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I want to show that, if every closed curve $\gamma$ on the connected surface $S$ satisfies $$ \int_{\gamma} \left(\frac{\tau}{\kappa}\right)ds =0 $$ where $\tau$ and $\kappa$ are the torsion and the curvature of the curve($\kappa>0$).

Then $S$ is a part of plane or sphere. I know how to prove the converse, but how can I deal with this?

Bob brant
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  • I don't see how @MichaelHardy addresses the quantifier "for all curves" $\gamma$, but surely the statement of the question should require that $S$ be connected (and that the curve have $\kappa>0$ everywhere). – Ted Shifrin Jul 01 '17 at 14:53
  • "If any curve $\gamma$ on a surfcae $S$ satisfies [etc.]" can be construed by normal English usage conventions as meaning "If there is any curve $\gamma$ on a surface $S$ that satisfies [etc.]". But I don't think that was intended. Just changing "any" to "every" would eliminate all ambiguity. – Michael Hardy Jul 01 '17 at 16:18
  • @TedShifrin : $\uparrow \qquad$ – Michael Hardy Jul 01 '17 at 16:23
  • @Michael: I would say "some" if I meant an unambiguous "there exists" :) But I realize that language is often the hardest part of mathematics. That said, this result is reminiscent of Scherrer's Theorem on total twist of closed curves, but I've not seen this one before. – Ted Shifrin Jul 01 '17 at 16:25
  • This is a very sloppily stated problem. We've already established that we need to assume $S$ is connected. But it's also the case that the assumption should be regarding all closed curves $\gamma$ (with $\kappa>0$). With those corrections, I now know how to prove the converse. – Ted Shifrin Jul 01 '17 at 16:29
  • @TedShifrin : I've changed the ambiguous "any" to "every". Omission of mathematical hypotheses seems at worst a lesser offense than that. Just which hypotheses would be appropriate can be discussed in answers and comments. – Michael Hardy Jul 01 '17 at 18:10
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    HINT for a solution: Compute the first variation of this integral (over normal variations in the surface)—which must obviously be dead $0$—and conclude that any curve must be a line of curvature. This means every point is umbilic, so, assuming connectedness, the surface is a (subset of a) plane or sphere. (If necessary, I can write out some of the details, but if I do so it'll be couched in the language of moving frames.) – Ted Shifrin Jul 01 '17 at 20:39

1 Answers1

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Im assuming $S\subset{\mathbb R}^3$.

If this integral is $=0$ for every curve $\gamma$ of nonzero curvature $\kappa$ then $\tau$ has to be $\equiv0$ along every such curve. This implies that every such curve on $S$ is in fact plane. I'd say that $S$ has to be part of a plane in this case.

Christian Blatter
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  • I truly believe the correct hypothesis is that the curve be closed. Otherwise the integral can be nonzero for (non-closed) curves on a sphere. – Ted Shifrin Jul 01 '17 at 18:25