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edited Apr 9, 2013 at 20:16

Rodrigo A. Pérez

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I just taught Leau's Flower Theorem today...

Your best bet to understand this map is to read Milnor's book. Click here for a copy of the original notes (which are less polished than the book). The chapter on parabolic points is perhaps the most pedestrian, but even that is clearer than other books!

To give you an idea of the flow of ideas, Milnor considers an analytic function $f$ with a parabolic fixed point at 0. He first assumes the multiplier is $\lambda = 1$, and proves the attraction/repulsion picture directly from the power series (call this result T1). Some corollaries follow, and then he describes the case $\lambda = {\rm e}^{2\pi{\rm i}p/q}$, which is the same as before, except that the petals are permuted by $f$ instead of determining independent basins of attraction. THIS IS YOUR MAP. The multiplier is $\lambda = {\rm e}^{2\pi{\rm i}/5}$, so the petals map counterclockwise onto each other.

The fifth iterate of your $f$ has a fixed point with multiplier 1 so each petal maps into itself. This is the situation where the Abel function makes sense. Milnor shows how to construct explicitly this coordinate change by refining the computations from the proof of T1. THIS IS THE ANSWER TO YOUR QUESTION.

I just taught Leau's Flower Theorem today...

Your best bet to understand this map is to read Milnor's book. The chapter on parabolic points is perhaps the most pedestrian, but even that is clearer than other books!

To give you an idea of the flow of ideas, Milnor considers an analytic function $f$ with a parabolic fixed point at 0. He first assumes the multiplier is $\lambda = 1$, and proves the attraction/repulsion picture directly from the power series (call this result T1). Some corollaries follow, and then he describes the case $\lambda = {\rm e}^{2\pi{\rm i}p/q}$, which is the same as before, except that the petals are permuted by $f$ instead of determining independent basins of attraction. THIS IS YOUR MAP. The multiplier is $\lambda = {\rm e}^{2\pi{\rm i}/5}$, so the petals map counterclockwise onto each other.

The fifth iterate of your $f$ has a fixed point with multiplier 1 so each petal maps into itself. This is the situation where the Abel function makes sense. Milnor shows how to construct explicitly this coordinate change by refining the computations from the proof of T1. THIS IS THE ANSWER TO YOUR QUESTION.

I just taught Leau's Flower Theorem today...

Your best bet to understand this map is to read Milnor's book. Click here for a copy of the original notes (which are less polished than the book). The chapter on parabolic points is perhaps the most pedestrian, but even that is clearer than other books!

To give you an idea of the flow of ideas, Milnor considers an analytic function $f$ with a parabolic fixed point at 0. He first assumes the multiplier is $\lambda = 1$, and proves the attraction/repulsion picture directly from the power series (call this result T1). Some corollaries follow, and then he describes the case $\lambda = {\rm e}^{2\pi{\rm i}p/q}$, which is the same as before, except that the petals are permuted by $f$ instead of determining independent basins of attraction. THIS IS YOUR MAP. The multiplier is $\lambda = {\rm e}^{2\pi{\rm i}/5}$, so the petals map counterclockwise onto each other.

The fifth iterate of your $f$ has a fixed point with multiplier 1 so each petal maps into itself. This is the situation where the Abel function makes sense. Milnor shows how to construct explicitly this coordinate change by refining the computations from the proof of T1. THIS IS THE ANSWER TO YOUR QUESTION.

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created Apr 9, 2013 at 3:54

Rodrigo A. Pérez

3.1k
2
41
47

I just taught Leau's Flower Theorem today...

Your best bet to understand this map is to read Milnor's book. The chapter on parabolic points is perhaps the most pedestrian, but even that is clearer than other books!

To give you an idea of the flow of ideas, Milnor considers an analytic function $f$ with a parabolic fixed point at 0. He first assumes the multiplier is $\lambda = 1$, and proves the attraction/repulsion picture directly from the power series (call this result T1). Some corollaries follow, and then he describes the case $\lambda = {\rm e}^{2\pi{\rm i}p/q}$, which is the same as before, except that the petals are permuted by $f$ instead of determining independent basins of attraction. THIS IS YOUR MAP. The multiplier is $\lambda = {\rm e}^{2\pi{\rm i}/5}$, so the petals map counterclockwise onto each other.

The fifth iterate of your $f$ has a fixed point with multiplier 1 so each petal maps into itself. This is the situation where the Abel function makes sense. Milnor shows how to construct explicitly this coordinate change by refining the computations from the proof of T1. THIS IS THE ANSWER TO YOUR QUESTION.