1,574 reputation
619
bio website cmcc.ufabc.edu.br/…
location Cotia, Brazil
age 36
visits member for 4 years, 5 months
seen 1 hour ago
I'm a professor at the Center of Mathematics, Computation and Cognition of the Federal University of the ABC, Santo André, Brazil.

Apr
30
answered constant rank theorem for banach spaces
Apr
8
comment Analog of Newlander–Nirenberg theorem for real analytic manifolds
After some thought, it seems to me that CR structures do not really address your point. Any smooth real hypersurface of $\mathbb{C}^n$ is a CR manifold, be it real analytic or not. This means that any intrinsic real analytic structure you put on it will fail to match the would-be one induced from the embedding if the embedding is not real analytic. There is an extensive discussion of CR manifolds, both intrinsic and embedded, I've found in the book of S. Dragomir and G. Tomassini, "Differential Geometry and Analysis on CR Manifolds" (Birkhäuser, 2006).
Mar
25
comment Stokes theorem with corners
This theorem is the one called the "Gauss-Green theorem" in Federer's book which you mentioned above. Have a look as well at the paper by M. Taylor, M. Mitrea and A. Vasy, "Lipschitz domains, domains with corners, and the Hodge Laplacian". Commun. PDE 30 (2005) 1445-1462, arXiv:math/0408438.
Mar
25
comment Stokes theorem with corners
One of the main basic results in Geometric Measure Theory is a version of the Stokes theorem for manifolds with a locally Lipschitz boundary, due to de Giorgi and Federer. It seems to me that such boundaries are general enough to include corners in the $\mathscr{C}^1$ sense. A proof of this theorem can be found in Federer's book and (I think) in Whitney's book "Geometric Integration Theory" quoted in Zurab Silagadze's answer below.
Feb
27
comment Foliations of Lorentzian manifolds by Spacelike Hypersurfaces
By the way, my former comment addresses the OP's first question. As for OP's second question, global hyperbolicity is also necessary, since the existence of a foliation by spacelike Cauchy hypersurfaces implies that such a hypersurface exists and hence the space-time is globally hyperbolic. In this case, Bernal and Sánchez have shown that one can even choose the function to have one of its level sets match the given hypersurface.
Feb
27
comment Foliations of Lorentzian manifolds by Spacelike Hypersurfaces
A weaker sufficient condition would be to require that the space-time is stably causal. In this case, you also have a real-valued smooth function with everywhere (say, past-directed) timelike gradient, hence all its level sets are regular and hence are spacelike hypersurfaces which foliate the space-time manifold. In this more general case, they are even allowed to change topology (in this case, the manifold must be disconnected, of course).
Feb
17
revised Integrals of pullbacks and the Inverse function theorem(s?)
Small amend to explanation
Feb
17
revised Integrals of pullbacks and the Inverse function theorem(s?)
Added explanation
Feb
17
revised Integrals of pullbacks and the Inverse function theorem(s?)
Added explanation of the role of Sard's theorem
Feb
16
revised Integrals of pullbacks and the Inverse function theorem(s?)
Added explanation
Feb
16
answered Integrals of pullbacks and the Inverse function theorem(s?)
Dec
8
comment Green's operator of elliptic differential operator
An implicit assumption in the question is that the base manifold $M$ should be compact, otherwise the operator cannot be Fredholm. Moreover, one only gets the tame estimates needed for Nash-Moser under this assumption. It is satisfied in the examples given, but it's not explicitly stated.
Nov
30
awarded  Yearling
Nov
15
awarded  Nice Answer
Oct
21
comment Is every Montel locally convex vector space compactly generated?
Oops, that was a typo (which unfortunately I can no longer edit out), sorry... Thanks!
Oct
20
comment Is every Montel locally convex vector space compactly generated?
I do not have the book of Frölicher and Kriegl at hand, but I guess these results are the ones you quoted, right?
Oct
20
comment Is every Montel locally convex vector space compactly generated?
Hmm... Come to think of it, this actually could be inferred indirectly from the discussion in the book of Kriegl-Michor (cited in the question) as well: Theorem 4.11 (3), pp. 39-40 states that the Kelley topology of a strict inductive limit of a sequence of Fréchet spaces coincides with its $c^\infty$ topology (i.e. the final topology induced by all smooth curves). If these Fréchet spaces are finite-dimensional and the sequence is strictly increasing, then Proposition 4.26 (iii), pp. 45 entails that the $c^\infty$ (and hence the Kelley) topology is not a vector space topology.
Oct
20
comment Is every Montel locally convex vector space compactly generated?
Strongly inaccessible cardinals are somewhat beyond my current mathematical knowledge, but of course your explanation makes clear why the continuum hypothesis is not needed regardless of that. Glad I learned something new... Thanks!
Oct
17
comment Is every Montel locally convex vector space compactly generated?
I'm actually thinking of the Mackey-Ulam theorem; moreover, Ulam originally showed that $\mathbb R$ does not admit a Ulam measure using the continuum hypothesis. Are there stronger recent results in this respect?
Oct
16
accepted Is every Montel locally convex vector space compactly generated?