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$exp_G(ad_X )(Y ) in Lie(H)$?

0

$begingroup$

Let $G le GL_N(Bbb R)$, G connected and $Hle G$, H closed and connected.

Let $mathcal{N} = {X ∈ Lie(G), ∀Y ∈ Lie(H), [X, Y ] ∈ Lie(H)}$

Let $X ∈ mathcal{N}, Y ∈ Lie(H)$. Prove that $Ad_{exp_G (X)} (Y ) ∈ Lie(H)$.

I am confused:

We know that $Ad_{exp_G (X)} (Y ) = exp_G(ad_X )(Y )$ with $ad_X in End(Lie(G))$ so by restriction to H: $ad_X in End(Lie(H))$.

Then composing with $exp_G$ we have $exp_G(ad_X )in End(H)$ so how comes $exp_G(ad_X )(Y ) in Lie(H)$?

I am surely missing something fundamental in Lie algebras and the exponential map.

Thank you for your help.

lie-groups lie-algebras

share|cite|improve this question

edited Mar 25 at 10:54

PerelMan

asked Mar 25 at 8:45

PerelManPerelMan

806414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check out my question here - math.meta.stackexchange.com/questions/29995/…
  $endgroup$
  – user619699
  Mar 25 at 8:48
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $G le GL_N(Bbb R)$, G connected and $Hle G$, H closed and connected.

Let $mathcal{N} = {X ∈ Lie(G), ∀Y ∈ Lie(H), [X, Y ] ∈ Lie(H)}$

Let $X ∈ mathcal{N}, Y ∈ Lie(H)$. Prove that $Ad_{exp_G (X)} (Y ) ∈ Lie(H)$.

I am confused:

We know that $Ad_{exp_G (X)} (Y ) = exp_G(ad_X )(Y )$ with $ad_X in End(Lie(G))$ so by restriction to H: $ad_X in End(Lie(H))$.

Then composing with $exp_G$ we have $exp_G(ad_X )in End(H)$ so how comes $exp_G(ad_X )(Y ) in Lie(H)$?

I am surely missing something fundamental in Lie algebras and the exponential map.

Thank you for your help.

lie-groups lie-algebras

share|cite|improve this question

edited Mar 25 at 10:54

PerelMan

asked Mar 25 at 8:45

PerelManPerelMan

806414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check out my question here - math.meta.stackexchange.com/questions/29995/…
  $endgroup$
  – user619699
  Mar 25 at 8:48
  

  
  
  
  

add a comment | 

$begingroup$

Let $G le GL_N(Bbb R)$, G connected and $Hle G$, H closed and connected.

Let $mathcal{N} = {X ∈ Lie(G), ∀Y ∈ Lie(H), [X, Y ] ∈ Lie(H)}$

Let $X ∈ mathcal{N}, Y ∈ Lie(H)$. Prove that $Ad_{exp_G (X)} (Y ) ∈ Lie(H)$.

I am confused:

We know that $Ad_{exp_G (X)} (Y ) = exp_G(ad_X )(Y )$ with $ad_X in End(Lie(G))$ so by restriction to H: $ad_X in End(Lie(H))$.

Then composing with $exp_G$ we have $exp_G(ad_X )in End(H)$ so how comes $exp_G(ad_X )(Y ) in Lie(H)$?

I am surely missing something fundamental in Lie algebras and the exponential map.

Thank you for your help.

lie-groups lie-algebras

share|cite|improve this question

edited Mar 25 at 10:54

PerelMan

asked Mar 25 at 8:45

PerelManPerelMan

806414

$endgroup$

share|cite|improve this question

edited Mar 25 at 10:54

PerelMan

asked Mar 25 at 8:45

PerelManPerelMan

806414

share|cite|improve this question

edited Mar 25 at 10:54

PerelMan

asked Mar 25 at 8:45

PerelManPerelMan

806414

asked Mar 25 at 8:45

PerelManPerelMan

806414

asked Mar 25 at 8:45

PerelManPerelMan

806414