# The Riemannian metric on manifold with boundary

$(M,g)$ is a Riemannian manifold with non-empty boundary and $DM$ is the double of $M$, is there a Riemannian metric $G$ on $DM$ such that $g=i^*G$? ($i$ is the inclusion from $M$ to $DM$)

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## 1 Answer

Generally there isn't a natural one, although metrics do exist. There is no such Riemann metrics if you demand $i^* G = g$ for both inclusions of $M$ in its double. For example, consider the flat disc $D^2$. Having a metric on the double $S^2$ that restricts to two flat discs would violate the Gauss-Bonnet theorem. But in the world of metric spaces, there are metrics on the double, they're just not Riemann metrics.

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But if we require only one inclusion such that $i^*G=g$, then it seems to me that we can use partition of unity to construct $G$ on $DM$. –  Hezudao Sep 19 '12 at 23:21
Yes, it just won't be a natural metric -- by that I mean it will depend on your choice of partition of unity. –  Ryan Budney Sep 20 '12 at 2:08
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