2,022 reputation
818
bio website homepages.ulb.ac.be/~samulisi
location Nantes, France
age
visits member for 3 years, 9 months
seen Dec 1 at 5:00

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


Mar
20
comment How much of an $n$-dimensional manifold can we embed into $\mathbb{R}^n$?
Another idea would be to take a Morse function with a unique minimum. Then, take the union of all the ascending manifolds from higher index critical points. This is of codimension 1 and thus of measure 0. I'm pretty sure the complement is the ascending manifold of the minimum and thus diffeomorphic to a ball. Oh, wait, that's the link that Jason gave. Sorry.
Mar
20
revised what are the holomorphic curves in $T^{*}S^3$ with boundary on the zero section?
added 1407 characters in body
Mar
20
comment what are the holomorphic curves in $T^{*}S^3$ with boundary on the zero section?
I recommend a nice survey article (though a little obsolete) by Hofer and Kriener. I'll update what I wrote with a few more details.
Mar
20
comment Vector field and integral curve
You know the total expression needs to be independent of the extension of $v$. However, you can extend $v$ however you want... what if you take an extension of $v$ that is invariant under the flow of $V$? Once you've seen what happens in that case, you can try to understand what happens more generally, though that case is sufficient to prove what you want.
Mar
20
answered what are the holomorphic curves in $T^{*}S^3$ with boundary on the zero section?
Mar
20
comment Vector field and integral curve
After you've extended the vector field, you want to expand the LHS as $= \nabla_V(d\phi_t \cdot v)|_{t=0} = \nabla_{d\phi_t \cdot v} V|_{t=0} + [V, d\phi_t \cdot v]|_{t=0}$ first using the definition of covariant derivative and then using the fact it is torsion-free. You now want to rewrite this last term as $L_V(d\phi_t v)$ and evaluate it.
Mar
19
comment Vector field and integral curve
Indeed, I was thinking of $v$ as being a local vector field. You can always extend your vector $v$ to a vector field and then show the result doesn't depend on your extension. (So, no, you can't prove $\nabla_V v = 0$, since that will depend on how you extend $v$.)
Mar
18
answered Vector field and integral curve
Mar
16
awarded  Yearling
Sep
17
awarded  Citizen Patrol
Sep
17
comment When is a $k$-form a $(p, q)$-form?
@MichaelAlbanese: that's how it always is with these things... impossible before you see it the right way and obvious afterwards.
Sep
16
answered Integrals on manifolds and pullbacks
Sep
16
comment Construction of cut-off function
Can you please verify what you mean instead of (3)? As I commented, it is impossible to satisfy it on the region where $2r < x < r$. The tricky part of proving that a cut-off function with these properties exists comes from the differential inequalities (3) and (4). One way to start is to see what functions are extreme cases of (3) and (4). This gives an idea of the type of behaviour the function must have and where the constraints have a chance to be problematic. Have you tried this? Let us know what you have tried to do.
Sep
16
answered When is a $k$-form a $(p, q)$-form?
Sep
16
comment Construction of cut-off function
Why are you interested in this question? Do you want an argument that such a function exists, or do you want an explicit formula? Also, is there a typo in condition (3) for the interval $-2r < x < -r$? On that interval, there must be points where $\phi'(x) > 0$.
Sep
11
comment Taking Differential Topology concurrently with Analysis
Nils's answer is the correct one. IMO, you really need to understand the implicit/inverse function theorem, but the professor is the best person to give you advice on what you should do.
Sep
10
answered Interpretation of Multilinear maps as tensors
Sep
7
comment $\{(x,y)\!\in\!\mathbb{B}^n; -\varepsilon\leq-\|x\|^2\!+\!\|y\|^2\leq\varepsilon\}\approx\mathbb{B}^k\!\times\!\mathbb{B}^{n-k}$
I was about to write a similar explanation, but was hesitating b/c of this corner smoothing issue. I don't know if you've noticed, Leon and LVK, but most topology books waive their hands at smooth handle attaching. It's not hard (as reading Max's explanation will make clear), but writing down every detail becomes a headache.
Sep
4
answered The connection in terms of local trivialization
Sep
2
comment The connection in terms of local trivialization
I'll try to write more later, but the key observation is that the map from $\pi^*E \to TE$ has image equal to the vertical tangent space (i.e. $\ker d\pi \subset TE$). For $A$ to split the sequence means exactly what you think: it is a projection to the vertical tangent space along what will now be defined to be the horizontal subspace. Let $\sigma$ be a section of $E$. The wiki definition has $\nabla \sigma \colon TX \to E$. Yours allows you to take $d\sigma \colon TX \to TE$ and then compose with $A$. IMO, wikipedia's more convenient for vector bundles, Taubes's generalizes better.