Area-preserving

Question

A diffeomorphism $\varphi: S \to \bar{S}$ is said to be area-preserving if the area of any region R \subset S is equal to the area of $\varphi(R)$. Prove that if $\varphi$ is area-preserving and conformal, then ϕ is an isometry.

I have two ideas for this excersise. First, i consider a parametrization $x: U \to R \subset S$ and $\bar{x}:\varphi \circ x : X(U) \to \varphi(X(U)) $ a parametrization for $\bar{S}$. We can note that if $\varphi$ is area-preserving then $A(R)=A(\varphi(R))$. Here, i can use the definition of Area with respect to its partial derivatives and then we have that $$A(R)=\int\int_{R}||x_{u} \times x_{v}||dudv$$ But for $\bar{S}$

$$A(\varphi(R))=\int\int ||(\varphi \circ x)_{u} \times (\varphi \circ x)_{v}||dudv$$

here i'm stuck, because i'dont know how to proceed with $A(\varphi(R))$ or how to describe the partial derivatives of the composition or use the hypotesis that $\varphi$ is conformal, i'm sure that conformal does not implies that coeficients of x and $\bar{x}$ are $\lambda$ - times of another, i.e, $\bar{E}=\lambda E, \bar{F}= \lambda F, \bar{G}= \lambda G$, so i don't know what to do here. Another idea for this is use the definition of area with respect to first fundamental form but here the only way to develop this idea is using the wrong assumption about coefficients that i was talking about. I appreciate some idea.

UPDATE

i want to be sure that this attempt is good. I will proof that $$d\varphi_{q}(x_{u})=\bar{x}_{u}$$

where $x_{u}$ and $\bar{x}_{u}$ are the partial derivatives of x and $\bar{x}$ respectively.

Now,for $(u_{0},v_{0}) \in U$ and $q=x(u_{0},v_{0}) $ and $\alpha$ a curve defined as $x(u,v_{0})=\alpha(u)$, therefore $\alpha'(u)=x_{u}(u,v_{0})$. Now, we have that

$$d\varphi_{q}(x_{u}(u_{0},v_{0}))=d\varphi_{q}(\alpha'(u_{0}))=(\varphi \circ \alpha)'(u_{0})$$

then we have that

$$d\varphi_{q}(x_{u}(u,v_{0}))=d\varphi_{q}(\alpha'(u))=(\varphi \circ \alpha)'(u)$$ for all $u$

but now, we can note that

$$(\varphi \circ \alpha)(u)=\varphi(\alpha(u))=\varphi(x(u,v_{0})=(\varphi \circ x)(u,v_{0})=\bar{x}(u,v_{0})$$

so, finally we have that $$(\varphi \circ \alpha)'(u,v_{0})=\bar{x}_{u}(u,v_{0})$$

and we can conclude that

$$(\varphi \circ \alpha)'=d\varphi_{q}(x_{u})=\bar{x}_{u}$$

this is enought or this proof is wrong?

Answer 1

Let’s say $(M,g)$ and $(\overline{M},\overline{g})$ are two Riemannian manifolds of the same dimension $n$, and $\phi:M\to \overline{M}$ is a diffeomorphism which is conformal and $n$-volume preserving (the volume being the Riemannian volume). We’ll show that $\phi$ is in fact an isometry.

First, by assumption of $\phi$ being conformal, it follows that there is a smooth positive function $f:M\to (0,\infty)$ such that $\phi^*(\overline{g})=f\cdot g$. Now, for any (nice enough $n$-dimensional) region $U\subset M$, we have the volume of $U$ equals the volume of $\phi[U]$. Let us evaluate the volume of $\phi[U]$: \begin{align} \text{vol}(\phi[U])&:=\int_{\phi[U]}1\,dV_{\overline{M},\overline{g}}\\ &=\int_U(1\circ \phi)\cdot dV_{M,\phi^*(\overline{g})}\tag{change of variables}\\ &=\int_U1\,dV_{M,fg}\\ &=\int_U1\cdot f^{n/2}\,dV_{M,g}, \end{align} where I have used the change of variables theorem on a Riemannian manifold, and in the final step, I used the fact that $dV_{M,fg}=f^{n/2}\,dV_{M,g}$ (this is clear from the local coordinate formula $dV_{M,g}=\sqrt{|\det g_{ij}|}\,|dx^1\wedge\cdots\wedge dx^n|$, so if you replace $g$ by $fg$, the determinant gets multiplied by $f^n$, so taking square root gives $f^{n/2}$).

