# Proposition about curves in $S^2$

Let $\gamma_1,\gamma_2:(a,b)\to S^2$ be unit speed curves in $S^2=\{\vec{v}\in\mathbb{R^3}:\vec{v}\cdot\vec{v}=1\}$. Then the following two statements are equivalent:

(1) There is a $3\times 3$ orthogonal matrix $M$ such that $\gamma_2\equiv M\gamma_1.$

(2) $\beta_j:(a,b)\to \mathbb{R}, \beta_j=\det(\gamma_j,\gamma_j',\gamma_j'')$ satisfy $\beta_1\equiv \beta_2$ or $\beta_1\equiv -\beta_2.$

I saw this proposition last night, but I still don't know how to prove it. Could you help me with it? Thanks in advance.

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Hint: The condition (2) means that the two curves have the same curvature functions. Now, since the curves are spherical, their torsion is actually determined by the curvature functions; this fact is stated as an exercise in essentially all textbooks on the differential geometry of curves. The fundamental theorem of curves —which tells you that the curvature and the torsion complete determine a curve up to rigid motions— and a little extra reasoning then shows (1) holds.

Going the other way is immediate.

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 I was wondering why the second condition mean that the two curves have the same curvatures. Generally, $k=\det(\gamma',\gamma'')/|\gamma'|^3.$ How is that condition related to this? – Vladimir May 7 '12 at 19:08 Differentiate $\gamma^2=1$ three times and you will find your answer (write $\gamma$ as a linear combination of $N$ and $B$!) – Blah May 8 '12 at 18:13 Could you show me how to write $\gamma$ as a linear combination of $N$ and $B$? – Vladimir May 8 '12 at 20:21 @JackWitt, $T$, $N$ and $B$ are an orthonormal basis of $\mathbb R^3$, so you can write $\gamma$ (and any other vector!) as a linear combination of those three. That's a start... :) – Mariano Suárez-Alvarez♦ May 8 '12 at 20:23 So... Differentiate $\gamma^2=1$ three times, I get $\gamma\gamma'''+3\gamma'\gamma''=0$. Suppose $\gamma=aT+bN+cB,$ then $(aT+bN+cB)T'+3TkN=0.$ What does it mean? – Vladimir May 8 '12 at 20:30
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1=>2 is easy, just compute (and use $|\det(M)|=1$) $$\det(M\gamma_1,M\gamma_1',M\gamma_1'')=\det(M)\det(\gamma_1,\gamma_1',\gamma_1'')$$

2=>1 is not so easy Fix $s_0\in (a,b)$. Since $\gamma_1(s_0)$ and $\gamma_2(s_0)$ have unit length, there is a $M\in O(3)$ such that $\gamma_2(s_0)=M\gamma_1(s_0)$. Now compare $\gamma_2$ and $M\gamma_1$ and use http://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas

Edit: better link: http://en.wikipedia.org/wiki/Fundamental_theorem_of_curves

Edit (sorry, this is not a full solution!): Since $\gamma(t)$ is parametrized by arclenth, the tangent is $$T(s)=\gamma'(s)$$ and the curvature $\kappa$ and the normal $N$ are given by $$\kappa(s)=|T'(s)|=|\gamma''(s)| \qquad T'(s)=\kappa(s) N(s)$$ The Binormal is $$B(s)=T(s)\times N(s) \quad\text{and the Torsion \tau is given by } B'(s)=\tau (s) N(s)$$ Now $\det(\gamma,\gamma',\gamma'')=\det(\gamma,T,\kappa N)=\kappa\det(\gamma,T,N)$ fixes $\kappa$

Another hint: Differentiate $\gamma(s)^2=1$ several times to fix $\tau$.

