image, and it turns out that the corresponding projective transformation is given by $G_P - 3I$, where $I$ is the identity matrix. Indeed, consider e.g. the first vertex of the pentagon and its lift $V_1$. Then the above formula gives $$(G_P - 3{ {I}})(V_1) = V_1 - \frac{V_{2}\wedge V_1 \wedge V_{4}}{V_{2}\wedge V_3 \wedge V_{4}}\,V_3 - \frac{V_{3}\wedge V_1 \wedge V_{5}}{V_{3}\wedge V_4 \wedge V_{5}}\,V_4.$$ Taking the wedge product of this expression with $V_2 \wedge V_4$ or $V_3 \wedge V_5$ we get zero. This means that $$(G_P - 3{ {I}})(V_1) \in \mathrm{span}(V_2, V_4) \cap \mathrm{span}(V_3, V_5),$$ so the corresponding point in the projective plane is the intersection of diagonals of the pentagon. Furthermore, since Glick’s operator is invariant under cyclic permutations, the same holds for all vertices, meaning that the operator $G_P - 3{ {I}}$ indeed takes a pentagon to its pentagram map image.

Likewise, the second iterate of the pentagram map on hexagons also leads to an equivalent hexagon, and the equivalence is again realized by $G_P - 3I$. Finally, notice that for quadrilaterals $G_P - 2I$ is a constant map onto the intersection of diagonals. These observations make us believe that the operator $G_P$ is per se an important object in 