Sam Lisi
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 Mar 10 comment symplectic surfaces in 4-manifolds Off the top of my head, I would guess that if the $\omega$ area of the homology class is positive, you can probably find a symplectic representative, but I haven't thought enough about this. Let me know if this modification of your question is of interest. Feb 18 comment Non-commutative symplectic geometry Do you have a reference for non-commutative symplectic geometry? I've never heard of this, and am curious. Dec 1 comment Is this some kind of trick question? Submanifold “proof” What is your definition of submanifold? This is indeed a simple exercise once you have stated the correct calculus theorem. Apr 6 comment Symplectic submanifolds in $\mathbb{R}^{4}$ @studiosus: your example isn't relevant to my claim because the punctured surface isn't a manifold with boundary. However, I think you are right that the actions on the boundary are relevant. I'll adjust my answer accordingly. Thank you. Apr 5 comment Symplectic submanifolds in $\mathbb{R}^{4}$ @studiosus: Thank you for the corrected reference. In this case, since the surfaces are compact with boundary, I think it's possible to make some minor modifications to Moser so that it works. In particular, I believe you can modify Moser's proof by hand to make the flow of the vector field he constructs be defined up to time 1. Since that's the piece where compactness is necessary, I think everything else should carry through. Apr 3 comment Symplectic submanifolds in $\mathbb{R}^{4}$ you didn't mention that you had cross-posted this to MO! Next time, be sure to mention it when you do, so we don't waste time duplicating effort. Mar 15 comment Is $f(x)+\sum_{p,i=1,…,m}\lambda_{p,i}x_{p,i}(x)$ globally defined? Where is this question coming from? This seems pretty unmotivated. Feb 17 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. Feb 16 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? Feb 16 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 $\partial_{x_i}$, $\partial_{y_i}$. Sep 24 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 $U' \setminus U$. Sep 24 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. Sep 18 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. Jun 4 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.) May 31 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? May 31 comment Horn and spindle tori Can you define what the spindle torus and horn torus are? May 29 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. May 24 comment Does this IVP have a unique solution for all $x \in \mathbb R$ @Tunococ: if you've cited a local uniqueness theorem, you've done all the necessary work to answer the question (see Artem's solution below). Ron, if you look at the example I give in my answer (with $y' = y^{2/3}$), you will see why I am somewhat concerned about how to provide a complete argument when you cross the singular points in your separation of variables. I'd think it was really cool if you could indeed provide the missing details in your argument to show global uniqueness -- if it worked, it would be a lower tech solution to the problem than any I know. May 23 comment Does this IVP have a unique solution for all $x \in \mathbb R$ OK, perhaps this is a question of taste -- nothing you wrote is false. I do, however, feel this is misleading because the uniqueness of global solutions to this IVP has nothing to do with its separability. This is why I asked above what theorems the OP knew about existence/uniqueness. May 23 comment Is there even a point in defining the notion of a 'metric' (as opposed to a metric space), etc.? To tack an example on to this idea, quite often one gets a set with some structure and then want to show it has even more structure. For instance, I may have a set of solutions to some equation, but then, exploiting something about the equation, I want to show that it is also a metric space. I guess I could say, "I will now find a metric space structure" on this set, but it is nicer to say "I will now find a metric". Of course, if I wanted to show that this set had a group structure... I'd have to say "group structure".