Decomposition matrices of S3 at p=2 and p=3

By [F2] and [F3], $M_k^{(2,1)}$ has $k$-basis $v_1,v_2,v_3$ and $S_k^{(2,1)}$ has $k$-basis $e_t=v_3-v_1$, $e_u=v_2-v_1$. Applying the orthonormal form ${\beta }_k$ of [F4], $\beta (e_t,e_t)=\beta (v_3-v_1,v_3-v_1)=1+1=2,\beta (e_u,e_u)=2,\beta (e_t,e_u)=\beta (v_3-v_1,v_2-v_1)=1,$ so $G_{(2,1)}=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$.

The unique $(3)$-tabloid $v$ is fixed by $S_3$, so $S_k^{(3)}=kv$ is the one-dimensional trivial module, and $G_{(3)}=(1)$ has rank $1$; over a field of characteristic $2$ the sign representation is trivial by [F8], and over a field of characteristic $3$ it is nontrivial. The column stabilizer of a $(1,1,1)$-tableau $w$ is all of $S_3$, and the single standard polytabloid is $e={\sum }_{\sigma \in S_3}\mathrm{sgn}(\sigma )\{\sigma \cdot w\};$ its six tabloid coefficients are $\pm 1$ at the six distinct orderings, so $e\mathrm{\ne }0$, and for $\tau \in S_3$ the substitution ${\sigma }^{\mathrm{\prime }}=\tau \sigma$ gives $\tau \cdot e={\sum }_{\sigma }\mathrm{sgn}(\sigma )\{\tau \sigma \cdot w\}=\mathrm{sgn}(\tau ){\sum }_{{\sigma }^{\mathrm{\prime }}}\mathrm{sgn}({\sigma }^{\mathrm{\prime }})\{{\sigma }^{\mathrm{\prime }}\cdot w\}=\mathrm{sgn}(\tau )e\mathrm{.}$ Hence $S_k^{(1,1,1)}=ke$ is the one-dimensional sign representation.

Reducing $G_{(2,1)}$ modulo $p$: for $p=2$ the reduction $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ has determinant $1$, hence rank $2$; for $p=3$ the reduction is nonzero with determinant $3\equiv 0$, hence rank $1$, its columns being proportional. By the dimension formula of [F4], ${\dim }_k{D}^{(2,1)}={\mathrm{rank}}_p(G_{(2,1)}<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p)=\{\begin{array}{ll}2,& p=2,\\ 1,& p=3.\end{array}$ Also ${\dim }_k{D}^{(3)}={\mathrm{rank}}_p(1)=1$ for both primes.

Let $p=2$. Since ${\dim }_k{D}^{(2,1)}=2={\dim }_k{S}_k^{(2,1)}$ by steps 2.1 and 1.1, the radical $R^{(2,1)}$ is zero and $S_k^{(2,1)}=D^{(2,1)}$ is simple by [F6]. Since ${\dim }_k{D}^{(3)}=1={\dim }_k{S}_k^{(3)}$, also $S_k^{(3)}=D^{(3)}$, the trivial module, and by [F8] and step 1.2 the sign module satisfies $S_k^{(1,1,1)}\cong S_k^{(3)}=D^{(3)}$. Therefore $d_{(3),(3)}=1$, $d_{(3),(2,1)}=0$; $d_{(2,1),(3)}=0$, $d_{(2,1),(2,1)}=1$; $d_{(1,1,1),(3)}=1$, $d_{(1,1,1),(2,1)}=0$, which is the matrix displayed for $p=2$.

Let $p=3$. Here $D^{(3)}$ is the trivial module of dimension $1$ by step 1.2 and step 2.1, and $D^{(2,1)}$ is a simple module of dimension $1$ that is not isomorphic to $D^{(3)}$ by [F6]. The sign module $S_k^{(1,1,1)}$ of step 1.2 is one-dimensional, hence simple by [F8], so it is isomorphic to $D^{(3)}$ or to $D^{(2,1)}$ by [F6]; it is nontrivial in characteristic $3$ by step 1.2, hence it is not $D^{(3)}$ and $S_k^{(1,1,1)}\cong D^{(2,1)}$, so $d_{(1,1,1),(3)}=0$ and $d_{(1,1,1),(2,1)}=1$. For $S_k^{(2,1)}$, by [F3] $w:=e_t+e_u=(v_3-v_1)+(v_2-v_1)=v_1+v_2+v_3-3v_1=v_1+v_2+v_3$ because $3v_1=0$ in characteristic $3$; here $w\mathrm{\ne }0$ and $\sigma \cdot w=w$ for every $\sigma \in S_3$ by [F2], so $kw$ is a one-dimensional trivial submodule of $S_k^{(2,1)}$, isomorphic to $D^{(3)}$. On the quotient $S_k^{(2,1)}\mathrm{/}kw$, which is one-dimensional, the transposition $(12)$ acts by $(12)\cdot e_t=v_3-v_2\equiv -e_t$, since $v_3-v_2+e_t=2v_3-v_1-v_2=3v_3-w\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}kw)$; the quotient therefore is a nontrivial one-dimensional simple module, hence isomorphic to $D^{(2,1)}$. Its dimension $2$ equals $1+1$, so the composition factors of $S_k^{(2,1)}$ are $D^{(3)}$ and $D^{(2,1)}$, each once: $d_{(2,1),(3)}=d_{(2,1),(2,1)}=1$. With $d_{(3),(3)}=1$ and $d_{(3),(2,1)}=0$ from $S_k^{(3)}=D^{(3)}$, this is the matrix displayed for $p=3$.

Both matrices satisfy the constraints of [F7]. The forced zero $d_{(3),(2,1)}=0$ holds in both, because $(2,1)⋭(3)$; the $p$-regular diagonal entries $d_{(3),(3)}=d_{(2,1),(2,1)}=1$ hold in both; and with the $p$-regular rows and columns in the decreasing lexicographic order $(3),(2,1)$ the leading block is $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ at $p=2$ and $\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ at $p=3$, lower unitriangular in both cases. The entries of the row $(1,1,1)$ lie in allowed positions since $(1,1,1)$ is dominated by every partition of $3$. The two matrices coincide with those recorded in the source reference for $S_3$, and the position $((2,1),(3))$ shows the characteristic dependence: $0$ at $p=2$ and $1$ at $p=3$.