The tangent space of a (compact) Lie group $G$ is given by its Lie algebra. Assuming for convenience that $G$ is connected, its Lie algebra $\frak{g}$ decomposes into a Cartan part, as well as positive and negative root spaces: $$ \frak{g} \simeq g_{-} \oplus \frak{g}_0 \oplus \frak{g}_+. $$ This gives a dual decomposition on the space of differential one-forms $$ \Omega^1 \simeq \Omega^1_- \oplus \Omega^1_0 \oplus \Omega^1_+. $$ This should extend to a $\mathbb{N}_{0}^3$-grading on the de Rham complex of $G$, if I am not missing something. Taking the differential into account, do we have some type of tri-complex here? (I am of course thinking of the analogy with complex manifolds.) If so, what is the geometric significance of all of this? Do there exist Riemannian metrics interacting well with this structure?
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3$\begingroup$ The decomposition only exists after tensoring with $\mathbb C$ and is not preserved by the differential - the annihilator of a subbundle generates a differential subalgebra iff sections of the subbundle are closed under Lie bracket. $\endgroup$– Bertram ArnoldCommented Nov 17, 2020 at 20:37
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1$\begingroup$ This is a grading in $\mathbf{Z}$, namely $\mathfrak{g}_{-1}\oplus\mathfrak{g}_0\oplus\mathfrak{g}_1$. On the Chevalley-Eilenberg complex this yields a $\mathbf{N}\times\mathbf{Z}$-grading, for which the differential shifts the grading by $(1,0)$ (or $(-1,0)$ in homology), and the homology inherits a $\mathbf{N}\times\mathbf{Z}$-grading. $\endgroup$– YCorCommented Nov 17, 2020 at 22:37
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