# On a curious map from the complex projective plane into $S^5$

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?

• Oh wow that would be cool, it would show that a codimension 1 embedding into Euclidean space doesn’t imply trivial normal bundle which also gives lots of other things like an embedding with no normal bundle - really shows how badly behaved topological manifolds are. – Connor Malin Feb 21 at 6:00
• I still don't think that the map ${\mathrm P}^2_{\mathbb C}\to S^5$ is injective. I may be confused again, but it seems that the map $P^1_{\mathbb C}\times P^1_{\mathbb C} \to \mbox{Sym}^2(\mathbb R^3)$ can be identified with the map $S^2\times S^2\to \mathbb R^6$ that sends $((x, y, z), (x', y', z'))$ to $(xx', yy', zz', xy'+x'y, xz'+x'z, yz'+y'z)$. This map identifies $(\bar u, \bar v)$ with $(-\bar u, -\bar v)$ and not just with $(\bar v, \bar u)$. – Gregory Arone Feb 21 at 16:08
• I strongly suspect that there is a cohomological obstruction to the existence of a topological embedding ${\mathbb P}^2_{\mathbb C} \hookrightarrow \mathbb R^5$ and even into $\mathbb R^6$. More specifically, I suspect that the van Kampen obstruction is not zero. – Gregory Arone Feb 21 at 16:21
• @GregoryArone, thank you so much for your comments! I have edited my post heavily in light of your comment that my map $f$ is not injective. – Malkoun Feb 21 at 17:03
• @GregoryArone: There is a topological obstruction to locally flat embedding of a closed 4-manifold X in $R^5$ or $R^6$. In either case, you would conclude that X is spin, and that the signature of X is 0. You can have a non-locally flat embedding in $R^6$ with a single non-flat point if X is spin; for instance a K3 surface has such an embedding. – Danny Ruberman Feb 21 at 19:55

## 1 Answer

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.

• Awesome answer. I just have a question please. Why is the Euler class of the $1$-dimensional sign representation equal to $u_1$? In particular, why is it nonzero? I forgot how to calculate Euler classes. – Malkoun Feb 22 at 15:19
• If $X$ is any simply connected space with a free action of $\Sigma_2$ then the (mod $2$) Euler class of the canonical line bundle over $X/_{\Sigma_2}$ is non-zero. To see this, think of the canonical map $X/_{\Sigma_2}\to \mathbb RP^\infty$. This map is induced by taking $\Sigma_2$ orbits of a $\Sigma_2$-equivariant map $X\to S^{\infty}$. By covering space theory the induced map of orbits is an isomorphism on $\pi_1$ and therefore it is an isomorphism on $H^1$. The Euler class is the image of the generator of $H^1(\mathbb RP^\infty; \mathbb F_2)$. – Gregory Arone Feb 22 at 15:44