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I've seen that the real projective plane $\mathbb{RP}^2$ can be embedded in $\mathbb{R}^4$. (see Wikipedia for example) Can the complex projective plane $\mathbb{CP}^2$ be embedded in $\mathbb{C}^4$. If so, what is the embedding?

zooby
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  • $\Bbb{CP}^2$ is $4$-dimensional, so embeds in $\Bbb R^8 \cong \Bbb C^4$ by the strong Whitney embedding. I wonder if there's a geometrically apparent way to do this. – Balarka Sen Sep 14 '16 at 16:44
  • But $\mathbb{R}^8 \neq \mathbb{C}^4$ For example $(x,y)\rightarrow (x,-y)$ is an allowed transform in $R^8$. But $z \rightarrow \overline{z}$ is not an allowed transform in $\mathbb{C}^4$ – zooby Sep 14 '16 at 20:31
  • I mean, they are homeomorphic. That is irrelevant of the complex structure $\Bbb C^4$ has. In case you wanted a holomorphic embedding (which you never wrote in your question), that's not possible due to what PVAL wrote. (aka $\Bbb{CP}^2$ is compact). – Balarka Sen Sep 14 '16 at 21:33
  • Thanks. I think I'm getting it. So a surface defined by $z_n\overline{z}n=1$ would be homeomorphic to the sphere $S{2n}$ but the complex projective space $\mathbb{CP}^2$ is not homeomorphic to say $\mathbb{RP}^4$ and is can be treated as its own interesting 4 dimensional manifold. – zooby Sep 14 '16 at 21:42

1 Answers1

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All closed holomorphic submanifolds of $\Bbb C^n$ are non-compact by the maximum modulus principle (restrict the harmonic function $\sum_{k=1}^n z_k\overline z_k$ to the manifold to show everything must have 0 modulus), so there is no complex differentiable embedding.

Looking at smooth embeddings we have:

$\Bbb CP^2$ cannot embed into $\Bbb R^5$ as it has nonzero signature by Rohklin's theorem.

To show $\Bbb CP^2$ smoothly embeds into $\Bbb R^7$ one can do the following. First write $\Bbb CP^2$ as the total space $\tau$ of the (anti)-tautological bundle over $\Bbb CP^1$ (which has boundary $S^3$) glued to a 4-ball along their common boundary. Note the definition of the tautological bundle

$$\tau=\{(x,v)\in \Bbb CP^1 \times \Bbb C^2 | v\in x, |v|\leq1\}$$

gives $\tau$ as a subset of $S^2 \times B^4 \subset B^3 \times B^4$. As every 3-sphere in the boundary of $B^3 \times B^4$ bounds a $4$-ball by a standard transversality argument, we can embed $\Bbb CP^2$ (smoothly after smoothing out the corners in this argument) in the double of $B^3 \times B^4=S^7$. By stereographic projection $\Bbb CP^2$ also embeds into $\Bbb R^7$.

I don't think $\Bbb CP^2$ can embed into $\Bbb R^6$ but I cannot find the nonembedding result in the literature (and I certainly have never gone through such a proof).

PVAL-inactive
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  • So does this mean that $\mathbb{CP}^2$ is equivalent to some sub- manifold of $S_7$? Is there some literature on this? It seems like quite a fundamental result of topology if its true? (I'm not quite sure what happens to the complex structure when it embeds into real space...) – zooby Sep 14 '16 at 20:42
  • @zooby none of the complex structures come along for the ride. In fact, it is a rather famous theorem that every smooth simply-connected 4-manifold smoothly embeds in $\Bbb R^7$, so one cannot hope to gain any topological facts about $\Bbb CP^2$ from this discussion. – PVAL-inactive Sep 14 '16 at 20:45
  • So does this mean $\mathbb{CP}^2$ embeds in $\mathbb{R}^7$ only if you remove the complex structure and treat it as a 4 dimensional real manifold? I'm confused because I just read that Smale and Spanier proved that $\mathbb{CP}^2$ can be embedded in $\mathbb{R}^7$. So is that technically true...? (Because I read also that complex manifolds must only admit holomorphic transformations.) – zooby Sep 14 '16 at 20:50
  • By the way would this surface in $\mathbb{R}^7$ still have isometry group of $SU(3)$? – zooby Sep 14 '16 at 21:46
  • Related question: Do you know if $\mathbb{CP}^4$ can be embedded in $\mathbb{R}^{10}$? –  Dec 03 '21 at 12:44
  • Sorry, do you know why we can not have this embedding? –  Dec 03 '21 at 12:45