Title: Part IB / Geometry
Tags: Abstract_Surfaces Fundamental_Forms_Curvature Surfaces Hyperbolic_Surfaces Geodesics Intro_Geometry

Q: Definition. An embedded polygonal surface is a compact subset \(\Sigma \subset \mathbb{R^2}\) equipped with a cover by finitely many compact subsets \(P_1 , \dots , P_k\) such that
[...]
A: i) each \(P_i\) is a planar polygon
ii) the intersection of any two \(P_i\) is either empty, a common edge, or a common vertex
iii) each edge lies in exactly two polygons

Q: Definition. The Euler characteristic of \(\Sigma\) is [...]
A: \(X(\Sigma) = V - E + F\)

Q: Definition. The defect at a vertex v is [...]
A: \[\Delta(v) = 2 \pi - \sum (\text{ interior angles meeting at v})\]

Q: Theorem. Polygonal Gauss-Bonnet. \[\sum_{\text{ vertices v} } \Delta (v) = 2 \pi X(\Sigma)\]
A: LHS = \(\sum_v \left ( 2 \pi - \sum \text{interior angles } \right ) = 2\pi V - \sum_v \sum_{\text{faces containing v}} (\text{angles in face at v})\) \[ = 2 \pi V - \sum_{\text{faces f}} \sum_{\text{v in f}} (\text{angle at v in f}) = 2\pi V - \sum_{\text{faces f}} \pi(\text{# of edges in F - 2})\] \[ = 2 \pi (V + F) - \pi \sum_f \text{# edges in f}\] the last term in exactly 2E so \[ = 2\pi X(\Sigma)\]

Q: Definition. A map \(f : U \to V\) is smooth if [...]  f is a diffeomorphism if [...]
A: it has all partial derivative of all orders.; it's smooth and has a smooth two-sided inverse.

Q: Theorem. Inverse Function Theorem. If \(f: U \to V\) is smooth and its derivative at \(p \in U\), denoted \(D_p f \in \mathcal{L}(\mathbb{R}^m, \mathbb{R^n})\), is invertible then there exist open neighbourhoods U' of p in U, V' of f(p) in V such that \(f|_{U'} : U' \to V'\) is a diffeomorphism. (note only possible if m=n).
A: Was proved in A+T but without higher differentiability of \(f^{-1}\). But we know \[D_q (f^{-1}) = \left ( D_{f^{-1}(q)} f \right )^{-1}\] by the chain rule. So the entires of the matrix \(D(f^{-1})\) can be expressed in terms of the entires of \(Df\) composed with \(f^{-1}\). So \(D(f^{-1})\) is differentiable (because Df and \(f^{-1}\) are)  so \(f^{-1}\) is twice differentiable, now induct.

Q: Definition. A smooth map \(f: U \to V\) is a local diffeomorphism at \(p \in U\) if [...]
A: \(D_p f\) is invertible.

Q: Definition. An embedded surface is a subset \(\Sigma \subset \mathbb{R}^3\) such that [...]
A: \(\forall p \in \Sigma\) there exists an open neighbourhood T of p in \(\mathbb{R}^3\) and a smooth function \(f: T \to \mathbb{R}\) such that
i) \(\Sigma \cap T = f^{-1}(0)\)
ii)\(D_p f \neq 0\)
Back Extra: We basically want the surface to have a 2 dimensional tangent plane at p.

Q: Definition. A (smooth 2-dimensional) parametrisation of \(\Sigma\) near a point p is an [...]
A: open neighbourhood U of p in \(\Sigma\), an open set V in \(\mathbb{R}^2\), and a homeomorphism \(\sigma: V \to U\) such that \(\sigma\) is smooth as a map to \(\mathbb{R}^3\) and \(\forall q \in V\) the map \(D_q (\sigma) \in \mathcal{L}(\mathbb{R}^2, \mathbb{R}^3)\) is injective.
Back Extra: We usually use (u,v) for coordinates on \(\mathbb{R}^2\) and subscripts for partial derivatives. Then \(D_q \sigma\) being injective means \(\sigma_u, \sigma_v\) are L.I at q.

Q: Lemma. \(\Sigma\) admits a parametrisation near p iff \(\exists\) open neighbourhood T of p in \(\mathbb{R}^3\) and a smooth function \(f: T \to \mathbb{R}\) such that \(\Sigma \cap T = f^{-1}(0)\) and \(D_p f \neq 0\)
A: Suppose such an f exists. By reordering the coordinates, WLOG \(f_z(p) \neq 0\). By Implicit Function Theorem(with k = 2, l = 1, coordinates ((x,y), z) ), \(f^{-1}(0)\) is locally a graph over the (x,y) plane. Explicitly there exists open sets \(A \subset \mathbb{R}^2, B \subset \mathbb{R}\), smooth \(F : A \to B\) such that \(p \in A \times B \subset T\) and \[\Sigma \cap (A \times B) = \{(x,y, F(x,y)) : (x,y) \ in A\}\] We get a parametrisation
Conversely, suppose \(\Sigma\) admits a parametrisation \(\sigma : V \to U\) near p.
Step 1: Reduce to a graph-type parametrisation \(\tau (u,v) = (u,v,F(u,v))\).
Let \(\pi : \mathbb{R}^3 \to \mathbb{R}^2\) denote projection to (x,y) plane. Want \(\tau\) such that \(\pi \circ \tau = \text{id}\).
By reordering coordinates, WLOG \(\sigma_u (p), \sigma_v(p), e_3\) forms a basis for \(\mathbb{R}^3\). This tells us that \(\pi \circ D_p (\sigma)\) is an isomorphism from \(\mathbb{R}^2 \to \mathbb{R}^2\) i.e. \(D_p(\pi \circ \sigma) = D_{\sigma(p)} \pi \circ D_p \sigma\)  is an isomorphism. So, shrinking U and V if necessary, \(\pi \circ \sigma\) is a diffeomorphism from V to \(\pi(U)\) which is an open set \(A \subset \mathbb{R}^2\). Then \[\tau : A \to U\]
defined by \(\sigma \circ (\pi \circ \sigma)^{-1}\) is the \(\tau\) that we want.
Step 2: Shrink the sets to make \[f(x,y,z) = z - F(x,y)\]
 work. Since U is open in \(\Sigma\), \(\exists\) open set T in \(\mathbb{R}^3\) such that \(U = \Sigma \cap T\). Then \[\Sigma \cap T = \{(x,y, F(x,y)) : (x,y) \in A\}\](\(U = \tau(A)\))
Now replace T with \(T \cap (A \times \mathbb{R})\). Then \(f(x,y,z) = z - F(x,y)\) is defined on all of T and satisfies \[\Sigma \cap T = f^{-1}(0), D_p f \neq 0\]
Corollary of this is that \(\Sigma\) is an embedded surface iff it can be covered by parametrisations.

Q: Definition. Let T be an open set in \(\mathbb{R}^{k + l}\) on which we take coordinates \(x \in \mathbb{R}^k, z \in \mathbb{R}^l\). Let's suppose that \(f: T \to \mathbb{R}^l\) is a smooth function and \(p = (a,b)\) is a point in \(f^{-1}(0)\).The set \(f^{-1}(0)\) is locally a graph (over x near p) if [...]
A: there exists an open neighbourhood A of a in \(\mathbb{R}^k\), open neighbourhood B of b in \(\mathbb{R}^l\), and a smooth map \(F: A \to B\) such that \(A \times B \subset T\) and \[f^{-1}(0) \cap (A \times B) = \{(x, F(x)) : x \in A\}\]
Back Extra: Intuitively, this means that the equation \(f(x,z) = 0\) implicitly defines z in terms of x, via \(z = F(x)\)

Q: Theorem. Implicit Function Theorem. If \[\det \left (\frac{\partial f_i}{\partial z_j} \big |_p \right ) \neq 0\] then f is locally a graph over x near p.
A: Define \(g : T \to \mathbb{R}^{k+l}\) by \[g(x,z) = (x, f(x,z))\]
So by given condition, \(D g\) is invertible. So by Inverse Function Theorem we can shrink T such that g gives a diffeomorphism from T to an open neighbourhood W of \(g(p) = (a,0)\) in \(\mathbb{R}^{k+l}\)
Note that the inverse \(h : W \to T\) is of the form \(h(u,v) = (u, H(u,v))\) for some smooth \(H : W \to \mathbb{R}^l\)
So its restriction to \[V = W \cap (\mathbb{R}^k \times \{0\})\] is of the form \[h(u,0) = (u,F(u))\] for smooth \(F : V \to \mathbb{R}^l\) (namely \(F(u) = H(u,0)\))
Pick small open neighbourhoods A of a, B of b such that \(A \times B \subset T\). By shrinking A, may assume \(A \subset F^{-1}(B)\). Then F is a smooth map \(A \to B\) and for \((x,z) \in A \times B\) we have \[(x,z) \in f^{-1}(0) \iff g(x,z) = (x,0) \iff (x,z) = h(x,0) \iff z = F(x) (\text{ since } h(x,0) = (x,F(x)) )\]
So \(f^{-1}(0) \cap (A \times B) = \{(x, F(x)) : x \in A\}\)

Q: Definition. The map [...] is the transition function from \(\sigma_1\) to \(\sigma_2\)
A: \[\sigma_2^{-1} \circ \sigma_1 : \sigma_1^{-1}(U_1 \cap U_2) \to \sigma_2^{-1}(U_1 \cap U_2)\]
Back Extra: It's a homeo between open sets
In fact its a diffeomorphism

Q: Definition. The tangent space to \(\Sigma\) at p is [...]
A: \[T_p \Sigma = \{ \dot{\gamma}(0) : \gamma \text{ is a smooth map } (-\epsilon, \epsilon) \to \Sigma \text{(for some } \epsilon > 0) \text{ such that } \gamma(0) = p\} \subset \mathbb{R}^3\]

