# Does $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carry an almost complex structure?

Recall the list of irreducible simply connected inner symmetric spaces of compact type in dimension $4k+2$:

1. Hermitian symmetric spaces (one can write them down explicitly);

2. Grassmannians of oriented real $p$-planes in $\mathbb{R}^{p+q}$, with $pq=4k+2$;

3. The 70-dimensional exceptional symmetric space $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$.

Hermitian symmetric spaces are of course complex manifolds, and real Grassmannians carrying almost complex structures were classified by P. Sankaran (PAMS, 1991) and Z. Tang (PAMS, 1994). Thus my question:

Does $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carry an almost complex structure?

Note that in dimension $4k$ it is known that an irreducible simply connected inner symmetric space of compact type carries an almost complex structure iff it is Hermitian symmetric.

Edit concerning the dimensions multiple of 4: It is on purpose that I did not mention our paper http://fr.arxiv.org/abs/1003.5172 when I first asked the question. Indeed, the method used in that paper to treat the $4k$-dimensional situation (an index-theoretical obstruction) seems uneffective in the $4k+2$-dimensional cases so I expected other methods to tackle this problem.

Let me nevertheless give a short account on how things work in dimension $4k$. The idea is to use the following criterion:

If a compact Riemannian manifold $(M,g)$ carries a self-dual vector bundle $E$ such that the index of the twisted Dirac operator on $\Sigma M \otimes E\otimes > TM^{\mathbb{C}}$ is odd, then $M$ has no almost complex structure (in fact $TM$ is not even stably isomorphic to a complex bundle).

Note that the complex spin bundle $\Sigma M$ and $E$ can be locally defined, only their tensor product has to be globally defined. On then checks that this criterion applies to all irreducible inner symmetric spaces of compact type which are neither Hermitian symmetric nor spheres.

Returning to the dimensions $4k+2$, it is easy to check that there exists a priori no bundle satisfying the above criterion. Nevertheless, one can show real Grassmannians and $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ do carry such a bundle $E$ which satisfies all conditions in the criterion except that it is not self-dual. I don't know if this can lead anywhere...

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A possibly irrelevant comment is that this is the scalar manifold in four-dimensional $N=8$ supergravity. – José Figueroa-O'Farrill Dec 11 '10 at 1:37
José: yes, thanks for the correction. When I have time I will edit my question to explain why our method to prove that symmetric spaces which are not Hermitian symmetric are not almost complex only works in dimension $4k$. Your comment about supergravity makes this space even more intriguing for me... – Andrei Moroianu Dec 11 '10 at 8:59

I'm sorry if this is obvious to everyone, but I thought that it was worth mentioning:

I don't know the answer to the question asked, but the answer to the easier question, "Does the space $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ carry an $\mathrm{E}_7$-invariant almost complex structure?", is 'no'.

To see this, consider the orthogonal complement of the subalgebra $\mathfrak{su}(8)$ in $\mathfrak{e}_7$, say $V = \mathfrak{su}(8)^\perp$. By inspecting the root diagram of $\mathfrak{e}_7$ (as described by Cartan's representation, not the usual one you see in most books), one sees that this real vector space of dimension $70$ has the property that its complexification is the irreducible complex representation $\Lambda^4(\mathbb{C}^8)$ of $\mathrm{SU}(8)$. Since the complexification of $V$ is irreducible under $\mathrm{SU}(8)$, the action of $\mathrm{SU}(8)$ does not preserve any complex structure on $V$.

Thus, the space $\mathrm{E}_7/(\mathrm{SU}_8/(\mathbb{Z}/2))$ does not have an $\mathrm{E}_7$-invariant almost complex structure.

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The following is mostly bogus, based on my overly quick reading and misunderstanding of the question:

No. $E_7 / (SU(8)/\mu_2)$ (where $E_7$ here denotes the compact real Lie group) is not a Hermitian symmetric space. You can tell, because there is no torus (copy of $U(1)$) in the center of $SU(8)$ (maybe Helgasson or Wolf is a good reference here).

The following comment does not apply, since the manifold has dimension $4k+2$ and not $4k$

Hence, by Gauduchon, Moroianu, Semmelmann (Inventiones Math., 2010), as you seem to know, this exceptional inner symmetric space carries no almost complex structure.

The following is correct:

I have no idea how to answer this question.

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@Marty: The paper you mention only treats dimensions multiple of 4, it does not apply to $E_7/SU_8$, whose dimension is 70. I am interested in this question since this space is the only irreducible symmetric space of compact type for which I don't know whether it has an almost complex structure. I hope that some people (maybe topologists) might have some hints on how to attack this problem. – Andrei Moroianu Dec 11 '10 at 8:52
Yup. I'm wrong. I tried to downvote, but I guess I can't downvote my own answer. Who knew... 70 is not a multiple of 4 after all :). I was kind of dismissive of the question - I should have read it more carefully. Apologies to Moroianu. – Marty Dec 11 '10 at 17:06
I apologize as well: I was focusing on the superficial. (Note that I have trivially edited Marty's answer so as to be able to remove my upvote.) – Pete L. Clark Dec 11 '10 at 17:32
Never mind, I appreciate your correctness, Pete and Marty. One last remark: the question I asked is certainly difficult, but as I (and my collaborators) have no further idea, I thought MO would be the right place to discuss it. If I was wrong, I apologise too. – Andrei Moroianu Dec 11 '10 at 21:39