Embedding surfaces inside small domains with minimal distortion

Abstract

Given two-dimensional Riemannian manifolds $\mathcal {M},\mathcal {N}$, we prove a lower bound on the distortion of embeddings $\mathcal {M}\rightarrow \mathcal {N}$, in terms of the areas’ discrepancy $V_{\mathcal {N}}/V_{\mathcal {M}}$, for a certain class of distortion functionals. For $V_{\mathcal {N}}/V_{\mathcal {M}} \ge 1/4$, homotheties, provided they exist, are the unique energy minimizing maps attaining the bound, while for $V_{\mathcal {N}}/V_{\mathcal {M}} \le 1/4$, there are non-homothetic minimizers. We characterize the maps attaining the bound, and construct explicit non-homothetic minimizers between disks. We then prove stability results for the two regimes. We end by analyzing other families of distortion functionals. In particular we characterize a family of functionals where no phase transition in the minimizers occurs; homotheties are the energy minimizers for all values of $V_{\mathcal {N}}/V_{\mathcal {M}}$, provided they exist.

Appendices

A Convexity results

Proof

[of Lemma 2.3] The strict convexity of F in $[a,\infty )$ implies that for $x>a\ge y$,

$$\begin{aligned} F\left( t x + (1-t) y\right) = tF(x) + (1-t)F(y) \Rightarrow t\in \{0,1\}. \end{aligned}$$

Denote $A = g^{-1}((a,\infty ))$. If $\mu (A)=0$ or $\mu (A^c)=0$, we are done: If $\mu (A)=1$, then $g>a$ a.e., so we are in the domain where F is strictly convex. If $\mu (A)=0$ then $g \le a$ a.e. - and the only case we need to check is when $\int _X g \in [a,\infty )$. We then must have $\int _X g =a$, so $g=a$ a.e.

We show that $0<\mu (A)<1$ cannot occur. Denote $x=\fint _A g$, and $y=\fint _{A^c} g$; then $y\le a<x$ and $\int _X g= \mu (A)x + \mu (A^c)y$. Thus, using the equality assumption and Jensen’s inequality,

$$\begin{aligned} F\left( \int _X g\right) = \int _X F\circ g = \int _A F\circ g + \int _{A^c} F\circ g \ge \mu (A)F(x) + \mu (A^c)F(y)\ge F\left( \int _X g\right) . \end{aligned}$$

Therefore equality holds, so the comment at the beginning of the proof implies that either $\mu (A)=1$ or $\mu (A^c)=1$. $\square$

The following lemma is a variant of Jensen inequality, when we have convexity only on a partial subset of our domain.

Lemma A.1

Let $F:(0,\infty ) \rightarrow [0,\infty )$. Assume that $F|_{(0,1]}$ is strictly decreasing and convex, with $F^{-1}(0)=\{1\}$.

Let $g:X \rightarrow (0,\infty )$ be a measurable function defined on a probability space X with $\int _X g \in (0,1]$. Then $F(\int _X g) \le \int _X F\circ g\,\,$ and if equality occurs $g(x) \in (0,1]$ a.e.  If $F|_{(0,1]}$ is strictly convex equality occurs if and only if g(x) is constant a.e.

Equivalently: F is convex at every point in [0, 1].

Proof

Set $F^*(x)=F(\min (x,1))$; $F^*\le F$ pointwise, and $F(x)=F^*(x)$ for $x\in [0,1]$. Since $\int _X g\in (0,1]$, $F\left( \int _X g\right) =F^*\left( \int _X g\right) .$ Since $F^*$ is convex

$$\begin{aligned} F\left( \int _X g\right) =F^*(\int _X g)\le \int _X F^*{\circ } g \le \int _X F{\circ } g \end{aligned}$$

as desired. If there is an equality, then we have $F^*{\circ } g=F {\circ } g$ a.e., so $g \in (0,1]$ a.e. If $F|_{(0,1]}$ is strictly convex then equality occurs if and only if g(x) is constant a.e. 

We prove that $F^*$ is convex: (Just draw it!:)

We need to show that $F^*(t x + (1-t) y) \le tF^*(x) + (1-t)F^*(y)$. If $t x + (1-t) y \ge 1$, then the LHS vanishes, so we are done. Thus, suppose that $t x + (1-t) y \le 1$. If x, y are both not greater than 1, then the assertion is just the convexity of $F|_{(0,1]}$.

