# Roots of a simple Lie algebra as elements of a Cartan subalgebra

Consider a Lie algebra $L$ over $\mathbb{C}$ and consider its Cartan decomposition: $$L = H \oplus L_1 \oplus ... \oplus L_k.$$ For each $1 \leqslant i \leqslant k$, take a non-zero element $e_i$ in a root subspace $L_i$. Since all $L_i$ are one-dimensional and $[H,L_i] = L_i$, for each $h\in H$ we have $$[h,e_i] = r_i(h) e_i,$$ where $r_i(h) \in \mathbb{C}$. Now $r_i$ is a linear map and $r_i$'s form a root system of $L$. Note that $r_i \in H^*$ (here $H^*$ denotes the dual space of $H$). In the book "Simple groups of Lie type", R.Carter very briefly describes a way to treat those roots as elements of $H$ using the Killing form. Hovewer, I do not get how to uniquely associate $r_i$ with some element of $H$. Could you please help me to understand this approach?

Upd. Just to put things straight: I do not understand how from the fact that the Killing form $(\cdot,\cdot)$ is non singular when restricted to the Cartan subgroup, it follows that for each $\varphi \in H^*$ there exists a unique $r_\varphi \in H$ such that $\varphi(h) = (r_\varphi, h)$.

For any (finite dimensional) vector space $V$, a non-degenerate bilinear form $(\cdot,\cdot)$ identifies $V$ with its dual $V^*$. Each element of $V$ corresponds to the linear functional $v^*=(v,?)$.
To see that this is an isomorphism, let $\{v_1,\ldots,v_n\}$ be an orthonormal basis with respect to $(\cdot,\cdot)$. Then, the linear functional $v_i^*=(v_i,?)$ is the (dual) basis for $V^*$: $$v_i^*(v_j)=(v_i,v_j)=\delta_{ij},$$ which is an easy exercise.
Now, the Killing form (restricted to $H$) is non-degenerate, so given an orthonormal basis $e_1,\ldots,e_n$ for $H$, and $\phi\in H^*$ we can write $$\phi=\sum_ia_ie_i^*$$ and $$r_\phi=\sum_ia_ie_i.$$