Apologies if this question is overly simple, I'm new to differential geometry.

Suppose I have two Riemannian manifolds $M_1$ and $M_2$, along with a diffeomorphism $f:M_1\to M_2$ between them. Let $d$ denote the geodesic distance on $M_1$. Let $B_r=\{x'\in M_1: d(x, x') \le r\}$ be a ball of radius $r$ centered around a point $x\in M_1$, and let $B_r'=\{x'\in M_2: d(f^{-1}(x'), x) \le r\}$, so that $B'$ is the image of $B$ under $f$. What (if anything) can I say about the volume of $B'$ as $r\to 0$? In particular, is it the case that there are constants $r_0>0,C>0$ such that $\text{Vol}(B_r') \ge C \text{Vol}(B_r)\forall r \le r_0$?


1 Answer 1


Yes, for small $r$ the volume ratio is determined by the determinant of the differential, as in the Euclidean case.

Fix orthonormal bases $e_i$ for $T_xM_1$ and $v_i$ for $T_{f(x)}M_2$ and let $A$ be the matrix of $Df$ in these frames; i.e. $A_{ij} = g_2(v_i, Df(e_j))$. Since $Df$ is a diffeomorphism, $A$ is invertible, and thus $|\det A| > 0$.

Then the precise result you're looking for is

$$\lim_{r\to 0}\frac{V(f(B_r))}{V(B_r)} = |\det A|.$$

In particular, for any $\epsilon > 0$ we can find an $r_0 >0$ such that $$(|\det A| - \epsilon)V(B_r) \le V(f(B_r)) \le (|\det A|+\epsilon) V(B_r)$$ whenever $r < r_0$.

We often write $|\det Df|$ instead of setting up $A$ and taking $|\det A|$, with the understanding that the determinant of a map between $n$-dimensional inner product spaces is taken in orthonormal bases for those inner products.

  • $\begingroup$ Thanks, I expected the answer would be something like that, but I wasn't sure how to formalize it. How standard is the notation for $\vert\text{det}Df\vert$? Would it be understandable to someone well-versed in differential geometry without first introducing the matrix $A$? $\endgroup$
    – wrvb
    May 7, 2016 at 19:31
  • $\begingroup$ @wrvb: Some would probably call it an abuse of notation, but I think most would understand what it means. The other option is to express it as the ratio $f^* d\mu_{g_2} / d\mu_{g_1}$ of volume forms. $\endgroup$ May 7, 2016 at 22:48

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