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I have the following problem: Find the canonical volume form of: $$i) R_f(a,b):={(f(z)cos(\phi),f(z)sin(\phi),z)|(\phi,z)\in\mathbb{R}\times(a,b)}$$ with $f\in C^{\infty}((a,b))$ and $f>0$ $$ii)W:={(rcos(\phi),rsin(\phi),\phi)|(\phi,r)\in \mathbb{R^2}}$$ $$iii) {(u,v,h(u,v)|(u,v)\in\mathbb{R^2}}$$ with $h\in C^{\infty}(\mathbb{R^2})$

We have the formula: The canonical volume formula $\omega_{M|U}$ for a riemanian manifold and the map (U,x) is: $$\omega_{M|U}=\sqrt{g}(dx^1\wedge,...,\wedge dx^n)$$ Let f(x,y) be a map of the manifold then $g_{ij}=<\partial xf,\partial yf>$. $g=det((g_{ij})_{i,j=1}^n)$

What exactly do I have to do? Do I have to find a map for each problem? If I know the map I can determine the g. But how do I find these maps? Can someone please help me?

Tobi92sr
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1 Answers1

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$\newcommand{\Brak}[1]{\left\langle #1\right\rangle}$For uniformity of notation, let $(u, v) \mapsto \Phi(u, v)$ be a regular parametrized surface, and define the "components of the first fundamental form" to be the functions $$ E = \Brak{\Phi_{u}, \Phi_{u}},\qquad F = \Brak{\Phi_{u}, \Phi_{v}},\qquad G = \Brak{\Phi_{v}, \Phi_{v}}. $$ The canonical volume form (i.e., the area form of the parametrized surface) is $$ \sqrt{EG - F^{2}}\, du \wedge dv. $$

Andrew D. Hwang
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  • Do I have to use that on for example i): with Φ(ϕ,z)=f(z)cos(ϕ),f(z)sin(ϕ),z)? – Tobi92sr Jun 14 '17 at 22:15
  • Yes, that's exactly right. – Andrew D. Hwang Jun 15 '17 at 00:21
  • So is the solution to i) is $g=(\partial_z f(z))^2*(f(z))^2$? – Tobi92sr Jun 15 '17 at 10:58
  • Not quite: If $(u, v) \leftrightarrow (z, \phi)$, then$$E = f'(z)^{2} + 1,\qquad F = 0,\qquad G = f(z)^{2},$$so $\sqrt{EG - F^{2}} = f(z) \sqrt{f'(z)^{2} + 1}$, and the volume form is this multiplied by $dz \wedge d\phi$. – Andrew D. Hwang Jun 15 '17 at 16:59
  • Thanks. I just forgot to type the +1 in E. Is the solution for ii) $$\sqrt{g}=\sqrt{r^2+1}$$ correct?

    And the solution for iii) should be: $$\sqrt{g}=\sqrt{(1+(\partial_uh(u,v))^2)*(1+(\partial_vh(u,v))^2)-(\partial_u\partial_vh^2(u,v))^2}=\sqrt{1+(\partial_uh(u,v))^2+(\partial_vh(u,v))^2}$$

     – Tobi92sr Jun 15 '17 at 17:16
  • Yes, those densities look good. (Again, though, the volume form needs to be multiplied by $du \wedge dv$, or its equivalent depending on the names of the parameters.) – Andrew D. Hwang Jun 15 '17 at 17:28
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    I know but I just wanted to check the $\sqrt{g}$. Thank you very much. – Tobi92sr Jun 15 '17 at 17:43