matgaio

Rio de Janeiro, Brazil

Mar
22
awarded Yearling
Mar
22
awarded Yearling
Oct
10
comment Radial geodesics in a graph of a function
Fine, now I've got this. Nice solution. Thanks a lot!
Oct
10
accepted Radial geodesics in a graph of a function
Oct
9
comment Radial geodesics in a graph of a function
Dear @Anthony Carapetis, you proved $\partial_\theta\in T_pS$, which doesn't imply (at least I can't see) that $\partial_\theta$ is a horizontal vector. Once $\partial_\theta\in T_pS$, it follows that $$0=\langle\partial_\theta,\nabla f-\partial_z\rangle=\langle\partial_\theta,\nabla f\rangle-\langle\partial_\theta,\partial_z\rangle,$$ but I can't see why is $\langle\partial_\theta,\partial_z\rangle=0$, once $\partial_\theta$ is not necessarily horizontal. Maybe it is just a silly question of mine. Thanks again.
Oct
2
asked Radial geodesics in a graph of a function
Sep
30
awarded Editor
Sep
30
revised Extending the hyperbolic splitting on $\Lambda$ to a neighborhood of $\Lambda$
minor corrections
May
8
awarded Caucus
Mar
24
comment Reference - Asymptotic geodesics on compact surfaces without conjugate points
Dear Anton, sorry for taking too long to unnacept the answer. I was really busy these days so I didn't see the changes on the status of the question. And thanks to Misha for pointing out the papers of Burns and Sullivan
Mar
21
comment Compact surfaces without conjugate points
Hi Daniel, thaks for the attention and for the pointing out about the covering. It seems to me that on the hyperbolic half plane, two vertical lines are asymptotic at the future and they deviate on the past, isn't it? At least I could see that they are asymptotic on the future, comparing the arcs joining the same height of both with the horizontal segments joining them.
Mar
20
asked Reference - Asymptotic geodesics on compact surfaces without conjugate points
Mar
20
asked Compact surfaces without conjugate points
Mar
20
accepted Surfaces without conjugate points
Mar
19
asked Surfaces without conjugate points
Feb
11
comment Manifold without conjugate points and positive curvature
I think the result is not exactly for the Ricci curvature an one point, but to the Ricci curvature calculated along a geodesic definer on all $\mathbb{R}$, but we still have that there is no manifold positively curved and without conjugate points.
Feb
10
comment Manifold without conjugate points and positive curvature
do you know if the inexistence of positively curved spaces without conjugate point was known before this paper?
Feb
8
comment frontier of class $C^{1}$.
The appendix of the Evans's book "Partial Differential Equations" have a definition of boundary of class $C^k$ which close to that you gave.
Feb
8
comment Approximation of Lipschitz functions on Riemannian manifolds
perhaps you can use an exaustion of the manifold by compact balls, approximate inside each of these balls and usa some sort of "cantor's diagonal argument"...
Sep
2
comment Geometric intuition behind the Lie bracket of vector fields
@AdamSaltz, this is a kind of common interpretation and proofs for this can be found in several textbooks on analysis on manifolds, for example Lee's "Smooth Manifolds" (a great book!). What amuses me on this is the fact that it is a really natural question to know when two vector fields commute (in the sense I gave before) and that the answer is a simple algebraic calculation.
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