Maximal curvature of a surface

How to compute curvature

Curvature of curves in the plane

Consider the problem of computing the point on a (smooth) real variety $V\subset \mathbb{R}^n$, where the curvature is maximal. For curves in the plane $ \mathbb{R}^2$ we can use the following formula for curvature at a point $p\in V = \{f(x_1,x_2)=0\}$.

$$\sigma(p) = \frac{v^T H v}{g^\frac{3}{2}}$$

where $v^T \,\nabla_p f(p) = 0$, $H$ is the Hessian of $f$ at $p$ and $g = \nabla_p f(p)^T\nabla_p f(p)$. The conditions for $\sigma(p)$ being maximal on $V$ are thus $v^T \, \nabla_p \sigma(p)=0$ and $f(p)=0$.

Thus, for maximizing $\sigma$ over

$$V =\{x_1^4 - x_1^2x_2^2 + x_2^4 - 4x_1^2 - 2x_2^2 - x_1 - 4x_2 + 1 = 0\}$$

we use the following code.

using HomotopyContinuation, DynamicPolynomials, LinearAlgebra
# using JLD2

@polyvar x[1:2]# initialize variables
f = x[1]^4 - x[1]^2*x[2]^2 + x[2]^4 - 4x[1]^2 - 2x[2]^2 - x[1] - 4x[2] + 1

∇ = differentiate(f, x) # the gradient
H = differentiate(∇, x) # the Hessian

g = ∇ ⋅ ∇
v = [-∇[2]; ∇[1]]
h = v' * H * v
dg = differentiate(g, x)
dh = differentiate(h, x)

F = [(g .* dh - ((3/2) * h).* dg) ⋅ v; f]

S = solve(F)

Then, S returns

julia> S
Result with 56 solutions
==================================
• 56 non-singular solutions (12 real)
• 0 singular solutions (0 real)
• 64 paths tracked
• random seed: 314288

Here is a picture of all solutions.

Curvature of general hypersurfaces

The definition of maximal curvature for points on general hypersurfaces is the maximal curvature of a geodesic through $p$:

$$\sigma(p) = \mathrm{max} \,\{\Vert \ddot{\gamma}(0)\Vert \mid \gamma \in V \text{ geodesic with\ } \gamma(0)=p, \Vert \dot{\gamma}(0)\Vert = 1\}$$

(curves with unit norm derivatives are called parametrized by arc-length).

Here, the equations are more complicated. The math for deriving them requires some knowledge on differential geometry. The theoretical part is at the end of this example. The reader who just wants to see code can execute the following script. It is written for the input data

$$V=\{x_1^2 + x_1x_2 + x_2^2 + x_1 - 3x_2 - 2x_3 + 2 = 0\}.$$

n = 3
@polyvar x[1:n] v[1:n] z[1:3]# initialize variables
f = x[1]^2 + x[1]*x[2] + x[2]^2  + x[1] - 3x[2]  -  2x[3] + 2# define f

∇f = differentiate(f, x) # the gradient
H = hcat([differentiate(∇f[i], x) for i in 1:n]...) # the Hessian

g = ∇f ⋅ ∇f
h = v ⋅ (H * v)
∇g = differentiate(g, x)
∇h = differentiate(h, x)
A = [(g .* ∇h - (3/2 * h) .* ∇g) ∇f H*v zeros(n); g.*(H*v) zeros(n) ∇f v]

# F is the system that is solved
F = [
    A * [1;z]
    v ⋅ v - 1;
    ∇f ⋅ v;
    f
]

S = solve(F)

# extract the real solutions
real_sols =  realsolutions(S)

# find the maximal σ
σ = map(p -> abs(h([x;v] => p[1:2n]) / g([x;v] => p[1:2n])^(3/2)), real_sols)
σ_max, i = findmax(σ)
p = real_sols[i][1:n]

The following animation was created with the @gif macro from the Plots.jl package.

The Algebraic Geometry of Curvature

Here is why the code above does what it is supposed to do.

