The exterior graded Swiss-Cheese Operad ΛVS2

Existence of the detS2$det^{S^2}$ map

“…Computing the determinant of the matrix 𝐴 1 ((𝑣 𝑖,𝑗 ) 1⩽𝑖<𝑗⩽4 ) one recovers the formula for 𝑑𝑒𝑡 𝑆 2 from [3] (see also [4]). A different approach to get the same result is to show that 𝑑𝑒𝑡(𝐴 1 ((𝑣 𝑖,𝑗 ) 1⩽𝑖<𝑗⩽4 )) satisfies the universality property of the map 𝑑𝑒𝑡 𝑆…”

“…Next we recall from [2,3] and [4] a few results about Λ 𝑆 2 𝑉 𝑑 and the 𝑑𝑒𝑡 𝑆 2 map. For every 𝑛 ⩾ 0 we define…”

“…Next we recall from [2, 3] and [4] a few results about $$\Lambda _{V_d}^{S^2}$$ and the $$det^{S^2}$$ map. For every $$n\geqslant 0$$ we define $$$$\begin{equation*} \Lambda _{V_d}^{S^2}[n]=\frac{\mathcal {T}^{S^2}_{V_d}[n]}{\mathcal {E}^{S^2}_{V_d}[n]}, \end{equation*}$$$$where $$\mathcal {T}^{S^2}_{V_d}[n]=V_d^{\otimes \frac{n(n-1)}{2}}$$, and $$\mathcal {E}^{S^2}_{V_d}[n]$$ is the subspace of $$\mathcal {T}^{S^2}_{V_d}[n]$$ generated by those elements $$\otimes _{1\leqslant i&lt;j\leqslant n}(v_{i,j})\in V_d^{\otimes \frac{n(n-1)}{2}}$$…”

“…And so, the corresponding columns of matrix $$A_1((v_{i,j})_{1\leqslant i&lt;j\leqslant 4})$$ are linearly dependent, which means that $$det(A_1((v_{i,j})_{1\leqslant i&lt;j\leqslant 4}))=0$$. Finally, one can check that if $$e_{1,2}=e_{1,4}=e_{2,3}=e_1$$, and $$e_{1,3}=e_{2,4}=e_{3,4}=e_2$$, then $$det(A_1((e_{i,j})_{1\leqslant i&lt; j\leqslant 4}))=1$$, which combined with the uniqueness of the map $$det^{S^2}$$ proved in [3] for $$d=2$$, it shows that …”

“…This conjecture is equivalent with the existence and uniqueness (up to a scalar) of a non‐trivial linear map $$det^{S^2}:V_d^{\otimes d(2d-1)}\rightarrow k$$ with the property that $$det(\otimes _{1\leqslant i&lt;j\leqslant 2d} (v_{i,j}))=0$$ if there exist $$1\leqslant x&lt;y&lt;z\leqslant 2d$$ such that $$v_{x,y}=v_{x,z}=v_{y,z}$$. The conjecture was checked to be true in the case $$d=2$$ [3] and $$d=3$$ [2].…”