# Representation of symmetric group as Cremona transformations

Question from me and a colleague:

Given a matrix $$$$U = \begin{bmatrix} U_{11} & U_{12} \\ U_{21} & U_{22} \end{bmatrix} \quad \text{with } U_{22} \neq 0,$$$$ denote its determinant by $$\Delta$$ and let $$$$\label{defineswap} \operatorname{sw}(2 \overline{2})(U) = \begin{bmatrix} \Delta/U_{22} & U_{12}/U_{22}\\ -U_{21}/U_{22} & 1/U_{22} \end{bmatrix},$$$$ a matrix with determinant $$U_{11}/U_{22}$$. (The notation is meant to suggest the word swap.") Suppose that the indeterminates $$U_{11}$$, $$U_{12}$$, $$U_{21}$$, $$U_{22}$$ are replaced by rational functions of four indeterminates $$z(1)$$, $$z(2)$$, $$z(\overline{1})$$, $$z(\overline{2})$$. Then these two equations are equivalent: $$$$U \begin{bmatrix} z(1) \\ z(2) \end{bmatrix} = \begin{bmatrix} z(\overline{1}) \\ z(\overline{2}) \end{bmatrix} \quad \Leftrightarrow \quad \operatorname{sw}(2 \overline{2})(U) \begin{bmatrix} z(1) \\ z(\overline{2}) \end{bmatrix} = \begin{bmatrix} z(\overline{1}) \\ z(2) \end{bmatrix}.$$$$

This seems to be the tip of an iceberg: there is a representation $$\operatorname{sw}: S_{4} \to \operatorname{Crem}(M_2)$$ of the symmetric group $$S_{4}$$ in the group of Cremona transformations of the space of $$2$$-by-$$2$$ matrices. We think of $$S_{4 }$$ as the group of permutations of the set $$\{1,2,\overline{1},\overline{2}\}$$, and for each such permutation $$\pi$$ we have $$$$\operatorname{sw}(\pi)(U) \begin{bmatrix} z(\pi(1)) \\ z(\pi(2)) \end{bmatrix} = \begin{bmatrix} z(\pi(\overline{1})) \\ z(\pi(\overline{2})) \end{bmatrix}.$$$$ If $$\pi$$ is an element of the subgroup $$S_2 \times S_2$$, then $$\operatorname{sw}(\pi)$$ is just the appropriate simultaneous permutation of rows and columns.

This leads us to ask whether there is a representation $$\operatorname{sw}: S_{2n} \to \operatorname{Crem}(M_n)$$ with similar properties. Suppose that $$U$$ is an $$n$$-by-$$n$$ matrix of indeterminates, and that its entries have been replaced by rational functions in the indeterminates $$z(1), z(2), \dots, z(n), z(\overline{1}), z(\overline{2}), \dots, z(\overline{n})$$ so as to satisfy this equation: $$$$U \begin{bmatrix} z(1) \\ z(1) \\ \vdots \\ z(n) \end{bmatrix} = \begin{bmatrix} z(\overline{1}) \\ z(\overline{2}) \\ \vdots \\ z(\overline{n}) \end{bmatrix}.$$$$ Think of $$S_{2n}$$ as the group of permutations of the set $$\{1,2,\dots,n,\overline{1},\overline{2},\dots,\overline{n}\}$$. We demand that for each such permutation $$\pi$$ we have $$$$\operatorname{sw}(\pi)(U) \begin{bmatrix} z(\pi(1)) \\ z(\pi(2)) \\ \vdots \\ z(\pi(n)) \end{bmatrix} = \begin{bmatrix} z(\pi(\overline{1})) \\ z(\pi(\overline{2})) \\ \vdots \\ z(\pi(\overline{n})) \end{bmatrix},$$$$ and we demand that if $$\pi$$ is an element of the subgroup $$S_n \times S_n$$, then $$\operatorname{sw}(\pi)$$ is the appropriate simultaneous permutation of rows and columns.