The Octahedron is the Complexification of the Tetrahedron (Part 3)

In this post, we show that the group of pseudo-projective transformations on $\mathbb{CP}^2$ preserving the set of 4 points of the tetrahedron is isomorphic to $S_4 \times \mathbb{Z}_2 \cong B_3$ . Pseudo-projective is the condition that complex projective subspaces are mapped to complex projective subspaces. In the case of $\mathbb{CP}^2$ , this would mean complex projective lines are mapped to complex projective lines.

Recall in part 2: we have shown that the group of complex projective transformations on $\mathbb{CP}^2$ preserving the set of 4 points of the tetrahedron is isomorphic to $S_4$ .

Thus, it seems natural to ask the question: does there exist a relation between pseudo-projective transformations and complex projective transformations? If so, what is it?

Turns out, any pseudo-projective transformation of the complex projective plane is either complex projective or becomes complex projective after (left or right) multiplication by the complex conjugation.

Note: we assume all transformations to be smooth.

Step 1: We first study the derivative of a pseudo-projective transformation at a point.

Given an affine chart, we can, WLOG, choose the point to be the pre-image of 0 (in this affine chart) under the transformation. This is possible by a suitable choice of coordinates.

To study the derivative of the transformation at this point, we must consider a neighbourhood of this point.

Choose an affine system of complex coordinates z, w in a neighbourhood around 0 (given by z = 0, w = 0). The complex lines passing through 0 are of the form w = pz, for some complex number p.

The derivative of the transformation at the point that is mapped to 0 is, by definition, a real $\mathbb{R}$ -linear transformation that maps $\mathbb{C}$ -lines to $\mathbb{C}$ -lines (Why?).

The fact that the derivative maps $\mathbb{C}$ -lines to $\mathbb{C}$ -lines follows from the definition of pseudo-projective transformations. For instance, denote the pseudo-projective transformation to be $F: \mathbb{CP}^2 \rightarrow \mathbb{CP}^2$ . We know that $F$ maps every complex projective line L to another complex projective line L’. Given a complex projective line L passing through p, a point in $\mathbb{CP}^2$ . Then, $T_p L \subset T_p \mathbb{CP}^2 \cong \mathbb{C}^2$ and $T_p L$ is one-dimensional. Thus, $T_p L \cong \mathbb{C}$ .

Since $F(L) = L'$ , $dF_p(L) \subset dF_{F(p)}(L')$ . Since F is a diffeomorphism, $dF_p(L) = dF_{F(p)}(L') \cong \mathbb{C}$ . Hence, $dF$ maps $\mathbb{C}$ -lines.

The derivative $dF_p: \mathbb{C}^2 \rightarrow \mathbb{C}^2$ is a $\mathbb{R}$ -linear transformation that maps complex lines to complex lines. We call these transformations that map complex subspaces to complex subspaces pseudo-complex.

Claim: Pseudo-complex $\mathbb{R}$ -linear transformation of $\mathbb{C}^2$ is either $\mathbb{C}$ -linear or a $\mathbb{C}$ -linear transformation by the complex conjugation.

Denote the transformation by T. Let the coordinates on $\mathbb{C}^2$ be (z, w). Let $x_1$ be a non-zero vector on the complex line, say L, that gets mapped to $w = 0$ under T. If we view L as an $\mathbb{R}$ -plane, then it is spanned by $x_1$ and $y_1 = i x_1$ .

Denote $X_1 = T(x_1)$ and $Y_1 = T(y_1)$ . Since T maps L to $w = 0$ , it follows that $X_1, Y_1$ span $w = 0$ when it is viewed as an $\mathbb{R}$ -plane.

$X_1, Y_1$ define a new complex structure $I_1$ on $w = 0$ :

$I_1 Y_1 = - X_1$ and $Y_1 = I_1 X_1$ .

Note: $I_1$ is in general different from the usual complex structure on $w = 0$ .

Let $e_1$ be a non-zero vector on $w = 0$ . Then, when viewed as an $\mathbb{R}$ -plane, $e_1$ and $f_1 = i e_1$ span $w = 0$ . Since $X_1, Y_1$ are both in $w = 0$ , there exist $a_1, b_1, c_1, d_1 \in \mathbb{R}$ such that:

$X_1 = a_1 e_1 + b_1 f_1$

$Y_1 = c_1 e_1 + d_1 f_1$ .

In the $e_1, f_1$ basis,

$I_1 = \frac{1}{a_1d_1 - b_1c_1} \begin{bmatrix} a_1b_1 + c_1d_1 & -(a_1^2 + c_1^2) \\ b_1^2 + d_1^2 & -(a_1b_1 + c_1d_1) \end{bmatrix}$ ,

whereas the usual complex structure in $e_1, f_1$ basis is given by:

$i = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ . $\square$

Similarly, we can choose a basis of vectors $e_2, f_2 = i e_2$ on $z = 0$ ; the complex structure $I_2$ induced by the transformation can be represented by $Y_2 = I_2 X_2$ and $I_2 Y_2 = - X_2$ (where $X_2, Y_2$ are the image of vectors $x_2, y_2 = ix_2$ on the pre-image of $z = 0$ ).

$x_1, x_2$ forms a complex basis for $\mathbb{C}^2$ . Thus, for any $v \in \mathbb{C}^2$ ,

$v = \alpha x_1 + \beta x_2$ , for some $\alpha, \beta \in \mathbb{C}$ .

If $\alpha \neq 0$ , $v = \alpha(x_1 + \frac{\beta}{\alpha} x_2)$ .

The complex line $L$ spanned by $v$ can be expressed as:

$L = \mathbb{C} v = \mathbb{C}(x_1 + \frac{\beta}{\alpha}x_2)$ .

