mathscinet? you need to access it from an institution, you can get temporary access for 30 days on your own machien if you access it from an institution and click some buttons ( I forgot which)
@TedShifrin given the fact that he is a von Neumann algebra guy who invented the Jones polynomial I already had the impression that he was super interdisciplinary, so a book on pointset topology was impressive but not completely out of left field
I harassed someone on main by suggesting that his answer sounded like a ChatGPT answer. He got more and more defensive and eventually flagged my comment. (By that point, I just deleted.)
@TedShifrin I don't have any good idea unfortunately. But the condition closed under convex combinations, so if you have solutions locally you can use a partition of unity to build global solutions
I don't see how to make that work. That was the first thing I thought of and started to type the answer, but writing $X=\sum\phi_k \frac{\partial}{\partial x^3_k}$ won't work (because it interacts badly with the "other" local forms of $\alpha$).
I wanted to find $X$ globally by looking at the Lie derivative of $\alpha$, but I don't see a nice invariance property of $\alpha$ since it is non-integrable.
I've been using a Riemannian metric, that I've been pushed to use, using the concept of an "area," because it's been easier to work with. I've run into a very daunting task, and that is to express that notion of distance on a surface that's not flat anymore.
And I think I need to read about differential 2-forms to solve this
@TedShifrin okay...I am just slightly confused because I calculated a Riemannian 1d metric, using integration. There's something called a Fisher metric that gives you a Riemannian metric for a given distribution, so that's what I did
it's a definite integral
just to briefly describe, the input is a distribution of say 1-parameter, you integrate over the support of the variable and so clearly you get a function of the parameter. That parameter, say $\theta \in \Bbb R$ is the new coordinate on the 1-manifold and the result of the integration is the Riemannian metric.
There seems to be a difference of opinion regarding the definition of the limit of a function. It seems like $0<|x-a|<\delta$ and $|x-a|<\delta$ are both used in some books. Which definition do you use or think is more appropriate? I am confused as to why the latter definition (with $|x-a|<\delta$) even makes sense.
@sunny I don't have any of my books handy, but I would define $\lim_{x\to a} f(x) = L$ to mean "for any $\varepsilon > 0$, there exists some $\delta > 0$ such that $0 < |x-a|<\delta$ implies that $|f(x) - L| < \varepsilon$."
I was suggesting that you are leaving out a lot of context from the authors who write $|x-a| < \delta$.
It is possible that they are leaving something out, but it is also possible that they are explicitly saying $x \ne a$ somewhere else in the definition.
to me it's cool that the 1-parameter family of metrics should satisfy that linear third order differential equation, as well as the distribution itself changing according to a diffusion equation. It's like Ricci flow but lower dimensions and linear instead of nonlinear (I guess it's not really like Ricci flow but it is type of geometric flow).
not coming from an AG background it just baffles me how someone could know enough about AG to be curious about that, and but not know enough to have some idea of the answer
didn't there used to be some true algebraic geometers on this chat? but it's been a while
ted cosplays as one, maybe he will have good examples
