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Vishesh

♦  DG Reading Group

Room for reading group working through Lee's "Intro to Smooth ...
Vishesh
Jan 3, 2016 13:39
Am I actually doing something sensible here?
Vishesh
Jan 3, 2016 13:38
that way as $s$ varies over $\mathbb{R}$, I can get the required answer
Vishesh
Jan 3, 2016 13:37
and I need to show that for distinct $s$ and $t$, the charts induced are distinct
Vishesh
Jan 3, 2016 13:37
Is there a painfully obvious way to do this??Because I am stuck for the moment on how to argue that. I am thinking varying $s$ will give me new structures
Vishesh
Jan 3, 2016 13:24
?
Vishesh
Jan 3, 2016 13:24
So compose a chart with $F_s$
Vishesh
Jan 3, 2016 13:16
So induction seems viable
Vishesh
Jan 3, 2016 13:15
Obviously for $s=1$, that map is the identity
Vishesh
Jan 3, 2016 13:14
Well disappointingly, I really didnt have any kind of intuition for this problem. So I was actually trying to sort of why that hint props up, by reverse engineering. Not getting anywhere. Can you just motivate the hint??
Vishesh
Jan 3, 2016 13:09
The hint is to use the homeomorphism $F_s(x) = |x|^{s-1}x$ from $\mathbb{B}^n$ to itself for $s>0$.
Vishesh
Jan 3, 2016 13:06
There is a hint. The hint sort of came out of left field for me.
Vishesh
Jan 3, 2016 13:05
Oh sure. The Problem is to show that if a topological manifold of dimension $n\geq 1$ has a smooth structure then it has uncountably many distinct such structures.
Vishesh
Jan 3, 2016 13:03
@Huy Would you mind if I discuss it with you?
Vishesh
Jan 3, 2016 13:02
I had a doubt about one of the problems. 1-6
Vishesh
Jan 3, 2016 13:01
Oh alright. Its heartening to see you here regularly though :)
Vishesh
Jan 3, 2016 12:54
Have people already started on the second chapter??Or will we discuss the solutions to the problems in the first one before starting up on the next one?
Vishesh
Dec 31, 2015 07:29
@Huy Thanks...yeah Its just that today is the last day for the 1st chapter as per the schedule...so I was wondering whether skipping would be okay
Vishesh
Dec 31, 2015 07:18
Do manifolds with boundary figure heavily even in a first course in DG??I am working on the problems but I skipped the manifolds with boundary part, is that risky??
Vishesh
Dec 22, 2015 18:10
Well, where I come from shockingly I have never had either a course on manifolds or Algebraic topology, but the only course offered was Riemannian Geometry.
Vishesh
Dec 22, 2015 18:05
Well you noticed it, thats all that matters. Yeah practice is the key. I have the same issue. At any rate if you are slow, you will still have me for company.
Vishesh
Dec 22, 2015 17:34
Hmm... I guess I am not doing a good job of explaining what I understood. To me it the only reason they seem to mention Proposition 2 is to let you know that you are not in the Euclidean space. Sort of like " this time its ok, but next one may not be so simple, so better understand what actually was done here". Crude way of putting it, I know :)
Vishesh
Dec 22, 2015 17:29
Why $\dfrac{\partial}{\partial x_1}|_g$?
Vishesh
Dec 22, 2015 17:28
I dont think the representation is the issue. It is the fact that $GL(n, \mathbb{R})$ is not exactly the Euclidean space so there is a chart (just the inclusion map) and so Proposition 2 is in action allowing to write the derivative in terms of the basis vectors.
Vishesh
Dec 22, 2015 17:22
Aargh...my LaTeX i pretty shoddy. I am not able to delete my previous comment or edit it either
Vishesh
Dec 22, 2015 17:17
Well what I meant was that Proposition 2 allows you to say $\dfrac{d}{dt}|_{t=0}c = \Sigma_{i=1}^{n} \. c(0) \dfrac{partial}{partial x_i}|_{x=0}$.
Vishesh
Dec 22, 2015 17:04
@Huy...yeah force of habit using the double dollar signs...Sorry
Vishesh
Dec 22, 2015 17:02
The $\mathbb{R}$ -linearity is to take $g$ out and allow $\dfrac{d}{dt}$ to operate on $c$ explicitly and Proposition 2 seems to be mentioned only to make sure that we identify the fact that $GL(n)$ is a open submanifold and hence a chart has come into play in the background
Vishesh
Dec 22, 2015 16:47
@RobertCardona... just a trivial point to add to the discussion about the left multiplication differential. It seems to me that invoking Proposition 2 to write $$ g \dfrac{d}{dt}|_{t=0}(c) = gc'(0) $$ is just cosmetic.
Vishesh
Dec 22, 2015 15:55
its fine now....yeah my laptop...might be also due the fact that there are a lot of windows open, some with MathJax.
Vishesh
Dec 22, 2015 15:54
Chrome
Vishesh
Dec 22, 2015 15:52
Yeah it does...though the rendering is really slow, sure it is just my browser.
Vishesh
Dec 22, 2015 15:50
@ Huy Thanks a lot . I should have been more observant.
 

