I start with some background, but people familiar with the subject may jump directly to the question.

Let $M^{4n}$ be a compact oriented smooth manifold. Recall that an almost hypercomplex structure on $M$ is a 3-dimensional sub-bundle $Q\subset End(TM)$ spanned by three endomorphisms $I$, $J$ and $K$ satisfying the quaternionic identities: $I^2=J^2=-Id$, $IJ=-JI=K$.

An almost quaternionic structure on $M$ is a 3-dimensional sub-bundle $Q\subset End(TM)$ which is locally spanned by three endomorphisms with the above property.

In both cases one may assume (by an averaging procedure) that $M$ is endowed with a Riemannian metric $g$ compatible with $Q$ in the sense that $Q\subset End^-(TM)$, i.e. $I$, $J$ and $K$ are almost Hermitian. Using this one sees that an almost hypercomplex or quaternionic structure corresponds to a reduction of the structure group of $M$ to $\mathrm{Sp}(n)$ or $\mathrm{Sp}(1)\mathrm{Sp}(n)$ respectively, but this is not relevant for the question below.

Notice that in dimension $4$ every manifold has an almost quaternionic structure (since $\mathrm{Sp}(1)\mathrm{Sp}(1)=\mathrm{SO}(4)$), but there are well-known obstructions to the existence of almost hypercomplex structures. For example $S^4$ is not even almost complex. Finally, here comes the question:

Are there any known topological obstructions to the existence of almost quaternionic structures on compact manifolds of dimension $4n$ for $n\ge 2$?

EDIT: Thomas Kragh has shown in his answer that there is no almost quaternionic structure on the sphere $S^{4n}$ for $n\ge 2$. I have found further obstructions in the litterature and summarized them in my answer below.

  • $\begingroup$ Isn't the case of $S^8$ (or $S^{2n}$ for $n \geq 4$) also excluded since they don't admit almost complex structures? $\endgroup$ Jan 18, 2011 at 12:07
  • $\begingroup$ No, this is the point: almost quaternionic does not imply almost complex (see the example of S^4). More generally, the quaternionic projective spaces $\mathbb{H}\mathrm{P}^n$ are quaternion-K\"ahler, but have no almost complex structure (Hirzebruch, 1953). $\endgroup$ Jan 18, 2011 at 12:16
  • $\begingroup$ What do you mean by $Sp(1)Sp(n)$? since $Sp(1)$ is a sub-Lie-group of $Sp(n)$ this is with the obvious definition simply $Sp(n)$ gain. $\endgroup$ Jan 18, 2011 at 19:20
  • $\begingroup$ $Sp(1) Sp(n)$ is shorthand notation in this context denoting the Lie group $Sp(1) \times Sp(n) / \{\pm 1\}$. $\endgroup$ Jan 18, 2011 at 19:28
  • $\begingroup$ @Thomas: this is standard notation,, although, of course slightly misleading. In fact $Sp(1)$ is obtained by right multiplication with unit quaternions on $\mathbb{H}^n$, while $Sp(n)$ is the centralizer of $Sp(1)$, and is given by left multiplication with matrices with quaternionic entries. The diagonal $Sp(1)\subset Sp(n)$ is of course different from the former $Sp(1)$! $\endgroup$ Jan 18, 2011 at 19:35

3 Answers 3


I know this is a bit late, but as you mentioned Cadek's and Vanzura's paper, I'd like to point out (selfishly?) that there's also my paper which uses a bit of their work and gives some integrality conditions on the existence of quaternionic structures on closed manifolds -- an example is stated below. I should emphasize that I really mean honest quaternionic not just almost quaternionic here, although the referee believed that the same should hold for only almost quaternionic structures too.

Theorem: Let M be an 8-dimensional compact quaternionic manifold with Pontryagin classes $p_1(TM)$ and $p_2(TM)$ and a fundamental class $[M]$. Then the following expressions are integers $$\biggl(\frac{143}{960}p_{1}^{2}-\frac{89}{240}p_{2}\biggr)[M], \quad \biggl( -\frac{17}{480}p_{1}^{2}+\frac{71}{120}p_{2}\biggr)[M].$$

  • 3
    $\begingroup$ Probably irrelevant comment: twice the first expression, minus the second, gives that $(p_1^2 - 4p_2) [M] \equiv 0 \pmod 3$. $\endgroup$ Aug 13, 2016 at 6:55
  • $\begingroup$ @Bruno Le Floch just to point out, that congruence holds in general on any closed smooth 8-manifold, by the integrality of the Hirzebruch L-genus. $\endgroup$ Jul 5, 2020 at 15:42

Thomas' proof for the fact that $S^{4n}$ has no almost quaternionic structure is correct, but I have found an alternative argument for this statement using the twistor space. Indeed, if $S^{4n}$ has an almost quaternionic structure $Q$, then the twistor space $S(Q)$ is an $S^2$-bundle over $S^{4n}$ whose total space has an almost complex structure. On the other hand, Theorem 1.4 in the following article by W. Sutherland shows that this can only happen for $n=1$.

Of course, the general question is still open. I do not know, in particular, whether the complex projective spaces $\mathbb{C}\rm{P}^{2n}$ have almost quaternionic structures for $n\ge 2$, although I strongly suspect they don't.

