Results 1 to 3 of 3

Math Help - lie algebra

  1. #1
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    lie algebra

    Who can solve this question?

    1. Let L be a complex Lie algebra . Show that L is nilpotent if and only if every 2- dimensional subalgebra of L is abelian. ( use the second version of engel's Theorem.)

    I can not get resolved
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Re: lie algebra

    Quote Originally Posted by vernal View Post
    Who can solve this question?

    1. Let L be a complex Lie algebra . Show that L is nilpotent if and only if every 2- dimensional subalgebra of L is abelian. ( use the second version of engel's Theorem.)

    I can not get resolved
    you should post questions like this in the algebra section if you want to get a faster response.

    anyway, suppose first that every 2-dimensional subalgebra of L is abelian and let a \in L. we only need to show that \text{ad}(a) is nilpotent because then we will be done by Engel's theoem. so suppose that \lambda \in \mathbb{C} is an eigenvalue of \text{ad}(a) and x \in L is an eigenvector corresponding to \lambda. then [a,x] = \lambda x. so the subalgebra generated by \{a,x\} would be at most 2-dimensional and thus, by our hypothesis, abelian (note that 1-dimensional algebras are always abelian). hence \lambda x = [a,x]= 0, which implies \lambda = 0. so every eigenvalue of \text{ad}(a) is zero and therefore \text{ad}(a) is nilpotent.
    conversely, suppose that L is nilpotent. so L^{n+1} = 0, for some integer n \geq 1. now, let A be a subalgebra of L and \dim_{\mathbb{C}} A = 2. suppose that A is not abelian. then there exist a,b  \in A such that [a,b] = \lambda a + \mu b \neq 0, for some \lambda, \mu \in \mathbb{C} and we may assume that \lambda \neq 0. so [[a,b],b] = \lambda [a,b] and [[[a,b],b],b] = \lambda^2[a,b] and, if we continue this process, we will eventually get \lambda^n[a,b] = 0 because L^{n+1}=0. but \lambda \neq 0 and [a,b ] \neq 0 and so \lambda^n [a,b] \neq 0. this contradiction proves that A is abelian.
    Last edited by NonCommAlg; December 18th 2011 at 02:37 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    Re: lie algebra

    Quote Originally Posted by NonCommAlg View Post
    you should post questions like this in the algebra section if you want to get a fater response.

    anyway, suppose first that every 2-dimesnional subalgebra of L is abelian and let a \in L. we only need to show that \text{ad}(a) is nilpotent because then we will be done by Engel's theoem. so suppose that x \in L is an eigenvector of L. then there exists and \lambda \in \mathbb{C} such that [a,x] = \lambda x. but then the subalgebra generated by \{a,x\} would be at most 2-dimensional and thus, by our hypothesis, abelian (note that 1-dimensional algebras are always abelian). hence \lambda x = [a,x]= 0, which implies \lambda = 0. so every eigenvalue of \text{ad}(a) is zero and therefore \text{ad}(a) is nilpotent.
    conversely, suppose that L is nilpotent. so L^{n+1} = 0, for some integer n \geq 1. now, let A be a subalgebra of L and \dim_{\mathbb{C}} A = 2. suppose that A is not abelian. then there exist a,b  \in A such that [a,b] = \lambda a + \mu b \neq 0, for some \lambda, \mu \in \mathbb{C} and we may assume that \lambda \neq 0. so [[a,b],b] = \lambda [a,b] and [[[a,b],b],b] = \lambda^2[a,b] and, if we continue this process, we will eventually get \lambda^n[a,b] = 0 because L^{n+1}=0. but \lambda \neq 0 and [a,b ] \neq 0 and so \lambda^n [a,b] \neq 0. this contradiction proves that A is abelian.
    Thank you very much..... tanks tanks tanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: February 4th 2011, 09:39 AM
  2. Replies: 2
    Last Post: December 6th 2010, 04:03 PM
  3. Algebra or Algebra 2 Equation Help Please?
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 12th 2010, 12:22 PM
  4. Replies: 0
    Last Post: April 24th 2010, 12:37 AM
  5. algebra help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 20th 2007, 11:33 AM

Search Tags


/mathhelpforum @mathhelpforum