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Jonathan Beardsley
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Let $E$ be a spectrum havehaving exactly two non-trivial homotopy groups, $\pi_k(E)=G$ and $\pi_j(E)=G'$ for $j>k\geq 0$, and having $k$-invariant $\alpha\colon\Sigma^{k}HG\to\Sigma^jHG'$. Also assume that $\Omega^\infty\alpha$ is null-homotopic.

Then for a space $X$ there is a group structure on $E^0(X)$ which, as a set, should be isomorphic to $H^k(X;G)\times H^j(X;G')$. This should correspond to homotopy classes of maps $X\to B^kG\times B^jG'$, and the group structure on this set should be the product group structure if the $k$-invariant $\alpha$ is trivial (i.e. the splitting $\Omega^\infty E\simeq B^kG\times B^jG'$ is one of infinite loop spaces).

We can model Eilenberg-MacLane spaces as topological Abelian groups, and I believe that there is going to be a group extension of $B^kG$ by $B^jG'$ determined by a 2-cocycle which is determined by the $k$-invariant $\alpha$ which will, in turn, determine the group structure of $E^0(X)$. Moreover, it seems like in nice cases, maybe when the $k$-invariant is easy to write down, one should basically be able to read off the cocycle formulaically and give the group structure on $E^0(X)$ as the pointwise multiplication on maps $X\to \Omega^\infty E$. Is this true, and if so, is it written down clearly anywhere? If it's not written down anywhere, what's an argument for it?

Let $E$ be a spectrum have exactly two non-trivial homotopy groups, $\pi_k(E)=G$ and $\pi_j(E)=G'$ for $j>k\geq 0$, and having $k$-invariant $\alpha\colon\Sigma^{k}HG\to\Sigma^jHG'$. Also assume that $\Omega^\infty\alpha$ is null-homotopic.

Then for a space $X$ there is a group structure on $E^0(X)$ which, as a set, should be isomorphic to $H^k(X;G)\times H^j(X;G')$. This should correspond to homotopy classes of maps $X\to B^kG\times B^jG'$, and the group structure on this set should be the product group structure if the $k$-invariant $\alpha$ is trivial (i.e. the splitting $\Omega^\infty E\simeq B^kG\times B^jG'$ is one of infinite loop spaces).

We can model Eilenberg-MacLane spaces as topological Abelian groups, and I believe that there is going to be a group extension of $B^kG$ by $B^jG'$ determined by a 2-cocycle which is determined by the $k$-invariant $\alpha$ which will, in turn, determine the group structure of $E^0(X)$. Moreover, it seems like in nice cases, maybe when the $k$-invariant is easy to write down, one should basically be able to read off the cocycle formulaically and give the group structure on $E^0(X)$ as the pointwise multiplication on maps $X\to \Omega^\infty E$. Is this true, and if so, is it written down clearly anywhere? If it's not written down anywhere, what's an argument for it?

Let $E$ be a spectrum having exactly two non-trivial homotopy groups, $\pi_k(E)=G$ and $\pi_j(E)=G'$ for $j>k\geq 0$, and having $k$-invariant $\alpha\colon\Sigma^{k}HG\to\Sigma^jHG'$. Also assume that $\Omega^\infty\alpha$ is null-homotopic.

Then for a space $X$ there is a group structure on $E^0(X)$ which, as a set, should be isomorphic to $H^k(X;G)\times H^j(X;G')$. This should correspond to homotopy classes of maps $X\to B^kG\times B^jG'$, and the group structure on this set should be the product group structure if the $k$-invariant $\alpha$ is trivial (i.e. the splitting $\Omega^\infty E\simeq B^kG\times B^jG'$ is one of infinite loop spaces).

We can model Eilenberg-MacLane spaces as topological Abelian groups, and I believe that there is going to be a group extension of $B^kG$ by $B^jG'$ determined by a 2-cocycle which is determined by the $k$-invariant $\alpha$ which will, in turn, determine the group structure of $E^0(X)$. Moreover, it seems like in nice cases, maybe when the $k$-invariant is easy to write down, one should basically be able to read off the cocycle formulaically and give the group structure on $E^0(X)$ as the pointwise multiplication on maps $X\to \Omega^\infty E$. Is this true, and if so, is it written down clearly anywhere? If it's not written down anywhere, what's an argument for it?

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Jonathan Beardsley
  • 10.5k
  • 1
  • 36
  • 85

Group structure on cohomology with coefficients in a spectral 2-type

Let $E$ be a spectrum have exactly two non-trivial homotopy groups, $\pi_k(E)=G$ and $\pi_j(E)=G'$ for $j>k\geq 0$, and having $k$-invariant $\alpha\colon\Sigma^{k}HG\to\Sigma^jHG'$. Also assume that $\Omega^\infty\alpha$ is null-homotopic.

Then for a space $X$ there is a group structure on $E^0(X)$ which, as a set, should be isomorphic to $H^k(X;G)\times H^j(X;G')$. This should correspond to homotopy classes of maps $X\to B^kG\times B^jG'$, and the group structure on this set should be the product group structure if the $k$-invariant $\alpha$ is trivial (i.e. the splitting $\Omega^\infty E\simeq B^kG\times B^jG'$ is one of infinite loop spaces).

We can model Eilenberg-MacLane spaces as topological Abelian groups, and I believe that there is going to be a group extension of $B^kG$ by $B^jG'$ determined by a 2-cocycle which is determined by the $k$-invariant $\alpha$ which will, in turn, determine the group structure of $E^0(X)$. Moreover, it seems like in nice cases, maybe when the $k$-invariant is easy to write down, one should basically be able to read off the cocycle formulaically and give the group structure on $E^0(X)$ as the pointwise multiplication on maps $X\to \Omega^\infty E$. Is this true, and if so, is it written down clearly anywhere? If it's not written down anywhere, what's an argument for it?