# 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?

• 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 $\mathfrak g$. – Mike Miller Aug 13 '20 at 16:11
• 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. – 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.