Suppose $\U0001d524$ is abelian, so the bracket is zero. Then ${\mathrm{ad}}_{X}=0$ for all $X$ , so $K(X,Y)=0$ for all $X$ and $Y$ . The Killing form was introduced to equip $\U0001d524$ with some interesting geometry, and defining the dot product of two vectors to be identically zero isn't an interesting geometry, so we want to focus on examples where this doesn't happen.
Dual Killing form
Killing form
Recall that any Lie algebra admits a symmetric bilinear form called the Killing form $$K(X,Y)=\mathrm{Tr}({\mathrm{ad}}_{X}{\mathrm{ad}}_{Y}).$$
A symmetric bilinear form $K$
is called
We say that $\U0001d524$
is
I'm being immoral here. The correct definition of semisimplicity is actually quite different from this, involving diagonalisability of certain matrices. It's then a theorem ("Cartan's criterion") that semisimplicity is equivalent to nondegeneracy of the Killing form. However, the way I've set things up, we don't need to know the usual definition of semisimplicity, but we do need to know about nondegeneracy of the Killing form.
Compact groups
Here is a theorem (which we won't prove; it will be an exercise) which gives a large class of examples of semisimple Lie algebras.
If $G$ is a compact group then the Killing form $K:\U0001d524\times \U0001d524\to \mathbf{R}$ is (realvalued and) negative semidefinite, i.e. $K(X,X)\le 0$ for all $X\in \U0001d524$ . Moreover $K(X,X)=0$ if and only if ${\mathrm{ad}}_{X}=0$ , i.e. if $X$ commutes with everything in the Lie algebra.
The
Recall that the centre $Z(G)$ of a group $G$ is the set of all elements $g$ which commute with every other element of the group. It's possible to have Lie groups with nontrivial centre but where the Lie algebra has trivial centre, for example $SU(2)$ has $Z(SU(2))=\{\pm I\}$ while $Z(\U0001d530\U0001d532(2))=\{0\}$ .
If $G$ is compact and $Z(\U0001d524)=0$ then $\U0001d524$ is semisimple.
Since $Z(\U0001d524)=0$ , $$ for all $X\in \U0001d524\setminus \{0\}$ , so $\U0001d524$ is semisimple.
Assume $Z(\U0001d524)=0$ . Let $T\subset G$ be a maximal torus. Let $\U0001d531$ be the Lie algebra of $T$ . Let $\U0001d525=\U0001d531\otimes \mathbf{C}$ and ${\U0001d525}_{\mathbf{R}}=i\U0001d531$ . Then ${K}_{{\U0001d525}_{\mathbf{R}}}$ is positive definite.
$$ , so $K(iX,iX)=K(X,X)>0$ (unless $X=0$ ).
Another name for a positivedefinite symmetric bilinear form $K:V\times V\to \mathbf{R}$
is a
The dual
Definition
Remember that our weight diagram lives in ${\U0001d525}_{\mathbf{R}}^{*}$ , not in ${\U0001d525}_{\mathbf{R}}$ .
If $K:V\times V\to \mathbf{C}$ is a nondegenerate symmetric bilinear form then ${V}^{*}$ inherits a symmetric bilinear form ${K}^{*}:{V}^{*}\times {V}^{*}\to \mathbf{C}$ .

There is an isomorphism $\mathrm{\u266f}:{V}^{*}\to V$ (its inverse is called $\mathrm{\u266d}:V\to {V}^{*}$ ) defined in the following way: $$K({\alpha}^{\mathrm{\u266f}},v)=\alpha (v)$$ for all $\alpha \in {V}^{*}$ and $v\in V$ .
This uniquely determines the map $\mathrm{\u266f}$ because $K$ is nondegenerate (it will be an exercise to see this).

Define ${K}^{*}(\alpha ,\beta )=K({\alpha}^{\mathrm{\u266f}},{\beta}^{\mathrm{\u266f}})$ .
More concretely...
