📂Geometry

Geodesic Curvature is Intrinsic

Theorem¹

The geodesic curvature $\kappa_{g}$ of a curve on a surface is intrinsic.

Description

In other words, $\kappa_{g}$ can be calculated solely using the coefficients of the Riemannian metric, without the unit normal $\mathbf{n}$. Of course, it can also be expressed using the extrinsic formula, as $\kappa \mathbf{N} = \mathbf{T}^{\prime} = \alpha^{\prime \prime} = \kappa_{n}\mathbf{n}+ \kappa_{g}\mathbf{S}$, so

$$ \begin{align*} \kappa_{g} =&\ \left\langle \mathbf{T}^{\prime}, \mathbf{S} \right\rangle \\ =&\ \left\langle \mathbf{T}^{\prime}, \mathbf{n} \times \mathbf{T} \right\rangle \\ =&\ \left[ \mathbf{T}^{\prime}, \mathbf{n}, \mathbf{T} \right] \\ =&\ \left[ \mathbf{n}, \mathbf{T}, \mathbf{T}^{\prime} \right] \\ =&\ \left\langle \mathbf{n}, \mathbf{T} \times \mathbf{T}^{\prime} \right\rangle \\ =&\ \left\langle \mathbf{n}, \mathbf{T} \times \kappa \mathbf{N} \right\rangle \\ =&\ \kappa \left\langle \mathbf{n}, \mathbf{T} \times \mathbf{N} \right\rangle \\ =&\ \kappa \left\langle \mathbf{n}, \mathbf{B} \right\rangle \\ =&\ \kappa \cos \theta \end{align*} $$

Here, $\left[ \cdot, \cdot, \cdot \right]$ signifies the scalar triple product. $\mathbf{B}$ is the binormal. $\theta$ is the angle between $\mathbf{n}$ and $\mathbf{B}$.

Proof

• $g_{ij}$: Coefficients of the Riemann metric
• $L_{ij} = \left\langle \mathbf{x}_{ij}, \mathbf{n} \right\rangle$: Coefficients of the second fundamental form
• $\Gamma_{ij}^{k} = \sum \limits_{l=1}^{2} \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk} = \left\langle \mathbf{x}_{ij}, \mathbf{x}_{l} \right\rangle g^{lk}$: Christoffel symbols

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Let $\mathbf{x} : U \to \mathbb{R}^{3}$ be the coordinate chart mapping, and the coordinates of $U$ be $(u_{1}, u_{2})$. Suppose a curve $\alpha (s) = \mathbf{x}\left( u_{1}(s), u_{2}(s) \right)$ is given on it. Then, $\alpha^{\prime \prime}$ is expressed as follows.

$$ \alpha^{\prime \prime}(s) = \kappa_{n}(s)\mathbf{n}(s) + \kappa_{g}(s)\mathbf{S}(s) $$

$\mathbf{n}$ is the unit normal, $\mathbf{S} = \mathbf{n} \times \mathbf{T}$. Also, $\mathbf{x}_{i}$, $\mathbf{x}_{ij}$ are respectively the first and second partial derivatives of $\mathbf{x}$.

$$ \mathbf{x}_{i} := \dfrac{\partial \mathbf{x}}{\partial u_{i}} \quad \text{and} \quad \mathbf{x}_{ij} := \dfrac{\partial^{2} \mathbf{x}}{\partial u_{j} \partial u_{i}} $$

• Part 1.

  First, let’s represent the scalar triple product of unit normals and the bases of the tangent plane as follows.

  $$ \epsilon _{ij} = \langle \mathbf{n}, \mathbf{x}_{i} \times \mathbf{x}_{j} \rangle = \left[ \mathbf{n}, \mathbf{x}_{i}, \mathbf{x}_{j} \right] $$

  Then, since $\mathbf{x}_{i} \times \mathbf{x}_{i} = \mathbf{0}$, the values ​​are as follows.

  $$ \epsilon_{11} = \epsilon_{22} = 0 $$

  Since $\mathbf{n}$ is perpendicular to $\mathbf{x}_{1} \times \mathbf{x}_{2}$ by definition,

  $$ \langle \mathbf{n}, \mathbf{x}_{1} \times \mathbf{x}_{2} \rangle = \left| \mathbf{n} \right| \left| \mathbf{x}_{1} \times \mathbf{x}_{2} \right| $$

  Here, since $\mathbf{n}$ is a unit vector and $g = \left| \mathbf{x}_{1} \times \mathbf{x}_{2} \right|^{2}$,

  $$ \epsilon_{12} = \langle \mathbf{n}, \mathbf{x}_{1} \times \mathbf{x}_{2} \rangle = -\langle \mathbf{n}, \mathbf{x}_{2} \times \mathbf{x}_{1} \rangle = -\epsilon_{21} = \sqrt{g} $$

• Part 2.

