# 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 Suppose that $$U$$ is an $$n$$-by-$$n$$ matrix of indeterminates, and that its entries have been replaced by rational functions in the indeterminates $$z(1), z(2), \dots, z(n), z(\overline{1}), z(\overline{2}), \dots, z(\overline{n})$$ so as to satisfy this equation: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.

Serge Cantat provided a definitive explanation. In the Grassmannian $$G(n,2n)$$, there is an open dense subset consisting of those $$n$$-planes that are the graphs of linear maps from $$\mathbb A^n$$ to $$\mathbb A^n$$. The symmetric group $$S^{2n}$$ acts on the Grassmannian, and if we restrict the action to this open set we see the desired representation.

To illustrate in the case $$n=2$$, here's how to obtain the matrix formula shown above. Represent a point of the Grassmannian $$G(2,4)$$ by a 4-by-2 matrix whose columns span the 2-plane. The matrix $$U$$ determines the point

$$\begin{equation*} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ U_{11} & U_{12} \\ U_{21} & U_{22} \end{bmatrix}. \end{equation*}$$

We now compute how the transposition $$(2 \overline{2})$$ acts: $$\begin{equation*} (2 \overline{2}) \cdot \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ U_{11} & U_{12} \\ U_{21} & U_{22} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ U_{11} & U_{12} \\ U_{21} & U_{22} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ U_{21} & U_{22} \\ U_{11} & U_{12} \\ 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \Delta/U_{22} & U_{12}/U_{22}\\ -U_{21}/U_{22} & 1/U_{22} \end{bmatrix}. \end{equation*}$$ (To obtain the last equation we have used column operations to restore an identity matrix at the top.) The bottom square matrix is exactly what we wrote above.

Alternatively, we may use the Plücker coordinates obtained from the six 2-by-2 subdeterminants: \begin{equation*} \begin{aligned} U \mapsto \, & [1:U_{12}:U_{22}:-U_{11}:-U_{21}:\Delta] \\ (2 \overline{2}) \cdot U \mapsto \, & [U_{22}:U_{12}:1:-\Delta:U_{21}:U_{11}] \\ & = \left[1: \frac{U_{12}}{U_{22}}: \frac{1}{U_{22}}: \frac{-\Delta}{U_{22}}: \frac{U_{21}}{U_{22}}: \frac{U_{11}}{U_{22}}: \right] \end{aligned} \end{equation*} and note that the point of projective space $$\mathbb P^5$$ we obtain is the image of the matrix in our formula.