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I am trying to understand the following argument in Humphreys

Arguement

I am good with the start of this proof and begin to get shakey here.

(1) $ad_T y(x)=-ay$ is itself an eigenvector of $ad_Ty$, of eigenvalue zero

I think this is because ad_t(y)(ad_T y(x))=ad_T y(-ay)=-a[y,y]=0.

(2) On the other hand, we can write $x$ as a linear combination of eigenvectors of $ad_T y$

I don't understand how why this is true. I know that $x\in T$ so $x$ is a sum of semisimple elements.

(3) after applying $ad_T y$ to $x$, al that is left is a combination of eigenvectors which belong to nonzero eigenvalues

Not understanding the previous step makes this hard to understand. Is this saying all the eigenvalues must be nonzero? or just that there must be some that are non zero? In eithercase why?

This contracdicts the previous conclusion

I do not see the contradiction but this is likey since I have not understood the previous two steps.

Could you please expand on (1), (2), and (3) ideally with comment on my confusions?

This is not a duplicate of this question as we are dealing with different aspects.

Mark Murray
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    For (2) Since y is semisimple $ad_T(y)$ is diagonalizable, i.e. there exists a basis of eigenvectors. So you can write any element as linear combination of these eigenvectors. Then (3) is just by definition of eigenvectors. – laura_b Jul 24 '20 at 13:10
  • Could you expand a little on being diagonalizable and there existing a basis of eigenvectors, (my linear algebra is very bad), is this saying that we will have a basis of eigenvectors for $T$ or is it that we have a basis of eigenvectors for $L$? I assume the correct answer is $T$ but I do not see why. – Mark Murray Jul 24 '20 at 13:26
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    As long as $ad_T(y):L\to L$ is diagonalizable, there exists a basis of eigenvectors on $L$. – laura_b Jul 24 '20 at 13:42

1 Answers1

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Since the field is algebraically closed, if $c_1,...,c_p$ are the eigenvalues of the characteristic polynomial of $ad_Ty$, $T=\sum Ker(ad_Ty-c_i)^{n_i}$, this implies that $x=a_1u_1+...+a_nu_n$ where $ad_Ty(u_i)=b_iu_i$

This implies that $[y,x]=-ay=a_1[y,u_1]+..+a_n[y,u_n]=a_1b_1u_1+..+a_nb_nu_n$, $-ay=a_{i_1}b_{i_1}u_{i_1}+..+a_{i_l}b_{i_l}u_{i_l}$ where $[y,u_{i_j}]\neq 0$

we deduce that $[y,-ay]=a_{i_1}b_{i_1}[y,u_{i_1}]+..+a_{i_l}b_{i_l}[y,u_{i_l}]=a_{i_1}b_{i_1}b_{i_1}u_{i_1}+..+a_{i_l}b_{i_l}b_{i_l}u_{i_l}=0$ this implies that $a_{i_j}b_{i_j}=0$ and $a_{i_j}=0$ since $[y,u_{i_j}]=b_{i_j}u_{i_j}\neq 0$.

This implies $x=a_1u_1+..+a_nu_n, a_i=0,$ if $[y,u_i]\neq 0 $ and $x\in kerad_Ty$ contradiction.

Tsemo Aristide
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