Fdtxjr

The meaning of the "at the of"

Running a subshell from the middle of the current command

What Happens when Passenger Refuses to Fly Boeing 737 Max?

What has been your most complicated TikZ drawing?

What does it mean when multiple 々 marks follow a 、?

Why would a jet engine that runs at temps excess of 2000°C burn when it crashes?

Can infringement of a trademark be pursued for using a company's name in a sentence?

Time travel short story where dinosaur doesn't taste like chicken

Do Bugbears' arms literally get longer when it's their turn?

What is the likely impact on flights of grounding an entire aircraft series?

Excess Zinc in garden soil

"However" used in a conditional clause?

Why doesn't the EU now just force the UK to choose between referendum and no-deal?

Is going from continuous data to categorical always wrong?

Is it true that real estate prices mainly go up?

Counter-example to the existence of left Bousfield localization of combinatorial model category

Co-worker team leader wants to inject the crap software product of his friends into our development. What should I say to our common boss?

Humans have energy, but not water. What happens?

What is the blue range indicating on this manifold pressure gauge?

Life insurance that covers only simultaneous/dual deaths

Can "semicircle" be used to refer to a part-circle that is not a exact half-circle?

Word for a person who has no opinion about whether god exists

Ban on all campaign finance?

How do anti-virus programs start at Windows boot?

“If $g$ is semisimple, It is not too hard to see that $H^2(g,a)=0$. With a little supplementary argument…”

Orthogonal subspace relative to the Killing formWhat's wrong with my proof that reductive Lie algebras are semisimple?Prove the Weyl's complete reducibility Theorem on finite-dimensional $mathfrakg-modules$ by Kostant's $mathfrakn$-cohomology resultDon't understand Levi decomposition theoremFinite-dimensional complex representations of nilpotent Lie algebras over a subfield of $BbbC$Semisimple algebraic group vs semisimple Lie algebraIs the theorem of complete reducibility or the abstract Jordan decomposition needed for structure theory of semisimple Lie algebras?On complex irreducible representations of Lie algebrasClassification results for solvable lie algebras.Necessity of diagonalizability of adjoint representation of cartan subalgebra in definition

1

$begingroup$

This is a statement made in Knapp, Lie groups, Lie algebras, Cohomology Chpt 4 last paragraph of Sec 2.

$H^i(g,a)$ is the $i-$th cohomology group of complex $Hom(wedge^i g,a)$ with $a$ abelian lie algebra.

"If $g$ is semisimple, it is not too hard to see that $H^2(g,a)=0$. With a little supplementary argument, it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra. The same theorem and proof apply over $R$ and a structure theorem for lie groups drops out. Results for $H^1$ then furnishes a uniqueness theorem"

$textbfQ1:$ Maybe this is too obvious for others. I do not see why obviously $H^2(g,a)=0$. Fix $pi:gto End(a)$ representation. I am using Brian, Hall's Lie algebra's semisimple meaning reductive and trivial center. I could see all semi simples are decomposed into simples and this is unique upto reordering. So given an extension $0to ato hto gto 0$. Now $h=aoplus_pi g$ as semi-direct product but the lie bracket between $a,g$ will be twisted by representation of $gto End(a)$. I do not see why all extensions are equivalent to $0to ato aoplus_pi gto g$.

$textbfQ2:$ Why it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra?

$textbfQ3:$ Why the same theorem and proof apply over $R$ and a structure theorem for lie groups drops out? Over $R$, I no longer have unitary representation which may not allow me to have decomposition of $g$ into simple ones.

$textbfQ4:$ What are results for $H^1$ furnishing a uniqueness theorem? What is the meaning of uniqueness theorem here?

abstract-algebra representation-theory lie-algebras homological-algebra

share|cite|improve this question

asked Mar 10 at 19:58

user45765user45765

2,6702724

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The result in Q1 is called "Whitehead's second lemma"; that might be a useful keyword.
  $endgroup$
  – Qiaochu Yuan
  Mar 10 at 20:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $mathfraka$ is not just an abelian Lie algebra (the question would make no sense, or would assume implicitly that the action is trivial), but is also endowed with a $mathfrakg$-action.
  $endgroup$
  – YCor
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Q2: This is Levi's theorem. There is a short proof of it using Whitehead's second Lemma from above (which can be found in several books, but is also a nice exercise).
  $endgroup$
  – Dietrich Burde
  2 days ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

This is a statement made in Knapp, Lie groups, Lie algebras, Cohomology Chpt 4 last paragraph of Sec 2.

