Recent content by Pond Dragon

One word: practice. Find a good book, and do exercises. As many as you can find time for.

Here's a picture:

I think you are referring to their universal covering groups...?

Yes, but I clearly spoke about looking for a Riemannian metric on \mathbb{R} such that the isometry group is trivial. Thus, I am clearly not talking about the usual metric on \mathbb{R}.

I have figured it out. The only Riemannian automorphism (or endo-isometry or...better terminology) of (\mathbb{R},(v_p,w_p)\mapsto e^{2p}v_pw_p) is the identity. Thus, the isometry group is trivial.

See Theorema Egregium. How does this relate?

First, yes. Sorry I misinterpreted; that was my fault. It does go both ways, you'd just have to take the inverse of the diffeomorphism. Being isometric is an equivalence relation, after all! Not just homeomorphic, though. Remember that an isometry must be a diffeomorphism.

Not quite. We're giving M the pullback metric, so the original metric that we pull back from is the one on N.

Instead of continuing to make snide comments, perhaps you could work to uphold PF's values by actually explaining? Note that we are talking about Riemannian manifolds with the manifold being \mathbb{R} and the metric being nonstandard. A map \varphi:(M,\mathrm{g})\to (N,\mathrm{g'}) is an...

This is precisely what I don't follow. (0,+\infty)\cong\mathbb{R} as smooth manifolds and, using a diffeomorphism such as \varphi:\mathbb{R}\to(0,+\infty),~t\mapsto e^t, (\mathbb{R},\phi^*\mathrm{g})\cong((0,+\infty),\mathrm{g}) as Riemannian manifolds. Don't isometric manifolds have isomorphic...

So, you don't know how to compute the product? It looks like you could just apply definitions.

Particularly beautiful? How about de Rham's theorem?

Is B the imaginary unit? Your question is unclear. I am particularly confused by your conjugation by "e^{-B\frac{\pi}{2}}," since I'm not sure how it is acting. Would you please clarify?

In the following stackexchange thread, the answerer says that there is a Riemannian metric on \mathbb{R} such that the isometry group is trivial. http://math.stackexchange.com/questions/492892/isometry-group-of-a-manifold This does not seem correct to me, and I cannot follow what he is...

Perhaps you can give us more information about what you're looking for?