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I am currently studying for an exam in differential geometry. There's a problem which I am not able to solve and do not even know where to start.

There is the curve β on the spherical surface. which radius is a.

If curve $\alpha$ is defined as $$\alpha(t)=\int \beta(t)\times\beta '(t)dt$$ show that torsion $\tau$ is $\frac{1}{a^2}$

By advice, to find binormal, I started to find $$\alpha'(t)\times\alpha''(t)$$ $$\alpha'(t)=\beta(t)\times\beta'(t)$$ $$\alpha''(t)=\beta(t)\times\beta''(t)$$ so $$\alpha'(t)\times\alpha''(t)=((\beta(t)\times\beta'(t))\cdot\beta''(t))\beta(t)$$ but how can I found $$\left|\alpha'(t)\times\alpha''(t)\right|$$?

And by knowing that binormal is $\pm\beta$ , how can I found a torsion? Maybe $\beta'(t)$ is tangent. And maybe $$\beta(t)\cdot\beta(t)=a^2$$ by derivative it, $$\beta'(t)\cdot\beta(t)=0$$ This is the whole thing that I can do. With that idea... still I can't find the way to solve...

eccentric
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    HINT: Show that the binormal is $\pm\beta$. – Ted Shifrin Sep 28 '20 at 01:42
  • sorry.... why binormal is $\pm \beta$? – eccentric Sep 28 '20 at 01:53
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    I'm telling you to figure that out. – Ted Shifrin Sep 28 '20 at 01:54
  • okay, I'll try it out more. – eccentric Sep 28 '20 at 01:56
  • More explicit hint: What direction is orthogonal to $\alpha’$ and $\alpha”$? – Ted Shifrin Sep 28 '20 at 01:58
  • $\alpha'$ $\alpha''$ is orthogonal in binormal. and I'm trying to get $\alpha''$ to get binormal of $a(t)$.

    $$\alpha''(t)=(\frac{d}{dt}\beta\times\frac{d}{dt}\beta)+(\beta\times\frac{d^2}{dt^2}\beta)=(\beta\times\beta'')$$ then $$\alpha'(t)\times\alpha''(t)=(\beta(t)\times\beta'(t))\times(\beta(t)\times\beta''(t))=((\beta(t)\times\beta'(t))\cdot\beta''(t))\beta(t)$$ Is $$(\beta(t)\times\beta'(t))\cdot\beta''(t))=1?$$ and I still don't know why torsion be like that.

     – eccentric Sep 28 '20 at 03:10
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    We don't even have an arclength parametrization. But you've established my hint. So what is $dB/ds$? – Ted Shifrin Sep 28 '20 at 05:01
  • Isn't it tangent? – eccentric Sep 28 '20 at 12:14
  • Frenet tells you that $dB/ds = -\tau N$. But you know $dB/dt = \beta'$. So how do you find $dB/ds$? You are correct that $\beta\cdot\beta' = 0$, which tells you that $|\alpha'| = |\beta\times\beta'| = a|\beta'|$ (why?). – Ted Shifrin Sep 28 '20 at 15:19
  • I really thanks for your help. I found that torsion is $(\beta(t)\times\beta'(t))\cdot\beta''(t) \over \left| \beta(t) \right|$ and $\left| \beta(t) \right|$ is a

    so $(\beta(t)\times\beta'(t))\cdot\beta''(t)$ have to be $ 1 \over a $

    why?

     – eccentric Sep 28 '20 at 15:25
  • Actually, I lied. Since $|\beta|=a$, we have $B = \pm\beta/a$. I'm not suggesting using that formula for torsion; I'm suggesting thinking it through from first principles. However, I don't believe that formula is correct. You need $\tau = \frac{\alpha'\cdot(\alpha''\times\alpha''')}{|\alpha'\times\alpha''|^2}$. So I definitely do not follow. Also, I just realized that the problem left out an important hypothesis: You need to know that $\beta,\beta',\beta''$ are linearly independent at each $t$. – Ted Shifrin Sep 28 '20 at 23:31

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