Connected Sum of Orientable Manifolds is Orientable

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  • #1
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Hi, All:
I am trying to show that the connected sum of orientable manifolds M,M' is orientable , i.e., can be given an orientation. I am using the perspective from Simplicial Homology.

Consider the perspective of simplicial homology, for orientable manifolds M,M', glued about cycles C,C' respectively. The idea is that we can use the original orientations and then select orientations on C,C', so that they cancel each other out when glued together, and then the remaining orientations on (M-C) and (M'-C') remain the same. Still, I guess I am assumming that manifolds are simplicial complexes; I don't know if we need any additional condition like, e.g., C^1 or higher.


Assume WOLG that M,M' are both connected: if an m-manifold M is orientable , this means

that the top cycle --call it m'-- can be assigned a coherent orientation, so that m' is a cycle -

-that does not bound, since m is the highest dimension--, i.e., the net boundary of m

cancels out , e.g., in the simplest case of a loop with boundary a-a=0. This means m', which

represents M itself, is a non-trivial cycle, which generates the top homology class. If your

coefficient ring is Z, then the top homology will be Z; consider going n-times about the

cycle. Now, the key is that the two orientable manifolds can be glued so that, at the circle of

gluing, the total boundary cancels out, and the resulting manifold M#M' is still orientable. As

a specific example, consider a square a,b,c,d, with arrows going all in the same direction, so

that the net boundary is : (b-a)+(c-b)+(d-c)+(a-d)=(b-b)+(a-a)+(c-c)+(d-d)=0 . Now glue a

second square a',b',c',d' along a common edge, (say (b,c) with (b',c')), but reverse the

orientation of the edge (b',c') in M when gluing, and notice how the simplex resulting from

the gluing also has net boundary zero.

Now, the key general point is that , at the cycle C where we collapse M with M', we change

the orientation of C in either M or M', so that, along the common cycle, where you are doing

the gluing, the respective boundaries cancel each other out, and the remaining orientations

of M-C and M-C' remain the same, so that M#M' is orientable.


Does this Work?
 

Answers and Replies

  • #2
lavinia
Science Advisor
Gold Member
3,259
642
Hi, All:
I am trying to show that the connected sum of orientable manifolds M,M' is orientable , i.e., can be given an orientation. I am using the perspective from Simplicial Homology.

Consider the perspective of simplicial homology, for orientable manifolds M,M', glued about cycles C,C' respectively. The idea is that we can use the original orientations and then select orientations on C,C', so that they cancel each other out when glued together, and then the remaining orientations on (M-C) and (M'-C') remain the same. Still, I guess I am assumming that manifolds are simplicial complexes; I don't know if we need any additional condition like, e.g., C^1 or higher.


Assume WOLG that M,M' are both connected: if an m-manifold M is orientable , this means

that the top cycle --call it m'-- can be assigned a coherent orientation, so that m' is a cycle -

-that does not bound, since m is the highest dimension--, i.e., the net boundary of m

cancels out , e.g., in the simplest case of a loop with boundary a-a=0. This means m', which

represents M itself, is a non-trivial cycle, which generates the top homology class. If your

coefficient ring is Z, then the top homology will be Z; consider going n-times about the

cycle. Now, the key is that the two orientable manifolds can be glued so that, at the circle of

gluing, the total boundary cancels out, and the resulting manifold M#M' is still orientable. As

a specific example, consider a square a,b,c,d, with arrows going all in the same direction, so

that the net boundary is : (b-a)+(c-b)+(d-c)+(a-d)=(b-b)+(a-a)+(c-c)+(d-d)=0 . Now glue a

second square a',b',c',d' along a common edge, (say (b,c) with (b',c')), but reverse the

orientation of the edge (b',c') in M when gluing, and notice how the simplex resulting from

the gluing also has net boundary zero.

Now, the key general point is that , at the cycle C where we collapse M with M', we change

the orientation of C in either M or M', so that, along the common cycle, where you are doing

the gluing, the respective boundaries cancel each other out, and the remaining orientations

of M-C and M-C' remain the same, so that M#M' is orientable.


Does this Work?

This seems right.

If the manifold is orientable then the boundary of the fundamental cycle minus an n-simplex is the boundary of that removed n-simplex with the induced orientation. In the connected sum the two boundaries of the two removed n-simplices cancel if you choose them to have complimentary orientations and so again you get a fundamental cycle.
 
  • #3
662
1
Thanks for your patience in going over a messy proof, Lavinia.
 

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