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Let $(M,g)$ be a compact manifold of dimension $n$ with boundary. If $\partial M$ is smooth then one has a control on the determinant of the Jacobian of the diffeomorphism in the collar theorem, i.e. for $\epsilon$ small enough the diffeomorphism $F:\partial M\times (0,\epsilon)\to M_\epsilon$ can be given by $F(x,t)=\exp_x(t\,{\bf n}(x))$. Moreover there exist a universal constant $C>0$ such that $C^{-1}\le|\det DF|\le C$ (one can take C=2 here). Let say that a family of manifolds satisfies a "Riemannian collar theorem" if for each manifold in the family there exists a collar diffeomorphism $F$ with a universal bounded on $|\det DF|$ as above. Does the family of manifolds with piecewise smooth boundary satisfies a "Riemannian collar theorem"? For example, assume that $\partial M=\cup_{i=1}^m N_i$, where $N_i$ are smooth $(n-1)$-manifolds, and each point of singularity lies on a transversal intersection of exactly two elements of $\{N_i\}$. Do we have a "Riemannian collar theorem" for such family of $(M,g)$? Probably not in general but can we say anything about when it does?

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  • $\begingroup$ I'm no expert, but there's a collar neighbourhood theorem for topological manifolds. Might be worth looking at the proof to see if you can adapt their ideas. $\endgroup$
    – ಠ_ಠ
    Commented Apr 20, 2017 at 11:03
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    $\begingroup$ It's not super clear to me what a collar neighborhood theorem means here. Does the unit square have a collar neighborhood? (Certainly I can't see the corner singularities on the interior.) One thing that is clearly true is that each smooth face has a collar neighborhood (this is obvious for the square). $\endgroup$
    – mme
    Commented Apr 20, 2017 at 11:33
  • $\begingroup$ @MikeMiller thanks for your remark. I agree that the question was very ambiguous. Please see the edited version. $\endgroup$
    – Math101
    Commented Apr 20, 2017 at 12:15
  • $\begingroup$ you probably want to mumble compact if you want a universal constant for one manifold. $\endgroup$
    – Thomas Rot
    Commented Apr 20, 2017 at 13:02
  • $\begingroup$ @ThomasRot Yes, indeed. I forgot to put the assumption. Also by a universal constant, I meant a constant independent of the family of manifolds under consideration. $\endgroup$
    – Math101
    Commented Apr 20, 2017 at 13:26

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