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It is obvious that not every homogeneous manifold is parallelizable (take for example the two-sphere $S^{2}$). In contrast, every Lie group $G$ is parallelizable, as you can construct a pointwise basis of left-invariant vector fields by acting with the adjoint action of the group onto himself. My question is, what fails if one tries to carry the same construction on an homogeneous space $M = G/H$? If I am not mistaken, one has a transitive action of $G$ on $M$, which could be used to generate left-invariant vector fields.

Thanks.

Bilateral
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1 Answers1

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If a group $G$ acts on itself by left multiplication and if $p$, $q$ are arbitrary points of $G$, there is exactly one element $g$ such that $gp = L_{g}(p) = q$. Consequently, a tangent vector at $p$ corresponds to a unique tangent vector at $q$ via the tangent map $dL_{g}$.

In general, however, the left $G$-action on $G/H$ has non-trivial stabilizers, and "left translation" of tangent vectors at $p$ is well-defined only up to the isotropy action (i.e., the action of $H$). Consequently, the $G$-action doesn't define single-valued vector fields unless $H$ acts trivially on tangent spaces, as for the action of $\mathbf{Z}^{n}$ on $(\mathbf{R}^{n}, +)$.

For $S^{2}$ viewed as $SO(3)/SO(2)$, the isotropy action is about as far from trivial as you can get (transitive on the set of unit vectors). :)

Andrew D. Hwang
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