I know that if $M$ is a manifold without boundary and $N$ a manifold with boudary, then $\partial(M\times N)=M\times \partial N$, but, if I have the product of two manifolds with boudary, it's known what would be the boundary of that? I suppose it could be $\partial{M}\times \partial{N}$, but when I think in $I\times I$ $\left(I=[0,1]\right)$, the corners doesn't look as something differentiable.
What is the boundary of two manifolds with boundary?
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differential-geometry
manifolds-with-boundary
2 Answers
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You're right that "corners" won't be differentiable. In fact, a product of two manifolds $M$ and $N$ with nonempty boundary does not have any natural structure of a differentiable manifold, because of the "corners" in $\partial M\times\partial M$. If you remove these corners, you do get a differentiable manifold $(M\times N)\setminus(\partial M\times\partial N)$ with boundary $(M\setminus \partial M)\times\partial N\cup\partial M\times(N\setminus\partial N)$.
(There is a notion of a "differentiable manifold with corners" that does allow such products. These have not just boundary points that look locally like $[0,\infty)\times\mathbb{R}^{n-1}$ but also higher codimension "corners" that look locally like $[0,\infty)^k\times\mathbb{R}^{n-k}$ for $k>1$.)
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0So quitting the intersection of the boundaries I get a differentiable manifold? – 2017-01-21
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1I'm not quite sure what you mean by that. But if you remove the points that are in $\partial M\times\partial N$ from $M\times N$, then yes, you do get a differentiable manifold with boundary. – 2017-01-21
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0Yes, I wrote it wrong, but you put what I needed to understand. Thanks! – 2017-01-21
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$$ \partial (M \times N) = (\partial M \times N) \cup (M \times \partial N).$$