1
$\begingroup$

The setup for my question is an embedded surface $\Sigma\to M$ in a smooth, compact 4-manifold $M$. Assuming one knows the induced metric $g_\Sigma$ on $\Sigma$ , I would like to know if there is a way to find the metric on a tubular neighborhood of $\Sigma$ without knowing the metric on $M$.

I was thinking that the tubular neighborhood is "just like" the normal bundle of the surface. If the normal bundle is $\pi:N\Sigma \to \Sigma$, then locally we have a trivialization like $\Sigma \times \pi^{-1}(\Sigma) \subset M$ and we should be able to put a product metric on the tubular neighborhood. If the normal bundle is trivial, does this splitting become global (meaning I can now write $\Sigma \times \pi^{-1}(\Sigma) = M$)? Can someone give some references on determining when the normal bundle is trivial?

  • 0
    The fibers have dimension equal to the codimension. But since the normal bundle is a bundle over $\Sigma$, the total dimension of $N\Sigma$ (as a manifold) is codimension + dimension = dimension of $M$.2011-06-21

1 Answers 1

1

You can't get the metric on $N\Sigma$ just by knowing it on $\Sigma$ since it could do anything in the normal directions. Concretely, take $\Sigma$ to be the $xy$ plane embedded in four space $M$ with coordinates $x,y,z,w$. Then for any metric like $dx^2 + dy^2 + e^{f(z,w)} (dz^2 + dw^2)$ a tubular neighborhood is just all of $M$ but there's no way we can recover the metric just by knowing it on $\Sigma$.

  • 3
    Sorry I guess I do not understand your question. I thought it was that if the induced metric on $\Sigma$ uniquely determines the induced metric on $N\Sigma$.2011-06-21