page2069 Lie Algebra Minute Matthias Lorentzen...mattegrisenforlag.com
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Lie Algebra Minute
Topic: `"The Kernel of" \ ad_f`.
Introduction: Every adjoint map ad_x in a Lie Algebra is a singular (non-invertible) linear transformation because x is always in the kernel. Now we complete the trio by examining `ad_f` `(x)= [f , x]`. Since we know that the Kernel is the set of vectors that "collapse" to zero , finding the kernel of `ad_f` will tell us which "twists" this operation ignores.
Exercise: Using the structure constants of `sl (2 , RR)` , construct the matrix for `ad_f` and identify its kernel.
Solution: We have the basis `beta = {e , f , h}` and the structure constants: `[h , e] = 2e` , `[h , f] = -2f` , `[h , h] = 0` , `[e , f] = h`.
1. Column 1: `ad_f \ (e) = [f , e] = -[e , f] = -h`.
- Coordinate: `([0] , [0] , [-1])`.
2. Column 2: `ad_f \ (f) = [f , f] = 0`.
- Coordinate: `([0] , [0] , [0])`.
3. Column 3: `ad_f \ (h) = [f , h] = -[h , f] = -(-2f) = 2f`.
- Coordinate: `([0] , [2] , [0])`.
Result: `Matrix \ M_f = ([0 , 0 , 0] , [0 , 0 , 2] , [-1 , 0 , 0]) `.
QED
The Kernel: Looking at the second column , we see that `ad_f \ (f) = 0`. Therefore , the vector f (or the coordinate `(0 , 1 , 0)`) is the basis of the Kernel of `ad_f`.
Question
We found that `e in ker(ad_e)` and `f in ker(ad_f)`. In the language of the vector Vector Space exercise , this means both `ad_e` and `ad_f` are "collapsing" maps. Based on the matrix `M_f` above , is `ad_f` an isomorphism of the vector space `sl(2 , RR)`
?
A) Yes , because it still maps three basis vectors in the target space.
B) No , because the kernel contains more than just the zero vector (it contains all multiples of `f`) , which means the map is not injective.