So, we may assume W.L.O.G that $0<x\le 1< y$, and $t x + (1-t) y \le 1$. The assumptions imply $1\ge t x + (1-t) y \ge t x + (1-t) 1$, hence

$$\begin{aligned} F\left( t x + (1-t) y\right) \le F\left( t x + (1-t) 1\right) \le tF(x) + (1-t)F(1) =tF(x), \end{aligned}$$

where the second inequality is due to the convexity of $F|_{(0,1]}$. Thus

$$\begin{aligned} F^*\left( t x + (1-t) y \right) =F\left( t x + (1-t) y\right) \le tF(x)= tF^*(x)=tF^*(x) + (1-t)F^*(y). \end{aligned}$$

$\square$

Proof

[of Lemma 5.10] We prove the claim under the assumption that F is convex on $(1-\epsilon ,1]$. The proof for the case of strict convexity is identical. Since $F|_{[1-\epsilon ,1]}$ is (strictly) convex we have

$$\begin{aligned} F(x) \ge T_y(x):=F(y)+F_-'(y) (x-y) \, \, \, \text { for every } \, \, x,y \in [1-\epsilon ,1]. \end{aligned}$$

(A.1)

Let $y \in [1-\epsilon ,1].$ Since F is decreasing on (0, 1], $F_-'(y) \le 0$, so $x \mapsto T_y(x)$ is decreasing. Since $T_y(1) \le F(1)= 0$, $T_y(x) \le 0 \le F(x)$ for every $x \ge 1$, so Inequality (A.1) holds for every $x \in [1-\epsilon ,\infty )$ and every $y \in [1-\epsilon ,1].$

Let $x \in (0,1-\epsilon ]$. For every $y <1$ sufficiently close to 1, we have

$$\begin{aligned} F(x) \ge F(1-\epsilon ) {\mathop {\ge }\limits ^{(1)}} F(y)+|F_-'(y)|\ge F(y)+F_-'(y) (x-y)= T_y(x), \end{aligned}$$

where inequality (1) follows from $\lim _{y \rightarrow 1^-} F_-'(y)=F_-'(1)=0$ together with $\lim _{y \rightarrow 1^-}F(y)=F(1)=0$. (We used the fact that the left derivative of a convex function is left-continuous.) Thus, we proved that for $y <1$ sufficiently close to 1, Inequality (A.1) holds for every $x \in (0,\infty )$, which means that $F=F|_{(0,\infty )}$ is convex at all such y. $\square$

B Proof of Lemma 3.4

Lemma 3.4 is concerned with the distance of a matrix to the set K. We will therefore need the following claim:

Proposition B.1

Let $A \in M_2$ with $\det A \ge 0$, and let $\sigma _1\le \sigma _2$ be its singular values. Then

$$\begin{aligned} {\text {dist}}^2(A,K)={\left\{ \begin{array}{ll} \frac{1}{2} \left( \sigma _1+\sigma _2-1\right) ^2, &{} \text { if }\, \sigma _2 \le \sigma _1 + 1 \\ \sigma _1^2+\left( \sigma _2-1 \right) ^2 , &{} \text { if }\,\sigma _2 \ge \sigma _1 + 1 \end{array}\right. } \end{aligned}$$

(B.1)

We first use Proposition B.1 for proving Lemma 3.4, then we prove it.

Proof

[Of Lemma 3.4] Let $\sigma _1,\sigma _2$ be the singular values of A. Then

$$\begin{aligned}&{\text {dist}}^2(A,{\text {SO}}_2)-(1-2\det A) \nonumber \\&\quad = (\sigma _1-1)^2+(\sigma _2-1)^2-(1-2\sigma _1 \sigma _2) \nonumber \\&\quad = \sigma _1^2+\sigma _2^2+1-2\sigma _1-2\sigma _2+2\sigma _1 \sigma _2 \nonumber \\&\quad =\left( \sigma _1+\sigma _2-1\right) ^2. \end{aligned}$$

(B.2)

The presence of the sum $\sigma _1+\sigma _2$ is no coincidence here! We represented the symmetric polynomial $P(\sigma _1,\sigma _2)=(\sigma _1-1)^2+(\sigma _2-1)^2$ as a polynomial in $\sigma _1 \sigma _2$ and $\sigma _1+\sigma _2$: $(x-1)^2+(y-1)^2=(1-2xy)+\left( x+y-1\right) ^2.$

Comment: Equation (B.2) implies that ${\text {dist}}^2(A,{\text {SO}}_2) \ge 1-2\det A$ and equality holds exactly when $\sigma _1+\sigma _2=1$. This gives another proof for Lemma 2.1 in the regime where $\det A \le 1/4$.

Equation (B.2) and Proposition B.1 imply that if $\sigma _2 \le \sigma _1 + 1$ then

$$\begin{aligned} {\text {dist}}^2(A,{\text {SO}}_2)=(1-2\det A) +2{\text {dist}}^2(A,K). \end{aligned}$$

Suppose that $\sigma _2 \ge \sigma _1 + 1$. Setting $x=\sigma _1, y=\sigma _2-1$, $x,y \ge 0$, hence

$$\begin{aligned} x^2+y^2 \le (x+y)^2 \le 2(x^2+y^2). \end{aligned}$$

Equations (B.2) and (B.1) imply that $(x+y)^2={\text {dist}}^2(A,{\text {SO}}_2)-(1-2\det A)$ and $x^2+y^2={\text {dist}}^2(A,K)$ which completes the proof. $\square$

1.1 B.1 Computing ${\text {dist}}(.,K)$

We prove Proposition B.1. By Definition 1.2 $K=\cup _{0\le s\le \frac{1}{4}}K_s$; we use the following lemma (that we prove below):

Lemma B.2

Let $0\le \sigma _1 \le \sigma _2$, and let $A \in M_2$ satisfy $\det A \ge 0$. Then

$$\begin{aligned} {\text {dist}}^2(A,K_{\sigma _1,\sigma _2})=\left( \sigma _1(A)-\sigma _1 \right) ^2+\left( \sigma _2(A)-\sigma _2 \right) ^2. \end{aligned}$$

(B.3)

Proof

[Of Proposition B.1]

Define $F:[0,\frac{1}{4}] \rightarrow [0,\infty )$ by $F(s)={\text {dist}}^2(A,K_s)$. Since $K=\cup _{0\le s\le \frac{1}{4}}K_s$,

$$\begin{aligned} {\text {dist}}^2(A,K)=\min _{0 \le s \le \frac{1}{4}}{\text {dist}}^2(A, K_s)=\min _{0 \le s \le \frac{1}{4}}F(s). \end{aligned}$$

We prove that $\min _{0 < s \le \frac{1}{4}} F(s)$ equals the RHS of Equation (B.1).

Given $0 \le s \le 1/4$, let $\sigma _1(s) \le \sigma _2(s)$ be the unique numbers satisfying

$$\begin{aligned} \sigma _1(s)\sigma _2(s)=s,\,\,\,\sigma _1(s)+\sigma _2(s)=1. \end{aligned}$$

Since $K_s=K_{\sigma _1(s),\sigma _2(s)}$, Lemma B.2 implies that

$$\begin{aligned} F(s)=\left( \sigma _1-\sigma _1(s) \right) ^2+\left( \sigma _2-\sigma _2(s) \right) ^2. \end{aligned}$$

Note that $\sigma _1(s)=\frac{1}{2} - \frac{\sqrt{1-4s}}{2},\sigma _2(s)=\frac{1}{2} + \frac{\sqrt{1-4s}}{2}$; thus $\sigma _i(s), F(s)$ are smooth functions of s on $[0,\frac{1}{4})$ and continuous on $[0,\frac{1}{4}]$. We shall use the following lemma, which we prove at the end of the current proof.

Lemma B.3

The function $F(s)={\text {dist}}^2(A,K_s)$ has a critical point $0\le s^* < \frac{1}{4}$ if and only if $\sigma _1<\sigma _2 \le \sigma _1+1$. When such a critical point exists, it is unique and satisfies $F(s^*)=\frac{1}{2} \left( \sigma _1+\sigma _2-1\right) ^2$.

