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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?

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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.

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