10
$\begingroup$

I'm interested in cohomology operations (in ordinary cohomology) $$H^i(-, G)\rightarrow H^{i+1}(-, H)\;,$$ that is, elements of $$H^{i+1}(K(G, i), H)\;.$$ I know that $K(G, 1)=BG$, so for $i=1$, those cohomology operations are in $H^2(BG, H)$, and therefore given by the Bocksteins of the corresponding central extensions of $G$ by $H$. Also, for $G=H=\mathbb{Z}_p$, the stable cohomology operations are given by the Steenrod algebra, and the only degree-1 elements are Bocksteins.

However, I don't know how it is for unstable cohomology operations, or groups other than $\mathbb{Z}_p$. I'm mostly interested in simple groups, such as finitely generated or $\mathbb{R}/\mathbb{Z}$.

Improve this question
$\endgroup$
1
  • 5
    $\begingroup$ We can make $K(G,i)$ by gluing to a point one $i$-cell for each generator of $G$, then one $i+1$-cell for each relation, and then additional cells to kill the remaining homotopy groups. The remaining homotopy groups are in degree $i+1$ and above, so we kill them by adding cells of dimension $i+2$ and above, which can't add any new cohomology classes in degree $i+1$. Then the cells we added in degree $i+1$ for the relations must give the Bocksteins. $\endgroup$
    – Will Sawin
    Commented Jan 28, 2022 at 20:46

1 Answer 1

Reset to default
16
$\begingroup$

Yes. For $i\ge1$ you can build $K(G,i)$ from the Moore space $M(G,i)$ by adding cells of dimension $\ge i+2$, so $H_i(K(G,i); Z) = G$ and $H_{i+1}(K(G,i); Z) = 0$. Hence $Ext(G, H) \cong H^{i+1}(K(G,i); H)$ by the UCT. The elements of $H^{i+1}(K(G,i); H)$ represent the cohomology operations $H^i(-;G) \to H^{i+1}(-;H)$, and the elements of $Ext(G,H)$ correspond to the Bockstein operations. The case $i=0$ may require special attention.

Improve this answer
$\endgroup$

You must log in to answer this question.

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