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A surface of revolution has the following first fundamental form $$ \begin{pmatrix} r^2(s)& 0\\ 0 &1\end{pmatrix}. $$

But does this first fundamental form imply that the surface is a surface of revolution?

Chappers
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gbox
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    Implicit in that notation for a surface of revolution is the assumption that $|r'(s)|\le 1$. If $r(s)=s$, we'll get the plane parametrized (as a surface of revolution) by polar coordinates. What if $r(s)=2s$? – Ted Shifrin Jul 24 '17 at 20:42

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In a word, no. Such a surface could be a cylinder over an arbitrary smooth plane curve, or could admit a one-parameter family of "helical" ambient isometries, see "Self-sliding" surfaces.

Further, as Ted notes, not every such metric embeds as a surface of rotation (even if, like Ted's "saddle cone", the metric embeds isometrically in Euclidean $3$-space).

You might be interested in A symplectic look at surfaces of revolution, l'Enseignement Mathématique 49 (2003), 157-172.

Andrew D. Hwang
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