Showing Local Diffeomorphism (Complete. Last Updated: August 27th)

This is a discussion of A.4 (Explicit Formula) of the appendix of Arnold’s PDE.

Goal: Show $Sym^n(\mathbb{RP}^2) \cong \mathbb{RP}^{2n}$ .

In Arnold’s proof, he split the proof into two parts.

First, he showed using the real algebraic theorem to show the existence of a homeomorphism between $Sym^n(\mathbb{RP}^2)$ and $\mathbb{RP}^{2n}$ . A detailed post about this section will be made soon.

Then, we must show this homeomorphism is a diffeomorphism (it suffices to show that it is a local diffeomorphism).

The map we construct is the following:

Given n lines and their Hermitian-orthogonal passing through 0 in $\mathbb{C}^2$ , we can represent them as:

$\{[z_1, w_1], [-\overline{w_1}, \overline{z_1}]\}, ..., \{[z_n, w_n], [-\overline{w_n}, \overline{z_n}]\}$ .

From each pair, we can choose a representation with a non-zero $w$ coordinate. Setting $w = 1$ (because we are in projective space), we obtain:

$[z_1, 1], ..., [z_n, 1]$ .

Key: Arnold said that given repeating pairs, we must choose the same representative. This is essential later to showing the Jacobian of our map is non-vanishing. Why?

Consider the following form:

$f(X, Y) = \prod_{k=1}^{n}(z_kX + Y) \prod_{k=1}^{n} (X - \overline{z_k}Y)$ (*)

Expanding (*) gives:

$f(X, Y) = f_0 X^{2n} + f_1 X^{2n-1} Y + ... + f_{2n} Y^{2n}$ ,

where (taken from Arnold’s PDE), the particular expressions are given by:

$f_0 = \sigma_n$

$f_1 = \sigma_{n-1} - \sigma_n \overline{\sigma_1}$

…

$f_n = 1 - \sigma_1 \overline{\sigma_1} + ... + (-1)^n \sigma_n \overline{\sigma_n}$

and $latex f_{2n-k} = (-1)^{n-k} \overline{f_k}

Claim: the map $(z_1, z_2, ..., z_n) \rightarrow (f_0, f_1, ..., f_2n)$ determines a unique ray in $\mathbb{R}^{2n+1}$ .

Proof via an example (can be generalized):

When n = 1, consider:

$F(X, Y) = (z_1 X + w_1 Y)(\overline{w_1} X - \overline{z_1} Y) = z_1\overline{w_1} X^2 + (w_1\overline{w_1} - z_1\overline{z_1})XY - \overline{z_1}w_1 Y^2$

From the above form, we obtain the map:

$\psi: [z_1, w_1] \rightarrow \{ |\lambda|^2 (z_1\overline{w_1}, w_1\overline{w_1} - z_1\overline{z_1}, -\overline{z_1}w_1): \lambda \in \mathbb{C} \backslash \{0\} \}$

this is because $[\lambda z_1, \lambda w_1] = [z_1, w_1]$ for any non-zero $\lambda \in \mathbb{C}$ .

By a change of coordinate, we have:

$\tilde{\psi}: [z_1, w_1] \rightarrow \{ |\lambda|^2 (\text{Re}(z_1\overline{w_1}), \text{Im}(z_1\overline{w_1}), w_1\overline{w_1} - z_1\overline{z_1}): \lambda \in \mathbb{C} \backslash \{0\} \} \subseteq \mathbb{R}^3$

which is a ray passing through the origin in $\mathbb{R}^3$ .

The ray corresponding to $[\overline{w_1}, -\overline{z_1}]$ is:

$[\overline{w_1}, -\overline{z_1}] \rightarrow \{ |\lambda|^2 (- \text{Re}(z_1\overline{w_1}), - \text{Im}(z_1\overline{w_1}), - w_1\overline{w_1} + z_1\overline{z_1}): \lambda \in \mathbb{C} \backslash \{0\} \}$

i.e. the antipodal ray.

Suppose WLOG $w_1 \neq 0$ (otherwise, we choose the Hermitian-orthogonal), then we can set $w_1 = 1$ . The map becomes:

$[z_1, w_1] = [z_1, 1] \rightarrow (\text{Re}(z_1), \text{Im}(z_1), 1 - z_1\overline{z_1}) = (\sigma_1, -\overline{\sigma_1}, 1 - \sigma_1 \overline{\sigma_1}) \subset \mathbb{R}^3$

this point corresponds uniquely to the ray we obtained before and this map (up to ordering of the coordinates) is the one that Arnold had described.

The coefficients of the polynomial $P(x, y) = \Pi_{k=1}^n (z_k x + y) \Pi_{k=1}^n (x - \overline{z_k}y)$ are the same as the coefficients of the polynomial $P(z) = (z_1 ... z_n) \Pi_{k=1}^n (z + \frac{1}{z_k}) \Pi_{k=1}^n (z - w_k)$ .

For a polynomial $a \Pi_{k=1}^n (x - \alpha_k)$ , the Jacobian of the map from the roots to the coefficients is $a^n \Pi_{i < j} (\alpha_i - \alpha_j)$ .

In our case of P(z), the determinant of the Jacobian is:

$(z_1 ... z_n) \Pi_{1 \leq i < j \leq 2n} (\alpha_i - \alpha_j)$

where $\alpha_i = -\frac{1}{z_i}, \alpha_{n+i} = -w_i$ for $1 \leq i \leq n$ .

But what would this determinant be if instead of the map $(\alpha_1, \alpha_2, ..., \alpha_n) \rightarrow (F_0, ... F_2n)$ , we wanted the map $(z_1, z_2, ..., w_n) \rightarrow (F_0, ..., F_2n)$ ?

We must account for the change of variable: $\alpha_i = -\frac{1}{z_i}, \alpha_{n+i} = -w_i$ .

The determinant of the Jacobian of this transformation is:

$|\frac{\partial{(\alpha_1, ..., \alpha_n)}{\partial{(z_1, ..., z_n, w_1, ..., w_n)}}|$

$= \Pi_{k=1}^n |\frac{d}{dz_k} (\frac{1}{z_k})| \Pi_{k=1}^n |\frac{d}{dw_k} (-w_k)|$

$= \Pi_{k=1}^n \frac{1}{z_k^2}$

Thus, by the chain rule, the determinant of the Jacobian of our map from $(z_1, ..., z_n, w_1, ..., w_n)$ to $(F_0, ..., F_2n)$ is:

$F_0^{2n} [\Pi_{1 \leq i < j \leq 2n} (\alpha_i - \alpha_j)] \cdot [\Pi_{k = 1}^n \frac{1}{z_k^2}]$ .

Notice that the determinant vanishes if and only if two roots $\alpha_i, \alpha_j$ coincide.

By our choice (no $z_i$ coincides with any $-\frac{1}{\overline{z_j}}$ , i.e. when $w_j = \overline{z_j}$ , $z_i \neq -\frac{1}{w_j}$ for any $1 \leq i, j \leq n$ ) and so, the product $\Pi_{i = 1}^n \Pi_{j = 1}^n (- \frac{1}{z_i} - w_j)$ does not vanish. Under a change of coordinate (i.e. $z_1, z_2, ..., w_n \rightarrow \sigma_1, \sigma_2, ... \tau_n$ ), the former part of the Jacobian ( $\Pi_{1 \leq i < j \leq n} (- \frac{1}{z_i} + \frac{1}{z_j})(w_i - w_j)$ ) is also guaranteed not to vanish.