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Let us define the following metric on $\mathbb{R}^n$: $$ g|_v(X, Y) := e^{-|v|^2} \langle X, Y\rangle,$$ where the brackets denote the standard scalar product.

How does the resulting manifold look like? If $n=2$, can it be isometrically embedded in $3$-space?

It has finite volume, but is it geodesically complete?

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    $\begingroup$ what is v in your expression? $\endgroup$ – Xipan Xiao Oct 14 '14 at 15:46
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It is not geodesically complete.

Any ray that begins at the origin is clearly the image of a geodesic. Since each one of those rays has a finite length with respect to $g$, geodesics are not defined at infinite time.

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