Proving A 2-Dimensional Manifold

Asked May 02 '20 at 20:00

Active May 02 '20 at 20:20

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I am stuck on how to prove that this is a 2D manifold. I would really appreciate a stp-by-step explanation on showing how:

Let $\gamma = (0,1)\times(0,2\pi)\rightarrow\mathbb{R}^3$ be given by $\gamma(u,v) = (u\cos(v), u\sin(v), v)$

Let $M = \{\gamma(u,v) \in\mathbb{R}^3: 0<u<1, 0< v < 2\pi\}$

How do we prove that $M$ is a smooth, 2D manifold with the information?

edited May 02 '20 at 20:08

Angina Seng

asked May 02 '20 at 20:00

UnevenMango

You want to show that $\gamma$ is a chart and that it covers $M$ given $u$ and $v$ – IAmNoOne May 02 '20 at 20:15
Try to imgaine what the surface is and calculate the Jacobian – IAmNoOne May 02 '20 at 20:20
Then show $\gamma$ also admits an inverse. – IAmNoOne May 02 '20 at 20:28
@IAmNoOne so the jacobian will be a 2 column by 3 row, correct? – UnevenMango May 02 '20 at 20:39
@IAmNoOne why wouldn't it be a 2 by 3, as we only have two variables and 3 equations? – UnevenMango May 02 '20 at 21:07
@IAmNoOne and from here, is it correct to check to see if the ranks of the subminors are 0? – UnevenMango May 02 '20 at 21:11
Sorry I said it too fast. It will be 3 by 2. To check the rank, just note the columns will be linearly independent. You will need the cross product to show it is nonzero. – IAmNoOne May 02 '20 at 21:38
@IAmNoOne what if we are to find $T_{C}M$, the tangent space at c=(1,0,0,2)? How are we to do this if there are only two variables? – UnevenMango May 02 '20 at 22:36
A tangent plane requires only two vectors. – IAmNoOne May 02 '20 at 22:44

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