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Kunneth -> Künneth, and other tidying

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edited Dec 8, 2022 at 18:15

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If $A_\ast$ has a KunnethKünneth theorem, then is $A$ a module over Morava $K$-theory?

Let$\newcommand\Spt{\mathit{Spt}}\newcommand\GrAb{\mathit{GrAb}}$Let $A$ be a ring spectrum. Suppose that $A$ has a KunnethKünneth theorem --— i.e. the homology theory $A_\ast : Spt \to GrAb$$A_\ast : \Spt \to \GrAb$ is a strong monoidal functor [1].

Question: Does it follow that $A$ is a module over Morava $K$-theory $K(h)$ for some prime $p$ and some $0 \leq h \leq \infty$?

An affirmative answer would be a variation on the theorem that every ring spectrum which is a field is a module over some $K(h)$. See e.g. Lurie's notes Lurie's notesUniqueness of Morava $K$-theory.

[1] I have learned here here that it's not so straightforward to say exactly what it means to "have a KunnethKünneth theorem" --— right now I'm just assuming that there is some way to make $A_\ast$ into a strong monoidal functor, but it seems in general there need not a canonical way to make $A_\ast$ into a lax monoidal functor, unless $A$ is homotopy commutative. I think for a start, I'd be happy with an answer which assumes that $A$ is homotopy commutative, and assumes that the strong monoidal structure comes from the lax monoidal structure which exists canonically in this case.

If $A_\ast$ has a Kunneth theorem, then is $A$ a module over Morava $K$-theory?

Let $A$ be a ring spectrum. Suppose that $A$ has a Kunneth theorem -- i.e. the homology theory $A_\ast : Spt \to GrAb$ is a strong monoidal functor [1].

Question: Does it follow that $A$ is a module over Morava $K$-theory $K(h)$ for some prime $p$ and some $0 \leq h \leq \infty$?

An affirmative answer would be a variation on the theorem that every ring spectrum which is a field is a module over some $K(h)$. See e.g. Lurie's notes.

[1] I have learned here that it's not so straightforward to say exactly what it means to "have a Kunneth theorem" -- right now I'm just assuming that there is some way to make $A_\ast$ into a strong monoidal functor, but it seems in general there need not a canonical way to make $A_\ast$ into a lax monoidal functor, unless $A$ is homotopy commutative. I think for a start, I'd be happy with an answer which assumes that $A$ is homotopy commutative, and assumes that the strong monoidal structure comes from the lax monoidal structure which exists canonically in this case.

If $A_\ast$ has a Künneth theorem, then is $A$ a module over Morava $K$-theory?

$\newcommand\Spt{\mathit{Spt}}\newcommand\GrAb{\mathit{GrAb}}$Let $A$ be a ring spectrum. Suppose that $A$ has a Künneth theorem — i.e. the homology theory $A_\ast : \Spt \to \GrAb$ is a strong monoidal functor [1].

Question: Does it follow that $A$ is a module over Morava $K$-theory $K(h)$ for some prime $p$ and some $0 \leq h \leq \infty$?

An affirmative answer would be a variation on the theorem that every ring spectrum which is a field is a module over some $K(h)$. See e.g. Lurie's notes Uniqueness of Morava $K$-theory.

[1] I have learned here that it's not so straightforward to say exactly what it means to "have a Künneth theorem" — right now I'm just assuming that there is some way to make $A_\ast$ into a strong monoidal functor, but it seems in general there need not a canonical way to make $A_\ast$ into a lax monoidal functor, unless $A$ is homotopy commutative. I think for a start, I'd be happy with an answer which assumes that $A$ is homotopy commutative, and assumes that the strong monoidal structure comes from the lax monoidal structure which exists canonically in this case.

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asked Dec 7, 2022 at 16:54

Tim Campion

If $A_\ast$ has a Kunneth theorem, then is $A$ a module over Morava $K$-theory?

Let $A$ be a ring spectrum. Suppose that $A$ has a Kunneth theorem -- i.e. the homology theory $A_\ast : Spt \to GrAb$ is a strong monoidal functor [1].

Question: Does it follow that $A$ is a module over Morava $K$-theory $K(h)$ for some prime $p$ and some $0 \leq h \leq \infty$?

An affirmative answer would be a variation on the theorem that every ring spectrum which is a field is a module over some $K(h)$. See e.g. Lurie's notes.