Geometric description of (simplicial)homologous cycles

[Bacle]

Hi, everyone:

I am trying to understand the geometric interpretation of two simplicial cycles being
homologous to each other.

Let C_k(X) be the k-th chain group in the simplicial complex X, and let c_k be
a chain in C_k(X)

The algebraic definition is clear: two k-cycles x=c_k and y=c_k' are homologous,
i.e., x~y , iff (def.) x-y is a boundary, i.e., if there is a cycle c_(k+1) in C_(k+1)(X)
with del(c_(k+1))= c_k-c_k' .

Still: how about geometrically: is there a nice geometric way of telling that two
cycles are homologous.?. I am having trouble translating the subtraction of cycles
into a geometric situation; it would seem like we could translate the expression
of c_k-c_k' is a boundary into saying that the curves c_k and -c_k' (i.e., c_k with
reversed orientation) are cobordant, in that there is a surface embedded in X--
the ambient complex--that is bounded by c_k and -c_k' .

Is this correct.?

Thanks.

 

[quasar987]

That's pretty much how I visualize it too... You should take a look at the little book "elementary concepts of topology" by Alexandroff. It is great for building geometrical understanding and intuition of homology.

 

[Bacle]

Thanks, Quasar :

I wonder if it makes sense to talk of a representing surface as generating homology,
in the way we could say a full loop generates the homotopy group of the circle, i.e:

1 loop --->1
2 loops--->2
...
...
n loops -->n

Any chances you know of examples of representing surfaces.?

Thanks.

 

[quasar987]

Mmh, how about this interpretation: If you have a a surface/loop topologically embedded in a space X, then a triangulation of that surface/loop induces in an obvious way a 2-cycle/1-cycle. Now, granted that two triangulations induce homologous cycles (i.e. cycles that differ by a boundary), then it makes sense to talk about the surface/loop as representing a homology class. (Namely, the one containing any and all cycles induced by a triangulation of said surface/loop.)

 

[Bacle]

Hi, Quasar, let me try again:

I think I understood that you actually meant that _if_ any two triangulations
induce homology cycles.

But, if the surface is oriented, then it is the only 2-cycle in X, right.?

Sorry if this sounds too lost.

Thanks.

 

[quasar987]

I forgot an "and if" somewhere. Let me try that again by making bold the parts I added or modified:

If you have a a surface/loop topologically embedded in a space X, then a triangulation of that surface/loop* induces in an obvious way a 2-chain/1-chain. Now, granted that this chain is a cycle (I mentioned in the other post that this will be the case iff the loop/surface is orientable. In particular, it will always be the case for loops because those are just circles up to homeomorphism) and that the two cycles induced by two different triangulations are homologous (i.e. cycles that differ by a boundary) (this is an assumption, I don't know if it holds. Zhentil, wofsy?) then it makes sense to talk about the surface/loop as representing a homology class. (Namely, the one containing any and all cycles induced by a triangulation of said surface/loop.)

*Note that any topological manifold of dimension 1, 2 or 3 admits a triangulation (See Lee Introduction to Topological Manifolds p.102ff)

 

[lavinia]

Two triangulations of the same space give the same homology groups. So if one triangulation of an embedded surface determines a cycle, any other will as well and the two cycles will be homologous.

If the manifold is orientable then a fundamental theorem says that is has a cycle in the top dimension that generates its top homology. If it is not orientable, the same theorem says that it has no top dimensional cycles and has top homology equal to zero.

Every manifold is orientable over Z/2 and thus has a mod 2 fundamental cycle. Any triangulation will give the mod 2 fundamental class.

Interestingly, some non-orientable manifolds - e.g. the Klein bottle - form the boundary of a one higher dimensional manifold. Yet they are not cycles over Z.