 the only semistandard tableaux which can occur in $\theta$ are those with a $2$ in each row, i.e. those of the form $$\young(1111233\star,2\star\star\star\star),\qquad
\young(111123\star\star,23\star\star\star)\quad\text{or}\quad
\young(11112\star\star\star,233\star\star).$$ Now the first and last of these three types can be ruled out using the same argument with ${\psi_{2,2}}$. So $\theta$ can only involve tableaux with a $2$ and a $3$ in each row; call these *usable* tableaux. Now we consider ${\psi_{d,1}}\circ\theta$ for $d\gs4$. Now for each usable tableau $T$, ${\psi_{d,1}}\circ{\hat\Theta_{T}}$ is either zero (if $d$ and $d+1$ occur in the same row in $T$) or a semistandard homomorphism. Furthermore, these semistandard homomorphisms ‘pair up’; for example, with $d=4$ we have $${\psi_{4,1}}\circ\,\young(11112356,23478)\,={\psi_{4,1}}\circ\,
\young(11112346,23578)\,=\,
\young(11112346,23478).$$ Since the semistandard tableau on the right can only arise in this way from the two semistandard tableaux on the left, these two semistandard homomorphisms must occur with equal coefficients in $\theta$. Now we observe that we can get from any usable tableau to any other by a sequence of steps in which we interchange the integers $d,d+1$ for various values of $d\gs4$. So if we apply the above argument for all $d\gs4$, we see that all usable tableaux occur with the same coefficient in $\theta$.

Homomorphisms from $S^\mu$ to $S^\la$
-------------------------------------

Now we consider homomorphisms from $S^\mu$ to $S^\la$, where $\la,\mu$ are as above. In view of Proposition \[cdhomdim1\], *we assume for the rest of this section that $3\ls v\ls a-1$*. It turns out that all such homomorphisms can be expressed as linear combinations of ${\hat\Theta_{A}}$ and ${\hat\Theta_{B}}$, where $A,B$ are the following $\mu$-tableaux of type $\la$: $$\begin{aligned}
A&=
{\text{\footnotesize$\gyoungx(1.2,;1_2{{\begin{tikzpicture}[baseline=0cm]\draw[thick,densely dotted](0,.13)--++(2*.25,0);\end{tikzpicture}}};1;2;2;2;3;4_2{{\b