take the following form: $$P_m = \left\{\left(\begin{pmatrix} *&* \\ *&*
        \end{pmatrix},\begin{pmatrix}
          *&*&0 \\ *&*&0 \\ *&*&*
        \end{pmatrix},
        \begin{pmatrix} *&0 \\ *&*
        \end{pmatrix}\right)\right\}, \qquad
    Z_m = \left\{\left(
        \begin{pmatrix}
          *&*&0\\ *&*&0
        \end{pmatrix}
        ,
        \begin{pmatrix}
          0&0 \\ 0&0 \\ *&0
        \end{pmatrix}
      \right)\right\}.$$

By Theorem \[thm:multidegree\] $$\begin{aligned}
    [Z_m] &= (x^{(1)}_1-x^{(2)}_3)(x^{(1)}_2-x^{(2)}_3)\
    (x^{(2)}_1-x^{(3)}_1)(x^{(2)}_1-x^{(3)}_2)
    (x^{(2)}_2-x^{(3)}_1)(x^{(2)}_2-x^{(3)}_2) (x^{(2)}_3-x^{(3)}_2) \\
    [\Omega] &=
    \partial_{x^{(2)}_1-x^{(2)}_2} \,\partial_{x^{(2)}_2-x^{(2)}_3}
    \,\partial_{x^{(3)}_1-x^{(3)}_2}   \, [Z_m]
  \end{aligned}$$ By Theorem \[thm:Kpoly\] $$\begin{aligned}
    [Z_m]^K_{\mathrm{Hom}}&=
    (1-e^{-x^{(1)}_1+x^{(2)}_3})(1-e^{-x^{(1)}_2+x^{(2)}_3})
    (1-e^{-x^{(2)}_1+x^{(3)}_1})(1-e^{-x^{(2)}_1+x^{(3)}_2}) \\
    &\,\,\,\,\,\,\,(1-e^{-x^{(2)}_2+x^{(3)}_1})(1-e^{-x^{(2)}_2+x^{(3)}_2})
    (1-e^{-x^{(2)}_3+x^{(3)}_2}) \\
    [\Omega]^K_{\mathrm{Hom}}&=
    d_{x^{(2)}_1-x^{(2)}_2}\, d_{x^{(2)}_2-x^{(2)}_3}\, 
    d_{x^{(3)}_1-x^{(3)}_2}\,     [Z_m]^K_{\mathrm{Hom}}.
  \end{aligned}$$ We shall work out the multidegree calculation explicitly. Let $a_i=x^{(1)}_i$, $b_i=x^{(2)}_i$, and $c_i=x^{(3)}_i$. We shall use the following properties of $\partial_\alpha$: $\partial_\alpha(f)=0$ if $r_\alpha(f)=f$, and $\partial_\alpha(fg)=\partial_\alpha(f)g+r_\alpha(f)
\partial_\alpha(g)$. In particular if $r_\alpha(f)=f$ then $\partial_\alpha(fg)=f\partial_\alpha(g)$.

Using the notation $\partial^a_i = \partial_{a_i-a_{i+1}}$ (and similarly for $b,c$) we have $$\begin{aligned}
  [\Omega] &= \partial^b_1 \partial^b_2 \partial^c_1
  (a_1-b_3)(a_2-b_3)(b_1-c_1)(b_1-c_2)(b_2-c_1)(b_2-c_2)(b_3-c_2) \\
  &= \partial^b_1 \partial^b_2
  (a_1-b_3)(a_2-b_3)(b_1-c_1)(b_1-c_2)(b_2-c_1)(b_2-c_2) \\
&= \partial^b_1 [ (a_2-b_3)(b_1-c_1)(b_1-c_2)(b_2-c_1)(b_2-c_2) \\
&+ (a_1-b_2)(b_1-c_1)(b_1-c_2)(b_2-c_1)(b_2-c_2) \\
&+ (a_1-b_2)(a_2-b_2)(b_1-c_1)(b_1-c_2)(b_2-c_2) \\
&+ (a_1-b_2)(a_2-b_2)(b_1-c_1)(b_1-c_2)(b_3-c_1) ] \\
&= [0 ]+ [(b_1-c_1)(b_1-c_2)(b_2-c_1)(b_2-c_2)] \\
&+ [(a_2-b_2)(b_1-c_1)(b_1-c_2)(b_2-c_2) \\
&+ (a_1-b_1)(b_1-c_1)(b_1-c_2)(b_2-c_2) \\
&+ (a_1-b_1)(a_2-b_1)(b_1-c_2)(b_2-c_2)] \\
&+ [ (a_2-b_2)(b_1-c_1)(b_1-c_2)(b_3-c_1) \\
&+ (a_1-b_1)(b_1-c_1)(b_1-c_2)(b_3-c_1) \\
&+ (a_1-b_1)(a_2-b_1)(b_1-c_2)(b_3-c_1) \\
&+ (a_1-b_1)(a_2-b_1)(b_2-c_1)(b_3-c_1) ].\end{aligned}$$ We check this against the component formula [@KMS Cor. 6.17], which is a sum over three minimal length lacing diagrams $$\begin{aligned}
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