# convolution of surface measures

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:

Ragozin

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

$$f(x,y)=x+y$$

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

Thank you very much.

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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