In the paper [2], the author defines the group $G$ to be generated by homeomorphisms of ${\bf S}^2\times{\bf S}^2$ that swap coordinates and/or map them to their antipodal point: $$G=\langle(y,x),(-x,y),(x,-y),(-x,-y),(-y,-x),(y,-x),(-y,x)\rangle$$

(the notation $(y,x)$ is to denote the map $f:(x,y)\mapsto(y,x)$.)

Now, the author proceeds to define two subgroups:

$$J=\langle(y,x)\rangle\qquad\text{and}\qquad H=\langle(y,x),(-x,-y),(-y,-x)\rangle.$$

We see that $J$ acts on ${\bf S}^2\times{\bf S}^2$ like $\mathfrak{S}_2$, which means that the quotient ${\bf S}^2\times{\bf S}^2/J$ is the Symmetric product $\mathrm{SP}^2({\bf S}^2)$. Taking the description of ${\bf CP}^n$ in terms of complex polynomials of degree less than $n$, up to scaling of a polynomial, I checked that $\mathrm{SP}^n({\bf S}^2)\cong{\bf CP}^n$ by the following homeomorphism: $$(z_1,\dots,z_n)\in\mathrm{SP}^n({\bf S}^2)\mapsto\prod_{z_k\neq\infty}(Z-z_k)\in{\bf C}_n[Z]/{\bf C}^\ast\cong{\bf CP}^n,$$ understanding ${\bf S}^2\cong{\bf C}\cup\{\infty\}$.

However, now, I have a problem to check the “explicit calculation” they mention in item (c) below the diagram. I have a map $${\bf S}^2\times{\bf S}^2/J\to{\bf S}^2\times{\bf S}^2/H$$ induced by the inclusion $J\subset H$. Supposedly, the quotient $H/J\cong{\bf Z}/2$ should act on ${\bf S}^2\times{\bf S}^2/J\cong{\bf CP}^2$ by complex conjugation. However, I couldn't do that computation myself. Maybe I am missing something obvious?

I tried saying that the quotient $H/J$ acting on ${\bf CP}^2$ is the same as $H$ acting on ${\bf S}^2\times{\bf S}^2$, up to order of the components. However, on each component, any homeomorphism of $H$ sends an element in ${\bf S}^2$ to its antipodal point. I couldn't relate this to complex conjugation, it seems to me that instead of sending $[z_0:z_1:z_2]$ to $[\overline{z_0}:\overline{z_1}:\overline{z_2}]$, it rather maps it to $[z_0:-z_1:-z_2]$... Could anyone give me a few hints towards proving the claim in the paper?

Another connected question is about proving that ${\bf CP}^2/conj$ is diffeomorphic to ${\bf S}^4$, which is done in [1]. Is there a simpler way to do it, knowing that we already have a homeomorphism, or is the method given by Kuiper the only one?

[1] N. Kuiper, The quotient space of CP(2) by complex conjugation is the 4-sphere, Math. Ann. 208 (1974), 175- 177.

[2] W. Massey, The quotient space of the complex projective plane under conjugation is a 4-sphere, Geom. Dedicata 2 (1973), 371-374. Link: https://www.maths.ed.ac.uk/~v1ranick/papers/massey5.pdf.


1 Answer 1


(The original version of this answer had the wrong map for complex conjugation on $\mathbb{C}\cup \{\infty\}$. Thanks to Grigory M for pointing it out. This new answer is substantially different in terms of its conclusion, but the work is mostly the same.)

The usual homeomorphism $\mathbb{C}P^1\times \mathbb{C}P^1/\sim \rightarrow \mathbb{C}P^2$ is given by $$([a_0:a_1],[b_0:b_1])\rightarrow [a_0b_0: a_0b_1 + a_1b_0: a_1b_1].$$

The identifcation of $S^2$ with $\mathbb{C}P^1$ identifies $[a_0:a_1]$ with $\frac{a_0}{a_1}\in \mathbb{C}\cup\{\infty\}\cong S^2$.

Under this identification, the antipodal map $x\mapsto -x$ for $S^2$ corresponds to the map of $\mathbb{C}\cup\{\infty\}$ given by $z\mapsto -1/\overline{z}$, which, in turn, corresponds to the map on $\mathbb{C}P^1$: $[a_0:a_1]\mapsto [\overline{a}_1:-\overline{a}_0]$.

So, the corresponding map on $\mathbb{C}P^2$ maps $[a_0b_0: a_0b_1 + a_1b_0: a_1b_1]$ to $[\overline{a}_1\overline{b}_1: -\overline{a}_1\overline{b}_0 - \overline{a}_0\overline{b}_1: \overline{a}_1 \overline{b}_1]$. In other words, it corresponds to swapping the first and last coordinates of $\mathbb{C}P^2$, negating the middle coordinate, and taking the complex conjuate of everything.

In other words, if $A = \begin{bmatrix}0 & 0 & 1\\ 0 & -1 & 0 \\ 1 & 0 & 0\end{bmatrix}$, then Massey's map is given by multiplication by $A$ followed by complex conjugation. Note that $A\in U(3)$, which is path connected, so $A$ is isotopic to the identity through diffeomorphisms. That is, $A$ lies in the identity component of $\operatorname{Diff}(\mathbb{C}P^2)$.

Thus, Massey's map is isotopic to complex conjugation.

  • 1
    $\begingroup$ I'd love for someone to point out my error! $\endgroup$ Jul 29, 2021 at 15:49
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    $\begingroup$ antipodal map is not $z\mapsto -1/z$ (which has fixed points!) but $z\mapsto -1/\bar z$ $\endgroup$
    – Grigory M
    Jan 9, 2022 at 9:23
  • $\begingroup$ @Grigory M: Oh, of course! I'll update this answer shortly. $\endgroup$ Jan 9, 2022 at 20:56

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