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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 $\Omega^*$?

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    $\begingroup$ 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. $\endgroup$
    – Mikhail Borovoi
    Commented Aug 9, 2019 at 2:30

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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 $\xi$.

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  • $\begingroup$ 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? $\endgroup$
    – Fofi Konstantopoulou
    Commented Aug 9, 2019 at 21:41
  • $\begingroup$ Why does this trivial action then extend to the tensor product $H^n(M,\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R}$? $\endgroup$
    – Fofi Konstantopoulou
    Commented Aug 9, 2019 at 21:42
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    $\begingroup$ 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. $\endgroup$
    – Mikhail Borovoi
    Commented Aug 10, 2019 at 1:00
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    $\begingroup$ 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 $H^n(M,\Bbb Z)\otimes_{\Bbb Z} \Bbb R$. $\endgroup$
    – Mikhail Borovoi
    Commented Aug 10, 2019 at 1:08
  • $\begingroup$ Great, thanks a lot! $\endgroup$
    – Fofi Konstantopoulou
    Commented Aug 10, 2019 at 15:35

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