TITLE: On a curious map from the complex projective plane into $S^5$
QUESTION [24 upvotes]: I have heavily edited the post (including the title), based on a comment by @GregoryArone that my map $f$ is not injective. In an earlier version of this post, I had thought to have constructed a smooth map from $\mathrm{P}^2_\mathbb{C}$ into $S^5$, which I thought was a topological embedding. Removing a point from $S^5$ and using stereographic projection, I had thought to have found a topological embedding of $\mathrm{P}^2_\mathbb{C}$ inside $S^5$, but I was mistaken. My map was actually not injective. However, this raised an interesting question.
Question set 1: does there exist a topological embedding of $\mathrm{P}^2_\mathbb{C}$ inside $\mathbb{R}^5$? Or is there maybe a topological obstruction to that?
In the second part of this post, I will describe a smooth map $f$ from $\mathrm{P}^2_{\mathbb{C}}$ into $S^5$ which is a double cover that is branched over a real slice of $\mathrm{P}^2_{\mathbb{C}}$, with respect to a real structure $\sigma$ on $\mathrm{P}^2$, defined by
$$ \sigma([z_0:z_1:z_2]) = [\bar{z}_2:-\bar{z}_1:\bar{z}_0].$$
Note that $\sigma$ is the real structure which is induced by the "antipodal map" $j$ on $P^1_\mathbb{C}$, defined by
$$j([u_0:u_1]) = [-\bar{u}_1:\bar{u}_0].$$
I will now describe how the map $f$ is defined. First, define the map $g: \mathrm{P}^1_{\mathbb{C}} \times \mathrm{P}^1_{\mathbb{C}} \to \mathrm{P}^2_{\mathbb{C}}$:
$$ ([u_0:u_1], [v_0:v_1]) \mapsto [2u_0v_0: u_0 v_1 + u_1 v_0: 2u_1v_1].$$ Then $g$ is holomorphic and onto. The symmetric group $S_2$ acts on the domain of $g$ by permuting the two factors, namely the $u$-point with the $v$-point, so to speak. The fibers of $g$ are actually the $S_2$ orbits in the domain of $g$.
The (extended) Hopf map $h$ is a smooth map from $\mathbb{C}^2$ onto $\mathbb{R}^3$, defined by
$$h(u_0,u_1) = \left( 2 \operatorname{Re}(u_0 \bar{u}_1), 2 \operatorname{Im}(u_0 \bar{u}_1), |u_0|^2 - |u_1|^2 \right).$$
The group $U(1)$ acts on the domain of $h$ by scalar multiplication, and the fibers of $h$ are the $U(1)$-orbits in the domain of $h$.
Then the map
$$ h \times h: \mathbb{C}^2 \times \mathbb{C}^2 \to \mathbb{R}^3 \times \mathbb{R}^3$$
followed by the map
$\operatorname{Sym}: \mathbb{R}^3 \times \mathbb{R}^3 \to S^2(\mathbb{R}^3)$ which maps $(x,y)$ to $x \odot y$, gives a map
$$k: \mathbb{C}^2 \times \mathbb{C}^2 \to S^2(\mathbb{R}^3),$$
where $k = \operatorname{Sym} \circ (h \times h)$. In turn, $k$ induces a smooth map
$$\tilde{k}: \mathrm{P}^1_\mathbb{C} \times \mathrm{P}^1_\mathbb{C} \to S^5,$$
where the latter is the unit sphere in $S^2(\mathbb{R}^3) \simeq \mathbb{R}^6$. Indeed, $k$ maps
$$(\mathbb{C}^2 \setminus \{ \mathbf{0} \}) \times (\mathbb{C}^2 \setminus \{ \mathbf{0} \}) \to S^2(\mathbb{R}^3) \setminus \{ \mathbf{0} \},$$
and the latter maps onto $S^5$ by the normalization map, with respect to the inner product on $S^2(\mathbb{R}^3)$ induced by the Euclidean inner product on $\mathbb{R}^3$. Note that this composed map
$$(\mathbb{C}^2 \setminus \{ \mathbf{0} \}) \times (\mathbb{C}^2 \setminus \{ \mathbf{0} \}) \to S^5$$
is invariant under rescaling each of the $2$ factors of its domain individually, and so induce a smooth map which we are denoting by $\tilde{k}$, from $\mathrm{P}^1_\mathbb{C} \times \mathrm{P}^1_\mathbb{C}$ into $S^5$.
The fibers of $\tilde{k}$ are actually of the form
$$(\mathbf{u}, \mathbf{v}), (j\mathbf{u}, j\mathbf{v}), (\mathbf{v}, \mathbf{u}), (j\mathbf{v}, j\mathbf{u})$$
where $\mathbf{u} = [u_0:u_1]$, $\mathbf{v} = [v_0:v_1]$ are points on $\mathrm{P}^1_\mathbb{C}$ and $j$ is the "antipodal map" which was previously defined. Note that $h(j\mathbf{u}) = -h(\mathbf{u})$.
We are now ready to define our map
$$f: \mathrm{P}^2_\mathbb{C} \to S^5.$$
Given a point $p \in \mathrm{P}^2_\mathbb{C}$, let $w \in g^{-1}(p)$ and define
$$f(p) = \tilde{k}(w).$$
Then $f$ is a well defined smooth map from $\mathrm{P}^2_\mathbb{C}$ into $S^5$, which is invariant under the real structure $\sigma$, which was previously defined. In fact, $f$ is a double cover onto its image (a codimension $1$ subset of $S^5$) which is branched over the real slice of $\mathrm{P}^2_\mathbb{C}$ with respect to $\sigma$. A generic fiber of $f$ is a pair of $\sigma$-conjugate points in $\mathrm{P}^2_\mathbb{C}$.
Note that if we think of the coordinates of $S^2(\mathbb{R}^3)$ as the components of a real $3$-by-$3$ symmetric matrix $A$, then it is not too difficult to see that $\tilde{k}$ maps $\mathrm{P}^1_\mathbb{C} \times \mathrm{P}^1_\mathbb{C}$ into the real quasi-affine variety
$$V = \{ \det(A) = 0 \} \cap \{ \operatorname{tr}(A^2) = 1 \} \cap 
\{ \operatorname{tr}(A)^2 \leq 1 \}.$$
In other words, these conditions ensure that the eigenvalues of $A$, which must be real, are of the form: $0$, $\lambda$, $\mu$ with $\lambda \mu \leq 0$ and $\lambda^2 + \mu^2 = 1$ (note that $\lambda$, or $\mu$, may be $0$).
Hence the image of $f$ is contained in $V$.
Question 2: is the image of $f$ equal to $V$? Edit: I think the image of $f$ is indeed $V$. Just note that it suffices to diagonalize $A$, and show that a diagonal matrix having $0$, $\lambda$ and $\mu$ as (real) eigenvalues and satisfying the previous conditions is in the image of $f$. And this is straightforward.
Finally, I suspect I am just rediscovering that the complex projective plane modulo complex conjugation is the $4$-sphere, except that instead of complex conjugation, I am using a different real structure. Indeed, this "folklore" result is proved and discussed for instance in

