It's 5:00 now, I'll see whether after sleeping I can get my brain clearer on this. I am interested in the results of exercise 5 and 6 and I also believe it will be very important for the later sections
Very few questions can not be answered by searching the web, and if repeat questions are not allowed. The site will slowly move from a forum to a data base. Is the site concerned about this?
@0celo7 If I had accepted that, I'd have to accept "let $(M, (U_\alpha, \varphi_\alpha), (TM, M, T_xM \cong \Bbb R^k, GL_n(\Bbb R^k)), (J:TM \to TM, J^2 = -I))$ be an almost complex manifold"
@ACuriousMind Good call on the protection of the sunset question.
I deleted the non-answers out of hand, but left the others to be buried in a hail of downvotes and deleted by community action.
@0celo7 It is very hard to make jokes that are both dry and subtle work over a text channel. Dry and over the top can work, as can subtle but pointed. Trying to have both adjectives at the same time is asking for trouble.
Heck, it is tough to make the combination work in person, when you get right down to it.
I am interested in knowing the degree of curvature of the earth's
surface. I am attempting to find a method which would allow me to
conceptualize and calculate the making of such a depiction on paper.
That is to say, if the earth curves from point A to point B, points
exactly 10 kilometers ap...
Isn't calculating e.g. how far one can see from the top of a ship until the curvature blocks your sight a completely standard exercise for applying the Pythagorean theorem?
I'm not sure if I am impressed by somebody gaming the system just to keep posting nonsensical answers. I guess he/she actually believes in what he/she is saying enough to want to keep saying it
The "young artist" bit sounds intriguing. I bet whatever help is given, if any, will end up as a diagram in one of the youtube videos or facebook group
A question was asked by someone in March 2015. It had 4 answers.
I added a (fifth) answer on 26th Sep, 2016 (18 months later) and the question was put on hold the next day.
Can someone please explain if this sounds normal, or my answer had a reaction.
The question is Entanglement, real or ...
Like, I definitely wouldn't have taken anything from that dude in my school who put a pack of apple juice on the heating and left it there for weeks to "ferment" it.
@ACuriousMind Ok, I am thoroughly confused by associated bundles. So we have the homeo $\pi^{-1}(U)\approx U\times G$ from the principal bundle. How the heck do we get $\pi^{-1}_E(U)\approx U\times F$ in the associated bundles?
Everyone pretty much says this is trivial, what am I missing?
@0celo7 Well, you define the projection $\pi_V : P\times_G V\to M$ from the associated bundle just by $\pi_V(p,v) = \pi(p)$ for $\pi : P\to M$ the projection of the principal bundle and checking it's well-defined on equivalence classes, right? On a trivializing $U$, you have that $\pi_V^{-1}(U) = \pi^{-1}(U)\times V/{\sim} = U\times V$ just by thinking about the equivalence relation. Since both $\pi^{-1}$ and quotienting out the relation are continuous, so is $\pi_V^{-1}$.
@DanielSank Interesting. Did you have any prior involvement with them or did you just "walk up" to them and suggest that?
Well, on $P\times V$, the relation is $(p,v)\sim (q,w) \iff \exists g\in G: q = pg \land w = \rho(g)^{-1}v$. On $U\times G\times V$, this means that $(x,g,v)\sim (y,h,w)\iff \exists k\in G: x=y\land h = gk \land w = \rho(k)^{-1}v$.
Since the action is free and transitive, to each $(x,g,v)$, $(x,1,\rho(g)v)$ is the only representant of its class with $(x,1)$ in the first two entries, and so specifying $x$ and $w\in V$ suffices to fix the equivalence class and no two such choices belong to the same class, hence the quotient is $U\times V$.
The representation of $G$ on $V$. I'm thinking of associated vector bundles, but you can replace it by a more general action on the fiber without any problems, I think