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| bio | website | homepages.ulb.ac.be/~samulisi |
|---|---|---|
| location | Nantes, France | |
| age | ||
| visits | member for | 2 years, 2 months |
| seen | 1 hour ago | |
| stats | profile views | 283 |
Postdoc working in symplectic/contact topology. I'm currently at the Université de Nantes.
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May 1 |
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An other question about Theorem 3.1 from Morse theory by Milnor let us continue this discussion in chat |
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May 1 |
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An other question about Theorem 3.1 from Morse theory by Milnor Do you understand what $\phi_t$ is? |
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Apr 30 |
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An other question about Theorem 3.1 from Morse theory by Milnor The tricky part is that $\phi_{b-a}$ maps $M^a$ to $M^b$. To check this, calculate what $f( \phi_{b-a}(x))$ is, if $f(x) \le a$. This is where you use the fact that your new vector field flows downwards so that the time $t$ flow changes the value of $f$ by $t$. |
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Apr 30 |
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An other question about Theorem 3.1 from Morse theory by Milnor The fact that it is a diffeomorphism onto its image follows just from the existence/uniqueness of ODE, the long-time existence (essentially by construction) and smoothness with respect to initial condition. I.e. this is standard ODE theory. |
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Apr 30 |
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An other question about Theorem 3.1 from Morse theory by Milnor The purpose of (1) is to make (2) obvious and to make it easy to write down a formula for $r_t$. Can you construct a simple example (e.g. for domains in $\mathbb{R}^2$) and see what he is doing? Can you say more about where you are stuck? |
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Apr 30 |
answered | Implicit Function Theorem and Rank Theorem Misunderstandings. |
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Apr 30 |
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Dynamical Systems. Bendixson's and Dulac-Bendixson's theorems. Great timing, I was just typing essentially the same thing. While this is clear from what you wrote, I want to emphasise that $U$ has positive area because $C$ is embedded. ($C$ is embedded by uniqueness of solutions to ODE.) |
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Apr 27 |
answered | Non-degenerate solutions to constant Hamiltonian flow |
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Apr 27 |
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Hamiltonian Isotopy in Symplectic geometry Can you give a bit more detail about what you are looking for? What do you mean by "visualize"? what makes a context "good"? Have you studied any classical physics where the Hamiltonian is giving the total energy of a conservative system? if so, this is the right visualization (at least on cotangent bundles). |
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Apr 27 |
answered | Lagrangian subspaces |
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Apr 16 |
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Area of flux homomorphism in symplectic topology I don't understand what you are trying to do, specifically. Are you trying to compute this explicitly? for that to make sense, it seems to me that you need some more explicit information. Or are you trying to do something else? this isn't clear to me. |
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Apr 13 |
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global vector fields in local coordinates In general you can't, for the reason you explained. What are you trying to do? |
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Apr 13 |
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Structure on manifolds I should add the question: have you studied surfaces in $\mathbb{R}^3$? This is the easiest scenario with a shape operator, and it might be worthwhile to understand this first. If you already know about this use of the shape operator, I can't tell you anything further. |
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Apr 13 |
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Structure on manifolds @Merri: I don't know the book you mention, and I know very little about uses of shape operators. However, I would guess that the key feature of a "shape operator" is that it is the differential of a "Gauss map". There are many situations in which you have something that looks like a Gauss map (but isn't the classical one), so maybe there are many non-standard shape operators. I don't really know. |
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Apr 11 |
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Can the chain rule be relaxed to allow one of the functions to not be defined on an open set? let us continue this discussion in chat |
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Apr 11 |
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Can the chain rule be relaxed to allow one of the functions to not be defined on an open set? I think you have some typos in your proposed modification of Theorem 3, or else I have no idea what you are asking for. In particular, I think you mean $f(E)$ instead of $F(E)$, but I still don't understand what exactly you are trying to accomplish. |
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Apr 11 |
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Can the chain rule be relaxed to allow one of the functions to not be defined on an open set? Observe that if the initial data is contained in the interior of the orthant ($\Omega$ above), then, for small time, the solution curve stays in $\Omega$, and thus you can use the standard chain rule. What's not at all clear is what behaviour you want to see at the boundary. |
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Apr 11 |
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Can the chain rule be relaxed to allow one of the functions to not be defined on an open set? I don't understand your motivation. Call $\Omega$ the (strictly) positive orthant. Then the non-negative orthant is $\overline \Omega$. If your initial data $x_0 \in \Omega$, then for small time, your solution curve $x(t)$ will live in $\Omega$. However, if you have initial data at the boundary $x_0 \in \partial \overline{\Omega}$, there is no reason for your integral curve to stay in $\overline \Omega$ even infinitesimally. In the former case, one expects $V(x(t))$ to make sense for short time, whereas in the latter, you don't. |
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Apr 11 |
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Symplectic geometry as a prequisite for Heegaard Floer homology Do you have someone in your department who can give you some guidance? there is a lot of symplectic geometry in both of these books that is not relevant to you and there are a number of technical points about holomorphic curves that are not explained in these books. A correct answer to your question, imo, requires more information about what you know and about what you are interested in doing. Unfortunately, even with this information, I wouldn't be qualified to give you a good answer, since I don't work in Heegaard Floer homology. |
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Apr 8 |
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Why is Cartan Formula just an avatar of Leibniz rule? Thank you for the bounty. I am grateful to you for the video. |
