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This is about something that I found intriguing while computing some geometric identities for the cylinder of radius $r$ which we denote by $C = \{(r\cos(\theta),r\sin(\theta),z) : \theta \in [0,2\pi), z \in \mathbb{R}\}$. We see that each point in the cylinder can be identified with a tuple $(\theta,z)$ which we take as coordinates.

A metric $g$ for $C$ can be induced by pulling back the Riemannian metric via the inclusion $\iota : C \rightarrow \mathbb{R}^3$. Then we see that $$g_C = \iota^*(g) = d(r\cos(\theta))^2 + d(r\sin(\theta))^2 + d(z)^2 = r^2d\theta^2 + dz^2.$$Then, $$g_C = \begin{pmatrix}r^2 & 0 \\ 0 & 1\end{pmatrix},$$so that $$\hat{g} = \begin{pmatrix} 1& 0 \\ 0&1/r^2\end{pmatrix}.$$From this we compute the gradient of $f : C \rightarrow \mathbb{R}$, $$\nabla f = (df)^\sharp = g^{ij} \frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^j} = \frac{\partial f}{\partial z}\frac{\partial}{\partial z}+\frac{1}{r^2}\frac{\partial f}{\partial \theta}\frac{\partial}{\partial \theta}.$$

From this we can also see the gradient in polar coordinates since $C \rightarrow \mathbb{R}^2 - \{0\} \cong S^1$ by $(\theta,z) \mapsto (\theta,R)$. Changing $z$ with $R$ gives the gradient in polar coordinates!

The Laplace operator for the cylinder reads $$ \Delta = -\frac{1}{\sqrt{\det{g}}}\frac{\partial}{\partial x^i}\left(g^{ij}\sqrt{\det{g}}\frac{\partial}{\partial x^j}\right) = -\frac{1}{r}\frac{\partial}{\partial \theta}\left(r\frac{\partial}{\partial \theta}\right) - \frac{1}{r}\frac{\partial}{\partial z}\left(\frac{1}{r}\frac{\partial}{\partial z}\right) = -\frac{\partial^2}{\partial \theta^2} - \frac{1}{r^2}\frac{\partial^2}{\partial z^2}.$$

Or is it that the coordinates change for the inverse matrix? i.e. $\hat{g}^{ij}$ considers first the $j$ coordinate? i.e. it should instead be $$-\frac{\partial^2}{\partial \theta^2} - \frac{1}{r^2}\frac{\partial^2}{\partial z^2} \rightarrow -\frac{\partial^2}{\partial z^2} - \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}.$$

Another question is that if this is in fact correct and if there's something more happening under the hood for the gradient to coincide with the Laplacian (up to a minus sign)?

Thanks in advance.

user57
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  • This is a bit confusing. On the cylinder $r$ is constant. So the metric is essentially the Euclidean metric - now for the coordinates $\theta$ and $z,.$ There is no curvature. Regarding your observation you make in bold letters: This looks to me like a formal coincidence. In polar coordinates for $\mathbb R^2$ the radius $r$ is a coordinate, unlike it is for the cylinder. – Kurt G. Dec 11 '23 at 17:17
  • Actually. The gradient in polar coordinates is $(\partial_r,\frac1{\color{red}r}\partial_\theta)$ not $\frac1{r^2},.$ And this is only true when we normalize the basis vectors to have unit lenght w.r.t the metric in polar coordinates. – Kurt G. Dec 11 '23 at 17:22
  • @KurtG. Yes, the notation made it confusing and I also got confused by it. What we would get according with the transformation $(\theta,z) \mapsto (\theta,R)$ is $\partial^R + 1/r^2\partial_\theta$. I will edit the map in the post. – user57 Dec 11 '23 at 17:39
  • @KurtG. According with Lee's book, the gradient in polar coordinates does have the factor $1/r^2$ on the $\theta$ coordinate. You can obtain this in exactly the same procedure. – user57 Dec 11 '23 at 17:41
  • That's correct. I forgot about that. The factor $\frac 1{r^2}$ appears when we express the gradient in the holonomic basis (aka coordinate basis $\partial_r,\partial_\theta,)$. The transformation $(\theta,z)\mapsto (\theta,r)$ is a formal coincidence. In the $\theta,z$ world of the cylinder $r$ is not a coordinate. It is merely a constant parameter of the cylinder. Then you identify that with $z$ and get an expression we are familiar with. Nothing special happening under the hood I think. – Kurt G. Dec 11 '23 at 17:48
  • @KurtG. Alright. What about the Laplacian, do you verify the result? – user57 Dec 11 '23 at 18:17
  • Why is there a minus sign? As I said: the metric on the cylinder is essentially the Euclidean metric tweaked by a constant. – Kurt G. Dec 11 '23 at 18:20
  • @KurtG. Indeed, then yes this must be correct. The minus is the convention in Lee's book. Thank you for your answer! – user57 Dec 11 '23 at 18:25

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