It can be shown that for hypersurfaces $V = \{f=0\}.$ the maximal geodesic curvature at $p$ is the spectral norm of the derivative of the Gauss map $G: V \to \mathbb{P}^{n-1}\mathbb{R},\, p\mapsto (\mathrm{T}_p V)^\perp$. The Gauss map sends a point $p$ to the normal space of $V$ at $p$. Since $V$ is of codimension $1$, the normal space is a line and lines are parametrized by the $(n-1)$-dimensional projective space $\mathbb{P}^{n-1}\mathbb{R}$. Summarizing:

$$\sigma(p) = \max_{v\in \mathrm{T}_p V,\, w\in \mathrm{T}_L \mathbb{P}^{n-1}\mathbb{R} \, \Vert v\Vert = \Vert w \Vert_L =1} \,\langle w, DG(p)v\rangle_L.$$

where $L=G(p)$ and $\langle \,,\,\rangle_L$ is the metric on $\mathrm{T}_L \mathbb{P}^{n-1}\mathbb{R}$. What is this metric? First, $V$ being embedded in $\mathbb{R}^n$ inherits the usual euclidean inner product $\langle\;,\,\rangle$. The inner product on the tangent space to $L$ is as follows: if $L$ is a line through $q\in \mathbb{R}^n$, then $\mathrm{T}_L \mathbb{P}^{n-1}\mathbb{R} \cong q^\perp$ and the inner product on $q^\perp$ is $\langle \;,\,\rangle_L = \frac{\langle\;,\,\rangle}{\langle q,q \rangle}$. It follows that,

$$\sigma(p) = \max_{v\in \mathrm{T}_p V,\, w\in q^\perp \, v^Tv = 1, \,w^Tw = q^T q} \,\frac{w^T \,DG(p) \,v}{q^Tq}$$ where $q$ is a point on $G(p)=(\mathrm{T}_p V)^\perp$.

A point on $(\mathrm{T}_p V)^\perp$ that can be computed easily from the input data is the gradient

$$\nabla_p f = \left(\frac{\partial f}{\partial x_1}(p),\ldots, \frac{\partial f}{\partial x_n}(p)\right)^T,$$

so that

$$\sigma(p) = \max_{v,w\in \nabla_p f^\perp,\, v^Tv = 1,\, w^Tw = \nabla_p f^T\,\nabla_p f} \,\frac{w^T \,DG(p)\, v}{\nabla_p f^T\,\nabla_p f}.$$

It remains to compute $DG(p)$. For this let $\pi : \mathbb{R}^n \to \mathbb{P}^{n-1}\mathbb{R}$ be the projection that sends $q\in \mathbb{R}^n$ to the line through $q$. Then, the Gauss map is written as $G(p) = \pi(\nabla_p f).$ Consequently, by the chain rule of differentiation:

$$DG(p) = D\pi(\nabla_p f) \, H$$ where $H = \begin{bmatrix} \frac{\partial \nabla}{\partial x_1} & \ldots & \frac{\partial \nabla}{\partial x_n}\end{bmatrix}$ is the Hessian of $f$.

One can show that $D\pi(\nabla_p f)$ is the orthogonal projection onto $\nabla_p f^\perp$. If $I_n$ denotes the $n\times n$ identity matrix: $D\pi(\nabla_p f) = I_n - \frac{\nabla_p f \nabla_p f^T}{\nabla_p f^T \nabla_p f}$. From this it is easy to see that $w^T\,D\pi(\nabla_p f) = w^T$ for all $w\in \nabla_p f^\perp$. Therefore, the following is an equation for $\sigma(p)$:

$$\sigma(p) = \max_{v, w\in \nabla_p f^\perp \, v^Tv = 1,\, w^Tw = \nabla_p f^T\,\nabla_p f} \,\frac{w^T \,H\, v}{\nabla_p f^T\,\nabla_p f}.$$

Diving $w$ by $\sqrt{g}$ for $g:=\nabla_p f^T\,\nabla_p f$ and using that $H$ is symmetric we have

$$\sigma(p) = \max_{v \in \nabla_p f^\perp \, v^Tv = 1} \,\frac{v^T \,H\, v}{g^\frac{3}{2}}.$$

(for curves in the plane this is the formula from above). Finally, the following formula for $\sigma$

$$\sigma = \max_{p\in V, v \in\nabla_p f^\perp \, v^Tv = 1} \,\frac{v^T \,H\, v}{g^\frac{3}{2}}$$

(actually, the last $\max$ is a $\sup$).

Writing $h = v^T H v$, $\nabla h = (\frac{\partial h}{\partial x_i})^{1\leq i\leq n}$ and $\nabla g = (\frac{\partial g}{\partial x_i})^{1\leq i\leq n}$ the corresponding critical equations of this are:

  • $A \begin{bmatrix} 1\\ z \end{bmatrix} = 0$ for $A= \begin{bmatrix}\nabla h \cdot g - \frac{3}{2} \cdot h \cdot \nabla g & \nabla f & Hv& 0 \\ Hv & 0 & \nabla f & v \end{bmatrix}$ and $z=(z_1,z_2,z_3)^T$.

  • $f=0$.

  • $v^Tv = 1$

  • $\nabla f^T v = 0$

These are the equations solved with the code above. It would be interesting to understand the degree of the equations for generic $f$.

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