$H^i(g,a)$ is the $i-$th cohomology group of complex $Hom(wedge^i g,a)$ with $a$ abelian lie algebra.

"If $g$ is semisimple, it is not too hard to see that $H^2(g,a)=0$. With a little supplementary argument, it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra. The same theorem and proof apply over $R$ and a structure theorem for lie groups drops out. Results for $H^1$ then furnishes a uniqueness theorem"

$textbfQ1:$ Maybe this is too obvious for others. I do not see why obviously $H^2(g,a)=0$. Fix $pi:gto End(a)$ representation. I am using Brian, Hall's Lie algebra's semisimple meaning reductive and trivial center. I could see all semi simples are decomposed into simples and this is unique upto reordering. So given an extension $0to ato hto gto 0$. Now $h=aoplus_pi g$ as semi-direct product but the lie bracket between $a,g$ will be twisted by representation of $gto End(a)$. I do not see why all extensions are equivalent to $0to ato aoplus_pi gto g$.

$textbfQ2:$ Why it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra?

$textbfQ3:$ Why the same theorem and proof apply over $R$ and a structure theorem for lie groups drops out? Over $R$, I no longer have unitary representation which may not allow me to have decomposition of $g$ into simple ones.

$textbfQ4:$ What are results for $H^1$ furnishing a uniqueness theorem? What is the meaning of uniqueness theorem here?

abstract-algebra representation-theory lie-algebras homological-algebra

share|cite|improve this question

asked Mar 10 at 19:58

user45765user45765

2,6702724

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The result in Q1 is called "Whitehead's second lemma"; that might be a useful keyword.
  $endgroup$
  – Qiaochu Yuan
  Mar 10 at 20:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $mathfraka$ is not just an abelian Lie algebra (the question would make no sense, or would assume implicitly that the action is trivial), but is also endowed with a $mathfrakg$-action.
  $endgroup$
  – YCor
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Q2: This is Levi's theorem. There is a short proof of it using Whitehead's second Lemma from above (which can be found in several books, but is also a nice exercise).
  $endgroup$
  – Dietrich Burde
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

This is a statement made in Knapp, Lie groups, Lie algebras, Cohomology Chpt 4 last paragraph of Sec 2.

$H^i(g,a)$ is the $i-$th cohomology group of complex $Hom(wedge^i g,a)$ with $a$ abelian lie algebra.

"If $g$ is semisimple, it is not too hard to see that $H^2(g,a)=0$. With a little supplementary argument, it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra. The same theorem and proof apply over $R$ and a structure theorem for lie groups drops out. Results for $H^1$ then furnishes a uniqueness theorem"

$textbfQ1:$ Maybe this is too obvious for others. I do not see why obviously $H^2(g,a)=0$. Fix $pi:gto End(a)$ representation. I am using Brian, Hall's Lie algebra's semisimple meaning reductive and trivial center. I could see all semi simples are decomposed into simples and this is unique upto reordering. So given an extension $0to ato hto gto 0$. Now $h=aoplus_pi g$ as semi-direct product but the lie bracket between $a,g$ will be twisted by representation of $gto End(a)$. I do not see why all extensions are equivalent to $0to ato aoplus_pi gto g$.

$textbfQ2:$ Why it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra?

$textbfQ3:$ Why the same theorem and proof apply over $R$ and a structure theorem for lie groups drops out? Over $R$, I no longer have unitary representation which may not allow me to have decomposition of $g$ into simple ones.

$textbfQ4:$ What are results for $H^1$ furnishing a uniqueness theorem? What is the meaning of uniqueness theorem here?

abstract-algebra representation-theory lie-algebras homological-algebra

share|cite|improve this question

asked Mar 10 at 19:58

user45765user45765

2,6702724

$endgroup$

share|cite|improve this question

asked Mar 10 at 19:58

user45765user45765

2,6702724

share|cite|improve this question

asked Mar 10 at 19:58

user45765user45765

2,6702724

asked Mar 10 at 19:58

user45765user45765

2,6702724

asked Mar 10 at 19:58

user45765user45765

2,6702724