# Is the E8 manifold homeomorphic to a CW complex?

Is the E8 manifold homeomorphic to a CW complex? (I know that it is not triangulable)

Edit: The E8 manifold is the unique compact (without boundary), simply connected topological 4-manifold, whose 2nd integer homology (equipped with its intersection form) is isomorphic to the $E_8$ lattice. A detailed description of this manifold can be found in an answer to this MSE question.

It appears to be an open problem to determine if every $n$-dimensional topological manifold admits structure of a CW complex. Positive answer (due to hard work of many people, most importantly - Kirby and Siebenmann for $n\ge 6$, Freedman and Quinn for $n=5$) is known in all dimensions except for $n=4$. Furthermore, all connected noncompact 4-dimensional manifolds are known to be triangulable (this is due to Quinn) and, hence, the answer is positive in this case as well.

Given that the E8-manifold is, in some sense, the simplest nonsmoothable 4-dimensional manifold, it makes sense to ask if it is homeomorphic to a CW complex.

• I don't understand why this was closed... What exactly is not clear? – Mariano Suárez-Álvarez May 19 '14 at 19:14
• @MarianoSuárez-Alvarez Apparently everything after the first line was only just added. But the first line itself is all that's necessary, so the question remains... – user98602 May 19 '14 at 19:17
• My guess is that the closers did not know what E8 is. The notion is standard in topology but might confuse somebody familiar only with, say, Lie groups. – Moishe Kohan May 19 '14 at 19:43
• Too bad it was closed. Given the tags, I thought that the question was clear enough. If I understand correctly, the answer isn't known!? – Lepanais May 20 '14 at 20:51
• @Lepanais: Yes, it is an open problem. – Moishe Kohan May 21 '14 at 17:31

Question. Is it true that a 4-dimensional manifold $X$ admits a CW-complex structure if and only if $X$ is smoothable?