9
$\begingroup$

Let $\mathbb{H}P^m$ be the $m$-th quaternionic projective space. What is the smallest integer $N$ such that there exists an embedding $$ \mathbb{H}P^2\longrightarrow \mathbb{R}^N? $$ Are there any references?

$\endgroup$
  • 2
    $\begingroup$ If I'm not mistaken, $12 \leq N \leq 16$, the lower bound coming from a standard calculation involving Stiefel-Whitney classes, and the upper bound coming from the Whitney embedding theorem. $\endgroup$ – Michael Albanese Jan 7 '16 at 5:26
  • 1
    $\begingroup$ S. Feder and D. M. Segal Proceedings of the American Mathematical Society Vol. 35, No. 2 (Oct., 1972), pp. 590-592 shows that $N>12$. $\endgroup$ – user83633 Jan 7 '16 at 9:19
14
$\begingroup$

I. M. James, Lectures on algebraic and differential topology, pp. 134–174, Lecture Notes in Math., Vol. 279, Springer, Berlin, 1972, Theorems 1.2 and 1.3 show that $$N=13.$$

$\endgroup$
  • 5
    $\begingroup$ The embedding can be geometrically described as follows. Let $V$ be the real vector space of $3\times3$ Hermitian quaternionic matrices with fixed (real) trace, say $1$. Note that $\dim V=14$. $\mathbb HP^2$ embeds into $V$ by mapping each quaternionic line in $\mathbb H^3$ to the matrix representing the corresponding orthogonal projection onto it. The image of the embedding in $V$ consists of the idempotent matrices, and it is also an orbit of the action of the group $Sp(3)$ on $V$ by conjugation. The image sits in the unit sphere of $V$, so it can be stereographically projected to $R^{13}$. $\endgroup$ – Claudio Gorodski Jan 13 '16 at 14:07

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.