Let $A$ be a homotopy ring spectrum. Then the homology theory $A_\ast : Spectra \to GrAb$ lifts to a homology theory valued in $GrMod(\pi_\ast A)$. If $A$ is homotopy commutative, then this functor $A_\ast : Spectra \to GrMod(\pi_\ast A)$ is lax monoidal. But it makes sense to ask for a lax monoidal structure on this functor even if $A$ is not homotopy commutative, so long as $\pi_\ast A$ is graded-commutative. And the answer can be yes --e.g. at the prime 2, Morava $K$-theory $K(n)$ is not homotopy commutative, but $K(n)_\ast$ is strong monoidal.

Question: Let $A$ be a homotopy ring spectrum such that $\pi_\ast A$ is graded-commutative. Then is the functor $A_\ast : Spectra \to GrMod_{\pi_\ast A}$ lax monoidal? Even lax symmetric monoidal?

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    $\begingroup$ I believe the answer is no for Morava K-theory at the prine 2 $\endgroup$ Dec 1, 2022 at 21:16
  • $\begingroup$ @MaximeRamzi wait really? $\endgroup$
    – Tim Campion
    Dec 1, 2022 at 22:28
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    $\begingroup$ See the discussion on page 15 of arxiv.org/pdf/2008.00330.pdf $\endgroup$
    – Drew Heard
    Dec 2, 2022 at 7:48
  • $\begingroup$ @DrewHeard : do you know if there's any chance of a "general"(-ish) statement of this form ? Like "if A is a homotopy ring spectrum such that blah, then $A_*$ can be made symmetric monoidal with values in some category of graded modules over some Hopf algebra ..." ? Anything that's not "$A$ is Morava $K$-theory" would be neat :D $\endgroup$ Dec 2, 2022 at 9:12
  • $\begingroup$ Like maybe some hypotheses that allow to control the failure of (homotopy) commutativity of $A$ in the homotopy groups $\endgroup$ Dec 2, 2022 at 9:20

1 Answer 1


As pointed out in the comments, the functor $A_*$ cannot in general be lax symmetric monoidal without making some alterations.

Here is an incomplete discussion of when $A_*$ can be lax monoidal.

The first observation is that, for any homotopy associative ring spectrum $A$, the functor $A_*$ naturally takes values in $\pi_* A$-bimodules. The left and right actions of $A$ on itself produce natural left and right actions of $A$ on $A \otimes X$ for any $X$, and lax monoidality of $\pi_*$ show that $A_* X$ is then a $\pi_* A$-bimodule.

The second observation is that this makes the functor $A_*$ a lax monoidal functor to the category of $\pi_* A$-bimodules. The homotopy associativity of $A$ ensures that the two composites $A \otimes A \otimes A \to A \otimes A \to A$ are homotopic, and so for any $X$ and $Y$ the two composites $$ (A \otimes X) \otimes A \otimes (A \otimes Y) \to (A \otimes X) \otimes (A \otimes Y) \to A \otimes (X \otimes Y) $$ are homotopic. The lax monoidality of $\pi_*$ then tells us that the two composites $$ A_* X \otimes \pi_* A \otimes A_* Y \to A_*(X \otimes Y) $$ are equal, establishing $\pi_* A$-bilinearity. However, tracking which maps actually appear on the spectrum level, this specifically uses the right $\pi_* A$-module structure on $A_* X$ and the left $\pi_* A$-module structure on $A_* Y$.

We now assume that the coefficient ring $\pi_* A$ is graded-commutative. We would like to show that, for any $X$, the left and right module structures "coincide": $a \cdot x = \pm x \cdot a$ for any $a \in \pi_* A$ and $x \in A_* X$ according to the Koszul sign rule. If we can do this, then bimodule bilinearity collapses to module bilinearity.

Suppose $x \in A_d X$ comes from a map $S^d \to A \otimes X$. We can express $A$ as a filtered hocolim of finite spectra $A_i$, and get a lift to a map $S^d \to A_i \otimes X$, with an adjoint map $DA_i \otimes S^d \to X$ using Spanier-Whitehead duality. This gives us a lift of $x$ to a factorization $$ S^d \to \xrightarrow{\eta} A \otimes (DA_i \otimes S^d) \to A \otimes X. $$ Therefore, it suffices to check this in the "universal" cases where $x \in A_* (DA_i)$ comes from the canonical map $\eta: S^0 \to A \otimes DA_i$.

This lifts to the limit over $i$, however: the canonical unit map $S^0 \to map(A,A)$ to the function spectrum. In these terms, we are asking if the canonical element $1 \in [A,A]$ is sent to the same element under post-multiplication by $a$ on either the left or the right; or equivalently, if "multiply by $a$ on the left" and "multiply by $a$ on the right" are always homotopic maps $A \to A$, rather than merely giving equal maps $\pi_* A \to \pi_* A$.

An example of this can be constructed from the following differential graded algebra. Start with the discrete ring $\Bbb Z[x]$. We first form a bimodule with a single generator $y$, annihilated by $x$ on the left and right; we form a second bimodule with a single generator $z$, annihilated by $x^2$ on the left and $x$ on the right (so it has a basis $\{z,xz\}$). Form the mapping cone $M$ of the map $z \mapsto 2y$: it has $H_0 = \Bbb Z/2$, generated by $y$, and $H_1 = \Bbb Z$, generated by $xz$. Let $A$ be the square-zero extension $\Bbb Z[x] \times M$ by this differential graded bimodule $M$. The coefficient ring is graded-commutative because almost all products are zero. However, in mod-$2$ homology we see the element $z$ which no longer commutes with the generator $x$ from $A_*$.

  • $\begingroup$ Ok thanks! At least we see that the condition "$\pi_\ast(A)$ is graded-commutative" is not sufficient to ensure that $A_\ast$ is lax monoidal, even non-symmetrically. Which heightens the mystery for me -- is $K(n)_\ast$ lax monoidal when $p = 2$, and if so why?... $\endgroup$
    – Tim Campion
    Dec 3, 2022 at 0:59
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    $\begingroup$ @TimCampion For K(n) you are fine. Another way to express this is that we are checking the composite $$A_* \otimes A_*^{op} \to \pi_*(A \otimes A^{op}) \to \pi_* end(A).$$ $\endgroup$ Dec 3, 2022 at 2:26
  • $\begingroup$ In the case of Morava K-theory, we have only one generator to check ($v_n$) and its two images in $\pi_*(K(n) \otimes K(n)^{op})$ are equal as a consequence of the formulas for the right unit for $BP_* BP$. (So, I believe, for genuinely interesting reasons!) $\endgroup$ Dec 3, 2022 at 2:30

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