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In Geometric Topology - Localization, Periodicity, and Galois Symmetry by Dennis Sullivan, we can read that there is a decomposition $$B\mathrm{SL}(\mathbb S_{(p)})\times K((\mathbf Z_{(p)})^\times)\xrightarrow{\simeq}B\mathrm{GL}(\mathbb S_{(p)})$$ induced by fiberwise join of associated bundles. I can guess that this decomposition is not $E_\infty$, for example for $p=2$, the first Stieffel Whitney class coming from $BO$ does not vanish under the application of Dyer Lashof operations.

We can find useful information about the action of the Dyer Lashof algebra by using the fact that $B\mathrm{SL}(\mathbb S_{(p)})\to B\mathrm{GL}(\mathbb S_{(p)})$ is an $E_\infty$-map and the (highy nontrivial) fact that $B\mathrm{SL}(\mathbb S_{(p)})$ is the $p$-localisation of $B\mathrm{SL}(\mathbb S)=BSF$, the homology of which is very well-known.

Is there any reference on the Dyer Lashof action on $H_1(B\mathrm{GL}(\mathbb S_{(p)}))=(\mathbf Z_{(p)})^\times$ ? Note that I am actually mostly interested in the case of $p$-completed spherical bundles, where all of the above discussion works but with $(p)$ replaced by $p$.

Probably, the case $p=2$ might be a bit different from the others... I could not work this calculation out myself.

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  • $\begingroup$ Do you mean $E_∞$-space structure? Since $BGL(\mathbb{S}_{(p)})$ is connected I don't see how it can be an $E_∞$-ring space.. $\endgroup$ Nov 18, 2020 at 16:02
  • $\begingroup$ Indeed, I modified the title thanks $\endgroup$
    – elidiot
    Nov 18, 2020 at 16:30

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$\newcommand{\QQ}{\widetilde{Q}}$ We can do the following computation at p=2 to figure out Dyer-Lashof operations in $GL_1(\Bbb S_{(2)})$. When we pass to the classifying space, decomposables under the circle product $\circ$ become zero; I don't know this decomposition well enough to do it here.

I'm going to write $\cdot$ / multiplication for the additive Pontrjagin product, $Q^n$ for the additive Dyer-Lashof operations, and $\QQ^n$ for the multiplicative ones, within the homology of the infinite loop space $\Omega^\infty \Bbb S_{(2)}$. I do not remember the odd-primary ones well enough. By restricting to unit components, this will give us the operations on $GL_1(\Bbb S_{(2)})$.

I'm also going to write $Q$ for $\sum Q^n$ and $\QQ$ for $\sum \QQ^n$ as formal series. The Cartan formula says $Q(x y) = Q(x) Q(y)$ and the mixed Cartan formula says, at least for representatives of elements in $\pi_0$ in degree zero, that $\QQ([a+b]) = \QQ[a] Q[ab] \QQ[b]$.

Because $[0]$ is the additive unit, $Q[0] = \QQ[0] = [0]$; because $[1]$ is the multiplicative unit, $\QQ[1] = [1]$.

The mixed Cartan formula then says $$\QQ[n+1] = \QQ[n] Q[n] \QQ[1] = \QQ[n] Q[1]^{n} [1].$$ Applying this inductively, we find that $\QQ[n] = [n]Q[1]^{\binom{n}{2}}$ for natural numbers $n$ (and in reverse, for all integers). Moreover, I believe that the standard relations can be used to show that this formula respects multiplication, and so it extends to be a valid formula for $\QQ[n]$ for all $n \in \Bbb Z_{(2)}$.

You can use this, for example, to show that $\QQ^1[-1]$ is essentially the same as $Q^1[1]$ in the classifying space, but $\QQ^2[-1]$ vanishes because it basically becomes decomposable under the multiplicative Pontrjagin product.

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  • $\begingroup$ Thank you for your answer. I will need some time to figure out if this enables me to make my calculation, and then I will accept the answer :) $\endgroup$
    – elidiot
    Nov 24, 2020 at 14:36
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    $\begingroup$ @elidiot You shouldn't feel obligated to accept the answer. I hope somebody else might be able to properly address what the indecomposables look like under the multiplicative Pontrjagin product. $\endgroup$ Nov 24, 2020 at 15:21

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