5
$\begingroup$

If $(N,h)$ is a Riemannian manifold and $F:M\to N$ is an immersion, we can trivially make $F$ into an isometric immersion by placing the induced metric on $M$. My question concerns the analogous procedure for submersions.

Let $f:M\to N$ be a submersion and let $g$ be a Riemannian metric on $M$. Let $\mathcal{V,H}$ be the vertical and horizontal distributions, respectively, i.e. the tangent space of $M$ at $p$ decomposes as $T_p M=\mathcal{V}_p\oplus \mathcal{H}_p$ for all $p\in M$. The restriction $\varphi_p:=(d\pi_p)|_{\mathcal{H}_p}:\mathcal{H}_p\to T_{f(p)}N$ of the differential is an isomorphism for all $p\in M$.

Is there a way to construct a metric on $N$ so that $f$ becomes a Riemannian submersion?

My idea is to define a metric $h$ on $N$ by $$h_{f(p)}(X,Y)=g_p(\varphi_p^{-1}(X),\varphi_p^{-1}(Y))?$$ Does this work?

$\endgroup$
2
  • $\begingroup$ Is $\mathcal H_p$ the $g$-orthogonal complement of $\mathcal V_p$? $\endgroup$ Jun 1, 2018 at 0:32
  • $\begingroup$ Yes. $\mathcal{V}_p=\mathrm{Ker}(d\pi_p)$ and $\mathcal{H}_p=\mathcal{V}_p^{\perp}$ with respect to $g$. $\endgroup$
    – smnas
    Jun 1, 2018 at 15:20

1 Answer 1

5
$\begingroup$

I think that in general, there is no metric on $N$ for which $\pi$ becomes a Riemannian subermsion. The issue is that you want $d\pi_p:\mathcal{H}_p\rightarrow T_{f(p)}$ to be an isometry for every $p$. In particular, if $p,q$ have $f(p)=f(q)$, then you need $(d\pi_q)^{-1} (d\pi_p):\mathcal{H}_p\rightarrow \mathcal{H}_q$ to be an isometry, and there is no reason it should be.

For an explicit example, let $M$ be the the cylinder $M = S^1\times \mathbb{R}$ with the natural projection map $\pi:M\rightarrow N:=S^1$. Suppose the metric on $M$ is a warped product metric of the form $\phi(t)d\theta^2 + dt^2$. Then the horizontal space at $p\in M$ is spanned by $\frac{\partial}{\partial \theta}$.

Then $M\rightarrow N$ is a Riemannian submersion for some metric on $N$ iff $\phi$ is constant.

(On the other hand, if the Riemannian metric on $N$ is assigned, I believe there is a metric on $M$ which makes $M\rightarrow N$ a Riemannian submersion. To do this, pick any random metric on $M$ in order to talk about $\mathcal{V}_p$ and $\mathcal{H}_p$. Then change this random metric on $\mathcal{H}_p$ by declaring $d\pi_p$ to be an isometry).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .