# Acting on CP(2) by conjugation

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.

(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.

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