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?

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    $\begingroup$ Just a note: $G_{\mathbb R}/H_{\mathbb R}$ is not necessarily the real points of anything. (Think of $\operatorname{SL}_2(\mathbb R)/\mu_2(\mathbb R)$, which is smaller than $(\operatorname{SL}_2/\mu_2)(\mathbb R) = \operatorname{PGL}_2(\mathbb R)$.) I would be suspicious that that sort of issue might cause problems, but I'm not sure. $\endgroup$
    – LSpice
    Dec 29, 2021 at 14:45
  • $\begingroup$ Try $G=GL_1$, $H=\{1\}$. $\endgroup$ Dec 29, 2021 at 19:02
  • $\begingroup$ I've added an assumption that the real points are compact. That is closer to the motivating example anyway. And it's really the case I'm most interested in. Thanks for the comments! $\endgroup$ Dec 29, 2021 at 19:25
  • $\begingroup$ So, in particular, $G(\mathbb{R})$, if compact, should be connected if $G$ is? $\endgroup$ Dec 29, 2021 at 21:46
  • $\begingroup$ @LaurentMoret-Bailly I believe that is correct. Certainly $ G(\mathbb{R}) $, if compact, is the maximal compact of $ G(\mathbb{C}) $. And the maximal compact of a connected lie group is always connected. See mathoverflow.net/questions/140622/… $\endgroup$ Dec 31, 2021 at 15:53

1 Answer 1


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

  • $\begingroup$ You have linked to your question. Did you mean to link to @JohannesHuisman's answer? $\endgroup$
    – LSpice
    Jan 27 at 4:10
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    $\begingroup$ Yes I mean his answer! Whenever I link I just copy paste the URL for the page. What's the trick for linking to a certain part of the page? $\endgroup$ Jan 27 at 13:20
  • $\begingroup$ Below each answer (and the question itself), one of the small text labels says 'share'; clicking on that will give you the link. For a comment, you instead click on the time stamp to get the link. $\endgroup$
    – LSpice
    Jan 27 at 13:29
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    $\begingroup$ Oh wow this is perfect MO is such a well designed site! Thank you! $\endgroup$ Jan 27 at 13:35
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    $\begingroup$ The consensus on that is muddied. See some discussion at How to contact a MathOverflow user?. $\endgroup$
    – LSpice
    Jan 27 at 13:43

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