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I have started reading about quaternionic and quaternionic Kähler manifolds. The most elegant definitions speak about the holonomy group of the manifold. However, it is possible to describe these type of manifolds using subbundles of $TM$. In particular, a manifold is called almost quaternionic if there exists a rank 3 vector subbundle $G\subset \operatorname{End}TM$ such that, locally it is spanned by 3 almost complex structures $IJ$ and $K$ verifying the quaternionic identity

$$IJK=-id_{TM}.$$

I imagine, this structures can be considered as the pre-image by the bundle trivialisations of the linear complex structures on $\mathbb H^n$, regarded as a real vector space. Hence, I have computed the complex structures on $\mathbb R^{4n}$ resulting of the operations $v\mapsto vi$, $v\mapsto vj$ and $v\mapsto vk$.

As you can see, I am considering $\mathbb H^n$ as a right module, because that is what Chevalley and others do. Then, considering a basis $\{(e_n),(f_n),(g_n),(h_n)\}_{n=1}^m$ of $\mathbb R^4$, I define $I:\mathbb R^{4n}\rightarrow \mathbb R^{4n}$ as

$$ \left\{ \begin{array}{l} Ie_n = (e_n)\cdot i = e_n i=f_n \\ If_n = (e_ni)\cdot i =-e_n \\ Ig_n = (e_nj)\cdot i = e_nji = -e_nk = -h_n\\ Ih_n=(e_nk)\cdot i = e_nki = e_nj= g_n , \end{array} \right. $$

so that $I$ is written

$$ \begin{pmatrix} 0 & -id_n & 0 & 0 \\ id_n & 0 & 0 & 0 \\ 0 & 0 & 0 & id_n \\ 0 & 0 & -id_n & 0 \\ \end{pmatrix}. $$

Similarly,

$$ J=\begin{pmatrix} 0 & 0 & -id_n & 0 \\ 0 & 0 & 0 & -id_n \\ id_n & 0 & 0 & 0 \\ 0 & id_n & 0 & 0 \\ \end{pmatrix} $$

and

$$ K=\begin{pmatrix} 0 & 0 & 0 & -id_n \\ 0 & 0 & id_n & 0\\ 0 & -id_n & 0 & 0 \\ id_n & 0 & 0 & 0 \\ \end{pmatrix}. $$

However, in this case $IJK=+id_{4n}$, while $KJI=-id_{4n}$. It seems logical because $KJI$ actin on $v$ corresponds to $v\cdot(ijk)$ and the other way round. And then it comes my question:

Why should I consider $\mathbb H$ acting on the right for the module structure but acting on the left for computing the complex structures $I,J$ and $K$? It seems me cheating.

EDIT:

This pdf is attached in a comment below this question. In page 35 it says:

More generally let $V=\mathbb H^n\cong =\mathbb R^{4n}$ be a "quaternionic vector space". We view it a sa left $\mathbb H$-module by left multiplication. $V$ has three complex structures $I,J,K$ given by left (component-wise) multiplication by $i,j,k$ respectively.

Let me emphasize the left multiplication. However, then the text says:

If we let $g\in GL_{4n}(\mathbb R)$ act on $q\in V$ as $qg^{−1}$ (where we view $q$ as a row vector), then the action of $GL_n(\mathbb H)$ commutes with the action of the complex structures, i.e. it is “quaternionic-linear”

But $I,J$ and $K$ are supposed to be in $GL_{4n}(\mathbb R)$, are not they? So I cannot understand how it is possible to consider them acting on the left first and then on the right.

P.S. As you probably have noticed, I have focused on almost quaternionic manifolds, since I am interested in the way complex structures have to be define. I speak neither the integrability of the complex structures nor the associated Kählerian forms $\omega_I,\omega_J,\omega_K$. In fact, what I really want is to understand how a(n almost) quaternionic (Kähler) structures works, trying to understand how it reduces the structure group.

Dog_69
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    It sounds like you're defining an almost quaternionic structure rather than a quaternionic structure. See wiki – jgon Jun 14 '19 at 00:15
  • @jgon Yes, you are right. I am ignoring the integrability of the complex structures in my question. I will edit the question to clarify this. The point is that I am interested in the definition of the complex structures (the order). Then I will worry about wheter they are integrable or not. – Dog_69 Jun 14 '19 at 07:36
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    I don't understand what the issue is. What's wrong with having $I$, $J$, $K$ act on the left and other transformations act on the right? If $\lambda$ is a quaternion, $v$ a quaternionic row vector, and $g$ an invertible quaternionic matrix, then $\lambda(vg^{-1})=(\lambda v)g^{-1}$, so the group action is "linear" over the quaternion scalar multiplication. If you view left-multiplication-by-$\lambda$ and right-multiplication-by-$g^{-1}$ as linear operators on real $4n$-dim space, then this basically says $I,J,K$ commute with the $4n\times4n$ matrices coming from quaternionic $n\times n$ ones. – anon Jun 21 '19 at 04:51
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    Previous comment is directed at your edit. For the question before that, I'm guessing because "complex structures" are usually defined with a linear operator acting on the left, and they didn't want to break that convention? (Even when getting them from a right $\Bbb H$-module structure.) I just wouldn't bother thinking about $\Bbb H$ acting on the left (except for the copy of $\Bbb H$ inside $M_n(\Bbb H)$ I guess) but I don't do manifold stuff generally. Though you can always convert a left $A$-module into a right $A^{\rm op}$-module and vice-versa, and $\Bbb H^{\rm op}\cong \Bbb H$. – anon Jun 21 '19 at 05:05
  • @runway44 First of all let me apologise for my delay replying your comments. I alsowant to thank your interest in my question. Regarding your first comment, especially the question ''what is wrong with having $I,J,K$ cating on the left and other transformations on the right?'' I say that $I,J,K$ should behave like other transformations, because they are elements of the general linear group. So I think $I,K,K$ must act on the same side that the rest of the elements. – Dog_69 Jun 25 '19 at 22:54
  • @runway44 And regarding your second comment, I think you are right. Reading an Alekseevsky's paper, it says something similar to you. He also says that $I,J,K$, define an left $\mathbb H$-module on a $4n$-dimensional real vector space. So they must be computed by letting the unit quaternions $i,j,k$ act on the left, even if we are considering $\mathbb H^n$ as a right $\mathbb H$-module. And you are also right: it seems it is not worth considering $\mathbb H^n$ as a left $\mathbb H$-module in general. Thanks again for your comments. – Dog_69 Jun 25 '19 at 22:59
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    No worries. The $I$, $J$, $K$ that act from the left are different from the $i$, $j$, $k$ acting on the right from the module structure (or vice-versa, whatever the convention is). While we usually have the scalar matrices to act the same as the scalars themselves, we no longer expect that when we're over a skew field and the matrices/scalars act from different sides. – anon Jun 26 '19 at 00:57

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