# Lecture 17—Discrete Curvature II (Variational Viewpoint)

In this lecture we wrap up our discussion of discrete curvature, and see how it all fits together into a single unified picture that connects the integral viewpoint, the variational viewpoint, and the Steiner formula. Along the way we’ll touch upon several of the major players in discrete differential geometry, including a discrete version of Gauss-Bonnet, Schläfli’s polyhedral formula, and the cotan Laplace operator—which will be the focus of our next set of lectures.

## 3 thoughts on “Lecture 17—Discrete Curvature II (Variational Viewpoint)”

1. Lqq88888 says:

Hmmmm~~~~, trying to get head around…
So why is the variation of energy $\delta E(f)$ a differential form? Isn’t it $\nabla_f(E)$?
And I don’t understand why do we get, say, $\int_M{H}dA$ from $HNdA$ by “$\int_M$” in the diagram.

1. Keenan says:

Good questions. There isn’t a formal writeup of what this diagram means in the notes yet, so some additional details are indeed in order. First, in the diagram, the integration really should be something like $\int_M \langle N, \cdot \rangle\ dA$. In other words, to go from a given vector quantity to the next scalar quantity in the sequence, we take the inner product with the normal (yielding a scalar) and then integrate. The only purpose here, really, is to point out that you’re going from one quantity to the next.

In terms of the variations, it’s often nice to write these out in terms of differential 2-forms rather than scalar functions. The reason is that the quantities $NdA$, $HNdA$, and $KNdA$ have nice simple expressions in terms of the immersion $f$ and the Gauss map $N$ (namely: $\frac{1}{2} df \wedge df$, $\frac{1}{2} df \wedge dN$, and $\frac{1}{2} dN \wedge dN$, respectively), whereas the corresponding 0-forms require uglier expressions. E.g., $H = \frac{1}{2} \langle N, \ast (df \wedge dN) \rangle$. Personally, I’d prefer not differentiating through all this extra “stuff” for no good reason!

To make it clear what I mean, let’s work through the derivation of the mean curvature normal as the gradient/first variation of surface area. As shorthand, I’ll adopt the notation
$\dot{\phi} := \frac{d}{dt}|_{t=0} \phi,$
\ie, a dot above a variable indicates the time derivative at time $t=0$. Consider, then, a family of immersions $f: M \to \mathbb{R}^3$ given by
$f(t) = f_0 + tuN_0$
where $f_0$ is some initial immersion, $N_0$ is its Gauss map, and $u: M \to \mathbb{R}$ is an arbitrary smooth function. One can easily show that
$df \wedge df = 2 N dA,$
where $dA$ is the area 2-form induced by $f$ and $N$ is its Gauss map. Then
$\begin{array}{rcl} \frac{1}{2}\frac{d}{dt}(df \wedge df) &=& d\dot{f} \wedge df \\ &=& N_0du \wedge df_0 + (1+t)u(dN_0 \wedge df_0) + tu(N_0 du \wedge dN_0) + tu^2(dN_0 \wedge dN_0), \end{array}$
hence
$\left.\frac{1}{2}\frac{d}{dt}(df \wedge df)\right|_{t=0} = N_0 du \wedge df_0 + u(dN_0 \wedge df_0).$
A standard calculation shows that $dN_0 \wedge df_0 = 2H_0 N_0 dA_0$, and if $X,Y$ are orthonormal with respect to the metric induced by $f_0$ then we also have
$\begin{array}{rcl} (N_0 du \wedge df_0)(X,Y) &=& N_0 du(X) \times df_0(Y) – N_0 du(Y) \times df_0(X) \\ &=& -du(X) df_0(X) – du(Y) df_0(Y) \\ &=& -df_0(du(X)X+du(Y)Y) \\ &=& -df_0(\nabla u), \end{array}$
i.e., $N_0 du \wedge df_0 = -df_0(\nabla u)dA_0$. Making these two substitutions, we get
$\left.d\dot{f} \wedge df\right|_{t=0} = (2 u H_0 N_0 – df_0(\nabla u)) dA_0.$
Finally, suppose we rewrite the relationship $df \wedge df = 2NdA$ as
$dA = \frac{1}{2}\langle N, df \wedge df \rangle.$
Then
$\left.2d\dot{A}\right|_{t=0} = \langle \dot{N}, df \wedge df \rangle + \langle N, 2d\dot{f} \wedge df \rangle.$
The first term vanishes because $df \wedge df$ is normal-valued, but $\dot{N}$ must always be tangential (since $N$ has unit length). For the second term, we get
$2\langle N, (2 u H_0 N_0 – df_0(\nabla u)) dA_0 \rangle.$
Here only the first term remains (since $df_0$ yields tangent vectors) and we get the formula
$\left. d\dot{A} \right|_{t=0} = 2uH_0 dA_0.$
In other words, if we move at speed $u$ along the normal direction, then the change in the area element will be proportional to twice the speed function $u$ times the mean curvature $H$.

Ok, so that gives us a sort of directional derivative of area with respect to some particular direction of motion. How do we get the gradient or first variation of surface area? We just look for the vector field $\nabla_f \text{area}$ that satisfies the relationship
$\langle\!\langle \nabla_f \text{area}, uN \rangle\!\rangle = \int_M 2u H_0\ dA_0$
for all $u: M \to \mathbb{R}$, where $\langle\!\langle U, V \rangle\!\rangle := \int_M \langle U, V \rangle\ dA_0$ denotes the $L^2$ inner product on vector fields on $f(M)$. In other words, the gradient is the thing that yields the directional derivative when you take its inner product with a given direction. But the only way to satisfy
$\int_M \langle \nabla_f \text{area}, uN \rangle\ dA_0 = \int_M 2u H_0\ dA_0$
is if
$\nabla_f \text{area} = 2H_0 N_0,$
as claimed. In other words, the direction we should move the surface to increase area as quickly as possible is in the normal direction, with speed equal to twice the current mean curvature.

1. Keenan says:

So in the end, you’re right—perhaps $dA$ shouldn’t show up in that diagram! Perhaps it would be better to omit the $dA$s, and write the downward arrows as $\int_M \langle N, \cdot \rangle\ dA$.