  Since $\mathbf{S} = \mathbf{n} \times \mathbf{T}$ and $\mathbf{S}$ are unit vectors,

  $$ \kappa _{g} = \left\langle k_{g} \mathbf{S}, \mathbf{S} \right\rangle = \left\langle k_{g} \mathbf{S}, \mathbf{n} \times \mathbf{T} \right\rangle = \left[ \kappa_{g}\mathbf{S}, \mathbf{n}, \mathbf{T} \right] $$

  Formula

  $$ \kappa_{g}\mathbf{S} = \sum \limits_{i=1}^{2} \left[ u_{k}^{\prime \prime} + \sum \limits_{i,j=1}^{2} \Gamma_{ij}^{k}u_{i}^{\prime}u_{j}^{\prime} \right] \mathbf {x}_{k} $$

  The tangent vector is $\mathbf{T} = \mathbf{x}_{l}u_{l}^{\prime}$, and according to the formula above, $\kappa_{g}$ is calculated as follows.

  $$ \begin{align*} \kappa_{g} =&\ \left\langle k_{g} \mathbf{S}, \mathbf{n} \times \mathbf{T} \right\rangle = \left[ \kappa_{g}\mathbf{S}, \mathbf{n}, \mathbf{T} \right] \\ =&\ \left\langle \sum \limits_{k=1}^{2} \left( u^{\prime \prime}_{k} + \sum\limits_{i=1}^{2} \sum\limits_{j=1}^{2} \Gamma_{ij}^{k} u_{i}^{\prime} u_{j}^{\prime} \right)\mathbf{x}_{k} , \mathbf{n} \times \mathbf{T} \right\rangle \\ =&\ \sum \limits_{k=1}^{2} \left( u^{\prime \prime}_{k} + \sum\limits_{i=1}^{2} \sum\limits_{j=1}^{2} \Gamma_{ij}^{k} u_{i}^{\prime} u_{j}^{\prime} \right) \left\langle \mathbf{x}_{k} , \mathbf{n} \times \mathbf{T} \right\rangle \\ =&\ \sum \limits_{k=1}^{2} \left( u^{\prime \prime}_{k} + \sum\limits_{i=1}^{2} \sum\limits_{j=1}^{2} \Gamma_{ij}^{k} u_{i}^{\prime} u_{j}^{\prime} \right) \left[\mathbf{x}_{k}, \mathbf{n}, \mathbf{T} \right] \\ =&\ \sum \limits_{k=1}^{2} \left( u^{\prime \prime}_{k} + \sum\limits_{i=1}^{2} \sum\limits_{j=1}^{2} \Gamma_{ij}^{k} u_{i}^{\prime} u_{j}^{\prime} \right) \left[\mathbf{n}, \mathbf{T}, \mathbf{x}_{k} \right] \\ =&\ \sum \limits_{k=1}^{2} \left( u^{\prime \prime}_{k} + \sum\limits_{i=1}^{2} \sum\limits_{j=1}^{2} \Gamma_{ij}^{k} u_{i}^{\prime} u_{j}^{\prime} \right) \left[\mathbf{n}, \mathbf{x}_{l}u_{l}^{\prime}, \mathbf{x}_{k} \right] \\ =&\ \sum \limits_{k=1}^{2} \left( u^{\prime \prime}_{k} + \sum\limits_{i=1}^{2} \sum\limits_{j=1}^{2} \Gamma_{ij}^{k} u_{i}^{\prime} u_{j}^{\prime} \right) u_{l}^{\prime} \left[\mathbf{n}, \mathbf{x}_{l}, \mathbf{x}_{k} \right] \\ =&\ \sum \limits_{k=1}^{2} \left( u^{\prime \prime}_{k} + \sum\limits_{i=1}^{2} \sum\limits_{j=1}^{2} \Gamma_{ij}^{k} u_{i}^{\prime} u_{j}^{\prime} \right) u_{l}^{\prime} \epsilon_{lk} \end{align*} $$

  Here, since $\Gamma_{ij}^{k}$ is intrinsic and, according to Part 1., $\epsilon_{lk}$ is also intrinsic, $\kappa_{g}$ is intrinsic.