Q: Lemma. If \(\sigma\) is a parametrisation of \(\Sigma\) near p and f is a smooth function near p s.t. \(D_p f \neq 0\) and \(\Sigma\) is locally \(f^{-1}(0)\) then \[\text{Im } D_{\sigma^{-1}(p)} \sigma \subset T_p \Sigma \subset \text{Ker } D_p f\]  Note LHS and RHS are 2-dimensional vector spaces so both '\(\subset\)' are '='. In particular \(T_p \Sigma\) is a 2-dimensional vector space with basis \(\sigma_u (p), \sigma_v(p)\)
A: Suppose \(w \in \text{Im } D_{\sigma^{-1}(p)} \sigma\).Then \(w = a\sigma_u + b\sigma_v\) for some \((a,b) \in \mathbb{R}^2\). Now define \[\gamma(t) = \sigma(\sigma^{-1}(p) + t(a,b))\]
By the chain rule \(\dot{\gamma}(0) = a\sigma_u + b \sigma_v = w\) so \(w \in T_p \Sigma\).
Now suppose \(w \in T_p \Sigma\). So \[w = \dot{\gamma}(0) \text{ for some } \gamma : (-\epsilon, \epsilon) \to \Sigma\]
Then have \[f \circ \gamma = 0\] so \(D_p f(\dot{\gamma}(0)) = 0\) by chain rule. Hence \(w = \dot{\gamma}(0) \in \text{Ker }D_p f\)

Q: Definition. A map \(F : \Sigma_1 \to \Sigma_2\) is smooth if [...]
A: it's continuous and for all parameterisations \(\sigma_1:V_1 \to U_1\), \(\sigma_2: V_2 \to U_2\) of \(\Sigma_1, \Sigma_2\) respectively the map \(\sigma_2^{-1} \circ F \circ \sigma_1\) is smooth.
Back Extra: \[\sigma_2^{-1} \circ F \circ \sigma_1 : \sigma_1^{-1}(U_1 \cap F^{-1}(U_2)) \to V_2\]
Note that \(\sigma_1^{-1}(U_1 \cap F^{-1}(U_2))\) and \(V_2\) are both open sets in \(\mathbb{R}^2\) so smoothness makes sense. In other words, F is smooth iff it's smooth when written in coordinates from parameterisations.
It's enough to check smoothness of the function for ccollections of parameterisations covering the surfaces, rather than all parameterisations. If it's smooth for some \(\sigma_1, \sigma_2\) and \(\hat{\sigma_1}\) is another parameterisations of \(\Sigma_1\) then an overlap between \(\sigma_1\) and \(\hat{\sigma_1}\) we have \[\sigma_2^{-1} \circ F \circ \hat{\sigma_1} = (\sigma_2^{-1} \circ F \circ \sigma_1)\circ(\sigma_1^{-1} \circ \hat{\sigma_1})\] and since the transition function is smooth so is this entire function.

Q: Definition. A diffeomorphism between surfaces is [...]
A: a smooth map with smooth two-sided inverse.
Back Extra: If \(F : \Sigma_1 \to \Sigma_2\) is smooth and \(D_pF \in L(T_p \Sigma_1, T_{F(p)} \Sigma_2)\) is invertible then F is a local diffeomorphism at p.

Q: Definition. A Gauss map for \(\Sigma\) on an open set U is [...]
A: a smooth map \(U \to \mathbb{R}^3\) such that \(\forall p \in U\), \(n(p) \perp T_p \Sigma\) and \(||n(p)|| = 1\).
Back Extra: Well usually view n as a month map \(U \to S^2\).

Q: Lemma. Existence and non-uniqueness. For any \(p \in \Sigma\) there exists an open neighbourhood U of p and a Gauss map \(n : U \to S^2\). If U is connected then n is unique up to a sign.
A: Choose a parameterisations \(\sigma : V \to U\) near p. Define \[n = \frac{\sigma_u \times \sigma_v}{|| \sigma_u \times \sigma_v ||}\] Suppose \(n_1, n_2\) are two Gauss maps on U, connected. Consider smooth map \(f : U \to \mathbb{R}\) given by \(\langle n_1 , n_2 \rangle\). This map lands in \(\pm 1\). Then \(f^{-1}((0 , \infty)), f^{-1}((-\infty, 0))\) is a partition of U into open sets. So by connectedness of U, one of these sets is empty. So either \(f \equiv 1\) or \(f \equiv -1\). Hence \(n_2 \equiv n_1\) or \(n_2 \equiv -n_1\).

Q: Definition. \(\Sigma\) is orientable if [...]
A: it admits a Gauss map defined on \(U = \Sigma\)
Back Extra: Otherwise \(\Sigma\) is non-orientable.

Q: Proposition. \(\Sigma\) is orientable iff it can be covered by parameterisations \(\sigma_{\alpha}\) such that each transition function \[\psi_{\alpha \beta} = \sigma_{\alpha}^{-1} \circ \sigma_{\beta}\] is orientation-preserving, meaning \[\det (D \psi_{\alpha \beta}) > 0\].
A: First note that if \(\sigma(u,v)\) and \(\tau(x,y)\) are overlapping parameterisations then \[\tau_x \times \tau_y = (u_x \sigma_u + v_x \sigma_v) \times (u_y \sigma_u + v_y \sigma_v) = \det \left ( \begin{matrix} u_x & u_y \\ v_x & v_y\end{matrix} \right) \sigma_u \times \sigma_v\] So \[\frac{\tau_x \times \tau_y}{||\tau_x \times \tau_y ||} = \text{sgn} (\det D\psi) \frac{\sigma_u \times \sigma_v}{||\sigma_u \times \sigma_v||}\] where \(\psi = \sigma^{-1} \circ \tau\).
"If direction": The local constructions \[\frac{(\sigma_{\alpha})_u \times (\sigma_{\alpha})_v}{|| (\sigma_{\alpha})_u \times (\sigma_{\alpha})_v}\] are consistent on overlaps so define a global Gauss map.
"Only if direction": Fix a global Gauss map n. Take nay cover of \(\Sigma\) by parameterisations \(\sigma_{\alpha}\)(WLOG domains are connected). If \[\frac{(\sigma_{\alpha})_u \times (\sigma_{\alpha})_v}{||(\sigma_{\alpha})_u \times (\sigma_{\alpha})_v||} = u\] then have \(\sigma_{\alpha}\) as is.
Otherwise replace \(\sigma_{\alpha}\) with \[\sigma_{\alpha} \circ ((u,v) \to (v,u))\] to fix it.
Now all transition functions are orientation preserving.

Q: Corollary. Orientability is diffeomorphism invariant, i.e. if there exists a diffeo \(F: \Sigma_1 \to \Sigma_2\) then \(\Sigma_1\) is orientable iff \(\Sigma_2\) is orientable
A: If \(\{\sigma_x\}\) is a collection of parametrisation covering \(\Sigma_1\) s.t. all transition function \(\psi\) have \(\det D \psi > 0\) then \(\{F \circ \sigma_x\}\) is a collection of parametrisations covering \(\Sigma_1\) with the same property.

Q: Definition. The First Fundamental Form (FFF) of \(\Sigma\), denoted by \(I\), is this family of inner products \(I_p\) on tangent spaces. Given a parametrisation about p we can express \(I_p\) as follows [...]
A: \[I_p(a\sigma_u + b \sigma_v, c\sigma_u + d\sigma_v) = Eac + F(ad + bc) + Gbd\]
Back Extra: \(E = \sigma_u \cdot \sigma_u, F = \sigma_u \cdot \sigma_v, G = \sigma_v \cdot \sigma_v\). So wrt basis \(\sigma_u, \sigma_v\), \(I_p\) has matrix \[\left ( \begin{matrix} E & F \\ F & G \end{matrix} \right )\]

Q: Definition. Write \(du^2, du \; dv, dv^2\) for the symmetric bilinear forms on \(T_p \Sigma\) with matrices, [...] w.r.t \(\sigma_u, \sigma_v\). Then \[I_p = [...]\]
A: \[\left ( \begin{matrix} 1 & 0 \\ 0 & 0\end{matrix}\right ), \left ( \begin{matrix} 0 & 1/2 \\ 1/2 & 0\end{matrix}\right ), \left ( \begin{matrix} 0 & 0 \\ 0 & 1\end{matrix}\right )\]; Edu^2 + 2F du \; dv + G dv^2

Q: Lemma. For any smooth path \(\gamma : [a,b] \to \Sigma\) there exists \(t_0 = a < t_1 < t_2 < \dots < t_k = b\) such that \(\forall i\) \(\gamma([t_{i-1}, t_i])\) is contained in the image of a parametrisation.
A: For all \(t \in [a,b]\) there exists a parametrisation \(\sigma : V \to U\) about \(\gamma(t)\). Then \(\gamma^{-1}(U)\) is an open neighbourhood of t in \([a,b]\). Let \(C_t\) be a connected sub-neighbourhood. The intervals \(C_t\) give an open cover of \([a,b]\) so can pass to a finite subcover by compactness, say \[(x_0, y_0) \cap [a,b], (x_1, y_1) \cap [a,b], \dots, (x_k,y_k) \cap [a,b]\] WLOG these satisfy \[x_0 < x_1 < \dots < x_k < b\] WLOG also \[a < y_0 < y_1 < \dots y_k\] this is possible because otherwise some interval would be a sub interval of another. Pick \(t_1 \in (x_1, y_0), t_2 \in ( x_2, y_1) \dots\) then \([t_{i-1},t_i] \subset (x_{i-1}, y_{i-1})\)

