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Suppose $M$ is a smooth submanifold of $\mathbb{R}^n$. Through some transversality tricks, I was able to prove that if $\dim M<n-2$, then $\mathbb{R}^n\setminus M$ is always connected and simply connected.

This motivates me to ask is it true that if $\dim M\geq n-2$, then $\mathbb{R}^n\setminus M$ is not connected and simply connected? Even looking at some small cases, if $\dim M=0$, say $M$ is a finite number of points, then although $\mathbb{R}^2\setminus M$ is connected, it is not simply connected since it retracts onto a wedge of circles. I was told does indeed follow from mod-$2$ intersection theory, but I don't see how to apply that here.

Clara
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  • Are you assuming $M$ is a compact submanifold? It seems that the people answering your question are assuming so. Does the non-compact submanifold $(e^t\cos t, e^t\sin t)$, $t\in\Bbb R$, separate $\Bbb R^2$? – Ted Shifrin Mar 19 '15 at 13:05

2 Answers2

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First of all: 1-connected = connected and trivial fundamental group, so "connected and 1-connected" is redundant.

Now if the codimension is 1, then the following statement holds: $M^{n-1} \subset N^n$ submanifolds both closed oriented, then $M$ is seperating (i.e. the complement is not connected) if and only if $i_*([M]) = 0\in H_{n-1}(N)$. To show this you basically use $mod2$ intersection theory by saying that if you can find a loop in $N$ that does intersect $M$ precisely odd times, then $M$ is not nullhomologous in the sense above. (Note that if $M$ is non-seperating you can always find a loop intersecting $M$ precisely once, since the normal bundle is trivial).

Apply that to the contractible space $\mathbb R^n$ and since homology is a homotopy functor, $M$ is nullhomologous in there. Hence the result.

Now for codimension 2 a similar argument yields that it is not 1-connected. However, the complement is 0-connected, since e.g. the 2-dimensional normal bundle minus the zero section is, or by another argument you could also use transversality for paths and $n-2+1 \neq n$.

Daniel Valenzuela
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  • you also could use Alexander Duality to show most of this, but usually a geometric light on the picture can be quite useful, since Alexander duality is not very intuitive. – Daniel Valenzuela Mar 18 '15 at 00:30
  • Thanks. Sorry, I hadn't heard of Alexander Duality until now. If $M$ has codim $2$, does it work something like $\tilde{H}_1(R^n\setminus M;Z/(2))=\tilde{H}^{n-2}(M;Z/(2))=Z/(2)$, so $H_1(R^n\setminus M;Z/(2))=Z/(2)\oplus Z/(2)$, so $\pi_1(R^n\setminus M)\neq 0$, hence $R^n\setminus M$ is not simply connected? (I dropped the blackboard font for $R$ and $Z$ to save characters.) – Clara Mar 18 '15 at 00:47
  • Almost: $\tilde H_1(R^n-M;Z_2) = H_1(R^n-M;Z_2)=Z_2$. The reduced homology only differs in dimension 0. But other than that your implications are correct. – Daniel Valenzuela Mar 18 '15 at 03:13
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In codimension 1, suppose that $\mathbb R^n\setminus M$ were path connected. Then, assuming the embedding is nice enough, you can find a loop $\gamma$ that intersects $M$ in a single point. Basically, you start on one side of the manifold and loop your way around to the the other side, since the complement is connected. However, looking at $M$ and $\gamma$ as representing mod 2 homology classes, the intersection number has to be zero, since they both represent the trivial class. This is a contradiction.

One can also use Alexander Duality to establish this, which will also tell you how to solve the codimension 2 case: the first cohomology of the complement is the top dimensional homology of $M$.

Cheerful Parsnip
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