3

Suppose we have two parametrized surfaces in $\mathbb{R}^3$: $$ X,Y:\mathbb{R}^2 \rightarrow \mathbb{R}^3 $$

The induced metric on either surface is the pullback of the Euclidean metric $\bar g$ due to the inclusion map: $$ g_X = X^\ast \bar g, \quad g_Y = Y^\ast \bar g $$

To check if $X$ and $Y$ are conformally equivalent, can we directly compare $g_X$ and $g_Y$, since they both are defined on $T\mathbb{R}^2$, or do we need to find out the a diffeomorphism $\varphi$ (if it exists) from $X$ to $Y$ and then compare $\varphi^\ast g_Y$ and $g_X$?

Addendum:

Thanks to This is much healthier for the clarifying comment. I have slightly edited my question above due to this.

The specific case I am interested in is of parallel surfaces: $$ Y(u,v) = X(u,v) + aN(u,v) $$

Let us assume that $X$ is convex, e.g. an ellipsoid, and $N$ is the "outward" normal, i.e. the surface curves away from $N$. Is it possible to determine for this case if the surfaces are conformal or not?

One can show that $$ g_Y = Q^T g_X Q $$ where $Q$ is the change-of-basis matrix from $\{X_u,X_v\}\ (\equiv X_\ast)$ to $\{Y_u,Y_v\}\ (\equiv Y_\ast)$. Does this help in any way in deciding conformality?

Community
  • 1
ushnishbasu
  • 81

1 Answers1

2

can we directly compare $g_X$ and $g_Y$, since they both are defined on $T\mathbb R^2$

Generally, one should think of parameter spaces as different copies of $\mathbb R^2$ floating somewhere in Platonic universe and not interacting with each other at all. But yes, you could check if $g_X$ is a scalar multiple of $g_Y$; if they are, the surfaces are conformally equivalent. Just realize that you are taking a blind shot at the problem when you do, because the identity map $\operatorname{id}:\mathbb R^2\to\mathbb R^2$ is just one of infinitely many diffeomorphisms between these surfaces, and not a distinguished one. If it's not conformal, that tells nothing.

do we need to find out the diffeomorphism $\varphi$ (if it exists) from $X$ to $Y$

There is no such thing as "the" diffeomorphism; given any two manifolds, you either have no diffeomorphisms between them, or you have uncountably many. Feast or famine. To show that the surfaces are not conformally equivalent, you'd have to prove that none of these uncountably many diffeomorphisms are conformal.

You may be interested in Is conformal equivalence the same as topological equivalence?

Community
  • 1