$\def\RR{\mathbb{R}}$Let $D$ be a closed disc, smoothly embedded in $\RR^3$. The Gauss-Bonnet theorem tells me that $\int \!\! \int_D K + \int_{\partial D} \kappa = 2 \pi$, where $K$ is the Gaussian curvature, and $\kappa$ is the geodesic curvature.

Let $g: D \to S^2$ be the Gauss map. Then $K = g^{\ast} \mathrm{Area}$, where $\mathrm{Area}$ is the standard area $2$-form on $S^2$. For simplicity, I'll assume that the Gauss map is injective on $D$. So $\int \!\! \int_D K = \int \! \! \int_{g(D)} \mathrm{Area}$.

Now, applying the Gauss-Bonnet theorem to $g(D)$, we get $\int \! \! \int_{g(D)} \mathrm{Area} + \int_{g(\partial D)} \kappa = 2 \pi$. So $\int_{\partial D} \kappa = \int_{g(\partial D)} \kappa$, where the two $\kappa$'s denote the respective Gaussian curvatures. For notational clarity, let $\kappa_1$ and $\kappa_2$ be the corresponding $1$-forms on $\partial D$ and $g(\partial D)$.

The simplest explanation would be if $\kappa_1 = g^{\ast} \kappa_2$. But I think this is false! In particular, this would suggest that if a portion of $\partial D$ were geodesic, so would be the same portion of $g(\partial D)$, and that would show that the Gauss map took geodesics to geodesics, which is not true.

My best guess is that there is some closed $1$-form $\eta$ on the tangent bundle $TS$ so that $\kappa_1 - g^{\ast} \kappa_2 = \eta|_{\partial D}$, where we have lifted $\partial D$ from $S$ to $TS$ by taking the velocity of a parametrization. The lifting of $\partial D$ to $TS$ is contractible in $TS$ (first shrink down to $S$ by dilating the tangent spaces, and then use that $\partial D$ is the boundary of $D$), so the integral of closed $1$-form on $\partial D$ is $0$.

Is my guess right? Where do I read about this? Or, if I'm wrong, is there a conceptual explanation of why $\int_{\partial D} \kappa_1 = \int_{g(\partial D)} \kappa_2$ which doesn't go through Gauss-Bonnet?


Great question. First of all, note that your preliminary assumption of injectivity of the Gauss map is sort of a red herring. By Gauss's definition (or the change of variables theorem), we'll have $\text{Area}(g(D)) = \int_D K\,dA$ in any event. But note that in the case of a flat disk, the Gauss map is constant, and so the $2\pi$ comes entirely from geodesic curvature for $D$, whereas there's no corresponding curvature for $g(\partial D)$.

In general, a Darboux frame ($e_1$ tangent to the curve, $e_2$ normal to the curve, still tangent to the surface) for $\partial D$ and a Darboux frame $e_1^*,e_2^*$ for $g(\partial D)$ will differ by rotation through some (varying) angle $\theta$. Then it's an easy calculation (standard for those of us in the moving frames world) that, writing $\omega_{12} = de_1\cdot e_2$, we have (i) $\omega_{12} = \kappa_1$ (in your notation), (ii) $\omega^*_{12} = \kappa_2$, and — the kicker — $\omega_{12}^* = \omega_{12} + d\theta$. By simple connectivity, $\int_{\partial D} d\theta = 0$, and this proves what you wanted.

EDIT: David, whenever you have two orthonormal frames $e_1,e_2$ and $e_1^*,e_2^*$ with $e_1^*=\cos\theta e_1+\sin\theta e_2$ and $e_2^*=-\sin\theta e_1+\cos\theta e_2$, set $\omega_{ij} = de_i\cdot e_j$ and $\omega_{ij}^* = de_i^*\cdot e_j^*$. Differentiating and dotting (noting that $de_i\cdot e_j=-de_j\cdot e_i$), we get \begin{align*} \omega_{12}^*=de_1^*\cdot e_2^* &= \big(\cos\theta\, de_1+\sin\theta\, de_2+(-\sin\theta e_1+\cos\theta e_2)d\theta\big)\cdot(-\sin\theta e_1+\cos\theta e_2) \\ &=de_1\cdot e_2 + d\theta = \omega_{12}+d\theta. \end{align*} Although it's only a brief introduction to this stuff, I have a short section on surface theory with differential forms in Chapter 3, Section 3 of my differential geometry notes.

  • $\begingroup$ I went and added the routine computation. You might be amused to see how it came out for someone who isn't in the moving frames world yet. $\endgroup$ – David E Speyer Apr 2 '15 at 2:23
  • $\begingroup$ Thanks for your edit and your notes! I guess the computation below is overly complicated because I am simultaneously checking that $\omega_{ij} = \kappa$ and the claim is true. I come at this having learned calculus with differential forms first, abstract Riemannian manifolds later and only now trying to teach myself the embedded story, so I can pass it on to students in the reverse order. $\endgroup$ – David E Speyer Apr 10 '15 at 14:14

$\def\RR{\mathbb{R}}$I finally managed to do the routine computation Ted Shifrin describes, and I'm writing up the details for the record.

