# Connected Sum of Orientable Manifolds is Orientable

• Bacle
In summary, the conversation discusses the proof of orientability of connected sums of orientable manifolds using the perspective of simplicial homology. The key idea is to select orientations on the cycles being glued together so that they cancel each other out, resulting in an orientable connected sum. This is demonstrated through an example with squares and a general point is made about changing the orientation of the cycle being glued to achieve orientability.
Bacle
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?

Bacle said:
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.

Thanks for your patience in going over a messy proof, Lavinia.

## 1. What is the definition of "Connected Sum of Orientable Manifolds"?

The connected sum of two orientable manifolds is a new manifold formed by removing a small open ball from each manifold and then connecting the resulting boundary spheres together. This can be visualized as gluing the two manifolds together along their boundaries.

## 2. Why is the connected sum of orientable manifolds orientable?

The connected sum of two orientable manifolds is orientable because it can be given a consistent orientation by orienting each of the two original manifolds and then extending this orientation to the boundary spheres that are connected together. This process ensures that the resulting manifold is still orientable.

## 3. Can the connected sum of non-orientable manifolds be orientable?

No, the connected sum of non-orientable manifolds can never be orientable. This is because the connected sum operation of two non-orientable manifolds will always result in a non-orientable manifold, regardless of the orientations chosen for the original manifolds.

## 4. Is the connected sum of orientable manifolds commutative?

Yes, the connected sum of orientable manifolds is commutative. This means that the order in which the manifolds are connected together does not affect the resulting manifold. This property is important in understanding the algebraic structure of manifolds.

## 5. Are there any limitations or conditions for the connected sum of orientable manifolds to be orientable?

The connected sum of orientable manifolds will always be orientable as long as the original manifolds are orientable and the connected sum is performed along a closed, smooth boundary. However, if the boundary of one or both of the original manifolds is not smooth, the resulting manifold may not be orientable.

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