Q: Lemma. \[d(p,q) = \text{inf} \{\text{length}(\gamma) : \gamma : [0,1] \to \Sigma \text{ smooth path in } \Sigma \text{ from p to q}\}\] is a metric
A: Positivity: Clearly \(d(p,q) = 0\) if \(p = q\)
Otherwise we have \[d(p,q) = \inf \{\text{length}(\gamma) : \gamma \text{ a path in } \Sigma \text{ from p to q}\}\] \[ \geq \inf \{\text{length}(\gamma) : \gamma \text{ a path in } \mathbb{R}^3 \text{ from p to q}\} = d_2(p,q) > 0\]
Symmetry: Obvious by reversing paths
Triangle Inequality: Suppose \(\gamma_1\) is a path p to q and \(\gamma_2\) is a path q to r. Let \(\psi : [0,1] \to [0,2]\) be a smooth function which is surjective, non-decreasing, equal to 1 on a neighbourhood of 1/2.
Now define \(\gamma_3 : [0,1] \to \Sigma\) by \[\gamma_3 (t) = \begin{cases} \gamma_1(\psi(t)) \; \; t\leq 1/2 \\ \gamma_2(\psi(t) - 1) \; \; t \geq 1.2 \end{cases}\]
This is a smooth path p to r. \[d(p,r) \leq \text{length}(\gamma_3) = \text{length}(\gamma_1) + \text{length}(\gamma_2)\] Now take infimum over \(\gamma_1\) and \(\gamma_2\) to get \[d(p,r) \leq d(p,q) \leq d(q,r)\]

Q: Proposition. The topology induced by d coincides with the topology induces by \(d_2\)(the subspace topology on \(\Sigma\))
A: We need to show that the "identity map" \((\Sigma, d) \to (\Sigma, d_2)\) and its inverse are continuous. Using sequences definition of continuity, STP that for points \(p_1, \dots , p_\infty\) in \(\Sigma\), \(p_n \to p_{\infty}\) w.r.t d iff \(p_n \to p_{\infty}\) w.r.t \(d_2\). i.e. \[d(p_n, p_{\infty}) \to 0 \iff d_2 (p_n, p_{\infty}) \to 0\]
We say that \(d \geq d_2\) so if \(d(p_n, p_{\infty}) \to 0\) then \(d_2(p_n, p_{\infty}) \to 0\)
Fix a parametrisation \(\sigma: V \to U\) about \(p_{\infty}\). Recall U is open in \(\Sigma\) with respect to the subspace topology, i.e. wrt \(d_2\). Since \(d_2(p_n, p_{\infty}) \to 0\) wrt \(d_2\) \(\exists N\) s.t. \(\forall n \geq N, p_n \in U\). Let \(q_i = \sigma^{-1}(p_i)\) for \(i = N, N+1, \dots , \infty\). Since \(\sigma\) is a homeomorphism,  \(q_n \to q_{\infty}\).
Since V is open, it contains a closed disc D of radius \(r > 0\) about \(q_{\infty}\). For n large, \(q_n \in D\). \(d(p_n, p_{\infty}) = \inf \{\text{length} (\gamma) : \gamma \text{ a path in } \Sigma \text{ from p to q}\} \leq \text{length}(\sigma \circ (\text{straight line segment } l_n \text{ from } q_n \text{ to } q_{\infty}))\) \[= \int_{l_n} \sqrt{E\dot{u}^2 + 2F\dot{u}\dot{v} + G \dot{v}^2} dt \leq \int_{l_n} \sqrt{(E+G)(\dot{u}^2 + \dot{v}^2)} dt \leq \left ( \max_{D} \sqrt{E + G} \right ) \int_{l_n} \sqrt{\dot{u}^2 + \dot{v}^2} dt \to 0\] as the second integral is \(d_2(q_n, q_{\infty})\)

Q: Definition. \[\text{Area}(U) = [...]\]
A: \int_V \sqrt{EG - F^2} du \; dv
Back Extra: Note that if \[\tilde{\sigma} : \tilde{V} \to U\] is another parametrisation of U, with transition function \(\psi  = \sigma^{-1} \circ \tilde{\sigma}\), then \(D \psi \) is the change of basis matrix between \(\sigma_u, \sigma_v\) and \(\tilde{\sigma}_{\tilde{u}}, \tilde{\sigma}_{\tilde{v}}\). So
Take det: \[\tilde{E}\tilde{G} - \tilde{F}^2 = (EG - F^2)(\det D \psi)^2\] So \[\int_{\tilde{V}} \sqrt{\tilde{E}\tilde{G} - \tilde{F}^2} d \tilde{u} \; d \tilde{v} = \int_{\tilde{V}} \sqrt{EG - F^2} \; |\det D \psi| d \tilde{u} \; d \tilde{v} = \int_V \sqrt{EG - F^2} du \; dv\] by change of variables. So the area is independent of the choice of parametrisation. For a general region we can compute area by breaking it into pieces covered by parametrisations.

Q: Definition. \[d(p,q) = [...]\]
A: \text{inf} \{\text{length}(\gamma) : \gamma : [0,1] \to \Sigma \text{ smooth path in } \Sigma \text{ from p to q}\}

Q: Definition. The pullback of \(I_2\) is [...]
A: the family \(H^* I_2\) of symmetric bilinear forms on tangent spaces to \(\Sigma_1\)

Q: Lemma. \(H^* I_2\) is smooth in the sense that if we pick a parametrisation \(\sigma(u,v)\) then we can locally write \(H^* I_2\) as \[Adu^2 + 2B du dv + C dv^2\] got unique functions A,B,C, and these functions are smooth.
A: We have \[A = (H^* I_2)( \sigma_u, \sigma_u) = I_2(DH( \sigma_u), DH( \sigma_u)) = I_2(H_u, H_u) =  H_u \cdot H_u \] where \(H_u\) is shorthand for \((H \circ \sigma)_u\). Similarly \(B = H_u \cdot H_v, C = H_v \cdot H_v\). So A,B,C all smooth.

Definition. The map H is;
i) A local isometry if \(H^* I_2 = I_1\), i.e. if it preserves all inner products(hence lengths and angles)
ii) Conformal if \(\exists\) positive smooth function \(\lambda: \Sigma_1 \to \mathbb{R}\) such that \(H^* I_2 = \lambda I_1\), i.e. H stretches lengths by \(\lambda\) but preserves angles
iii) Area preserving if \[\sqrt{AC - B^2} = \sqrt{E_1 G_1 = F_1^2}\] i.e. if it preserves area of infinitesimal parallelograms
if) An isometry if it's a bijective local isometry.

Q: Lemma. (i) H is a local isometry iff conformal and area preserving
(ii) If H is a local isometry, or conformal, or are preserving then it's a local diffeomorphism.
A: (i) If H is a local isometry then \(A = E_1, B = F_1, C = G_1\) so \(\sqrt{AC - B^2} = \sqrt{E_1 G_1 - F_1^2}\) so area preserving. And conformal with \(\lambda = 1\)/
Conversely, if conformal then \(A = \lambda E_1, B = \lambda F_1, C = \lambda G_1\) for some positive function \(\lambda\). If also area preserving then \[\lambda \sqrt{E_1 G_1 - F_1^2} = \sqrt{AC - B^2} = \sqrt{E_1 G_1 - F_1^2}\] so \(\lambda =1\). Hence a local isometry.
(ii)Suppose H is conformal. Then \[H^* I_2 = \lambda I_1\] for some smooth \(\lambda > 0\). In particular \((H^* I_2)_p\) is an inner product \(\forall p \in \Sigma_1\). So \(\forall p, \forall w \in T_p \Sigma_1 \) except 0, \( (H^* I_2)_p (w,w) > 0 \). So \[(I_2)_{H(p)} (D_p H(w), D_p H(w)) > 0\] so \(D_p H(w) \neq 0\). So \(D_p H\) injective, hence an isomorphism.
Suppose instead that H is area preserving, then \[\det \left ( \begin{matrix} H_u \cdot H_u & H_u \cdot H_v \\ H_u \cdot H_v & H_v \cdot H_c \end{matrix} \right ) = AC - B^2 > 0\] So \(H_u, H_v\) are linearly independent. So \(DH\) is injective, hence an isomorphism.

Q: Definition. \(II_p\) is the symmetric bilinear form on \(T_p \Sigma\) given by [...] Given a choice of Gauss map n on an open set \(U \in \Sigma\) the second fundamental form, \(II\), is [...]
A: the Hessian of f.; the family of symmetric bilinear forms on tangent spaces to U given by \(II_p\) at each point \(p \in U\)
Back Extra: Remark: The sign of \(II\) depends on the sign on n.

