Carlo Mantegazza
|
Registered User
|
I am an analysis researcher at the Scuola Normale Superiore di Pisa
|
|
Feb 25 |
comment |
Topology of ${\mathbb R}^n$ @Sam/Wlodzimierz - After your (Sam) comment on contractibility, it seems to me that the point is exactly that... the problem can also be reduced to show that a compact $(n-1)$-dimensional submanifold embedded in ${\mathbb R}^n$ cannot be retracted (in itself) on some ${\mathbb S}^m$, with $m\leq n-2$ ($m$ is actually the dimension of the factor N minus 1). I am wondering whether the fact that the hypersurface "disconnects" ${\mathbb R}^n$ and the "final" sphere ${\mathbb S}^m$ instead does not, can be used (at least in the differential case) to get a conclusion without using (co)-homology. |
|
Feb 22 |
comment |
Topology of ${\mathbb R}^n$ no problem Zev |
|
Feb 22 |
comment |
Topology of ${\mathbb R}^n$ I had in mind something in this spirit. I was just wondering if it was possible something easier, the question came from one of my undergraduate students. |
|
Feb 22 |
asked | Topology of ${\mathbb R}^n$ |
|
Feb 5 |
comment |
analysis of the regularity using Hormander condition Possibly, you should say in what spaces you are looking for uniqueness/regularity/continuous dependence |
|
Jan 29 |
comment |
Vitali sets of full outer measure yes, exactly . |
|
Jan 29 |
comment |
Vitali sets of full outer measure Nice answer Robert! Thanks |
|
Jan 29 |
asked | Vitali sets of full outer measure |
|
Jan 25 |
awarded | ● Commentator |
|
Jan 25 |
comment |
Cutlocus and conjugate points Very interesting reference, nice paper, thanks! |
|
Jan 24 |
comment |
Cutlocus and conjugate points Thanks, nice answer, I was really too optimistic... possibly, also $S^2\times S^2$ is a counterexample. Anyway, do you know a reference where I can find the description of the possible cut loci of the symmetric spaces? They should have dimension less than $n-1$ ($n$ is the dimension of the manifold). I am now asking myself if the property $Cut_p=Conj_p$ holds for $every$ point $p$ of the manifold, what can be said about this latter? |
|
Jan 24 |
awarded | ● Nice Answer |
|
Jan 24 |
comment |
Fattening of totally convex sets thanks, perfect! |
|
Jan 24 |
asked | Cutlocus and conjugate points |
|
Jan 23 |
comment |
Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure? Yes I got it. I wrote this before Ramiro posted the reference to the very nice result of Cichon. Anyway, I am now wondering how to construct a Vitali set with full outer measure in R. |
|
Jan 23 |
answered | Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure? |
|
Jan 23 |
accepted | Examples on small cut radius of totally convex set in non-negatively curved manifold |
|
Jan 21 |
revised |
2x2 subdeterminants of a matrix corrected the mistakes due to the invariance under change of sign of A |
|
Jan 21 |
comment |
How to show that x-y is Lebesgue-Lebesgue measurable T is Lipschitz, hence zero-sets go to zero-sets |
|
Jan 21 |
answered | Examples on small cut radius of totally convex set in non-negatively curved manifold |
|
Jan 20 |
asked | Fattening of totally convex sets |
|
Jan 18 |
awarded | ● Critic |
|
Jan 18 |
comment |
Examples on small cut radius of totally convex set in non-negatively curved manifold What about a small geodesic ball around the north pole and the south pole in a sphere? |
|
Jan 18 |
accepted | Totally Geodesic Submanifolds |
|
Jan 17 |
answered | level set of function f coincides with that of |▽f|,M splits? |
|
Jan 17 |
comment |
Totally Geodesic Submanifolds Ciao Pietro!$ $ |
|
Jan 17 |
awarded | ● Nice Question |
|
Jan 17 |
comment |
2x2 subdeterminants of a matrix Sorry $N\geq3$ here above. |
|
Jan 17 |
awarded | ● Supporter |
|
Jan 17 |
awarded | ● Teacher |
|
Jan 17 |
answered | Totally Geodesic Submanifolds |
|
Jan 16 |
awarded | ● Scholar |
|
Jan 16 |
comment |
2x2 subdeterminants of a matrix Thanks, very interesting reference. Just a comment on my question: it comes from determining explicitely the unique (up to the sign) second fundamental form $A$ of a hypersurface of dimension $N>3$ in $R^{N+1}$. As $Riem=A*A$ (where $*$ is the Kulkarni-Nomizu product of two symmetric 2-forms) all the 2x2 determinants of the matrix associated to $A$ in an orthonormal basis are given by the sectional curvatures, hence they are all "intrinsic", so also $A$, up to the sign. |
|
Jan 16 |
comment |
2x2 subdeterminants of a matrix You're right, I should have written A=B or A=-B. The same for the last sentence, you can get the determinant of A up to the sign. Sorry for the inaccuracy. |
|
Jan 16 |
awarded | ● Student |
|
Jan 16 |
awarded | ● Editor |
|
Jan 16 |
revised |
2x2 subdeterminants of a matrix edited body |
|
Jan 16 |
asked | 2x2 subdeterminants of a matrix |