It turns out that the Gauss map is a bit of a red herring. Rather, let $X$ and $Y$ be two surfaces in $\RR^3$ and let $f: \RR \to X$ and $g: \RR \to Y$ be two curves which have the following interesting property: For all $t$, the tangent planes $T_{f(t)}X$ and $T_{g(t)}Y$ are parallel. Let $\theta(t)$ be the angle between $f'(t)$ and $g'(t)$. Then $d \theta$ is the difference between the geodesic curvature $1$-forms. (This is a slightly vague statement that will be made better below.)

In particular, $Y$ could be $S^2$, and $g$ could be the composition of $f$ with the Gauss map $X \to S^2$, but this case isn't special in any way.

First, I need to explain how to write the geodesic curvature as a $1$-form. This is surely well known to experts, but most of the books I've been looking at define it as a number, and I had to convert to $1$-forms before I understood what was going on.

Let $C \subset X \subset \RR^3$ be a curve in an oriented surface in $3$-space. For $x \in X$, let $n_x$ be the unit normal vector. We define a $1$-form $\kappa_{C/X}$ on $C$ as follows: Let $\phi: \RR \to C$ be a parametrization of $C$ (or a segment of $C$). Let $\phi'(t) = |\phi'(t)| u(t)$, so $u(t)$ is a unit vector. Write $u'(t)$ for the derivative of $u(t)$ with respect to $t$. We define $\kappa_{C/X}$ so that $\phi^{\ast} \kappa = \det(u(t)\ u'(t)\ n_{u(t)}) dt$. Clearly, if $\phi$ is the arc length parametrization, then $\kappa$ is $(\mbox{geodesic curvature}) d (\mbox{arc length})$. But I claim that this definition works for any parametrization.

The reason is as follows: Suppose we change to another parametrization, with parameter $s$. Then $u$ and $n$ don't change. We have $du/dt = (du/ds) (ds/dt)$. But $dt = (ds/dt)^{-1} ds$. Since $\det$ is linear, these cancel.

I also need

Lemma Let $a$, $b$ and $c$ be three vectors in $\RR^3$ such that $a$ is a unit vector and $b$ is normal to $a$. Then $(a \times b) \cdot (a \times c) = b \cdot c$.

Proof We may assume that $a = (0,0,1)$. Then $b = (b_1, b_2, 0)$ and $c=(c_1, c_2, c_3)$. The claimed result is $(b_2, -b_1,0) \cdot (c_2, -c_1,0) = (b_1, b_2, 0) \cdot (c_1,c_2,c_3)$, which is obvious. $\square$

Now, let $X$, $f$, $Y$, $g$ and $\theta$ be as above. We write $n(t)$ for the common normal to $T_{f(t)} X$ and $T_{g(t)} Y$. Write $f(t) = |f(t)| u(t)$ and $g(t) = |g(t)| v(t)$, so $u$ and $v$ are unit vectors.

From our previous computations about geodesic curvature, we want to show $$\frac{d \theta(t)}{dt} = \det(v(t), v'(t), n(t)) - \det(u(t), u'(t), n(t))$$ where prime is differentiation with respect to $t$. (Note that it was important to work out the formula for geodesic curvature without assuming we have an arc length parametrization, since it is unlikely that both $f(t)$ and $g(t)$ are unit speed.)

Now, since $u$ and $v$ are unit vectors, $\theta = \cos^{-1}(u \cdot v)$. We have $$\frac{d \theta}{dt} = - \ \frac{d (u \cdot v)/dt}{\sin \theta} = - \ \frac{u' \cdot v + u \cdot v'}{\sin \theta}$$ so we must show that $$u' \cdot v + u \cdot v' = - \det(v, v', (\sin \theta) n) + \det(u, u', (\sin \theta) n). \quad (\ast)$$

But $u$ and $v$ are unit vectors and $n$ is normal to both of them, so $(\sin \theta) n = u \times v$ and we can write the right hand side of $(\ast)$ as $$\det(v, v', v \times u) + \det(u, u', u \times v) = (v \times v') \cdot (v \times u) + (u \times u') \cdot (u \times v).$$ We must show that $$u \cdot v' + u' \cdot v = (v \times v') \cdot (v \times u) + (u \times u') \cdot (u \times v).$$

The result now follows from the Lemma, using $(a,b,c) = (v, v', u)$ in the first case and $(u, u', v)$ in the second. (Since $u$ and $v$ are unit length, we have $u \perp u'$ and $v \perp v'$, as the lemma requires.) $\square$


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