Next, we claim that if $\sigma _1<\sigma _2 \le \sigma _1+1$, then $\min _{0 \le s \le \frac{1}{4}} F(s)=F(s^*)$. (One can prove that F is convex, so any critical point is a global minimum, but we won’t do that.) The possible candidates for minimum points are interior critical points $s \in (0,1/4)$, and the endpoints $0,1/4$. Thus, we need to show that

$$\begin{aligned} F(0) \ge \frac{1}{2} \left( \sigma _1+\sigma _2-1\right) ^2, F\left( \frac{1}{4}\right) \ge \frac{1}{2} \left( \sigma _1+\sigma _2-1\right) ^2. \end{aligned}$$

Since $\sigma _1(\frac{1}{4})=\sigma _2(\frac{1}{4})=\frac{1}{2}$, $F( \frac{1}{4})=\sigma _1^2+\sigma _2^2+\frac{1}{2}-\sigma _1-\sigma _2$; thus

$$\begin{aligned}&F\left( \frac{1}{4}\right) \ge \frac{1}{2} \left( \sigma _1+\sigma _2-1\right) ^2 \\&\quad \iff \sigma _1^2+\sigma _2^2+1-2\sigma _1-2\sigma _2+2\sigma _1\sigma _2 \le 2\sigma _1^2+2\sigma _2^2+1-2\sigma _1-2\sigma _2 \\&\quad \iff 0 \le \sigma _1^2+\sigma _2^2-2\sigma _1\sigma _2=(\sigma _1-\sigma _2)^2. \end{aligned}$$

Since $\sigma _1(0)=0,\sigma _2(0)=1$, $F(0)=\sigma _1^2+\sigma _2^2-2\sigma _2+1.$ Thus,

$$\begin{aligned}&F( 0) \ge \frac{1}{2} \left( \sigma _1+\sigma _2-1\right) ^2 \\&\quad \iff \sigma _1^2+\sigma _2^2+1-2\sigma _1-2\sigma _2+2\sigma _1\sigma _2 \le 2\sigma _1^2+2\sigma _2^2-4\sigma _2+2 \\&\quad \iff -2\sigma _1-2\sigma _2+2\sigma _1\sigma _2 \le \sigma _1^2+\sigma _2^2-4\sigma _2+1 \\&\quad \iff 2(\sigma _2-\sigma _1) \le (\sigma _2-\sigma _1)^2+1, \end{aligned}$$

which always holds since $2x \le x^2+1$ holds for every real x.

If $\sigma _1=\sigma _2$ or $\sigma _2 > \sigma _1 + 1$, then F has no critical points. In these cases all is left to do is to compare F(0) and $F( 1/4)$. A direct computation shows that $F(0) \le F( \frac{1}{4})$ iff $\sigma _2 \ge \sigma _1 + 1/2$, from which the conclusion follows. $\square$

Proof

[Of Lemma B.3]

$$\begin{aligned} 2F'(s)=-\left( \sigma _1-\sigma _1(s)\right) \sigma _1'(s)-\left( \sigma _2-\sigma _2(s)\right) \sigma _2'(s). \end{aligned}$$

Since $\sigma _1(s)+\sigma _2(s)=1 \Rightarrow \sigma _1'(s)=-\sigma _2'(s)$, we get

$$\begin{aligned} 2F'(s)=\sigma _1'(s)\left( \Delta \sigma - \Delta \sigma (s)\right) , \end{aligned}$$

where $\Delta \sigma :=\sigma _2-\sigma _1, \Delta \sigma (s):=\sigma _2(s)-\sigma _1(s)$.

$\sigma _1'(s)>0$ so $F'(s)=0$ if and only if $\Delta \sigma = \Delta \sigma (s)$. Since $s \rightarrow \Delta \sigma (s)=\sqrt{1-4s}$ is strictly decreasing, the uniqueness of the critical point is established. We now prove existence. The condition $\sigma _1 < \sigma _2 \le \sigma _1+1$ is necessary: $0 \le \sigma _i(s) \le 1$ implies that if $F'(s^*)=0$ then $\sigma _2-\sigma _1= \Delta \sigma (s^*) \le 1$. Furthermore, since $s^* < \frac{1}{4}$, $\Delta \sigma (s^*)>0$, the equality $\Delta \sigma = \Delta \sigma (s^*)$ implies that $\sigma _1<\sigma _2$.

