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I have a question and I'm not exactly sure if I'm on the right track. It isn't homework, just a curiosity I'm following:

Consider a right circular cylinder with fixed radius of 1. This is parameterized by $\phi = [cos(\theta), sin(\theta), z] = \phi(\theta,z)$. I want to define the metric (tensor?) on the cylinder as a manifold. I believe I'm doing something wrong here but I can't figure out what. If I understand it correctly, the metric is:

$g = J^T J$

And the Jacobian is:

$$ \left[ \matrix { -sin(\theta)&0 \\ cos(\theta)&0 \\ 0&1 } \right] $$

Then the metric is just $g = I_{2x2}$. However, from another answer, it seems that this should be a 3x3 matrix. Does it have something to do with the fact that I want the metric on the manifold itself, without embedding it into Euclidean space? Or am I just totally off base?

Michael Stachowsky
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1 Answers1

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Just apply the formula for the metric and substitute $r=1$: $$ J=\frac {\partial (x,y,z)}{\partial (r,\theta,z)}=\begin{pmatrix} \cos {\theta} & \sin {\theta} & 0 \\ -r\sin {\theta} & r\cos {\theta} & 0 \\ 0 & 0 & 1 \end {pmatrix} $$ Then you will get the euclidean metric: $$ g=\begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {pmatrix} $$

SrJaimito
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  • I'm still a bit uncertain. If I understand it correctly, a curve defined on my cylinder is parameterized by only two variables, $\theta$ and $z$. For instance, a helix on the cylinder may be given as $r(\theta(t),z(t)) = [2\pi t, t]$. The 3x3 metric won't work here. I'm not sure what mistake I've made, though, since I agree with you as well... – Michael Stachowsky May 06 '16 at 15:58
  • I should also mention that using my 2x2 metric and determining the arc length of that helix on the interval $t \in [0,1]$, I do in fact get the correct answer of $4\pi^2 + 1$, which is the identical answer to considering the helix in Euclidean space $r(t) = [cos(2 \pi t), sin(2 \pi t), t]$. So perhaps it does matter that I've chosen not to embed the cylinder in $R^3$? – Michael Stachowsky May 06 '16 at 16:01