Is there an embedding of the complex projective plane in complex space? 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?
 A: 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).