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 Thank you. But I'm sorry that I don't understand your last step--"Now compare $\gamma_2$ and $M\gamma_1$ and use...". Could you explain more about it? – Vladimir May 6 '12 at 20:26 Since $\gamma(t)$ and $\gamma'(t)$ have unit length and are orthogonal, what does the equality of the determinants really tell you? And there is a theorem that two curves having the same curvature and torsion are congruent. – Blah May 6 '12 at 20:50 Excuse me, I still don't understand... I don't know what the equality of the determinants "really" tells me, and I can't see the connection between your first step and the second step of the second part of the proof... – Vladimir May 6 '12 at 22:33 Why can that equation fix $k$? Do $\gamma_1$ and $\gamma_2$ share the same $\det(\tau,T,N)$? Why? Besides, we have $\tau=\det(\gamma',\gamma'',\gamma''')/|\gamma'\times\gamma''|^2.$ How does this fix $\tau$? And, you get $\gamma_2(s_0)=M\gamma_1(s_0)$, how to use this? Are we going to show that for any $s_0$, they share the same $M$? Quite confusing... If you could provide a outline of the proof, I'll be very grateful. Thanks. – Vladimir May 7 '12 at 15:55

Here is another way to prove the proposition. Curves will be called $c$ and $d$ for short. I will assume (without real loss of generality) that $(a,b)=(-1,1)$. Furthermore, upon switching (if needed) to the pair of curves $(c,-d)$, we can assume $\beta_1=\beta_2$ and call this function simply $\beta$. As a last simplification, you can further assume $c(0)=d(0)$ and $c^.(0)=d^.(0)$ : just switch to the couple of curves $(c, P\cdot d)$ for the appropriate rotation $P$.

Since $c$ and $c^.$ are curves on the sphere, you get upon differentiation that $c\perp c^.,~c^.\perp c^{..}$ and $\langle c, c^{..}\rangle=-1$. Consider now the basis $C(t)=(c(t),c^.(t),c(t)\wedge c^.(t))$ : this is a direct orthonormal basis of your space, and thus the determinant defining $\beta$ can be calculated in this basis. Also what precedes shows that $c^{..}=-c+(\cdots)c\wedge c^.$ which because of the determinant calculations is actually $$c^{..}=-c+\beta\cdot c\wedge c^{.}.$$ Exactly the same holds for $d$.

We also set $(e,f,g)=(c(0),c^.(0),c(0)\wedge c^.(0)=C(0)=D(0)$.

Now consider the family of rotations $R_c(t)$ defined by sending the vectors of $C(0)$ to those of $C(t)$ in order of appearance : by definition, for all $t$, we have $c(t)=R_c(t)\cdot c(0)$. Now differentiate this family with respect to $t$ : the resulting map has $$\frac{dR_c}{dt}(t):\lbrace \begin{array}{ll} e\mapsto & c^.(t) \\ f\mapsto & c^{..}(t)=-c(t)+\beta(t)c(t)\wedge c^.(t) \\ g\mapsto & c(t)\wedge c^{..}(t)=\beta(t) c(t)\wedge(c(t)\wedge c^.(t))=-\beta(t)c^.(t) \end{array}$$ In other words, $$\frac{dR_c(t)}{dt}=R_c(t)\times \tau(t)$$ where $$\mathrm{Mat}(\tau(t);(e,f,g))= \left(\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & -\beta(t) \\ 0 & \beta(t) & 0 \end{array}\right)$$

Exactly the same equations stand if we do things with the analoguoulsy defined $R_d(t)$. We can now show that $R_c=R_d$ and thus for all $t,~c(t)=R_c(t)\cdot e=R_d(t)\cdot e=d(t)$. You can either say that $R_c(0)=R_d(0)=\mathrm{id}$ and they are both solutions to the same first order linear differential equation, and by the Cauchy Lipschitz theorem they (are globally defined and) coincide, or you can differentiate $R_c(t)(R_d(t))^{-1}=R_c(t)R_d(t)^*$ by hand and get $$\frac{dR_c(t)R_d(t)^*}{dt}=R_c(t)\tau(t)\times R_d(t)^*+R_c(t)\times (-\tau(t)R_d(t)^*)=0,$$ where we have made use of the skew symmetry of $\tau(t)$ this shows equality for all $t$.

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 Thanks! What do you mean by "$\text{Mat}(\tau(t);(e,f,g))$"? – Vladimir May 7 '12 at 18:03 For all $,~\tau(t)$ is a linear map. This just means the matrix of $\tau(t)$ in the (direct orthonormal) basis $(e,f,g)$. – Olivier Bégassat May 7 '12 at 18:12 ... Did you just award me $150$ points for this answer???!?! Thank you very much :D I'm glad I was of assistance! – Olivier Bégassat May 9 '12 at 21:19