# What kind of isometry? A metric tensor "respects" the foliation?

1. May 24, 2014

### center o bass

Suppose we have a foliation of leaves (hypersurfaces) with codimension one of some Riemannian manifold $M$ with metric $g$. For any point $p$ in $M$ we can then find some flat coordinate chart $(U,\phi) = (U, (x^\mu, y))$ such that setting $y$ to a constant locally labels each leaf (hypersurface) uniquely.

Let the dimension of the manifold $M$ be $n$ and let latin indicies run from $1$ to $n$ and greek run from $1$ to $n-1$.

In the chart we can introduce a coordinate basis and express the metric as $g=g_{ab}dx^a \otimes dx^b$. Now suppose that
$$\frac{\partial}{\partial y} g_{ab} = 0$$
in _any_ flat chart for the foliation.

Since the $y$-curves defined by $x^\mu$ to constant are only defined within each flat chart for the foliation they are not globally defined (they might be disconnected and be at completely different angles from their respective leaves). And hence $frac{\partial}{\partial y}$ is not a globally defined killing vector.

But what are this kind of isometry then kalled? Somehow the metric $g$ respects the foliation. Are there any equivalent definitions of this situation?

2. Jun 26, 2014

### Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Jun 27, 2014

### Geometry_dude

What you're doing is a bit weird, in the sense that $y$ is not a coordinate, but a label for the leaves, which can be considered a global parameter. The leaves $\Lambda_y$ need not admit a global chart. A better notation for the chart would be
$$(U_y \subseteq \Lambda_y, (x^i_y))$$
Corresponding to the foliation you have a distribution locally spanned by the pushforward of the coordinate frames to $M$. If $\frac{d}{d y}$ commutes with this distribution, then locally you can find an (adapted) chart on $M$ s.t. the y becomes a coordinate and the other coordinates coincide with the ones on the leaves. This distinction should be made.
In the former case, the equation you actually wrote down is
$$\frac{d}{d y} g \downharpoonright_{\Lambda_y} = 0 \, ,$$
which is trivial. In the latter case you wrote
$$\mathcal{L}_{\frac{d}{d y}} g = 0 \, ,$$
which is exactly the condition for it to be a Killing vector field and might not be true in general. If it is, the metric is preserved by the flow of the vector field $\frac{d}{d y}$ and, again assuming the foliation exists, it is enough to study one of the leaves and then the manifold $M$ is the orbit of the action of the flow on the leaf. Note that this implies that your manifold is diffeomorphic to $Q \times \mathbb R$ or
$Q \times S^1$.

Last edited: Jun 27, 2014