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You can find below the proof due to G.M; Tuynman, of the third Lie theorem.

The proof is similar to using Ado's theorem, but requires an 'advanced' result:

the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

http://ifile.it/hy0q139

I posted a related question in math.stackexchange.com

http://math.stackexchange.com/questions/56899/elementary-proof-of-the-third-lie-theoremhttps://math.stackexchange.com/questions/56899/elementary-proof-of-the-third-lie-theorem

You can find below the proof due to G.M; Tuynman, of the third Lie theorem.

The proof is similar to using Ado's theorem, but requires an 'advanced' result:

the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

http://ifile.it/hy0q139

I posted a related question in math.stackexchange.com

http://math.stackexchange.com/questions/56899/elementary-proof-of-the-third-lie-theorem

You can find below the proof due to G.M; Tuynman, of the third Lie theorem.

The proof is similar to using Ado's theorem, but requires an 'advanced' result:

the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

http://ifile.it/hy0q139

I posted a related question in math.stackexchange.com

https://math.stackexchange.com/questions/56899/elementary-proof-of-the-third-lie-theorem

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amine
amine
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You can find below the proof due to G.M; Tuynman, of the third Lie theorem.

The proof is similar to using theAdo's theorem of Ado, but requires an 'advanced' result:

the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

http://ifile.it/hy0q139

In the Godbillon book, the Künneth theorem that identifies the cohomology of product manifolds with the tensor product of their cohomologies: $$K : H(M)\otimes H(N)\rightarrow H(M\times N)$$ If $M$ is compact, $K$ is an isomorphism. There are two things I can not understand, among others:posted a related question in math.stackexchange.com

  1. Godbillon consider $d=K^{-1}\circ\mu^\star$, where $\mu$ is the multiplication on the Lie group $G$, and $K$ is invertible, the compactness is it essential for $K$ to be an isomorphism?!

  2. Godbillon theorem shows in his page 202, with $d$, if $G$ is connected then the smallest integer $q>0$ such that $H^q(G)\neq 0$ is odd. First, $G$ must be compact? Why $H^1(G)=0$? This is not a consequence of the above theorem where compactness is essential!

http://math.stackexchange.com/questions/56899/elementary-proof-of-the-third-lie-theorem

You can find below the proof due to G.M; Tuynman, of the third Lie theorem.

The proof is similar to using the theorem of Ado, but requires an 'advanced' result:

the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

http://ifile.it/hy0q139

In the Godbillon book, the Künneth theorem that identifies the cohomology of product manifolds with the tensor product of their cohomologies: $$K : H(M)\otimes H(N)\rightarrow H(M\times N)$$ If $M$ is compact, $K$ is an isomorphism. There are two things I can not understand, among others:

  1. Godbillon consider $d=K^{-1}\circ\mu^\star$, where $\mu$ is the multiplication on the Lie group $G$, and $K$ is invertible, the compactness is it essential for $K$ to be an isomorphism?!

  2. Godbillon theorem shows in his page 202, with $d$, if $G$ is connected then the smallest integer $q>0$ such that $H^q(G)\neq 0$ is odd. First, $G$ must be compact? Why $H^1(G)=0$? This is not a consequence of the above theorem where compactness is essential!

You can find below the proof due to G.M; Tuynman, of the third Lie theorem.

The proof is similar to using Ado's theorem, but requires an 'advanced' result:

the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

http://ifile.it/hy0q139

I posted a related question in math.stackexchange.com

http://math.stackexchange.com/questions/56899/elementary-proof-of-the-third-lie-theorem

Source Link
amine
amine
  • 513
  • 5
  • 14

You can find below the proof due to G.M; Tuynman, of the third Lie theorem.

The proof is similar to using the theorem of Ado, but requires an 'advanced' result:

the fact that for a simply connected Lie group $G$ not only the first de Rham cohomology space $H^1(G)=\{0\}$ but also $H^2(G)=\{0\}$.

http://ifile.it/hy0q139

In the Godbillon book, the Künneth theorem that identifies the cohomology of product manifolds with the tensor product of their cohomologies: $$K : H(M)\otimes H(N)\rightarrow H(M\times N)$$ If $M$ is compact, $K$ is an isomorphism. There are two things I can not understand, among others:

  1. Godbillon consider $d=K^{-1}\circ\mu^\star$, where $\mu$ is the multiplication on the Lie group $G$, and $K$ is invertible, the compactness is it essential for $K$ to be an isomorphism?!

  2. Godbillon theorem shows in his page 202, with $d$, if $G$ is connected then the smallest integer $q>0$ such that $H^q(G)\neq 0$ is odd. First, $G$ must be compact? Why $H^1(G)=0$? This is not a consequence of the above theorem where compactness is essential!