line $(v_{i+1},v_{i+2})$ (i.e. the line on the right).

These corner invariants are defined on almost the entire space ${\mathcal{P}}_n$ of twisted $n$-gons. Furthermore, these numbers are invariant under projective transformations and hence descend to the space ${\mathcal{P}}_n \,/\, {{\mathbb P}\mathrm{GL}}({\mathbb P}^2)$ of projective equivalence classes of twisted polygons. As shown in [@schwartz2008discrete], the functions $x_1, \dots, x_n, y_1, \dots, y_n$ constitute a coordinate system on an open dense subset of ${\mathcal{P}}_n \,/\, {{\mathbb P}\mathrm{GL}}({\mathbb P}^2)$. This in particular allows one to express the pentagram map, viewed as a transformation of ${\mathcal{P}}_n \,/\, {{\mathbb P}\mathrm{GL}}({\mathbb P}^2)$, in terms of the corner invariants.

(v1) at (-4,2.5) ; (v2) at (-3,1) ; (v3) at (0,0) ; (v4) at (2,1) ; (v5) at (3,2.5) ;

(v1)–(v2); (v5)–(v4); (v4) – (v3); (v2) – (v3); (i1) at (-.5,-2.75) ; (i2) at (-1.75,-.875) ; (i3) at (1.0909,-.363636) ;

(v2) – (i1); (v4) – (i1);

(v3) – (i2); (v3) – (i3);

If we are given a twisted $n$-gon with corner invariants $(x_i,y_i)$, then the corner invariants $(x_i,y_i)$ of its image under the pentagram are given by $$\begin{aligned}
\label{eq3}
    x'_i=x_i\frac{1-x_{i-1} y_{i-1}}{1-x_{i+1} y_{i+1}} \qquad
    y'_i=y_{i+1} \frac{1-x_{i+2}y_{i+2}}{1-x_iy_i}. \end{aligned}$$ These formulas assume a specific labeling of vertices of the pentagram map image. 