The unshaded part must exist, whereas the shaded part is optional with variable size. $$T = \raisebox{-0.5\height}{\ \begin{texdraw}
 \fontsize{6}{6}\selectfont
 \textref h:C v:C
 \drawdim em
 \setunitscale 1.35
 \move(14.5 0)\lvec(1 0)\lvec(1 -1)\lvec(14.5 -1)\lvec(14.5 0)
 \ifill f:0.8
 \move(0 0)\lvec(-2.5 0)\lvec(-2.5 -2)\lvec(0 -2)\lvec(0 0)
 \ifill f:0.8
 \move(14.5 0)\lvec(-3.5 0)
 \move(14.5 -1)\lvec(-3.5 -1)
 \move(0 -2)\lvec(-3.5 -2)
 \move(-3.5 0)\lvec(-3.5 -2)
 \htext(-3 -0.5){$1$}
 \htext(-0.55 -0.5){$1$}
 \htext(-1.7 -0.5){$\cdots$}

 \move(0 0)\rlvec(0 -2)
 \htext(0.5 -0.5){$1$}
 \move(1 0)\rlvec(0 -1)
 \htext(2.25 -0.5){$2\!\cdots\!2$}
 \move(3.5 0)\rlvec(0 -1)
 \htext(4.75 -0.5){$3\!\cdots\!3$}
 \move(6 0)\rlvec(0 -1)
 \htext(6.5 -0.5){$0$}
 \move(7 0)\rlvec(0 -1)
 \htext(8.25 -0.5){$\bar{3}\!\cdots\!\bar{3}$}
 \move(9.5 0)\rlvec(0 -1)
 \htext(10.75 -0.5){$\bar{2}\!\cdots\!\bar{2}$}
 \move(12 0)\rlvec(0 -1)
 \htext(13.25 -0.5){$\bar{1}\!\cdots\!\bar{1}$}
 \move(14.5 0)\rlvec(0 -1)
 \htext(-1.25 -1.5){$3\!\cdots\!3$}
 \move(-2.5 -0.9)\rlvec(0 -1.1)
 \htext(-3 -1.5){$2$}
\end{texdraw}}.$$ The element $T_{\infty}$ is $$T_{\infty} = \raisebox{-0.5\height}{\ \begin{texdraw}
 \fontsize{6}{6}\selectfont
 \textref h:C v:C
 \drawdim em
 \setunitscale 1.35
 \move(0 0)\lvec(2 0)\lvec(2 -1)\lvec(0 -1)
 \move(1 0)\lvec(1 -2)\lvec(0 -2)\lvec(0 0)
 \htext(0.5 -0.5){$1$}
 \htext(1.5 -0.5){$1$}
 \htext(0.5 -1.5){$2$}
\end{texdraw}}\,.$$

We recall the action of Kashiwara operators $\fit$, $\eit$ $(i\in I)$ on marginally large tableaux $T\in\T(\infty)$.

1.  We first read the boxes in the tableau $T$ through the *far eastern reading* and write down the boxes in *tensor product form.* That is, we read through each column from top to bottom starting from the rightmost column, continuing to the left, and 