Isometry from warped product onto the base.

Question

Let $B$ and $F$ be semi-Riemannian manifolds with metric tensors $g_B$ and $g_F$, and consider the warped product $B \times_f F$ by a smooth map $f: B \to \Bbb R$, with metric tensor: $$g = \pi^\ast g_B + (f \circ \pi)^2 \sigma^\ast g_F,$$where $\pi$ and $\sigma$ are the projections on $B$ and $F$.

O'Neill says in page $205$ that for every $q \in F$, $\pi\big|_{B \times q}$ is an isometry onto $B$. I'm am having trouble checking that.

My idea was to follow my nose and check that $\left(\pi\big|_{B \times q}\right)^\ast g_B = g$. For this, let $p \in B$ be any point, and take $(v,w),(v',w') \in T_{(p,q)}(B \times_f F)$. We have: $$\begin{align} \left(\pi\big|_{B \times q}\right)^\ast (g_B)_{(p,q)}((v,w),(v',w')) &= (g_B)_{\pi\big|_{B \times q}(p,q)}\left({\rm d}\left(\pi\big|_{B \times q}\right)_{(p,q)}(v,w),{\rm d}\left(\pi\big|_{B \times q}\right)_{(p,q)}(v',w')\right) \\ &= (g_B)_p(v,v'),\end{align}$$ but I don't see how or why we must have $(g_F)_q(w,w') = 0$ for the desired identity to hold. Help?

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Edit: I realize now that if I take $(v,w) \in T_{(p,q)}(B \times q)$, then we must necessarily have $w = 0$. My question is now conceptual: why restricting $\pi$ to $B \times q$ means that I should restrict my tangent vectors only to $T_{(p,q)}(B \times q)$ instead of $T_{(p,q)}(B \times_f F)$?

Answer 1

To check that a map $f:M\to N$ is an isometry, one does the following:

1. Pick a point $p\in M$ and two vectors $u,v\in T_pM$
2. Push the vectors over to $T_{f(p)}N$
3. Check that $(f^*u,f^*v)_{T_{f(p)}N} = (u,v)_{T_p(M)}$.

In your case, $M=B\times \{q\}$. What is the tangent space of this manifold? Well, multiplication by a singleton doesn't really change a manifold: it's naturally identified with $B$, and in particular has the same dimension. In the product notation, the elements of its tangent space are $(v,0)$.