Let \(\Phi:\R^m\to\R^n\) be a smooth map. Recall S09P01 and S09P02. Define the pullback \(\Phi^*\omega\) when \(\omega\) is a \(0\)-form (i.e., a function), a \(1\)-form, and a \(2\)-form on \(\R^n\).
What is the cross product of two vectors \(u\) and \(v\)? We denote this “\(u \times v\).”
For a vector field \(F\) and a parameterized surface \(\Phi:D\to\R^3\), define the flux integral \(\displaystyle\iint_\Phi F\cdot d\vec{S}\) by finding a \(2\)-form \(\omega_F\) on \(\R^3\) built from the components of \(F\) so that \(\displaystyle\iint_\Phi F\cdot d\vec{S} = \iint_D \Phi^*\omega_F\).
The torus with radii \(0<r<R\) is parametrized by \[ \Phi(\theta,\phi)=\bigl((R+r\cos\phi)\cos\theta,\;(R+r\cos\phi)\sin\theta,\;r\sin\phi\bigr) \] for \((\theta,\phi)\in[0,2\pi]\times[0,2\pi]\). Compute the surface area; then compute the flux of \(F(x,y,z)=(x,y,0)\) through the torus, oriented via the outward normal.
Recall the angle form from S08P09. Define the solid angle form on \(\R^3\setminus\{0\}\) by \[ \Omega = \frac{x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy}{(x^2+y^2+z^2)^{3/2}}. \] Compute \(\displaystyle\iint_{S^2}\Omega\) where \(S^2\) is the unit sphere with outward orientation.
The helicoid may be parameterized by \(\Psi(u,v)=(\sinh u \sin v,-\sinh u \cos v,v)\). Compute the three pullbacks \(\Phi^*(dx\wedge dy)\) and \(\Phi^*(dy\wedge dz)\) and \(\Phi^*(dz\wedge dx)\).
Write each as \(A\,du\wedge dv\), \(B\,du\wedge dv\), \(C\,du\wedge dv\) respectively, and verify that \(|\Phi_u\times\Phi_v|^2=A^2+B^2+C^2\).
For two 1-forms \(\alpha\) and \(\beta\), define \[ (\alpha\wedge\beta)(v,w) = \det\begin{pmatrix} \alpha(v) & \alpha(w) \\ \beta(v) & \beta(w) \end{pmatrix} = \alpha(v)\,\beta(w) - \alpha(w)\,\beta(v). \] For 1-forms \(\omega\) and \(\eta\) on \(\R^n\), and a smooth \(\Phi:\R^m\to\R^n\), prove \(\Phi^*(\omega\wedge\eta)=(\Phi^*\omega)\wedge(\Phi^*\eta)\).
Let \(\Phi:\R^m\to\R^n\) be smooth. Prove that pullback commutes with the exterior derivative (recall S09P03), i.e., prove that \(d(\Phi^*\omega)=\Phi^*(d\omega)\) for a \(k\)-form \(\omega\).
Check this explicitly using the helicoid from S10P06 and \(\omega=xz\,dy-y\,dz\).
Let \(\Phi:D\to\R^3\) be a regular parametrized surface. Write \[ E=\Phi_u\cdot\Phi_u,\qquad F=\Phi_u\cdot\Phi_v,\qquad G=\Phi_v\cdot\Phi_v \] for the coefficients of the first fundamental form. Show that \(|\Phi_u\times\Phi_v|=\sqrt{EG-F^2}\).
For a graph \(z=h(x,y)\) over a domain \(D\subset\R^2\), deduce that the surface area equals \[ \iint_D\sqrt{1+h_x^2+h_y^2}\,dx\,dy. \] Why is this always \(\ge\operatorname{area}(D)\), with equality iff \(h\) is constant?
Show that the solid angle form \(\Omega\) from S10P05 is closed, i.e., \(d\Omega=0\) on \(\R^3\setminus\{0\}\).
The catenoid is the surface of revolution obtained by revolving \(x=\cosh z\) about the \(z\)-axis, parametrized by \[ \Phi(u,v)=(\cosh u\cos v,\cosh u\sin v,u), \qquad (u,v)\in\R\times[0,2\pi]. \] The helicoid from S10P06 is \(\Psi(u,v)=(\sinh u \sin v,-\sinh u \cos v,v)\). Consider the one-parameter family \[ \Phi_t(u,v)=\cos t \cdot \Phi(u,v) + \sin t \cdot \Psi(u,v) \text{ for \(t\in[0,\pi/2]\).} \] Note that \(\Phi_0\) is the catenoid and \(\Phi_{\pi/2}\) is the helicoid, so this deformation transforms one into the other.
Compute the first fundamental form coefficients \(E_t,F_t,G_t\) for each \(t\). Show that \(E_t G_t - F_t^2\) is independent of \(t\).
What does this say about areas?
Let \(\Phi_1:D_1\to\R^3\) and \(\Phi_2:D_2\to\R^3\) be regular parametrizations with the same image, i.e., \(\Phi_1(D_1)=\Phi_2(D_2)\).
Then a \(2\)-form \(\omega\) on \(\R^3\), \( \displaystyle\iint_{D_1}\Phi_1^*\omega = \iint_{D_2}\Phi_2^*\omega. \) \hfill(Hint: review S09P15.)
Let \(\Phi:D_1\to\R^3\) be a regular parametrized surface and let \(\psi:D_2\to D_1\) be a \(C^1\) diffeomorphism. Consider \(\Psi=\Phi\circ\psi\) and suppose \(\det D\psi > 0\) everywhere.
Then S10P07 shows \(\displaystyle\iint_{D_2}\Psi^*\,\omega=\displaystyle\iint_{D_1}\Phi^*\omega\).
Let \(\Phi:D\to\R^3\) be a regular parametrized surface, and let \(\pi:\R^3\to\R^2\) denote the projection \(\pi(x,y,z)=(x,y)\). Then \(\Area(\Phi(D))\ge\Area\bigl(\pi(\Phi(D))\bigr)\).