6
$\begingroup$

Hi! Probably this is an easy question, but i can't see the answer.

Let $X$ be a a smooth real manifold with $\dim(X)=d$ and $M,N\subset X$ two smooth submanifolds with $\dim(M)=m$ and $\dim(N)=n$. The submanifolds $M,N$ intersect but not transversely.

What can i say about connected components of $M\cap N$? More precisely, is it possible to find three manifolds $X,M,N$ as above such that a connected component of $M\cap N$ is not a manifold? Or a connected component that is not smooth? (In all the examples i thought, connected components of $M\cap N$ were smooth)

Thank you in advance.

$\endgroup$

2 Answers 2

7
$\begingroup$

Let $M$ be any manifold, and let $Z$ be a closed subset of $M$. Suppose there exists a smooth function $f:M\to\mathbb{R}$ with $f^{-1}\{0\}=Z$. We can then take $X=M\times\mathbb{R}$ and identify $M$ with $M\times\{0\}$ and put $N=\{(m,f(m)):m\in M\}$. Then $M$ and $N$ are embedded submanifolds of $X$ with $M\cap N=Z$.

Moreover, I think it is true that such a function $f$ exists for every closed subset $Z$, no matter how wild or fractal. I don't remember the argument in detail, but if I recall correctly it is not too hard. One issue is to patch together things done locally using a partition of unity, and another is to express $f$ as a countable sum of nonnegative smooth functions $f_n$ which need to be rescaled aggressively to force the higher derivatives of the sum to converge.

$\endgroup$
1
  • 2
    $\begingroup$ It's a Lemma due to Whitney - one of the useful consequences of the existence of partitions of unity on smooth manifolds. $\endgroup$
    – Qfwfq
    Aug 21, 2011 at 17:30
7
$\begingroup$

You could try the real quadric surface $x^2+y^2-z^2=1$ and the hyperplane $x=1$. They intersect along a union of two (intersecting) lines ($z=\pm y$, $x=1$).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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