Q: Lemma. Given a parametrisation \(\sigma(u,v)\) near p, we have \[II = L du^2 + 2M du \; dv + N dv^2\] where \(L = \sigma_{uu} \cdot n\), \(M = \sigma_{uv} \cdot n\), \(N = \sigma_{vv} \cdot n\). In particular, L,M,N are smooth so \(II\) is smooth
A: Fix \((a,b) \in \mathbb{R}^2\) and consider the curve \[\gamma(t) = \sigma(\sigma^{-1}(p) + t(a,b))\] in \(\Sigma\) through p. Let \(\Gamma = H \circ \gamma\) - a curve in \(T_p \Sigma\) through 0.
By definition of \(II_p\), Taylor's theorem (with Peano remainder) \[f(\Gamma (t)) = f(\Gamma (0)) + t D_0 f(\dot{\Gamma}(0)) + \frac{t^2}{2} II_p (\dot{\Gamma}(0), \dot{\Gamma}(0)) + o(t^2) = \frac{t^2}{2} II_p (\dot{\Gamma}(0), \dot{\Gamma}(0)) + o(t^2)\] On the other hand \[f(\Gamma(t)) = (H^{-1}(\Gamma (t)) -p) \cdot n(p) = (\gamma(t) - p) \cdot n(p) = \left [ (\gamma(0) - p) + t\dot{\gamma}(0) + \frac{t^2}{2} \ddot{\gamma}(0) + o(t^2)\right ] \cdot n(p)\]
But \(\gamma(0) - p = 0\), \(\dot{\gamma}(0) \perp n(p)\) so \[f(\Gamma(t)) = \frac{t^2}{2} \ddot{\gamma}(0) \cdot n(p) + o(t^2)\]
Comparing the two expressions for \(f(\Gamma(t))\), we get \[II_p (\dot{\Gamma}(0), \dot{\Gamma}(0)) = \ddot{\gamma}(0) \cdot n(p)\] Note that \(\dot{\Gamma}(0) = D_p H(\dot{\gamma}(0)) = \dot{\gamma}(0)\) since \(D_p H = \text{id}\) so this equal \(a \sigma_u + b \sigma_v\) and \(\ddot{\gamma}(0) = a^2 \sigma_{uu} + 2ab \sigma_{uv} + b^2 \sigma_{vv}\)
Plug in: \[II_p (a\sigma_u + b\sigma_v, a\sigma_u + b\sigma_v) = (a^2 \sigma_{uu} + 2ab \sigma_{uv} + b^2\sigma_{vv}) \cdot n(p)\] Equate coefficients to get \[L = \sigma_{uu} \cdot n(p), M = \sigma_{uv} \cdot n(p), N = \sigma_{vv} \cdot n(p)\]

Q: Definition. The Gaussian curvature of \(\Sigma\) at p, denoted \(\kappa(p)\), is [...]
A: \[\det (D_p n)\]
Back Extra: It is the factor by which n scales the areas of infinitesimal parallelograms. Note if we reverse the sign of n, \(\kappa\) is unchanged since \(\det (-D_p n) = (-1)^2 \det (D_p n) = \kappa (p)\) so \(\kappa\) is independent of the choice of n, hence can be defined globally on any embedded surface even if non-orientable.

Q: Proposition. \(D_p n\) is self-adjoint (wrt \(I_p\)) i.e. the bilinear form \[I_p ( -, D_p n(-))\] is symmetric. In fact, it is \(- II_p\)
A: WTS for all \(w,z \in T_p \Sigma\) \[I_p (w, D_p n(z)) = I_p (D_p n(w), z)\] Enough to check \(w,z \in \{\sigma_u, \sigma_v\}\)
Trivial if \(w =z \). So just need to check \(I_p (\sigma_u, D_p n(\sigma_v)) = I_p(\sigma_v, D_p n(\sigma_u))\)
LHS = \(\sigma_u \cdot D_p n(\sigma_v) = \sigma_u \cdot (n \circ \sigma)_v = \sigma_u \cdot n_v = (\sigma_u \cdot n)_v - \sigma_{uv} \cdot n = -M\)
Similarly RHS = \(\sigma_v \cdot n_u = (\sigma_v \cdot n)_u - \sigma_{vu} \cdot n = -M\)
So \(D_p n\) is self-adjoint. Similarly \(I_p (\sigma_u, D_p n(\sigma_u)) = \sigma_u \cdot n_u = -\sigma_{uu} \cdot n = -L\)
\(I_p (\sigma_v, D_p n(\sigma_v)) = -N\) so the components of \(I_p(-, D_pn(-))\) agree with the components of \(- II_p\)

Q: Corollary. We can use matrix form to get \[\kappa = [...]\]
A: \frac{LN - M^2}{EG - F^2}
Back Extra: Take determinants

Q: Definition. The point p is elliptic/ parabolic/ hyperbolic if \(\kappa(p)\) is [...] \(\Sigma\) is flat if [...]
A: positive/zero/negative.; \(\kappa \equiv 0\)

Q: Lemma. If p is an elliptic point then \(\exists\) neighbourhood of p in \(\Sigma\) s.t. \(U / p\) lies strictly on one side of the affine tangent plane. If p is hyperbolic then \(\forall\)neighbourhoods of p, \(\exists p_{\pm} \in U\) on strictly opposite sides of the affine tangent plane
A: Choose parametrisation \(\sigma\) and Gauss map n near p. WLOG \(\sigma(u,0) = p\). Then \[\sigma(u,v) \cdot n(p) = p \cdot n(p) + \frac{1}{2} (u^2 \sigma_{uu} \cdot n(p) + 2uv \sigma_{uv} \cdot n(p) + v^2 \sigma_{vv} \cdot n(p)) + o(||(u,v)||^2)\] So \((\sigma(u,v) - p) \cdot n(p) = \frac{1}{2}(u \; v) \left ( \begin{matrix} L & M \\ M & N \end{matrix} \right ) \left ( \begin{matrix} u \\ v \end{matrix} \right ) + o(||(u,v)||^2)\) SO if \(V_{\pm} \subset \mathbb{R}^2\) are sub spaces on which \(\left ( \begin{matrix} L & M \\ M & N \end{matrix} \right )\) is positive/negative definite then on \(V_{+} \ \{0\}\).
\(\sigma(u,v)\) lies on positive n(p) side of affine tangent plane  and \(V_{-} \ \{0\}\) on the negative side. If p elliptic then \(V_+ = \mathbb{R}^2\) or \(V_- = \mathbb{R}^2\). If p is hyperbolic then they are both one-dimensional.

Q: Definition. A smooth triangle in \(\Sigma\) compact is [...] The face/edges/vertices of the triangle are the image of T/its edges/its vertices under f. A smooth triangulation of \(\Sigma\) is [...] The Euler characteristic \(\chi\) of a smooth triangulation is [...]
A: a pair \((T,f)\) where \(T \subset \mathbb{R}^2\) is a closed triangle, f is a diffeo \(V \to U\) where V is an open neighbourhood of \(T\) in \(\mathbb{R}^2\) and U is an open set in \(\Sigma\).; a finite collection of smooth triangles s.t. faces cover \(\Sigma\), intersection of two distinct faces is empty or common vertex or common edge.; \(V - E + F\).

Q: Theorem. Gauss-Bonnet for closed surfaces. For any smooth triangulation of \(\Sigma\) [...]
A: \[\int_{\Sigma} K dA = 2\pi \chi\]
Back Extra: See part II DG, idea is to cut \(\Sigma\) into triangles and think about parallel transport around the boundary of each one.

Q: Corollary. \(\chi\) is independent of the choice of triangulation. And \(\int_{\Sigma} K dA\) is a diffeomorphism invariant, i.e. if \(\exists\) diffeomorphism \(F: \Sigma_1 \to \Sigma_2\) then [...]
A: \[\int_{\Sigma_1} K_{\Sigma_1} dA = \int_{\Sigma_2} K_{\Sigma_2} dA\]
Back Extra: [given a triangulation of \(\Sigma_1\), apply F to get a triangulation of \(\Sigma_2\) with same value of \(\chi\).]

Q: Definition. A path of vectors along \(\gamma\) is [...] The path is parallel if [...]
A: a smooth map \(V:[0,1] \to \mathbb{R}^3\) such that \(\forall t \; V(t) \in T_{\gamma(t)}\Sigma\).; \(\forall t\) \[\dot{V}(t) \perp T_{\gamma(t)} \Sigma\]

Q: Lemma. If \(V,W\) are parallel along \(\gamma\) then \(V \cdot W\) is a constant. So length of parallel paths of vectors, and (unsigned) angles between them are constant
A: \[\frac{d}{dt}(V \cdot W) = \dot{V} \cdot W + V \cdot \dot{W} = 0 + 0\] since \(\dot{V} \perp T \Sigma\) and \(W \in T \Sigma\) and vice versa

Q: Definition. A geodesic in \(\Sigma\) is [...]
A geodesic is complete if [...]
A: a path \(\gamma : I \to \Sigma\) in \(\Sigma\) defined on an open or closed interval such that \(\forall t \in I\) \[\ddot{\gamma} (t) \perp T_{\gamma(t)} \Sigma\] and \(\gamma\) is non-constant; it can be extended to a geodesic defined on \(\mathbb{R}\)
Back Extra: \(\dot{\gamma}\) is a parallel path of vectors along \(\gamma\)

Corollary. Geodesic equations. A curve \(\gamma(t) = \sigma(u(t), v(t))\) is a geodesic (ie. \(\dot{\gamma}\) parallel along \(\gamma\)) iff

Q: Proposition. Being a geodesic is a local-isometry invariant. I.e. if \(F : \Sigma_1 \to \Sigma_2\) is a local isometry then a curve \(\gamma\) in \(\Sigma_1\) is a geodesic in \(\Sigma_1\) iff \(F \circ \gamma\) is a geodesic in \(\Sigma_2\)
A: Proof on sheet 3, essentially because E,F,G invariant

Q: Lemma. Geodesic equations v2. A curve \(\gamma(t) = \sigma(u(t), v(t))\) is a geodesic iff \[(E \dot{u} + F \dot{v})^{\cdot} = \frac{1}{2}E_u \dot{u}^2 + F_u \dot{u}\dot{v} + \frac{1}{2}G_u \dot{v}^2\] \[(F \dot{u} + G \dot{v})^{\cdot} = \frac{1}{2} E_v \dot{u}^2 + F_v \dot{u} \dot{v} + \frac{1}{2}G_v \dot{v}^2\]
A: Can expand out and compare with v1. For a more conceptual proof see below.

