bio | website | |
---|---|---|
location | Brazil | |
age | ||
visits | member for | 5 years, 5 months |
seen | Jul 18 at 0:05 | |
stats | profile views | 715 |
I'm a graduat student at the Instituto de Matemática Pura e Aplicada (IMPA), Rio de Janeiro, Brazil.
email: [the shortest string of letters containing my first name and last name as substrings]@gmail.com
Apr 9 |
revised |
Why worry about the axiom of choice?
added 2 characters in body |
Apr 9 |
revised |
Why worry about the axiom of choice?
edited body |
Feb 20 |
awarded | Yearling |
Feb 2 |
comment |
Examples of common false beliefs in mathematics
A Finsler metric (symmetric or not) can always be bounded between two Riemannian metrics, and induces the same topology that the manifold already has before considering any metric. This topology is metrisable. |
Oct 14 |
revised |
Does for every vector field there always exist a volume form for which the vector field is a homothety?
added 15 characters in body |
Oct 14 |
revised |
Does for every vector field there always exist a volume form for which the vector field is a homothety?
added 40 characters in body |
Oct 14 |
answered | Does for every vector field there always exist a volume form for which the vector field is a homothety? |
Oct 14 |
revised |
Manifolds admitting flat connections
edited body |
Oct 14 |
revised |
Manifolds admitting flat connections
added 8 characters in body |
Oct 14 |
revised |
Manifolds admitting flat connections
added 8 characters in body |
Oct 13 |
comment |
Manifolds admitting flat connections
I corrected the "symmetric flat" thing. I like the idea of using Cartan-Hadamard to prove that spheres don't admit torsionfree flat connections. This saves the task of proving that only toruses admit free $\mathbb R^n$ actions, which is something I didn't do in my answer. |
Oct 13 |
revised |
Manifolds admitting flat connections
added 164 characters in body |
Oct 12 |
answered | Manifolds admitting flat connections |
Sep 24 |
awarded | Autobiographer |
Sep 12 |
comment |
Does anyone know an intuitive proof of the Birkhoff ergodic theorem?
If $f\equiv 1$ and $T=I$ we have $M_T'\equiv+\infty$, so the recurrence formula is $+\infty + f=+\infty$, which is useless. Am I missing something? |
Aug 23 |
awarded | Nice Answer |
Aug 12 |
revised |
real symmetric matrix has real eigenvalues - elementary proof
added 391 characters in body |
Aug 12 |
revised |
real symmetric matrix has real eigenvalues - elementary proof
added 391 characters in body |
Aug 12 |
revised |
real symmetric matrix has real eigenvalues - elementary proof
added 391 characters in body |
Aug 6 |
revised |
real symmetric matrix has real eigenvalues - elementary proof
added 1 character in body |