# Finding a metric on a tubular neighborhood of an embedded surface

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?

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Do you really mean $\Sigma \times N \Sigma$? The dimension of $N\Sigma$ as a manifold is equal to the dimension of $M$. –  Eric O. Korman Jun 21 '11 at 15:23
shouldn't the normal bundle of a surface have dimension equal to the codimension of the surface? Since $N\Sigma=TM_{i(\Sigma)}/T\Sigma$? Perhaps I mean the fibers of the normal bundle have (in this case) dimension 2? –  levitopher Jun 21 '11 at 16:24
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 $M$. –  Eric O. Korman Jun 21 '11 at 17:09

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 $\Sigma$.
I understand we cannot find THE metric $g$ on $M$, but the normal bundle should have a description independent of the metric right? I mean we have an exact sequence $0\to T\Sigma \to TM_{i(\Sigma)} \to N\Sigma \to 0$ which does not reference the metric. In your example I think you assume a trivial normal bundle so you can put A metric (not THE metric $g$) on all of $M$? –  levitopher Jun 21 '11 at 16:21
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 $N\Sigma$. –  Eric O. Korman Jun 21 '11 at 17:10