Q: Definition. The energy of \(\gamma\) is [...]
A: \[\text{Energy}(\gamma) = \int_{t_0}^{t_1} \dot{\gamma}(t) \cdot \dot{\gamma}(t) dt = \int_{t_0}^{t_1} I_{\gamma(t)}(\dot{\gamma}(t), \dot{\gamma}(t)) dt\]
Back Extra: Analogously to length, if \(\gamma\) lies in a parametrisation then \[\text{Energy}(\gamma) = \int_{t_0}^{t_1} E \dot{u}^2 + 2F\dot{u}\dot{v} G \dot{v}^2 dt\] for a general curve, can break into finitely many pieces contained in parametrisations and sum their energies.
However length is independent of parametrisation. Energy is not.

Q: Lemma. Energy is "differentiable with respect to \(\gamma\)" in the following sense: if \(\Gamma: (-\epsilon, \epsilon) \times [t_0, t_1] \to \Sigma\)  is a smooth map with \(\Gamma (0,t) = \gamma(t) \forall t\) then the map \( \mathcal{E}: (-\epsilon, \epsilon) \to \mathbb{R}\) given by \[ \mathcal{E}(s) = \text{Energy} ( \Gamma (s, -)) \] is differentiable at s = 0 with \[\mathcal{E}'(0) = 2 \int_{t_0}^{t_1} \dot{\gamma}(t) \cdot \Gamma_{st}(0,t) dt\]
A: Morality differentiate things through \(\int\). For rigorous proof see Sheet 3.

Q: Proposition. \(\gamma\) is a geodesic iff it's a stationary point of energy among paths with fixed end points. i.e. iff \(\forall \Gamma\) as above with \(\Gamma (s,t_0) = \gamma(t_0)\) and \(\Gamma(s,t_1) = \gamma(t_1) \; \forall s\)  we have \(\mathcal{E}'(0) = 0\)
A: Consider a variation \(\Gamma\) with fixed end-points. By previous lemma \[\mathcal{E}'(0) = 2 \int_{t_0}^{t_1} \dot{\gamma}(t) \cdot \Gamma_{st}(0,t) dt = 2[\dot{\gamma}(t) \cdot \Gamma_s (0,t)]_{t_0}^{t_1} - 2 \int_{t_0}^{t_1} \ddot{\gamma}(t) \cdot \Gamma_2 (0,t) dt\] Since end-points are fixed \[\Gamma_s (s,t_0) = 0 \; \forall s, \Gamma_s (s,t_1) = 0 \; \forall s\] So boundary term vanishes and therefore \[\mathcal{E}'(0) = -2 \int_{t_0}^{t_1} \ddot{\gamma}(t) \cdot \Gamma_s (0,t) dt\] Suppose \(\gamma\) is a geodesic then \[\ddot{\gamma}(t) \perp T_{\gamma(t)}\Sigma \] and \(\Gamma_2 (0,t) \in T_{\gamma(t)}\Sigma\) so the integrated is identically 0, so \(\gamma\) is a stationary point.
Conversely suppose \(\gamma\) not a geodesic. Then \(\exists T \in (t_0, t_1)\) such that \(\ddot{\gamma}(T) \not \perp T_{\gamma(t)} \Sigma\). By working in parametrisation about \(\gamma(T)\) and using smooth bump functions, can construct \(\delta > 0\) and a variation \(\Gamma(s,t)\) s.t.
- For \(t \not \in (T-\delta, T+\delta )\), \(\Gamma(s,t) = \gamma(t) \; \forall s\)
- For \(t \in (T - \delta, T + \delta)\), \(\Gamma_s (0,t) \cdot \ddot{\gamma}(t) > 0\)
Then \[\int_{t_0}^{t_1} \ddot{\gamma}(t) \cdot \Gamma_s (0,t) > 0\] so \(\mathcal{E}'(0)<0\) so \(\gamma\) not a stationary point.

Q: Lemma. \(\gamma\) is a geodesic iff it's a stationary point of length (with fixed end points) and has constant speed.
A: Let \(K = \sqrt{L} = \) speed of geodesic.
So Energy = \(\int_{t_0}^{t_1} L dt\) and Length = \(\int_{t_0}^{t_1} K dt\)
Know \(\gamma\) is a geodesic iff it satisfies \(L_u - \dot{(L_{\dot{u}})} = 0 \) and \(L_v - \dot{(L_{\dot{v}})} = 0\)
Similarly \(\gamma\) is a stationary point of length with fixed end-points iff \(K_u - \dot{(K_{\dot{u}})} = K_v - \dot{(K_{\dot{v}})} = 0\)
Using \(L = K^2\), have \[L_u - \dot{(L_{\dot{u}})} = 2K K_u - \dot{(2KK_{\dot{u}})} = 2KK_u - 2\dot{K}K_{\dot{u}} - 2K\dot{(K_{\dot{u}})} = 2K(K_u - \dot{(K_{\dot{u}})})\] Similarly with v in place of u. So if \(\gamma\) is a stationary point of length with constant speed then \(K_u - \dot{(K_{\dot{u}})} = 0\) and \(\dot{K} = 0\). So \(L_u - \dot{(L_{\dot{u}})} = 0\).
Conversely, if \(\gamma\) is geodesic then \(\dot{K} = 0\) and \(L_u - \dot{(L_{\dot{u}})} = 0\) so \(K_u - \dot{(K_{\dot{u}})} = 0\).

Q: Proposition. A geodesic \(\gamma\) is locally length minimising, in the sense that \(\forall T \; \exists \delta > 0\) s.t. if \(\tau_0, \tau_1 \in (T-\delta, T+\delta)\) then \(d(\gamma(\tau_0), \gamma_(\tau_1) = \text{Length}(\gamma |_{[\tau_0, \tau_1]})\)
A: Proof outside the course

Q: Proposition. (Version of Noether's theorem). If FFF is independent of u (in some parametrisation) then \(E \dot{u} + F \dot{v}\) is conserved along geodesics.
A: By geodesic equations v2 \[\dot{(E\dot{u} + F \dot{v})} = \frac{1}{2}E_u \dot{u}^2 + F_u \dot{u}\dot{v} + \frac{1}{2} G_u \dot{v}^2\] and \(E_u, F_u , G_u = 0\).

Q: Definition. X is locally homeomorphic to \(\mathbb{R}^2\) if [...] The maps \(\psi\) are called [...] and a collection of them whose domains cover X is an [...]
A: \(\forall p \in X \; \exists\) open neighbourhood U of p in X, open set \(V \subset \mathbb{R}^2\), homeomorphism \(\psi : U \to V\).; charts,; atlas.

Q: Definition. X is second-countable if [...]
A: there exists a countable basis for the topology.
Back Extra: i.e. \(\exists\) countable collection \(\{U_1, U_2, \dots\}\) of open sets such that every open set can be written asa union of \(U_i\)'s

Q: Definition. An abstract topological surface is [...]
A: a topological space \(\Sigma\) which is locally homeomorphic to \(\mathbb{R}^2\), Hausdorff, and second-countable.

Q: Lemma. For a space X locally homeomorphic to \(\mathbb{R}^2\):
(i) X is connected iff it's path connected
(ii) X is second-countable iff it's Lindelöf(every open cover has a countable subcover) iff it admits a countable atlas.
(iii)X is Hausdorff iff it's regular (if \(C \subset X\) is closed and \(p \in X \setminus C\) \(\exists\) disjoint open sets U,V in X s.t. \(C \subset U, p \in V\))
A: Sheet 3 for proof.

Q: Definition. An action of a group G on X is free and proper if
[...]
A: (i) \(\forall p \in X\) \(\exists\) open neighbourhood U of p such that \(\forall g \neq e\), \(g U \cap U = \emptyset\)
(ii) If \(p_1, p_2 \in X\) lie in different orbits \(\exists\) open neighbourhoods \(U_i\) of \(p_i\) such that \(\forall g \in G\), \(g U_1 \cap U_2 = \emptyset\)
Back Extra: Actions on topological spaces are always by homeomorphisms, i.e. \(\forall g \in G\) the map \(x \mapsto gx\) is a homeomorphism

Q: Proposition. If \(\Sigma\) is an abstract topological surface with a free and proper action of G then \(\Sigma / G\) is an abstract topological surface (with quotient topology)
A: Let \(q : \Sigma \to \Sigma / G\) be the quotient map. Recall that \(W \subset \Sigma / G\) is open iff \(q^{-1}(W)\) is open in \(\Sigma\).
Given \(p \in \Sigma\), \(\exists\) chart \(\psi ' : U' \to V'\) about p. Also \(\exists\) open neighbourhood U'' s.t. \(\forall g \neq e\), \(g U '' \cap U'' = \emptyset\).
Let \(U = U' \cap U''\), \(\psi = \psi'|_U : U \to V\), \(V = \psi'(U' \cap U'')\)
Get a chart \(\psi : U \to V\) about p s.t. \(\forall g \neq e\), \(g U \cap U = \emptyset\).
Claim that \(q(U)\) is open in \(\Sigma / G\) and \(q |_U : U \to q(U)\) is a homeo. Then \[\psi'' \circ (q|_U)^{-1} : q(u) \to V\] is a chart on \(\Sigma / G\) about \(q(p)\).
- \(q|_U\) is surjective onto \(q(U)\)
- If \(p_1, p_2 \in U\) satisfy \(q(p_1) = q(p_2)\) then \(\exists g\) s.t. \(g p_1 = p_2\). Then \(g U \cap U \neq \emptyset \). So \(g = e\) and for \(gp_1 \in gU\) and \(p_2 in U\), \(p_2 = gp_1 = p_1\) so \(q |_U\) is injective.
- q is continuous so \(q |_U\) is continuous.
- Given open \(T \subset \Sigma\), we have \(q^{-1}(q(T)) = \bigcup_{g \in G} gT\) union of open sets, hence open. So q is an open map hence \(q|_U\) is open and \[q |_U : U \to q(U)\] is a homeomorphism(it's a bijection and is continuous and open)
Remains to show that \(\Sigma / G\) is Hausdorff and second-countable.
- Given \(p_1, p_2 \in \Sigma\) in different orbits, to show \(\Sigma / G\) is Hausdorff we need to find disjoint open neighbourhoods of \(q(p_1), q(p_2)\) in \(\Sigma / G\).
Since the action is free and proper \(\exists\) open neighbourhoods \(U_1, U_2\) of \(p_1, p_2\) in \(\Sigma\) such that \(\forall g \in G\), \(gU_1 \cap U_2 = \emptyset\). Then \(q(U_1), q(U_2)\) are disjoint, and they're open since q is an open map.
- Since \(\Sigma\) is second countable, it's Linedlöf(every open cover has a countable subcover). A continuous image of a Linedlöf space is Lindelöf. So \(\Sigma / G\) is Linedlöf, so \(\Sigma / G\) is second-coutable.

