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I have the following problem: Let $A, B\subset R^3$, $A$ is homeomorphic to a ball, while $B$ is a standard Euclidean ball. Can it happen that the fundamental group of $A\setminus B$ is a perfect group? I am interested in answers for $A$ and $B$ both closed and open, so in fact this is 4 questions.

I am aware of disturbing examples like the infinite grope or the complement of the Alexander's horned sphere, but I still strongly believe that the answer should be no.

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In the generic case the fundamental group is trivial, hence perfect. But I guess you are asking if the fundamental group of (some component of) $A\setminus B$ can be a non-trivial perfect group. –  Mark Grant Aug 27 '12 at 11:23
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Where does this question come from? WHy should perfection be hard to attain? –  Igor Rivin Aug 27 '12 at 14:34
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@Igor: If you use, say, PL balls, then perfection is equivalent to triviality of the fundamental group: A compact 3-manifold with boundary different from a union of spheres, always has nontrivial $H_1$. On the other hand, if you remove a doubly wild arc from an open ball, the complement has perfect nontrivial $\pi_1$: However, this is an example for $B-A$, not $A-B$ as requested. –  Misha Aug 27 '12 at 21:49
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No, one is enough, thanks a lot for your answer. The source of the problem is metric topology, in particular questions of local connectivity of certain sets. The standard question is: taking a homotopy trivial arc in A, that is contained in a small ball, can you glue a disk into it that is contained in a ball twice the size? And the same for the complement of A and arcs outside a ball - how much space contracting the arc might take. This is phrased in homotopy terms, and I wondered if there is an obstruction for rephrasing it homologically. –  Pawel Goldstein Aug 28 '12 at 9:04
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I still wonder, however, if one can prove that no component has nontrivial perfect $\pi_1$... By the way, Misha, could you give a reference to the result you mention, on non-triviality of $H_1$ for compact 3-manifolds? It seems something one should know; thanks in advance. –  Pawel Goldstein Aug 28 '12 at 9:20
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2 Answers 2

up vote 9 down vote accepted

Consider a smooth properly embedded surface $P\subset \mathbb{R}^3$. Then $\mathbb{R}^3= X\cup Y$, where $X\cap Y=P$ and $X, Y$ are properly embedded submanifolds with $\partial X=\partial Y=P$. By Mayer-Vietoris, we have an exact sequence $0=H_2(\mathbb{R}^3)\to H_1(P)\to H_1(X)\oplus H_1(Y)\to H_1(\mathbb{R}^3)=0$, so we see that $H_1(X)\oplus H_1(Y)\cong H_1(P) \neq 0$ unless $P$ is a union of smoothly properly embedded planes and spheres. Therefore at least one component of $\mathbb{R}^3\backslash P$ does not have perfect fundamental group, or else $P$ is a union of planes and spheres (since $P$ is smoothly properly embedded, there's a nice collar neighborhood, so $H_i(X)\cong H_i(int(X))$, and same for $Y$). In the case that $P$ is a union of properly embedded planes and 2-spheres, by Seifert-Van Kampen's theorem, each $\pi_1(X,x)$ and $\pi_1(Y,y)$, for $x\in X, y\in Y$ injects into $\pi_1(\mathbb{R}^3)$, so is trivial.

Let's apply this to your situation. I'll consider the case of $int(A)\backslash B$, since there's not issue of local connectivity for an open set. Then $P=\partial B \cap int(A)\subset int(A)\cong \mathbb{R}^3$ is a properly embedded smooth surface, so we have either there is a component of $int(A)\backslash B$ which does not have perfect fundamental group, or $P$ is a union of properly embedded planes in $int(A)$ (or a sphere), in which case each complementary region has trivial fundamental group. I think this answers at least one interpretation of your question.

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Great, that's exactly what I hoped for, and the answer proved easier than I feared. –  Pawel Goldstein Aug 28 '12 at 8:55
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Let me deal with the following very special case: the closure of $B$ is contained in the internal part of $A$, and $A$ is bounded. In this case, using that $\partial A$ is compact one can show that there exists a standard open ball $B'$ which has the same center as $B$ but a strictly larger radius, and is such that the closure of $B'$ is contained in the internal part of $A$. Then, one may compute the fundamental group of $A$ by applying Van Kampen Theorem to the open covering $B'$, $A\setminus B$. Since $A$ and $(A\setminus B)\cap B'=B'\setminus B$ are simply connected, we easily get that $A\setminus B$ is simply connected too.

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That is of course true, but this case is rather trivial - all you get is a 3-d annulus "wrinkled" outside. Thanks anyway. –  Pawel Goldstein Aug 28 '12 at 9:23
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