Root space of lie algebra - understanding the definition

Question

I am taking a first course in "Semi-simple Lie algebras" at the moment. In the last lecture we were introduced to the concept of root decomposition, which I find extremely confusing. In the lecture we defined the following

Definition - Let $\mathfrak{g}$ be a complex semi-simple finite dimensional Lie algebra. A sub-algebra $\frak{h}\subseteq \frak{g}$ is called a toral sub-algebra if for all $h\in \frak{h}$ the operator $ad_{h} \in \frak{gl(g)}$ is diagonalizable. A toral sub algebra $\frak{h} \subseteq \frak{g}$ is called maximal toral sub algebra if it is not properly contained in any other toral subalgebra.

and we proved that every toral sub-algebra is abelian (i.e $\left[ \frak{h}, \frak{h} \right] =\{0\}$). Then my professor said the following things:

1. $\left[ \mathfrak{h}, \mathfrak{h} \right] =\{0\}$ therefore $\left[ ad(\mathfrak{h}), ad(\mathfrak{h}) \right] =\{0\}$.
2. $ad(\mathfrak{h}) \subseteq \mathfrak{gl(g)}$ is an abelian Lie algebra of diagonalizable operators on $\mathfrak{g}$, therefore it is simultaneously diagonalizable. This means that there is a basis $\mathcal{B} \subseteq \mathfrak{g}$ consisting of common eigenvectors for all $ad(\mathfrak{h})$.
3. Notice that every common eigenvector $\xi \in \mathfrak{g}$ defines a functional $ \alpha \in \mathfrak{h}^{\ast}$ by $$ad_{h}(\xi) = \left[h, \xi \right] = \overbrace{\alpha(h)}^{\in \mathbb{C}} \cdot \xi$$ where $\alpha(h)$ is the eigenvalue of $ad_{h}$ corresponding to the common eigenvector $\xi$ (which obviously depends on the choice of $h\in \mathfrak{h}$).

Up to this point, I feel that I understand and agree with everything that has been said. The problem is that then out of the blue he said: "therefore there exists a finite set $\Phi \subseteq \left(\mathfrak{h}^{\ast} \backslash \{0\} \right)$ such that $$\left(\bigstar \right) \;\;\;\; \mathfrak{g} \cong \mathfrak{g}_{0} \oplus \left( \bigoplus_{\alpha \in \Phi} \mathfrak{g}_{\alpha}\right)$$ where $\mathfrak{g}_{\alpha}$ are defined as follows $$\mathfrak{g}_{\alpha} := \left\{ x\in \mathfrak{g} \mid \forall h\in \mathfrak{h} \left( ad_{h}(x) = [h,x] = \alpha (h) x \right) \right\}$$ Then he said (defined) that the functionals $\alpha \in \Phi$ are called roots of $\mathfrak{g}$ with respect to $\mathfrak{h}$ and, $\mathfrak{g}_{\alpha}$ are called the root-spaces.
There are two things that I aim to understand with this question:

1. The definition of a root-space (for some reason I am not able to comprehend the fact that $\alpha$ can vary for different $h_{1},h_{2} \in \mathfrak{h}$). In particular I don't understand what is the purpose of this definition in the sense that, how can a root space be anything but the whole of $\frak{g}$?
2. Why $\left(\bigstar \right)$ is true.

Thanks in advance for any insightful comments.

Answer 1

Consider, for instance, $\mathfrak g=\mathfrak{sl}(2,\Bbb C)$ and$$\mathfrak h=\left\{\begin{bmatrix}x&0\\0&-x\end{bmatrix}\,\middle|\,x\in\Bbb C\right\}.$$Let $H=\left[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\right]$ (and note that $\mathfrak h=\Bbb CH$), $X=\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$, and $Y=\left[\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right]$. It is easy to check that $[H,X]=2X$, that $[H,Y]=-2Y$ and that $[X,Y]=H$. But then, if $\alpha\colon\mathfrak h\longrightarrow\Bbb C$ is the only linear form such that $\alpha(H)=2$, then\begin{align}\mathfrak{sl}(2,\Bbb C)_\alpha&=\{Z\in\mathfrak{sl}(2,\Bbb C)\mid(\forall W\in\mathfrak h):[W,Z]=\alpha(W)Z\}\\&=\{Z\in\mathfrak{sl}(2,\Bbb C)\mid[H,Z]=2Z\}\text{ (since $\mathfrak h=\Bbb CH$)}\\&=\Bbb CX\end{align}and, by a similar argument, $\mathfrak{sl}(2,\Bbb C)_{-\alpha}=\Bbb CY$. Finally,\begin{align}\mathfrak{sl}(2,\Bbb C)_0&=\{Z\in\mathfrak{sl}(2,\Bbb C)\mid(\forall W\in\mathfrak h):[W,Z]=0\}\\&=\{Z\in\mathfrak{sl}(2,\Bbb C)|[Z,H]=0\}\\&=\Bbb CH.\end{align}So, clearly we have $\mathfrak{sl}(2,\Bbb C)=\mathfrak{sl}(2,\Bbb C)_\alpha\oplus\mathfrak{sl}(2,\Bbb C)_0\oplus\mathfrak{sl}(2,\Bbb C)_{-\alpha}$.

I hope that this clarifies the meaning of the root spaces.

Finally, we have $\mathfrak g=\mathfrak g_0\oplus\left(\bigoplus_{\alpha\in\Phi}\mathfrak g_\alpha\right)$ because, for each $h\in\mathfrak h$, $\operatorname{ad}(h)$ is diagonalizable and because if you have a family of diagonalizable endomorphisms of a vector space such that each two of them commute, then they are simultaneously diagonalizable.

Answer 2

The eigenvalue does not depend on the vector within a root space. However, it depends (linearly) on the element of the torus. In other words, it only depends on the operator.

Say, if $h_1,h_2,\ldots,h_n$ span the maximal torus, and $x$ is a simultaneous eigenvector belonging to eigenvalue $\lambda_i$ of $ad(h_i)$, then for an arbitrary element of the torus $$h=a_1h_1+\cdots+a_nh_n$$ $x$ is an eigenvector of $ad(h)$ belonging to the eigenvalue $\lambda=a_1\lambda_1+a_2\lambda_2+\cdots+a_n\lambda_n$. In such a case the root is the linear function

$$\alpha:\mathfrak{h}\to\Bbb{C}, h\mapsto a_1\lambda_1+a_2\lambda_2+\cdots+a_n\lambda_n.$$

The number of distinct roots is finite, because each and every one of the operators $ad(h_i)\in\mathfrak{gl}(\mathfrak{g})$ can only have finitely many eigenvalues. As always, the linear transformation $\alpha$ is uniquely determined if we know its values on a basis, here the set of numbers $\lambda_i=\alpha(h_i), i=1,2,\ldots,n$.

The collection of vectors $x\in\mathfrak{g}$ such that $ad(h_i)(x)=\lambda_ix$ for all $i=1,2,\ldots,n$ is easily seen to be a subspace of $\mathfrak{g}$. Use the subspace criterion to verify this if you don't see it right away. Those subspaces are the root spaces (assuming at least one of the $\lambda_i$s is non-zero).