Is there a metric on $\mathbb{R}^2$ such that the unit circle is a geodesic?

Question

It is well known that all 1D spaces are flat, and furthermore that all paths in 1D spaces with unit tangent vectors are geodesics. In particular the space $S^1$ is flat, and the closed loop traversing the space once is a geodesic.

I've been attempting to embed $S^1$ as the unit circle in $\mathbb{R}^2$ so that the unit tangent vector field $\vec V = [1]$ on $S^1$, which is parallel transported about $S^1$, remains parallel transported about the unit circle in $\mathbb{R}^2$, and furthermore that the path $l(t) = [t]$ which is a geodesic in $S^1$ has its $\mathbb{R}^2$ embedding $l(t) = [\cos (t), \sin (t)]$ also a geodesic.

However, I haven't been able to make the terms work out. I've been attempting to transform the metric tensor and the vector field from $S^1$ to the higher-dimensional space $\mathbb{R}^2$ using the partial derivatives of the coordinate transformations, and from the derived $\mathbb{R}^2$ metric tensor finding the Christoffel symbols, then checking that $V$ and $l$ are parallel transported and a geodesic respectively using the parallel transport condition $<\vec U, \nabla \vec V > = 0$ and the geodesic condition $<\vec U, \nabla \vec U> = 0$.

So, is there a metric on $\mathbb{R}^2$ so that the path $l(t) = [\cos(t), \sin(t)]$ is a geodesic, and if so, what is it? (I'm ready for the answer, since I've burned through all the options that seem evidently available to me).

Answer 1

There are many possibilities.

For example, you can use a stereographic projection to map $\mathbb R^2$ to a sphere of unit diameter minus a point, such that the unit circle maps to the equator. Then pull the metric on the sphere back to $\mathbb R^2$. This gives a nice conformal metric: $$ ds^2 = \frac{dx^2+dy^2}{(x^2+y^2+1)^2} $$

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Intuitively I would expect that whenever you have $ds^2=f(x^2+y^2)^2\cdot(dx^2+dy^2)$ for some function $f:\mathbb R_{\ge 0}\to\mathbb R_{>0}$ that falls off to $0$ "sufficiently fast", there will be some circle around the origin that is a geodesic. With some luck, this might allow you to choose an $f$ of a form that makes your subsequent computations simpler.

Answer 2

This may not quite satisfy you, but here's one approach. $\Bbb R^2-\{0\}$ is diffeomorphic to the cylinder $x^2+y^2=1$. Indeed, we can map in polar coordinates by $$f(r,\theta) = (\cos\theta,\sin\theta,\log r).$$ Any circle on the cylinder is a geodesic when we use the induced metric coming from $\Bbb R^3$. And any such circle pulls back to a circle centered at the origin in the punctured plane. Now pull back this metric back by $g$ to get the metric on the punctured plane: $$g^*(dx\otimes dx + dy\otimes dy + dz\otimes dz) = \dfrac1{r^2}dr\otimes dr + d\theta\otimes d\theta.$$ (If you insist, you can translate this into cartesian coordinates on the punctured plane.)

Since you're interested in only the unit circle, we can in fact use a simple bump function argument to combine this metric on $r>1/2$, say, and the Euclidean metric on $r<1/4$, thereby obtaining a metric on the entire plane with the desired property. If you want to do this and don't know how, let me know.