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Qiaochu Yuan
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The Lie bracket on rational homotopy has a relatively simple conceptual explanation: if $X$ is simply connected, then the rational homology $H_{\bullet}(\Omega X, \mathbb{Q})$ of its loop space is a connected graded Hopf algebra, and (under some mild hypotheses, maybe?) must therefore be the universal enveloping algebra of its Lie algebra of primitive elements, which turns out to be the rational homotopy $\pi_{\bullet}(\Omega X, \mathbb{Q})$ of $\Omega X$. This is of course the rational homotopy of $X$ but shifted by a degree, and that shift is responsible for the degree shift of the bracket. (These brackets vanish for rational spectra.)

Roughly speaking this rational homology is the "group algebra" of $\Omega X$ and so what this suggests is that the homotopy Lie algebra is the "Lie algebra of $\Omega X$" in a suitable sense. Like all good Lie algebra structures it is fundamentally attached to a group, which in this case is the loop space.

Now, I understand even less about chromatic homotopy than I do about rational homotopy, but my understanding is that the formal group structures appearing there are most simply understood as coming from Chern class computations on complex oriented cohomology theories, so fundamentally the group structure here is tensor product of line bundles. There is a graded Hopf algebra that can be written down here but it is (I think) $E^{\bullet}(BU(1))$ where $E$ is complex oriented, whose comultiplication (which encodes the formal group law) comes from the group structure on $BU(1)$ representing tensor product of line bundles. This has quite a different flavor from the loop space group structure above so, at least based on my limited background, the two don't seem all that related to me. To make a basic observation, the formal group laws appearing in the chromatic picture are all (commutative and) $1$-dimensional, whereas the Lie algebras appearing in rational homotopy are decidedly not.

I feel like I should say something here about Snaith's theorem which feels relevant but I'm not sure what to say.

Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741