I've heard somewhere that a compact submanifold of a manifold gives a homology class of the whole manifold and vice versa. Can somebody explain this to me?
Compact submanifold represents a homology class
-
0I don't know a lot about this, but in one direction it seems that one should take the inclusion $i\colon N \to M$ and push forward the [fundamental class](http://en.wikipedia.org/wiki/Fundamental_class) $[N]$ of $N$. In the other direction, maybe you could use [Poincaré duality](http://en.wikipedia.org/wiki/Poincaré_duality)? This last bit I'm not so sure of. – 2011-11-12
1 Answers
For starters, you have the definition of a cycle wrong. A cycle is a fomal sum of maps $\Delta^n\rightarrow X$ whose boundary is 0. It represents something nontrivial in homology iff it's not a boundary.
Now, given a compact submanifold $M$, triangluate it. Then this gives a formal sum of maps $f:\Delta^n\rightarrow M$. The boundary of this will be $0$, basically because $M$ has no boudnary. Composing this with the inclusion map gives a formal sum of maps into the big manifold which also has boundary $0$, i.e., gives a cycle.
The converse, interestingly, is not true. That is, there are homology classes which are not representable by smoothly immersed compact submanifolds, though it is true that given any class, some multiple of it is representable by an immersed compact submanifold. These results (if I haven't misstated them) were proved by Thom, but I'm not sure where.
Edit Per Dylan's suggestion, here are links to a discussion about this on Math Over flow https://mathoverflow.net/questions/1489/cohomology-and-fundamental-classes and a copy of Thom's paper: http://www.springerlink.com/content/e83tm550v40p8850/ .
-
0@Qiaochu: Steve D. nailed it - I'm thinking of the smooth category. – 2011-11-13