1,944 reputation
816
bio website homepages.ulb.ac.be/~samulisi
location Nantes, France
age
visits member for 3 years, 1 month
seen Apr 9 at 21:20

Postdoc working in symplectic/contact topology. I'm currently at the Université de Nantes.


Aug
28
revised Isoperimetric Area Function
fixed $\L$ to $\mathcal L$.
Aug
28
comment Isoperimetric Area Function
I may have fixed your $\L$. Let me know if I have chosen the wrong interpretation of what you meant. So you know for next time, I used \mathcal L
Aug
28
suggested suggested edit on Isoperimetric Area Function
Aug
20
comment existence of nonlinear second order ODE boundary value problem
It's not clear to me how standard existence results apply to this boundary value problem. For instance, if you allow the constants to have arbitrary sign, the boundary value problem has completely different features. Can you please clarify what you had in mind?
Aug
20
answered existence of nonlinear second order ODE boundary value problem
Aug
9
awarded  Necromancer
Jul
17
comment understanding this differential operator on a tensor product
Question 1 is clear once you expand $z = x+iy$ and $\bar z = x - iy$. Then, the metric you wrote down becomes $h ( dx^2 + dy^2 - i dx \wedge dy)$, which is precisely $h$ times the standard Hermitian metric. The real part of this is a Riemannian metric. I think that's where your confusion is from.
Jul
16
comment A “correct” hierarchical scoring scheme?
@PhD: you use "correct" twice in your description of the properties you want your scoring scheme to have. What does "correct" mean? Another way of wording this question: if I gave you a different scoring scheme from the one you proposed, what concrete things about it would enable you to determine that it was better.
Jun
26
comment Why do mathematicians use single-letter variables?
however, this fantastic formulation of the chain rule is hilarious.
Jun
26
comment Why do mathematicians use single-letter variables?
This is a great answer to the question.
Jun
21
comment Riemannian metric making a given function harmonic
This is essentially the idea I described in my comment above. I guess you have to choose $\alpha$ preserved by the monodromy? I'd like some more details if you have figured them out.
Jun
19
comment Riemannian metric making a given function harmonic
Thank you, Willie and Leonid, for confirming my guess and for the reference.
Jun
18
comment Riemannian metric making a given function harmonic
I'm not sure at the moment of how to get this to work in general, but if $f$ is a surjective submersion (what I called a geometer's fibration), then $M$ locally looks like a product, $I \times S$, where $S$ is a level set of $f$, and locally we have $(t, p) \mapsto t$. In this local picture, it suffices to have a metric for which the interval is orthogonal to $S$ and $\partial_t$ has constant length. More generally, I would then want to construct the metric fibrewise, so the fibres are orthogonal to $\nabla f$. I haven't thought about this enough to complete the idea.
Jun
18
comment Riemannian metric making a given function harmonic
@WillieWong: I suspect that what Kutluhan and Taubes mean is that the closed 1-form $df$ is also co-closed (i.e. $\star df = 0$). My best guess is that this is the same as asking for $f$ to be a harmonic map $M \to S^1$, but I am not sure. (While $f$ is manifold valued, you can think of $df$ as being an honest $1$-form by identifying $T S^1 = \mathbb{R} \times S^1$.)
Jun
18
comment Riemannian metric making a given function harmonic
when you say it fibres over the circle, do you have a geometer's fibration or a topologist's fibration in mind? i.e. is it a fibre bundle with diffeomorphic fibres (equivalently, $df$ of maximal rank)?
Jun
16
awarded  Revival
Jun
16
revised Gradient nonzero extensions of a vector field on the circle
explained the error (pressed enter too soon on previous edit)
Jun
16
comment Gradient nonzero extensions of a vector field on the circle
Thanks for giving a nice counterexample to what I claimed (in step 1). I'll edit my answer to point out where the problem shows up.
Jun
16
comment Gradient nonzero extensions of a vector field on the circle
@Leonid Kovalev: thanks for reopening this discussion... I had forgotten about it, and see that one of the points I didn't discuss in enough detail is actually false.
May
15
awarded  Nice Answer