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?

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    $\begingroup$ 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. $\endgroup$ Feb 21, 2021 at 6:00
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    $\begingroup$ 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 $(\bar v, \bar u)$. $\endgroup$ Feb 21, 2021 at 16:08
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    $\begingroup$ 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. $\endgroup$ Feb 21, 2021 at 16:21
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    $\begingroup$ @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. $\endgroup$
    – Malkoun
    Feb 21, 2021 at 17:03
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    $\begingroup$ @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. $\endgroup$ Feb 21, 2021 at 19:55

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Malkoun
    Feb 22, 2021 at 15:19
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    $\begingroup$ 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 $H^1(\mathbb RP^\infty; \mathbb F_2)$. $\endgroup$ Feb 22, 2021 at 15:44

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