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This homework set is out of 40 points with a possibility of extra credit.

If \(M \subseteq \mathbb{R}^n\) is a smooth manifold and \(\mathbf{p} \in M\), define \(N_p M\) to be all elements of \(\mathbb{R}^n_{\mathbf{p}}\) which are orthogonal to every element of \(T_p M\). This is the vector space of normal vectors to \(M\) at \(\mathbf{p}\). Define the normal bundle of \(M\) to be \(N M := \bigcup_{\mathbf{p} \in M} N_{\mathbf{p}} M\). If \(I \subseteq \mathbb{R}\) is an interval and \(\gamma:I \to M\) is continuous, a lift of \(\gamma\) to \(\Gamma \subseteq \mathbb{R}^n \times \mathbb{R}^n\) is a continuous function \(\widetilde{\gamma} : I \to \Gamma\) such that for all \(t \in I\), \(\widetilde{\gamma}(t) \in \Gamma \cap \mathbb{R}^n_{\gamma(t)}\). A lifting \(\widetilde{\gamma}\) has unit length if \(\|\widetilde{\gamma}(t)\| = 1\) for all \(t \in I\).
Remark: If \(\mathbf{u}_p,\mathbf{v}_p \in \mathbb{R}^n_p\), then \(\mathbf{u}_{\mathbf{p}} \cdot \mathbf{v}_{\mathbf{p}} = \mathbf{u} \cdot \mathbf{v}\) and \(\|\mathbf{u}_{\mathbf{p}} \| = \| \mathbf{u} \|\) (i.e. the dot product and norm on \(\mathbb{R}^n_{\mathbf{p}}\) come from viewing it as isomorphic to \(\mathbb{R}^n\), not from the inclusion \(\mathbb{R}^n_{\mathbf{p}} \subseteq \mathbb{R}^{2n}\)).

1. (8+2 points) Suppose that \(S \subseteq \mathbb{R}^n\) is a smooth surface, \(\gamma:(a,b) \to S\) is a regular parametrized curve with range \(C\). Show that if \(n = 3\), then \(\gamma\) has at most two unit length lifts to \(N S\) and in general \(\gamma\) has at most two unit length lifts to \(N C \cap T S\).
   Hint: If \(\widetilde{\gamma}_1\) and \(\widetilde{\gamma}_2\) are unit length lifts of \(\gamma\) to \(NS\), consider the function \(f(t) = \widetilde{\gamma}_1(t) \cdot \widetilde{\gamma}_2(t)\). What are its possible values? Show that any such function must be constant. A similar idea applies to the second part.
   Remark: In fact in each case there are always exactly two such lifts of \(\gamma\). A unit length lift of \(\gamma\) to \(NS\) amounts to a selection of a "side" of the surface for each point on \(\gamma\). A unit length lift of \(\gamma\) to \(NC \cap TS\) is a choice of "side" of the curve within the surface.
   
   Solution Observe that if \(n=3\) then for any \(\mathbf{p} \in S\), \(N_{\mathbf{p}} S\) is a 1-dimensional vector space since \(\mathbb{R}^3_{\mathbf{p}}\) is the direct sum of \(T_{\mathbf{p}} S\) and \(N_{\mathbf{p}} S\). In particular, there are exactly two unit vectors in \(N_{\mathbf{p}} S\). If \(\tilde \gamma_1\) and \(\tilde \gamma_2\) are lifts of \(\gamma_1\) and \(\gamma_2\), then \(\tilde \gamma_1 (t) \cdot \tilde \gamma_2(t) \in \{1,-1\}\). Since \(\tilde \gamma_1 \cdot \tilde \gamma_2\) is continuous, the Intremediate Value Theorem implies that it must be constant. This implies \(\tilde \gamma_2 = \pm \tilde \gamma_1\) (here and below multiplication by \(-1\) is only on the vector component of \(\tilde \gamma_1\), not the base point). In particular there are exactly two lifts of a curve \(\gamma\) to \(N_{\mathbf{p}} S\) if there are any lifts at all. If \(n\) is arbitrary, notice that \(T_{\mathbf{p}} S\) is the direct sum of \(T_{\mathbf{p}} C\) and \(N_{\mathbf{p}} C \cap T_{\mathbf{p}} S\). In particular, \(N_{\mathbf{p}} C \cap T_{\mathbf{p}} S\) has dimension 1 and contains exactly two unit vectors. The proof that curves in \(S\) have at most 2 lifts to \(NC \cap TS\) now proceeds in exactly the same way as in the previous case.


