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Let $L$ be a semisimple Lie algebra. I am trying to understand root space decomposition of $L$ on my own. Since $L$ is semisimple, $L$ possesses an abelian maximal toral subalgebra i.e. an abelian subalgebra $H$ which is $\text {ad}$-semisimple, known as Cartan subalgebra. But then $\text {ad} (H)$ is a commuting family of semisimple operators on $L$ and hence they are simultaneously diagonalizable. So there exists a basis $\{e_1, \cdots, e_n \}$ of $L$ and $\lambda_i \in H^{\ast}$ corresponding to each basis element $e_i$ such that $$[h, e_i] = \lambda_i (h) e_i$$ for all $h \in H.$ Define $$L_{\lambda_i} : = \left \{x \in L\ |\ [h, x] = \lambda_i (h) x\ \text {for all}\ h \in H \right \}.$$

Then it's clear that $L = \sum\limits_{i = 1}^{n} L_{\lambda_i}.$ But I can't see why the sum is direct. First of all how can conclude that all the $\lambda_i$'s are distinct? Because if for some $i \neq j$ we have $\lambda_i = \lambda_j$ then clearly $L_{\lambda_i} = L_{\lambda_j}.$ But then the sum won't be direct. So at first we have to somehow show that all $\lambda_i$'s are distinct. If they are distinct they are all one dimensional.

In particular, if the sum is direct then $H$ is also one-dimensional. Is it always the case?

Could anyone please answer the questions? Also please let me know where I am going wrong if there is any.

Thanks for your time.

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  • $\begingroup$ The most difficult item to show is $\dim L_{\lambda_i}=1$ $\endgroup$
    – kabenyuk
    Feb 28, 2023 at 12:10
  • $\begingroup$ @kabenyuk$:$ How to show that the sum is direct or equivalrntly $\dim L_{\lambda_i} = 1$ for all $i\ $? Also there exists $i \in \{1, \cdots, n \}$ such that $\lambda_i = 0$ which correspond to the centralizer of $H$ in $L$ and it is well known that this centralizer is $H$ itself. So if the sum is direct doesn't it imply that $H$ has to be one dimensional? $\endgroup$ Feb 28, 2023 at 12:43
  • $\begingroup$ I didn't write, but $L_0=H$ and the dimension of $H$ can be as large as you like. The proof that $\dim L_\lambda=1$ for non-zero $\lambda$ uses some facts about representations of the algebra $sl_2$. It is written in many books. I don't know which book you are reading. $\endgroup$
    – kabenyuk
    Feb 28, 2023 at 12:57
  • $\begingroup$ Check out Dietrich Burde's answer here, it might help you. $\endgroup$
    – kabenyuk
    Feb 28, 2023 at 13:05
  • $\begingroup$ @kabenyuk$:$ In the linked answer the author wrote that $\sum\limits_{i = 1}^{n} L_{\lambda_i}$ is direct because the eigenvectors in different eigenspaces are linearly independent. Here $L_{\lambda_i}$ is not like an eigenspace as eigenvalues keep on changing as we vary the elements of $H.$ I don't understand what exactly he tried to mean. $\endgroup$ Feb 28, 2023 at 13:47

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With your definitions / setup / notations, the $\lambda_i$ are not yet distinct in general: Because if we choose the basis $e_1, ..., e_n$ so that the first $r$ elements $e_1, ... e_r$ are a basis of $H$, which is its own $0$-space a.k.a. centralizer, then $\lambda_1 = ... = \lambda_r = 0 \in H^*$.

The correct way to define the root space decomposition starting from your notation is to define $R:= \{\lambda_i: 1\le i \le n\} \color{red}{\setminus \{0\}}$, and write

$$L = L_0 \oplus \bigoplus_{\alpha \in R} L_{\alpha}$$

Note that I'm kind of cheating here because I have now made the $\alpha$ mutually distinct by definition. From there, the standard way to proceed is to note $H=L_0$ and then show that each of the "proper" root spaces $L_\alpha$ ($\alpha \in R)$ has dimension $1$ -- which is more involved than one might think, cf. Why are root spaces of root decomposition of semisimple Lie algebra 1 dimensional? and When is the Lie bracket of two root spaces nonzero? including links from there, also discussion in https://math.stackexchange.com/a/4583911/96384.

Once one has that, of course it is clear in hindsight that the $\color{red}{\text{nonzero}}$ $\lambda_i$, i.e. in my above notation the $\lambda_i$ with $r+1 \le i \le n$, were mutually distinct.

After that comes the real fun in showing that $R$ is a root system in a natural way.

I also advertise my own answer here for better understanding of the root space decomposition by seeing it in a basic but telling matrix example. That should clear things up.

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  • $\begingroup$ How do you know that the sum $L_0 + \sum\limits_{\alpha \in R} L_{\alpha}$ is direct? How do you even know that $L_{\alpha}$'s are all distinct for $\alpha \in R.$ $\endgroup$ Feb 28, 2023 at 19:52
  • $\begingroup$ In order to show that the sum is direct I think it's enough to show that the elements of $L_{\alpha}$'s are linearly independent. If for any $\alpha \neq (0)$ we can show that either $L_{\alpha} = (0)$ or $1$-dimensional then that would imply the sum is indeed direct. $\endgroup$ Feb 28, 2023 at 20:20
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    $\begingroup$ @AnilBagchi. You don't want to prove 1-dimensional before you see the sum is direct. The sum is direct because that is what it means for $\mathfrak{h}$ to be simultaneously diagonalisable $\endgroup$
    – Callum
    Mar 2, 2023 at 15:51
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    $\begingroup$ Exactly. Anil, if you already have your basis $\{h_1, ..., h_m, e_1, ..., e_{d-m} \}$, then order it further so that $e_1, ..., e_{i_1}$ belong to (and hence span) $L_{\alpha_1}$, $e_{i_1+1}, ..., e_{i_2}$ belong to (hence span) $L_{\alpha_2}$ etc. Then it is obvious that the sum of those $L_\alpha$ is direct, they are disjoint spans of mutually disjoint subsets of your basis vectors. And yes, in hindsight, once you have proved they are each one-dimensional, that means $i_j = j$ for all $1 \le j \le d-m$, there are $d-m$ roots, and the root system is as you describe. $\endgroup$ Mar 3, 2023 at 2:28
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    $\begingroup$ @AnilBagchi. Did you read my comment? $\endgroup$ Mar 3, 2023 at 15:51

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