22a - Lecture 22 Theorem 1(Engel If each element of g is...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Lecture 22 Theorem 1. (Engel) If each element of g is ad-nilpotent, then g is nilpotent. Proof of a particular case : Suppose each x g satisfies ad 2 x = 0. That is, for all x, y g , [ x, [ x, y ]] = 0. Claim g 2 = 0, provided char F 6 = 3. That is, [ x, [ y, z ]] = 0 for all x, y, z g . Given w, z g = [ z, [ z, w ]] = 0. Replace z by x + y . 0 = [ x + y, [ x + y, w ]] = [ x, [ x, w ]] + [ x, [ y, w ]] + [ y, [ x, w ]] + [ y, [ y, w ]] = 0 + [ x, [ y, w ]] + [ y, [ x, w ]] + 0 [ x, [ y, w ]] is skew symmetric in x, y Now by Jacobi, 0 = [ w, [ x, y ]] + [ x, [ y, w ]] + [ y, [ w, x ]] = [ w, [ x, y ]] - [ x, [ w, y ]] - [ w, [ y, x ]] = [ w, [ x, y ]] + [ w, [ x, y ]] + [ w, [ x, y ]] = 3[ w, [ x, y ]] g 2 = 0 provided char F 6 = 3. Remark : In this case it turns out that g 3 = 0. To prove Engel, we use Theorem 2. Let V be a finite dimensional vector space and g gl ( V ) a subalgebra. If each element of g is nilpotent as a linear map V V , then T z g Ker ( z ) 6 = 0 . 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This theorem will be proved later. Proof of Theorem = Engel : Suppose each element of g is ad-nilpotent. Apply the theroem to ad ( g ) gl ( g ), where ad ( g ) = { ad z : z g } . By the theorem, some z 6 = 0 lies in the kernel of each ad w so that Z ( g ) 3 z is nonzero. Now let g /Z ( g
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern