Geodesic curvature

In Riemannian geometry, the geodesic curvature $$k_g$$ of a curve $$\gamma$$ measures how far the curve is from being a geodesic. For example, for 1D curves on a 2D surface embedded in 3D space, it is the curvature of the curve projected onto the surface's tangent plane. More generally, in a given manifold $$\bar{M}$$, the geodesic curvature is just the usual curvature of $$\gamma$$ (see below). However, when the curve $$\gamma$$ is restricted to lie on a submanifold $$M$$ of $$\bar{M}$$ (e.g. for curves on surfaces), geodesic curvature refers to the curvature of $$\gamma$$ in $$M$$ and it is different in general from the curvature of $$\gamma$$ in the ambient manifold $$\bar{M}$$. The (ambient) curvature $$k$$ of $$\gamma$$ depends on two factors: the curvature of the submanifold $$M$$ in the direction of $$\gamma$$ (the normal curvature $$k_n$$), which depends only on the direction of the curve, and the curvature of $$\gamma$$ seen in $$M$$ (the geodesic curvature $$k_g$$), which is a second order quantity. The relation between these is $$k = \sqrt{k_g^2+k_n^2}$$. In particular geodesics on $$M$$ have zero geodesic curvature (they are "straight"), so that $$k=k_n$$, which explains why they appear to be curved in ambient space whenever the submanifold is.

Definition
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. Conversely the normal curvature is the norm of the projection of $$DT/ds$$ on the normal bundle to the submanifold 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 $$dT/ds$$.

If $$\gamma$$ is unit-speed, i.e. $$\|\gamma'(s)\|=1$$, and $$N$$ designates the unit normal field of $$M$$ along $$\gamma$$, the geodesic curvature is given by

k_g = \gamma''(s) \cdot \Big( N( \gamma(s)) \times \gamma'(s) \Big) = \left[ \frac{\mathrm{d}^2 \gamma(s)}{\mathrm{d}s^2} , N(\gamma(s)), \frac{\mathrm{d}\gamma(s)}{\mathrm{d}s} \right]\,, $$ where the square brackets denote the scalar triple product.

Example
Let $$M$$ be the unit sphere $$S^2$$ in three-dimensional Euclidean space. The normal curvature of $$S^2$$ is identically 1, independently of the direction considered. Great circles have curvature $$k=1$$, so they have zero geodesic curvature, and are therefore geodesics. Smaller circles of radius $$r$$ will have curvature $$1/r$$ and geodesic curvature $$k_g = \frac{\sqrt{1-r^2}}{r}$$.

Some results involving geodesic curvature

 * The geodesic curvature is none other than the usual curvature of the curve when computed intrinsically in the submanifold $$M$$. It does not depend on the way the submanifold $$M$$ sits in $$\bar{M}$$.
 * Geodesics of $$M$$ have zero geodesic curvature, which is equivalent to saying that $$DT/ds$$ is orthogonal to the tangent space to $$M$$.
 * On the other hand the normal curvature depends strongly on how the submanifold lies in the ambient space, but marginally on the curve: $$k_n$$ only depends on the point on the submanifold and the direction $$T$$, but not on $$DT/ds$$.
 * In general Riemannian geometry, the derivative is computed using the Levi-Civita connection $$\bar{\nabla}$$ of the ambient manifold: $$DT/ds = \bar{\nabla}_T T$$. It splits into a tangent part and a normal part to the submanifold: $$\bar{\nabla}_T T = \nabla_T T + (\bar{\nabla}_T T)^\perp$$. The tangent part is the usual derivative $$\nabla_T T$$ in $$M$$ (it is a particular case of Gauss equation in the Gauss-Codazzi equations), while the normal part is $$\mathrm{I\!I}(T,T)$$, where $$\mathrm{I\!I}$$ denotes the second fundamental form.
 * The Gauss–Bonnet theorem.