Sam Lisi
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 Feb17 comment Derivative of a function involving inverse of a matrix-lifting a diffeomorphism to a symplectomorphism I haven't given this enough thought for this to count as an answer... in particular, I'm ignoring the specific details of your construction of $f$. The general fact is that a diffeomorphism $f \colon X \to X$ induces a symplectic diffeomorphism of $T^*X \to T^*X$, and $T^*X$ has a canonical symplectic structure. If all is right in the world, you have merely restated this fact in a special case. Feb17 answered Second Hirzebruch surface as Delzant space associated to trapezoid Feb16 comment On the definition/notation for pseudoholomorphic curves Now, consider $u \colon D \to M$. Look at some point $z \in D$, and let $p = u(z) \in M$. Then, we take a coordinate chart around $p$ as in my previous comment. We can then think of $u \colon D \to \mathbb{R}^{2n}$. The differential $du \colon T_z D \to T_{u(z)} \mathbb{R}^{2n}$. Now, $J$ is an endomorphism of $T_{u(z)}\mathbb{R}^{2n}$, and this matrix is then $J( u(z ))$. Does this make sense? Feb16 comment On the definition/notation for pseudoholomorphic curves A key point: $M$ is not a complex manifold. This means that the coordinate chart we choose is just a real coordinate chart. Thus, we should think of $z = (x_1, y_1, x_2, y_2, \dots, x_n, y_n)$. At each point in this coordinate chart, the tangent space is spanned by $\partial_{x_i}, \partial_{y_i}$. This gives a trivialization of the tangent space over each point in our coordinate chart. Now, finally, $J(x_1, y_1, \dots, x_n, y_n)$ is a matrix that tells you what $J$ does to the local vector fields $\partial_{x_i}$, $\partial_{y_i}$. Feb14 answered On the definition/notation for pseudoholomorphic curves Feb10 answered About symplectic embedding Feb10 answered Why is the dividing set nonempty when a convex surface has Legendrian boundary? Sep24 comment Isotopy between two open disks on a surface You're definitely on the right track. Not sure if this is the kind of hint you are looking for, but I would consider also a $U'$ that is a $2\epsilon$ neighbourhood of $P$. Then, you can find an isotopy that does nothing on $U$ and sends the annulus $A \setminus U$ to $U' \setminus U$. Sep24 revised What is nonhomogeneous linear mapping? added more text from Milnor, making the quotation's context clearer. Sep24 suggested approved edit on What is nonhomogeneous linear mapping? Sep24 comment What is nonhomogeneous linear mapping? @studiosus: can you put that as an answer? The continuation of the text makes this clearer. I have taken the liberty of editing it in to the question, though it will wait for peer review. Sep18 comment Natural diffeomorphism between $T\mathbb{S}^n\times \mathbb{R}$ and $\mathbb{S}^n\times\mathbb{R}^{n+1}$ @OhMyGod: yes, that's the idea. The sphere sits inside of $\mathbb{R}^{n+1}$ (as the unit sphere, say). The tangent bundle of $\mathbb{R}^{n+1}$ (which is trivial), restricted to the sphere, can also be seen as the tangent bundle of the sphere summed with its normal bundle. Sep17 answered is any hamiltonian system with just one degree of freedom completely integrable? Sep16 answered Can the system $\partial_x f(x,y) = \dot{y}$, $\partial_y f(x,y) = \dot{x}$ be related to some Hamiltonian system? Sep16 answered Understanding the definition and meaning of cotangent space Jul25 answered symplectic strucutre Jun4 comment Prove that the circle $S^1$ is not the boundary of any compact manifold with boundary in $\mathbb R^2-{(0,0)}$ This point is clear from the question already, but I want to emphasize that when you say "the circle $S^1$", you mean the unit circle in $\mathbb R^2$. There are plenty of other embedded circles in $\mathbb R^2 \setminus \{ 0 \}$ that do bound disks (e.g. take the boundary of the disk of radius 2 centred at $(0, 500)$). In this problem, there is no ambiguity, but it is often useful to keep track of what you mean exactly. (In particular, after you have figured this out, a good exercise is to see the difference between my circle and your circle.) May31 comment Horn and spindle tori This "slice" idea is not the right approach. You can easily construct a submanifold of $\mathbb{R}^3$ so that a coordinate slice (e.g. $z = const$) is not a submanifold for some suitable choice of constant. (Simple example: take a standard torus and choose a slice that gives you a figure 8). In your horn and spindle tori, you have points of self-intersection. Look at the neighbourhood of one of them. Does it look like a neighbourhood in a Euclidean space? May31 comment Horn and spindle tori Can you define what the spindle torus and horn torus are? May29 comment Chern class of tautological line bundle After you have done the computation this way, I think it would be instructive to see how Milnor-Stasheff do it. I learned a lot from working through that section of their book.