To prove sufficiency, note that for every $0 < r \le 1$, there exists $s \in [0,\frac{1}{4})$ satisfying $\Delta \sigma (s)=r$. Now, let $s^*$ be the critical point. Since $\sigma _1(s^*)+\sigma _2(s^*)=1$,

$$\begin{aligned} -2\sigma _1(s^*)=-1+\sigma _2(s^*)-\sigma _1(s^*)=\sigma _2-\sigma _1-1, \end{aligned}$$

thus

$$\begin{aligned} \sigma _1-\sigma _1(s^*)=\frac{1}{2} \left( 2\sigma _1-2\sigma _1(s^*)\right) =\frac{1}{2} \left( 2\sigma _1+\sigma _2-\sigma _1-1\right) =\frac{1}{2} \left( \sigma _1+\sigma _2-1\right) , \end{aligned}$$

so

$$\begin{aligned} F(s^*)=2\left( \sigma _1-\sigma _1(s^*) \right) ^2=\frac{1}{2} \left( \sigma _1+\sigma _2-1\right) ^2, \end{aligned}$$

where in the first equality we have used the implication $F'(s^*)=0 \Rightarrow \sigma _2-\sigma _2(s^*) = \sigma _1-\sigma _1(s^*)$. This completes the proof. $\square$

Proof

[Of Lemma B.2]

Given $X \in K_{\sigma _1,\sigma _2}$, we have

$$\begin{aligned} |A-X|^2=|A|^2 + |X|^2 -2\langle A,X\rangle . \end{aligned}$$

Since |A| and $|X|=\sigma _1^2+\sigma _2^2$ are constant, we need to maximize $X \rightarrow \langle A,X\rangle$ over $X \in K_{\sigma _1,\sigma _2}$. By Von Neumann’s trace inequality,

$$\begin{aligned} \langle A,X\rangle ={\text {tr}}(A^TX) \le \sigma _1(A)\sigma _1+\sigma _2(A)\sigma _2. \end{aligned}$$

It remains to show that this upper bound is realized by some $X \in K_{\sigma _1,\sigma _2}$. Using the bi-${\text {SO}}_2$ invariance, we may assume that $A={\text {diag}}(\sigma _1(A),\sigma _2(A))$ is positive semidefinite and diagonal; taking $X={\text {diag}}(\sigma _1,\sigma _2)$ then realizes the bound. Thus

$$\begin{aligned} {\text {dist}}^2(A,K_{\sigma _1,\sigma _2})&=\sigma _1(A)^2+\sigma _2(A)^2+\sigma _1^2+\sigma _2^2-2\sigma _1(A)\sigma _1-2\sigma _2(A)\sigma _2 \\&=\left( \sigma _1(A)-\sigma _1 \right) ^2+\left( \sigma _2(A)-\sigma _2 \right) ^2. \end{aligned}$$

Comment: In the reduction of the problem to the diagonal positive semidefinite case, we explicitly use the assumption that $\det A \ge 0$. Indeed, let $A=U\Sigma V^T$ be the SVD of A. If $\det A>0$, then either both $U,V \in {\text {SO}}_2$ or both $U,V \in {\text {O}}^{-}_2$. In the latter case, we can multiply by ${\text {diag}}\left( -1,1\right)$ from both sides of $\Sigma$ to make them in ${\text {SO}}_2$. A similar argument works when $\det A=0$. $\square$

C Estimating ${\text {dist}}(.,{\text {CO}}_2)$

Proof

[Of Lemma 3.5]

First, expand

$$\begin{aligned}&{\text {dist}}^2(A,{\text {SO}}_2)-2(\sqrt{\det A}-1)^2 \nonumber \\&\quad = (\sigma _1-1)^2+(\sigma _2-1)^2-2(\sqrt{\sigma _1 \sigma _2}-1)^2 \nonumber \\&\quad = \left( \sigma _1^2+\sigma _2^2-2\sigma _1 \sigma _2 \right) -2\sigma _1-2\sigma _2+4\sqrt{\sigma _1 \sigma _2} \nonumber \\&\quad =\left( \sigma _2-\sigma _1\right) ^2-2\left( \sqrt{\sigma }_2-\sqrt{\sigma }_1\right) ^2 \le \left( \sigma _2-\sigma _1\right) ^2=2{\text {dist}}^2(A,{\text {CO}}_2). \end{aligned}$$

(C.1)

Now,

$$\begin{aligned} \left( \sigma _2-\sigma _1\right) ^2-2\left( \sqrt{\sigma }_2-\sqrt{\sigma }_1\right) ^2 = \left( \sqrt{\sigma }_2-\sqrt{\sigma }_1\right) ^2 \left( \left( \sqrt{\sigma }_2+\sqrt{\sigma }_1\right) ^2-2\right) . \end{aligned}$$

(C.2)

