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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 $R^n$.

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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.

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