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Let $M$ be a manifold with boundary $\partial M$. Form the manifold $M'$ by attaching a half-infinite cylinder $\partial M\times[0,\infty)$ to $M$ along its boundary. In other words, $$M'=M\cup_{\partial M}\partial M\times[0,\infty),$$ where we identify $\partial M\sim\partial M\times\{0\}$.

Let $g$ be a Riemannian metric on $M$.

Question: Does there always exist a Riemannian metric $g'$ on $M'$ such that the restriction of $g'$ to $M$ is equal to $g$?

Comment added later: Now that I think about it, perhaps the required property is built into the definition of smoothness of $g$ at the boundary, namely that it's extendible slightly beyond the boundary of the chart into some open neighborhood; one then uses a partition of unity to get a metric on all of $M'$.

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  • $\begingroup$ Just a quick thought: One big problem here seems to be whether $M'$ admits a differentiable structure - which is possibly linked to the regularity of $\partial M$. For instance if $M=\mathbb{R}_{\geq 0}\times \mathbb{R}_{\geq 0}$, i.e. the quarter space in $\mathbb{R}^2$, then the glued manifold $M'$ might look like $\partial \mathbb{R}_{\geq 0}^3$, i.e. the boundary of an octant in $\mathbb{R}^3$. $\endgroup$
    – dennis_s
    Sep 25 '20 at 14:31
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Take a collar neighborhood of $\partial M$. This is given by a flow $\phi:\partial M \times[0,\infty)\rightarrow M$, generated by vector field $X$ such that $\phi_{*}(\partial_{t})_{(p,t)}=X_{\phi(p,t)}$. Define $g'$ on $M'$ as follows:

$$g' = \begin{cases} g & p\in M\\ \phi_{t}^{*}g_{\phi_{t}(p)} & (p,t)\in\partial M \times[0,\infty) \end{cases}$$

Since $\phi(p,0) = id$, at all points in $\partial M$, $g'$ restricts to g as desired.

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  • $\begingroup$ I don't think that $\partial M \times [1,\infty)$ is diffeomorphic to $M$ (for example, take the closed disk $D^2$, it is not diffeomorphic to $\partial D^2 \times \mathbb{R}_+ = \mathbb{S}^1 \times \mathbb{R}_+$). But is is diffeomorphic to an open subset of $M$ and it would be sufficient I guess. $\endgroup$
    – Didier
    Sep 25 '20 at 14:46
  • $\begingroup$ Ah, yes,I see that it wasn't what I wanted to say $\endgroup$ Sep 25 '20 at 14:48
  • $\begingroup$ @DIdier_ fixed it $\endgroup$ Sep 25 '20 at 14:51
  • $\begingroup$ The metric on $M'$ obtained in this way doesn't seem to restrict to the original metric on $M$ though. $\endgroup$
    – geometricK
    Sep 25 '20 at 15:19
  • $\begingroup$ @geometricK Is that more helpful? $\endgroup$ Sep 25 '20 at 16:07

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