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For all $p \in \mathbb{R} \smallsetminus \lbrace 0,1 \rbrace$, let $\displaystyle M(p)=\frac{1}{p^{2}(p-1)^{2}}$. Then, let $g_{p} \, : \, (u,v) \, \longmapsto \, uM(p)v$.

I am not sure about the following assertions :

  • Let $M=\mathbb{R} \smallsetminus \lbrace 0,1 \rbrace$. $\big( g_{p} \big)_{p \in M}$ defines a Riemannian metric on $M$. Hence, $M$ equipped with this metric is a Riemannian manifold.
  • $M$ cannot be geodesically complete for this metric because, according to Hopf-Rinow theorem, $(M,\vert \cdot \vert)$ would be a complete metric space (which it is not).

I think both are true but I'd feel better if someone could validate or invalidate these assertions.

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You have modified the standard metric by a conformal factor function which is always positive so this certainly defines a Riemannian metric.

On the other hand, the metric resembles a hyperbolic metric on the y-axis near the poles of the conformal factor, so it is actually complete "near" these points. Whether or not it is complete "at infinity" I leave to you as an exercise (hint: integrate).

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  • $\begingroup$ Thank you for the answer but what about the Hopf-Rinow theorem ? If $M$ equipped with this metric was complete, then $M$ as a metric space would be complete as well, right? $\endgroup$ – jibounet Jul 31 '14 at 12:45
  • $\begingroup$ Whether or not it is incomplete depends on whether you run off to infinity in finite time. This is consistent with Hopf-Rinow. $\endgroup$ – Mikhail Katz Jul 31 '14 at 12:51
  • $\begingroup$ I have written the differential equation for geodesics and solved it. I might be mistaken but the geodesics but not all the geodesics are defined on $\mathbb{R}$. Therefore, $M$ would be incomplete. $\endgroup$ – jibounet Jul 31 '14 at 13:00
  • $\begingroup$ Here you don't even need the geodesic equation. Integrate the conformal factor to show that you run off to infinity in finite time. $\endgroup$ – Mikhail Katz Jul 31 '14 at 14:04
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    $\begingroup$ Ok sorry, I misunderstood "arclength". If the curve is parametrized by arclength, its length is equal to the length of the interval on which it is defined (say $t \, \mapsto \, \tilde{\gamma}$ (with $a \leq t \leq b$) is the curve parametrized by arclength) then $L(\tilde{\gamma}) = b-a$. $\endgroup$ – jibounet Jul 31 '14 at 16:15

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