An Extension of Viète’s Theorem

Viète’s theorem tells us that $Sym^n(\mathbb{CP}^1) \cong \mathbb{CP}^{2n}$ , i.e. there is a diffeomorphism between the space of sets of unordered n lines passing through the origin of $\mathbb{C}^2$ and the space of lines passing through the origin of $\mathbb{C}^{2n+1}$ .

In his PDE book, Arnold claimed that $Sym^n(\mathbb{RP}^2) \cong \mathbb{RP}^{2n}$ and gave a proof based upon some corollaries and theorems obtained while proving Maxwell’s theorem. He also said that Viète’s theorem is the quaternion version of $Sym^n(\mathbb{RP}^2) \cong \mathbb{RP}^{2n}$ ; this will be the main focus of this post. The proof that Arnold gave tends more towards algebraic geometry (I will give a thorough examination of this later in this post). In the first part of this post, I will share a proof that I have in mind, which is really an extension of Viète’s theorem. By constructing multi-valued maps between $Sym^n(\mathbb{CP}^1)$ and $Sym^n(\mathbb{RP}^2)$ , and $\mathbb{CP}^n$ and $\mathbb{RP}^2n$ , respectively, we will lift the diffeomorphism between $Sym^n(\mathbb{CP}^1)$ and $\mathbb{CP}^n$ to one between $Sym^n(\mathbb{RP}^2)$ and $\mathbb{RP}^{2n}$ .

We start by constructing the $2^n$ -sheeted covering from $\mathbb{CP}^n$ to $\mathbb{RP}^{2n}$ (**); we will also show that $\mathbb{CP}^2/B(2) \cong \mathbb{RP}^4$ .

We first look at two examples: when n = 1 and n = 2.

The maps from $\mathbb{CP}^n \rightarrow \mathbb{RP}^{2n}$ are constructed from the elementary symmetric forms.

When n = 1, we construct $r: \mathbb{CP}^1 \rightarrow \mathbb{RP}^2$ in the following method:

Let $[z, w] \in \mathbb{CP}^1$ be arbitrary. This corresponds to the line $zx + wy = 0$ in $\mathbb{C}^2$ . Its Hermitian-orthogonal is given by $\bar{w}x - \bar{z}x = 0$ , and the corresponding point in $\mathbb{CP}^1$ is $[\bar{w}, -\bar{z}]$ .

We look at $f_{[z, w]} = (zx + wy)(\bar{w}x - \bar{z}y) = z\bar{w}x^2 + (w\bar{w} - z\bar{z})xy - \bar{z}wy^2$ .

Using the coefficients of the above polynomial, for each point $[z, w] \in \mathbb{CP}^1$ , we obtain a unique ray (scaling by a positive constant) in $\mathbb{R}^3$ :

$[z, w] \rightarrow \{(\lambda \cdot Re(z\bar{w}), \lambda \cdot Im(z\bar{w}), \lambda \cdot (w\bar{w} - z\bar{z})) : \lambda \in \mathbb{R}_{+}\}$ .

Notice that the Hermitian-orthogonal $[\bar{w}, -\bar{z}]$ does not get mapped to the same ray; it gets mapped to the antipodal ray in $\mathbb{R}^3$ :

$f_{[\bar{w}, -\bar{z}]} = (\bar{w}x - \bar{z}y)(-zx - wy) = -z\bar{w}x^2 - (w\bar{w} - z\bar{z})xy + \bar{z}wy^2$

The space of rays in $\mathbb{R}^3$ can be identified as $S^2$ in the obvious way (each ray corresponds to a point on the sphere).

Thus, the map $[z, w] \rightarrow [\bar{w}, -\bar{z}]$ (which is the antiholomorphic involution map on $\mathbb{CP}^1$ , discussed in a previous post) under this map gives rise to the antipodal map on $S^2$ .

It follows that by identifying $[z, w]$ and $[\bar{w}, -\bar{z}]$ as one point (we are now looking at the space $\mathbb{CP}^1/B(1)$ ), we have the following bijection induced from the above bijection (it can be shown that is a diffeomorphism) from $\mathbb{CP}^1 \rightarrow S^2$ :

${\ [z, w], [\bar{w}, -\bar{z}]} \rightarrow [Re(z\bar{w}), Im(z\bar), w\bar{w} - z\bar{z}] \in \mathbb{RP}^2$ .

This map can again be shown to be a diffeomorphism.

Thus, we have $\mathbb{CP}^1/B(1) \cong \mathbb{RP}^2$ .

Now, let’s look at when n = 2. We will construct a $2^2$ -sheeted covering from $\mathbb{CP}^2 \rightarrow \mathbb{RP}^4$ .

Since $Sym^n(\mathbb{CP}^1) \cong \mathbb{CP}^n$ , a point in $\mathbb{CP}^2$ can be identified as an unordered pair of points in $\mathbb{CP}^1$ , i.e. $(z_1, w_1), (z_2, w_2)$ .

These two vectors together with their Hermitian-orthogonals give us:

$(z_1, w_1), (\bar{w_1}, -\bar{z_1}), (z_2, w_2), (\bar{w_2}, -\bar{z_2})$ .

