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I am rather confused about elements of the method of moving frames and how to apply the method of moving frames.

$S^2$ is a homegeneous space for the Lie group $SO(3)$.

We want to construct a lift $\tilde{\alpha}:U\subset\mathbb{R} \to SO(3)$ of a curve $\alpha: U \subset \mathbb{R} \to S^2$ suitable to our curve geometry, and then compute the pullback of the Maurer-Cartan form to obtain invariants.

Identifying the frame bundle with $SO(3)$, I get the following underdetermined lift: $$ (\alpha(t), e_1(t), e_2(t)). $$

Now, since $\langle \alpha,\alpha \rangle =1$, we have $\langle \alpha', \alpha \rangle =0$, so if we choose (these choices are, I believe, $SO(3)$ invariant) \begin{align} e_1 &= \frac{\alpha'}{|\alpha'|} \\ e_2 &= \frac{e_1' - \langle e_1',\alpha \rangle \alpha}{| e_1' - \langle e_1',\alpha \rangle \alpha|} \end{align} we then pull back the Maurer-Cartan form and use the structure equations to compute the following two invariants:

$$ | \alpha'| \ \text{and} \ \Bigg| \Bigg(\frac{\alpha'}{|\alpha'|} \Bigg)' - |\alpha'| \alpha \Bigg| .$$

Where, if anywhere, am I going wrong?

Andrew Whelan
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1 Answers1

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Yes, assuming your curve is regular (has nowhere-vanishing derivative), you may as well assume it's arclength-parametrized (as always in the theoretical curve game), so $\alpha'$ will be a unit vector. Then, quite simply, you should get your frame by taking $e_0 = \alpha$, $e_1=\alpha'$, and $e_2=e_0\times e_1$. Now, what are the structure equations going to be?

Ted Shifrin
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  • Well, using the fact that we only have three potential non-zero entries for the Maurer-Cartan form, $de_i = e_j \omega^j_i$ is what we should use. After some work I think we get just the only non-constant-coefficient form of the pullback as $\omega^2_1 = \langle \alpha'' , \alpha \times \alpha' \rangle$. Is this correct? – Andrew Whelan Jul 12 '17 at 19:48
  • You're missing a $ds$, of course. Well, what about $\omega_0^1$ and $\omega_1^0$? – Ted Shifrin Jul 12 '17 at 19:52
  • Ah, I meant the coefficients of these forms. Would these not just be $\pm ds$? – Andrew Whelan Jul 12 '17 at 20:01
  • Yes. Sure. Note that $\omega_1^0=-\omega_0^1=-ds$ is just the fact that the normal curvature of any curve on the unit sphere is $-1$ (assuming outward-pointing normal). – Ted Shifrin Jul 12 '17 at 20:06
  • Ah okay, thanks! I was just assuming the invariants might be a little more interesting on $S^2$ as compared to $\mathbb{R}^2$. Actually, are they not pretty much the same, since we can write the coefficients of $\omega^2_1$ as $ \langle \alpha , \alpha' \times \alpha'' \rangle = |\alpha' \times \alpha'' | = \kappa(s)$? – Andrew Whelan Jul 12 '17 at 20:08
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    Be careful. That's $\kappa_g(s)$ (intrinsic curvature). And we already said $\kappa_n = -1$. So $\kappa$ for the curve as a space curve is different. But I think you're thinking just intrinsically. If you want more invariants for curves, you need higher codimension (whether in $\Bbb R^n$, $S^n$, or $\Bbb H^n$). – Ted Shifrin Jul 12 '17 at 20:13
  • Excellent, thanks - always appreciate your comments. I'll give the $S^3$ case a go to compare. – Andrew Whelan Jul 12 '17 at 20:35