Producing finite objects by forcing!
It is a trivial fact that forcing can not produce finite sets of ground model objects. However there are situations,
where we can use forcing to prove the existence of finite objects with some properties. For example consider the following
result…
Mohammad Golshani
- 31,966
48
votes
4 answers
Fermat's last theorem over larger fields
Fermat's last theorem implies that the number of solutions of $x^5 + y^5 = 1$ over $\mathbb{Q}$ is finite.
Is the number of solutions of $x^5 + y^5 = 1$ over $\mathbb{Q}^{\text{ab}}$ finite?
Here $\mathbb{Q}^{\text{ab}}$ is the maximal abelian…
Pablo
- 11,229
48
votes
4 answers
Central limit theorem via maximal entropy
Let $\rho(x)$ be a probability density function on $\mathbb{R}$ with prescribed variance $\sigma^2$, so that:
$$\int_\mathbb{R} \rho(x)\, dx = 1$$
and
$$\int_\mathbb{R} x^2 \rho(x), dx = \sigma^2$$
Fact: the density function which maximizes the…
Paul Siegel
- 28,772
48
votes
4 answers
Is Schauder's conjecture resolved?
Schauder's conjecture: "Every continuous function, from a nonempty
compact and convex set in a (Hausdorff) topological vector space into
itself, has a fixed point." [Problem 54 in The Scottish Book]
I wonder whether this conjecture is resolved. I…
57319
- 603
48
votes
2 answers
Geometric interpretation of the half-derivative?
For $f(x)=x$, the half-derivative of $f$ is
$$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x = 2 \sqrt{\frac{x}{\pi}} \;.$$
Is there some geometric interpretation of (Q1) this specific derivative, and, (Q2) of the half-derivative more generally? I have…
Joseph O'Rourke
- 149,182
- 34
- 342
- 933
48
votes
2 answers
Why should I care about topological modular forms?
There seems to be a lot of recent activity concerning topological modular forms (TMF), which I gather is an extraordinary cohomology theory constructed from the classical theory of modular forms on the moduli space of elliptic curves. I gather that…
Doug P
- 481
48
votes
1 answer
When does $A^A=2^A$ without the axiom of choice?
Assuming the axiom of choice the following argument is simple, for infinite $A$ it holds: $$2\lt A\leq2^A\implies 2^A\leq A^A\leq 2^{A\times A}=2^A.$$
However without the axiom of choice this doesn't have to be true anymore. For example if $A$ is an…
Asaf Karagila
- 38,140
48
votes
3 answers
Pullback measures
Why do all measure theory textbooks present the concept of push-forward measure, but never the concept of pull-back measure? Doesn't the latter exist?
It's true that the naive treatment of such a concept would sometimes lead to contradictions. For…
Alex M.
- 5,217
48
votes
7 answers
What's the use of a complete measure?
A complete measure space is one in which any subset of a measure-zero set is measurable.
For what reasons would I want a complete measure space? The only reason I can think of is in the context of probability theory: using complete probability…
Tom E
- 481
- 1
- 5
- 3
48
votes
1 answer
A function whose fixed points are the primes
If $a(n) = (\text{largest proper divisor of } n)$, let $f:\mathbb{N} \setminus \{ 0,1\} \to \mathbb{N}$ be defined by $f(n) = n+a(n)-1$. For instance, $f(100)=100+50-1=149$. Clearly the fixed points of $f$ are the primes.
Is every number…
Rodrigo A. Pérez
- 3,071
48
votes
8 answers
Who wrote up Banach's thesis?
Sometime ago I read somewhere (and I don't remember where it was) that Stefan Banach--a highly creative and great mathematician--did not always write down his ideas.
Allegedly, he did not write his own thesis (but of course, all the mathematics in…
Andreas Loos
- 331
48
votes
8 answers
A sudden smiley? :-)
This is a vague question, and I will no doubt be (properly!) chastised for posing it.
I would like to generate a set $S$ of points in $\mathbb{R}^3$—$|S|$ finite or infinite—which
has the property that, viewing $S$ under orthogonal projection along…
Joseph O'Rourke
- 149,182
- 34
- 342
- 933
48
votes
2 answers
Isomorphic general linear groups implies isomorphic fields?
Suppose $n > 1$ is a natural number. Suppose $K$ and $L$ are fields such that the general linear groups of degree $n$ over them are isomorphic, i.e., $GL(n,K) \cong GL(n,L)$ as groups. Is it necessarily true that $K \cong L$?
I'm also interested in…
Vipul Naik
- 7,240
48
votes
5 answers
Theories of Noncommutative Geometry
[I have rewritten this post in a way which I hope will remain faithful to the questioner and make it seem more acceptable to the community. I have also voted to reopen it. -- PLC]
There are many ways to approach noncommutative geometry.
What are…
Anweshi
- 7,272
47
votes
2 answers
Local structure of rational varieties
I've been asked this question by a colleague who's not an algebraic geometer; we both feel that the answer should be "no", but I don't have a clue how to prove it.
Here's the question:
let $X$ be a smooth rational variety (over the complex numbers,…
rita
- 6,213