In fact, $\U0001d525={\U0001d524}_{0}$ .
sl(2,C) subalgebras, 1
Review of SU(3)
When we studied the representation theory of $G=SU(3)$ (or of its Lie algebra $\U0001d524=\U0001d530\U0001d529(3,\mathbf{C})$ , one of the key pictures we studied was the root diagram: the weight diagram of its adjoint representation. This was a hexagonal configuration of roots at ${L}_{i}{L}_{j}$ with $i,j\in \{1,2,3\}$ , $i\ne j$ .
We'll write ${\U0001d524}_{{L}_{i}{L}_{j}}=\mathbf{C}\cdot {E}_{ij}$ for the root space corresponding to the root ${L}_{i}{L}_{j}$ and ${\U0001d524}_{0}$ for the weight space corresponding to $0$ . Recall that $${\U0001d524}_{0}=\mathbf{C}\cdot {H}_{13}\oplus \mathbf{C}\cdot {H}_{23}=\{\left(\begin{array}{ccc}\hfill {\theta}_{1}\hfill & \hfill \hfill & \hfill \hfill \\ \hfill \hfill & \hfill {\theta}_{2}\hfill & \hfill \hfill \\ \hfill \hfill & \hfill \hfill & \hfill {\theta}_{1}{\theta}_{2}\hfill \end{array}\right):{\theta}_{1},{\theta}_{2}\in \mathbf{C}\}.$$
This Lie algebra has the nice property that if we pick one root ${L}_{i}{L}_{j}$ and its opposite root ${L}_{j}{L}_{i}$ ), and we pick generators ${E}_{ij}\in {\U0001d524}_{{L}_{i}{L}_{j}}$ and ${E}_{ji}\in {\U0001d524}_{{L}_{j}{L}_{i}}$ then the three elements $${E}_{ij},{E}_{ji},{H}_{ij}=[{E}_{ij},{E}_{ji}]$$ span a Lie subalgebra isomorphic to $\U0001d530\U0001d529(2,\mathbf{C})$ . Our study of the structure of $\U0001d530\U0001d529(3,\mathbf{C})$ representations used the three $\U0001d530\U0001d529(2,\mathbf{C})$ subalgebras we obtained this way: for example, it gave us the Weyl symmetry group. This will generalise.
Roots in general
The setup
Suppose $K$ is a compact matrix group with Lie algebra $\U0001d528$ . Let's write $\U0001d524$ for the complexification $\U0001d528\otimes \mathbf{C}$ . Inside $K$ we have a maximal torus $T$ with Lie algebra $\U0001d531$ . Let's write $\U0001d525=\U0001d531\otimes \mathbf{C}\subset \U0001d524$ . The adjoint representation is a map $\mathrm{Ad}:K\to GL(\U0001d528)$ , defined by $$g\mapsto {\mathrm{Ad}}_{g},{\mathrm{Ad}}_{g}(X)=gX{g}^{1}.$$ The same formula defines a complex representation, also written $\mathrm{Ad}:K\to GL(\U0001d524)$ , if we allow $X$ to live in $\U0001d524$ . By taking the derivative we get the Lie algebra representation $$\mathrm{ad}:\U0001d528\to \U0001d524\U0001d529(\U0001d524)$$ and its complexification $${\mathrm{ad}}^{\mathbf{C}}:\U0001d524\to \U0001d524\U0001d529(\U0001d524).$$
The root diagram
Because we have a maximal torus, we get a weight space decomposition $$\U0001d524=\oplus {\U0001d524}_{\alpha}$$ where $${\U0001d524}_{\alpha}=\{X\in \U0001d524:{\mathrm{ad}}_{H}X=\alpha (H)X\forall H\in \U0001d525\}.$$ The direct sum is happening over a finite set of weights. Which weights occur in this direct sum?

