In great generality a Lie group mod its maximal compact subgroup is contractible (for example this is true for all connected Lie groups). Whenever this is true then the Lie group $ D $ is diffeomorphic to a cartesian product of its maximal compact $ K $ with a contractible piece $ D/K $. So (again assuming $ G $ is connected or semisimple or some other sufficient condition to guarantee $ D/K $ contractible) if $ D $ is dimension $ 2n $ and the maximal compact subgroup is dimension $ n $ then we have a diffeomorphism
$$
D \cong K \times (D/K)
$$
where $ D/K $ is a contractible piece of dimension equal to $ K $. So $ D $ is diffeomorphic to a trivial $ \dim(K) $ dimensional real vector bundle over $ K $. On the other hand, since $ K $ is a Lie group it must be parallelizable. That is, the tangent bundle of $ K $ is a trivial $ \dim(K) $ dimensional real vector bundle over $ K $. In other words, we have that the Lie group $ D $ is diffeomorphic to the tangent bundle of its maximal compact subgroup,
$$
D \cong T(K).
$$
A particular case of this is when $ G $ is a linear algebraic group whose real points $ G\_\mathbb{R} $ are compact. A compact group is always the maximal compact subgroup of its complexification (this follows from the fact that for subgroups of a complex linear algebraic group maximal compact is equivalent to compact plus Zariski dense). In other words, $ G\_\mathbb{R} $ is the ($ n $ dimensional) maximal compact subgroup of the ( $ 2n $ real dimensional) group $ G\_\mathbb{C} $. So, taking $ D=G\_\mathbb{C} $ and $ K=G\_\mathbb{R} $ in the argument above, we have that the complex points are diffeomorphic to the tangent bundle of the real points,
$$
G\_\mathbb{C} \cong T(G\_{\mathbb{R}}).
$$
Again this argument requires that the real points are compact. For example $ G\_\mathbb{R} $ compact implies $ G\_\mathbb{C} $ reductive group and so by Iwasawa decomposition for reductive groups we have that $ G\_\mathbb{C} $ mod its maximal compact is contractible.


**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} $?**


$\DeclareMathOperator\SO{SO}$Motivation:
$$
 \SO\_{n+1}(\mathbb{C})/\SO\_{n}(\mathbb{C}) 
$$
is diffeomorphic to the tangent bundle of the $ n $ sphere
$$
 S^n= \SO\_{n+1}(\mathbb{R})/\SO\_{n}(\mathbb{R}).
$$
More Motivation: When $ H $ is trivial the result follows from the discussion above.


$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SU{SU}\DeclareMathOperator\SO{SO}$Comment on the non-compact case: Without the compactness assumption this is false. For example we can take $ H $ trivial and $ G=\SL\_n $ (and $ n\geq 2 $ ). Then $ G\_\mathbb{C}=\SL\_n(\mathbb{C}) $ is homotopy equivalent to $ \SU\_n $ while the tangent bundle of $ G\_\mathbb{R}=\SL\_n(\mathbb{R}) $ is homotopy equivalent to $ \SL\_n(\mathbb{R}) $ which is homotopy equivalent to $ \SO\_n(\mathbb{R}) $. But $ \SU\_n $ and $ \SO\_n(\mathbb{R}) $ are not homotopy equivalent. Thus $ \SL\_n(\mathbb{C}) $ and the tangent bundle $ T(\SL\_n(\mathbb{R})) $ are not homotopy equivalent so they certainly are not diffeomorphic.


Possible counter examples to consider next:
The Steifel manifold $ O\_4/O\_2 $ or the Grassmanian $ O\_4/(O\_2 \times O\_2) $. What do the tangent bundles look like? Are they diffeomorphic to the complex points?



Is the manifold of complex points of a quotient of compact groups just the tangent bundle?
This fact about the tangent space being the complexification is known to be true if $ G\_\mathbb{R}/H\_\mathbb{R} $ is a compact symmetric space. In other words, if $ H\_\mathbb{R} $ is the fixed points of an involution of $ G\_\mathbb{R} $. This includes, for example, all the spheres written as
$$
S^n \cong SO\_{n+1}(\mathbb{R})/SO\_n(\mathbb{R})
$$
This is taken directly from <https://mathoverflow.net/a/414174/387190>