Q: Definition. Given an action of a group G on a space X, a fundamental domain for the action is [...]
A: a connected, closed set \(D \subset X\) s.t.
(i) D contains at least 1 point in every orbit
(ii)The interior of D (the union of all open sets in X contained in D) contains at most 1 point from each orbit.
Back Extra: On D / interior,  there may be multiple points from the same orbit, and we usually indicate how these get identified in the quotient.

Q: Definition. An atlas \(\{\psi_{\alpha} : U_{\alpha} \to V_{\alpha}\}\) is smooth if [...] Two smooth atlases are equivalent if [...] This is an equivalent relation and an equivalence class is a smooth structure on \(\Sigma\). An abstract smooth surface is [...]
A: \(\forall \alpha, \beta\) the transition function \[\psi_{\beta} \circ \psi_{\alpha}^{-1} : \psi_{\alpha}(U_{\alpha} \cap U_{\beta}) \to \psi_{\beta}(U_{\alpha} \cap U_{\beta})\] is smooth (as a map between open subsets of \(\mathbb{R}^2\)).; their union is smooth.; an abstract topological surface equipped with a choice of smooth structure.
Back Extra: The smooth structure is part of the data but we don't usually notate it explicitly.

Q: Definition. An abstract smooth surface \(\Sigma\) is orientable if [...]
A: it admits an atlas (representing its smooth structure) such that transition functions \(\psi\) satisfy \[\det (D \psi) > 0\]

Q: Definition. A map \(F : \Sigma_1 \to \Sigma_2\) between smooth surfaces is smooth if [...]. A diffeomorphism is [...] F is a local diffeomorphism if [...]
A: it's continuous and for all charts \(\psi_1, \psi_2\) the map \[\psi_2 \circ F \circ \psi_1^{-1} : \psi_1 (U_1 \cap F^{-1}(U_2)) \to V_2\] is smooth; a smooth map with smooth two-sided inverse.; it's smooth and \(D(\psi_2 \circ F \circ \psi_1^{-1})\) is invertible at each point \((\forall \psi_1, \psi_2)\)
Back Extra: We need F to be continuous so \(F^{-1}(U_2)\) is open and so the map is between open sets.

Q: Lemma. If \(\Sigma\) is an abstract smooth surface and G acts on \(\Sigma\) by diffeomorphism then the charts defined previously define a smooth atlas on \(\Sigma / G\), making it into an abstract smooth surface.
A: Recall that charts on \(\Sigma / G\) were defined by taking a small chart \(\psi : U \to V\) on \(\Sigma(gU \cap U = \emptyset \; \forall g \neq e)\) s.t. \(q|_U : U \to q(U)\) is a homeo (\(q : \Sigma \to \Sigma / G\) quotient map) and then constructing \[\psi \circ q|_U^{-1}\]
We need to show transition functions \((\psi_2 \circ q|_{U_2}^{-1}) \circ (\psi_1 \circ q|_{U_1}^{-1})^{-1}\) are smooth. This is \[\psi_2 \circ (q|_{U_2}^{-1} \circ q|_{U_1}) \circ \psi_1^{-1}\]
By definition, this is smooth iff \[T := q|_{U_2}^{-1} \circ q|_{U_1}\] is smooth (as a map between open subsets of \(\Sigma\))
WTS T smooth. Take \(p \in W_1\). Since \(q(T(p)) = q(p)\), we have \(\exists g \in G\) s.t. \(T(p) = gp\). Since T maps to \(W_2\) we have \(gp \in W_2\), so \(p \in g^{-1}W_2\). Hence \(Z := W_1 \cap g^{-1}W_2\) is an open neighbourhood of p in \(\Sigma\). We claim that on Z, \[T(p') = gp'\] so T is smooth since the action of g is smooth. Hence T is smooth near p, but p is arbitrary so T is smooth.
To prove the claim, consider \(p' \in Z\). Know \(q(T|_{p'}) = qp'\) so \(\exists g'\) s.t. \(T(p') = g'p'\). WTS \(g' = g\). Notice that
(i) \(g'p' \in W_2\) since T maps to \(W_2\)
(ii) \(g'p' = g' g^{-1}W_2\) since \(p' \in g^{-1}W_2\).
So \(W_2 \cap g' g^{-1}W_2 \neq \emptyset\). But \(W_2 \subset U_2\) and recall \(U_2\) was chosen such that \(\forall h\neq e \; U_2 \cap hU_2 = \emptyset\). So \(g' g^{-1} = e\) so \(g' = g\) so \(T(p') = gp'\).

Q: Definition. A smooth family of symmetric bilinear forms on \(\Sigma\) is:
[...]
If u,v are the coordinates associated to \(\varphi\) then we write the family as \(Adu^2 + 2B du dv + C dv^2\). A Riemannian metric on \(\Sigma\) is [...] A Riemannian surface is [...]
A: For each chart \(\varphi : U \to V\) it's a triple of smooth functions \(A,B,C : V \to \mathbb{R}\) such that at each point \(p \in U_1 \cap U_2\) we have
where \(\psi = \varphi_2 \circ \varphi_1^{-1}\).; a smooth family of symmetric bilinear forms \(Edu^2 + 2F du dv + G dv^2\) such that at each point \(\left ( \begin{matrix} E & F \\ F & G \end{matrix} \right )\) is positive definite.; a smooth surface equipped with a Riemannian metric g.

Q: Definition. The length of a smooth path \(\gamma = [a,b] \to (\Sigma, g)\) contained in a chart \(\varphi : U \to V\) is [...] For a general path, break into pieces contained in charts and define its length to be the sum of the corresponding integrals. The area of a region R is defined by breaking into finitely many pieces contained in charts and summing the integrals of [...] on each piece.
A: \[\int_a^b \sqrt{E \dot{u}^2 + 2F \dot{u} \dot {v} + G \dot{v}^2} dt\]; \(\sqrt{EG - F^2}\)
Back Extra: Transformation law guarantees this is independent of \(\varphi\).

Q: Definition. A non-constant smooth curve \(\gamma : I \to \Sigma\) is a geodesic if [...] Being a geodesic is local-isometric invariant.
A: it satisfies the geodesic equations. Equivalently if it's a stationary point of Energy among curves with fixed end-points.

Definition. If \[H : (\Sigma_1 ,g_1) \to (\Sigma_2, g_2)\] is a smooth map between Riemannian surfaces, the pullback \(H^* g_2\) is the smooth family of symmetric bilinear forms on \(\Sigma_1\) defined as follows. Given a chart \(\varphi_1 : U_1 \to V_1\) on \(\Sigma_1\) and a point \(p \in U_1\), pick a chart \(\varphi_2 : U_2 \to V_2\) on \(\Sigma_2\) about H(p) and let \(E_2, F_2, G_2\) be the components of \(g_2\) wrt \(\varphi_2\). We define the components of \(A,B,C\) of \(H^* g_2\) wrt \(\varphi_1\) near p by

where \(\psi = \varphi_2 \circ H \circ \varphi_1^{-1}\).
These satisfy the correct transformation law by the chain rule (exercise). The map H is conformal if \(\exists\) smooth function \(\lambda : \Sigma to (0,\infty)\) such that \(H^* g_2 = \lambda g_1\), it's a local isometry if \(H^* g_2 = g_1\), an isometry if it's also a diffeomorphism, area-preserving if \(\sqrt{AC - B^2} = \sqrt{E_1 G_1 - F_1^2}\)

Q: Lemma. If a group G acts freely and properly by isometries on \((\Sigma, g)\) then there exists a unique Riemannian metric on \(\Sigma / G\) s.t. \(q : \Sigma \to \Sigma / G\) is a local isometry.
A: Recall charts on \(\Sigma / G\) were defined by taking small charts \(\varphi : U \to V\) on \(\Sigma\) such that \(q(U)\) is open in \(\Sigma / G\) and \(q |_U : U \to q(U)\) is a homeo, and then using \(\varphi \circ q|_U^{-1} : q(U) \to V\).
If g has components \(E,F,G\) wrt \(\varphi\) then we defined \(\overline{g}\) on \(\Sigma / G\) to have components \(E,F,G\) wrt \(\varphi \circ q|_U^{-1}\).
These are positive-definite and the unique choice making q a local isometry. RTP they transform correctly.
Well, an overlap between \(\varphi_1 \circ q|_{U_1}^{-1}\) and \(\varphi_2 \circ q|_{U_2}^{-1}\), the components are related by
where \(\psi = \varphi_2 \circ \varphi_1^{-1}\)(by transformation law). WTS same holds with \[\overline{\psi} := (\varphi_2 \circ q|_{U_2}^{-1}) \circ (\varphi_1 \circ q|_{U_1}^{-1})^{-1}\] in place of \(\psi\). Notice \[\psi = \overline{\psi} \circ (\varphi_1 \circ q |_{U_1}^{-1} \circ q|_{U_2} \circ \varphi_1^{-1})\] If we call the part in brackets \(\chi\). Then it suffices, by chain rule, to show that
This is equivalent to \(T = q|_{U_1}^{-1} \circ q |_{U_2}\) being an isometry wrt \(\varphi_1\). We saw before that locally T is given by the action of a fixed \(g \in G\), which is an isometry by assumption

Q: Theorem. Gauss-Bonnet for abstract closed surfaces. If \(\Sigma\) is a Riemannian surface then for any smooth triangulation we have \[\int_{\Sigma} K dA = 2\pi \chi\]
A: Proof outside the course.

