I know that if two manifolds intersect transversally then their intersection is a manifold. But I was trying to construct an example where the intersection is not a manifold. But I still do not see how intersection of two manifolds cannot be a manifold. It would be great, if any one has any counterexample giving intersection as $xy=0$ or any other set which is not a manifold .

  • $\begingroup$ I think there's an example like this in Guillemin and Pollack. Let one of the manifolds be the $xy$-plane and the other something which lies entirely in the upper half-space, such that the points where $z=0$ is precisely $xy=0$. $\endgroup$
    – user98602
    Oct 5, 2015 at 4:16

1 Answer 1


Let $M_1 \subset \Bbb R^3$ be the $xy$-plane and $M_2 = \{(x,y,z) | (xy)^2 = z\}$. Then $M_2 \cap M_1 = \{(x,y,z) | xy = 0, z = 0\}$, as you desire. $M_2$ is a manifold by the regular value theorem, because the map $f: \Bbb R^3 \to \Bbb R$, $f(x,y,z) = z-(xy)^2$ is a submersion.

Transversality breaks rather wildly, as you notice: in fact, the tangent planes of $M_2$ at points where $z=0$ are the $xy$-plane.

Guillemin and Pollack have some nice pictures of non-transverse intersections in their book. Something with this result is probably in there.

You can see a picture of $M_2$ on WolframAlpha here.

E: In fact, let $X = f^{-1}(0)$ be the zero set of a smooth function $\Bbb R^2 \to \Bbb R$. Then the exact same construction works to build $X$ as the intersection of two submanifolds of $\Bbb R^3$; let $M_2 = \{(x,y,z) | z = f(x,y)^2\}$; the map $f(x,y) - z$ is still a submersion.

And, in fact, any closed subset of $\Bbb R^n$ is the zero set of a smooth function, in particular, say, the Cantor set or the Koch snowflake. So you can make these intersections pretty wild!

  • $\begingroup$ thanks a lot for the nice example $\endgroup$
    – happymath
    Oct 5, 2015 at 4:26
  • $\begingroup$ @happymath: I added some extra comments that might be worth reading. $\endgroup$
    – user98602
    Oct 5, 2015 at 4:28
  • 1
    $\begingroup$ This is really nice this simple construction seems to yield a lot of counterexamples. But I am wondering if $z=xy$ is also a counter example $\endgroup$
    – happymath
    Oct 5, 2015 at 4:36
  • $\begingroup$ @happymath: Yeah, that works too. I dunno why I squared, didn't really matter. $\endgroup$
    – user98602
    Oct 5, 2015 at 5:00

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