Let $E={\mathbb C}^3$ with coordinate basis $\{e_1,e_2,e_3\}$, $F={\mathbb C}^4$ with coordinate basis $\{f_1,\ldots,f_4\}$ and $H={\mathbb C}^2$ with coordinate basis $\{h_1,h_2\}$. Denote $R$ the symmetric algebra $Sym((E\oplus F)\otimes H)$; $R$ is a polynomial ring with variables $x_{i,j} = e_i\otimes h_j$ and $y_{i,j} = f_i\otimes h_j$. We take the variables $x_{i,j}$ to have degree $(1,0)$ and the variables $y_{i,j}$ to have degree $(0,1)$. Let $G$ be a $2\times 7$ generic matrix of variables in $R$ and $I$ the ideal generated by the $2\times 2$ minors of $G$. The minimal free resolution of $I$ is an example of an Eagon-Northcott complex (see Eisenbud - Commutative Algebra, Appendix A2.6).
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Notice how the first three columns of $G$ involve only the $x_{i,j}$ variables while the other columns involve only the $y_{i,j}$ variables. In fact, $G$ is the matrix of the map $$\phi : E\otimes R(-1,0) \oplus F\otimes R(0,-1) \rightarrow H^* \otimes R, e_i \otimes 1 \mapsto \sum_{k=1}^2 h_k^* \otimes x_{i,k}, f_j \otimes 1 \mapsto \sum_{k=1}^2 h_k^* \otimes y_{j,k}$$ with respect to the bases $\{e_1\otimes 1,\ldots,e_3\otimes 1,f_1\otimes 1,\ldots,f_4\otimes 1\}$ of the domain and $\{h_1^*\otimes 1,h_2^*\otimes 1\}$ of the codomain.
The ring $R$ carries a degree compatible action of $SL_3 ({\mathbb C}) \times SL_4 ({\mathbb C}) \times SL_2 ({\mathbb C})$. Notice how the $SL_3 ({\mathbb C})$ factor acts non trivially on $E$, i.e., on the variables $x_{i,j}$, and trivially on $F$, i.e., on the variables $y_{i,j}$. The opposite holds for the $SL_4 ({\mathbb C})$ factor. The map $\phi$ is $G$-equivariant and so the ideal generated by the $2\times 2$ of $\phi$ inherits the $G$-action on $R$.
The weight of $x_{i,j} = e_i\otimes h_j$ is obtained by concatenating the weight of $e_i$ with the trivial weight {0,0,0} for the action of $SL_4 ({\mathbb C})$ and then with the weight of $h_j$. Similarly the weight of $y_{i,j} = f_i\otimes h_j$ is obtained by concatenating the trivial weight {0,0} for the action of $SL_3 ({\mathbb C})$ with the weight of $f_i$ and then with the weight of $h_j$. As in Example 2, we automate the process as illustrated below and then we attach the list of weights to the ring $R$.
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Now we decompose the resolution of $I$.
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We show what part of the resolution looks like in terms of Schur functors: $$R \leftarrow (\wedge^2 F \otimes R(0,-2)) \oplus (E \otimes F \otimes R(-1,-1)) \oplus (\wedge^2 E \otimes R(-2,0)) \leftarrow (\wedge^3 F \otimes H \otimes R(0,-3)) \oplus (E \otimes \wedge^2 F \otimes H \otimes R(-1,-2)) \oplus (\wedge^2 E \otimes F \otimes H \otimes R(-2,-1)) \oplus (H \otimes R(-3,0)) \leftarrow \ldots$$
Finally we decompose some graded components in the quotient ring $R/I$. Unlike with $\ZZ$-gradings, when working in the multigraded setting it is not possible to decompose a range of degrees but only one multidegree at a time.
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We have $Q_{(2,0)} = S_2 E \otimes S_2 H$, $Q_{(1,1)} = E \otimes F \otimes S_2 H$ and $Q_{(0,2)} = S_2 F \otimes S_2 H$.