4
$\begingroup$

I have been told that the following is an open problem in mathematics, but I am pretty sure that experts in the topic surely know the answer.

Is the manifold

$$M=\frac{E_{7(7)}}{SU(7)}\times \mathbb{R}^{+}$$

a (pseudo)Kähler-Hodge manifold? Just to be clear, by this I mean if $M$ can be equipped with a (pseudo)Kähler structure, namely a complex structure $J$, a symplectic structure $\omega$ and a compatible metric $g$ (of indefinite signature) such that $\omega$ is integral, that is, $[\omega] \in H^{2}(M,\mathbb{Z})$, together with a hermitian line-bundle $\mathcal{L}\to M$ such that its first Chern class is

$c_{1}(\mathcal{L}) = [\omega]$

$E_{7(7)}$ refers to the split real form of $E_{7}$.

Thanks.

Edit: I have been required to provide the reference that states this as an open problem. Here it is:

http://arxiv.org/abs/0804.1362

(see section 3.3) That reference also explains the relevance of this problem in describing the moduli space of M-theory flux compactifications on seven-dimensional manifolds.

$\endgroup$

1 Answer 1

-1
$\begingroup$

If i am not wrong the following is true, G compact connected Lie group, H the centralizer of a torus in G, then G/H is a projective manifold, does this answer your question??

$\endgroup$
1
  • 1
    $\begingroup$ $E_{7(7)}$ is not compact. $\endgroup$
    – Bilateral
    Feb 16, 2015 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.