The multiplication on $THH$ of finite fields

Let $k$ be a finite field, $THH(k)$ its topological Hochschild homology spectrum. For essentially formal reasons, we know that it's an $E_\infty$-algebra over the Eilenberg-Mac Lane spectrum $Hk$, and so can be modeled by an $E_\infty$-dga over $k$.

Question: Is this $E_\infty$-structure equivalent to a strictly commutative multiplication?

We know that the homotopy algebra $\pi_* THH(k)$ is a polynomial ring on a generator in degree $2$, so in particular it cannot be equivalent to a simplicial commutative $k$-algebra (for then it would support a divided-power structure relative to its ideal of positive-degree elements.) So in a sense a positive answer to the question above would say that $THH(k)$ is "as commutative as it could possibly be" given its homotopy algebra.

This is not the case, and you can use Dyer-Lashof operations to see so. In the following I'll show this for $k = \Bbb F_2$ because that's the easiest case to compute. Bokstedt proved that $THH_*(\Bbb F_2) = \Bbb F_2[\sigma]$ where $|\sigma| = 2$. If it came from a commutative DGA, then the only nonzero Dyer-Lashof operations would be the squaring operations (because the operad action would factor through the projection down to the bottom class). We'll show that $Q^6 \sigma = \sigma^4$.
The description of $THH(k)$ by the cyclic bar construction makes it the derived smash product $k \wedge_{k \wedge^{\Bbb L} k^{op}} k$. The associated Kunneth spectral sequence comes from a simplicial commutative $k$-algebra, which gives the associated spectral sequence compatible Dyer-Lashof operations. It sits inside the Bokstedt spectral sequence (this is how Bokstedt computed the hidden extensions to show $THH_*(\Bbb F_2) = \Bbb F_2[\sigma]$). In particular, this spectral sequence takes the form of a Tor-spectral sequence $$Tor_{A_*} (\Bbb F_2, \Bbb F_2) \Rightarrow THH_*(\Bbb F_2)$$ where $A_*$ is the dual Steenrod algebra $\Bbb F_2[\xi_i]$. The Tor-groups form an exterior algebra generated by $I/I^2$, where $I$ is the augmentation ideal of the dual Steenrod algebra. The exterior algebra is therefore generated by $[\xi_i]$.
However, Steinberger's calculations from the $H_\infty$ book tell us that $$Q^6 \xi_1 =\xi_1^7 + \xi_1^4 \xi_2 + \xi_1 \xi_2^2 + \xi_3,$$ and therefore in $THH(k)$ we have an identity $$Q^6 [\xi_1] = [\xi_3]$$ because everything else is decomposable. The spectral sequence degenerates and so $Q^6$ on the generator in degree two hits the only nonzero thing in degree 8.
• Thanks, Tyler, this is very useful. I'm a little confused by your statement that "The associated Kunneth spectral sequence comes from a simplicial commutative $k$-algebra," though. I believe that it comes from a simplicial $E_\infty$-algebra over $k \wedge^\mathbb{L} k^{op}$, but if I'm not mistaken that thing is not a simplicial commutative $k$-algebra. Perhaps I'm misunderstanding what you mean there? – Aaron Royer May 22 '15 at 20:08
• @Aaron Ah, that's because I meant an $Hk$-algebra, meaning over the Eilenberg-Mac Lane spectrum. Sorry. – Tyler Lawson May 22 '15 at 20:32
• It might be worth noting that "compatible Dyer-Lashof operations" is a phrase that should be interpreted with care. For example, you do not seem to mean that there are operations on the $E_2$-page that behave like the DL operations but rather that there is a structure on the filtration that gives rise to a theory of operations whose algebraic behavior may be unpredictable. – Sean Tilson May 25 '15 at 14:07