type of fermion and from derivatives (in the fermionic ${{\mathfrak{sl}}}(2)$ subsector) [@Belitsky:2005bu]. The states we check, given in Table \[tab:comparefield\], are in complete agreement with these previous computations. Therefore, we conclude that we have found the correct solution for ${\delta{\mathfrak{D}}}_4$. Since our comparison includes length-two states, we find confirmation that no additional wrapping terms are needed.

$\begin{array}{|l|l|l|}\hline
{{\mathfrak{D}}}_0&(R, L, B) &{\bigl({\delta{\mathfrak{D}}}_2,{\delta{\mathfrak{D}}}_4\bigr)}^P\\\hline
4  &(2,2,0)& \begin{array}{l} (6,-12)^+ \end{array} \\
\hline 
5  &(3,3,0)& \begin{array}{l} (4,-6)^- \end{array} \\
\hline
6  &(2,2,0)& \begin{array}{l} (\frac{25}{3},-\frac{925}{54})^+ \end{array} \\
\hline
6 & (3, 3, 0)& \begin{array}{l} (\frac{15}{2}, -\frac{225}{16})^\pm \end{array} \\ \hline
6  &(4,4,0)&\begin{array}{l} (2.76393,-2.90983)^+ \\ (7.23607,-14.0902)^+ \end{array} \\ \hline
7 & (5, 5, 0)& \begin{array}{l} (2, -\frac{3}{2})^- \\ (6, -\frac{21}{2})^- \end{array} \\ \hline 
7.5 & (0, 3, \pm \frac{3}{2})& \begin{array}{l} (10, -\frac{245}{12})^\pm \end{array} \\ \hline 
8  &(2,2,0)& \begin{array}{l} (\frac{49}{5},-\frac{45619}{2250})^+ \end{array} \\
\hline 
8 & (6, 6, 0)& \begin{array}{l} (1.50604, -0.830063)^+ \\ (4.89008, -7.30622)^+ \\ (7.60388, -14.8637)^+ \end{array} \\ \hline
9  &(6,8,\pm 1)& \begin{array}{l} (1.17157,-0.4895952)^- \\ (4, -5)^- \\ (6.82843, -12.5104)^- \end{array} \\
\hline 
9.5 & (0, 3, \pm \frac{3}{2})& \begin{array}{l} (\frac{133}{12},-\frac{131117}{5760})^\pm \end{array} \\ \hline 
10  &(7,9,\pm 1)& \begin{array}{l} (0.935822,-0.304865)^+ \\ (3.30540, - 3.44381)^+ \\ (6, -10)^+ \\  (7.75877, -15.2513)^+ \end{array} \\ \hline 
10.5 & (0, 3, \pm \frac{3}{2})& 