# Construct a Riemannian metric on the target manifold so that a surjective submersion becomes a Riemannian submersion

Suppose we have a Riemannian manifold $(M^m, g_M)$ and a smooth manifold $N^n$ (with dimensions $m > n$). And suppose we have a smooth surjective submersion $F: M^m \to N^n$.

Question: Is there a natural way to construct a Riemannian metric $g_N$ on $N^n$ such that $F$ can actually be "upgraded" to a Riemannian submersion?

In most cases, there is no Riemannian metric on $N$ which makes $F$ a Riemannian submersion. However, if there is such a metric, it is unique. To shed more light on this matter, let us try to construct the desired metric.
Let $p\in N$, and let $q\in F^{-1}(p)$. Let $H_q$ denote the orthogonal complement of $\ker dF_q$ in $T_qM$. Then $dF_q$ restricts to an isomorphism of vector spaces $$dF_q|_{H_q}:H_q\xrightarrow{\sim}T_pN,$$ and there is a unique inner product on $T_pN$ which makes the above isomorphism an isometry. This explains the uniqueness part in the first paragraph. Now, if we repeat this argument for some $q'\in F^{-1}(p)$ other than $q$, there is no reason to expect we get the same inner product on $T_pN$, and this is why the desired Riemannian metric does not exist usually.
• (+1) It may be worth mentioning that there's an important but very special case where everything does work out: a $G$-invariant Riemannian metric on the total space $P$ of a principal $G$-bundle $P \to B$ canonically induces a unique Riemannian metric on the base space $B$ that makes the quotient map $P \to B$ a Riemannian submersion. – Branimir Ćaćić Dec 22 '17 at 23:37