# proving that a curve with constant curvature contained in a sphere its a circle

i have a curve $\alpha:I\rightarrow \mathbb{R}^3$ such that his curvature $k$ is constantand $\alpha$ is entirely contained in a sphere, i must prove that this curve is a cirlce.

My try:

I need to prove that $\alpha$ has zero torsion, i supose the sphere with center at origin and radius $r>0$, so i got $|T^{'}(s)|=k$ for all $s$ where $T$ is the tangent line of $\alpha$ and $|\alpha(s)|^2=r^2$ for all $s$, i wanted to prove that the binormal $B$ of $\alpha$ is contants, but i derivatite these two equation a lot and i couln't conclude anything, anyone can help?

• what do you know about $\alpha \cdot \alpha \; ?$ It appears you are switching between $\alpha$ and $\gamma$ – Will Jagy Jan 21 '18 at 0:02
• that is equal to the radius of the sphere. P.S. sorry, i edited – Eduardo Silva Jan 21 '18 at 0:04
• no. That is the square root. I want you to tell me about the derivative of $\alpha \cdot \alpha$ by the variable $s$ – Will Jagy Jan 21 '18 at 0:05
• thats right, but i tried this as you can see in my question, so how this is going to work? the derivatives will give me that $\alpha$ and its tangents are perpendicular – Eduardo Silva Jan 21 '18 at 0:07
• How do you write $\alpha$ in terms of $T,N,B \; ?$ – Will Jagy Jan 21 '18 at 0:09

thanks to the coments of @willjagy i could conclude my demonstrations, first i find by derivating $|\alpha(s)|^2$ that $<T,\alpha>=0$, derivating again and using Frenet Formulas i get

$k<N,\alpha>+1=0$ and one more time:

$k^{'}<N,\alpha>-k\tau<B,\alpha>=0$, where $\tau$ is the torsion of the curve, so, as $k$ is contant and positive:

$\tau<B,\alpha>=0$ $\forall s\in I$,

and then at any cases $\tau=0$ or $<B,\alpha>=0$ i can conclude the $\alpha$ is a circle, for the first i get $\alpha$ in a great circle of the sphere and for the second an arbitrary cirlce with radius less or equal than $r$.

• not quite. If we introduce two functions of $s,$ call them $p$ and $q,$ and say $\alpha = p N + q B,$ differentiate and collect the coefficients of $T,$ then of $N,$ then of $B,$ the conclusions are, given constant $k,$ that both $p$ and $q$ are constant, $p$ is nonzero, so $\tau$ is zero. – Will Jagy Jan 21 '18 at 2:26
• If you think geometrically, two things are wrong here. First, $N$ points inward and $\alpha$ points outward. Second, the only time this equation can hold is for a great circle (the intersection of the sphere with a plane through the origin — the center). – Ted Shifrin Jan 21 '18 at 5:04
• By your coments i tried again and reformulate my argumentation, any gap now? – Eduardo Silva Jan 21 '18 at 15:09