7
$\begingroup$

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$Let $ g $ be the round metric on the sphere $ S^n $. Since $ S^n $ is compact the isometry group $ \Iso(S^n,g) $ is also compact. And every compact group can be realized as the real points of some (reductive) linear algebraic group. Indeed, $ \Iso(S^n,g) = \O_{n+1}(\mathbb{R}) $. The complex points of this group are $ \O_{n+1}(\mathbb{C}) $. And $ \O_{n+1}(\mathbb{C}) $ acts transitively on the tangent bundle of the sphere $ T(S^n) $.

Does this generalize from the round sphere to other compact homogeneous Riemannian manifolds?

In other words, Let $ (M,g) $ be a compact Riemannian homogeneous space. Then $ \Iso(M,g) $ is a compact Lie group. So there exists some (reductive) linear algebraic group whose real points are isomorphic to $ \Iso(M,g) $. The question is, does there always exist a linear algebraic group $ G $ such that the real points of $ G $ are isomorphic to the isometry group $$ G_\mathbb{R} \cong \Iso(M,g) $$ and, in addition, the complex points of $ G $ act (transitively, smoothly) on the tangent bundle $ T(M) $?

Note that this question is equivalent to a question which does not a priori in involve any geometry:

The manifold $ G_\mathbb{C}/H_\mathbb{C} $ is the tangent bundle to $ G_\mathbb{R}/H_\mathbb{R} $ where $ G_\mathbb{R}$, $H_\mathbb{R} $ are compact real forms of $ G_\mathbb{C}$, $H_\mathbb{C} $

Note that while the action of $ G_\mathbb{R} $ is by isometries, the action of $ G_\mathbb{C} $ on $ T(M) $ can only be by isometries if $ M $ is parallelizable. So in particular the action of $ \O_{n+1}(\mathbb{C}) $ on $ T(S^n) $ can only be by isometries in the cases $ n=1,3,7 $.

$\endgroup$
3
  • 1
    $\begingroup$ Did you intend to consider the action on the projectivized tangent bundle? Obviously the isometry group preserves the zero section of the tangent bundle, and this orbit is disjoint from the other orbit. $\endgroup$ Commented Jan 2, 2022 at 0:03
  • $\begingroup$ MathJax note: leaving a blank line after your \DeclareMathOperators will produce a blank in the output document, at least unless you include the blank line within the $ $ "environment". I have edited to remove it. $\endgroup$
    – LSpice
    Commented Jan 2, 2022 at 0:15
  • $\begingroup$ @JasonStarr no I'm just interested in the regular tangent bundle $\endgroup$ Commented Jan 18, 2022 at 15:25

1 Answer 1

3
+50
$\begingroup$

Maybe these arguments are of interest to you. It is known that for any compact symmetric space $M$ the tangent bundle $TM$ possesses a canonical structure of a complex manifold. Multiplication by $-1$ on $TM$ is an antiholomorphic involution; its set of fixed points is $M$, when identified with the zero section of $TM$ (see e.g. Thm 2.5a of Szőke, R.: Complex structures on tangent bundles of Riemannian manifolds, Math. Ann. 291 (1991), 409-428). It follows that the action of the Lie group $G_{\mathbf R}$ on $M$ extends to an action on the complex manifold $TM$ by holomorphic automorphisms. One proves that it extends to a holomorphic action of the complexification $G_{\mathbf C}$ of $G_{\mathbf R}$ on $TM$. This can be done by considering the induced morphism from the real Lie algebra ${\mathfrak g}_{\mathbf R}$ of $G_{\mathbf R}$ into the complex Lie algebra ${\mathcal A}(TM)$ of holomorphic vector fields on $TM$. It naturally extends to the complexification ${\mathfrak g}_{\mathbf C}$ of ${\mathfrak g}_{\mathbf R}$, which gives rise to a holomorphic action of the complexification $G_{\mathbf C}^0$ of the neutral component $G_{\mathbf R}^0$ of the real Lie group $G_{\mathbf R}$. The latter action can easily be extended to a holomorphic action of $G_{\mathbf C}$ on the complex manifold $TM$. It is not hard to see that the action is transitive (for more details see e.g. Thm C of Szőke, R.: Automorphisms of certain Stein manifolds, Math. Z. 219 (1995), 357-385)

$\endgroup$
1
  • $\begingroup$ Great answer! I'm still interested in generic Riemannian homogeneous manifolds but I'm really excited to see this result for symmetric spaces. Since the result that originally inspired me to ask about this was just tangent bundles of sphere, which are symmetric spaces, this answer is certainly sufficient to earn the bounty. Let me know if you have any thoughts on the non symmetric case, for example, a way to identify the tangent space of a Stiefel manifold with $ SO_n(\mathbb{C})/SO_{n-k}(\mathbb{C}) $. $\endgroup$ Commented Jan 19, 2022 at 0:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.