Connected sum of topological manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:23:15Z http://mathoverflow.net/feeds/question/121571 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121571/connected-sum-of-topological-manifolds Connected sum of topological manifolds ACL 2013-02-12T08:59:55Z 2013-02-13T21:06:57Z <p>A definition of the connected sum of two $n$-manifolds $M$ and $M'$ begins by considering two $n$-balls $B$ in $M$, $B'$ in $M'$, and glueing the varieties $M\setminus \mathring B$ and $M'\setminus \mathring B'$ along their boundary (an $(n-1)$-sphere) by an orientation-reversing homeomorphism. The construction depends a priori on these various choices, but it is asserted at many places of the litterature (Lee's book on topological manifolds for example, as well as <a href="http://en.wikipedia.org/wiki/Connected_sum" rel="nofollow">Wikipedia</a>) that the result does not depend on these choices.</p> <p>In the differentiable case, a reference is given to a theorem of Palais (<em>Natural operations on differential forms</em>, Thm. 5.5) which asserts — roughly — that two embedding of $n$-balls differ by a global diffeomorphism which is isotopic to identity.</p> <p>Are the details of this independence written somewhere in the litterature, both in the continuous and the smooth case?</p> http://mathoverflow.net/questions/121571/connected-sum-of-topological-manifolds/121574#121574 Answer by Bruno Martelli for Connected sum of topological manifolds Bruno Martelli 2013-02-12T09:48:40Z 2013-02-12T09:55:18Z <p>In the smooth setting, I would suggest you to read the <a href="http://www.maths.ed.ac.uk/~aar/papers/kervmiln.pdf" rel="nofollow">paper of Kervaire and Milnor</a> on homotopy spheres. </p> <p>The paper explains clearly that connected sum must be defined with some care in higher dimensions: if you choose any orientation-reversing diffeomorphism of the two $(n-1)$-spheres, then the resulting smooth manifold is not determined (low-dimensional topologists don't care about that, because problems arise only when $n\geq 7$).</p> <p>In fact, every exotic sphere is obtained by gluing two $n$-discs via some diffeomorphism of their boundaries, and this would imply that every exotic sphere is the connected sum of two $n$-spheres! Indeed, to define unambiguously the connected sum you need to choose a particular isotopy class of diffeomorphisms of the two boundary $(n-1)$-spheres, and luckily this can be done by taking one that extends to the two removed discs. This is clearly explained in the paper of Milnor and Kervaire. </p> <p>With that requirement, connected sum is well-defined and produces a unique result (up to diffeomorphism). The independence of the result is obtained using a <a href="http://www.jstor.org/stable/2032968" rel="nofollow">theorem of Palais</a> and <a href="http://archive.numdam.org/article/BSMF_1961__89__227_0.pdf" rel="nofollow">Cerf</a> that states that two smooth embeddings of the $n$-disc in a connected $n$-manifold are always related by a diffeomorphism of the manifold.</p> http://mathoverflow.net/questions/121571/connected-sum-of-topological-manifolds/121635#121635 Answer by Lee Mosher for Connected sum of topological manifolds Lee Mosher 2013-02-12T19:17:12Z 2013-02-13T21:06:57Z <p>In the topological category the proof that connected sum is well-defined depends on the Annulus Theorem, first proved by Kirby; the necessity of the Annulus Theorem is seen from Bruno Martelli's answer. So you are not likely to find a proof before Kirby's paper. Perhaps someone jotted a proof down, maybe someone who thought about the Annulus Theorem when it was still a conjecture, and realized that well-definedness of connected sum was a good application. But, I do not know. </p> <p>Anyway, the proof is straightforward once you have the Annulus Theorem. Here's a sketch.</p> <p>There's a couple of missing hypotheses. One must assume $M,M'$ are connected. One must also assume $M,M'$ are orientable. And one must assume the balls $B,B'$ are "nicely embedded"; at the minimum, assume that the boundary spheres $S,S'$ are locally bicollared, which implies globally bicollared by Brown's theorem. This rules out nastiness like an Alexander horned ball.</p> <p>Now one shows that the connected sum is independent of the choice of gluing map $S \to S'$. This follows from the fact that any two homeomorphisms $S \to S'$ which agree on orientations are isotopic: once that is known, one absorbs the isotopy into the collar neighborhoods. Proving this fact may already require the Annulus Theorem.</p> <p>For the rest, it suffices to prove that for any two nicely embedded balls $B_1,B_2 \subset M$ there exists an orientation preserving homemeorphism of $M$ taking $B_1$ to $B_2$, in fact an ambient isotopy. Using the boundary bicollaring, we may assume $B_1,B_2$ are contained respectively in open balls $U_1,U_2$, which are centered on points $p_1,p_2$ in some coordinate chart. We can also assume that $p_1=p_2$, because there is an ambient isotopy of $M$ taking $p_1$ to $p_2$: connect $p_1$ to $p_2$ by a path, cover the path by finitely many charts, and concatenate a sequence of ambient isotopies supported in these finitely many charts, moving $p_1$ along the path step by step to $p_2$. We can also replace $B_1$ by an arbitrarily small subball in $U_1$ centered at $p_1$, and similarly for $B_2$; this is straightforward to check using an ambient isotopy supported in the coordinate charts for $U_1$ and $U_2$. In particular, we can assume $B_1$ is contained in the interior of $B_2$.</p> <p>Now apply the annulus theorem: the difference $B_2 \setminus B_1$ is homeomorphic to a sphere crossed with an interval. Using this, one can then ambiently isotope $B_2$ to $B_1$.</p> http://mathoverflow.net/questions/121571/connected-sum-of-topological-manifolds/121642#121642 Answer by Igor Belegradek for Connected sum of topological manifolds Igor Belegradek 2013-02-12T20:05:08Z 2013-02-12T22:24:27Z <p>I just taught this in my undergraduate topology class (in the topological category), and in fact I usually give a version of this as a homework exercise (with hints). </p> <p>The idea is to define connected sum along the <i>coordinate balls</i>, i.e. the balls that are mapped via coorinate chats to standard balls in $\mathbb R^n$. For such balls the argument is easy, and no annulus theorem is needed. For notational convinience let's write a coordinate ball as $B_x$ where $x$ is the point in $M$ that cooresponds to the center of the corresponding ball in $\mathbb R^n$. Fix $y\in M$ and consider the subset $Y$ of $M$ consisting of points $x$ such that there is a homeomorphism of $M$ taking $B_x$ to $B_y$. It is easy to show that $Y$ is open and closed (the point is that given two metric balls in $\mathbb R^n$ there is a homeomorphism that maps one ball into the other one, and is the identity outside a compact set; such a homeomorphism can be constructed with bare hands, and this is where the work is). Thus if $M$ is connected, then $Y=M$. The same argument shows that any two coordinate balls are ambiently isotopic. </p>