The zero-loop perfect-matching standard module is simple

For a parameter variable $b$, let ${\varphi }_b$ be the link-state form on the fixed basis of $W_{2p,0}(\mathbb{C},b)$. If $a,c$ are perfect matchings, their overlay is a disjoint union of $C(a,c)\ge 1$ alternating cycles, counting a shared arc as a two-edge cycle; therefore ${\varphi }_b(a,c)=b^{C(a,c)}$. Define the bilinear form ${\varphi }^{\mathrm{\prime }}(a,c)={\frac{d}{db}{\varphi }_b(a,c)\mid }_{b=0}\mathrm{.}$ It follows that ${\varphi }^{\mathrm{\prime }}(a,c)=1$ when $C(a,c)=1$ and $0$ when $C(a,c)>1$.

Delete site $2p$ and its incident arc from a perfect matching, and declare its partner to be the single defect. This is a bijection from perfect matchings on $2p$ sites to the link-state basis of $W_{2p-1,1}$: the inverse joins the unique defect to the appended site $2p$. For a pair $a,c$, deleting the two arcs incident to site $2p$ turns exactly the overlay cycle containing that site into the open path between the defects and leaves the other $C(a,c)-1$ cycles unchanged. Hence the ordinary one-defect Gram value of the images is $b^{C(a,c)-1}$, whose value at $b=0$ equals ${\varphi }^{\mathrm{\prime }}(a,c)$. Thus the matrix of ${\varphi }^{\mathrm{\prime }}$ is exactly the Gram matrix of $W_{2p-1,1}(\mathbb{C},0)$.

For $W_{2p-1,1}$ the arc count is $p-1$, the defect count is $1$, and $q^2=-1$ has order $\kappa =2$. Since $\kappa \mid (1+1)$, this is a critical label. By [F4], its Gram form is nondegenerate; step 2.1 then shows that ${\varphi }^{\mathrm{\prime }}$ is nondegenerate.

For two link states $x,y$ on $2p$ sites with two defects, define $D_{x,y}$ to be the Temperley–Lieb diagram whose bottom arcs are those of $x$, whose top arcs are those of $y$, and whose two unmatched bottom sites are joined to the two unmatched top sites in left-to-right order by propagating strands. This matching is planar and is an allowed basis diagram. Let $x^{\mathrm{\prime }}$ and $y^{\mathrm{\prime }}$ be the perfect matchings obtained by joining the two defects of $x$ and $y$, respectively. If $z$ is a basis perfect matching and $C(y^{\mathrm{\prime }},z)$ is the number of cycles in the overlay of $y^{\mathrm{\prime }}$ and $z$, stacking $z$ above $D_{x,y}$ leaves the bottom matching $x^{\mathrm{\prime }}$ and removes exactly $C(y^{\mathrm{\prime }},z)-1$ closed cycles. Therefore $z\cdot D_{x,y}={\delta }^{C(y^{\mathrm{\prime }},z)-1}{x}^{\mathrm{\prime }}$, where the coefficient at $\delta =0$ is defined as $1$ when $C(y^{\mathrm{\prime }},z)=1$ and $0$ when $C(y^{\mathrm{\prime }},z)>1$. This coefficient is exactly ${\varphi }^{\mathrm{\prime }}(y^{\mathrm{\prime }},z)$, so $z\cdot D_{x,y}={\varphi }^{\mathrm{\prime }}(y^{\mathrm{\prime }},z)x^{\mathrm{\prime }}$. The same equality holds for every $z$ by linearity.

Let $z\mathrm{\ne }0$. Nondegeneracy of ${\varphi }^{\mathrm{\prime }}$ gives a nonzero functional ${\varphi }^{\mathrm{\prime }}(-,z)$, so at least one perfect-matching basis vector $y^{\mathrm{\prime }}$ has ${\varphi }^{\mathrm{\prime }}(y^{\mathrm{\prime }},z)\mathrm{\ne }0$. For any target basis matching $x^{\mathrm{\prime }}$, cut from $x^{\mathrm{\prime }}$ and $y^{\mathrm{\prime }}$ the arcs incident to site $2p$ to obtain two-defect link states $x,y$. By step 3.2, $z\cdot D_{x,y}$ is a nonzero scalar multiple of $x^{\mathrm{\prime }}$. Thus the submodule generated by $z$ contains every basis vector and equals $W_{2p,0}$. Every nonzero submodule contains such a $z$, so the module is simple.