How to verify that an element in the root lattice is an imaginary root of a non-hyperbolic root system? In my research I encounter some elements in a root lattice and I would like to verify that these elements are imaginary roots. Consider the root system $J_{6, 11}$ with the following Dynkin diagram:
\begin{align}
\circ - \circ - \circ - \circ - \circ - & \circ -  \circ - \circ - \circ - \circ \\
& \ | \\
& \ \bullet
\end{align}
where the $\circ$'s corresponds to the simple roots $\alpha_1, \ldots, \alpha_{10}$ (from left to right) and $\bullet$ corresponds to the simple root $\alpha_{11}$.
Let $\gamma = \alpha_1+2\alpha_2+3\alpha_3+4\alpha_4+5\alpha_5+6\alpha_6+5\alpha_7+4\alpha_8+3\alpha_9+2\alpha_{10}+2\alpha_{11}$. I would like to verify that $\gamma$ is an imaginary root in $J_{6,11}$. Are there some method to do this? Thank you very much.
 A: Denote
$$K = \{ \alpha\in Q_+\setminus\{0\} \mid \langle \alpha,\alpha_i^\vee \rangle \leqslant 0 \text{ for all $i$ and $\operatorname{supp}(\alpha)$ is connected} \}.$$
Here $Q_+$ is the positive part of the root lattice and $\operatorname{supp}(\alpha)$ is the support of $\alpha$, that is, the subdiagram of the Dynkin diagram corresponding to the simple roots having non-zero coefficient in $\alpha$.
Then Lemma 5.3 in "Infinite dimensional Lie algebras" by V. Kac states that $K\subset \Delta_+^\mathrm{im}$ (the set of positive imaginary roots), and since $\Delta_+^\mathrm{im}$ is $W$-invariant, $WK\subseteq\Delta_+^\mathrm{im}$ (in fact, Theorem 5.4 shows that they are equal).
Now for the root $\gamma$ you mention. Using simple reflections $s_1,\ldots,s_{10}$, one can transform $\gamma$ to the following element of the root lattice:
$$\gamma' = \alpha_2+2\alpha_3+3\alpha_4+4\alpha_5+5\alpha_6+4\alpha_7+3\alpha_8+2\alpha_9+\alpha_{10}+2\alpha_{11}$$
(this is the lowest height element in the $W(\langle\alpha_1,\ldots,\alpha_{10}\rangle)$-orbit of $\gamma$). Then
$$\langle\gamma',\alpha_1^\vee\rangle = \langle\gamma',\alpha_{11}^\vee\rangle=-1 \quad \text{and} \quad \langle\gamma',\alpha_i^\vee\rangle=0 \quad \text{for} \quad i=2,\ldots,10,$$
so $\gamma'\in K$ and hence $\gamma$ is an imaginary root.
