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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$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:

Let $ G $ be a linear algebraic group and $ H $ a linear algebraic subgroup. Suppose the real points $ G_\mathbb{R}$, $H_\mathbb{R} $ are compact. Consider the manifold of complex points $$ M_\mathbb{C}=G_\mathbb{C}/H_\mathbb{C}. $$ Then is the tangent bundle of $$ M_\mathbb{R}=G_\mathbb{R}/H_\mathbb{R} $$ diffeomorphic to $ M_\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 $.

Also note that this result is essentially claimed by Nicolas Tholozan in the comments on Is a quotient of real linear algebraic groups always a Cartesian product of compact and contractible factors? I am really just looking for a proof or reference to verify his claim

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$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 $.

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$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 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 $.

Also note that this result is essentially claimed by Nicolas Tholozan in the comments on Is a quotient of real linear algebraic groups always a Cartesian product of compact and contractible factors? I am really just looking for a proof or reference to verify his claim

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$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:

Let $ G $ be a linear algebraic group and $ H $ a linear algebraic subgroup. Suppose the real points $ G_\mathbb{R}$, $H_\mathbb{R} $ are compact. Consider the manifold of complex points $$ M_\mathbb{C}=G_\mathbb{C}/H_\mathbb{C}. $$ Then is the tangent bundle of $$ M_\mathbb{R}=G_\mathbb{R}/H_\mathbb{R} $$ diffeomorphic to $ M_\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 $.

Also note that this result is essentially claimed by Nicolas Tholozan in the comments on Is a quotient of real linear algebraic groups always a Cartesian product of compact and contractible factors? I am really just looking for a proof or reference to verify his claim

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$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 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 $.

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$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 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 $.

Also note that this result is essentially claimed by Nicolas Tholozan in the comments on Is a quotient of real linear algebraic groups always a Cartesian product of compact and contractible factors? I am really just looking for a proof or reference to verify his claim