Michael Atiyah, Jurgen Berndt, Projective planes, Severi varieties and spheres, Surveys in Differential Geometry VIII, Papers in Honor of Calabi, Lawson, Siu and Uhlenbeck, International Press (2003) pp.1-27. doi:10.4310/SDG.2003.v8.n1.a1,
arXiv:math/0206135.

Question 3: is the image of $f$ diffeomorphic to $S^4$? If so, then it would provide yet another proof of the previous folklore result. A related question is whether or not $V$ is diffeomorphic to $S^4$.
Edit: I think I can build a diffeomorphism from $S^4$ onto $V$. Think of $S^4$ as the set
$$W = \{ B \,|\, \text{$B$ real symmetric $3$-by-$3$, } \operatorname{tr}(B) = 0 \text{ and } \operatorname{tr}(B^2) = 1 \}.$$
It is not too difficult to see that $W$ is diffeomorphic to $S^4$. Define a map from $W$ into $V$, by
$$ B \mapsto \frac{B - \lambda_2(B) I}{\lVert B - \lambda_2(B)I \rVert},$$
where $\lambda_1(B) \leq \lambda_2(B) \leq \lambda_3(B)$ are the $3$ eigenvalues of $B$. I think that this map is perhaps a diffeomorphism from $W$ onto $V$. However, I am not sure about its smoothness when $2$ eigenvalues of $B$ collide. Can someone comment on that please?

REPLY [28 votes]: I think I can prove the following
Claim There is no topological embedding of $\mathbb CP^2$ into $\mathbb R^6$.
The proof uses the van Kampen obstruction. Let me review the idea. Suppose there is a topological embedding $f\colon \mathbb CP^2\hookrightarrow\mathbb R^6$. Then $f$ induces a $\Sigma_2$-equivariant map of deleted squares
$$
f^2_\Delta\colon \mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2 \to \mathbb R^6\times \mathbb R^6\setminus \mathbb R^6.
$$
Let $\widetilde S^5$ denote the $5$-dimensional sphere with the antipodal action of $\Sigma_2$. There is a $\Sigma_2$-equivariant map (in fact a homotopy equivalence)
$$
\mathbb R^6\times \mathbb R^6\setminus \mathbb R^6 \xrightarrow{\simeq} \widetilde S^5.
$$
It follows that a topological embedding $f$ would induce a $\Sigma_2$-eqivariant map
$$
\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2 \to \widetilde S^5.
$$
So to prove that there is no topological embedding, it is enough to prove that there is no such map. An equivariant map like this is essentially the same things as a nowhere vanishing section of the vector bundle
$$
(\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2)\times_{\Sigma_2} {\widehat {\mathbb R}}^6 \to (\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2)_{\Sigma_2}.
$$
Here $\widehat {\mathbb R}^6$ is the $6$-dimensional sign representation of $\Sigma_2$. The Euler class of this vector bundle is an obstruction to the existence of a section, and therefore to the existence of a topological embedding. This is the van Kampen obstruction.
It remains to prove that the Euler  class is non-zero. All cohomology groups will be taken with mod 2 coefficients. The Euler class is an element of $H^6\left((\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2)_{\Sigma_2}\right)$. The cohomology ring of $\left(\mathbb CP^2\times \mathbb CP^2\setminus \mathbb CP^2\right)_{\Sigma_2}$ was calculated in the following paper
Samuel Feder, The reduced symmetric product of projective spaces and the generalized Whitney theorem, Illinois J. Math. 16 (1972), 323–329  https://doi.org/10.1215/ijm/1256052288
If I am parsing the result of this paper correctly, the cohomology ring is generated by two elements $u_1, x_2$, subject to just the relations $x_2^3=0$ and $u_1^3=u_1x_2$. It follows that as a vector space, the cohomology has following basis
$$
1, u_1, u_1^2, x_2, u_1x_2=u_1^3, u_1^2x_2=u_1^4, x_2^2, u_1x_2^2=u_1^5, u_1^2x_2^2=u_1^6.
$$
The main point is that $u_1^6\ne 0$. Clearly $u_1$ is the Euler class of the $1$-dimensional sign representation, so $u_1^6$ is the Euler class of the $6$-dimensional sign representation.