Is every manifold triangulable?


by quasar987
Tags: manifold, triangulable
quasar987
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Sep22-10, 06:44 AM
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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?
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lavinia
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Sep22-10, 07:20 AM
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Quote Quote by quasar987 View Post
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?
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
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Sep22-10, 07:32 AM
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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
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Sep22-10, 07:58 AM
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Is every manifold triangulable?


Quote Quote by quasar987 View Post
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?
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
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Sep23-10, 11:24 PM
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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
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Oct5-10, 02:39 PM
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Quote Quote by lavinia View Post
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.
In Whitney's geometric integration p.124, he credits S.S. Cains (1934) with Whitehead (1940) giving an improvement of the proof in "On Cą complexes, Annals of Math. 41"


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