Sam Lisi
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 Sep 11 comment Taking Differential Topology concurrently with Analysis Nils's answer is the correct one. IMO, you really need to understand the implicit/inverse function theorem, but the professor is the best person to give you advice on what you should do. Sep 10 answered Interpretation of Multilinear maps as tensors Sep 7 comment $\{(x,y)\!\in\!\mathbb{B}^n; -\varepsilon\leq-\|x\|^2\!+\!\|y\|^2\leq\varepsilon\}\approx\mathbb{B}^k\!\times\!\mathbb{B}^{n-k}$ I was about to write a similar explanation, but was hesitating b/c of this corner smoothing issue. I don't know if you've noticed, Leon and LVK, but most topology books waive their hands at smooth handle attaching. It's not hard (as reading Max's explanation will make clear), but writing down every detail becomes a headache. Sep 4 answered The connection in terms of local trivialization Sep 2 comment The connection in terms of local trivialization I'll try to write more later, but the key observation is that the map from $\pi^*E \to TE$ has image equal to the vertical tangent space (i.e. $\ker d\pi \subset TE$). For $A$ to split the sequence means exactly what you think: it is a projection to the vertical tangent space along what will now be defined to be the horizontal subspace. Let $\sigma$ be a section of $E$. The wiki definition has $\nabla \sigma \colon TX \to E$. Yours allows you to take $d\sigma \colon TX \to TE$ and then compose with $A$. IMO, wikipedia's more convenient for vector bundles, Taubes's generalizes better. Aug 31 answered What am I losing if I decide to perform all math by computer? Aug 30 comment What am I losing if I decide to perform all math by computer? I don't understand the question. What do you mean by "solve mathematical problems everyday"? what kind of problems? what constitutes a solution? Aug 20 comment existence of nonlinear second order ODE boundary value problem It's not clear to me how standard existence results apply to this boundary value problem. For instance, if you allow the constants to have arbitrary sign, the boundary value problem has completely different features. Can you please clarify what you had in mind? Aug 20 answered existence of nonlinear second order ODE boundary value problem Aug 9 awarded Necromancer Jul 17 comment understanding this differential operator on a tensor product Question 1 is clear once you expand $z = x+iy$ and $\bar z = x - iy$. Then, the metric you wrote down becomes $h ( dx^2 + dy^2 - i dx \wedge dy)$, which is precisely $h$ times the standard Hermitian metric. The real part of this is a Riemannian metric. I think that's where your confusion is from. Jul 16 comment A “correct” hierarchical scoring scheme? @PhD: you use "correct" twice in your description of the properties you want your scoring scheme to have. What does "correct" mean? Another way of wording this question: if I gave you a different scoring scheme from the one you proposed, what concrete things about it would enable you to determine that it was better. Jun 26 comment Why do mathematicians use single-letter variables? however, this fantastic formulation of the chain rule is hilarious. Jun 26 comment Why do mathematicians use single-letter variables? This is a great answer to the question. Jun 21 comment Riemannian metric making a given function harmonic This is essentially the idea I described in my comment above. I guess you have to choose $\alpha$ preserved by the monodromy? I'd like some more details if you have figured them out. Jun 19 comment Riemannian metric making a given function harmonic Thank you, Willie and Leonid, for confirming my guess and for the reference. Jun 18 comment Riemannian metric making a given function harmonic I'm not sure at the moment of how to get this to work in general, but if $f$ is a surjective submersion (what I called a geometer's fibration), then $M$ locally looks like a product, $I \times S$, where $S$ is a level set of $f$, and locally we have $(t, p) \mapsto t$. In this local picture, it suffices to have a metric for which the interval is orthogonal to $S$ and $\partial_t$ has constant length. More generally, I would then want to construct the metric fibrewise, so the fibres are orthogonal to $\nabla f$. I haven't thought about this enough to complete the idea. Jun 18 comment Riemannian metric making a given function harmonic @WillieWong: I suspect that what Kutluhan and Taubes mean is that the closed 1-form $df$ is also co-closed (i.e. $\star df = 0$). My best guess is that this is the same as asking for $f$ to be a harmonic map $M \to S^1$, but I am not sure. (While $f$ is manifold valued, you can think of $df$ as being an honest $1$-form by identifying $T S^1 = \mathbb{R} \times S^1$.) Jun 18 comment Riemannian metric making a given function harmonic when you say it fibres over the circle, do you have a geometer's fibration or a topologist's fibration in mind? i.e. is it a fibre bundle with diffeomorphic fibres (equivalently, $df$ of maximal rank)? Jun 16 awarded Revival