EDIT: In fact there are topological obstructions! The first one (which I should have been aware of) is that the second Stiefel-Whitney class of an almost quaternionic manifold of real dimension $8n$ vanishes. This was first noticed by Marchiafava and Romani in 1975, then by Salamon in 1982, and thus rules out the complex projective spaces $\mathbb{C}\rm{P}^{4n}$.

Moreover, in dimension 8, Cadek and Vanzura not only have found further obstructions (e.g. $4p_2(M)=p_1^2(M)+8e(M)$), but they also gave sufficient topological conditions for the existence of a $\rm{Sp}(1)\rm{Sp}(2)$ structure on 8-dimensional manifolds. Their article is available here.


It seems to me that if I understood the comments to my comment correctly that the map

$$\mathrm{Sp}(1) \times \mathrm{Sp}(n) \to \mathrm{SO}(4n)$$

induced by right unit quarternionic multiplaction on $\mathbb{H}^n$ of the left factor and right matrix multiplication on $\mathbb{H}^n$ of the left factor has kernel $\{ \pm 1\}$. Since the source is simply connected it must lift to the spin group. So we have a map

$$\mathrm{Sp}(1) \times \mathrm{Sp}(n) \to \mathrm{Spin}(4n)$$.

Covering the map

$$\mathrm{Sp}(1)\mathrm{Sp}(n) \to \mathrm{SO}(4n)$$

Since the covering fiber is $\mathbb{Z}/2\mathbb{Z}$ and we can check that after taking the functor $B$ both fibers are $K(\mathbb{Z}/2\mathbb{Z},1)$-spaces we see that

$$\begin{matrix} B(\mathrm{Sp}(1)\times \mathrm{Sp}(n)) & \longrightarrow & B(\mathrm{Spin}(4n)) \\\ \downarrow && \downarrow \\\ B(\mathrm{Sp}(1)\mathrm{Sp}(n)) & \longrightarrow & B(\mathrm{SO}(4n)) \end{matrix} $$

is homotopy cartesian.

So if $M$ is spinable and has an almost Quarternionic structure it means that its classifying map lifts to $B(\mathrm{Sp}(1) \times \mathrm{Sp}(n))$

Edit: The conclusion (which is now removed) was wrong, but at least it seems to simplify the picture when $M$ is spin.

Added: For spheres $S^{4n}$ we may use the above on the $4n$th homotopy group and deloop. This implies that if we had a quartenionic structure on $S^{4n}$ we would have that the image of the map

$$\pi_{4n-1}(\mathrm{Sp}(1)\times \mathrm{Sp}(n) ) \to \pi_{4n-1} (\mathrm{SO}(4n))$$

contains the image of the map $\mathbb{Z} \cong \pi_{4n-1}(\Omega S^{4n}) \to \pi_{4n-1}(\mathrm{SO}(4n)) \cong \mathbb{Z}\times \mathbb{Z}$ (*) induced by the delooping of the classifying map for the tangent bundle of $S^{4n}$.

We know that not having an almost hypercomplex structure implies that the image of $\pi_{4n-1}(\mathrm{Sp}(n)) \to \pi_{4n-1} (\mathrm{SO}(4n))$ never contains this image, and since $\pi_{4n-1}(\mathrm{Sp}(1))$ is torsion for $n\geq 2$ the above map can not do so either for $n\geq 2$.

(*) $\pi_{4n-1}(\mathrm{SO}(4n)) \cong \mathbb{Z}\times\mathbb{Z}$ follows WHEN $n\geq 4$ from the paper

Barratt, M. G.; Mahowald, M. E. The metastable homotopy of O(n). Bull. Amer. Math. Soc. 70 1964 775-760.

I think this is true in general. Indeed, it is true for $n=1$ where the above is not a contradiction because there $\pi_3(\mathrm{Sp}(1))\cong \mathbb{Z}$. Andrei pointed out in a comment that this is also true for $n=1,2$.

  • $\begingroup$ Hope my addition makes the point clear - and that there are no errors :). $\endgroup$ Jan 18, 2011 at 23:20
  • $\begingroup$ Ups - My reference for $\pi_{4n-1}(SO(4n))$ only works for $n\geq 4$. So have not settled $S^8$ and $S^12$. $\endgroup$ Jan 18, 2011 at 23:44
  • $\begingroup$ I put this into the answer, and also added the extra stable $\math{Z}$ in $\pi_{4n-1}(SO(4n))$ which I had forgotten (I replaced $\mathbb{Z}$ with $\mathbb{Z}\times\mathbb{Z}$. Sorry for the numerous edits. $\endgroup$ Jan 19, 2011 at 0:28
  • $\begingroup$ I agree that your proof works for all spheres, including for $S^8$ and $S^{12}$, using the fact that your relation (*) actually holds for all $n\ge 2$. I have nevertheless found a shorter proof based on Sutherland's theorem about almost complex structures on sphere bundles over spheres. I wrote this proof in a separate answer. $\endgroup$ Jan 19, 2011 at 13:14
  • $\begingroup$ I realized that non-surjectivity of the map on $\pi_{4n}$ is detected by the Euler class. That is - the Euler class evaluates to 0 on the image. So any manifold with non-trivial euler class restricted to $\pi_{4n}$ cannot have a quaternionic structure for $n\geq 2$. This of course does not help with $\mathbb{C}P^{2n}$. $\endgroup$ Jan 21, 2011 at 16:44

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