1
$\begingroup$

Why the boundary of a contractible, simply connected 2 dimensional simplicial manifold is connected? The conclusion is false for simplicial complexes if you consider the two cones $CS^1$ with their top points identified. That is a contractible simply connected simplicial complex with two $S^1$ as its boundary, but it is not a simplicial manifold. The neighborhood around the identified point is not homeomorphic to $R^n$.

$\endgroup$
1
$\begingroup$

If the boundary wasn't connected, you could draw a path from one component to the other. You could make this path a properly-embedded arc in the manifold. By design, removal of this arc from the manifold would not disconnect it (because there's a path from one side of the arc to the other, say in one of the boundary circles).

The Thom class of the normal bundle of this arc is then a non-trivial element of $H^1$ of the manifold $M$, so the manifold can't be disconnected. Said another way, given a loop in your manifold, you could take the mod-2 intersection number with this arc. That's a non-zero functional $Hom(H_1 M, \mathbb Z_2)$.

Edit: here is another construction. A little tubular neighbourhood of the arc $A$ I describe above is homeomorphic to $A \times [-1,1]$, where $(\partial A) \times [-1,1]$ correspond to the points in $\partial M$. Take the composite $A \times [-1,1] \to [-1,1] \to S^1$ where the first map is projection onto the $[-1,1]$ factor and the second map is $[-1,1] \to S^1$ given by $x \longmapsto e^{\pi i x}$. Extend this map $A \times [-1,1] \to S^1$ to a continuous function $M \to S^1$ by sending all the points outside of $A \times [-1,1]$ to $1 \in S^1$.

The fact that our arc does not separate $M$ tells us that this map $M \to S^1$ is non-trivial on the fundamental group.

$\endgroup$
2
  • $\begingroup$ "so the manifold can't be disconnected" should read "so the manifold can't be simply connected", I guess. $\endgroup$
    – user53153
    Feb 18 '13 at 6:37
  • $\begingroup$ No, it says what I meant. Removal of the arc from the manifold results in a connected manifold. $\endgroup$ Feb 19 '13 at 6:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.