Computing volume of Riemannian manifolds and its $n$-sheeted covering

Question

I have a question-

It is given that $f: M \to N$ is an $n$-sheeted covering map and a local isometry then I have to show that volume$(M) = n$ volume$(N)$, where $M$ and $N$ are Riemannian manifolds.

Let $p$ be an element in $M$, and let $(U,(x_1,...x_n))$ be a coordinate chart around $p$. Now $f(p) \in N$, let $V$ be an open set in $N$ and $(V,(y_1,...y_n))$ be a coordinate chart around $f(p)$ in $N$.

Now, looking at the expression of the volume of a region $R $ in manifold, $$ vol(R) = \int_{R} \sqrt{\det(g_{ij})} dx_{1}...dx_{n} $$

it seems like we have to use local isometry and covering space definition at some point to express $g_{ij}$, Riemannian-metric of $M$ in terms of Riemannian-metric $h_{ij}$ of $N$....I've been trying to play around with the definitions, but unsuccessfully!

Can someone give a hint on where to proceed from here?

Answer 1

The basic idea is the following.

For each point $p\in N$, the fiber $f^{-1}(p)$ is a set of $n$ points in $M$. Each such point has a neighbourhood such that the restriction of $f$ to it is an isometry. Now take the intersection of the images of these sets, which will be some open neighbourhood $p \in U \subset N$. By its definition, the restriction of $f$ to each connected component of $f^{-1}(U)$ will be an isometry. I claim you can prove the following (let $\nu$ denote the volume form on $N$ corresponding to the metric):

Lemma

$$Vol(f^{-1}(U)) = \int_{f^{-1}(U)}f^*(\nu) = n\cdot \int_{U}\nu = n\cdot Vol(U)$$

In fact, we have the following slightly stronger statement:

Lemma For any function $\psi$ and a volume form $\nu$,

$$\int_{f^{-1}(U)}f^*(\psi\nu) = n\cdot \int_{U}\psi\nu $$

This lemma is a local version of the statement you want. We used the local isometry property here. We would now like to make it global. How do you typically achieve such a thing in differential geometry? One common strategy is to use a partition of unity.

Take a cover of $N$ by open sets $U_\alpha$, such that for each $U_\alpha$, the restriction of $f$ to each connected component of $f^{-1}(U_\alpha)$ is an isometry. (I claim you can do this, but you should check carefully that it is true).

Take a partition of unity $\{ \psi\}$ on $N$ that is subordinate to your chosen open cover (I omit indices). The volume form for our Riemannian metric, $\nu$, is equal to

$$ \nu = \sum \psi \cdot \nu.$$

(where the sum is over functions in our partition). Now

$$ Vol(M) = \int_M f^*\nu = \int_M f^*\left(\sum \psi \cdot \nu\right) = \sum\int_M f^* \left(\psi \cdot \nu\right) = \sum\int_{f^{-1}(U_\alpha)} f^* (\psi \cdot \nu)$$

(if your volume is infinite there might be some convergence issues around commuting the sum and integral, but this case can probably be handled separately). By our lemma,

$$ \int_{f^{-1}(U_\alpha)} f^* \psi \cdot \nu = n \cdot \int_{U_\alpha} \psi \cdot \nu $$

Putting this together with the fact that $\psi$ is a partition of unity, we continue from above:

$$\sum\int_{f^{-1}(U_\alpha)} f^* (\psi \cdot \nu) = n\sum \int_{U_\alpha} \psi \cdot \nu = n \int_{M} \sum\psi \cdot \nu =n \int_{M} \nu = n Vol(N)$$

Answer 2

Here's my attempt to solve this question thanks to the hints by AnonymousCoward, I'm sure there are some major flaws and I'd like to have them pointed out by someone.

Let $\{U_{\alpha}\}_{{\alpha}\in I}$ be a locally finite covering (by co-ordinate charts) of $N$ and $\{f_{\alpha}\}_{ \alpha \in I}$ be a smooth partition of unity. Let these ${U_{\alpha}}$s be open balls (WHY?)

Now, let $x \in N$. As $f: M \mapsto N$ is a covering map, there exists an open neighborhood $U_{\alpha}$ containing $x$ such that $f^{-1}(U_{\alpha}) = \cup_{k=1}^{n} V_{k} $ where $V_k \cap V_l = \phi$ for $k \neq l$ and $V_k \subset M$ is an open ball (again why?)

Also, $ restr{f}{V_{k}} : V_{k} \mapsto U_{\alpha}$ for all $k=1,...,n$ is a homeomorphism. And as $f$ is given to be a local isometry, so $restr{f}{V_{k}}$ is an isometry, and therefore by definition if $g_{ij} $ is metric component of Riemannian-metric on $N$ and $h_{ij}$ is metric component of Riemannian-metric on $M$ then we must have $g_{ij} = h_{ij}$ for all $i,j = 1,...n$.

And so, $vol(U_{\alpha})$ which depends on metric components $g_{ij}$s is the same as $vol(V_{k})$

Now what can I do??