So far, we have calculated the volume of $\phi[U]$ as an integral over $U$. By hypothesis, this integral is equal to the volume of $U$, i.e \begin{align} \int_Uf^{n/2}\,dV_{M,g}&=\int_{U}1\,dV_{M,g}, \end{align} and this equality is supposed to hold for all nice enough regions $U$ (e.g all Borel sets $U$). If two integrals are equal for “all nice regions”, then the two integrands must be identical; this is essentially the fundamental lemma of calculus of variations (in $n$-dimensions). Thus, $f^{n/2}=1$ identically on $M$, and hence $f=1$. Thus proves $\phi$ is an isometry.

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Edit: Specialized version

Parametrize the surface $S\subset\Bbb{R}^3$ as you have, $x:U\subset\Bbb{R}^2\to x[U]\subset S$. Your “first fundamental form” gives you a $2\times 2$ symmetric matrix of inner products \begin{align} [g_{ij}]&=\begin{pmatrix} \langle\frac{\partial x}{\partial u},\frac{\partial x}{\partial u}\rangle & \langle\frac{\partial x}{\partial v},\frac{\partial x}{\partial u}\rangle\\ \langle\frac{\partial x}{\partial u},\frac{\partial x}{\partial v}\rangle& \langle\frac{\partial x}{\partial v},\frac{\partial x}{\partial v}\rangle \end{pmatrix} \equiv \begin{pmatrix} E&F\\ F&G \end{pmatrix}, \end{align} where $E,F,G:U\to\Bbb{R}$ are smooth functions (such that the determinant $EG-F^2$ is nowhere vanishing). Now, since $\phi$ is a diffeomorphism, you get a parametrization $\overline{x}=\phi\circ x$ of $\overline{S}$. With respect to this parametrization, as in your update, the chain rule tells you that for each $p\in U$, $\frac{\partial \overline{x}}{\partial u}(p)=d\phi_{x(p)}\left(\frac{\partial x}{\partial u}(p)\right)$, and likewise with $v$. In words, the tangent vector $\frac{\partial x}{\partial u}(p)\in T_{x(p)}S$ is mapped via $d\phi_{x(p)}:T_{x(p)}S\to T_{\phi(x(p))}\overline{S}=T_{\overline{x}(p)}\overline{S}$ to the tangent vector $\frac{\partial \overline{x}}{\partial u}(p)\in T_{\overline{x}(p)}\overline{S}$. So, we have \begin{align} \left\langle \frac{\partial \overline{x}}{\partial u^i}(p),\frac{\partial \overline{x}}{\partial u^j}(p)\right\rangle_{\overline{x}(p)} &= \left\langle d\phi_{x(p)}\left(\frac{\partial x}{\partial u^i}(p)\right), d\phi_{x(p)}\left(\frac{\partial x}{\partial u^i}(p)\right)\right\rangle_{\phi(x(p))}\\ &= \lambda^2(p) \left\langle \frac{\partial x}{\partial u^i}(p),\frac{\partial x}{\partial u^j}(p)\right\rangle_{x(p)}. \end{align} The first line is by the computation we just made, and the second line is the definition of conformality. Here, for convenience, I have set $(u^1,u^2)=(u,v)$. This is saying precisely that \begin{align} \begin{pmatrix} \overline{E}&\overline{F}\\ \overline{F}&\overline{G} \end{pmatrix} &=\lambda^2 \begin{pmatrix} E&F\\ F&G \end{pmatrix}. \end{align} Finally, the norm of the cross product is $\|x_u\times x_v\|=\sqrt{EG-F^2}$. Put all of this together, and try to mimic what I said above.

(btw what I call $\lambda^2$ here (since this is how doCarmo writes it) is what I called $f$ in my original answer).