Are invariant forms on homogeneous spaces necessarily closed? 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?
 A: 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.
