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Let $\mathfrak{g}$ be a finite dimensional real Lie algebra, and $\mathfrak{g}^* $ be the dual vector space. We have the standard Cartan-Eilenberg complex $(\wedge^{\cdot} \mathfrak{g}^* ,\text{d}_{CE})$ where the differential is given by the dual of the Lie bracket.

Now assume that we have a Lie subalgebra $\mathfrak{k}\subset \mathfrak{g}$ and a $\mathfrak{k}$ invariant complement $\mathfrak{p}\subset \mathfrak{g}$. That is: $$ \mathfrak{g}=\mathfrak{k}\oplus\mathfrak{p} $$ as vector spaces, $[\mathfrak{k},\mathfrak{k}]\subset \mathfrak{k}$ and $[\mathfrak{k},\mathfrak{p}]\subset \mathfrak{p}$. Remember that in general $[\mathfrak{p},\mathfrak{p}]\nsubseteq$$\mathfrak{p}$.

We can construct $\wedge^{\cdot} \mathfrak{p}^* $ and define a derivative on it as follows. For any $ \theta \in \wedge^k \mathfrak{p}^* $, use the split $ \mathfrak{g}^* =\mathfrak{k}^* \oplus\mathfrak{p}^* $ we can treat $\theta $ as an element in $\wedge^k \mathfrak{g}^* $, take the Cartan-Eilenberg differential, we get $\text{d}_{CE}\theta \in \wedge^{k+1} \mathfrak{g}^* $. Again use $ \mathfrak{g}^* =\mathfrak{k}^* \oplus\mathfrak{p}^* $ we get $$ \wedge^{k+1} \mathfrak{g}^* = \bigoplus_{i=0}^{k+1} \wedge^i \mathfrak{k}^* \otimes \wedge^{k+1-i} \mathfrak{p}^* $$ Let $\text{pr}$ denote the projection onto the $i=0$ component of the above decomposition. Then our derivation $\overline{\text{d}}$ is defined to be $\text{pr}\circ \text{d}_{CE}: \wedge^{\cdot} \mathfrak{p}^* \rightarrow \wedge^{\cdot+1}\mathfrak{p}^* $.

It is not difficult to check that $\overline{\text{d}}$ is a derivation but $\overline{\text{d}}\circ \overline{\text{d}}\neq 0$. It measures the failure of $[\mathfrak{p},\mathfrak{p}]\subseteq\mathfrak{p}$.

My question is: what does this $\overline{\text{d}}\circ \overline{\text{d}}$ stand for? Is it a kind of curvature? Is there any nice expression of it? In particular, does it have anything to do with the connection on the principle bundle $K\rightarrow G \rightarrow G/K$?

I have the impression that Kostant has done some work of this kind but I just cannot find them.

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    $\begingroup$ Take a look at Chapter X in the third volume of "Connections, curvature and cohomology" by Greub, Halperin and Vanstone. You will find that indeed what you have can be interpreted as a curvature of an algebraic connection. $\endgroup$ Commented Aug 8, 2012 at 14:11

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