# Coinvariant representative of homogeneous space cohomology

Given a compact homogeneous space $$M = K/L$$, consider its de Rham complex $$(\Omega^*,d)$$. Will every cohomology class $$[\omega] \in ker(d)/im(d)$$ contain a representative $$\nu$$ which is invariant with respect to the left $$K$$-action on $$\Omega^*$$?

• Yes, assuming that $K$ is a connected compact Lie group. Indeed, $K$ acts on the integral cohomology trivially, because $K$ is connected. Hence $K$ acts trivially on the de Rham cohomology. Furthermore, since $K$ is compact, it preserves a Riemann metric on $M$, hence it acts on harmonic forms, and acts trivially, because it acts trivially on the de Rham cohomology. Such a harmonic form is the canonical $K$-invariant representative of a cohomology class. – Mikhail Borovoi Aug 9 at 2:30

Yes, assuming that $$K$$ is a connected compact Lie group.

Indeed, fix $$n$$ such that $$0\le n\le d={\rm dim}(M)$$. The group $$K$$ acts on the integral cohomology group $$H^n(M,\Bbb Z)$$ trivially, because $$K$$ is connected, while $$H^n(M,\Bbb Z)$$ is discrete. Therefore, $$K$$ acts trivially on the de Rham cohomology group

$$H^n_{\rm dR}(M,\Bbb R)=H^n(M,\Bbb R)=H^n(M,\Bbb Z)\otimes_{\Bbb Z} \Bbb R.$$

Consider the point $$x=e\cdot L\in K/L=M$$ with stabilizer $$L$$. Then $$L$$ acts on the tangent space $$T_x(M)$$. Since $$L$$ is compact, it preserves a positive definite quadratic form on $$T_x(M)$$. It follows that $$M$$ admits a $$K$$-invariant Riemannian metric $$g$$. Therefore, $$K$$ acts on the vector space $${\mathcal H}^n(M,g)$$ of harmonic differential $$n$$-forms on $$M$$ with respect to $$g$$.

Since $$M$$ is compact, by a theorem of Hodge a cohomology class $$\xi\in H^n_{\rm dR}(M,\Bbb R)$$ is represented by a unique harmonic form $$\omega\in {\mathcal H}^n(M,g)$$; see for instance the book by Jürgen Joost "Riemannian Geometry and Geometric Analysis", Theorem 2.2.1. Since $$K$$ acts trivially on $$H^n_{\rm dR}(M,\Bbb R)$$, it acts trivially on $${\mathcal H}^n(M,g)$$. Thus the harmonic differential form $$\omega$$ is a $$K$$-invariant representative of $$\xi$$.

• Why must $K$ act trivially on $H^n(M,\mathbb{Z})$? Is this a general result about discrete sets with an action of a compact connected Lie group? – Fofi Konstantopoulou Aug 9 at 21:41
• Why does this trivial action then extend to the tensor product $H^n(M,\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R}$? – Fofi Konstantopoulou Aug 9 at 21:42
• Let $G$ be a connected topological group acting continuously on a discrete topological space $X$. Let $x\in X$. Consider the map $$\phi_x\colon G\to X, \quad g\mapsto g\cdot x.$$ The map $\phi_x$ is continuous, hence the image $G\cdot x=\phi_x(G)$ of the connected group $G$ is a connected subset of $X$, containing $x$. Since $X$ is discrete, we conclude that $G\cdot x=\{x\}$. Thus $G$ acts on $X$ trivially. – Mikhail Borovoi Aug 10 at 1:00
• Concerning $H^n(M,\Bbb R)$: You need some definitions of the functors $H^n(M,\Bbb Z)$ and of $H^n(M,\Bbb R)$. Then you will see immediately that $$H^n(M,\Bbb R)=H^n(M,\Bbb Z)\otimes_{\Bbb Z} \Bbb R$$ and that ${\rm Aut}(M)$ acts on $H^n(M,\Bbb R)$ via the first factor of $H^n(M,\Bbb Z)\otimes_{\Bbb Z} \Bbb R$. – Mikhail Borovoi Aug 10 at 1:08
• Great, thanks a lot! – Fofi Konstantopoulou Aug 10 at 15:35