Local isometric embedding right inverse to a Riemannian submersion

Question

Let $M$ and $N$ be Riemannian manifolds such that $\pi:M\to N$ is a surjective Riemannian submersion, i.e. for each $x\in M$, $$\langle \pi_{*x}(v),\pi_{*x}(w) \rangle_{\pi(x)} = \langle p(v), p(w) \rangle_x$$ where $p: T_xM = \text{Ker}(\pi_{*x}) \oplus \text{Ker}(\pi_{*x})^\perp \to \text{Ker}(\pi_{*x})^\perp$ is the orthogonal projection. For instance $M = G$ is a Lie group with a right-invariant Riemannian metric, $N= G/H$ is a homogeneous space with the metric induced by $G$, and $\pi: G\to G/H$ is the projection.

By the implicit function theorem we know that for $x\in M$, there is a neighbourhood $U$ of $\pi(x)$ and a smooth function $f:U \to M$ such that $f(\pi(x)) = x$ and $\pi\circ f = id_U$. My question is, can $U$ and $f$ be chosen such that $f$ is an isometric embedding? If yes, how so?

Some ideas: For $x\in M$, $$\pi_{*x}|_{\text{Ker}(\pi_{*x})^\perp}: \text{Ker}(\pi_{*x})^\perp \to T_{\pi(x)}N$$ is an isometry. So asking for $f$ to be an isometric embedding is equivalent to asking that for all $x'\in U$, the map $$Df: T_{\pi(x')}N \to T_{x'}M = \text{Ker}(\pi_{*x'}) \oplus \text{Ker}(\pi_{*x'})^\perp$$ vanishes on the $\text{Ker}(\pi_{*x'})$ coordinate. I can see that the implicit function theorem allows us choose $f$ so that $Df|_x$ satisfies this, but I don't know whether it is possible to do so locally around $x$.

Intuition: For $\pi: \mathbb{R^2}\to \mathbb{R}$, projection onto the first coordinate, there is a map $f:\mathbb{R}\to \mathbb{R^2}$ such that $\pi \circ f = id_\mathbb R$. For example we can take $f(x) = (x, g(x))$ for any function $g:\mathbb R \to \mathbb R$. However $f(x) = (x,a)$ for constant $a\in \mathbb R$ have the special property that they are isometric embeddings. So I'm interested to know if we can always do this locally in the general setting above, and if so how.

Any help is appreciated!

Answer 1

You can do so if and only if the horizontal distribution $x \mapsto \ker(\pi_*)^\perp$ is integrable. This is equivalent to the vanishing of O'Neill's $A$-tensor for the Riemannian submersion: $$ \langle \nabla_X Y, U \rangle = 0, $$ for all vector fields $X,Y,U$ with $X,Y$ horizontal (i.e. tangent to $\ker( \pi_*)^\perp$, $U$ vertical (i.e. tangent to $\ker(\pi_*)$). Here $\nabla$ is the Levi-Civita connection of $M$.

Since this tensor is skew-symmetric in $X,Y$, this is always satisfied for instance when $\dim N = 1$.

See e.g. Besse, "Einstein manifolds", Chapter 9.