Timeline for Which homotopy 2-types are H-spaces?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 23, 2022 at 11:24 | vote | accept | Blazej | ||
| Jan 8, 2022 at 23:58 | comment | added | Fernando Muro | A non-zero $\beta$ is the non-trivial element of $H^3(K(\mathbb{Z}/2,1), \mathbb{Z}/4)\cong \mathbb{Z}/2$. The corresponding 2-type is the loop space of the Postnikov piece with $k$-invariant the generator of $H^4(K(\mathbb{Z}/2,2), \mathbb{Z}/4)\cong \mathbb{Z}/4$. | |
| Jan 7, 2022 at 19:16 | comment | added | Tyler Lawson | @mme If the groups involved (even just G) are finitely generated, yes, absolutely; a direct sum decomposition gives you a matrix decomposition of the primitives, and you can reduce to the case where G is cyclic. I'm worried (perhaps needlessly?) about eg the possibility of lim^i terms if G is not finitely generated. | |
| Jan 7, 2022 at 19:08 | comment | added | mme | @Chris I'm just making a linear algebra claim: if $\bigoplus_i G_i \to \bigoplus_j A_j$ is linear, then it can be computed as the action of a matrix of homomorphisms $G_i \to A_j$, and I think this is just the claim that finite direct sums are biproducts. If all of these terms are zero, then so is their direct sum. So additivity is crucial. (I have now used the terms "additive", "group homomorphism", and "linear" synonymously, which might add to confusion.) | |
| Jan 7, 2022 at 18:51 | comment | added | Chris Schommer-Pries | @mme "If beta is nonzero [then] one of these is non-zero". Is that correct? I think that is right for the A's, but for the G's aren't there "cross terms"? For example if A=R is a ring and G = R x R xR, there is an operation corresponding to 3-fold cup product. Its restriction to any one factor of R of G will be zero. However that beta is not additive, so maybe this is okay for the additive betas? | |
| Jan 7, 2022 at 17:48 | comment | added | mme | I think there are none. Split A and G into cyclic summands. If beta is a natural group homomorphism, so are the operations obtained by including summands into G and projecting to summands in A. If beta is nonzero one of these is nonzero. So we can reduce to the case that G and A are both cyclic of prime or infinite order, where one can explicitly compute all cohomology operations and see that the only additive one is zero. Have I made an error here? | |
| Jan 7, 2022 at 14:51 | comment | added | Tyler Lawson | Using a free resolution $0 \to R \to F \to G \to 0$ to get a fiber sequence $K(F,1) \to K(G,1) \to K(R,2)$ we can use the Serre spectral sequence. There are two potential nonzero contributions to $H^3$, but the first is from $H^3(K(F,1);A)$ and we already said that has no primitives. That reduces to checking if there are any primitive elements in the cokernel of $$ H^2(K(F,1); A) \to H^2(K(R,2); H^1(K(F,1); A)) $$ which is isomorphic to a map $$ Hom(\Lambda^2 F, A) \to Hom(R \otimes F, A). $$ This is about where I ran out of steam... | |
| Jan 7, 2022 at 14:38 | comment | added | Tyler Lawson | @ChrisSchommer-Pries That I don't know. As you noted above, the most natural source of additive cohomology operations does not give any. If $G$ is free then there are also no such cohomology operations. | |
| Jan 6, 2022 at 22:16 | comment | added | Chris Schommer-Pries | Are there non-zero beta which satisfy this condition? | |
| Jan 6, 2022 at 19:30 | history | answered | Tyler Lawson | CC BY-SA 4.0 |