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A condition for the intersection of two smooth manifolds to be a smooth manifold is that they intersect transversally. Is this only an obstruction because of the smooth structure?

Question: Is the intersection of two topological manifolds always a topological manifold? If not, are there conditions which can be added to the manifold(Not the smooth+transversal intersection) so that it is true?

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    $\begingroup$ There is a topological transversality theory, but it's quite hard. See Kirby-Siebenmann, "Foundational Essays on Topological Manifolds". $\endgroup$
    – user98602
    Aug 12, 2015 at 17:13
  • $\begingroup$ Thank you very much. I will check it out. $\endgroup$
    – joseph123
    Aug 12, 2015 at 17:56

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Actually, the intersection of two topological manifolds in a Euclidean space can be almost anything. Here's an example to show how bad things can get. Let $C$ be any closed subset of $\mathbb R^n$ whatsoever, and let $f\colon \mathbb R^n \to \mathbb R$ be the function $$ f(x) = \operatorname{dist}(x,C). $$ Thus $f$ is continuous, and $f(x)=0$ if and only if $x\in C$.

Let $M\subset\mathbb R^{n+1}$ be the graph of $f$, and let $N\subset\mathbb R^{n+1}$ be the graph of the zero function (i.e., $N = \mathbb R^n\times \{0\}$). Then $M$ and $N$ are both topological submanifolds of $\mathbb R^{n+1}$, and $M\cap N = C \times \{0\}$.

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  • $\begingroup$ Thank you very much for the example. $\endgroup$
    – joseph123
    Aug 12, 2015 at 17:57
  • $\begingroup$ If I may ask another question: Are there any known conditions ( apart from smooth + transversality) that guarantee that the intersection is a topological manifold? $\endgroup$
    – joseph123
    Aug 12, 2015 at 18:11
  • $\begingroup$ None that I know of. I don't expect you'll find anything better than the Kirby-Siebenmann topological transversality theory mentioned by @MikeMiller. $\endgroup$
    – Jack Lee
    Aug 12, 2015 at 20:55
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No, the general intersection of topological manifolds need not be another topological manifold.

For instance, suppose I have two two-dimensional manifolds $M_1$ and $M_2$. Here $M_1$ is the $xy$-plane in $\mathbb{R}^3$, and $M_2$ is the union of the unit sphere $S_a$ with center $a=(0,0,1/2)$ and of the unit sphere $S_b$ with center $b=(3,0,1)$. If you want $M_2$ to be connected, you can connected $S_a$ to $S_b$ via a tube that doesn't intersect the $xy$-plane.

As you can see, the intersection $M_1\cap M_2$ consists of a circle in the $xy$-plane (in $\mathbb{R}^3$) with center $(0,0,0)$ together with the point $(3,0,0)$. It is some object that is both one- and two-dimensional, which cannot be a manifold.

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    $\begingroup$ Is this the only obstruction? If the intersection turned out to be equi-dimension(in some sense), would it then be a manifold? $\endgroup$
    – joseph123
    Aug 12, 2015 at 11:35
  • $\begingroup$ I reckon you also need some conditions to ensure that the topologies of the two manifolds agree on the overlap. $\endgroup$
    – GFR
    Aug 12, 2015 at 12:20
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    $\begingroup$ @joseph123 Of course it depends on the definition of equi-dimension. As you know, if $M_1,M_2\subset X$ are submanifolds of some manifold $X$, and if $M_1$ and $M_2$ intersect transversely, then $M_1\cap M_2$ is again a manifold. I guess what you are looking for is a set $S$, such that everyone can agree on the dimension of $S$, but something disqualifies $S$ from being a manifold. One example of this might be if $S$ is the Cantor set, which looks pretty $0$-dimensional, but it is not a manifold. The real problem is then to find $M_1$ and $M_2$, such that $M_1\cap M_2 = S$ (if possible). $\endgroup$
    – Mankind
    Aug 12, 2015 at 14:55
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    $\begingroup$ @joseph123: Is this the only obstruction? If the intersection turned out to be equi-dimension(in some sense), would it then be a manifold? No. For example, consider the intersection of two congruent squares having a side in common. The intersection is a closed line segment, which is a manifold-with-boundary but not a manifold. $\endgroup$
    – user13618
    Jan 28, 2017 at 3:09

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