1
$\begingroup$

I plan to give a talk on Steenrod algebra in a student seminar. but there are some questions that I didn't find an answer to, and it seems to me that I'm missing something:

  1. if the algebra of cohomology operations $\mathcal{A}\left(E\right)$ is isomorphic to the cohomology algebra $E^*\left(E\right)$, why it isn't (graded) commutative?

  2. the right and left units (source & target homomorphism) on the dual $E_*\left(E\right)$ are induced by the morphisms of spectra $\mathbb{S}\wedge E\to E\wedge E$ and $E\wedge\mathbb{S}\to E\wedge E$. why they are usually distict? and why they coincide in the classical cases ($H\mathbb{F}_p$)?

and another related question:

how by "correcting" the non-commutativity of cup product (in spaces or chain complexes) the stability is automaticly "corrected"?

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ The multiplication on $E^{\ast}(E)$ is not the cup product, it's composition. $\endgroup$
    – Qiaochu Yuan
    Commented Aug 25, 2014 at 17:14

1 Answer 1

Reset to default
8
$\begingroup$

For part 2: The two "units" are very nearly always distinct. You should regard it as an accident that they coincide for $H\mathbb{F}_p$.

In fact, if $k$ is a field which is not a prime field, then the two units are distinct. In this case, $$ \pi_0 (Hk\wedge Hk) \approx k\otimes_{\mathbb{Z}} k, $$ and in homotopy the units are the two evident maps of $k$ into the factors of the tensor product. If $k$ is not $\mathbb{F}_p$ or $\mathbb{Q}$, then $k\otimes k\not\approx k$, so the units are different.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ I think it is worth while to note that even in the case of $E=H\mathbb{F}_p$, the two maps of spectra $E\rightarrow E\wedge E$ are different, although they induce the same map in homotopy. $\endgroup$
    – user43326
    Commented Aug 26, 2014 at 15:02

You must log in to answer this question.

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