Intuition for warped product manifold.

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.

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 $[0,2\pi]/_\sim = \Bbb S^1$).
• 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? – BigbearZzz May 29 '18 at 18:21