3
$\begingroup$

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.

$\endgroup$

2 Answers 2

6
$\begingroup$

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 $k>1$.)

$\endgroup$
3
  • $\begingroup$ So quitting the intersection of the boundaries I get a differentiable manifold? $\endgroup$
    – iam_agf
    Jan 21, 2017 at 0:16
  • 1
    $\begingroup$ I'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. $\endgroup$ Jan 21, 2017 at 2:35
  • $\begingroup$ Yes, I wrote it wrong, but you put what I needed to understand. Thanks! $\endgroup$
    – iam_agf
    Jan 21, 2017 at 4:23
2
$\begingroup$

$$ \partial (M \times N) = (\partial M \times N) \cup (M \times \partial N).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.