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It's known, from Ravenel's green book, as well as other sources, that we build formal group laws over a ring from n-buds, where an n-bud is essentially a truncated formal group law (sometimes called a formal group law n-chunk). It is also known that a map of ring spectra $X(n)\to E$ where $X(n)$ is the Thom spectrum associated to the map $\Omega SU(n)\to \Omega SU\simeq BU$, determines an n-bud over $E_\ast$. These spectra are also important in the proof of the celebrated Nilpotence Theorem of chromatic homotopy theory. My question is a rather general one, but begins with wondering why we use $\Omega SU(n)$ rather than $BU(n)$? How different is the former filtration of $BU$ from the latter? Moreover, if we take the Thom spectrum of the inclusion $BU(n)\to BU$, do we still get the $X(n)$? If so, why do Devinatz, Hopkins and Smith use $\Omega SU(n)$ at all? I have a number of questions surrounding this issue, but I guess that's probably enough for now.

Thanks as usual!

-Jon

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    $\begingroup$ The $\Omega SU(n)$ filtration is a filtration by loop maps. Their Thom spectra inherit a product from the loop structure which you use to make their homotopy groups into a ring. You do not have this structure for $BU(n)$. $\endgroup$
    – Justin Noel
    Commented Oct 31, 2013 at 15:32
  • $\begingroup$ For reference: the relationship between n buds and X(n) is in Ravenel’s green book $\endgroup$
    – Tim Campion
    Commented May 25, 2019 at 20:48

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