@Ryan Let $X$ be a normed space. $C\subseteq X$ is called a set of uniqueness if for every $x\in X$ there is at most one $c\in C$ with $d(x,C)=d(x,c)$. $C\subseteq X$ is called a set of existence if for every $x\in X$ there is some $c\in C$ with $d(x,c)=d(x,C)$. $C\subseteq X$ is called a Chebyshev set if it is bot a set of uniqueness and a set of existence.
Let $X$ be a normed space, then $X$ is strictly convex iff every nonempty convex set in $X$ is a set of uniqueness iff every nonempty closed convex set in $X$ is a set of uniqueness
Sup there, I invite all of you to math chat with built-in LaTeX to talk about various branch of mathematics, atm we're talkin here about graph theory and doin some proofs https://hack.chat/?math It's completely anonymous chat without registration, come and visit
@Secret what are your thoughts on consciousness? Do you think matter and mind are equivalent representations of reality and there is some isomorphism that links the two representations of reality
You just want to take a parametrisation: $$\Phi(\theta,\phi)=(\cos(\theta)\sin(\phi),\sin(\theta)\sin(\phi),\cos(\phi)),$$ and then you want to compute $\Phi^*(d\omega)$?
@RyanUnger Not with that now I guess. I just experienced that computing the christoffel symbols and then the components of the levi-civita connection, and then the riemann curvature tensor in that order isn't so bad
As I said, @F.White, I don't know what @anakhro is looking for. If we have the $1$-form as the restriction of a $1$-form on $\Bbb R^3$ in the first place, then we're just working in the situation where $x\,dx+y\,dy+z\,dz=0$ and this relation can allow us to simplify.
I have an area form $\Omega$ and a vector field $u$ on $S^2$ (everything written in Cartesian coordinates), and I want to compute $d(\iota_u\Omega)$. The part with $d$ of course doesn't work when I treat it just like the exterior derivative in R^3 (thanks Balarka) and so I need to either convert to spherical coordinates, or figure out a way to calculate the exterior derivative of $\iota_u\Omega$ on $S^2$ without this change.
Not to bombard you with questions Ted (although I am) If we let $S$ be the surface given by rotating the curve $(f(v),0,g(v))$ for $v_0<v<v_1$ around the $z$-axis, and we assume the parametrisation is an immersion: $$\Phi:U\subset \Bbb R^2\to S\subset \Bbb R^3,\qquad \Phi(u,v)=(f(v)\cos(u),f(v)\sin(u),g(v)),$$
where $U=\{(u,v)\in\Bbb R^2\mid u_0<u<u_1,\quad v_0<v<v_1\}$
Then one can deduce the pulled back metric on this manifold is:
$$g=f(v)^2du^2+(f'(v)^2+g'(v)^2)dv^2$$
is there a smart way to compute the Gaussian curvature?
Ahh I'll have to look at that, the only procedure I know at the moment is to compute it as sectional curvature: $$\frac{\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle}{\|\partial_1\wedge\partial_2\|}$$
I just computed by hand $R(\partial_1,\partial_2)\partial_2 = \nabla_1\nabla_2\partial_2 - \nabla_2\nabla_1\partial_2$ in terms of christoffel symbols, so I suppose implicitly I have it in terms of christoffel symbols
It's takes a little bit over a page though, which would suck in a short exam
I guess I probably wouldn't be given such an arbitrary surface though in an exam (hopefully :))
@TedShifrin I'd probably apply the vector field $u$ to $f(x,y,z) = x^2 + y^2 + z^2$. And yes, I am fairly certain that I calculated the interior product correctly.
OK. There's a nice formula, as I said, which you should check once in your life. (I always assigned that mess of algebra to my undergrad class to turn in.) It is $$K=-\frac1{2\sqrt{EG}}\Big((G/\sqrt{EG})_u+(E/\sqrt{EG})_v\Big),$$ where $E=g_{11}$ and $G=g_{22}$.
@TedShifrin I remember using this in one of my exams. The professor was shocked that I didn't take any of the two sheets he had given to find the Gaussian Curvature.
I gave my students the relevant formulas on their exams/final, because the point is not to memorize ridiculous formulas. If you know differential forms, everything is super easy, but the classical approach is full of formulas.
Yeah, it is. Sadly we have some people here who still believe in making us memorize. One had asked us to memorize all kinds of wonky probability distributions
Hi guys, Consider a function $f$ ($\mathbb{R}\to \mathbb{R}$) that has an Oblique asymptote in $y=2x-1$. I want to find the limit $\lim_{x\to\infty}\sin\left(\frac{2}{x}\right)f(x)$. How can I do it? I know from the Oblique asymptote definition that $lim_{x\to\infty} (f(x)-2x)=-1$, but how does it help me?
Quick question, what's the usually recommended text to learn about spectral sequences? I know of a few, Lecture Notes in Algebraic Topology by Kirk and Davis, Hatcher's incomplete SSAT, A User's Guide to Spectral Sequences and Weibel's Homological Algebra
Ahh I see, I think I read in Kirk and Davis' notes that you could also use them to give quick proofs of some big theorems, in fact, I think I remember seeing somewhere that you can prove the Thom isomorphism theorem with a certain spectral sequence
Requires more than spectral sequences technically speaking, you need the rationalization construction, but essentially spectral sequence arguments.
You also need Serre's theorem but that's also essentially a spectral sequence argument.
@Perturbative Ya ok but Thom isomorphism theorem is very easy to see by hand
if you're looking for lo-fi computations, a quick corollary of spectral sequences is the Gysin sequence, with which you can prove groups acting freely on spheres have periodic cohomology
They're slowly becoming essential to my toolbox, although I still need a lot of practice to get used to them
If $E$ is an orientable rank $k$ vector bundle on $M$ then the map $H^{n+k}_{vc}(E) \to H^n(M)$ from the vertically supported cohomology of $E$ to the base is given by evaluating on the fiberwise orientation class, basically
If you're working deRham, it's integration along the fibers
A disjoint number of things. I was reading about h-principles and stratified spaces but got sidetracked and started learning about localizations of spaces etc
Great. A diagonalizable (affine) group scheme over a field F is one whose representing Hopf algebra F[H], the group algebra over F for some abelian group H. (The comultiplication, inversion and identity maps are what you'd expect they are - I can elaborate if need be)
A beautiful fact is that there is an equivalence of categories between groups of MT over F and abelian groups with a continuous action (by homomorphisms) of Gal(F_sep/F)
@anakhro What does "K is orientable" even mean? A circle is an orientable manifold. I am saying the following is true: Every knot in M has trivial normal bundle $\iff$ M is an orientable manifold (regardless of dimension)
On the other hand, given an abelian group H with the corresponding T := Gal(F_sep/F) action, one takes the group scheme represented by the Hopf algebra (F_sep[H])^{T} (that is, take T-fixed points)
Lol I get that feeling. I've spent a lot of time trying to understand the technical details of general scheme theory, and only now after years am starting to feel like I understand the basics. It is a lot of baggage
Right, because given any oriented rank 2 bundle you can build an oriented manifold in which has a knot with exactly that normal bundle; just take the double of it's disk bundle
Hi , how do I show that z^4 -z^3 -4z +1 has no roots for |z|>=2 ? Sorry for no latex
I know about Rouche theorem and all of that stuff
Actually, I need to show that there are exactly three roots in the area { 1 < |z| < 2 }. But it is easy to show there is only 1 root in the { |z| <= 1 } and since there is 4 in total ...
@Ultradark I think they probably be representing some notions of reality but I have no reason to believe they are isomorphic (otherwise we would have detected ghosts and such with science already)
