# Geometric description of (simplicial)homologous cycles

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
Homework Helper
Gold Member
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.

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
Homework Helper
Gold Member
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.)

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
Homework Helper
Gold Member
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