To make this a little less abstract, let's write out what this means in terms of matrices. Pick a basis ${e}_{1},\mathrm{\dots},{e}_{n}$ of $V$ and define a matrix $K$ whose $i,j$ entry is ${K}_{ij}=K({e}_{i},{e}_{j})$ . Then $$K(v,w)=K(\sum _{i}{v}_{i}{e}_{i},\sum _{j}{w}_{j}{e}_{j})=\sum _{i,j}{v}_{i}{w}_{j}{K}_{ij}.$$
Given $\alpha \in {V}^{*}$ , we can think of $\alpha $ as a row vector $\left(\begin{array}{ccc}\hfill {\alpha}_{1}\hfill & \hfill \mathrm{\cdots}\hfill & \hfill {\alpha}_{n}\hfill \end{array}\right)$ and we want to produce a column vector ${\alpha}^{\mathrm{\u266f}}=\left(\begin{array}{c}\hfill {\alpha}_{1}^{\mathrm{\u266f}}\hfill \\ \hfill \mathrm{\vdots}\hfill \\ \hfill {\alpha}_{n}^{\mathrm{\u266f}}\hfill \end{array}\right)$ such that $K({\alpha}^{\mathrm{\u266f}},v)=\alpha (v)$ for all $v\in V$ .
We have $$\alpha (v)=\sum {\alpha}_{j}{v}_{j}\mathit{\hspace{1em}}K({\alpha}^{\mathrm{\u266f}},v)=\sum {K}_{ij}{\alpha}_{i}^{\mathrm{\u266f}}{v}_{j}.$$ Now taking coefficients of ${v}_{j}$ on both sides we get $${\alpha}_{j}=\sum _{i}{K}_{ij}{\alpha}_{i}^{\mathrm{\u266f}}.$$
Since ${K}_{ij}$ is symmetric, this implies ${\alpha}_{j}={\sum}_{i}{K}_{ji}{\alpha}_{i}^{\mathrm{\u266f}}$ , which we can write as $${\alpha}^{T}=K{\alpha}^{\mathrm{\u266f}}.$$ Hence $${\alpha}^{\mathrm{\u266f}}={K}^{1}{\alpha}^{T}.$$
This only makes sense if ${K}_{ij}$ is invertible, and this turns out to be equivalent to nondegeneracy. The flat map is then given by ${v}^{\mathrm{\u266d}}={(Kv)}^{T}$ , which makes sense even if $K$ is degenerate.
Finally, let's define ${K}_{ij}^{*}$ by ${K}^{*}(\alpha ,\beta )=\sum {K}_{ij}^{*}{\alpha}_{i}{\beta}_{j}$ . Since ${K}^{*}(\alpha ,\beta ):=K({\alpha}^{\mathrm{\u266f}},{\beta}^{\mathrm{\u266f}})$ , we have $$\sum {K}_{ij}^{*}{\alpha}_{i}{\beta}_{j}=\sum {K}_{ij}{K}_{ip}^{1}{\alpha}_{p}{K}_{jq}^{1}{\beta}_{q}=\sum {K}_{pq}^{1}{\alpha}_{p}{\beta}_{q}.$$
Therefore ${K}_{ij}^{*}={K}_{ij}^{1}$ . So if we want to compute the matrix of the dual symmetric bilinear form ${K}^{*}$ (with respect to the dual basis), we first compute the matrix of $K$ , then we invert it.
Example
For $SU(3)$ , we calculated $$K({H}_{13},{H}_{13})=K({H}_{23},{H}_{23})=12$$ and $$K({H}_{13},{H}_{23})=6$$ for the Killing form on ${\U0001d525}_{\mathbf{R}}$ (spanned by ${e}_{1}={H}_{13}$ and ${e}_{2}={H}_{23}$ ). So, as a matrix ${K}_{ij}=K({e}_{i},{e}_{j})$ , we get $$K=\left(\begin{array}{cc}\hfill 12\hfill & \hfill 6\hfill \\ \hfill 6\hfill & \hfill 12\hfill \end{array}\right).$$
On the dual space ${\U0001d525}_{\mathbf{R}}^{*}$ , we therefore get $${K}^{*}=\frac{1}{108}\left(\begin{array}{cc}\hfill 12\hfill & \hfill 6\hfill \\ \hfill 6\hfill & \hfill 12\hfill \end{array}\right).$$ Let ${e}_{1}^{\mathrm{\u266d}}$ and ${e}_{2}^{\mathrm{\u266d}}$ be the dual basis of ${\U0001d525}_{\mathbf{R}}^{*}$ . Then $${e}_{i}^{\mathrm{\u266d}}=1/3,{e}_{1}^{\mathrm{\u266d}}\cdot {e}_{2}^{\mathrm{\u266d}}=6/108,$$ so if $\varphi $ is the angle between these two vectors then $\mathrm{cos}\varphi =6\times 9/108=1/2$ , and $\varphi $ is $120$ degrees.
This is why I was drawing my $k$ and $\mathrm{\ell}$ axes at 120 degrees to one another in $SU(3)$ weight diagrams.