6
$\begingroup$

For a topological group G, we have a Pontryagin product in homology by multiplying representative cycles. This gives the homology the structure of an associative graded algebra. Am I correct in thinking we can prove this by seeing a topological group as an algebra over the Ass-operad and applying homology everywhere?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

You can do that. If an operad O acts on a space X, then the structure maps O(n) x Xn -> X induce homology operations H*O(n) ⊗ H*(X)⊗n -> H*(X). In particular, any path component in O(2) produces a multiplication on H*X, if it's in the same path component as its own image under the symmetric group action it's commutative, if the two composites of it are in the same path component of O(3) it's associative, et cetera.

In particular if O is the associative operad (so O(n) are discrete) then this structure reduces to the Pontrjagin ring structure.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.