Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that given a curve $r(t)$ in an euclidean space and $s(t)$ the arc length on the curve we can build a $TNB$ reference frame, using the Frenet formulas: $$T=\frac{dr(t)}{ds}$$ $$N=\frac{\frac{dT}{ds}}{||\frac{dT}{ds}\|}$$ $$B=T\times N$$ My question is: given a generic $3\times 3$ metric tensor $g_{\mu \nu}$ defining an element: $$ds^2=g_{\mu \nu}dx^{\mu}dx^{\nu}$$ is it still possible to define a $TNB$ reference frame as in the euclidean geometry?

share|cite|improve this question
up vote 3 down vote accepted

Yes, it's not very difficult actualy if you know a bit of Riemannian geometry.

Take a smooth curve $c : [0,l] \rightarrow M$ where $M$ is a 3-manifold with $g$ a Riemannian metric.

We can assume it's parametrised by arc-length, then $c'$ is always of norm $1$. We have $T$

You can define what is $\frac{d}{dt} c'(t)$, using the fact that $c'$ can always be extended locally to a smooth vector field $\tilde c$, and define $\frac{d}{dt} c'(t) = D_{c'(t)} \tilde c'(t)$ where $D$ is the Levi-Civita connexion. You just need to check that it doesn't depends on the extension $\tilde c$. Assuming that it's nowhere zero, $B =\frac{\frac{d}{dt} c'(t)}{||\frac{d}{dt} c'(t) ||} $.

Now define $N$ is easy. Depending on the fact that $M$ is oriented, take one of the two vector such that $(T,N,B)$ is in each point an orthonormal basis for $T_pM$, and the choice is unique if you want it to be direct.

Note that this construction is local, but can be extended at the whole segment.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.