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I have found the upper half-space coordinates on hyperbolic space $\mathbb H^d$ highly technically convenient. In these coordinates the metric takes the form $$g = f(x)((dx^1)^2 + \cdots + (dx^{d - 1})^2 + dy^2).$$ So the metric is a rescaling of euclidean metric written in Cartesian coordinates (if we take $f(x) = (x^1)^{-2}$), and have the added bonus that $\partial_y$ is a Killing field. For the project I've been working on recently, neither of these facts are terribly useful by themselves, but together they are highly powerful.

But I do not understand the mechanism by which such coordinates arise. Right now it just seems like some weird miracle that happens in constant nonpositive curvature. The obvious necessary conditions are that the Cotton ($d = 3$) or Weyl ($d \geq 4$) tensor vanishes, and that the local isometry group of $g$ is nontrivial (so you can pick a local Killing field). But I would be surprised if these conditions are sufficient.

What are necessary and sufficient conditions to write the metric in the form $$g = f(x)((dx^1)^2 + \cdots + (dx^{d - 1})^2 + dy^2)?$$

Here are some preliminary thoughts. I think that you can get a similar metric on $\mathbb S^2$ by writing out the scalar curvature for a metric $e^{2\varphi}(dx^2 + dy^2)$ where $\partial_y \varphi = 0$: $$R = -2e^{-2\varphi} \partial_x^2 \varphi$$ which is an ODE for $\varphi$ that can be solved using local wellposedness for ODE. Actually, feeding the ODE into a computer algebra system for $R = 2$, I think you can take $$e^{2\varphi} = 1 - \tanh^2 x.$$

This method clearly doesn't work for $\mathbb S^3$ since it's not enough to impose constant scalar curvature, you also have to impose that the traceless Ricci tensor vanishes. So you get a somewhat overdetermined-looking system of PDE. For $\mathbb S^d$, $d \geq 4$, you must also impose that the Weyl tensor vanishes...

By the way, I know that sometimes these questions about "when can I write the metric in such and such form" can be deceptively easy. If this question is more suited to MathOverflow, please let me know.

Aidan Backus
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  • To begin with, $\partial_y$ is not a Killing field on the hyperbolic space (in the upper half-space model). The only coordinate fields which are Killing fields will be $\partial_{x_i}, i=1,...,d-1$. But maybe your question is about necessary and sufficient conditions for a Riemannian metric to be locally conformally flat? The answer is usually given in terms of vanishing of the Weyl tensor in dimensions $\ge 4$ and the Schouten tensor in dimension $3$. – Moishe Kohan May 23 '22 at 01:56
  • Oh, my apologies, I should’ve been clearer: the indexes that I used for the entries in the hyperbolic metric were unconventional. As I said the metric takes the form f(x)(dx^2+dy^2) so that y rather than x is the killing direction. – Aidan Backus May 23 '22 at 02:43
  • I can only repeat: If you consider the standard upper-half plane model of the hyperbolic plane, ${(x,y): y>0}, ds^2= y^{-2}(dx^2+dy^2)$, then $\partial_y$ is not a Killing field. – Moishe Kohan May 23 '22 at 15:35
  • I can only repeat: we are taking $f(x) = (x^1)^2$, so $\partial_y$ is a Killing field. – Aidan Backus May 23 '22 at 15:44
  • Then edit your question accordingly, since the formula $f(x)=(x^1)^2$ does not appear anywhere in the original post. – Moishe Kohan May 23 '22 at 15:46
  • You will have to forgive me for assuming that the expression f(x), rather than f(x, y), was quite clear already. – Aidan Backus May 23 '22 at 15:50
  • Could you explain your rationale for your notation? Clearly the "vertical" direction in the upper half-space model is different from the other horizontal directions, so it makes sense to give the vertical direction a variable name that distinguishes it from the others. In your notation, $y$ is a horizontal direction and therefore is equivalent to the coordinates $x^2, \dots, x^n$ and, $x^1$ is the vertical direction and therefore, despite appearances, different from the other coordinates $x^2, \dots, x^n$? – Deane May 23 '22 at 16:09
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    Sure; I now see that it unfortunately probably obfuscated things, but I wanted to privilege the single direction y, as I'm mainly interested in conditions for the existence of one Killing coordinate (with the conformal condition), rather than d - 1 Killing coordinates. – Aidan Backus May 23 '22 at 18:39

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