2
$\begingroup$

In Pugh's Real Mathematical Analysis he writes:

The quotient vector space $$H^k(U) = Z^k(U)/B^k(U)$$ is called the $k$th de Rham cohomology group of $U$. Its members are the "cohomology classes" of $U$.

While of course there's a forgetful functor $\operatorname{Vect}_\mathbb{R}\to\operatorname{Grp}$, and we don't really need the scalars to do homology, it seems like a bit of a fraud to call something a group that we've constructed as a vector space, and whose most interesting property is its dimension.

In these notes on homology a $k$-chain is defined as a member of the free abelian group $Q_k$ of finite linear combinations of $k$-cubes. But it then talks about subspaces, rank, and bases as if $Q_k$ really was a vector space after all.

Why do we call these vector spaces groups?

$\endgroup$
2
  • 1
    $\begingroup$ The 'notes on homology' talk about singular homology, which is different from de Rham cohomology. The former is only an abelian group, and the latter is an $\mathbb R$-vector space. 'Rank' is the abelian group version of 'dimension', and free abelian groups still have bases. $\endgroup$
    – Remy
    Commented Mar 25, 2018 at 6:18
  • $\begingroup$ Ah, okay. Shows how much I know about free groups I guess. $\endgroup$ Commented Mar 25, 2018 at 6:25

1 Answer 1

6
$\begingroup$

It's just traditional. The first kinds of (co)homology objects to be studied were abelian groups (specifically, simplicial homology groups), and the name stuck in some other contexts even when there is more structure than just a group.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .