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Apr 13, 2010 at 14:17 comment added Paul @Don: It depends on what you mean by "self-intersection" and whether you allow disconnected submanifolds. eg if you take twice the first generator of H1(T), then 2x\cdot 2x=0 but the curve t\mapsto (e^{4\pi i t},1) has self-intersection 1. But 2 parallel copies of x is an embedded, disconnected manifold that represents 2x
Apr 13, 2010 at 13:46 comment added Tim Perutz About your last sentence: as Paul notes, classes of codim 1 or 2 are always representable. In the codim one case, to double the homology class, compose a map to $S^1$ with the double covering of $S^1$ by itself.
Apr 13, 2010 at 6:36 history answered Don Stanley CC BY-SA 2.5