3
$\begingroup$

Let $M$ be a smooth manifold in $\mathbb{R}^n$. If Lebesgue measure of $M$ is zero i.e $l(M)=0$, does it mean that volume of manifold is also zero i.e $Vol(M)=0$? Are they the same thing (volume and Lebesgue measure).

$\endgroup$
5
$\begingroup$

No, that doesn't mean $\mathrm{Vol}(M)=0$. The volume of an oriented Riemannian manifold $M$ is an intrinsic quantity defined by $$\mathrm{Vol}(M)=\int_MdV,$$ where $dV$ is the Riemannian volume form of $M$. Hence, for example, if $M\subseteq\Bbb R^3$ is a surface, then $\mathrm{Vol}(M)$ is define as the area of $M$, which is in general non-zero, while its Lebesgue measure in $\Bbb R^3$ is zero. For instance, if $S^2_r\subseteq\Bbb R^3$ is the sphere of radius $r$, then $$\mathrm{Vol}(S^2_r)=4\pi r^2,\quad l_{\Bbb R^3}(S^2_r)=0.$$

However, if $M\subseteq\Bbb R^n$ is an open submanifold of $\Bbb R^n$, then the Riemannian volume form of $M$ is the restriction of that of $\Bbb R^n$ and hence we have $$\mathrm{Vol}(M)=\int_MdV=\int_M dx^1\cdots dx^n=l(M),$$ so the two quantities are the same.

Conclusion: $\mathrm{Vol}(M)$ is an intrinsic quantity, while $l(M)$ is an extrinsic quantity that depends on the particular embedding. Hence, they are in general not equal.

Additional Remark: To prove that if $M\subseteq\Bbb R^n$ is a smooth manifold of dimension $<n$, then $l(M)=0$, we can use Sard's Theorem. The inclusion $\iota:M\hookrightarrow \Bbb R^n$ has critical values everywhere because $\dim M<n$, so $\iota(M)=M$ has measure $0$ in $\Bbb R^n$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Cool :)! Thank you for editing +1 $\endgroup$ – ClassicStyle Jul 18 '15 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.