Q: Definition. A smooth triangle (T,f) is a geodesic triangle if [...] Similarly can define smooth polygons and geodesic polygons.
A: the edges are geodesics, up to reparametrisation.

Q: Theorem. Gauss-Bonnet for geodesic polygons. If \((T,f)\) is a geodesic n-gon in \(\Sigma\) then \[\int_{f(T)} K dA = 2\pi - \sum \text{ exterior angles}\] Equivalently \[\sum \text{ interior angles} = (n-2)\pi + \int_{f(T)} K dA\]
A: The (n-2)\(\pi\) is where the Euclidean answer comes from, 0 curvature.

Q: Theorem. Every compact orientable abstract smooth surface is diffeomorphic to one of the following:
A: Proof outside the course. These are surfaces of genus g = 0,1,2… and denoted \(\Sigma_g\)

Q: Proposition. The Euler characteristic of \(\Sigma_g\) is \(2 - 2g\).
A: (\(\chi\) independent of choice of triangulation by Gauss-Bonnet.)
Proof by induction on g.
\(\Sigma_0 = S^2\), \(\Sigma_1 = T^2\) compute \(\chi = 2\), \(\chi = 0\) by hand.
For the inductive step fix triangulations of \(\Sigma_1, \Sigma_g\).
Delete the interior of a face from from each, and glue together the resulting triangular boundaries.
We get a surface diffeomorphic to \(\Sigma_{g+1}\), with a triangulation such that \[V = V_1 + V_2 - 3\] \[E = E_1 + E_2 - 3\] \[F = F_1 + F_2 - 2\] Hence \[\chi(\Sigma_{g+1}) = (V_1 + V_2 - 3) - (E_1 + E_2 - 3) + (F_1 + F_2 - 2) = (V_1 - E_1 + F_1) + (V_2 - E_2 + F_2) - 2 = 2 - 2(g+1)\]

Q: Corollary. If \(\Sigma_g\) admits a Riemannian metric of constant curvature c then
If \(g = 0\) then \(c > 0\)
If \(g = 1\) then \(c = 0\)
If \(g \geq 2\) then \(c < 0\)
A: By Gauss Bonnet we have \[c \times \text{Area} (\Sigma_g) = \int_{\Sigma g} K dA = 2\pi \chi = 2\pi (2 - 2g)\] So c must have the same sign as \(1-g\)

Q: Lemma. (i)\(\varphi\) lies in \(M_{\mathbb{R}}\) iff it can be written with \(a,b,c,d \in \mathbb{R}\). This corresponds to the image of \(PGL(2, \mathbb{R})\) in \(PGL(2, \mathbb{C})\)
(ii) \(\varphi\) lies in \(M_H\) iff it can be written with \(a,b,c,d \in \mathbb{R}\) and \(ad - bd > 0\). This corresponds to the image of \(PSL(2, \mathbb{R})\) in \(PGL(2,\mathbb{C})\)
A: (i)If \(a,b,c,d \in \mathbb{R}\) then it is clear that \(\varphi \in M_{\mathbb{R}}\). Conversely suppose \(\phi \in M_{\mathbb{R}}\). We'll show \(\varphi = \psi_1^{-1} \psi_2\) where \(\psi_1, \psi_2\) are real.
- If \(\varphi(\infty) = \infty\), let \(\psi_1 = \text{id}\). Otherwise, let \[\psi_1 (z) = - \frac{1}{z - \varphi(\infty)}\] Notice \(\psi_1\) is real and \(\psi_i \circ \varphi\) sends \(\infty \to \infty\).
- So \(\psi_2 := \psi_1 \circ \varphi\) preserves \(\mathbb{R}\) and sends \(\infty \to \infty\). So \(\psi_2\) is of the form \[z \mapsto \frac{az+b}{d}\] WLOG \(d = 1\)
- So \(\psi_2 (z) = az+b\) preserves \(\mathbb{R}\) so \(a,b \in \mathbb{R}\)
(ii) A map \(\varphi \in M_{\mathbb{R}}\) is in \(M_H\) iff it sends \(i\) to the upper half-plane. For \(a,b,c,d \in \mathbb{R}\) we have \[\varphi(i) = \frac{ai+b}{ci +d} = \frac{(ai +b)(d - ci)}{c^2 + d^2} \in \text{ upper half-plane iff } ad-bc > 0\] If \(ad - bc = 1\) then \(ad - bc > 0\). So image of \(PSL(2, \mathbb{R})\) lies in \(M_H\). Conversely, if \(a,b,c,d \in \mathbb{R}\) and \(ad- bc > 0\) then we can rescale a,b,c,d by \(\frac{1}{\sqrt{ad - bc}}\) to make \(ad - bc = 1\) so \(z \mapsto \frac{az + b}{cz + d}\) is in the image of \(PSL(2, \mathbb{R})\)

Q: Corollary. \(M_H\) is generated by \[z \mapsto az \; \; a \in \mathbb{R}_{ > 0}\] \[z \mapsto z + c \; \; c \in \mathbb{R}\] \[z \mapsto \frac{-1}{z} \]
A: Take \(\varphi \in M_H\). Above we wrote \(\varphi = \psi_1^{-1} \circ \psi_2\) where \[\psi_1(z) = - \frac{1}{z - \varphi(\infty)} \; \; (\text{ or } \psi_1 = \text{id})\]
\[\psi_2 (z) = az +b\]
Notice \(\psi_1\) is generated by second and third types of element. We saw \(a,b \in \mathbb{R}\) and if \(\varphi\) preserves H then \(a > 0\) so \(\psi_2\) is generated by the first two types of element.

Q: Lemma. We have \[M_D = \{z \mapsto e^{i\alpha} \frac{z-w}{1-\overline{w}z}\} : \alpha \in \mathbb{R}/ 2\pi \mathbb{Z}, w \in D\]
A: Calculate directly or write down a \(\psi\) s.t. \(\psi(D) = H\) and use previous description of \(M_H\)

Q: Definition. The (Poincaré) upper half-plane model for the hyperbolic plane is [...]
A: H equipped with Riemannian metric \[g_H = \frac{dx^2 + dy^2}{y^2}\]
Back Extra: This is an orthogonal chart \((F = 0)\) so \[K = -\frac{1}{2 \sqrt{EG}} (\frac{\partial}{\partial x} \frac{G_x}{\sqrt{EG}} + \frac{\partial}{\partial y} \frac{E_y}{\sqrt{EG}}) = \frac{-y^2}{2} \cdot \frac{\partial}{\partial y}(y^2 \cdot \frac{-2}{y^3}) = y^2 \frac{\partial}{\partial y}(\frac{1}{y}) = -1\]

Q: Lemma. The action of \(M_H\) on H is via isometries of \(g_H\).
A: Suffices to check for \[z \mapsto az\] \[z \mapsto z + c\] \[z \mapsto -\frac{1}{z}\]
For \(z \mapsto az\): \(dx,dy,y\) all get scaled by a
For \(z \mapsto z + c\) : \(dx,dy,y\) all unchanged
For \(z \mapsto -\frac{1}{z}\) is \[x + iy \mapsto - \frac{1}{x+iy} = \frac{-x+iy}{x^2 + y^2}\] So \(\varphi(x,y) = (\frac{-x}{x^2 + y^2}, \frac{y}{x^2 + y^2} ) \)
Notice : \(\varphi^* dx = d(\frac{-x}{x^2 + y^2}) = \frac{(x^2 - y^2) dx + 2xy dy}{(x^2 +y^2)^2}\) and similarly \(\varphi^* dy = \frac{(x^2 - y^2)dy - 2xy dx}{(x^2 + y^2)^2}\) So \[\varphi^* g_H = \frac{1}{(\frac{y}{x^2 + y^2})^2}\left [ (\frac{(x^2 - y^2)dx + 2xydy}{(x^2 + y^2)^2})^2 + (\frac{(x^2 - y^2)dy - 2xy dx}{(x^2 + y^2)^2})^2\right ]= \frac{dx^2 + dy^2}{y^2} = g_H\]

Q: Definition. The Poincaré disc model for the hyperbolic plane is defined to be [...]
A: D equipped with Riemannian metric \(g_D := \psi^* g_H\) for any \(\psi \in \mathcal{M}\) such that \(\psi(D) = H\)
Back Extra: The unique metric on D such that \(\psi : (D, g_D) \to (G, g_H)\) is an isometry.

