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The Product Neighborhood Theorem states that if $N\subseteq M$ is a smooth compact submanifold without boundary of codimension $n$ and there is a trivialization of the normal bundle (wrt. some smooth metric) of $N$ in $M$, then some neighborhood of $N$ is diffeomorphic to $N\times \mathbb{R}^n$ s.t. the diffeomorphism takes the framing vectors to the canonical basis of $\mathbb{R}^n$.

Is the same true if $N,M$ are smooth manifolds with boundary and $N$ is a neat submanifold of $M$?

For the proof, in most texts I found the idea to map $(n,x)\in N\times\mathbb{R}^n$ to $\varphi(1)$ for a geodesics $\varphi$ starting in $n$ with tangent vector $\epsilon x$ for $\epsilon$ small enough. This probably doesn't work on the boundary: for example, if $M$ is the closed unit 2-disc and $N:=\{0\}\times [-1,1]$ with the horizontal framing $(1,0)$ then I don't see how to generalize the above construction. I can hardly have a geodesics in $M$ starting in $(0,1)$ in the $(1,0)$ direction.


Remark: What confuses me is that the boundary-version of PNT is used in the proof of the Thom-Pontryiagin construction -- namely, that $[M,S^n]$ is isomorphic to the framed cobordism group $\Omega^{fr}_{n;M}$ (a framed cobordism has boundary) -- but in all books, I have only seen the statement of PNT for the boundaryless case.

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  • $\begingroup$ Not to be nitpicky, but your $M$ is not a manifold with boundary. It has corners. And you're supposed to be allowing only $N$ to have boundary? But clearly your tubular neighborhood should only allow appropriate "inward-pointing" normal directions. (Sounds like a normal cone in geometric measure theory ...) $\endgroup$ – Ted Shifrin Aug 1 '14 at 21:33
  • $\begingroup$ Ted: thanks for your remark: I changed the $M$ in my example to be a unit ball.. $\endgroup$ – Peter Franek Aug 1 '14 at 21:57
  • $\begingroup$ Peter, I'm still confused. Are you allowing both $M$ and $N$ to have boundary, or just $N$? $\endgroup$ – Ted Shifrin Aug 1 '14 at 21:57
  • $\begingroup$ Let say both, $N$ is neat in $M$ and $\partial N\subseteq \partial M$. $\endgroup$ – Peter Franek Aug 1 '14 at 21:58
  • $\begingroup$ OK, that's not how the question reads :) $\endgroup$ – Ted Shifrin Aug 1 '14 at 21:59
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I believe I've found the answer in this text and more details in Kosinski's differential manifolds, p. 53. In a nutshell: for a neat submanifold $N$ of $M$, one can define the normal bundle abstractly as the quotient $(T_N M)/TN$. Any choice of Riemannian metric induces an "orthogonal" representation of the normal bundle.

Choosing a vector field $X$ on $\partial M$ in the "inwards" direction such that $X_{\partial N}\subseteq TN$ induces a line bundle $\nu$ over $\partial M$ and a decomposition $T_{\partial M}M=T(\partial M)\oplus \nu$ which induces an isomorphism $$T_{\partial N}M/T_{\partial N}N\simeq T_{\partial N}\partial M/T(\partial N),$$ so the notion of the normal bundle of $N$ in $M$ and $\partial N$ in $\partial M$ are compatible.

The key idea to get a tubular neighborhood diffeomorphic to the normal bundle is to choose a metric on $\partial M$, extend it to a product metric on a collar neighborhood of $\partial M$ that is induced by the vector field $X$, and then extend it to a smooth metric on whole $M$. Then normal vectors to $N$ on $\partial N$ coincide with normal vectors to $\partial N$ in $\partial M$, geodesics in $\partial M$ stay in $\partial M$, and the "geodesic" construction of tubular neighborhood works up to the boundary. If the normal bundle is trivial, it reduces to the answer to my question.

(Note that in Kosinski, the notation of $N$ and $M$ is reversed.) I hope that it makes sense and I understood the key points of Freeds concise text correctly.

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  • $\begingroup$ This was so nice, thanks. $\endgroup$ – Behnam Esmayli Aug 18 '16 at 18:55

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