Commit d431d3ef authored by Felix Hilsky's avatar Felix Hilsky
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chapter 3.1 started

parent 228dec2f
......@@ -186,6 +186,48 @@ Let $`Q ∈ W^{1,p}(Ω, 𝒬), 1 ≤ p ≤ ∞`$ be non-orientable. Then there e
- In chapter 3, orienting a line field is translated into lifting $`Q`$ from $`ℝP^2`$ to covering space $`𝕊^2`$.
- Classical algebraic topology tools only consider continous maps, but we have $`W^{1,p}(Ω)`$ maps and want to preserve this regularity
### Continous line fields
Chapter 3.1 deals with the lifting of continous line fields.
Similar to "normal" algebraic topology.
*Lemma 3*:
- $`Q : [t_1, t_2] → 𝒬`$ continous path can be lifted in two ways (depending on choice of start).
- If $`Q`$ stays close ($`≤ ε s`$ for $`0 < ε < \sqrt{2}`$) to some $`\overline{Q} ∈ 𝒬`$, then the lifted $`n`$ stays on the same half-circle as the start value ($`|n(t) - \overline m| ≤ ε`$)
*Proof notes for Lemma 3*: First calculation:
```math
| n ⊗ n - m ⊗ m |^2 = Σ_{i,j} | n_i n_j - m_i m_j |^2 = Σ_{i,j} ( n_i n_j - m_i m_j )^2
```
```math
= Σ_{i,j} (n_i n_j)^2 + (m_i m_j)^2 - 2 (n_i n_j m_i m_j)
= (\underbrace{Σ_i n_i^2}_{=1} (\underbrace{Σ_j n_j^2}_{=1})) + (\underbrace{Σ_i m_i^2}_{=1} (\underbrace{Σ_j m_j^2}_{=1})) - 2 (\underbrace{Σ n_i m_i}_{=n · m} (\underbrace{Σ_j n_j m_j}_{= n·m}))
```
```math
= 1 + 1 - 2 (n·m)^2 = 2 (1 - (n·m)^2)
```
Define $`n`$ arbitrary (possibly non-continous) as $`Q(t) = s(n(t) ⊗ n(t) - \frac13\operatorname{Id})`$. Construct $`n^+`$ by starting at $`n(t_1)`$, then staying on the same side of the circle as $`n^+(t_1)`$ as long ("$`δ`$") as it is possible to prove. That makes it continous. Then start over at $`n^+(t_1 + δ)`$. This process is finite since a continous path into a bounded set on a compact domain is uniformly continous.
There are only two continous liftings because otherwise we would have a jump by 180° where it changes from $`n^+`$ to $`n^-`$.
Given that $`Q`$ stays close to some $`\overline{Q}`$, calculation shows that $`n^+ · m`$ can never be 0, so must be always $`>0`$.
---
*Proposition 5*: $`Ω`$ simply connected, $`Q`$ continous. Then there exists a continous lifting $`n: P(n) = Q`$.
*Proof*: See algebraic topology: q4-q23 Proposition 1.33, currently p. 70
---
*Theorem 1*: If $`Q: G := Ω \setminus \bigcup_{i = 1}^N \overline{ω_i} → 𝒬_2`$ is orientable at all $`∂ω_i`$, then $`Q`$ is orientable.
*Proof sketch*:
- For $`f:[0,1] → \overline{Ω}`$ define $`f^*`$ by replacing parts of $`f`$ in ω_i by (shorter) parts on $`∂ω_i`$. $`f^*`$ is continous (with proof in Lemma 4, but really?).
- *Lemma 5*:
1) $`\overline{Ω}`$ is path-connected and simply connected. Proof: Schoenflies theorem: Jordan curve creates homeomorphism to unit disk.
2) $`\overline{G}`$ is path-connected. Proof: see previous bullet point.
3) $`\overline{G}`$ is locally path-connected: for every point $`x ∈ \overline{G}`$ and every radius $`ε`$, there is a ball $`B_{δ}(x)`$, such that all points in $`B_{δ}(x) ∩ \overline{G}`$ are path-connected to each other with paths within $`B_{ε}(x) ∩ \overline{G}`$ is path-connected. Proof: Use homeomorphism from Schoenflies theorem, build small balls and a straight line in unitdisk.
- *Lemma 6*:
## Questions
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