Given a warped product, I want to compute the ditance of any two points. First get the equation for the geodesic, then compute the length of the geodesic.

Consider the two-dimensional surface $$ g=dr^2+e^{2r}d\theta^2, \quad r\in (-\infty,\infty), \quad \theta \in [0,2\pi] $$ Let $s\to (r(s),\theta(s))$ be a geodesic. Using the formula for the geodesic, I can get $$ r''(s)=C^2 e^{-2r} $$ for some constant $C$. But it seems we can't get the explicit form for $C(s)$, so I don't know how to compute the distance.

Given two points $x_1=(r_1,\theta_1), x_2=(r_2,\theta_2)$, is there a formula for $d(x_1,x_2)$ w.r.t $r_1, r_2, \theta_1, \theta_2$?

The second question is : for a warped product $g_M=dr^2+\varphi^2 g_N$, Is there a distance formula for $x_1=(r_1,y_1), x_2=(r_2,y_2)$ w.r.t. $r_1, r_2, d_N(y_1,y_2)$?


First, the metric that you write down has Gauss curvature $K\equiv-1$, so if you want to compute distances in this metric, you should write it in more standard coordinates and then use the known distance formulae for the hyperbolic plane. Thus, if you set $(x,y) = (\theta,e^{-r})$, you'll find that $$ \mathrm{d}r^2 + e^{2r}\mathrm{d}\theta^2 = \frac{\mathrm{d}x^2+\mathrm{d}y^2}{y^2} $$ and the right hand side is the Poincaré metric on the upper half plane. Many books have the formula for distance between points in the Poincaré metric. Of course, if you really want $\theta$ to be $2\pi$-periodic (i.e., the points $(r,\theta)=(0,0)$ and $(r,\theta)=(0,2\pi)$ are to be regarded as the same), then you'll have to modify the Poincaré distance formula a little bit to get the true distance.

Second, in the more general case, there is no known way to write down the distance formula for a general warped metric, even in the simplest case: a general surface of revolution. Even in dimension $2$, the set of Riemannian metrics for which an explicit distance formula is known is very small. For one example, see my answer to this question. There are a few others (some of which have only discrete symmetries), but, as far as I know, there is no general theory on how to classify them.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.