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I want to understand the following question:

Let $\delta$ be the Euclidean metric in $\mathbb{R}^2$. Is there any criteria for smooth function $u$ such that $(\mathbb{R}^2, e^{2u}\delta)$ can be compactified to a compact closed Riemannian surface?

For example, if $u(x,y)=\ln(\frac{2}{1+x^2+y^2})$, then $(\mathbb{R}^2, e^{2u}\delta)$ is the unit sphere minus the north pole. Hence $\mathbb{R}^2$ can be compactified under such metric.

I'm wondering is there a general criteria?

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    $\begingroup$ The meaning of "can be compactified" is not clear. The surface with a metric of your type is always conformally equivalent to the plane. The plane, as a Riemann surface can be compactified to the sphere (and to nothing else). But probably you mean compactification not of the Riemann surface but of the metric space. Could you explain the precise meaning of "a metric space can be compactified"? $\endgroup$
    – Alexandre Eremenko
    Commented Jul 6, 2020 at 12:39
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    $\begingroup$ For exmple, the plane with original, Euclidean metric "can be compactified" or cannot? $\endgroup$
    – Alexandre Eremenko
    Commented Jul 6, 2020 at 12:41

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As a Riemann surface, the plane can have only one compactification, the sphere. Introducing a factor to the metric does not change the conformal structure. So whatever factor you put there, the resulting Riemann surface remains conformally equivalent to the plane and thus have one and only one compactification, namely the sphere. In other words, possible compactifications of a Riemann surface do not depend on the metric in the given conformal class.

Perhaps you are thinking of a metric completion, instead of compactification: metric completion means that you add to your metric space the infinite elements corresponding to equivalence classes of Cauchy sequences. If you are asking for which metrics this completion will add exactly one point at infinity, then the criterion is the following: for every $\epsilon>0$ there exists a compact such that for $z,w$ outside this compact, the distance between $z$ and $w$ is less than $\epsilon$.

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  • $\begingroup$ Sorry I didn't clarify in my question. I know that if the criteria is satisfied, then $\mathbb{R}\cup \{\infty\}$ is a complete metric space, under the metric $g=e^{2u}\delta$. Also, I know that if $\int_{\mathbb{R}^2}e^{2u}dx<\infty$, then the criteria is satisfied. My true question is that, how to make sure such metric is still a Riemannian metric at $\infty$? Or how to choose a local chart in a neighborhood of $\infty$ such that it is a smooth extension of the original metric defined only on $\mathbb{R}^2$? $\endgroup$
    – student
    Commented Jul 6, 2020 at 14:26
  • $\begingroup$ Choose the chart $1/z$ and the condition is that your $e^u$ written in this chart is smooth at $0$. $\endgroup$
    – Alexandre Eremenko
    Commented Jul 6, 2020 at 19:54
  • $\begingroup$ I see. Thank you very much! $\endgroup$
    – student
    Commented Jul 8, 2020 at 14:39
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I answer the following question (which I think is the intention): Under which condition does there exist a compact Riemannian manifold $(M,g)$ and a chart $\phi=(x,y)\colon U\subset M\to \mathbb R^2$ such that the metric $$g_{\mid U}=\phi^*{e^{2u}\delta}?$$

Because every Riemannian metric induces a Riemann surface structure, $M$ must be diffeomorphic to the 2-sphere, see the answer of Alexandre Eremenko.

Then, you can use (conformal) coordiantes $(\tilde x,\tilde y)=\tfrac{1}{x^2+y^2}(x,-y)$ centered at infinity $\infty$ to compute wether for the given function $u$ the $(0,2)$-tensor extends smoothly as a Riemannian metric to $\infty.$ This is a standard exercise in a first course on differential geometry.

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  • $\begingroup$ Thank you very much for your post. Yes, the question you posted is really what I want to understand. I have trouble in understanding why the metric completion under the metric $g=e^{2u}\delta$ implies that such completed metric is also Riemannian, or why it is also a smooth extension of the original metric defined only on $\mathbb{R}^2$? For example, given $e^{2u}=1/(x^2+4y^2)$, what are such $M$, $U$ and $\phi$? Sorry I have a very poor basis of Riemannian geometry. $\endgroup$
    – student
    Commented Jul 6, 2020 at 14:29
  • $\begingroup$ Sorry for the typo in the previous comment. It should be $e^{2u}=1/(1+x^2+4y^2)$. Could you give me a reference on how to use the conformal coordinates you gave to proceed computation? $\endgroup$
    – student
    Commented Jul 6, 2020 at 14:48
  • $\begingroup$ You have $\phi^*\delta=\tfrac{1}{(\tilde x^2+\tilde y^2)^2} \delta$. Therefore you need that $e^{2u\circ\phi(\tilde x,\tilde y)} \tfrac{1}{(\tilde x^2+\tilde y^2)^2}$ extends smoothly to $0$ and that the value of this function at $0$ is bigger than 0. $\endgroup$
    – Sebastian
    Commented Jul 6, 2020 at 20:42
  • $\begingroup$ Thank you very much! This makes a lot of sense. $\endgroup$
    – student
    Commented Jul 8, 2020 at 14:38

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