I have found two different definitions for integrable quaternionic structure in the literature, and I need to know if they agree with one another.

One definition that I have found (from Differential Geometry of Lightlike Submanifolds - Duggal, Sahin) is that for an almost quaternion manifold, integrable quaternionic structure implies that there exists coordinates in each coordinate neighborbood whereby the three almost complex structures $J_1,J_2,J_3$ (that define the almost quaternionic structure) take the form

$$ J_1= \left[\begin{array}{cccc} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1\\ 0 & 0 & 1 & 0 \end{array}\right] $$ $$ J_2= \left[\begin{array}{cccc} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \end{array}\right] $$

$$ J_3= \left[\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{array}\right], $$ where $1$ is the identity matrix.

On the other hand, from this article, I have found a definition where integrable quaternionic structure implies that the Nijenhuis tensor for each of the almost complex structures given above vanish.

Do these definitions agree?


These two 'definitions' do not agree. Also, you should be careful about your choice of sources. Most differential geometers use the terminology 'almost quaternionic' to mean that the structure group of a $4n$-manifold $M$ has been reduced to a subgroup of $\mathrm{GL}(n,\mathbb{H}){\cdot}\mathrm{Sp}(1)\subset\mathrm{GL}(4n,\mathbb{R})$. So, for example, $\mathbb{HP}^n$ is a(n integrable) quaternionic manifold in this sense, even though it does not have any almost complex structures. When the structure group has been reduced to $\mathrm{GL}(n,\mathbb{H})\subset\mathrm{GL}(4n,\mathbb{R})$, i.e., when there are $3$ anti-commuting almost complex structures defined on $M^{4n}$, then one says that $M$ is almost hypercomplex. From your question, I assume that this latter term is what you mean by almost quaternionic.

In any case, the condition that there exist local coordinates in which the three almost complex structures have constant coefficients is the condition that the hypercomplex structure be locally flat, which is much more restrictive than the second condition, which is that each of the $J_i$ be an integrable almost complex structure (i.e., that its Nijnhuis tensor vanish). For example, even in the much more restrictive case of hyperHermitian structures (i.e., when the structure group is $\mathrm{Sp}(n)\subset\mathrm{GL}(n,\mathbb{H})\subset\mathrm{GL}(4n,\mathbb{R})$) flatness is much more restrictive than the condition that the three almost complex structures be integrable.

Any hyperKähler manifold is integrable in the second sense, but it is not integrable in the first sense unless the metric is actually flat.

For another kind of example, see The Hypercomplex Structure of $SU(3)$.

  • $\begingroup$ Sorry, I think there is a typo in your last sentence. $\endgroup$ – Mtheorist Sep 23 '17 at 9:50
  • $\begingroup$ Thank you for your answer. Can't we obtain the three almost complex structures in the first sense on a hyperkaehler manifold with non-flat metric, via the hyperkaehler quotient? In this (projecteuclid.org/download/pdf_1/euclid.jmsj/1227107863) reference, Nakajima and Gocho show that when taking such a quotient of a manifold, $Y$, the three almost complex structures on $Y$ descend to the three almost complex structures on the quotient, $M$. Then wouldn't it be true that if the almost complex structures on $Y$ take the above form with constant coefficients, the same should hold for $M$? $\endgroup$ – Mtheorist Sep 23 '17 at 10:05
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    $\begingroup$ @Mtheorist: No, that is not true. For example, one can construct the Eguchi-Hanson metric (which is hyperKähler) on $M=TS^2$ by a hyperKähler quotient of $Y=\mathbb{H}^2$ by a standard $S^1$-action. However, there do not exist local coordinates on $M$ in which the three complex structures have constant coefficients: The hyperKähler structure is not flat, and its Levi-Civita connection leaves the complex structures parallel, so it must agree with the Obata connection. If there were coordinates in which the three complex structures had constant coefficients, the Obata connection would be flat. $\endgroup$ – Robert Bryant Sep 23 '17 at 10:49
  • $\begingroup$ Sorry, I am a little confused. In a local patch, $\nabla_{\nu} {(J_i)^{\lambda}}_{\mu}=\partial_{\nu} {(J_i)^{\lambda}}_{\mu}+{\Gamma^{\lambda}}_{\nu \kappa}{(J_i)^{\kappa}}_{\mu}-{\Gamma^{\kappa}}_{\nu \mu}{(J_i)^{\lambda}}_{\kappa}$. The complex structures being parallel and having constant coefficients would imply that ${\Gamma^{\lambda}}_{\nu \kappa}{(J_i)^{\kappa}}_{\mu}={\Gamma^{\kappa}}_{\nu \mu}{(J_i)^{\lambda}}_{\kappa}$. It is not obvious to me how this implies flatness of the connection. $\endgroup$ – Mtheorist Sep 23 '17 at 18:07
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    $\begingroup$ @Mtheorist: Maybe not obvious, but it does follow. You have to use the torsion-free condition, i.e., $\Gamma^\kappa_{\mu\nu}=\Gamma^\kappa_{\nu\mu}$. At, say, the origin of your coordinates $x=0$, you'll then find that the quadratic map $\Phi:\mathbb{R}^{4n}\to\mathbb{R}^{4n}$ defined by $\Phi^\kappa(x) = \Gamma^\kappa_{\mu\nu}(0)x^\mu x^\nu$ must have its differential satisfy $$\Phi'(x)(J_iy)=J_i \bigl(\Phi'(x)(y)\bigr)$$ for all $i=1,2,3$, i.e., $\Phi$ is quaternion differentiable, which forces $\Phi(x)\equiv0$, since, as is well-known, quaternion differentiable maps must be affine linear. $\endgroup$ – Robert Bryant Sep 23 '17 at 18:47

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