Ochanine proved in this paper that for genera taking values in $\mathbb{Q}$-algebras, vanishing on even-dimensional projective bundles is equivalent to being an elliptic genus (i.e. a specialization of Euler's formal group law). Does the implication "vanishing on even-dim projective bundles => specialization of Euler" still apply integrally, up to a finite set of primes (probably 2 and 3)? Just from skimming through it, it seems the argument would still go through over $\mathbb{Z}[\frac{1}{6}]$, but I'm not certain.

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    $\begingroup$ I'm not sure, but you might want to take a look at Hirzebruch-Berger-Jung. $\endgroup$ – skd Oct 18 '18 at 4:09
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    $\begingroup$ Ochanine is using that we can rationally generate $MSO_*$ by $\mathbb{CP}^2$ and the Milnor manifolds $H_{2,3}$ and $H_{3,2j}$. But I do not think that they are an integral generating set. The crucial point seems to to find a generating set of $MSO_*[\frac16]$, where all but two generators are even-dimensional projective bundles. $\endgroup$ – Lennart Meier Oct 18 '18 at 10:09
  • $\begingroup$ yeah very true - but if the target $R$ is torsion-free, say, and we only care about the composite genus $MU_* \to MSO_* \to R$ so that there's no zero-divisors, i thought maybe this wouldn't matter since rational generators determine the image completely. is this true? $\endgroup$ – xir Oct 18 '18 at 15:58
  • $\begingroup$ If the target $R$ of your genus is torsionfree, the rational statement gives you the statement you want also for $R$ itself. Btw: $MU_*$ surjects onto $MSO_*/2\text{-torsion}$, i.e. if $R$ does not have $2$-torsion, the composite genus $MU_* \to MSO_* \to R$ determines the original one. $\endgroup$ – Lennart Meier Oct 19 '18 at 10:33
  • $\begingroup$ ah of course. thanks for the tip on torsion, i wanted to say something like that but wasn't sure what the torsion in $MSO_*$ actually looked like. so in general i guess this is some kind of lifting problem for genera. $\endgroup$ – xir Oct 19 '18 at 16:11

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