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Published online 1 May 2009 | Nature | doi:10.1038/news.2009.427
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Hidden riddle of shapes solved
Mathematicians crack the Kervaire invariant problem.
A team of three has solved a 45-year-old problem in the mathematics of topology.
The Kervaire invariant problem is "one of the major outstanding problems in algebraic and geometric topology" says fellow mathematician Nick Kuhn, at the University of Virginia in Charlottesville.
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For a donut and a coffee cup to be topologically equivalent, said cup must have a closed circuit handle enclosing a void, as does the donut. The example pictured is invalid. Substitute something by Wedgewood. Are a hat band and a Moebius strip topologically equivalent? (Given the parity of the former's generating functions, is its apparent chirality only an artifact of construction?)
Yes, Uncle Al, I was certainly a bit confused by the statement 'a donut and a coffee cup are equivalent' after seeing the misleading picture given. Perhaps if the donut biter had bitten all the way through the ring it would have been more accurate!
Not just "a conference in Edinburgh" but the Atiyah80 conference celebrating the 80th birthday of Sir Michael Atiyah, http://www.maths.ed.ac.uk/~aar/atiyah80.htm
The picture should show an ordinary mug not a disposable one.
Apologies for the misleading picture, which has now been replaced. Thanks!
A nice animation of a donut turning into a mug (and back) can be found in the Topology wikipedia article, or directly here: http://upload.wikimedia.org/wikipedia/commons/2/26/Mug_and_Torus_morph.gif
You say "That is in line with what mathematicians intuitively expected," but that is not correct. In the 70s many mathematicians tried to prove the opposite statement, that manifolds with Kervaire invariant one exist in all dimensions allowed by Browder's theorem. I do not know of anyone, myself included, who expected the answer that we found.