 Differential Geometry

Discussions about differential geometry in an informal spirit
Vishesh
Dec 31, 2015 09:40
You will have to tell me what is the new definition and the old one??If the new one is tangent vector as an operator and the old one as an arrow sticking out of the sphere, then a bit of imagination will help you see how they match. Use the charts on the sphere.
Vishesh
Dec 31, 2015 09:36
just take the example of a sphere or a cylinder, that will help
Vishesh
Dec 31, 2015 09:35
if you are interested in tangent vectors to the surface.
Vishesh
Dec 31, 2015 09:34
and the domain of that function is the manifold itself, so in your case, the domain of the smooth function is the surface itself not $\mathbb{R}^3$
Vishesh
Dec 31, 2015 09:33
yes
Vishesh
Dec 31, 2015 09:33
I meant LATEX in chat
Vishesh
Dec 31, 2015 09:32
To the right of your screen, there is a link to chat guidelines, read that.
Vishesh
Dec 31, 2015 09:30
Yes you can take the directional derivative for that function. Just type you equations between dollar signs, that will bring MathJax into play.
Vishesh
Dec 31, 2015 09:28
You mentioned Lee's book, go through the tangent vectors part again, you can see that the idea is that you cannot look at tangent vectors to a manifold as arrows as in Euclidean spaces because that automatically assumes that you are talking about an ambient space. Your surface is a submanifold, so the confusion, however when defining tangent vectors in general, it is better not to think of the surface as inside $\mathbb{R}^3$, but as a stand alone manifold.
Vishesh
Dec 31, 2015 09:20
so the domain is the surface
Vishesh
Dec 31, 2015 09:19
the function is not defined on $\mathbb{R}^3$, but on the surface itself
Vishesh
Dec 31, 2015 09:19
No, if you are looking to define tangent vectors for a surface in $\mathbb{R}^3$, then
Vishesh
Dec 31, 2015 09:18
Well in that case, yes a surface is a "special manifold". What I mean was that for Euclidean spaces the many definitions of tangent vectors all coincide, so it is a special case. But in general that does not happen, like for surfaces.
Vishesh
Dec 31, 2015 09:15
What exactly do you mean by "special manifold"?
Vishesh
Dec 31, 2015 09:14
I mean Euclidean spaces are the special manifolds
Vishesh
Dec 31, 2015 09:14
Can you just elaborate a bit??I did not understand what exactly is confusing you??From what I could gather looking at your question, I think the problem comes with thinking manifolds as embedded inside some larger Euclidean space, which distracts from the true definition of tangent vectors.

Yes, as you have stated that in case of Euclidean space, the tangent space and the manifold itself(i.e. $\mathbb{R^n$) coincide. But that is the special case, so in a way Euclidean manifold is the "special manifold".
Vishesh
Dec 31, 2015 09:12
I saw the question. You are not sure how to define a smooth function on an embedded surface in $\mathbb{R}^3$??
Vishesh
Dec 31, 2015 09:05
I don't know how much help I can be. But you can tell me the problem...