- #1
Bacle
- 662
- 1
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.
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.