2. (10 points) Suppose that \(S \subseteq \mathbb{R}^n\) is a smooth, orientable surface. Show that if \(n=3\), any unit length lift of a regular closed curve to \(NS\) is closed. For +5 points extra credit show that in general any unit length lift of a regular closed curve \(C\) to \(NC \cap TS\) is closed.
   Remark: Such a lift of a closed curve not being closed amounts to traveling around the surface, returning to the starting point but in a way such that the chosen "side" of the surface or curve has changed. These are useful tools in demonstrating that a surface is not orientable - just show that an appropriate lift of a closed curve isn't closed.
   
   Solution Suppose that \(\mathbf{N}\) is an orientation of \(S\). If \(\gamma:[a,b] \to S\) is any closed curve, then \(\tilde \gamma(t) := (\gamma(t),\mathbf{N}(\gamma(t)))\) is a lift of \(\gamma\) to \(S\). Since \(\gamma(a) = \gamma(b)\), \(\mathbf{N}(\gamma(a)) =\mathbf{N}(\gamma(b))\) and therefore \(\tilde \gamma\) is closed as well. Notice that \(-\tilde \gamma\) is also a lift of \(\gamma\) which is closed. By Problem 1, every lift of \(\gamma\) to \(N S\) is either \(\tilde \gamma\) or \(-\tilde \gamma\) and hence is closed.


3. (10 points) Let \(S\) be the range of \[ \sigma(x,y) := (\cos (x) \cos (y), \sin (x) \cos (y), \cos (x/2) \sin (y), \sin (x/2) \sin (y)). \] Show that \(S\) is a smooth surface.
   Remark: This is a Klein bottle. Observe that \(\sigma(x,y+2\pi) = \sigma(x,y)\) and \(\sigma(x+2 \pi,2 \pi-y) = \sigma(x,y)\).
   
   Solution It suffices to show that \(\sigma\) is a local diffeomorphism. First notice that \(\sigma\) is one-to-one when restricted to each of the following open sets: \((0,2\pi) \times (0,2 \pi)\), \((-\pi,\pi) \times (0,2\pi)\), \((0,2\pi) \times (-\pi,\pi)\), and \((-\pi,\pi) \times (-\pi,\pi)\). Moreover, the image of each of these open sets under \(\sigma\) is open (since its complement in \(S\) is closed).
   Next we must show that \(\sigma\) is regular. \[ \sigma_x = (-\sin(x)\cos(y),\cos(x)\cos(y),-\frac{1}{2}\sin(x/2)\sin(y),\frac{1}{2}\cos(x/2)\sin(y) ) \] \[ \sigma_y = (-\cos(x)\sin(y),-\sin(x)\sin(y),\cos(x/2)\cos(y),\sin(x/2)\cos(y) ) \] If we restrict to the first two coordinates and take the determinant, it is \(\sin(y)\cos(y) = \frac{1}{2} \sin (2y)\) which is only \(0\) if \(y = n \pi\). If \(y=n \pi\), then we have: \[ \sigma_x = (\pm \sin(x), \pm \cos(x),0,0) \qquad \sigma_y = (0,0,\pm \cos(x/2),\pm \sin(x/2)) \] It follows that \(\sigma_x\) and \(\sigma_y\) are always linearly independent (notice that \(\sigma_x\) and \(\sigma_y\) are never 0).


4. (10 points) Show that the surface \(S\) in exercise 3 is not orientable.
   Hint: Use the extra credit component of exercise 2 with the curve \(\gamma(t) = \sigma(t,0) = (\cos(t),\sin(t),0,0)\).
   
   Solution Define \(\gamma:[0,2\pi] \to S\) as above noting that \(\dot \gamma(t) = (-\sin(t),\cos(t),0,0)\). In particular \(\gamma(0)=\gamma(2\pi)\) and \(\dot \gamma(0) = \dot \gamma(2\pi)\). hence \(\gamma\) is closed. If \(y=0\), then we have that \[ \sigma_x(x,0) = (-\sin (x), \cos (x), 0,0) \qquad \sigma_y(x,0) = (0,0,\cos(x/2),\sin(x/2)). \] Notice that \(\sigma_x(t,0) = \dot \gamma(t)\), \(\sigma_y(t,0) \cdot \dot \gamma(t) = 0\), and \(\|\sigma_y(t,0)\| = 1\). Thus \(\sigma_y(t,0)\) is a continuuous unit normal to \(\gamma\) and \(\tilde \gamma (t) = (\gamma(t),\sigma_y(t,0))\) is a lift of \(\gamma\) to \(N C \cap TS\). Since \(\sigma_y(2\pi,0) = (0,0,-1,0) \ne (0,0,1,0) = \sigma_y(0,0)\), \(\tilde \gamma:[0,2\pi] \to NC \cap TS\) is not closed. Thus \(S\) is not orientable by Problem 2.