bio | website | perso-math.univ-mlv.fr/users/… |
---|---|---|
location | Créteil | |
age | 30 | |
visits | member for | 4 years, 10 months |
seen | 1 min ago | |
stats | profile views | 1,423 |
Now at "Université Paris Est Créteil". I work in Riemannian geometry, and more specifically on Ricci flow.
Jun 11 |
answered | Gauss-Bonnet formula for 2-dimensional Alexandrov spaces |
Jun 11 |
comment |
Gauss-Bonnet formula for 2-dimensional Alexandrov spaces
Existence is known, I'm currently writing an answer about it. |
Jun 11 |
comment |
Gauss-Bonnet formula for 2-dimensional Alexandrov spaces
Actually in this context, just requiring that the curvature measure is not zero and nonnegative should be enough (which is weaker than what you require, there could be dirac masses in the curvature, think about the surface of a cube for instance). |
Jun 11 |
comment |
Gauss-Bonnet formula for 2-dimensional Alexandrov spaces
For question 2, you need to require that the curvature is not everywhere 0 to exclude the torus. |
May 1 |
awarded | Nice Answer |
Apr 30 |
awarded | Popular Question |
Apr 1 |
comment |
Shortest paths in Alexandrov spaces
@valeri why don't you make this an answer ? |
Apr 1 |
comment |
Thales Style Level Sets
I can write down an explicit formula in finite time for plane polygons, but it will be ugly... |
Mar 31 |
comment |
Thales Style Level Sets
Except the original motivation, is it that important to restrict the question to plane 2-dimensional figures ? (Actually the only sets for which I can answer the question are balls !) |
Mar 22 |
comment |
A version of isotone projection cones
What about the projection from $\mathbb{R}^2$ to $C=\{y=-x\}$, with $a=(0,2)$ and $b=(0,0)$ ? It seems to prove that the general statement you were looking for is false. |
Feb 26 |
comment |
Obstruction to the existence of global isometries on a constant-curvature Riemannian manifold
Even in the case where $M$ is simply-connected $(M,g)$ need not be an open subset of a complete constant curvature manifold: consider for instance the universal cover of an open subset of the plane with several holes. |
Feb 23 |
comment |
Existence and uniqueness of a quasi-linear pde system on a surface
May I inquire where this equation come from ? Can it be written in a more geometric fashion ? Does it encode some special properties of the 1-form $I_\alpha dx^\alpha$ ? If you manage to present your question in this way, this would help us giving you more precise answers. |
Dec 9 |
comment |
Surjectivity of “nice maps” from local properties
For your example from analysis, don't you need to assume that your map $f$ is proper ? It seems to me that the inclusion map from an open subset of $\mathbb{R}^d$ to $\mathbb{R}^d$ is a counterexample. |
Dec 3 |
revised |
Frobenius Condition for a specific first order pde
added 104 characters in body |
Dec 3 |
comment |
Frobenius Condition for a specific first order pde
My bad. This case is integrable actually. |
Dec 2 |
revised |
metric on ${\bf SPD}_n({\mathbb R})$
added 36 characters in body |
Dec 2 |
answered | metric on ${\bf SPD}_n({\mathbb R})$ |
Dec 2 |
revised |
Frobenius Condition for a specific first order pde
Added the fact that the solution needs to be non constant. |
Dec 2 |
comment |
Frobenius Condition for a specific first order pde
Ok, I did the computation: if $u$ solves the problem, so does any $f(u)$ for $f$ a smooth function from $\mathbb{R}$ to $\mathbb{R}$. So I'll take $u(x,y,z)=z^2-y^2$ instead of yours. Then $\nabla\phi_1=\partial_x$ and $\nabla\phi_2=yz\partial_x+xz\partial_y+xy\partial_z$, while $\nabla u= 2z\partial_z-2y\partial_y$. On can check that the inner products between the gradients don't vanish everywhere. |
Nov 29 |
comment |
Frobenius Condition for a specific first order pde
I'll have a look but I'm skeptical. |