By the AM-GM inequality, if $\det A \ge 1/4$ then

$$\begin{aligned} \left( \sqrt{\sigma }_2+\sqrt{\sigma }_1\right) ^2 \ge 4\sqrt{ \sigma _1\sigma _2} \ge 2, \end{aligned}$$

which implies

$$\begin{aligned} \left( \sqrt{\sigma }_2+\sqrt{\sigma }_1\right) ^2-2 \ge \left( \sqrt{\sigma }_2+\sqrt{\sigma }_1\right) ^2 - 4\sqrt{ \sigma _1\sigma _2}= \left( \sqrt{\sigma }_2-\sqrt{\sigma }_1\right) ^2. \end{aligned}$$

(C.3)

Equations (C.2), (C.3) complete the proof. $\square$

D The Euler–Lagrange equation of ${\text {dist}}^2(.,{\text {SO}})$

In this section we prove that the Euler–Lagrange equation of the functional of $E_2$ is

$$\begin{aligned} \delta \left( d\phi -O(d\phi )\right) =0, \end{aligned}$$

where $O(d\phi )$ is the orthogonal polar factor of $d\phi$. (The derivation of the EL equations for $p \ne 2$ follows from the special case of $p=2$.)

For brevity, we show the derivation only for the Euclidean case where $\mathcal {M}=\Omega \subseteq \mathbb {R}^n, \mathcal {N}=\mathbb {R}^n$ are endowed with the usual flat metrics; the general Riemannian case follows in a similar fashion.

Let ${\text {GL}}_n^+$ be the group of real $n \times n$ matrices having positive determinant, and let $O:{\text {GL}}_n^+\rightarrow {\text {SO}}_n$ map $A\in {\text {GL}}_n^+$ into its orthogonal polar factor, i.e.

$$\begin{aligned} O(A)=A\left( \sqrt{A^TA}\right) ^{-1}. \end{aligned}$$

$\sqrt{A^TA}$ denotes the unique symmetric positive-definite square root of $A^TA$. The map O is smooth. We use the following observation:

Lemma D.1

Let $A \in {\text {GL}}_n^+$. Given $B\in M_n$ write $\dot{O}=dO_A(B)$. Then for every $B \in M_n$,

$$\begin{aligned} \langle \dot{O},O\rangle =\langle \dot{O},A\rangle =0. \end{aligned}$$

Proof

The equality $\langle \dot{O},O\rangle =0$ follows from differentiating $\langle O,O\rangle =n$. Now,

$$\begin{aligned} \dot{O} \in T_{O}{\text {SO}}_n=OT_{{\text {Id}}}{\text {SO}}_n=O\text {skew} \end{aligned}$$

implies that $\dot{O}=OS$ for some $S \in \text {skew}$. Thus,

$$\begin{aligned} \langle \dot{O},A\rangle = \langle OS,OP\rangle = \langle S,P\rangle =0, \end{aligned}$$

where the last equality follows from the fact that the spaces of symmetric matrices and skew-symmetric matrices are orthogonal. $\square$

Derivation of the EL equation Recall that

$$\begin{aligned} E_2(\phi )=\int _{\Omega } {\text {dist}}^2\left( d\phi ,{\text {SO}}_n\right) =\int _{\Omega }|d\phi -O(d\phi )|^2. \end{aligned}$$

Let $\phi \in C^2(\Omega ,\mathbb {R}^n)$, and let $\phi _t=\phi +tV$ for some $C^2$ vector field $V:\Omega \rightarrow \mathbb {R}^n$. Then,

$$\begin{aligned} \frac{1}{2}\left. \frac{d}{dt} E\left( \phi _t \right) \right| _{t=0}&= \int _{\Omega } \left. \langle \nabla _{\frac{\partial }{\partial t}} d\phi _t - \nabla _{\frac{\partial }{\partial t}} O(d\phi _t) , d\phi _t-O(d\phi _t)\rangle \right| _{t=0} dx \nonumber \\&=\int _{\Omega } \langle \left. \nabla _{ \frac{\partial }{\partial t} } d\phi _t \right| _{t=0} , d\phi -Q(d\phi )\rangle dx=\int _{\Omega } \langle \nabla V , d\phi -Q(d\phi )\rangle dx \nonumber \\&=-\int _{\Omega } \langle V , {\text {div}}\left( d\phi -Q(d\phi )\right) \rangle dx. \end{aligned}$$

(D.1)

The passage from the first to the second line relied upon the fact that $\nabla _{\frac{\partial }{\partial t}} O(d\phi _t)$ is orthogonal to $d\phi _t,O(d\phi _t)$, which essentially follows from Lemma D.1.

