I'm working on the following problem from Lee's "Introduction to Smooth Manifolds":
Let $D \subseteq \mathbb R^3$ be the surface obtained by revolving the circle $(r-2)^2 + z^2 = 1$ around the z-axis, with the induced Riemannian metric from $\mathbb R^3$, and let $C \subseteq D$ be the “inner circle” defined by $C = \{(x,y,z) : z=0, \, x^2 + y^2 = 1\}$. Show that $C$ is calibrated, and therefore is the shortest curve in its homology class.
In this case, a calibration of a Riemannian manifold $M$ is a closed $p$-form $\omega$ on $M$ so that $\omega(v_1, \ldots, v_p) \leq 1$ for every orthonormal set $\{v_1, \ldots, v_p\}$, and a Riemannian submanifold $S \subseteq M$ is calibrated if there is a calibration $\omega$ so that $\iota_S^*\omega$ is the induced Riemannian volume form on $S$.
So I need to find a 1-form $\omega \in \Omega^1(D)$ for which $\omega(v) \leq 1$ for every unit tangent vector $v\in TD$, and $\iota_C^* \omega$ is the induced Riemannian volume form on $C$. Let $F(\theta, t) = \big((2t+\cos t)\cos\theta, (2+\cos t)\sin\theta, \sin t\big)$; then $F : [0,2\pi]^2 \to \mathbb R^2$ parametrizes $D$. My original thought was to let $\omega$ be the 1-form $(\overline F_\theta)^\flat$, i.e. $(\overline F_\theta)^\flat(v) = g_D (v, \overline F_\theta)$ for every $v \in TD$, where $g_D$ is the induced Riemannian metric on $D$ and $\overline F_\theta = F_\theta / |F_\theta|$ is the normalization of the tangent vector $F_\theta$. It seems clear to me in this case that $\iota^*_C \omega$ is the induced volume form on $C$, and that $\omega(v) \leq 1$ for unit tangent vectors $v$. However, I have two concerns:
- I'm not sure how to show $\omega = (\overline F_\theta)^\flat$ is closed, and
- if I could show the above, why would a similar argument not apply to the outer circle of $D$, which is clearly not minimizing in its homology class?
I would appreciate any thoughts or clarifications.
