

Triangulation Conjecture Disproved - digital55
https://www.quantamagazine.org/20150113-a-proof-that-some-spaces-cant-be-cut/

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tunesmith
Are there historical examples of conjectures widely believed to be true and
then proven to be false?

I suppose Gödel's incompleteness theorems and Hilbert's second problem is
along those lines, but I'm not sure Hilbert's problem was regarded as a
conjecture.

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jonnybgood
There is the two sentence counterexample to Euler's conjecture.

[http://www.openculture.com/2015/04/shortest-known-paper-
in-a...](http://www.openculture.com/2015/04/shortest-known-paper-in-a-serious-
math-journal.html)

~~~
tunesmith
Thanks, that's awesome!

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Xcelerate
Pretty cool. I was actually wondering the other day if, for a given manifold,
there is a general algorithm to randomly choose a point on it such that each
point has an equal probability of being chosen.

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panic
This isn't possible for some manifolds. In particular, you can't choose a
point uniformly from an infinite plane. To see why, split the plane into an
infinite number of equally-sized (say, one by one) squares. Each square should
have an equal probability that one of its points is chosen. Call this
probability _s_.

Now let's look at the entire plane. If you pick a point, it will surely be on
the plane. That means the probability that you choose a point on the plane is
equal to one.

Putting the squares together gives you the whole plane, so you'd expect that
summing their overall probabilities _s_ would give you the overall probability
of the plane, which is one. But there are an infinite number of squares, and
there's no real number _s_ for which the sequence _s_ , _s_ \+ _s_ , _s_ \+
_s_ \+ _s_ , ... converges to one.

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enupten
[http://planetmath.org/anysigmafinitemeasureisequivalenttoapr...](http://planetmath.org/anysigmafinitemeasureisequivalenttoaprobabilitymeasure)

~~~
apoklasdjasd
Note that the OP desired that "each point has an equal probability of being
chosen". The infinite plane being infinite, this requirement does not formally
make sense, but a natural way to interpret it is to ask for invariance under
isometries (or translation-invariance). As the post that you replied to
indicates, this is not possible for the Euclidean plane. The measure
constructed by the lemma that you cite does not have this property.

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Retric
Seems odd that they did not provide a counter example. My suspicion there is
an incorrect assumption for non fractal structures.

PS: This is really one of those cases where a paper does not mean much. You
really need to dig through
[http://arxiv.org/abs/1303.2354](http://arxiv.org/abs/1303.2354) and find the
flaw as there is no peer review. Though it supposedly is going to show up in
Journal of the AMS so it's not junk and probably worth diging into.

~~~
panic
Here is the actual paper:
[http://arxiv.org/pdf/1303.2354v4.pdf](http://arxiv.org/pdf/1303.2354v4.pdf).
It looks like this is their family of counter examples:

 _A specific non-orientable five-dimensional manifold M5 with Sq1 ∆(M) ̸= 0 is
constructed in [25]. Hence M5 is non-triangulable. To get non-orientable
examples of non-triangulable manifolds in dimensions n > 5 we can take the
product of M5 with the torus Tn−5. To get an orientable example in dimension 6
we can consider the non-orientable S1-bundle over M given by M ̃ ×Z/2 S1,
where M ̃ → M is the oriented double cover; the total space of this bundle is
orientable. To get orientable examples in dimensions n > 6, we can then take
products with Tn−6._

~~~
Grue3
Is there a reason why previously found 4-dimensional counterexample couldn't
have been used as the basis of induction?

