Conformal Equivalence of two Riemann metrics

Question

I'm reading a paper and encountered a concept of conformal equivalence between two Riemannian metrics on a differentiable $2$-manifold $M$ :

Two Riemannian metric $g$ and $f$ are conformally equivalent if

$$ f = e^{2u}g$$

for some smooth function $u:M \rightarrow R$

I don't have much knowledge about topology so please understand if my question sounds stupid.

Question: so can u be "any function?" I mean if I set an arbitrary smooth function u then can I make conformally equivalent metric from a given metric?

Also, it seems like the conformal equivalence relation is an equivalence relation (I mean obviously from the name of it) but I don't really see how this could be hold.

$f$ is obviously not $e^{2u}f$.

Thank you in advance.

Answer 1

Short answer till you ask for more details.
(1) yes it is true for arbitrary differentiable functions $u$, we just require the factor to be positive and the exponential do it.
(2) if you to get an equivelence relation defined by $$f \ \ \text{is related to}\ \ g \ \ \ \text{if and only if}\ \ \ \ f=e^{2u}g$$ It seems to be true:

• If $f$ is related to $g$ then $g$ is related to $f$ using $-u$ instead of $u$
• $f$ is related to itself using $u=0$,

and so on. I hope it helps out.

Answer 2

$u$ should actually be any scalar field. As you've said, $u:M\to \mathbb{R}$. "$f$ is obviously not $e^{2u}f$." - this is not right. Conformally equivalent metrics have a conformal transformation, so $f\neq e^{2u}f$, but instead $f\to e^{2u}f$. A conformal transformation preserves angles. In Riemannian geometry, two Riemannian metrics $g$ and $h$ on smooth manifold $M$ are called conformally equivalent if $g=qh$ for some positive scalar field $q$ on $M$. The field $q$ is called the conformal factor, and is usually written as $e^{nu}$, with constant $n$. When there is a conformal equivalence between two metrics $f$, $g$, since $g=fe^{2u}$, the transformation $f\to e^{2u}f$ is a conformal transformation.