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In Humphreys, given a finite dimensional semisimple Lie algebra $L$ and a maximal toral subalgebra $H$, $$L_\alpha := \{x\in L|[hx] = \alpha(h)x\;\forall h\in H\}$$ Then since $ad_L\;H$ is a commuting family of semisimple endormorphisms of $L$ thus can be simultaneously diagonalized. Thus $$L = \bigoplus_{\alpha \in H^*} L_\alpha$$where only finitely many summands are nonzero.

Later we see in fact $L_0 = C_L(H) = H$, and let $\Phi$ be the set of $\alpha$ where $LK_\alpha\ne0$, then $$L = H\bigoplus_{\alpha\in \Phi} L_\alpha$$

So far so good, but I'm confused about a particular set of $\{f_i\}\subset H^*$ that I think will do the job. Fix a basis $\{e_1,\cdots,e_n\}$ of $L$ where $ad_L\;H$ are simultaneously diagonalized. Define $f_i\in H^*$ to be the function that takes $h\in H$ and returns the $i$th diagonal entry of the matrix form of $ad\;h$. Then I think $L = \bigoplus_{i =1}^n L_{f_i}$ will be enough because each $L_{f_i}$ contains the subspace spanned by the $i$th basis vector. (It's possible that some of the $f_i$ concide, then we only count it in the sum once).

Why is this not good? Also, I know $H\subset L_0$, but can't see how $L_0$ sits in $L$ directly in this decomposition $L = \bigoplus_{i =1}^n L_{f_i}$.

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    $\begingroup$ Yeah. You do get such a decomposition. If you calculate an example you will see that some of the $f_i$ vanish identically. The interesting game is about finding the sets of the non-zero functions $f_i$ that arise and their relations. $\endgroup$ Jan 14, 2014 at 6:50
  • $\begingroup$ Well, you will see that $L_\alpha$ is 1-dimensional whenever $\alpha\neq0$. OTOH $L_0=H$ often has higher dimension. $\endgroup$ Jan 14, 2014 at 11:44
  • $\begingroup$ @JyrkiLahtonen Since some $f_i$ vanish identically, that means there is a component where all $ad\;h$ are 0, plus $C_L(H) = H$, which means for any basis of $H$, it can be extended into a basis of $L$ for which $ad\;H$ simultaneously diagonalizes, correct? $\endgroup$
    – mez
    Jan 15, 2014 at 12:40

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You do have that decomposition, because you just said ``Let's take a basis of eigenvectors and for each eigenvector consider the eigenspace that contains it''. The problem is that this does not give you any explicit information about the algebra itself. You can see $L_0$ inside, if you know what $e_i$ lie in $H$. Just look at $L_{f_i}$ in that case.

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  • $\begingroup$ I'm not saying $\alpha\in \Phi$ must be equal to one of $f_i$, I am saying $\bigoplus_i L_{f_i}$ seems to me a decomposition of $L$, because it contains all the basis vectors of $L$ for the particular basis that makes $ad\;H$ simultaneously diagonalized. $\endgroup$
    – mez
    Jan 14, 2014 at 1:32
  • $\begingroup$ Changed the answer. Maybe it is helpful now $\endgroup$
    – user68061
    Jan 14, 2014 at 5:51

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