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Wikipedia article on geodesic curvature https://en.wikipedia.org/wiki/Geodesic_curvature says the following :

Consider a curve $\gamma$ in a manifold $\bar{M}$, parametrized by arclength, with unit tangent vector $T=d\gamma/ds$. Its curvature is the norm of the covariant derivative of $T : k=||DT/ds||$. If $\gamma$ lies on $M$, the geodesic curvature is the norm of the projection of the covariant derivative $DT/ds$ on the tangent space to the submanifold and the normal curvature is the norm of the projection of the covariant derivative $DT/ds$ on the normal bundle at the point considered. If the ambient manifold is the euclidean space $\mathbb{R}^n$, then the covariant derivative $DT/ds$ is just the usual derivative $dT/ds$.

I am studying geodesic/normal curvature of curves on surfaces in $\mathbb{R}^3$.

Let $\sigma:U\rightarrow \mathbb{R}^3$ be a parametrized surface, $\gamma:I\rightarrow\mathbb{R}^3$ be a parametrized curve on $\sigma$ i.e., $\gamma=\sigma\circ \mu$ where $\mu:I\rightarrow U$ given by $\mu(s)=(x_1(s),x_2(s))$ is a regular smooth parametrized curve.

For this, we have defined $\hat{t}(s)=\gamma'(s),\hat{m}(s)=N(\mu(s))$ and $\hat{u}(s)=\hat{m}(s)\times\hat{t}(s)$ where $$\hat{N}(u,v)=\frac{\sigma_u\times\sigma_v}{||\sigma_u\times \sigma_v||}$$

We have then seen that $\{\hat{t}(s),\hat{m}(s),\hat{u}(s)\}$ forms a coordinate system and any vector perpendicular to one of these three is in the span of remaining two vectors. As $\gamma''(s)=\hat{t}'(s)$ is perpendicular to $\hat{t}(s)$ we need to have $\gamma''(s)$ as linear combination of $\hat{m}(s)$ and $\hat{u}(s)$ so there exists $k_g(s),k_n(s)\in \mathbb{R}$ such that $$\gamma''(s)=k_g(s)\hat{u}(s)+k_n(s)\hat{m}(s).$$ $k_g(s)$ defined as above is called the geodesic curvature of $\gamma$ at $s$ and $k_n(s)$ defined as above is called the normal curvature of $\gamma$ at $s$.

Now, I would like to relate this to what wikipedia article says about geodesic curvature. Projection of the covariant derivative $\gamma''(s)$ on the normal bundle i.e., $||\gamma''(s)\cdot \hat{N}(\mu(s))||=||\gamma''(s)\cdot \hat{m}(s)||=k_n(s)$ is the normal curvature. Projection of the covariant derivative $\gamma''(s)$ on the tangent space ? It is not really clear which tangent space are they talking about. Any explanation on this is useful.

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Since your curve has unit speed (i.e., is parametrized by arc length), the equation $$ \frac{dT}{ds} = \gamma''(s) = k_{g}(s) \hat{u}(s) + k_{n}(s) \hat{m}(s) $$ is the decomposition of the accereration into tangential and normal components at $\gamma(s)$. That is, the projection of $\gamma''(s)$ on the tangent plane $T_{\gamma(s)}M$ is $\gamma''(s)$ minus the normal component: $$ \gamma''(s) - k_{n}(s) \hat{m}(s) = k_{g}(s) \hat{u}(s). $$

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  • $\begingroup$ Thanks for your answer. Any reference regarding this would be helpful. So, $\hat{u}(s)$ is the tangent vector(space) they are talking about? But $\hat{u}(s)=\hat{m}(s)\times \hat{t}(s)$.. Please say more about this. $\endgroup$ – user312648 May 22 '17 at 5:50
  • $\begingroup$ The "tangent space" is the tangent plane $T_{\gamma(s)}M$, spanned by $\hat{t}$ and $\hat{u}$. (The acceleration $\gamma''(s)$ has zero tangential component because the curve is unit-speed: $1 = \|\gamma'\|^{2} = \gamma' \cdot \gamma'$, so the product rule gives $0 = 2\gamma' \cdot \gamma''$.) Not sure what type of reference you're seeking; everything in this answer comes straight from your post...? $\endgroup$ – Andrew D. Hwang May 22 '17 at 10:02

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