# Product-like metric on a pseudo-Riemmanian manifold foliated by Lie group orbits

Suppose we have an $n$-dimensional pseudo-Riemmanian manifold $(M,g)$ on which a connected Lie group $G$ acts isometrically (I am most interested in the Lorentzian case if it matters). Suppose that the orbits of $G$ foliate $M$.

Are there general conditions such that we can choose local flat coordinates $(x^a,y^i)$, with $x^a$ ($a,b=1,2,\cdots,k$) parameterizing the leaves of the foliations, and with the leaves of the foliation determined by $y^i = c^i$ for constants $c^i$ ($i,j=k+1,\cdots,n$), such that the metric is locally a warped product of the form $$ds^2 = g_{ij}(y)dy^i\,dy^j + f(y)\tilde{g}_{ab}(x)dx^a\,dx^b?$$

Some motivation for the question. In general relativity, metrics of such types are often studied. For example, the Schwarzchild metric $$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1} + r^2\left(d\theta^2 + \sin^2\theta\ d\phi^2\right),$$ among others, is of the above type.

In general, one can show that a (spatially) spherically symmetric (Lorentzian) metric takes the form $$ds^2 = -A(r,t) dt^2 + B(r,t) dr^2 + C(r,t)dr\,dt + D(r,t)\left(d\theta^2 + \sin^2\theta\ d\phi^2\right).$$ I would be very happy to see a rigorous derivation of the above fact. All of the arguments I've seen so far are by physicists and are largely heuristic. Why can we choose coordinates $(t,r,\theta,\phi)$ such that the metric takes the above form?

I am also interested in generalizations of the above fact, which is the special case of the metric in a manifold foliated by (space-like) $SO(3)$ orbits. Can it be generalized to other Lie groups? If so, how? And under what conditions?

• This is already false when $M=R^3$, $g_{ij}=\delta_{ij}$ and $G\cong {\mathbb R}$ acting via skew motions. – Moishe Kohan Jul 30 '16 at 13:49
• @studiosus Right, that's why I'm asking for conditions for which it's true. But a skew motion isn't an isometry of $g_{ij}=\delta_{ij}$ right? – EuYu Jul 30 '16 at 14:39
• It is an isometry of course! Then I do not know what kind of conditions you are looking for. – Moishe Kohan Jul 30 '16 at 14:41
• @studiosus Is a skew motion really an isometry? You mean a "shear" when you say skew motion right? Unless my geometry is really out of date, that doesn't seem correct. – EuYu Jul 30 '16 at 14:42
• No, a skew motion is not a shear. It is a composition of a translation with rotation along the axis of the translation. If you take 1-parameter group of such motion, the orbits will be the above axis as well as some helixes. – Moishe Kohan Jul 30 '16 at 14:47