$\alpha =0$ occurs, in other words ${\U0001d524}_{0}\ne 0$ . This is because $\U0001d525$ is an abelian Lie algebra, so ${\mathrm{ad}}_{H}({H}^{\prime})=0$ for all $H,{H}^{\prime}\in \U0001d525$ . Therefore $\U0001d525\subset {\U0001d524}_{0}$ (just like in the $SU(3)$ example).
Suppose $Z\in {\U0001d524}_{0}$ , i.e. $Z\in \U0001d524$ and ${\mathrm{ad}}_{H}Z=0$ for all $H\in \U0001d525$ . Then $[H,Z]=0$ for all $H\in \U0001d525$ . In particular, $[H,Z]=0$ for all $H\in \U0001d531\subset \U0001d525$ .
Let's write $Z=X+iY$ where $X,Y\in \U0001d528$ (so they are the real and imaginary parts of $Z$ ). Then $0=[H,Z]=[H,X]+i[H,Y]$ for all $H\in \U0001d531$ , therefore both real and imaginary parts must vanish, and $[H,X]=[H,Y]=0$ . This means that $X,Y\in \U0001d528$ commute with all elements in $\U0001d531$ .
If either $X$ or $Y$ is not contained in $\U0001d531$ then either $\U0001d531\oplus \mathbf{C}\cdot X$ or $\U0001d531\oplus \mathbf{C}\cdot Y$ will be an abelian subalgebra ${\U0001d531}^{\prime}$ strictly containing $\U0001d531$ . Then $\overline{\mathrm{exp}({\U0001d531}^{\prime})}$ is a torus strictly containing $T$ , which contradicts the assumption that $T$ is a maximal torus.
Which other weights occur?
Any nonzero weight of the adjoint representation is called a
Action of root vectors

If $X\in {\U0001d524}_{\alpha}$ and $Y\in {\U0001d524}_{\beta}$ then $[X,Y]\in {\U0001d524}_{\alpha +\beta}$ . (We won't prove this: it's an exercise, very similar to the corresponding results for $\U0001d530\U0001d529(2,\mathbf{C})$ and $\U0001d530\U0001d529(3,\mathbf{C})$ , like "$X$ moves things to the right and $Y$ moves things to the left".)

If $X\in {\U0001d524}_{\alpha}$ and $Y\in {\U0001d524}_{\beta}$ then $K(X,Y)=0$ unless $\alpha +\beta =0$ .

If $\U0001d524$ is semisimple then $\alpha \in R$ if and only if $\alpha \in R$ .
(2). Pick a basis of $\U0001d524$ consisting of root vectors. We'll compute the matrix of ${\mathrm{ad}}_{X}{\mathrm{ad}}_{Y}$ with respect to this basis and then take the trace to find $K(X,Y)$ . Suppose $Z\in {\U0001d524}_{\gamma}$ is one of our basis vectors. Where does $Z$ go under ${\mathrm{ad}}_{X}{\mathrm{ad}}_{Y}$ ? By part (a), ${\mathrm{ad}}_{X}{\mathrm{ad}}_{Y}(Z)=[X,[Y,Z]]\in {\U0001d524}_{\alpha +\beta +\gamma}$ .
Let's write ${\mathrm{ad}}_{X}{\mathrm{ad}}_{Y}$ as a block matrix with respect to the splitting into weight spaces (i.e. the $i,j$ block in the matrix is the matrix of the map ${\U0001d524}_{{\lambda}_{i}}\to {\U0001d524}_{{\lambda}_{j}}$ ).
The diagonal blocks encode the maps ${\U0001d524}_{\lambda}\to {\U0001d524}_{\lambda}$ . But if $\alpha +\beta \ne 0$ then ${\mathrm{ad}}_{X}{\mathrm{ad}}_{Y}$ sends ${\U0001d524}_{\lambda}$ to ${\U0001d524}_{\lambda +\alpha +\beta}\ne {\U0001d524}_{\lambda}$ , so there are no nonzero diagonal blocks if $\alpha +\beta \ne 0$ . Therefore the trace of this matrix vanishes and $K(X,Y)=0$ .
(3) If $\U0001d524$ is semisimple then the Killing form is nondegenerate, so for all nonzero $X$ there exists $Y$ such that $K(X,Y)\ne 0$ . In particular, if $X\in {\U0001d524}_{\alpha}$ then there exists $Y$ such that $K(X,Y)\ne 0$ .
Take the components of $Y={\sum}_{\lambda}{Y}_{\lambda}$ with respect to the weight space splitting. Then $K(X,{Y}_{\lambda})=0$ unless $\lambda =\alpha $ . Therefore, if $K(X,Y)\ne 0$ , we must have ${Y}_{\alpha}\ne 0$ , so ${\U0001d524}_{\alpha}\ne 0$ and $\alpha \in R$ .
The trick for extracting the $\U0001d530\U0001d529(2,\mathbf{C})$ subalgebras will be the following theorem which we'll prove next time.
If $X\in {\U0001d524}_{\alpha}$ , $Y\in {\U0001d524}_{\alpha}$ are nonzero then $X$ , $Y$ and $H=[X,Y]$ will span a subalgebra isomorphic to $\U0001d530\U0001d529(2,\mathbf{C})$ .
Preclass exercise
True or false: If we use a submaximal torus, is it still true that $\U0001d525={\U0001d524}_{0}$ ? If so, why? If not, what do we get instead?