- chapter 4: different cases of $`Q`$ orientable on boundary $`⇒`$/$`⇔`$/$`⇐`$ on $`Ω`$. Sometimes $`Q: → 𝒬_2`$, sometimes $`Q`$ continous (on boundary)

### Chapter 2 Propositions

(choose better subsection title!)

*Proposition 2*: $`Q ∈ W^{1,p}(Ω, 𝒬)`$ ($`1 \leq p \leq ∞`$) can have only two orientations.

*Proof*:

In one dimension weakly differentiable means that $`f(x) = f(a) + ∫_a^x f'(y) dy`$.

(Which is in turn equivalent to absolutely continous.)

In several dimensions this is true in each direction but only for almost all $`x_2`$, $`x_3`$.

So: $`n`$, $`m`$ have representatives such that for almost everywhere $`x_2`$, $`x_3`$: $`n`$, $`m=τn`$ are absolutely continous, so also $`τ = τ·1 = τn·n = m·n`$.

Absolute continuity implies continuity, so $`τ`$ has a representative that is for those almost all $`x_2, x_3`$, constant.

Now use this in Fubini on a small ball in $`Ω`$ around $`y`$:

(first =: Fubini, second =: $`τ`$ is constant along $`x_1`$ for almost all $`x_2, x_3`$, third =: first integrated over $`x_1`$ direction, this is $`± φ(x_{1, \max}, x_2, x_3) - (± φ(x_{1, \min}, x_2, x_3)) = 0 - 0 = 0`$, then integrated over $`x_2`$, $`x_3`$, it stays 0)

Hence $`∇τ = 0`$ (since $`x_2`$, $`x_3`$ are analogous) weakly but this means that $`τ`$ is constant on $`B_{ε}(y)`$ for every $`y ∈ Ω`$ and therefore constant on $`Ω`$. Done.

---

Trace operator is introduced by Evans but with continuity estimate for $`\operatorname{tr} : W^{1,p} → L^p`$. Here we need the stronger statement that $`\operatorname{tr}`$ maps into the [Sobolev-Slobodeckij space](https://en.wikipedia.org/wiki/Trace_operator#Characterization_using_Sobolev%E2%80%93Slobodeckij_spaces) $`W^{1-1/p, p}`$.

*Proposition 3*:

*Proof sketch*:

## Energy functionals

- specific energy functionals are regarded (p. 11/503) because they were looked at before and Oseen-Frank was successful for them

- conversion between energy functionals possible

## Questions

- What is the third index $`k`$ for $`Q`$ on page 3 (= 495)?

- Probably it's a derivative of $`Q_{ij}`$ in direction $`k`$.

- And why is it written as a product instead of a square$`^2`$?

- To make spectral representation of $`Q`$ fit to $`Q = s(n \otimes n - \tfrac13 \operatorname{Id})`$, we need $`s = -3 λ_1 = -3 λ_2, n = ê_3`$. How can $`s`$ then depend on $`a, b, c`$ (see formula (3) on page 3 (=495))? If that's the case, $`Q`$ is further reduced to the choice of $`n`$ but that's what we assumed as our model at the beginning. So apparently that's fine.

- in Proposition 1 it should be $`\det Q = +\frac{2s^3}{27}`$ since the $`-`$ exists in exactly two eigenvalues: $`\det Q = λ_1 λ_2 λ_3 = (-\frac s3)(-\frac s3)(+\frac {2s}3) = +\frac{2s^3}{27}`$. Correct?

## Errors in the paper

p.5 (=497) Proposition 1 (iii): sign of $`\operatorname{det} Q`$ is wrong. Must be positive, since exactly two of the eigenvalues are negative

p. 7 (=499) Proposition 2: it says $`Q(x) ∈ W^{1,p}(Ω, 𝒬)`$ but $`(x)`$ must be erased ($`Q(x) ∈ 𝒬`$)

Also the statement of the proposition is trivial and not what the colloquial formulation says: it should be "... with $`P(n) = P(m) = Q`$ and $`n ≠ m`$ [as $`W^{1,p}`$-generalized functions, so on a non-0-Lebesgue set], we have $`m = -n`$ almost everywhere in $`Ω`$. If $`n ≠ m`$ almost everywhere, it follows straight away that $`m = -n`$ almost everywhere since pointwise $`n(x) = ± m(x)`$

- Gute Quelle lesen zum $`\operatorname{tr}`$-Operator an Rändern inkl. $`W^{1-\frac1p,p}(Ω, 𝕊^2)`$ bzw. $`W^{1-\frac{1p},p}(∂Ω, 𝒬)`$ (s. unter Def. 1 auf S. 7 (=499) und Proposition 3 von B&Z).