In Bott and Tu's Differential Forms in Algebraic Topology, the authors show using Pontrjagin classes that $\mathbb CP^4$ cannot be smoothly embedded in $\mathbb R^k$ when $k\le 11$. The obvious question arises: can $\mathbb CP^4$ be embedded in $\mathbb R^{12}$?

The only result I know in this direction is the Whitney embedding theorem, which says that a smooth $m$-dimensional manifold can be embedded in $\mathbb R^{2m}$. That is clearly not good enough here, as $\mathbb CP^4$ has dimension $8$.


1 Answer 1


No, see theorem 1.3 here http://www.lehigh.edu/~dmd1/CPcrabb4.pdf, and the reference given there. Here $\alpha (n)$ denotes the number of $1$'s in the binary expression of $n$.

In this case $4$ has binary expansion $100$, so the first case of theorem 1.3 implies $\Bbb CP^4$ cannot even immerse into $\Bbb R^{14}$. The Whitney immersion theorem implies that this is optimal. In fact any compact orientable $n$-manifold embeds in $\Bbb R^{2n-1}$ (according to wikipedia this is due to Haefliger and Hirsch (for $n>4$) but I do not know the specific references off hand).

  • $\begingroup$ Excellent, thank you. $\endgroup$
    – Potato
    Dec 19, 2015 at 0:34

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