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This homework set is out of 40 points.

1. (10 points) Suppose that \(S \subseteq \mathbb{R}^3\) is a smooth surface. Show that for every point \(\mathbf{p} \in S\) there is an open set \(U \subseteq \mathbb{R}^3\) containing \(\mathbf{p}\) such that \(S \cap U\) is an orientable smooth surface.
   
   Solution Given \(\mathbf{p} \in S\), let \(U\) be an open set containing \(\mathbf{p}\) such that for some open \(V \subseteq \mathbb{R}^2\), there is a diffeomorphism \(\sigma: V \to S \cap U\). For each \(\mathbf{q} \in S \cap U\), define \[ \mathbf{N}_{\mathbf{q}} := \frac{\sigma_x \times \sigma_y}{\| \sigma_x \times \sigma_y\|} \] where \(\sigma_x\) and \(\sigma_y\) are evaluated at \(\sigma^{-1}(\mathbf{q})\). Since \(\sigma\) is regular, \(\sigma_x \times \sigma_y\) is nonzero and \(\mathbf{N}\) is defined and continuous on \(S \cap U\).


2. (10 points) Show that \[ S:=\{(x,y,z) \in \mathbb{R}^3 \mid x^2 + y^2 = z^2 \textrm{ and } z \geq 0\} \] is not a smooth surface. Hint: Use the previous exercise at the point \((0,0,0)\) then argue that any assignment \(\mathbf{N}\) of normal vectors to \(S \cap U\) can't be continuous at \((0,0,0)\) by showing that \({\displaystyle \lim_{(x,y,z) \to (0,0,0)} \mathbf{N}(x,y,z)}\) does not exists.
   
   Solution Suppose for contradiction that \(S\) is smooth. By the previous exercise, there is an open \(U\) containing \((0,0,0)\) such that \(S \cap U\) has an orientation \(\mathbf{N}\). Observe that for all \((x,y,z) \ne (0,0,0)\) which are on \(S\), \[ \mathbf{N}(x,y,z) = \pm \frac{1}{\sqrt{x^2 + y^2 + z^2}} (x,y,-z) = \pm\frac{\sqrt{2}}{2}(\frac{x}{|z|},\frac{y}{|z|},-1) \] By replacing the orientation with its negative if necessary, we may assume that \(\mathbf{N}(1,0,1) = \frac{\sqrt{2}}{2} (1,0,-1)\). By continuity of \(\mathbf{N}\cdot\mathbf{k}\) and Intermediate Value Theorem, \(\mathbf{N} \cdot \mathbf{k} = -\frac{\sqrt{2}}{2}\) on \(S \cap U\). But now \[ \lim_{t \to 0^+} \mathbf{N}(t,0,t) = \frac{\sqrt{2}}{2}(1,0,-1) \ne \frac{\sqrt{2}}{2} (-1,0,-1) = \lim_{t \to 0^-} \mathbf{N}(t,0,t) \]


3. (10 points) Suppose that \(U \subseteq \mathbb{R}^3\) is open and \(f:U \to \mathbb{R}\) has \(q\) as a regular value. Show that \[ S:=\{\mathbf{p} \in \mathbb{R}^3 \mid f(\mathbf{p}) = q\} \] is an orientable surface by explicitly defining an orientation.
   
   Solution Define \(\mathbf{N}\) on \(S\) by \[ \mathbf{N} = \frac{\nabla f }{\|\nabla f \|} \] Since \(f\) is regular on \(S\), \(\mathbf{N}\) is continuous. We need to show that if \(\mathbf{v}_{\mathbf{p}} \in T_{\mathbf{p}} S\), then \(\mathbf{N}_{\mathbf{p}} \cdot \mathbf{v}_{\mathbf{p}} = 0\). Let \(\gamma:(-\epsilon,\epsilon) \to S\) be a smooth curve such that \(\gamma(0) = \mathbf{p}\) and \(\dot \gamma(0)= \mathbf{v}\). Observe that \[ \frac{d}{dt} f(\gamma(t)) = \nabla f (\gamma(t)) \cdot \dot \gamma(t) \] Since \(f(\gamma(t)) = q\) is constant, it follows that \(\nabla f(\gamma(t)) \cdot \dot \gamma(t) = 0\). If we take \(t =0\) we obtain \(\nabla f(\mathbf{p}) \cdot \mathbf{v}=0\). Since \(\mathbf{N}\) is parallel to \(\nabla f(\mathbf{p})\), this finishes the proof.


4. (10 points) Calculate the first fundamental form of the surface parametrized by \( \sigma(u,v) := (\cos (u) \cosh (v), \sin(u) \cosh (v), v) \).
   
   Solution \[ \sigma_u = (-\sin(u) \cosh(v), \cos(u) \cosh(v),0) \qquad \sigma_v = (\cos(u)\sinh(v),\sin(u)\sinh(v),1) \] Which gives \[ E= \sin^2(u) \cosh^2(v) + \cos^2(u) \cosh^2(v) = \cosh^2(v) \] \[ F=0 \] \[ G=\cos^2(u)\sinh^2(v) + \sin^2(u) \sinh^2(v) + 1 = \sinh^2(v) + 1 = \cosh^2(v) \] The first fundamental form is \[ \cosh^2(v) \ du^2 + \cosh^2(v) \ dv^2 \]