Hi, everyone:(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Geometric description of (simplicial)homologous cycles

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