$\gdef\sl{\mathfrak{sl}}$ $\gdef\hf{\mathfrak{h}}$ $\gdef\asl{\widehat{\mathfrak{sl}}}$ $\gdef\ahf{\widehat{\mathfrak{h}}}$

As a vector space we have $$\asl_2 = sl_2[t,t^{-1}] \oplus \C K \oplus \C d$$ As Lie algebra, we have: $$ [X t^m, Y t^n] = [X, Y] t^{m+n} + m \delta_{m+n=0} K \quad \forall X, Y \in sl_2 $$ $$ [d, X t^m] = m X t^m, \quad d= t \d_t $$

The affine Cartan (where coroot lives) is $$ \ahf = \hf \oplus \C K \oplus \C d$$ The dual affine Cartan (where root, weights lives) is $$ \widehat{\mathfrak{h}} = \mathfrak{h}^* \oplus \C K^* \oplus \C \delta, \quad \delta:= d^*$$

To talk about root system, we need to first list all roots $\hat \Phi$, which is the (non-zero) eigenvalues (living in $\ahf^*$) of how stuff in $\ahf$ acts on $\asl_2$.

For example, eigenvalue for $e$ inside $\hf^*$ is $\alpha$, and the coroot $\alpha^\vee \in \hf$ is defined to be $\alpha^\vee = [e,f]$, which is just usually denoted $h$.

We will see that $\hat \Phi \cup \{0\} = (\Phi \cup \{0\}) + \Z \delta \In \ahf^*$. Here are the positive roots (the usual choice for Borel) $$ \hat \Phi_+ = \{ \alpha + n \delta, n \geq 0 \} \sqcup \{ n \delta, n > 0\} \sqcup \{-\alpha + n \delta, n > 0\}. $$ Finally, we can talk about simple roots, which are primitive elements in positive roots, they are $$ \alpha_1 = \alpha, \quad \alpha_0 = -\alpha + \delta. $$ The corresponding elements $e_{\alpha_i} \in \mathfrak{g}_{\alpha_i}$ are $$ e_1 = e, \quad e_0 = f \otimes t, \quad f_1 = f, \quad f_0 = e \otimes t^{-1} $$ Now, there is no mistake or typo that I mistook an f with an e. One should think of, $t$ is SO positive that, it overcomes the negativity of $f$ or whatsoever.

Finally, we can get to the coroot, which I will $\alpha_0^\vee, \alpha_1^\vee$. $$ \alpha_0^\vee = h_0 = [e_0, f_0] = [f \otimes t, e \otimes t^{-1}] = [f,e] + 1 \delta_{1-1=0} K = K - h $$ and $$ \alpha_1^\vee = h_1 = [e_1, f_1] = [e,f] = h. $$

Sanity check, does $(e_0, h_0, f_0)$ satisfies an $sl_2$-triple condition? $$ [h_0, e_0] = [K - h, f\otimes t] = [-h, f \otimes t] = 2 f \otimes t = 2 e_0. $$

Fix an integral level $K = k > 0$. All integrable representation is given by highest weight $v$, we determine $v$ by its two weights $$ \alpha_0^\vee v = a_0 v, \quad \alpha_1^\vee v = a_1 v $$ where since $K = \alpha_0^\vee + \alpha_1^\vee$, and $K v = k v$, so $a_0 + a_1 = k$. By the definition of integrable representation, the finite $\sl_2$ acts locally finitely. In particular, we need to have $a_1 \geq 0$. Can we have $a_0 < 0$? Somehow