Quotient of $\mathbb{C}^2$ by the action of $S_2$ - contradiction?

Question

Question: Is there not a contradiction between (1) the answer of Qiaochu Yuan to this question and (2) the answer of Georges Elencwajg to that question in the case $n=2$?

In (1) the question is about the quotient $Q$ of $\mathbb{C}^n$ by the group action of $\mathbb{Z}_n$ via cyclically permuting the coordinates. In the case $n=2$, the answer boils down to $Q$ being homeomorphic to $\mathbb{C} \times A := \mathbb{C} \times \{z \in \mathbb{C} : 0 \leq \text{arg}(z) < \pi \}$. This results from the action being conjugate to the $\mathbb{Z}_2$-action given by flipping the second coordinate and leaving the first fixed; i.e. by $(z_1, z_2) \mapsto (z_1, - z_2)$ and the quotient of $\mathbb{C}$ by $z \mapsto -z$ being $A$.

In (2) the question is about continuity of the solutions of a polynomial over $\mathbb{C}$ in terms of its coefficients. Let $p(z) = \sum_{k=0}^{n-1} c_k z^k = \prod_{k=0}^{n-1} (a_k - z)$ be a polynomial over $\mathbb{C}$. Then the map $v: \mathbb{C}^n \rightarrow \mathbb{C}^n; (a_0, \ldots, a_{n-1}) \mapsto (c_0, \ldots, c_{n-1})$ is polynomial and invariant under the $S_n$-action on $\mathbb{C}^n$ given by permuting the components. Hence $v$ descends to a map $w:\mathbb{C}^n/S_n \rightarrow \mathbb{C}^n$ and the answer claims that this is a homeomorphism.

But in the case $n=2$, by (1) and $\mathbb{Z}_2 = S_2$ there cannot be such a homeomorphism; e.g. since $\mathbb{C} \times A$ is not a manifold.

---

What am I missing here?

Thank you very much in advance for any input.

Answer 1

I don't see anywhere in Qiaochu's answer where it is claimed that $\Bbb C^2/S_2$ is not a manifold. Which is good, because $\Bbb C^2/S_2$ is a manifold. For instance, by a classical theorem of Bourbaki (mathSE source), if $E\subset M\times M$ is the graph of an equivalence relation such that $E$ is a closed submanifold of $M\times M$ and the first projection $E\to M$ is a submersion, then the quotient space $M/E$ inherits from $M$ a smooth manifold structure.

In the case where we leave the $S_2$ action as the coordinate swap, the equivalence relation $E\subset \Bbb C^2\times \Bbb C^2$ corresponding to identifying a point and its image under the group action by $S_2$ can be written as the common vanishing locus of $z_1-w_2$ and $z_2-w_1$, or a 2-plane in 4-space. Then $E$ verifies the hypotheses is a closed submanifold of $\Bbb C^2\times\Bbb C^2$ and the first projection is a submersion, so $\Bbb C^2/S_2$ is a manifold. In the case where we conjugate the $S_2$ action to the action by $(z_1,z_2)\mapsto (z_1,-z_2)$, the equivalence relation $E\subset\Bbb C^2\times\Bbb C^2$ is the common vanishing locus of $z_1-w_1$ and $z_2+w_2$, which is again verifies the hypotheses of the theorem.

---

This same theorem actually proves that your $A$ when viewed as $\Bbb C/S_2$ is a manifold by taking $E\subset\Bbb C\times\Bbb C$ as the vanishing locus of $z_1+w_1$. (This can also be seen intuitively by looking at CW complex structure on $\Bbb C$ with cells the origin, the positive real axis, the negative real axis, and the upper and lower half-planes - the quotient by this is the CW complex with one zero-cell, one one-cell, and one two-cell attached just like they are if you were creating a CW complex structure on $\Bbb C$ with the origin, the positive real axis, and the entire space as your cells. More formally, the squaring map and the principal branch of the square root define homeomorphisms between $A$ and $\Bbb C$.)