Take a compact homogeneous space $G/K$, and a left $G$-invariant differential $k$-form $\omega \in \Omega^k(G/K)$. Will $\omega$ necessarily be closed? Might it even be harmonic when $G/K$ is endowed with a Riemannian metric?

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    $\begingroup$ No. See Bredon's book "geometry and topology" for a good discussion of the complex of invariant differential forms. For $K = 1$ you recover the Chevalley-Eilenberg complex of $\mathfrak g$. $\endgroup$ – Mike Miller Aug 13 '20 at 16:11
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    $\begingroup$ They don't have to be closed, but each closed form differs from a closed invariant form by an exact (not necessarily invariant) form, so the de Rham cohomology of forms is the same as that of invariant forms, assuming $G$ is compact. $\endgroup$ – Ben McKay Aug 13 '20 at 16:59

Note that the answer depends on the pair $(G,K)$.

For example, if $K=\{e\}$, then one is asking whether the ring of left-invariant forms on $G$ consists only of closed forms. This only happens when $G$ is abelian.

On the other hand, if $M=G/K$ is a compact Riemannian symmetric space and $G$ is the identity component of the isometry group of $M$, then, indeed, every $G$-invariant form is closed and, in fact, the ring of $G$-invariant forms on $M$ is equal to the space of harmonic forms on $M$. This is a well-known result, but for a short proof, one can consult this note by Michael E. Taylor.

For example, when $M=\mathbb{CP}^n$ endowed with its Fubini-Study metric, one has $G = \mathrm{SU}(n{+}1)/\mathbb{Z}_{n+1}$, and the only $G$-invariant forms are (linear combinations of) powers of the Kähler form $\omega$.

As another example, if $K$ is compact and $M = (K\times K)/\Delta$, where $\Delta = \{ (k,k)\ |\ k\in K \}$, then the $(K\times K)$-invariant forms on $M$ are simply the bi-invariant forms on $K$, which are all closed.


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