Similar to the case of when n=1, the corresponding symmetric forms can be obtained as the coefficients of the following polynomial:

$f = (z_1x + w_1y)(\bar{w_1}x - \bar{z_1}y)(z_2x + w_2y)(\bar{w_2}x - \bar{z_2}y)$ .

Expanding the LHS gives:

$f$

$= z_1 \bar{w_1} z_2 \bar{w_2} x^4 + [z_1 \bar{w_1}(w_2 \bar{w_2} - z_2 \bar{z_2}) + z_2\bar{w_2}(-z_1\bar{z_1} + w_1\bar{w_1})]x^3y + [-z_1\bar{w_1}\bar{z_2}w_2 + (w_1\bar{w_1} - z_1\bar{z_1})(w_2\bar{w_2} - z_2\bar{z_2}) - \bar{z_1}w_1 z_2 \bar{w_2})x^2y^2 - \overline{[z_1 \bar{w_1}(w_2 \bar{w_2} - z_2 \bar{z_2}) + z_2\bar{w_2}(-z_1\bar{z_1} + w_1\bar{w_1})]}xy^3$ $- \overline{[z_1\bar{w_1}z_2\bar{w_2}]}y^4$

$\equiv f_0x^4 + f_1x^3y + f_2x^2y^2 - \bar{f_1}xy^3 - \bar{f_0}y^4$ .

Notice $f_2 \in \mathbb{R}$ .

And so we have a map from $Sym^2(\mathbb{CP}^1)$ to $\mathbb{R}^5$ :

$[z_1, w_1], [z_2, w_2] \rightarrow (Re(f_0), Im(f_0), Re(f_1), Im(f_1), f_2)$ .

Thus, we have a map from $Sym^2(\mathbb{CP}^1)$ to $S^4$ .

This is a 2:1 map; let’s call it $r$ . Notice that $[\bar{w_1}, - \bar{z_1}], [\bar{w_2}, -\bar{z_2}]$ gets mapped to the same ray, i.e. the antiholomorphic involution in $Sym^2(\mathbb{CP}^1) \cong \mathbb{CP}^n$ under $r$ induces the identity map on $S^4$ .

In his book, Arnold claimed that this antiholomorphic involution under $r$ is the same as the complex conjugation map (in this case, they leave the resulting image unchanged in $S^4$ ); this is because each factor of this map is a symmetric form of degree n in terms of $z_1, ..., z_n, w_1, ..., w_n$ . Hence, each factor is invariant under permutations of $z_1, ..., z_n, w_1, ..., w_n$ and when n is even, the -1’s cancel out.

Hence, we have a bijection from $\mathbb{CP^2}/conj. \rightarrow S^4$ (showing that this bijection is smooth tells us that $\mathbb{CP^2}/conj \cong S^4$ .

Now, in $\mathbb{CP}^2$ , if we identify {{ $[z_1, w_1], [z_2, w_2]$ }}, {{ $[\bar{w_1}, -\bar{z_1}], [z_2, w_2]$ }}, {{ $[z_1, w_1], [\bar{w_2}, \bar{z_2}]$ }}, {{ $[\bar{w_1}, -\bar{z_1}], [\bar{w_2}, \bar{z_2}]$ }} as one point (i.e. we consider the space $Sym^2(\mathbb{CP}^1)/B(2)$ ), then we have a bijection to the space of lines in $\mathbb{R}^5$ . This is because the points {{ $[\bar{w_1}, -\bar{z_1}], [z_2, w_2]$ }}, {{ $[z_1, w_1], [\bar{w_2}, \bar{z_2}]$ }}, under the map $r$ , are both mapped to the antipodal ray corresponding to $r([z_1, w_1], [z_2, w_2])$ .

Thus, we have $\mathbb{CP^2}/B(2) \cong \mathbb{RP}^4$ .

I will not show the general result for $\mathbb{CP}^n/B(n) \cong \mathbb{RP}^{2n}$ in this post.

From the construction we did for n = 2, the diffeomorphism is really between $Sym^n(\mathbb{CP}^n)/B(n)$ and $\mathbb{RP}^{2n}$ .

We now focus on (*), i.e. finding a $2^n$ -sheeted covering from $Sym^n(\mathbb{CP}^n)$ and $Sym^{2n}(\mathbb{RP}^2)$ .

Given a point {{ $[z_1, w_1], [z_2, w_2], ..., [z_n, w_n]$ }} $\in Sym^n(\mathbb{CP}^n)$ , my thinking is that we can take the diffeomorphism we constructed for $\mathbb{CP}^1 \rightarrow \mathbb{RP}^2$ to map each $[z_i, w_i]$ for $1 \leq i \leq n$ to a point in $\mathbb{RP}^2$ :

$[z_i, w_i] \rightarrow [Re(z_i\bar{w_i}, Im(z_i\bar{w_i}), z_i\bar{z_i}-w_i\bar{w_i}] \in \mathbb{RP}^2$ .

For each factor of $\mathbb{CP}^1$ , this map is a 2:1 cover (recall the above for n = 1), and so, putting this together as a map from $Sym^n(\mathbb{CP}^1) \rightarrow Sym^n(\mathbb{RP}^2)$ , we have a $2^n :1$ cover.

An Extension of Viète’s Theorem

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