math q

[twelwe]

what are the imp... impl... oh for fucks sake, what does this mean?

[Arrhythmia]

what are the imp... impl... oh for fucks sake, what does this mean?

i 'unno

[TrumpTrain]

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.

can you skip 1 & 2 and just formally show 3 :p I love proof by contradiction