I know that Ciprian Manolescu has settled the triangulation conjecture in the negative: Not all manifolds can be triangulated. I've only read secondary literature on this result, which did not detail in which dimensions it is known that there exist untriangulable manifolds. My understanding is that Freedman and Casson showed there exist counterexamples in dimension $4$.

Q. Does Manolescu's proof provide counterexamples in all dimensions $\ge 5$?

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    $\begingroup$ If I believe the abstract of this article, "non-triangulable manifolds exist in each dimension >4". $\endgroup$
    – Gro-Tsen
    Mar 11, 2017 at 1:13

1 Answer 1


In dimensions up to three, every manifold is triangulable (this is classical). In dimension 4, there are simply connected non-triangulable manifolds (such as the E8 manifold); in fact, a closed 4-manifold is triangulable if and only if it's smoothable. (Pick a triangulation; the links of vertices are always both homology and homotopy spheres, thus by the Poincare conjecture are $S^3$, so this gives a PL structure, and thus is smoothable. This is probably true more generally than closed manifolds, but I'm feeling a little paranoid.)

The crucial papers pre-Manolescu are, firstly, Galewski-Stern and Matumoto's thesis (which I don't have a link to): these prove that every closed manifold if dimension $n \geq 5$ are triangulable if and only if there is a homology 3-sphere $Y$ such that $Y \# Y$ bounds a homology ball, and the Rokhlin invariant $\mu(Y) = 1 \in \Bbb Z/2\Bbb Z$. This is what Manolescu disproved (there is no such 3-manifold); the Galewski-Stern paper clarifies what dimensions their theorem is proved for after you start dropping assumptions about compactness and the boundary.

But importantly for your question is Galewski and Stern's sequel paper. Their theorem 2.1 (plus Manolescu's result) implies that there are non-triangulable manifolds in every dimension $n \geq 5$. However, all orientable 5-dimensional manifolds are triangulable. In dimensions at least 6, though, you can use their construction to produce non-triangulable orientable manifolds.

EDIT: I can't describe this better than Ciprian's own Lectures on the Triangulation Conjecture; Chapter 2 outlines the geography of triangulable manifolds, giving progressively more detail about the older results until he gives a sketch of his construction.

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    $\begingroup$ Does bordism/cobordism classify the cobordant of ONLY triangulated manifolds, or does it classify the cobordant of non-triangulated manifolds as well? $\endgroup$ Sep 26, 2018 at 14:55
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    $\begingroup$ @annieheart You can define "bordism groups" for pretty much any kind of manifolds you like. Normally one means smooth manifolds. Certainly you can make sense of it for non-triangulable manifolds, but these are necessarily non-smoothable. So you would want to know about $\Omega_*^{\text{Top}}$; there is some study of this in a book by Madsen and Milgram (classifying spaces for...). You might find this answer useful. $\endgroup$
    – mme
    Sep 26, 2018 at 18:03

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