A complex manifold admits an almost complex structure, $J$, which satisfies $$ {J^i}_j{J^j}_k={\delta^i}_k, $$ and a Hermitian metric, $g$, which satisfies $$ g_{st}{J^s}_i{J^t}_j=g_{ij}. \tag{1} $$ A hyperkahler manifold, on the other hand, admits an almost quaternionic structure which satisfies $$ {(J_u)^i}_j{(J_v)^j}_k={\delta^i}_k{\delta}_{uv}+\varepsilon_{uvz}{(J_z)^i}_k, $$ where $u,v,z=1,2,3$. What is the generalization of (1) for a hyperkahler manifold? In other words, $$\tag{2} g_{st}{(J_u)^s}_i{(J_v)^t}_j=? $$ If the metric is Hermitian w.r.t. each complex structure, then the the RHS of (2) must contain $g_{ij}\delta_{uv}$. But are there any other terms on the RHS of (2)? How does this change when the manifold is quaternionicKahler instead of hyperkahler?

1$\begingroup$ Look up hyperkähler geometry and quaternionkähler geometry, and the distinction between the two. I did not want to give you straight the answer, because there are some subtleties involved, and you should be careful about them. $\endgroup$ – Malkoun Aug 2 '17 at 10:21

$\begingroup$ @Malkoun Thank you for pointing that out, I am mostly interested in hyperkahler manifolds, though don't mind knowing the distinction in the quaternionicKahler case. I have edited the question appropriately. $\endgroup$ – Mtheorist Aug 2 '17 at 10:33

1$\begingroup$ In short, to answer your question directly for the hyperkähler case, it is enough to require that each $J_u$ is $g$orthogonal, for each value of $u$ (between 1 and 3). Let us write $I = J_1$, $J = J_2$ and $K = J_3$, which is a notation close to that of the quaternions. Then for instance $g(Iv,Jw) = g(Iv,IKw) = g(v,Kw)$, since $I$ is $g$orthogonal, so all other relations can be deduced using $g$orthogonality, and the equation for the products of the $J_u$ that you wrote down. But please read my answer below as well. $\endgroup$ – Malkoun Aug 2 '17 at 11:25

1$\begingroup$ As another remark, the equation you are interested in is the same for quaternionkähler manifolds too, but you need to be careful that, while the metric $g$ is globally defined, the almost complex structures $I$, $J$ and $K$ (or $J_u$ in your notation) are only locally defined, and there is no canonical choice for them, in the sense that an $SO(3)$rotated choice of such local $I$, $J$ and $K$ is an equally valid choice. $\endgroup$ – Malkoun Aug 2 '17 at 11:34
Here are some small facts about Quaternionic Geometry. First let us leave metrics aside, for the moment. There are 2 analogues of complex manifolds in the quaternionic world, namely triholomorphic manifolds and quaternionic manifolds. They are very different. Triholomorphic manifolds admit a 2sphere of complex structures. When I write complex structure, I mean integrable almost complex structure. On the other hand, quaternionic manifolds are manifolds admitting an almost quaternionic structure satisfying an integrability condition. They may not admit a 2sphere of almost complex structures globally, though they do locally. But these almost complex structures need not be integrable.
As an example of a triholomorphic manifold, think of $\mathbb{H}^n$ for instance. An example of a quaternionic manifold would be $\mathbb{H}P^n$, which does not even admit a globally defined almost complex structure.
The metric analogues of these 2 classes of manifolds, would be hyperkähler manifolds and quaternionkähler manifolds. Their definitions can be found in many places (wikipedia for instance, or Besse's Einstein manifolds, or books or lecture notes by say, Salamon for instance).
I hope this will point you in the right direction. Note that hyperkähler manifolds have holonomy which is a subgroup of $Sp(k)$, where the dimension of the manifold is $4k$, while the holonomy of a quaternionkähler manifold of dimension $4k$ is a subgroup of $Sp(k)\times_{\mathbb{Z}_2} Sp(1)$.
Edit: I will answer your question directly, for hyperkähler manifolds, where the $J_u$ are integrable and globally defined. If $u,v,w$ is a cyclic permutation of $1,2,3$, then:
$g_{st}(J_u)^s_{\,i}(J_v)^t_{\,j} = g_{st}(J_u)^s_{\,i}(J_u)^t_{\,k}(J_w)^k_{\,j} = g_{ik}(J_w)^k_{\,j}$
So the general answer is:
$g_{st}(J_u)^s_{\,i}(J_v)^t_{\,j} = g_{ij} \delta_{uv}  \epsilon_{uvw}g_{ik}(J_w)^k_{\,j}$
where summation over repeated indices was implicitly assumed.
Edit: the OP asked me in the comment about the case where $J_u$ is denoted by $(J_u)_i^{\,s}$ say (the "transpose" convention, so to speak). We repeat the first calculation here, with this alternate convention.
If $u,v,w$ is a cyclic permutation of $1,2,3$, then:
$g_{st}(J_u)^{\,s}_i(J_v)^{\,t}_j = g_{st}(J_u)^{\,s}_i(J_w)^{\,k}_j(J_u)^{\,t}_k = g_{ik}(J_w)^{\,k}_j$
So the general answer with this alternate convention is:
$g_{st}(J_u)^{\,s}_i(J_v)^{\,t}_j = g_{ij} \delta_{uv} + \epsilon_{uvw}g_{ik}(J_w)^{\,k}_j$

$\begingroup$ It seems that some references use the convention where each complex structure is represented as ${(J_u)_i}^s$ instead of ${(J_u)^s}_i$. It seems that when we use this convention, the second term in the answer you gave has a plus sign in front of it instead of a minus sign. Is this correct? $\endgroup$ – Mtheorist Aug 3 '17 at 11:38

$\begingroup$ I just saw your question now, $3$ years later! You would like to have relations similar to the quaternions, such as $IJ=K$ or, if you prefer, $I_1 I_2 = I_3$ and cyclic permutations, together with the relations $I_1^2 = I_1$ and so on. By the way, suppose you have $I_1$, $I_2$ and $I_3$ satisfying the usual relations, and would like to construct new almost complex structures, say $I'_1$, $I'_2$ and $I'_3$ satisfying $I'_1 I'_2 = I'_3$ and cyclic, just take $I'_1 = I_1$, $I'_2 = I_2$ and $I'_3 = I_3$ for example. So this sign is thus not so important. I do like to get signs right though. $\endgroup$ – Malkoun May 12 '20 at 20:30