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Let $Ell^*(X)$ be the elliptic cohomology theory (associated to a given elliptic curve $E$) of a nice space $X$.

Recall the Landweber-Ravenel-Stong construction: $MU^*(X) \otimes_{MU^*} R \simeq Ell^*(X)$, where $R \simeq Ell^*$.

How does the Landweber-Ravenel-Stong construction of $Ell^*(-)$ tell us how to actually compute $Ell^*(X)$ with the Atiyah-Hirzebruch spectral sequence?

Let $L$ be the Lazard ring, and $R$ is flat over $L$. It seems like the Landweber-Ravenel-Stong construction tells us that:

  1. Given a map $L \to R$, $$MU^* \simeq L \longrightarrow R \simeq Ell^*$$ This gives us a map between their Atiyah-Hirzebruch SS? $$MU^*-AHSS(X) \longrightarrow Ell^*-AHSS(X)$$
  2. The formal group law associated to our elliptic curve $\hat{E} \simeq \text{Spf } Ell^*(\mathbb{CP}^\infty)$.

It was told that $Ell^*(X)$ is computationally simpler than computing $MU^*(X)$. But we seem to need to know $MU^*(X)$ before we can compute $Ell^*(X)$, so I must be missing something!

Do we need to know the AHSS of $MU^*(X)$ before we can compute the AHSS of $Ell^*(X)$?

Edit wrt computing the AHSS of $Ell^*(X)$: It was explained to me that we don't need $MU^*(X)$, we need $Ell^*(pt)$ and $H^*(X)$ (plus differentials) to compute $Ell^*(X)$. We know $MU^*(pt)$ and thus our choice of elliptic curve (or better, it's ground ring) determines $Ell^*(pt)$.

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    $\begingroup$ As the spectrum $Ell$ is Bousfield equivalent to $K(0)\vee K(1)\vee K(2)$ (with an implicit localization at a prime), we have $Ell_*(K(A,n)) = 0$ for $A$ finite abelian and $n\geq 3$ (see math.rochester.edu/people/faculty/doug/mypapers/emspace.pdf). We certainly do not have to compute $MU_*(K(A,n))$ for this. $\endgroup$
    – Lennart Meier
    Commented Mar 28, 2015 at 23:32

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The AHSS is unlikely to be a good method for computing either $MU^*(X)$ or $Ell^*(X)$ except in cases where the AHSS collapses for easy reasons (the $E^2$ term is torsion free, or concentrated in even total degree). Even in those cases, it is usually desirable and possible to use other methods (usually Chern classes) to produce some generators and relations for $MU^*(X)$, and then use the collapsing AHSS to show that no more generators or relations are required. In these cases one can compute both $MU^*(X)$ and $Ell^*(X)$.

The basic case where $Ell^*(X)$ is easier than $MU^*(X)$ is the case $X=BG$, where $G$ is a finite group. Even here, the group $Ell^*(X)$ itself is not so easy. It is better to consider the $I_2$-adic completion of $Ell$, which we could call $E$. Then $E$ is essentially a version of Morava $E$-theory, and $E^*(BG)$ is a finitely generated module over the Noetherian ring $E^*(\text{point})$, which is often a free module. Everything can be computed quite explicitly in the case where $G$ is abelian. Even if $G$ is nonabelian one can use HKR character theory, Chern classes of representations, transfers and power operations. The AHSS is almost never useful in this context.

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  • $\begingroup$ How is the $I_2$-adic completion of $Ell$ essentially a version of Morava $E$-theory? Do you have a recommended source to read more about this? $\endgroup$
    – Catherine Ray
    Commented Mar 28, 2015 at 21:40
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    $\begingroup$ @fractalcows As both theories are Landweber exact, the question is completely algebraic. $I_2$-adic completion corresponds to completing (the moduli stack of elliptic curves with $\Gamma_0(2)$-structure) at the locus where $(p,v_1)$ vanishes, i.e. the characteristic-$p$-points $k$ corresponding to supersingular elliptic curves. By Serre-Tate theory, this corresponds to the universal deformation of the formal groups of the supersingular elliptic curves. So if you have only one supersingular elliptic curve (e.g. for $p\leq 7$), you get $E(k,\Gamma)^{hG}$, where $G$ is the automorphism group... $\endgroup$
    – Lennart Meier
    Commented Mar 28, 2015 at 23:18
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    $\begingroup$ of the supersingular elliptic curve with $\Gamma_0(2)$-structure and $\Gamma$ its formal group law. (As all formal group laws over $\mathbb{F}_{p^n}$ of the same height are isomorphic, you can often pretend that $\Gamma$ is the Honda formal group law so that $E(k,\Gamma) = E_2$.) Some of this is explained in math.mit.edu/~mbehrens/papers/buildTMF.pdf Section 4. $\endgroup$
    – Lennart Meier
    Commented Mar 28, 2015 at 23:19

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