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Jordan algebra and quaternionic projective space
One more question. I can understand why is the quaternionic projective space the space of elemenents of $J_{n}(H)$ of rank 1, because the quaternionic projective space is the space of 1-dimensional subspaces of $H^{n}$. But, why the space of elements of $J_{n}(H)$ of trace 1?
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Jordan algebra and quaternionic projective space
Is the condition $\overline{A}^{T}=A$ here because these matrices preserve $H$-inner product? Is there any connection between $Sp(n)$ and and $J_{n}(H)$?
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
Yes, I just want that condition $JT(M) = T(M)$ and not $KTM\bot TM$ and $LTM\bot TM$
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
Now I am also confused about parallel condition and integrable condition for J, I have to check it. So, if an almost complex structure is parallel then it is integrable, because of Nijenhuis tensor?
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
is $R^{2} = span\{x,y\}$. With $\gamma=i\Omega (.,\tau .)$ we define a Hermitian form. Now, we calculate $\phi^{*}\gamma = \frac{\partial\phi_{1}}{\partial p} \frac{\partial \overline{\phi}_{2}}{\partial\overline{p}}(-2dq_{1}\wedge dq_{2})$. Here $C=span\{p\}$ and $p = q_{1} + i q_{2}$. Is then $\phi$ totally complex if $\frac{\partial\phi}{\partial q_{1}}=0$ ?
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
I don't have the option to delete my last two comments, so I am writing them again.
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
* Here $C=span\{p}$ and $p = q_{1} + i q_{2}$. Is then $\phi$ totally complex if $\frac{\partial\phi}{\partial q_{1}}=0$?
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
is $R^{2} = span\{x,y\}$. With $\gamma = i\Omega(.,\tau .)$ we define a Hermitian form. Now, we calculate $\phi^{*}\gamma = \frac{\partial\phi_{1}}{\partial p}\frac{\partial\overline{\phi}_{2}}{\partial{\overline{p}}}(-2dq_{1}\wedge dq_{2}). Here $C=span{p}$ and $p=q_{1}+iq_{2}$. Is then $\phi$ totally complex if $\frac{\partial\phi}{\partial q_{1}}=0$?
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
@Robert: I don't think than J should be integrable. I wonder does the condition 1) from my previous comment and the condition J(T(M))=T(M) necessarily imply that $K(T(M))\bot T(M)$ and $L(T(M))\bot T(M)$, for other two sections. So, holomorphic submanifold should satisfy just the condition 1) and J(T(M)) = T(M). But, the other two conditions for K and L don't have to be satisfied. Let's take on simple example: let $\phi:C\to C^{2}$ be an immersion. $\Omega = dz\wedge dw$ is standard complex symplectic form. $z=x+iu$; $w=y+iv$. Let $\tau$ be the standard real structure and its fixed point set
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
@Nina: Thank you for the references!
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Totally complex submanfiolds in quaternionic Kaehlerian manifolds
I was thinking on this: On an ambient manifold M we have 3 almost complex structures, J, K and L. And if for a submanifold N these conditions are satisfied: 1) $\nabla_{X}J=0$, for $X\in T(N)$, 2) $J(T(M))=T(M)$, $K(T(M))\bot T(M)$, $L(T(M))\bot T(M)$, then we say that N is totally complex submanifold. Or maybe I misunderstood it.
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Quaternion Kaehlerian manifold, definition
Is there some reference where I can find the examples of totally complex submanifolds in, for example, complex space $C^{n}$? And also holomorphic submanifolds. I am trying to find some example of submanifold that is holomorphic but not totally complex in complex space. Since that is not my field, I am not quite sure how will I define almost complex structures $K$ and $L$. They should be orthogonal to naturally defined structure $J$ which is parallel, and map tangent bundle of submanifold to normal bundle of submanifold.
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