Are there topological obstructions to the existence of almost quaternionic structures on compact manifolds?

Question

$\DeclareMathOperator\End{End}\newcommand\Id{\mathrm{Id}}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SO{SO}$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 $\Sp(n)$ or $\Sp(1)\Sp(n)$ respectively, but this is not relevant for the question below.

Notice that in dimension $4$ every manifold has an almost quaternionic structure (since $\Sp(1)\Sp(1)=\SO(4)$), 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 $n\ge 2$?

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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 literature and summarized them in my answer below.

Answer 1

I know this is a bit late, but as you mentioned Čadek's and Vanžura's paper, I'd like to point out (selfishly?) that there's also my paper Indices of quaternionic complexes 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 $$\biggl(\frac{143}{960}p_{1}^{2}-\frac{89}{240}p_{2}\biggr)[M], \quad \biggl( -\frac{17}{480}p_{1}^{2}+\frac{71}{120}p_{2}\biggr)[M].$$

Answer 2

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 A Note on Almost Complex and Weakly Complex Structures 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 $\mathbb{C}{\operatorname{P}^{2n}}$ have almost quaternionic structures for $n\ge 2$, although I strongly suspect they don't.

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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 $\mathbb{C}{\operatorname{P}^{4n}}$.

Moreover, in dimension 8, Čadek and Vanžura 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 $\operatorname{Sp}(1)\operatorname{Sp}(2)$ structure on 8-dimensional manifolds. Their article is Almost quaternionic structures on eight-manifolds.

Answer 3

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 $n=1,2$.