Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Thanks for reading my question. Assume we are given two compact (possibly with boundary) $(n-1)$-dimensional $C^{\infty}$ manifolds $M_1$ and $M_2$ embedded in $\mathbb{R}^n$, with induced measure $d\sigma_1$ and $d\sigma_2$. For example two $S^{n-1}$ embedded in $\mathbb{R}^n$. Then whether $d\sigma_1\ast d\sigma_2$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}^n$?

I know there is an old result due to Ragozin here:


which deals with analytic submanifolds. So I wonder if his results can be extended to smooth manifolds as well. The only place where the analyticness is used is to show that

$$M_1\times M_2\overset{f}{\rightarrow}\mathbb{R}^n$$

defined by


is an analytic map, so its critical set, not critical value, is of null measure.

Thank you very much.

share|cite|improve this question
I forgot to add a comment that it seems easy to see the convolution is absolutely continuous if the manifolds $M_1$ or $M_2$ does not contain any "flat" part, say they have no open subset with vanishing curvature. – Peng Aug 27 '13 at 16:58
I reviewed the Ragozin argument again, and it seems that as long as each of $M_1$ and $M_2$ contains a flat part and especially these two parts are parallel, then $d\sigma_1\ast d\sigma_2$ will definitely contain a singular part near the center of these two pieces. Am I right? – Peng Aug 27 '13 at 18:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.