This is a continuation of part 1.

In this post, we will show that we consider the 4 points of the tetrahedron in \mathbb{CP}^2, the group of complex projective transformations that preserve this set of 4 points is isomorphic S_4 (the full symmetry group of the tetrahedron).

The natural homomorphism from the transformation group to the permutation group of the 4 points is the action itself. We will now show that this homomorphism is an isomorphism.

To do this, it is easier to consider extending a permutation of points to a transformation. Through this construction, we will find a projective transformation permuting the 4 points in an arbitrary way.

Given the 4 points of the tetrahedron in \mathbb{CP}^2, there correspond 6 different lines passing through the points. Since complex projective transformation maps complex lines to complex lines, we can choose such a transformation that sends one of the 6 lines to the line at infinity. Given homogeneous coordinates [X : Y : Z], the affine chart we consider is Z \neq 0 and so, the line at infinity is given Z = 0. Denote P_1 and P_2 to be the two points that lie on this line at infinity. Choose an affine coordinate on the affine chart such that the x-direction is given by P_1 and the y-direction is given by P_2. Then, P_1 and P_2 are expressed as [1:0:0] and [0:1:0], respectively.

By a suitable translation, we can set one of the remaining 2 points, say P_3, to be at the origin on the affine chart. Thus, P_3 becomes [0:0:1].

There remains one non-zero finite point, P_4. By a suitable homothety, we can assume P_4 to be [1:1:1].

The order of the 4 points in this construction is arbitrary. Thus, there always exists a projective transformation that maps the 4 points, in any order, to the 4 points [1:0:0], [0:1:0], [0:0:1], [1:1:1].

It follows that the image of the above homomorphism from the group of projective transformations preserving the set of 4 points to the permutation group of the 4 points is surjective.

We claim this homomorphism is also injective. It suffices to show that if the order of the 4 points is fixed, then the projective transformation is also fixed.

Claim: the projective transformation fixing the 4 points [1:0:0], [0:1:0], [0:0:1], [1:1:1] is trivial.

Since this projective transformation maps [1:0:0] and [0:1:0] to themselves, it preserves the line at infinity. Hence, it is an affine transformation. In affine coordinates, this transformation can be expressed as (x, y) \mapsto (ax+by+c, dx+ey+f) for some a, b, c, d, e, f \in \mathbb{C}. Since [0:0:1] is preserved, it follows that the transformation is linear, i.e. c = 0 = f. Since the direction of the coordinate axes x and y are preserved, b = d = 0. Since [1:1:1] is preserved, a = e = 1. This proves the claim.

Hence, the homomorphism is an isomorphism. The group of projective transformations the preserve the set of 4 points of the tetrahedron in \mathbb{CP}^2 is indeed isomorphic to the permutation group of the 4 points, S_4.

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