Rolling out pie dough

Question

I was rolling out a pie crust tonight. I would like to produce a perfect circle, but part way through the border of my crust was a rather different closed curve. I am aware of the Riemann mapping theorem which says I can map my existing crust to a circle with an angle preserving map, but that may stretch and shrink locally, which will make the crust an uneven thickness. I don't care about preserving angles. If we assume my crust is uniformly thick now, I want to map it to an equal area disk preserving local areas so the thickness stays uniform. Is there a theorem that says I can?

Answer 1

I'm going to assume your pizza has smooth boundary.

Let $(M_i, \omega_i)$ be compact oriented smooth manifolds, possibly with boundary, equipped with a volume form. Suppose $M_1$ is oriented diffeomorphic to $M_2$, and they have the same volume $\int_{M_i} \omega_i.$ Then there is a diffeomorphism $f: M_1 \to M_2$ with $f^*\omega_2 = \omega_1$.

We can prove this by a trick of Moser's. First, reduce to the case that $M_1 = M_2$. Then we want to show there's a self-diffeomorphism of $M$ such that $f^* \omega_1 = \omega_0$. Because the volume forms give the same orientation of $M$, they are (smoothly) homotopic as volume forms; indeed, as volume forms in the same cohomology class: there is a smooth path $\omega_t$ of volume forms connecting them, with $[\omega_0] = [\omega_t]$. What I would like is to find a flow $f_t$ of a time-varying vector field $X_t$ such that $f_t^* \omega_t = \omega_0$; taking derivatives, if this were true, $$0 = \frac{d}{dt} (f_t^*\omega_t) = f_t^* (\mathcal L_{X_t} \omega_t) + f_t^*\left(\frac{d}{dt}\omega_t\right) = f_t^* \left(d\iota_{X_t} \omega_t+ \frac{d}{dt} \omega_t\right).$$ (Note that for the flow to make sense on a manifold with boundary, $X_t$ should always be tangent to the boundary, on the boundary.) Going backwards, if we can find a vector field $X_t$ with $d\iota_{X_t} \omega_t = - \frac{d}{dt} \omega_t$, then $X_t$ generates the desired flow. Because $[\frac{d}{dt} \omega_t] = 0$, $\frac{d}{dt} \omega_t = d\alpha_t$, where we choose $\alpha_t$ to vary smoothly in $t$. The nondegeneracy of the volume form is precisely what allows me to solve $\iota_{X_t} \omega_t = -\alpha_t$ uniquely for all $t$.

There is only one problem in the case with boundary: forcing $X_t$ to be tangent to the boundary. If it solves the above equation, this is precisely the same as saying that $\alpha_t\big|_{\partial M} = 0$. We can choose $\alpha_t$ to satisfy $d\alpha_t = \frac{d}{dt}\omega_t$ and $\alpha_t\big|_{\partial M} = \eta_t$ for any choice of $\eta_t$ on the boundary with zero volume (this is essentially the statement that $\frac{d}{dt} \omega_t$ is zero in the relative cohomology group $H^n(M,\partial M)$ - true precisely because $\int \omega_t$ is constant), this proving that we can choose an appropriate $\alpha_t$, thus obtaining an appropriate $X_t$, thus proving the theorem.

This in hand, your desired theorem follows from Schoenflies' theorem that the closed region bounded by any smooth circle in the plane is diffeomorphic to a disc.