E Additional proofs

Lemma E.1

The function F defined in (1.9) is well-defined and continuous; it is strictly decreasing on (0, 1] and strictly increasing on $[1,\infty )$.

Proof

First, suppose that $s \le 1$; then the minimum is obtained at a point (a, b) where both $a,b \le 1$. Indeed, if $a>1$ (and so $b <s \le 1$), we can replace a by 1 and b by s to get the same product with both numbers closer to 1. Thus, it suffices to show that the minimum exists when $0<a,b \le 1$. Now, $b \le 1 \Rightarrow s=ab \le a$, and similarly $b \ge s$. So, the problem reduces to proving existence of a minimum over the compact set $\{ (a,b) \in [s,1]^2 \, | \, ab=s\}$. Since f was assumed continuous we are done.

Next, we prove that F is strictly decreasing on (0, 1]. Indeed, let $0< s_1 < s_2 \le 1$, and suppose that $F(s_1)=f(a)+f(b)$, for some $(a,b) \in [s_1,1]^2, ab=s_1$. Choose a smooth path (a(t), b(t)) from (a, b) to (1, 1), where a(t), b(t) are both strictly increasing. Then for $t>0$

$$\begin{aligned} F\left( a(t)b(t)\right) \le f(a(t))+f(b(t)) < f(a)+f(b)=F(s_1). \end{aligned}$$

Since a(t)b(t) movies continuously from $s_1$ to 1, it hits $s_2$ at some time $t>0$, which establishes the claim. Finally, a symmetric argument shows that if $s \ge 1$, then the minimum is obtained in $\{ (a,b) \in [1,s]^2 \, | \, ab=s\}$, and that F is strictly increasing on $[1,\infty )$. Proving F is continuous is routine and we omit it. $\square$

Proof

[of Proposition 5.6] Suppose that $(\sqrt{s}_n,\sqrt{s}_n)$ is a minimizer of (1.9) for some sequence $s_n \in (0,1)$ which converges to zero. We prove that f diverges to $\infty$ at zero. Recasting everything in terms of $g(x)=f(e^x)$, we get

$$\begin{aligned} g\left( \frac{x + y}{2}\right) \le \frac{g(x) + g(y)}{2} \, \,\, \, \text { whenever } \, \, x,y \le 0 \, \, \, \text { and } x+y=\lambda _n, \end{aligned}$$

where $\lambda _n=\log s_n \rightarrow -\infty$. Choose a subsequence $(\lambda _{n_k})$ such that $\lambda _{n_k} < 2 \lambda _{n_{k-1}}$ for all k. Choosing $x=0$ and $y= \lambda _{n_k}$ in the condition above and the monotonicity of g give

$$\begin{aligned} g(\lambda _{n_{k-1}}) \le g\left( \frac{1}{2} \lambda _{n_k}\right) \le \frac{1}{2} g(\lambda _{n_k}) \end{aligned}$$

so

$$\begin{aligned} g(\lambda _{n_k}) \ge 2 g(\lambda _{n_{k-1}}) \ge 2^2 g(\lambda _{n_{k-2}}) \ge \cdots \ge 2^{k-1} g(\lambda _{n_{1}}) \, . \end{aligned}$$

$\square$

Proof

[Of Lemma 2.1] Since the problem is bi-${\text {SO}}_2$-invariant, using SVD we can assume that $A={\text {diag}}(a,b)$ is diagonal with $a,b>0$. We need to compute

$$\begin{aligned} F(s)=\min _{a,b \in \mathbb {R}^+,ab=s} (a-1)^2+(b-1)^2. \end{aligned}$$

Using Lagrange’s multiplier, there exists $\lambda$ such that $\left( a-1,b-1\right) =\lambda (b,a)$. Thus $a(1-a)=b(1-b)$ which implies $a=b$ or $a=1-b$. In the latter case $s=ab=b(1-b)$. Since $a=1-b,b,s$ are positive, we must have $0<b<1,0<s\le \frac{1}{4}$. (since $\max _{0<b<1} b(1-b)=\frac{1}{4}$). We then have

$$\begin{aligned} (a-1)^2+(b-1)^2 =b^2+(b-1)^2=2b(b-1)+1=1-2s. \end{aligned}$$

• If $s \ge 1/4$ then there is only one critical point $(a,b)=(\sqrt{s},\sqrt{s})$, so the minimum is obtained exactly when $a=b$ and $F(s)=2(\sqrt{s}-1)^2.$

• If $s \le 1/4$, there are up to 3 to critical points: $(\sqrt{s}, \sqrt{s}),(x,1-x),(1-x,x)$ when $0<x<1$ satisfies $x(1-x)=s$. (For $s=1/4$ they all merge into a single point. For $s<1/4$ these are 3 distinct points.)