Q: Lemma. \[g_D = \frac{4(du^2 + dv^2)}{(1-u^2 - v^2)^2}\] with respect to standard coords on D.
A: Fix \(\psi(z) = \frac{z+i}{iz+1}\) and let \(g_D = Edu^2 + 2Fdudv + G dv^2\).
We know \(\psi\) is homomorphism, hence conformal, and know \(g_H\) is proportional to \(dx^2 + dy^2\) so \(E=G, F=0\).
We know rotations \(z \mapsto e^{i \alpha}z\) lie in \(M_D\) so are isometries of \(g_D\), so \(g_D\) is rotationally invariant. Hence \(E = G = \lambda(r)\) for some positive smooth \(\lambda\).
Want to show \[\lambda(r) = \frac{4}{(1-s^2)^2}\]
Let \(\gamma = it\), i.e. \(u(t) = 0, v(t) = t\)
We have \[g_D(\dot{\gamma}(t), \dot{\gamma}(t)) = E \dot{u}^2 + 2F\dot{u}\dot{v} + G\dot{v}^2 = G = \lambda(t)\]
Since \(\psi\) is an isometry, we have \[g_H(\dot{\psi \circ \gamma})(t), (\dot{\psi \circ \gamma})(t)) = \lambda(t)\] But \(\psi \circ \gamma(t) = \frac{it + i}{1-t} = i \frac{1+t}{1-t}\) i.e. \(x(t) = 0\), \(y(t) = \frac{1+t}{1-t}\) So LHS of equation is \[\frac{\dot{x}^2 + \dot{y}^2}{y^2} = \left (\frac{1-t}{1+t} \right )^2 \left ( \frac{1}{1-t} + \frac{1+t}{(1-t)^2} \right )^2 = \left ( \frac{1-t}{1+t} \right )^2 \left ( \frac{2}{(1-t)^2} \right )^2 = \frac{4}{(1-t)^2 (1+t)^2} = \frac{4}{(1-t^2)^2}\] so \(\lambda(t) = \frac{4}{(1-t^2)^2}\)

Q: Lemma. Vertical lines in H are geodesics, and diameters in D are geodesics (wrt \(g_H\) and \(g_D\) respectively, when parametrised at constant speed)
A: Fix \(b > a > 0\) and \(x_0\), suppose \(\gamma\) is a path H from \((x_0, a)\) to \((x_0, b)\)
We have \[\text{Length}(\gamma) = \int_{t_0}^{t_1} \sqrt{\frac{\dot{x}^2 + \dot{y}^2}{y^2}}dt \geq \int_{t_0}^{t_1} \sqrt{\frac{\dot{y}^2}{y^2}}dt \] with equality iff \(x = x_0\) \[= \int_{t_0}^{t_1} \frac{|\dot{y}|}{y} dt \geq \int_{t_0}^{t_1} \frac{\dot{y}}{y} dt\] with equality iff \(\dot{y} \geq 0\) \[ = \int_a^b \frac{1}{y}dy = \log \frac{b}{a}\] So \(\text{Length}(\gamma)\) is minimised iff \(x = x_0\) and y non-decreasing, i.e. iff \(\gamma\) is a reparametrisation of vertical line segment. So vertical line is a geodesic.
Apply \(\psi : \frac{z-i}{1-iz}\), which sends H to D and y-axis to vertical diameter. It's an isometry so vertical diameter is a geodesic. Rotations of D are isometries so all diameters are geodesics.

Q: Proposition. In H a path is a geodesic iff it's an arc oof a vertical line or a circle with centre on the x-axis. In D a path is a geodesic iff it's an arc of a circline orthogonal to the unit circle.
A: First consider D. Suppose \(\Gamma\)is a circline meeting unit circle orthogonally at \(a_1, a_3\). Pick \(a_2\) on the arc of unit circle anticlockwise from \(a_1\) to \(a_3\)
Let \(\psi\) be the unique Möbius map sending \((a_1, a_2,a_3)\) to \((1, i, -1)\). This lies in \(M_D\) (it sends unit circle to itself and preserves its orientation), and sends \(\Gamma\) to a circline meeting the unit circle orthogonally at \(\pm 1\). So \(\psi(\Gamma)\) is the diameter through \(\pm 1\). We saw that this is a geodesic, so \(\Gamma\) is a geodesic since \(\psi\) is an isometry.
Conversely, given a geodesic \(\gamma\) in D from p to q, let \(\psi \in M_D\) sending p to 0. (eg \(z \mapsto \frac{z-p}{1-\overline{p}z}\)). Then \(\psi \circ \gamma\) is a geodesic through the origin, so by uniqueness of geodesics with given starting point and direction \(\psi \circ \gamma\) is a diameter. So \(\gamma\) is an arc of \(\psi^{-1} \text{(diameter) = circline orthogonal to unit circle}\).
To get the result for H, note that vertical lines and circles centred on x-axis are precisely circlines orthogonal to \(\mathbb{R}_{\infty}\)

Q: Definition. A hyperbolic line in H or D is [...] Two hyperbolic lines are parallel if [...] They're ultraparallel if [...]
A: the image of a geodesic \(\gamma : \mathbb{R} \to H\) or D, i.e. a circline orthogonal to the boundary.; their closures in \(\mathbb{C}_{\infty}\) meet on the boundary.; the closures are disjoint.

Q: Definition. A hyperbolic polygon is [...] An ideal hyperbolic polygon is [...]
A: a geodesic polygon in the hyperbolic plane.; a region bounded by a cycle of parallel hyperbolic lines.

Q: Theorem. Gauss-Bonnet for hyperbolic polygons. Given a hyperbolic n-gon \[\sum \text{interior angles} = (n-2)\pi - \text{area}\]
A: Proof on sheet 4, application of Green's theorem.

Q: Definition. Given a Riemannian surface(assumed connected) (\(\Sigma, g\)), let \(\text{Isom}(\Sigma , g)\) denote the group of isometries. If \(\Sigma\) is orientable, with \(\{\varphi_{\alpha}\}\) on atlas such that all transition functions \(\psi\) have \(\det D \psi > 0\), then a diffeomorphism \(\chi\) of \(\Sigma\) is orientation-preserving if [...]
Let \(\text{Isom}^+ (\Sigma , g)\) denote the group of orientation-preserving isometries.
A: \[\forall \alpha, \beta \; \det D(\varphi_{\alpha} \circ \chi \circ \varphi_{\beta}^{-1}) > 0\]

Q: Lemma. (i) The definition of orientation-preserving is independent of choices and the map \[\text{Isom}(\Sigma, g) \to \{ \pm 1 \}\] \[\varphi \mapsto \begin{cases} 1 & \text{if } \varphi \text{ is orientation-preserving} \\ -1 & \text{otherwise} \end{cases} \] is a group homomorphism, with kernel \(\text{Isom}^+ (\Sigma, g)\)
(ii) If \(G \subset \text{Isom}(\Sigma, g)\) acts freely and properly then \(\Sigma / G\) is orientable iff \(\Sigma\) orientable and \(G \subset \text{Isom}^+ (\Sigma, g)\)
A: Exercise

Q: Lemma. If \(\varphi \in \text{Isom}(D, g_D)\) and \(\varphi(0) = 0\) and \(D_0 \varphi = \text{id}\) then \(\varphi = \text{id}\).
A: Given \(p \in D\) let \(\gamma\) be the unique geodesic 0 to p, say parametrised so \(\varphi(1) = p\). Consider \(\tilde{\gamma} := \varphi \circ \gamma\)
- This is a geodesic
- \(\tilde{\gamma}(0) = \varphi(\gamma(0)) = \varphi(0) = 0\)
- \(\dot{\tilde{\gamma}}(0) = D_0 \varphi(\dot{\gamma}(0)) = \dot{\gamma}(0)\) since \(D_0 \varphi = \text{id}\)
So by uniqueness of geodesics \[\tilde{\gamma} = \gamma\]
So \(\varphi(p) = \varphi(\gamma(1)) = \tilde{\gamma}(1) = \gamma(1) = p\)

Q: Proposition. \(\text{Isom}^+(D, g_D) = M_D \) whilst \(\text{Isom}(D, g_D)\) is generated by \(M_D\) and \((u,v) \mapsto (u,-v)\), i.e. \(z \mapsto \hat{z}\)
A: We know \(\text{Isom}^+\) contains \(M_D\). Conversely, given \(\varphi \in \text{Isom}^+\) we'll show it can be reduced to id by composing with elements of \(M_D\).
Let \(\psi \in M_D\) by \[\psi(z) = \frac{z - \varphi(0)}{1 - \overline{\varphi(0)} z}\] Replacing \(\varphi\) with \(\psi = \varphi\) we may assume \(\varphi(0) = 0\). By composing with a rotation (\(\in M_D\)) we may assume \(D_0 \varphi\) fixes \(\sigma_u\). Since \(\varphi\) is an isometry, \(D_0 \varphi\) must send \(\sigma_v\) to a vector fo the same length orthogonal to \(D_0 \varphi(\sigma_u) = \sigma_u\). So \(D_0 \varphi = \pm \sigma_v\). Since \(\varphi\) is orientation-preserving we must have \(+ \sigma_v\) so \(D_0 \varphi = \text{id}\). Hence \(\varphi = \text{id}\) by previous lemma.
Now consider \(\text{Isom}\). Easy to check that \(z \mapsto \overline{z}\) lies in \(\text{Isom}\). Conversely suppose \(\varphi \in \text{Isom}\). Applying the above argument we can reduce to \(\varphi(0)= 0\), \(D_0 \varphi (\sigma_u) = \sigma_u \), and \(D_0 \varphi(\sigma_v) = \pm \sigma_v\). If \(D_0 \varphi (\sigma_v) = \sigma_v\) then \(\varphi = \text{id}\) as above. Otherwise compose with \(z \mapsto \overline{z}\) then \(\varphi = \text{id}\).

Q: Lemma. A non-identity Möbius map \(D \to D\) or \(H \to H\) is elliptic iff as a map \(D \to D\) it's conjugate to a rotation. It's parabolic or hyperbolic iff on H it's conjugate to \(z \mapsto z+1\) or \(z \mapsto az, a \in (1,\infty)\) respectively.
A: Sheet 4