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Let's recall that, a group $G$ being given, two subgroups $A,B\subset G$ are called permutable iff $AB=BA$ for the Minkowski law. It is straightforward to see that $(A,B)$ are permutable iff $AB$ is a subgroup of $G$.

Let now $G$ be a finite dimensional Lie group (real to begin with) and suppose that if $A,B\subset G$ are Lie subgroups (then closed by Cartan's theorem), one can easily show that, providing $G=AB$ and if it is a factorization (means that the decomposition is unique or, equivalently, $A\cap B=\{1_G\}$), then $Lie(A)\cap Lie(B)=\{0\}$

My questions are the following

  • Q1) If $G=AB$ is a factorization ($A\cap B=\{1_G\}$), do we have $Lie(A)\oplus Lie(B)=Lie(G)$ ?
  • Q2) Does the result hold if we just have $G=AB$ (and still $Lie(A)\cap Lie(B)=\{0\}$) without supposing that $A\cap B=\{1_G\}$ ?
  • Q3) What are the references about these questions ?

What was (and is still) not clear to me is why the tangent map of the multiplication $$ A\times B\rightarrow G $$ must be surjective. More precisely, the fact that it is a factorization provides a section $$s: G\rightarrow A\times B$$ (in fact an inverse bijection) but I am stuck in proving that it is C^1 (continuous would do, I think). Maybe the action of $A\times B$ on itself by $(a,b)*(x,y)=(ax,yb^{-1})$ could be exploited but I do not see how.

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  • $\begingroup$ You're right, of course. I will think further and delete my answer in the meantime. $\endgroup$ Jul 19, 2015 at 22:52
  • $\begingroup$ The property ${\rm{Lie}}(A)\cap{\rm{Lie}}(B)=0$ when $A\cap B=1$ is valid without hypotheses on $AB$ inside $G$. More generally, ${\rm{Lie}}(A)\cap {\rm{Lie}}(B)={\rm{Lie}}(A\cap B)$ for any two closed subgroups $A$ and $B$ of a Lie group $G$. (I haven't seen this in a textbook in that generality, but perhaps it is in a standard reference? It is not hard to prove, but requires some care since (i) $A \cap B$ might not be connected and (ii) when ${\rm{Lie}}(A)\cap{\rm{Lie}}(B)$ is exponentiated to a connected Lie subgroup $H$ of $G$ we don't know a-priori that $H$ has the subspace topology.) $\endgroup$
    – grghxy
    Jul 20, 2015 at 3:17
  • $\begingroup$ @grghxy OK, ${\rm{Lie}}(A)\cap {\rm{Lie}}(B)={\rm{Lie}}(A\cap B)$ is not hard to prove using one-parameter subgroups. I have a look to your answer later. $\endgroup$ Jul 20, 2015 at 5:27

2 Answers 2

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As desired in the question, the action of $A \times B$ on $G$ is the key point. This is a transitive action, and it is a general fact in the theory of actions of (separable) Lie groups on manifolds (see Bourbaki, Lie Groups & Lie Algebras, Ch. III, no. 1.7, Proposition 14) that if a Lie group $H$ with at most countably many connected components (equivalently, a countable base for its topology) acts transitively on a smooth manifold $X$ then for any $x_0 \in X$ with stabilizer $H_{x_0} \subset H$ the natural orbit map $H/H_{x_0} \rightarrow X$ is a diffeomorphism. (We need a countability hypothesis on $H$ to rule out the case where $X$ is a positive-dimensional connected Lie group and $H$ is the underlying discrete group acting by left-translation.) In particular, if the action is simply transitive then the orbit map $H \rightarrow X$ is an $C^{\infty}$-isomorphism.

Now consider a Lie group $G$ with only countably many connected components and $A, B$ closed Lie subgroups of $G$ (so they each have only countably many connected components) such that the multiplication map $A \times B \rightarrow G$ is surjective. For the Lie subgroup $H := A \cap B$ setting $X=G$ with the indicated transitive action, we get that the natural map $A \times B \rightarrow G$ defined by $(a,b) \mapsto ab^{-1}$ is a $C^{\infty}$-submersion (in fact, a quotient map modulo $A \cap B$), and a $C^{\infty}$-isomorphism when $A\cap B = 1$. In particular, in all such cases ${\rm{Lie}}(A) + {\rm{Lie}}(B) = {\rm{Lie}}(G)$, and the sum is direct if and only if $A \cap B$ is discrete. (The generalization to Lie subgroups that might not have the subspace topology is left to the interested reader!)

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  • $\begingroup$ OK, thank you very much, I had a look to your answer (it took some time to digest it though), but I think it is a nice application of the "Lie orbital Bourbaki". $\endgroup$ Jul 20, 2015 at 16:09
  • $\begingroup$ @DuchampGérardH.E.: Without a countability hypothesis on $G$ how do you prove that this map between connected components is surjective? (I had considered to make such an argument but didn't see how to link the connectedness to the purely algebraic hypothesis $AB=G$.) I won't be surprised if there is a short and easy justification, but I didn't see it after thinking for 10 seconds and so decided not to dwell more on it. :) $\endgroup$
    – grghxy
    Jul 20, 2015 at 18:26
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Let $i_A : A \to G$ and $i_B : B \to G$ be the embeddings and let $\mu : G \times G \to G$ be the group multiplication. Then the composition $$ A \times B \stackrel{i_A\times i_B}{\longrightarrow} G \times G \stackrel{\mu}{\longrightarrow} G $$ is a smooth map which is surjective and sends the identity to the identity. (It is not, however, a group homomorphism.)

In a previous, incorrect, version of this answer, I had claimed that the tangent map at the identity was surjective. This is of course false. This was used only to arrive at an inequality: $$ \dim\mathrm{Lie}(G) \leq \dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B)~. $$

In fact, this inequality follows from Sard's theorem and the fact that the map $A \times B \to G$ is surjective, for if $\dim G > \dim A + \dim B$ then the image of the map $A \times B \to G$ would have measure zero.

Having established the above inequality, we now establish the reverse inequality. To see this simply notice that if $\mathrm{Lie}(A) \cap \mathrm{Lie}(B) = 0$, then the fact that $\mathrm{Lie}(A)$ and $\mathrm{Lie}(B)$ are subspaces of $\mathrm{Lie}(G)$ implies $$ \dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B) \leq \dim\mathrm{Lie}(G)~. $$ So the answers to Q1 and Q2 are both true (as vector spaces not as Lie algebras).

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  • $\begingroup$ Of course, all this is straightforward once one knows that the multiplication $$A\times B\rightarrow G$$ is a submersion ... which is exactly what I need. $\endgroup$ Jul 19, 2015 at 22:32
  • $\begingroup$ @JoséFigueora Yes, the key point was of course (I say "of course" after interaction however) Sard's theorem which here says that "somewhere" the action is a submersion. And the situation is the same everywhere due to the transitivity of the action of $A\times B$ $\endgroup$ Jul 20, 2015 at 16:17

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