In linear algebra, a Vandermonde matrix, named after Alexandre-Théophile Vandermonde, is a matrix with the terms of a geometric progression in each row, i.e., an m × n matrix
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Gaussian curvature is the sectional curvature of a surface @Akiva. Sectional curvature tells you the curvature of a "2 dim slice" of a higher dim manifold
I guess they could also mean scalar curvature depending on the context, that strikes me as the "simplest" of the big curvatures people talk about in beginning Riemannian stuff
You can study constant scalar curvature things but those are much less constrained. Every closed manifold has a constant -1 scalar curvature metric except S^1, S^2, RP^2, T^2, K^2
I asked him if there were similar analogues to how we could get all these nice constraints from looking at bounds on Ricci for the other curvatures, and he remarked to me that the theorems you get from restricting scalar don't determine as much
Isn't it nontrivial that every hyperbolic 3-fold is a quotient of H^3? I guess you make the pi_1 act on H^3 by Milnor-Svarc and then nuke it because hyperbolic 3-manifolds are classified by fundamental groups (???) by geometrization (???) or something
I mean like I've seen the definition of spin structure, but what I meant by finding out what it actually means I mean viscerally by doing problems so that the definition isnt just words
I wasn't trying to claim I'm unsure how to prove this above. Like Akiva says you just pass to the universal cover and use the classification of complete simply conn
Yes. That's covered in every first differential geometry course. You'll do it soonish :)
@EricSilva Are you aware of Kazdan-Warner's results?
@BalarkaSen I feel like, if you prove that epsilon-balls in your thing "look like" epsilon-balls in H^3, you could use simple connectedness to show that your thing is H^3
I'm not buying that you can turn this into a proof.
1) Let $\Sigma$ be a closed surface and $f$ a smooth function. Then there is a metric with curvature f iff it fits with Gauss-Bonnet: It has to be somewhere positive if it's a sphere, everywhere zero or somewhere pos and somewhere
I think I had a question to parse out of Akiva's attempt but I am failing
like, $M$ and $N$ be two manifolds such that for any pair $p, q$ in $M, N$ resp there exists balls $B_p$ and $B_q$ around which are isometric, does that say anything at all about $M$ and $N$?
The really interesting thing is that scalar curvature is unconstrained in higher dimensions by theorems like GB. In fact KW prove that manifolds come in 3 classes. 1) All functions are sc functions. 2) The sc functions are those that take on the value 0 somewhere. 3) they're the functions that are negative somewhere
2) corresponds to having a zero but not positive scalar curvature metric
3) corresponds to having no zero scalar curvature metric
No, none. You are describing space forms, in this case the hyperbolic plane and 3-space. You might want to look at Cheeger and Ebin, Comparison Theorems in Riemannian Geometry.
Right, Theorem 1.37 on page 41, simply connected manifolds of the same dimension and constant (sectional) curvature $K...
I have a question regarding finding extrema of the following function: $f(x,y)=(y −1)(x^2 − y)^2$. Whilst searching for stationary points, I also stumble upon x^2=y satisfying D_x f = 0 =D_y f. How to interpreted this?
Nope, Semiclassic. Only my getting discouraged by the number of people asking undergrad diff geo without showing effort — and by their apparently using texts that don't do stuff they should do (or the student isn't reading the text).
No, none. You are describing space forms, in this case the hyperbolic plane and 3-space. You might want to look at Cheeger and Ebin, Comparison Theorems in Riemannian Geometry.
Right, Theorem 1.37 on page 41, simply connected manifolds of the same dimension and constant (sectional) curvature $K...
