# Kunneth Theorem and localisation

I have two, somehow related questions, and I would be very grateful if you point out at some references if the answer is known. If these are too elementary or not research level, then please feel free to move it mathexchange. I am not an expert in this, so I thought it is worth asking here.

According to Ravenel's orange book, the Nilpotence Theorem implies that $H\mathbb{F}$ and $K(n)$ are essentially'' the only homology theories for which the Kunneth isomorphism holds; I suppose this latter means that $$E_*(X\times Y)\simeq E_*(X)\otimes_{E_*(pt)}E_*(Y).$$ So, my first question is whether this conclusion is immediate or whether it is not that trivial and a proof is recorded somewhere? I have not tried to prove it.

Second, I wonder if there is a spectrum $E$ so that $$\pi_*L_E(S^0\wedge S^0)\simeq\pi_*L_ES^0\otimes\pi_*L_ES^0$$ where the tensor product is over some suitable coefficient ring, $L_E$ is the Bousfield localisation with respect to $E$, and $S^0$ is the sphere spectrum. Or, equivalently on the category of spaces $$\pi^s_*L_E(X\times Y)\simeq\pi^s_*L_EX\otimes\pi^s_*L_EY$$ again with tensor product over say $\pi_*L_ES^0$. A first guess is $E=H\mathbb{F}$ and $K(n)$, but these do not seem that immediate to me either.

Edit. Perhaps it is better to replace $\times$ with $\wedge$, and ask for an isomorphism $$\pi^s_*L_E(X\wedge Y)\simeq\pi^s_*L_EX\otimes_{\pi_*L_ES^0}\pi^s_*L_EY$$ as also in Neil's update to his answer.

The motivating example for the second question is the following. Consider a compact Lie group and an embedding $1^{\times n}<G^{\times n}$. Then, the transfer map of Becker-Schultz-Mann-Miller-Miller provides a transfer map (upon choosing a suitable twisting bundle) $$\Sigma^{n\dim(\mathfrak{g})}BG^{\wedge n}_+\to (B1^{\times n})_+=S^0.$$ Now, one may hope that in homotopy the image of $$\pi_*(\Sigma^{n\dim(\mathfrak{g})}BG^{\wedge n}_+)\to\pi_* B1^{\times n}_+=\pi_*S^0=\pi_*^s$$ falls into the submodule of decomposable elements in $\pi_*^s$. Here, $\mathfrak{g}$ is the Lie algebra of $G$. This, however, is not true, which I think is mainly because of the lack of a Kunneth theorem. So, one hope is to find some $E$ where homotopy of $E$-localised spaces/spectra admits some Kunneth theorem. For this reason, I am interested in a Kunneth theorem and not a Kunneth spectral sequence.

Equivalently, if there is a way to describe the submodule of decomposable elements in $\pi_*^s$ then I would be happy.

• The statement you want regarding homotopy of localizations is probably not going to be true. Recall that $S^0\wedge S^0=S^0$. I guess then you would replace this by a pair $X$ and $Y$, this is also unlikely to be true. I think what you should look at is the Kunneth spectral seqeuence which exists in general. – Sean Tilson May 5 '17 at 11:04
• @SeanTilson I know that such a spectral sequence exists, but I am interested in the Kunneth theorem. I have edited the question, and put more explanation on the motivation. – user51223 May 5 '17 at 11:34
• Right, so what you are asking for is conditions under which it might collapse, and these are homological in nature, because of the spectral sequence. This is why I suggest looking at the KSS. I don't know how the E-local KSS works, but that might address your other questions. In general, I am not sure how you would go about proving the Kunneth theorem without using the KSS. It seems like the easiest way to do such a thing. – Sean Tilson May 5 '17 at 11:58
• And maybe you want to replace the cartesian product with a smash product, otherwise you would tensoor over something else since $S^0$ is not the monoidal unit. – Sean Tilson May 5 '17 at 11:59

The first statement is not true as stated. Firstly, if $E$ and $F$ have Künneth theorems then so does $E\times F$. (Here $E\times F$ is the same as $E\vee F$, but calling it $E\times F$ makes the ring structure more visible.) Secondly, suppose that $E$ has a Künneth theorem and $A_*$ is an algebra over $E_*$ that is flat as an $E_*$-module. Then the functor $X\mapsto A_*\otimes_{E_*}E_*(X)$ is a multiplicative homology theory that also has a Künneth theorem. By applying these constructions to the $K(n)$'s (including $K(0)=H\mathbb{Q}$ and $K(\infty)=H\mathbb{F}_p$) we obtain many examples. It would not surprise me if this gives all examples, but I do not think that this has been written down. One might need to worry about infinite products.

