4
$\begingroup$

I am reading about wave equations in manifold and encountered the term warped product manifold. More specifically, in my case it is defined as follows,

$$N:=[0,\phi^*) \times_g \mathbb S^{k-1}$$ where $\phi^* \in \Bbb R\cup\{+\infty\}$ and $g:\Bbb R\to\Bbb R$ is an odd smooth function such that $g(0)=0, g'(0)=1$. On $N$ we have the ''polar'' coordinates $(\phi,\chi) \in [0,\phi^*)\times \mathbb S^{k-1}$. In these coordinates the metric of $N$ takes the form $$ d\phi^2 + g^2(\phi)d\chi^2 $$ where $d\chi^2$ is the standard metric of $\mathbb S^{k-1}\hookrightarrow \Bbb R^k$.

How should I think of $N$ is this case? More importantly, what is a warped product manifold in general?

Any help is very appreciated.

$\endgroup$
2
$\begingroup$

If $(B,g_B)$ and $(F,g_F)$ are pseudo-Riemannian manifolds and $f\colon B \to \Bbb R$ is a smooth positive function, the warped product of $(B,g_B)$ and $(F,g_F)$ is the pseudo-Riemannian manifold $B\times_fF = (B\times F, g)$, where $$g \doteq \pi_B^*g_B + (f\circ \pi_B)^2\pi_F^\ast g_F,$$with $\pi_B$ and $\pi_F$ being the projections onto $B$ and $F$.

It is similar to a direct product, in the sense that all the leaves $B\times q$ are isometric to $B$, while the fibers $p\times F$ are homothetic to $F$, with factor $1/f(p)$. That is, $f$ is warping all the leaves, hence the name. You can look at page 204 of Barret O'Neill's Semi-Riemannian Geometry with applications to Relativity for some results about warped products.

In your specific case, you can see that metric tensor as something coming from a revolution surface. As a simple example, consider a curve $\alpha\colon I \to \Bbb R^3$ parametrized by arc-length, written in the form $\alpha(s) = (\varphi(s),0,\psi(s))$, with $\varphi(s) > 0$. This last condition says that the curve does not meet the revolution axis. Then the revolution surface generated by $\alpha$ can be parametrized by the map $X\colon I\times \left[0,2\pi\right] \to \Bbb R^3$ given by $$X(s,\theta) = (\varphi(s)\cos \theta, \varphi(s)\sin \theta, \psi(s)),$$and the metric tensor of $\Bbb R^3$ induced in the image of $X$ is $$g = {\rm d}s^2 + \varphi(s)^2{\rm d}\theta^2,$$whence the revolution surface is isometric to $I\times_\varphi \Bbb S^1$ (here $I$ and $\Bbb S^1$ are equipped with ${\rm d}s^2$ and ${\rm d}\theta^2$, respectively, and since things are $2\pi$-periodic in $\theta$ they pass from $[0,2\pi]$ to $[0,2\pi]/_\sim = \Bbb S^1$).

$\endgroup$
  • $\begingroup$ Thank you for your answer, it is very helpful. I have one question, if you wouldn't mind. I follow the reference you gave and it seems that O'Neill's book uses the opposite terminology as you were using here, i.e. he calls $B\times q$ the leaves and $p\times F$ the fibers. Is it the case that the standard usage of the terms has changed? $\endgroup$ – BigbearZzz May 29 '18 at 18:21
  • 1
    $\begingroup$ No, what you said is correct, it was my mistake. I'll fix it in the answer now. $\endgroup$ – Ivo Terek May 29 '18 at 18:36

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.