The twelve D4 positive roots

Write ${\beta }_1,\dots ,{\beta }_{12}$ for the vectors in the displayed order, and put $R=\{\pm {\beta }_1,\dots ,\pm {\beta }_{12}\}$. For a vector with coordinates $(c_1,c_2,c_3,c_4)$, [F1] gives $c_i^{\mathrm{\prime }}=-c_i+{\sum }_{j\sim i}{c}_j$ for the coordinate changed by $s_i$, leaving the others fixed; $j\sim i$ means that $j$ and $i$ are joined in the D4 diagram. Direct substitution gives the following images of ${\beta }_k$, where $-j$ denotes $-{\beta }_j$ and $j$ denotes ${\beta }_j$. In the order $s_1,s_2,s_3,s_4$, the rows for $k=1,\dots ,12$ are $(-1,5,1,1)$, $(5,-2,6,7)$, $(3,6,-3,3)$, $(4,7,4,-4)$, $(2,1,8,9)$, $(8,3,2,10)$, $(9,4,10,2)$, $(6,8,5,11)$, $(7,9,11,5)$, $(11,10,7,6)$, $(10,12,9,8)$, and $(12,11,12,12)$. Thus each $s_i$ preserves $R$: the table covers positive entries and linearity covers their negatives. Since $R$ contains every simple root, [F2] gives $\mathrm{\Phi }\subseteq R$.

Conversely, ${\beta }_1,\dots ,{\beta }_4$ are simple roots. The table gives ${\beta }_5=s_2{\beta }_1$, ${\beta }_6=s_2{\beta }_3$, ${\beta }_7=s_2{\beta }_4$, ${\beta }_8=s_3{\beta }_5$, ${\beta }_9=s_4{\beta }_5$, ${\beta }_{10}=s_4{\beta }_6$, ${\beta }_{11}=s_4{\beta }_8$, and ${\beta }_{12}=s_2{\beta }_{11}$. Hence all ${\beta }_k$ are roots. Each negative $-{\beta }_k$ is also a root: for ${\beta }_k=w{\alpha }_i$, it equals $w{s}_i{\alpha }_i$ by [F1]. Therefore $R\subseteq \mathrm{\Phi }$. The twelve ${\beta }_k$ have nonnegative, nonzero coordinate tuples, whereas their negatives have a negative coordinate, so the stated positive roots are exactly those twelve.