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Let $G$ be a connected Lie Group and $K<G$ a maximal compact subgroup.

Denote by $\Omega^q(G/K)^G$ the $G$-invariant real-valued $q$-forms on the manifold $G/K$, i.e. those forms $\omega$ s.t. $g^*\omega=\omega$, where $g$ denotes the left translation mapping $hK$ to $(gh)K$.

Evaluation at the identity $eK$ yields an isomorphism $$\Omega^q(G/K)^G\cong\hom_{\mathfrak k}(\Lambda^q\mathfrak{g/k},\mathbb{R})$$

where $\mathfrak g$ and $\mathfrak k$ denote the Lie Algebras of $G$ and $K$.

I understand that a $G$-invariant form is determined by its values at the identity and so the isomorphism is given by mapping $\omega$ to $\omega_{eK}$.

But what does $\hom_{\mathfrak k}$ mean? Maybe $\mathfrak k$-equivariant? But what is the $\mathfrak k$-action and and why isn't it just $\mathbb R$-linear homomorphisms?

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1 Answer 1

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It should be $\operatorname{Hom}(\Lambda^q(\mathfrak{g}/\mathfrak{k}),\mathbb{R})^K$, invariant under $K$. It isn't just the $\mathbb{R}$-linear stuff. Imagine $G$ is the rotation group of the sphere, $K$ the subgroup fixing the north pole. Then to be invariant under $G$, you need to invariant under all transformations of $G$, in particular under all transformations in $K$. But these turn the north pole around, so you need to be invariant under the action of $K$ on the tangent plane at the north pole. If $K$ is connected, then $K$-invariance is the same as being null for the $\mathfrak{k}$-action, so you can write it as $\operatorname{Hom}(\Lambda^q(\mathfrak{g}/\mathfrak{k}),\mathbb{R})^{\mathfrak{k}}$.

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Very helpful, thanks. Could you maybe state the $\mathfrak k$-action explicitly? –  Peter Aug 28 '13 at 18:13
Get $G$ to act on its Lie algebra $\mathfrak{g}$ by the adjoint action. Restrict this from $G$ to $K$ to get an action of $K$ on $\mathfrak{g}$, which preserves $\mathfrak{k}$, and so induces an action on the quotient $\mathfrak{g}/\mathfrak{k}$. Call this action $\rho$. Every action of $K$ induces an action of $\mathfrak{k}$ by differentiation: define $\rho(A)$ for $A \in \mathfrak{k}$ by $e^{\rho(A)}=\rho(e^A)$. –  Ben McKay Aug 29 '13 at 6:26

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