Sam Lisi
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 Apr13 comment 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. Apr13 comment 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. Apr11 comment Can the chain rule be relaxed to allow one of the functions to not be defined on an open set? Apr11 comment 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. Apr11 comment 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. Apr11 comment 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. Apr11 comment Symplectic geometry as a prerequisite 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. Apr8 comment Why is Cartan Formula just an avatar of Leibniz rule? Thank you for the bounty. I am grateful to you for the video. Apr8 comment Vector fields as section of tangent bundle and what is $T_xM$ for you? I think this is what jerrysciencemath wanted to know. Apr8 answered Structure on manifolds Apr8 comment Structure on manifolds I don't understand what your second question means. Can you give some more details of what you are reading and/or looking to have clarified? Apr7 comment Why is Cartan Formula just an avatar of Leibniz rule? @DamienL: I've written some details. I hope this helps explain that identity. Apr7 revised Why is Cartan Formula just an avatar of Leibniz rule? added a proof of the key identity Apr4 comment Floquet's Theory, Hills Equation Have you looked at the wikipedia page? en.wikipedia.org/wiki/Floquet_multiplier (Of course, to apply what they write, you will need to convert your second order equation to a first order system.) Apr4 answered Why is Cartan Formula just an avatar of Leibniz rule? Apr1 answered Algebraic interpretation of Lyapunov functions Mar29 comment Why is Cartan Formula just an avatar of Leibniz rule? Thanks for the link to the video! Also, what Martin said is the answer. Mar29 comment What is the sense of the equality sign =? Very nice explanation. I can go delete mine now. Mar27 comment Winding number of vector field on surface It's not more general... what I described above really uses the fact that the structure group of the tangent bundle to an oriented surface can be reduced to U(1). This is not true for a general manifold. In some sense, what I describe can be generalized to a discussion of Maslov classes of Lagrangians in symplectic manifolds... but that's much fancier than what you are interested in. Mar26 answered Winding number of vector field on surface