Wanted: metric in which the volume of euclidean space is finite

Question

With the motivation to give a small mini-course with rudiments of Riemannian geometry to students of the end of the graduation in mathematics. I aspire to show how much the environment changes according to the way I choose to measure distances. For example, let $\mathbb{R}^ 2$ be the canonical metric. Using the definition of volume, it follows that vol of $\mathbb{R}^{2}$ "is" infinite. Now let $g = (dx^ {2} + dy ^ {2}).y ^ {- 2} $(metric of Lobachveski), again using the volume definition, I have gained $vol (\mathbb{R }^ {2}) = \infty$. which does not give finite volume. how could I find a metric for $\mathbb{R} ^ 2$ whose volume of this manifold was finite?

Answer 1

If we express our metric in polar coordinates, it becomes a bit easier to see what to do. Let's suppose that our desired metric is related to the Euclidean metric by a conformal transformation of the form: $$ ds^2 = \Omega^2(r) (dr^2 + r^2 d\theta^2). $$ Then the volume spanned by the coordinate range $r = [0,\infty)$ and $\theta = [0, 2 \pi]$ is $$ V = \iint \sqrt{g} \, d^2x = \int_0^{2 \pi} \int_0^\infty \Omega^2(r) r \, dr \, d \theta = 2 \pi \int_0^\infty \Omega^2(r) r \, dr. $$ So for any function $\Omega(r)$ for which this integral converges, the volume of the above coordinate range will be finite.

In particular, $\Omega(r)$ must fall off faster than $1/r$ as $r \to \infty$ for this integrand to be finite, so let's try $$ \Omega(r) = \frac{1}{1+r^2}, $$ which corresponds in the coordinates $x$ and $y$ to $$ ds^2 = \frac{1}{(1 + x^2 + y^2)^2}(dx^2 + dy^2). $$ Then we have $$ V = 2 \pi \int_0^\infty \frac{r dr}{(1+r^2)^2} = \pi. $$