Root space decomposition of a semisimple Lie algebra.

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.

• The most difficult item to show is $\dim L_{\lambda_i}=1$ Feb 28, 2023 at 12:10
• @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? Feb 28, 2023 at 12:43
• 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. Feb 28, 2023 at 12:57
• Check out Dietrich Burde's answer here, it might help you. Feb 28, 2023 at 13:05
• @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. Feb 28, 2023 at 13:47

1 Answer

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.

• 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 $\alpha \in R.$ Feb 28, 2023 at 19:52
• 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. Feb 28, 2023 at 20:20
• @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 Mar 2, 2023 at 15:51
• 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. Mar 3, 2023 at 2:28
• @AnilBagchi. Did you read my comment? Mar 3, 2023 at 15:51