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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 quaternionic-Kahler instead of hyperkahler?

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    $\begingroup$ Look up hyperkähler geometry and quaternion-kähler 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 quaternionic-Kahler case. I have edited the question appropriately. $\endgroup$ – Mtheorist Aug 2 '17 at 10:33
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    $\begingroup$ 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
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    $\begingroup$ As another remark, the equation you are interested in is the same for quaternion-kähler 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
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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 2-sphere 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 2-sphere 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 quaternion-kähler 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 quaternion-kähler 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:

$g_{st}(J_u)^{\,s}_i(J_v)^{\,t}_j = g_{ij} \delta_{uv} + \epsilon_{uvw}g_{ik}(J_w)^{\,k}_j$

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  • $\begingroup$ 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

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