# Represntative Surfaces/Curves in homology. Examples.?

• Bacle
In summary: I could be wrong about that) so that its 2-chain becomes a cycle. But this cannot happen if S is nonorientable.In summary, if M is orientable, then its top homology is Z. If M is not orientable, then the only n-cycle that has a chance of being a cycle is not a cycle and thus H_n(M;Z)=0.
Bacle
Hi, everyone:

1) Examples of representative surfaces or curves:

Please let me know if this is a correct definition of a surface representing

H_2(M;Z):

Let M be an orientable m-manifold, Z the integers; m>2 . Let S be an orientable

surface embedded in M. Then H_2(S;Z)=Z . We then say that S represents

H_2(M;Z)~Z if the homomorphism h: Z-->H_2(M;Z) sends 1 --as a generator of Z --

to the homology class of S. specifically, if we have h(1)=alpha ; alpha a homology

class, then there is an embedding i:S-->M , with [i(S)] =alpha.

If this is correct. Anyone know of examples of representative curves or surfaces.?

2)An argument for why non-orientable manifolds have top homology zero, and

for why orientable manifolds have top homology class Z.?.

I have no clue on this one. I know homology zero means that all cycles

are boundaries, but I don't see how this is equivalent to not being orientable.

How about this for 2: Specialize to simplicial homology (and thus, to triangulable manifolds). It is more or less obvious that in an n-manifold M the only n-chain in M that has a chance of being a cycle will be some integer multiple of the chain induced by a triangulation of M (in the sense of my reply to your question 1). Now, when does the n-chain induced by a triangulation is a cycle? Precisely when M is orientable. To get the flavor of this, consider the 2-sphere (orientable) and pick a simple triangulation. It is easy to see that the vertices of the triangles can be ordered so that this 2-chain is cycle. The same procedure works for the 2-torus. Ok, so it works in this case. Next consider the Mobius bundle (i.e. the Mobius strip, but infinitely long in order to remove the boundary... more like a "Mobius cylinder") Triangulate it in a simple way and start ordering the vertices of the triangles in order to make the induced 2-chain into a cycle. You will see that this is impossible. Exactly the same procedure can be applied to the Klein bottle and the projective plane.

So, if M is orientable, then in the language and notation of my reply to your question 1, the only nontrivial cycle is represented by M itself, and H_n(M;Z)={k[M]:k\in Z}. (Clearly [M] is not a boundary as there are no (n+1)-chains in M for dimensional reasons.) If M is not orientable, then the only n-cycle that has a chance of being a cycle is not a cycle and thus H_n(M;Z)=0.

Last edited:
Why must H_2(M;Z) be isomorphic to Z? Once this is false, the rest doesn't make sense.

Zhentil wrote:

"Why must H_2(M;Z) be isomorphic to Z? Once this is false, the rest doesn't make sense. "

If M is orientable, its homology over Z is Z. I don't know if this is a definition or a result,

but I know it to be true.

Bacle said:
Zhentil wrote:

"Why must H_2(M;Z) be isomorphic to Z? Once this is false, the rest doesn't make sense. "

If M is orientable, its homology over Z is Z. I don't know if this is a definition or a result,

but I know it to be true.
Its top homology is Z. You specified that the dimension was greater than 2.

Btw, it shouldn't be hard to see that H_2(S^3,Z)=0, since the image of any two-simplex that isn't surjective lands in R^3, and hence retracts to a point.

Zhentil wrote, in part:

"Its top homology is Z. You specified that the dimension was greater than 2. "

Your right. Now I am confused; I don't know if this is an intrinsic orientation of

the surface, or if it is some sort of orientation relative to, and compatible with,

that of the space in which the surface is embedded.

Continuing my previous, and using the helpful comment by Quasar --correct me if
I am wrong , Q * -- if the embedded surface S were orientable (leaving aside for
now issues of the relation between the orientability of S, and that of M) , then
it would be possible to triangulate it ( I think we only need C^1 to be able to
triangulate a manifold, sure that smoothness is way more than enough), and
to triangulate it and orient it in such a way that it is a cycle, and , within S,
it would be the only 2-cycle, again, within S, it would not be a boundary, since
there are no 3-chains in dimension 2. Then the cycle S is a 2-cycle which is
not a boundary, and the only cycles in dimension 2 are integral multiples of Z,
(and, again repeating Q's argument.) none of these cycles would be a boundary,
so that H_2(S;Z)=Z.

## 1. What is a representative surface/curve in homology?

A representative surface/curve in homology is a geometric object that represents a homology class. In other words, it is a surface or curve that is used to represent a group of topologically equivalent objects in a specific homology class. This allows for the classification and study of topological spaces.

## 2. How are representative surfaces/curves chosen?

Representative surfaces/curves are chosen based on their ability to accurately represent a specific homology class. They should possess certain topological properties, such as being connected and closed, and should also be relatively simple and easy to visualize.

## 3. Can you provide an example of a representative surface/curve in homology?

One example of a representative surface in homology is the sphere, which represents the first homology class in two-dimensional spaces. This is because all closed and connected surfaces in two dimensions can be continuously deformed into a sphere.

## 4. How do representative surfaces/curves aid in homology calculations?

Representative surfaces/curves provide a way to simplify and visualize complex topological spaces, making it easier to calculate their homology. By representing a homology class with a single surface/curve, the properties and relationships of that class can be studied and used in calculations.

## 5. Can representative surfaces/curves be used in other branches of mathematics?

Yes, representative surfaces/curves can also be used in other branches of mathematics, such as algebraic topology and differential geometry. In these fields, they are used to study the properties and relationships of topological spaces and their higher-dimensional counterparts.

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