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The irrational torus flow is free with dense orbits

Statement

Let αRQ. The formula

t(z,w)=(e2πitz,e2πiαtw)

defines a smooth free left action of R on T2=S1×S1, and every orbit is dense. Its orbit map through (1,1),

i(t)=(e2πit,e2πiαt),

is an injective immersion and a Lie-group homomorphism.

Facts & Assumptions

Given: An irrational real number α, the additive Lie group R, and the usual torus Lie group T2.

[F1]

A smooth left action is jointly smooth and satisfies the identity and associativity laws. Smooth left actions of Lie groups.

[F2]

Among N+1 points placed in N sets, two points lie in the same set. The pigeonhole principle on N.

Proof

technique · construct small irrational rotations by the finite pigeonhole principle
1.1

The displayed map is jointly smooth. Its phase factors satisfy e2πi(s+t)=e2πise2πit and likewise with α(s+t), so 0 acts as the identity and (s+t)x=s(tx); hence it is a smooth left action by [F1]. If t fixes any (z,w), then e2πit=e2πiαt=1, so tZ and αtZ. Irrationality forces t=0, proving freeness.

givenF1algebraconstruct
1.2

The integer rotation orbit {nα+Z:nZ} is dense in R/Z. Indeed, given ε>0, choose an integer N>1/ε and partition [0,1) into N half-open intervals of length 1/N. Applying [F2] to the N+1 fractional parts of 0,α,,Nα gives integers 0r<sN with (sr)α<1/N<ε. Replacing q=sr by q if necessary makes δ={qα} satisfy 0<δ<ε. For any u[0,1), the integer m=u/δ satisfies 0umδ<δ and mδ<1, so the orbit point mqα+Z=mδ+Z is within ε of u+Z.

F2algebra
2.1

Fix a source point (z,w) and a target (e2πiaz,e2πibw). Parameters t=a+n, nZ, put the first coordinate exactly at e2πiaz, while their second-coordinate phases are e2πiαae2πinα. Step 1.2 makes the latter dense in S1, so some such parameters put the action point arbitrarily close to the target. Hence every orbit is dense.

step 1.1step 1.2algebra
3.1

The orbit map i is a homomorphism by the phase calculation in step 1.1. It is injective because i(t)=(1,1) implies t=0 by the freeness calculation. Its differential sends 1T0R to the nonzero tangent vector 2πi(1,α); translating the homomorphism identity shows its differential is injective everywhere. Thus i is an injective immersion. All choices above are finite or single choices, so no choice principle is used.

step 1.1step 2.1algebradischarge-construct: flow and winding verified

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