The decomposition matrix of S3 in characteristic two

By [L1], only the values at $1$ and $(123)$ matter. The ordinary trivial and sign characters both restrict to $(1,1)$ on those classes, because a $3$-cycle is even. The ordinary standard character restricts to $(2,-1)$.

Let $k$ be an algebraic closure of ${\mathbb{F}}_2$ and let $W=\{(x_1,x_2,x_3)\in k^3:x_1+x_2+x_3=0\}$ with the permutation action of $S_3$. It is the reduction of the integral sum-zero lattice for the ordinary standard representation. A $3$-cycle acts on $W$ with the two distinct nontrivial cubic-root eigenvalues; a transposition interchanges their eigenlines. Thus no line in $W$ is $S_3$-invariant, so $W$ is simple. By [F2] its Brauer character is the restricted ordinary standard character $(2,-1)$, while the trivial module gives $(1,1)$. These two characters are independent on the two $2$-regular classes, so [L2] shows that they are the full irreducible Brauer basis. Consequently [F1] writes the restricted ordinary characters as $1^0={\phi }_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}},{\mathrm{sgn}}^0={\phi }_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}},{\mathrm{std}}^0={\phi }_{\mathrm{s}\mathrm{t}\mathrm{d}}\mathrm{.}$

Reading off the coefficients yields the displayed decomposition matrix.