Is every manifold triangulable?

quasar987

In Lee's Intro to topological manifolds, p.105, it is written that every manifold of dimension 3 or below is triangulable. But in dimension 4, threre are known examples of non triangulable manifolds. In dimensions greater than four, the answer is unknown.

But in Bott-Tu p.190, it is written that every manifold admits a triangulation.

Which is right?

lavinia

I would guess that Bott and Tu mean every smooth manifold since their book is about differential topology. It is a theorem of Whitehead, I believe, that every smooth manifold has a smooth triangulation.

Eynstone

It seems that the authors probably have different definitions of triangulation.
In my opinion, the problem boils down to whether every interval isomorphic to some interval in R^n is triangulable & hence the second statement looks good.

lavinia

R. ]Kirby and L. C. Siebenmann, On the triangulation of manifolds and the hauptvermutung,
Bull. Amer. Math. Soc., 75 (1969), 742-749.

This paper is said to have an example of a non-triangulable 6 manifold

mathwonk

it seems indeed to depend on your definition of triangulation. in gneral the answer may be unknown.

consult: the history of topology by ioan mackenzie (or ask ron stern)

http://books.google.com/books?id=7iR...ble%3F&f=false

quasar987

In Whitney's geometric integration p.124, he credits S.S. Cains (1934) with Whitehead (1940) giving an improvement of the proof in "On C¹ complexes, Annals of Math. 41"