Motivation: According to the "chromatic" picture of stable homotopy, we should think of the moduli stack $M_{FG}$ of formal groups as a "good approximation" to the stable homotopy category (more precisely, we should think of the category of quasi-coherent sheaves on the stack $M_{FG}$ as the "good approximation"). Every finite complex $X$ defines a module $MU_* X$ over the complex bordism ring $MU_*$, which is the Lazard ring classifying a universal formal group law; this is also a comodule over $MU_\ast MU$, which corresponds to strict automorphisms of a formal group law. Taking account of the grading lets one say "formal group" instead of "formal group law."

Let's say for instance that we have a two cell complex, with cells far away in dimensions from one another. Then this approach means that we get a two-dimensional vector bundle on $M_{FG}$. I'm wondering if we can understand these algebro-geometrically. For instance, line bundles on $M_{FG}$ can be understood (the Picard group is $\mathbb{Z}$, generated by the Lie algebra $\omega$ of a formal group), and perhaps 2-dimensional vector bundles are not too far off.

Here's the specific situation I have in mind. Let $\nu: S^3 \to S^0$ be the second (stable) Hopf map, which generates the 3-stem. Then the cofiber of $\nu$ is the desuspension $\Sigma^{-4} \mathbb{HP}^2$, and this (as an even, two cell complex) has free $MU$-homology. The homology of this corresponds to some vector bundle on $M_{FG}$, which is an extension of $\mathcal{O}$ and $\omega^4$. Alternatively, $\nu$ is detected in the 1-line of the ANSS as a class in $\mathrm{Ext}^1_{M_{FG}}(\omega^4, \mathcal{O})$, which has order twelve, I think. Is there a description of this class purely in terms of formal groups? I'd be interested as well in the pull-back of this bundle to $M_{1,1}$ (i.e., under the map $M_{1,1} \to M_{FG}$ sending an elliptic curve to its formal group.) More concretely, this describes cooperations in elliptic homology on $\mathbb{HP}^2$.

(As a simpler example, we can do this for $\eta$, the first Hopf map. Then we are looking at $\Sigma^{-2} \mathbb{CP}^2$, and the relevant bundle on $M_{FG}$ is the next order version of the Lie algebra. Given an even-periodic homology theory $E$, then $\widetilde{E}_0(\mathbb{CP}^2)$ is dual to $\widetilde{E}^0(\mathbb{CP}^2)$, which corresponds to functions on the formal group of $E$ mod functions which vanish to degree 3 or higher at the origin. $\widetilde{E}_0(\mathbb{CP}^2)$ is dual to this.)

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    $\begingroup$ Stably, and away from the prime 2, one can write $\mathbb{HP}^\infty$ as the Borel construction $E\mathbb{Z}/2 \times_{\mathbb{Z}/2} \mathbb{CP}^\infty$ for the action of complex conjugation on $\mathbb{CP}^\infty$. So perhaps you can do some sort of Galois-equivariant version of your analysis above for $\mathbb{CP}^2$? $\endgroup$ – Craig Westerland Jul 15 '12 at 16:21
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    $\begingroup$ Of course, the prime 2 is pretty important when talking about $\nu$... $\endgroup$ – Craig Westerland Jul 15 '12 at 16:23
  • $\begingroup$ The first sentence of the second paragraph contains an error, I think you mean two cell complex. The dimension of a cell complex is usually defined to be the dimension of the top cell, if it is finite. $\endgroup$ – Sean Tilson Jul 15 '12 at 16:53
  • $\begingroup$ @Sean: Thanks. Yes, that's what I meant to say. $\endgroup$ – Akhil Mathew Jul 15 '12 at 20:05
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    $\begingroup$ Away from $2 = | \mathbb{Z} / 2|$, the LHSSS for the fibration $E\mathbb{Z}/2 \times_{\mathbb{Z}/2} \mathbb{CP}^\infty \to B\mathbb{Z} / 2$, given by $H^*(\mathbb{Z} / 2, H^*(\mathbb{CP}^\infty))$ collapses at the $E_2$-term, and contains only 0-dimensional group cohomology (i.e., invariants). The generator of $\mathbb{Z} / 2$ acts on a a generator $x \in H^2(\mathbb{CP}^\infty)$ by multiplying by $-1$, and on the rest through ring homomorphisms. So the invariant subring is that which is generated by $x^2$, which happens to be the image of $H^*(\mathbb{HP}^\infty)$ under the natural map. $\endgroup$ – Craig Westerland Jul 16 '12 at 2:45

The answer seems to be yes: given an even-periodic cohomology theory $E$, then $E^*(\mathbb{HP}^\infty)$ can be described as even functions on the formal group of $E$ (just as $E^*(\mathbb{CP}^\infty)$ corresponds to functions on the formal group of $E$).

In fact, there is a natural map $\mathbb{CP}^\infty \to \mathbb{HP}^\infty$ (sending a complex line to the associated quaternionic line in some $\mathbb{H}^\infty$, or as the map $BU(1) \to B( \mathrm{Sp}(1))$ from the inclusion $U(1) \to \mathrm{Sp}(1)$) and the pull-back on ordinary cohomology has the following property: if $(-1): \mathbb{CP}^\infty \to \mathbb{CP}^\infty$ is the natural involution, then the cohomology of $\mathbb{HP}^\infty$ is identified with the invariant elements of $\mathbb{CP}^\infty$. The same holds for an even-periodic cohomology theory with $E^0(\ast)$-torsion-free at least, by the following logic. The map $$\mathbb{CP}^\infty \stackrel{(-1)}{\to} \mathbb{CP}^\infty \to \mathbb{HP}^\infty$$ is homotopic to the plain inclusion (the inclusion $U(1) \to \mathrm{Sp}(1)$ and $U(1) \stackrel{z \mapsto \overline{z}} U(1) \to \mathrm{Sp}(1)$ are homotopic as group-homomorphisms---in fact, they are conjugate by $j$). So $E^\ast(\mathbb{HP}^\infty) \to E^*(\mathbb{CP}^\infty)$ is a monomorphism into the subset of even functions on the formal group, and counting dimensions shows that it is an isomorphism (note that the AHSS degenerates).

So, it follows that $E^\ast(\mathbb{HP}^2)$ can be identified with even functions on the formal group, modulo functions that vanish to order six and higher. More interestingly, though, the fact that $12$ times the associated extension is trivial (that is, that there is a $\mathbb{Z}/12$ in the ANSS at that point corresponding to $\eta$---there is a $\mathbb{Z}/2$ above it because $\pi_3(S^0) = \mathbb{Z}/24$) can be seen algebro-geometrically, at least when one pulls the extension back to $M_{1,1}$. See M. Hopkins's ICM address for a discussion of this.

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