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Hey all. The setup for my question is an embedded surface $\Sigma \hookrightarrow M$ 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 so locally we have something like $\Sigma \times N\Sigma \subset M$ and we should be able to put a product metric on the tubular neighborhood. Does that sound correct?

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I'm not sure I understand the question. M has a Riemannian metric, which induces a Riemannian metric on any (smooth) submanifold of M. So both $\Sigma$ and its tubular neighborhood have induced metrics. Are you asking for a description of the induced metric on the tubular neighborhood? – Dan Ramras Jun 3 '11 at 18:58
Yeah I'm sorry I will edit that, I would like to describe the metric on the tubular neighborhood without having to know the metric on the manifold $M$. – cduston Jun 3 '11 at 19:15
If the normal bundle is trivial, then you can certainly put a product metric on it. However, in your first paragraph you wrote that you want a way to find "the" metric on a tubular neighborhood--the only sense I can make of that is that you want to find the metric induced from $M$. (Otherwise, there are many different metrics on the tubular neighborhood, and there's no reason to prefer one over another.) If you're asking whether the induced metric on the tubular neighborhood is determined by the induced metric on $\Sigma$, the answer is decidedly no. – Jack Lee Jun 3 '11 at 21:36
Ok, so my question must reduce to determining if the normal bundle is trivial; but I actually cannot find much information on this. I suppose it must be connected to the vanishing of some characteristic classes, but could someone direct me a little better? – cduston Jun 21 '11 at 0:56
Even if the normal bundle is trivial topologically or smoothly, it won't necessarily be metrically a product. So characteristic classes are not going to help you. – Ben McKay Jul 6 '11 at 11:28

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