24
$\begingroup$

$\DeclareMathOperator\End{End}\newcommand\Id{\mathrm{Id}}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SO{SO}$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 $n\ge 2$?


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.

$\endgroup$
7
  • $\begingroup$ 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

9
$\begingroup$

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 $$\biggl(\frac{143}{960}p_{1}^{2}-\frac{89}{240}p_{2}\biggr)[M], \quad \biggl( -\frac{17}{480}p_{1}^{2}+\frac{71}{120}p_{2}\biggr)[M].$$

$\endgroup$
2
  • 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
7
$\begingroup$

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.


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.

$\endgroup$
6
$\begingroup$

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 $n=1,2$.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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