Sam Lisi
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 May 7 comment Figure $\infty$ is immersion of circle @HoseynHeydari: This falls under the heading of things Gowers refers to as "Just Do It" proofs. (See gowers.wordpress.com/2008/08/16/just-do-it-proofs) To get you started: what is the definition of an immersion? Can you obtain a figure X as an immersion from the disjoint union of two intervals? OK, now you are done. May 7 comment Figure $\infty$ is immersion of circle @EricO.Korman: I think you are overlooking the fact that he takes the figure $\infty$ to be an immersed circle. It is the image of an injective immersion of $\mathbb{R}$. May 7 awarded Revival May 7 awarded Caucus May 7 answered When does a vector field admit orthogonal fields? May 7 answered An alternative description of the first Stiefel-Whitney class May 4 comment When does a vector field admit orthogonal fields? @Yvoz How does this last observation of yours not answer your question? It seems like a characterization of when such a vector field exists. What are you looking for if that isn't the answer? May 3 comment When does a vector field admit orthogonal fields? Observe that you don't need orthogonality for the second vector field -- you just need pointwise linearly independent and nowhere vanishing. Then, since $\mathbb{R}^3$ is oriented, you get a third vector field free. You are therefore trying to find a trivialization of the tangent bundle of the quotient of $U$ by the flow of $X$. In general, this quotient will be a horrible space and won't have a tangent bundle, but maybe something can be said anyway. (Note that in your two examples, the quotients are $\mathbb{R}^2$ are $S^2$ respectively, and in particular very nice.) May 3 comment An alternative description of the first Stiefel-Whitney class I was not thinking about Steenrod squares etc. I really was thinking about their chapter on obstruction theory, Chapter 12. I guess M&S refer back to Steenrod's book, which I haven't studied as carefully as I should have. On page 143, they explain how you can reconstruct the obstruction classes from the $w_i$, at least for small $i$. May 3 comment An alternative description of the first Stiefel-Whitney class @Mehdi : yes, that's what I was thinking about (to your first comment). I'll try to write something soon, but the short version of what I am thinking about is that the $\mathbb{Z}/2$ that shows up in the sequence $1 \to SO(n) \to O(n) \to \mathbb{Z}/2 \to 1$. May 3 comment An alternative description of the first Stiefel-Whitney class What definition(s) of $w_1$ do you know? If you have been studying Milnor-Stasheff, take a look at the obstruction theory chapter. May 1 answered Question about symplectic tranformations May 1 comment An other question about Theorem 3.1 from Morse theory by Milnor May 1 comment An other question about Theorem 3.1 from Morse theory by Milnor Do you understand what $\phi_t$ is? Apr 30 comment 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 $t$. Apr 30 comment 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. Apr 30 comment 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? Apr 30 answered Implicit Function Theorem and Rank Theorem Misunderstandings. Apr 30 comment 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.) Apr 27 answered Non-degenerate solutions to constant Hamiltonian flow