There is a related statement that is certainly true. Let $F$ be a ring spectrum such that $F_*$ is a graded field (ie every nonzero homogeneous element is invertible, and $F_*\neq 0$). This implies that every graded $F_*$-module is free, and thus that $F$ has a Künneth theorem. It also implies that $F\wedge X$ is always a wedge of (possibly suspended) copies of $F$. This includes the possibility that $F\wedge X$ could be zero, of course. This means that $F\wedge K(n)$ is a wedge of suspensions of $F$ and also a wedge of suspensions of $K(n)$. It follows from the Nilpotence Theorem that there exists a prime $p$ and a number $n\in [0,\infty]$ with $F\wedge K(n)\neq 0$. One can deduce from this that $F$ itself is a wedge of suspensions of $K(n)$. This still does not force $F$ to be isomorphic to $K(n)$, however; it could be $K(n)$ tensored with a finite field, or with a root of $v_n$ adjoined.

UPDATE: for the second question, I do not believe that the natural map $$\pi_*(L_EX)\otimes_{\pi_*(L_ES)}\pi_*(L_EY) \to \pi_*L_E(X\wedge Y)$$ is a natural isomorphism unless $E$ is zero (in which case $L_E=0$) or $E=E\mathbb{Q}\neq 0$ (in which case $L_E$ is rationalization). However, I cannot see a proof of this at the moment.

• I corrected the first statement. The statement in Ravenel's book, on page 8 of the Orange Book, is that $H\mathbb{F}$ and $K(n)$ are ''essentially'' the only homology theories for which Kunneth theorem holds. I suppose the word essentially'' therein then needs more explanation, and your arguments almost provide a proof. – user51223 May 5 '17 at 11:26
• Would perhaps the natural map fail to be monic in general, which then will prevent the map from being isomorphism?! – user51223 May 5 '17 at 18:12
• It seemed to me that it might be possible to prove non-existence of such a spectrum if we use the destabilisation $\Omega^\infty$ and if we knew a couple of things. We know that (1) At the prime $2$, use $\theta_i\in{_2\pi_{2^{i+1}-2}^s}$ elements which we know are not decomposable for $i>3$, (2) The functors $\Omega^\infty$ and $L_E$ commute, (3) $\theta_i$ maps nontrivially under $h:{_2\pi_*^s}\to H_*Q_0S^0$ with $Q=\Omega^\infty\Sigma^\infty$, (4) the image of $h$ when restricted to decomposable elements is given by $h(\theta_i)$ for $i=1,2,3$. – user51223 May 7 '17 at 7:01
• So, if we knew that $h(\theta_i)$ maps nontrivially under $H_*QS^0\to H_*L_EQS^0$ then it seems to me that we get a contradiction to the existence of $E$. But, I am not sure how much we can say about this. – user51223 May 7 '17 at 7:03

Jeff Strom correctly pointed to Hopkins-Smith II for the Kunneth Theorem question: the correct statement is that any E with a perfect Kunneth Theorem is (additively) a wedge of copies of $K(n)$ for some n, where $K(0) = H\mathbb Q$ and $K(\infty) = H\mathbb F_p$. There is also a discussion of this in Hopkin's Durham 1985 proceeding paper, which also has, if I remember right, an influential appendix in which Mike is relating the closely related Thick Subcategory Theorem to classical ring theoretic settings. He was wondering if the `identifying the prime fields theorem" as above was equivalent to the Nilpotence Theorem, and not just a consequence.

Regarding smashing the $n$--fold transfer: if $G$ contains the subgroup $\mathbb Z/p$, then this map has Adams filtration at least $n$, so the image in homotopy will be elements of Adams filtration at least $n$. This is a decomposablity statement of sorts since a map has Adams filtration $n$ if it factors as $n$ maps each of which is zero in mod p homology.

• For the second part, I am interested in a decomposition in terms of elements of $\pi_*^s$. I would be happy with a factorisation implied by the Adams filtration, if I could get some hand on the connectivity of the spectra that appear in the middle. – user51223 May 6 '17 at 7:20
• Getting any sort of interesting control over such connectivity, in general, would be a major major result in homotopy theory. Don't hold your breath on that one. – Nicholas Kuhn May 6 '17 at 11:51
• Well, for that such a connectivity result is probably non-existing at the moment, then I was interested in the decomposition that I have asked for. – user51223 May 6 '17 at 17:57
• For a while, I have been wondering that whether or not a factorisation, with suitable connectivity properties, could be obtained using Todabracket like constructions, say something like higher brackets defined by Joel Cohen and the algorithm provided by him, despite the indeterminacy problem that we may face!? – user51223 May 7 '17 at 8:33

For the first statement, I think you want Prop. 1.8 in "Nilpotence and Periodicity II" by Hopkins and Smith.