The complete weight-one decomposition matrix for S3 at p=3

The labels and their cores. The hook-length diagrams of the three partitions of $3$ are $3,2,1$ for $(3)$; $3,1$ in the first row and $1$ in the second for $(2,1)$; and $3,2,1$ vertically for $(1,1,1)$. In each case there is exactly one hook of length $3$ and removing its rim hook leaves the empty partition, so all three partitions have $3$-core $\mathrm{\varnothing }$; conversely any partition of $3$ with core $\mathrm{\varnothing }$ is one of these, since a partition whose core has size $3$ is its own core, and $\mathrm{\varnothing }$ has size $0=3-3$. Dominance orders them as $(3)⊳(2,1)⊳(1,1,1)$.

Compute the Weyl matrix directly. First work over ${\mathbb{F}}_3$ with lift $R={\mathbb{Z}}_{(3)}$, $\widehat{q}=1$, so the only nonzero hook valuation at rank three is $v([3{]}_1)=1$. The conjugate beta sequence of $(3)$ is $(3,2,1)$. Its hook pairs $(3,2)$ and $(3,1)$ give the balanced sequences $(5,0,1)$ and $(4,2,0)$, of sorted shapes $(1,1,1)$ with sign $-1$ and $(2,1)$ with sign $+1$, respectively. The other pair has coefficient zero. Thus its positive sum is $[W^{(2,1)}]-[W^{(1,1,1)}]$. For $(2,1)$, beta $(4,2,0)$ and its only nonzero pair $(3,1)$ give $(5,1,0)$, shape $(1,1,1)$ with sign $+1$, so its positive sum is $[W^{(1,1,1)}]$. The column shape has no row pair and zero positive sum, hence $W^{(1,1,1)}=L^{(1,1,1)}$. Positivity and $W(1)=\mathrm{rad}W$ then show that $W^{(2,1)}(1)=L^{(1,1,1)}$ and every higher submodule is zero, giving $[W^{(2,1)}]=[L^{(2,1)}]+[L^{(1,1,1)}]$. Substitute into the first sum to get $[L^{(2,1)}]$; the same argument gives $[W^{(3)}]=[L^{(3)}]+[L^{(2,1)}]$. This proves exactly the displayed complete Weyl matrix. Scalar extension to any original field $F$ of characteristic three preserves the Weyl modules/forms and the composition series, since all Schur simple heads are absolutely simple by [F2].

Determine the specific corner images directly. The group row symmetrizer $x_{(3)}={\sum }_{w\in S_3}w$ satisfies $x_{(3)}^2=6x_{(3)}$, so its rank-one Murphy form is zero in characteristic three. In the rank-one column cell the generator is $1$ and its square coefficient is $1$; the higher ideal contains $1+T_1$ and $1+T_2$ (the latter is a conjugate of the former), so this cell and its nonzero form quotient are the sign module. For the middle cell put $x=1+T_1$, $c_0=x$ and $c_1=x{T}_2$ modulo the higher row ideal. Its row symmetrizer satisfies $x_{(3)}=x+x{T}_2+x{T}_2{T}_1$, so $c_1{T}_1=-c_0-c_1$ while $c_0{T}_1=c_0$; and $T_2^2=1$ gives $c_0{T}_2=c_1$, $c_1{T}_2=c_0$. Thus its generator matrices are $M_1=\left(\begin{array}{cc}1& 0\\ -1& -1\end{array}\right)$ and $M_2=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. The independent cell products $x^2=2x$, $x{T}_{2}x=x_{(3)}-x\equiv -x$ and $x{T}_2^{2}x=2x$ give $G=\left(\begin{array}{cc}2& -1\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 2& 2\end{array}\right)$ over $F$. Its radical is the line $(1,-1)$; modulo it both generators act by $1$, so $D_{\mathrm{M}}^{(2,1)}$ is the nonzero trivial module. These two simple images are distinct since $1\mathrm{\ne }-1$. The exact corner/form identity [F3] therefore kills precisely $L^{(3)}$ among the three labels. Projecting the Weyl matrix just proved gives all displayed Murphy Specht rows over the original $F$, without any general finite-$e$ nonzero-label criterion.

Dipper-James rows and the direct group computation. Conjugating the partitions $(3)\leftrightarrow (1,1,1)$, $(2,1)\leftrightarrow (2,1)$ and applying the signed $\theta$-pullback to every Hecke module, the rows of step 3.1 become $[S^{(1,1,1)}]=[D^{(2,1)}]$, $[S^{(2,1)}]=[D^{(2,1)}]+[D^{(3)}]$ and $[S^{(3)}]=[D^{(3)}]$ in the $3$-regular labels, that is, the matrix $\left(\begin{array}{cc}1& 0\\ 1& 1\\ 0& 1\end{array}\right)$ with rows $(3),(2,1),(1,1,1)$ and columns $(3),(2,1)$. For the independent direct check, construct the tabloid module over $k$ using [F5]. Every middle-shape polytabloid is a difference $v_i-v_j$, and these span the two independent vectors $e_t,e_u$ displayed in [F5], so they are a basis of $S_k^{(2,1)}$. A one-row polytabloid is its unique tabloid, giving the trivial module; a one-column polytabloid is the nonzero alternating sum of all six distinct tabloids, giving the sign module. Put $w:=e_t+e_u=v_3-v_1+v_2-v_1=v_1+v_2+v_3$ in characteristic $3$: then $w\mathrm{\ne }0$ spans a one-dimensional submodule, since $\sigma \cdot v_i=v_{\sigma (i)}$ gives $\sigma \cdot w=w$ for every $\sigma \in S_3$, so it realises the trivial module; and in the quotient, $(12)\cdot e_t+e_t=(v_3-v_2)+(v_3-v_1)=2v_3-v_1-v_2=3v_3-w\equiv 0$ shows that $(12)$ acts by $-1$. All transpositions are conjugate and act on a one-dimensional quotient, so they all act by $-1$ and the quotient is sign. The orthonormal tabloid form has middle Gram matrix $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ in $(e_t,e_u)$; its characteristic-three kernel is $k(e_t+e_u)=kw$. Thus the James quotient $D^{(2,1)}$ is precisely this sign quotient, while the row form has norm $1$ and $D^{(3)}$ is trivial. With $S^{(3)}=\mathrm{triv}$, $S^{(1,1,1)}=\mathrm{sgn}$ by [F5], and the just-computed $D^{(3)}=\mathrm{triv}$, $D^{(2,1)}=\mathrm{sgn}$ (the complete list is [F4]), the direct rows of $S^{(3)},S^{(2,1)},S^{(1,1,1)}$ are $(1,0),(1,1),(0,1)$, matching the matrix obtained from the Schur functor entry by entry.