Revisions to Permutable (Lie) subgroups

Added some hints of how to solve the question and where I am stuck

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edited Jul 19, 2015 at 22:04

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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.

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).

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.

Added some hints of how to solve the question and where I am stuck

Source Link

edited Jul 19, 2015 at 21:58

Duchamp Gérard H. E.

3.7k
34
40

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).

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 ?

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).

Formatting the first block

Source Link

edited Jul 19, 2015 at 15:37

Duchamp Gérard H. E.

3.7k
34
40

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 $Lie(A)\cap Lie(B)=\{0\}$

Let now $G$ be a finite dimensional Lie group (real to begin with) and suppose that if $A,B\subset G$My questions are Lie subgroups (then closed by Cartan's theorem), one can easily show that,the following

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 ($G=AB;\ A\cap B=\{1_G\}$$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 ?

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 ($G=AB;\ A\cap B=\{1_G\}$), 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 ?

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 ?

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asked Jul 19, 2015 at 14:35

Duchamp Gérard H. E.

3.7k
34
40

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