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We know from the classy work of Joyce that "any compact Lie group becomes hypercomplex after it is multiplied by a sufficiently big torus". The quote comes from the Wikipedia page.

I am asking if it is known that these hypercomplex manifolds are hyper-Kaehler (or hyper-Hermitian) or not.

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The answer is already 'no' for the simplest case:
$$ S^1\times \mathrm{SU}(2) = (\mathbb{H}{\setminus}\{0\})/\mathbb{Z}, $$ which is clearly hypercomplex, but cannot even be Kähler, much less hyperKähler.

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  • $\begingroup$ Can we put on it something weaker, like an Hermitian structure, or an almost-symplectic? $\endgroup$
    – Tomasz Köner
    Commented May 21, 2016 at 12:39
  • $\begingroup$ @TomaszKöner: Yes, since these are only topological questions. In fact, any complex manifolds supports almost-symplectic structures and Hermitian structures. This is essentially because $\mathrm{GL}(n,\mathbb{C})$ and $\mathrm{Sp}(n,\mathbb{R})\subset \mathrm{SL}(2n,\mathbb{R})$ each contain $\mathrm{U}(n)$ as a maximal compact subgroup. (Hence, the latter is a deformation retract of each of the two former (non-compact) groups.) $\endgroup$
    – Robert Bryant
    Commented May 26, 2016 at 21:44

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