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Show that a curve takes values in a sphere if and only if the curvature $k_n$ and $k_g$ of a parallel frame satisfy the equation of a line in the plane. How can the radius of the sphere can be read off this equation?

I already tried to differentiate $\Vert\gamma-a\Vert=r²$ and I am familiar with the structure equations. I also read this question Prove that a curve is spherical iff it satisfies the relation but it didn't help me either. Can anyone help me?

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hardy
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  • It's really not appropriate to call the curvatures $k_n$ and $k_g$ when you're talking about a parallel or Bishop frame. These terms only make sense when you're talking about a curve lying on a surface. See exercise 10 on p. 33 of my differential geometry text. – Ted Shifrin Mar 26 '17 at 06:56
  • But in this case the curve lies on a surface - the surface of the sphere. Apart from that I just quoted hte exercises from my differential geometry course. I'll look at your text, thx! – hardy Mar 26 '17 at 07:33
  • Oh, duh, sorry. But that only makes sense for the only if direction. So the point is that we can take (although we needn't) $\mathbf N_1 = \gamma$. The result holds for any Bishop frame. – Ted Shifrin Mar 26 '17 at 07:41
  • Thx a lot. I think I can solve it now - I'll write it down later. By the way: your text is great! – hardy Mar 26 '17 at 08:19
  • Okay I've shown that the curvatures satisfy the equation of a line. I still have a problem with the other direction - my thought was that I just take the distance from curve to a fixed point and differentiate it, if equals zero then the distance is constant and this point is the centre of the sphere. – hardy Mar 27 '17 at 10:33
  • The problem is that you have no idea what point to use. The point is that $\gamma$ will lie on a sphere centered at $P$ if and only if $(\gamma-P)\cdot T = 0$ (why?). Supposing $ak_1+bk_2+1=0$, what can you say about $\alpha = \gamma - aN_1 - bN_2$? – Ted Shifrin Mar 27 '17 at 15:17
  • @why?: because $(\gamma - P)\cdot(\gamma - P)=R²$ if I differentiate it I get $2(\gamma -P) \cdot T = 0$ – hardy Mar 27 '17 at 19:10
  • Of course: And conversely by a well-known fact about integrating $0$. But this is used everywhere in differential geometry. – Ted Shifrin Mar 27 '17 at 19:11
  • Now I would differentiate $\alpha = \gamma - aN_1 - bN_2$ and multiple it with $T$ and then I can substitute $N_1' \cdot T = -k_1$ and $N_2' \cdot T = -k_2$ and use the equation for the curvatures from above. – hardy Mar 27 '17 at 19:27
  • Finally I get the equation $\alpha' \cdot T = \gamma' \cdot T - 1$, hence $\alpha' \cdot T = 0$ where $\alpha = \gamma - P$. Therefore the curve $\gamma$ has to lie on a sphere. – hardy Mar 27 '17 at 19:55
  • Where did $P$ come from? – Ted Shifrin Mar 27 '17 at 20:09
  • There is a combination of $N_1$ and $N_2$ which gives us $P=aN_1 + bN_2$. Tell me if I'm wrong. – hardy Mar 27 '17 at 20:13
  • No, you're not doing things right. The point of introducing $\alpha$ is to get $\alpha'=0$. – Ted Shifrin Mar 27 '17 at 20:15
  • Am I right to the point where I followed that $\alpha \cdot T = 0$? – hardy Mar 27 '17 at 20:36
  • Okay, I think I have another approach: I choose $a$ and $b$ so that $\alpha'=0$ therefore it is constant. Now I show that $\alpha$ is in fact the centre of the sphere by differentiating: $(\alpha-\gamma) \cdot (\alpha-\gamma)=0$ and with this I know that $\gamma$ lies on the sphere with the centre $\alpha$. – hardy Mar 27 '17 at 21:32
  • Right. :) Now you have it!! – Ted Shifrin Mar 27 '17 at 22:40
  • Thank you! You helped me a lot. – hardy Mar 27 '17 at 23:07

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