Write $\frac{\beta}{\alpha} = s + it$ for some $s, t \in \mathbb{R}$ . Then,

$v = x_1 + sx_2 + itx_2 = x_1 + sx_2 + ty_2$ and

$iv = ix_1 + isx_2 - tx_2 = y_1 + sy_2 - tx_2$ .

Under the transformation, we have:

$X = T(v) = T(x_1 + sx_2 + ty_2) = X_1 + sX_2 + tY_2$ and

$Y = T(iv) = T(y_1 + sy_2 - tx_2) = Y_1 + sY_2 - tX_2$ .

Since T maps complex lines to complex lines, $\text{span} \{ X, iX \} = \text{span} \{ X, Y \} = T(L)$ .

If $\mathbb{C}^2$ is viewed as $\mathbb{R}^4$ and $T(L)$ as one of its 2-dimensional real subspace, then $X, iX, Y$ are $\mathbb{R}$ -linearly dependent in $\mathbb{R}^4$ for any coefficients $s, t \in \mathbb{R}$ .

Note: Since $X, Y$ can be expressed uniquely in terms of $X_1, X_2, Y_1, Y_2$ , the induced complex structure $I_1, I_2$ on $z = 0, w = 0$ determines the induced complex structure on all other complex lines.

Substituting the expressions in terms of $e_1, e_2, f_1, f_2$ for $X_1, Y_1, X_2, Y_2$ from above, we find that the corresponding coefficient matrix is given by (if we consider $e_1, f_1, se_2, sf_2, te_2, tf_2$ as variables for the columns, respectively):

$\begin{bmatrix} a_1 & b_1 & a_2 & b_2 & c_2 & d_2 \\ -b_1 & a_1 & -b_2 & a_2 & -d_2 & c_2 \\ c_1 & d_1 & c_2 & d_2 & -a_2 & -b_2 \end{bmatrix}$

Key: the projections of $X, iX, Y$ to any 3-dimensional hyperspace are still linearly dependent.

For instance, setting $s = 1, t = 0$ , we obtain the projection of the 3 vectors to the hyperspace spanned by $e_1, f_1, e_2$ . Thus, the minor in the columns (1, 2, 3) vanishes.

Through this reasoning, it is easy to see that all 20 minors vanish. We thus obtain a system of equations and find (Arnold provided detailed calculations in his paper):

$d_1 = -a_1, c_1 = b_1, c_2 = b_2, d_2 = -a_2$ (1)or

$d_1 = a_1, c_1 = -b_1, c_2 = - b_2, d_2 = a_2$ (2).

If (1) occurs, then $i, I_1, I_2$ are the same. Our pseudo-complex real linear transformation preserves the usual complex structure. Hence, it is a complex linear transformation.

If (2) occurs, then $I_1, I_2 = -I$ . Thus, composing the transformation by a complex conjugation gives us a complex linear transformation. This proves our claim. $\square$

Now that we have discussed the derivative of the transformation at every point, let’s look at the pseudo-projective transformation on the entire $\mathbb{CP}^2$ .

Claim: Any pseudo-projective transformation of the complex projective plane is either complex projective or becomes complex projective after multiplication by the complex conjugation.

Since the complex projective plane is connected, from the previous claim, we can deduce that either the derivative is complex linear or anti-complex (i.e. complex linear after multiplication by complex conjugation) at every point. This is because the difference between complex and anti-complex lies in the difference between the orientations of any complex line defined by the original complex structure and the induced complex structure. Orientation is locally constant. By connectedness, orientation is fixed on the entire complex projective plane.

In the first case, the pseudo-projective transformation is complex smooth and maps complex lines to complex lines. By definition, it is complex projective.

In the second case, if we multiply the pseudo-projective transformation by complex conjugation, we see that the product is complex smooth and maps complex lines to complex lines. Hence, the product is complex projective. This proves the claim.

Returning back to our initial topic: the group of pseudo-projective transformations that permute the set of 4 points of the tetrahedron is isomorphic to $S_4 \times \mathbb{Z}_2$ .

Recall, from the previous post, we can assume the 4 points of the tetrahedron to be [0:1:0], [1:0:0], [0:0:1], [1:1:1]. These 4 points are all real, and so they are fixed by complex conjugation.

We know that the group of complex projective transformations that permute the 4 points is isomorphic to $S_4$ .

Thus, we have at least 48 pseudo-projective transformations that preserve the 4 points (these are the 24 complex projective transformations and the 24 products by conjugation).

We now show that these are all the pseudo-projective transformations that permute the 4 points.

Let B be a pseudo-projective transformation that permute the 4 points. Then, there exists a complex projective transformation A such that $BA^{-1}$ fixes the 4 points.

If $BA^{-1}$ is complex projective, then by the previous post, we know that it must be the identity.

Otherwise, $BA^{-1} \cdot \text{conj}$ is complex projective, and hence, the identity. It follows that $B = \text{conj} \cdot A$ .

Thus, B belongs to one of the 48 pseudo-projective transformations described above.

Note: $\text{conj}$ commutes with the 48 pseudo-projective transformations.

If $A$ is a complex projective transformation, then $\text{conj} \cdot A = A \cdot \text{conj}$ . This is because the 4 points are real and so $\text{conj} \cdot A$ and $A \cdot \text{conj}$ both give the same permutation to the 4 points. Hence, by uniqueness, $\text{conj} \cdot A = A \cdot \text{conj}$ .

The same argument follows if $A$ is not complex projective.

Hence, the group of 48 pseudo-projective transformations is isomorphic to $S_4 \times \mathbb{Z}_2$ . $\square$

The Octahedron is the Complexification of the Tetrahedron (Part 3)

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