curve of constant curvature on unit sphere is planar curve?

Question

I've studied differential geometry and get this question.

I'd like to verify following statement.

curve of constant curvature on unit sphere is planar curve

I've struggled with Frenet-Serret frame, differentiating, differentiating, differentiating, .....

BUT I didn't get something yet..

Could you give me some hint, please?

$$ $$ Ah!! FIRST OF ALL,

I'd like to know whether the statement is true or false.

Thanks in advance.

Answer 1

You want to show that the torsion is zero. Note that If $\mathbf{r}$ is the unit radius vector to the curve then $$\mathbf{t}=\frac{d \mathbf{r}}{ds}$$ Now $\mathbf{t} \cdot \mathbf{r}=0$ so on differentiating we get

$$\frac{d \mathbf{t}}{ds}\mathbf{r}+\mathbf{t}\frac{d \mathbf{r}}{ds}=0$$ or $$\kappa \mathbf{n}\cdot \mathbf{r}+ \mathbf{t}\cdot \mathbf{t}=0$$ (Note that $\kappa\neq 0$, since otherwise we have a straight line.) So $$\mathbf{n}\cdot \mathbf{r}=-\frac{1}{\kappa}$$ differentiating this and using that $\kappa$ is constant we get

$$(-\kappa \mathbf{t}+\tau \mathbf{b})\cdot \mathbf{r}+\mathbf{n}\cdot \mathbf{t}=0$$ which simplifies to $$\tau \mathbf{b}\cdot \mathbf{r}=0$$ If $\tau=0$ we are done so assume

$$\mathbf{b}\cdot \mathbf{r}=0$$

and differentiate which gives

$$-\tau \mathbf{n}\cdot \mathbf{r}+\mathbf{b}\cdot \mathbf{t}=0$$

$$-\tau \mathbf{n}\cdot \mathbf{r}=0$$ But $\mathbf{n}\cdot \mathbf{r}=-\frac{1}{\kappa}$ so again $\tau=0$.