Pandemonium Purger
Posts: 1341
Joined: Monday, 24th October 2011, 06:13
If it doesn't fit anywhere else, it belongs here. Also, come here if you just need to get hammered.
Pandemonium Purger
Posts: 1341
Joined: Monday, 24th October 2011, 06:13
Tartarus Sorceror
Posts: 1891
Joined: Monday, 1st April 2013, 04:41
Location: Toronto, Canada
twelwe wrote:what are the imp... impl... oh for fucks sake, what does this mean?
Halls Hopper
Posts: 80
Joined: Sunday, 8th January 2017, 05:05
Arrhythmia wrote:TrumpTrain wrote:Arrhythmia wrote:Here's a fun one I proved today in algebraic topology.
Let f be a continuous real function defined on S^2 (thought of as unit vectors in R^3). Then there exist three mutually orthogonal vectors, v_1, v_2, and v_3 \in R^3 such that f(v_1) = f(v_2) = f(v_3).
I'd like to see this proved. I have no idea where to start this.
Essentially:
1. Show every continuous function SO(3) \rightarrow R^3 - \{(x, x, x) : x \in R\} is null-homotopic by looking at the fundamental groups (\pi_1(SO(3)) = C_2, \pi_1(R^3 - \{(x, x, x) : x \in R\} = Z (deformation retractable to the circle)) so the only homomorphism between them is the trivial one).
2. Represent orthogonal triples of vectors in S^2 as elements of SO(3).
3. Assume towards contradiction that there is a continuous function f which doesn't satisfy the problem's conditions. Then (its representation in SO(3)) must map to R^3 - \{(x, x, x) : x \in R\} so it's null-homotopic, but you can come up with a path in S^2 which won't be null-homotopic because its image (under the SO(3) representation) will wind around the pole.
I could write it up more formally, if you want.
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