A Intution behind the definition of extrinsic curvature

1. Jun 15, 2017

Afonso Campos

Forgive me for asking a rather silly question, but I have thinking about the following definition of the extrinsic curvature $\mathcal{K}_{ij}$ of a sub-manifold (say, a boundary $\partial M$ of a manifold $M$):

$$\mathcal{K}_{ij} \equiv \frac{1}{2}\mathcal{L}_{n}h_{ij} = \nabla_{(i}n_{j)},$$

where $n$ is the inward-pointing unit normal to $\partial M$ and $h_{ij}$ is the induced metric on $\partial M$.

Is there an intuitive way to understand why the above must be the definition of the extrinsic curvature of a submanifold?

2. Jun 16, 2017

Without having any experience in this area, I think I can still offer an input that might be relevant: In curvilinear motion, there is the relation $\frac{d \hat{T}}{ds}= \kappa \hat{N}$ which is essentially one of the Frenet equations. This can easily be demonstrated in two dimensions by letting $\hat{T}=cos(\phi) \hat{i}+sin(\phi) \hat{j}$ and taking the derivative. ( The curvature $\kappa=\frac{d \phi}{ds}$.) Note that $\hat{N}=\frac{d \hat{T}}{d \phi}=-sin(\phi) \hat{i}+cos(\phi) \hat{j}$. $\\$ In your manifold problem, the rate of change of the vector normal to the surface is used to define the curvature, but its rate of change is the same as that of a unit tangent vector for a curve along the surface.

Last edited: Jun 16, 2017
3. Jun 18, 2017

lavinia

Given a curve parameterized by arc length with unit tangent $T$ the covariant derivative $∇_{T}T$ equals $kN$ where $k$ is the curvature and $N$ is the unit normal.

By the Product Rule $0 = T⋅<T,N> = <∇_{T}T,N> + <T,∇_{T}N> = k + <T,∇_{T}N>$ so the tangential component of the covariant derivative of the unit normal is negative the curvature of the curve. The upshot is that differentiating the unit normal to the curve is a way to compute its curvature.

On a manifold of codimension 1 with a well defined unit normal vector field - e.g. if the manifold is the boundary of a one higher dimensional manifold - then $∇_{T}T$ splits into a tangential and a normal component. The normal component is the normal curvature $k_{n}$ multiplied by the unit normal. It is the curvature of the curve that is extrinsic to the manifold while the tangential component is the intrinsic or "geodesic" curvature of the curve.

Since $<T,N> = 0$ one again has $0 = T⋅<T,N> =<∇_{T}T,N> + <T,∇_{T}N> = k_{n} + <T,∇_{T}N>$ so differentiating the unit normal is a way to compute the normal curvature of the curve. Note that if the curve is a geodesic on the submanifold, its geodesic curvature is zero so the normal curvature and the curvature are the same.

Generally speaking $∇_{T}N$ will not be parallel to $T$ so one gets a linear map $L(T) = ∇_{T}N$ of the tangent space into itself. But one can show that this map is symmetric so that it has n eigenvectors $T_{i}$. For these vectors the $∇_{T_{i}}N$ are parallel to $T_{i}$ and equal $-k_{i}T_{i}$ where $k_{i}$ is the normal curvature. These $k_{i}$ are called principal curvatures.

Now restrict to the case where the manifold is embedded in $R^{n+1}$. and consider the normal map $X: M× R→R^{n}$ defined by $(p,s) →p + sN_{p}$

$dX = Id + sdN$ so if $T$ is an eigenvector of $L$ then $dX(T) = T - skT$ so for $s= 1/k$ the point $p + sN$ is a singular point of the map $X$. At this point nearby normal lines "intersect" and $p + 1/kN$ is called a focal point of the normal map. One thinks of the normal lines as light rays reflecting off of the manifold and focussing at $p + 1/kN$. The $n$ different reciprocals of the curvatures $1/k_{j}$ are called the "radii of curvature".

Since $L$ is symmetric it is completely described by its eigenvectors and these in turn determine the focal points of the normal map. So extrinsic curvature may be thought of as being determined by n directions at each point p along which the normal lines focus at the radii of curvature along the unit normal.

Notes:

- For a circle the radii of curvature are constant and are equal to the radius of the circle. All normal lines focus perfectly at the center. Classically, the radius of curvature of a plane curve was thought of as the center of a circle and nearby normals were thought to approximate the normals to the circle. The curve itself is always tangent to this circle.

- If one parameterizes a neighborhood of the point $p$ on the manifold $M$ in Euclidean space $(t_1,...,t_{n}) →M(t_1, ..., t_{n})$ in such a way that the metric matrix $g_{ij}$ is the identity matrix at $p$ then taking inner products with the equations $∂X/∂t_{i} = ∂M/∂t_{i}+s∂N/∂t_{i}$ one gets the matrix $g_{ij} + s∂N/∂t_{i}⋅∂M/∂t_{j}$ and it is clear that this matrix is singular when $-1/s$ is an eigenvalue of the symmetric matrix $∂N/∂t_{i}⋅∂M/∂t_{j}$.

- If the manifold is embedded in codimension $k$ the by choosing $k$ unit normal vector fields one can perform the same type analysis.

Last edited: Jun 19, 2017