  To decide which point is the global minimizer, we need to compare the values of the objective function at the critical points, which are $2(\sqrt{s}-1)^2,1-2s.$ Since

  $$\begin{aligned} 1-2s \le 2(\sqrt{s}-1)^2 \iff (2\sqrt{s}-1)^2 \ge 0, \end{aligned}$$

  $F(s)=1-2s$ for $s \le \frac{1}{4}$.

$\square$

Proof

[of Lemma 4.2] By Proposition 4.1,

$$\begin{aligned} d\phi _x-O(d\phi _x)=\alpha {\text {Cof}}d\phi _x \end{aligned}$$

if and only if $d\phi _x \in K$ and $\alpha =-1$ or $d\phi _x$ is conformal and $\sigma _i(d\phi _x)=\frac{1}{1-\alpha }$. In both cases $\sigma _1(d\phi _x)+\sigma _2(d\phi _x)=\frac{2}{1-\alpha }$ is constant. If $\alpha =-1$, then $d\phi \in K$, and if $\alpha \ne -1$ then $d\phi \in {\text {CO}}_2$ with $\sigma _i(d\phi )=\frac{1}{1-\alpha }$, i.e. $\phi$ is a homothety.

Now, define $H(\phi )={\text {dist}}^{p-2}\left( d\phi ,{\text {SO}}(\mathfrak {g},\phi ^*\mathfrak {h})\right)$. Let $p \ne 2$ and suppose that

$$\begin{aligned} H(\phi )\left( d\phi -O(d\phi )\right) =\alpha {\text {Cof}}d\phi . \end{aligned}$$

Since we assumed $J\phi >0$, ${\text {Cof}}d\phi$ is invertible, hence $H(\phi ) \ne 0$, and

$$\begin{aligned} d\phi -O(d\phi )=\frac{\alpha }{H(\phi )} {\text {Cof}}d\phi . \end{aligned}$$

Proposition 4.1 implies that either $H(\phi )(x)=-\alpha$ and $d\phi _x \in K$, or that $d\phi _x$ is conformal and

$$\begin{aligned} \sigma (d\phi _x)=\frac{1}{1-\alpha /H(\phi )(x)}=\frac{1}{1-\frac{\alpha }{(\sqrt{2}|\sigma (d\phi _x)-1|)^{p-2}}}. \end{aligned}$$

Thus $\sigma (d\phi _x)$ is a solution for the equation

$$\begin{aligned} \sigma =\frac{1}{1-\frac{\beta }{|\sigma -1|^{p-2}}}, \end{aligned}$$

which has a finite number of solutions $\{\alpha _1,\dots ,\alpha _k\}$.

We showed that for every $x\in \mathcal {M}$, $d\phi _x \in K$, or $d\phi _x$ is conformal with $\sigma (d\phi _x) \in \{\alpha _1,\dots ,\alpha _k\}$. Thus $x \mapsto \sigma _1(d\phi _x)+\sigma _2(d\phi _x)$ is a continuous function on $\mathcal {M}$, which takes values in a finite set $\{1,2\alpha _1,\dots ,2\alpha _k\}$, hence it must be constant. Thus, $d\phi \in K$ or $\phi$ is a homothety with $\sigma (d\phi ) \in \{\alpha _1,\dots ,\alpha _k\}$. If $d\phi \in K$, then $H(\phi )=-\alpha$ is constant. Thus

$$\begin{aligned} \sigma _1(d\phi )+ \sigma _2(d\phi )=1, \left( \sigma _1(d\phi )-1\right) ^2+\left( \sigma _2(d\phi )-1\right) ^2 \end{aligned}$$

are constants, which implies that $\sigma _1(d\phi ), \sigma _2(d\phi )$ are constants as required. $\square$