Let $S$ be a smooth surface in $ℝ^3$ with normal $n$.

Define the local first fundamental form $A$ and the local second fundamental form $B$ in a smooth local parametrization $F$ for $S$. How do $A, B$ transform under changes of local parametrizations $F, \widetilde{F}$?
State a formula for the Gaussian curvature $K$ in terms of $A, B$. Show that $K$ does not depend on the choice of local parametrization $F$.

For $F=F(x, y)$, abbreviate $∂_1 F=∂_x F, ∂_2 F=∂_y F$. Then $A_{i j}=∂_i F ⋅ ∂_j F$ and $B_{i j}=n ⋅ ∂_{i j} F$ (using the dot product in $ℝ^3$ ).
Let $τ=\widetilde{F}^{-1} ∘ F$ be the transition map. Then $A=D τ^T ∘ \widetilde{A} ∘ D τ$ and $B=D τ^T ∘ \widetilde{B} ∘ D τ$ where $\widetilde{A}, \widetilde{B}$ are the forms in the parametrization $\widetilde{F}$, and $T$ means transpose.
$K=\det B / \det A=\det(D τ^T ∘ \widetilde{B} ∘ D τ) / \det(D τ^T ∘ \widetilde{A} ∘ D τ)=[(\det D τ)^2 \det \widetilde{B}] /[(\det D τ)^2 \det \widetilde{A}]=\det \widetilde{B} / \det \widetilde{A}$

State a criterion involving linear algebra for a smooth map $F: U → S$ to be a smooth local parametrization, where $U$ is an open subset of $ℝ^2$. The helicoid is the surface $H ⊂ ℝ^3$ parametrized by \[ F: ℝ^2 → H, F(x, y)=(y \cos x, y \sin x, x) . \] Prove that $H$ is a smooth surface.
Compute the first and second fundamental forms for $H$, and compute its Gaussian curvature.
What is the local first fundamental form for $H$ after the change of coordinates $x=\widetilde{x}$, $y=\sinh \tilde{y}$?

$F$ is a smooth parametrization iff $∂_x F, ∂_y F$ are linearly independent for all $(x, y) ∈ U$.
$F(x, y)=(y \cos x, y \sin x, x)$
$∂_x F=(-y \sin x, y \cos x, 1)$ and $∂_y F=(\cos x, \sin x, 0)$
Linear independence: $λ ∂_x F+μ ∂_y F=0$, then the third coordinate forces $λ=0$, but then also $μ=0$ since $∂_y F≠0$ ($\sin x, \cos x$ do not simultaneously vanish). \[ ∂_x F ⋅ ∂_x F=y^2+1,   ∂_x F ⋅ ∂_y F=0,   ∂_y F ⋅ ∂_y F=1 \] First fundamental form: $A=\pmatrix{1+y^2 & 0 \\ 0 & 1}$ (equivalently $I=(1+y^2) d x^2+d y^2$). \[ ∂_x F × ∂_y F=(-\sin x, \cos x,-y) \] Normal: $n(x, y)=\frac{1}{\sqrt{1+y^2}}(-\sin x, \cos x,-y)$ \begin{aligned} & ∂_{x x} F=(-y \cos x,-y \sin x, 0),   ∂_{x y} F=(-\sin x, \cos x, 0),   ∂_{y y} F=0 \text {, } \\ & n ⋅ ∂_{x x} F=0, n ⋅ ∂_{x y} F=\frac{1}{\sqrt{1+y^2}}, n ⋅ ∂_{y y} F=0 \end{aligned} Second fundamental form: $B=\pmatrix{0 & \frac{1}{\sqrt{1+y^2}} \\ \frac{1}{\sqrt{1+y^2}} & 0}$ (equivalently $\text{II}=\frac{2}{\sqrt{1+y^2}} d x d y$).
Gaussian curvature: $K=\frac{\det B}{\det A}=\frac{-\frac{1}{1+y^2}}{1+y^2}=-\frac{1}{(1+y^2)^2}$. \begin{gathered} (x, y)=τ^{-1}(\widetilde{x}, \widetilde{y})=(\widetilde{x}, \sinh \widetilde{y}), D τ^{-1}=\pmatrix{ 1 & 0 \\ 0 & \cosh \widetilde{y} }\\ \widetilde{A}=\pmatrix{ 1 & 0 \\ 0 & \cosh \widetilde{y} }\pmatrix{ 1+\sinh^2 \widetilde{y} & 0 \\ 0 & 1 }\pmatrix{ 1 & 0 \\ 0 & \cosh \widetilde{y} }=\pmatrix{ \cosh^2 \widetilde{y} & 0 \\ 0 & \cosh^2 \widetilde{y} } . \end{gathered}

What does it mean for two surfaces to be isometric? State a criterion involving first fundamental forms for two surfaces to be locally isometric.

Two surfaces are isometric if there is a diffeomorphism $φ: S_1 → S_2$ which preserves lengths of curves, $L(φ(γ))=L(γ)$ for all curves $γ ⊂ S_1$.
Two surfaces are locally isometric near $p_1, p_2$ iff there are local parametrizations $F_1: U →S_1, F_2: U → S_2$ defined on the same open set $U ⊂ ℝ^3$, with $p_1 ∈ F_1(U), p_2 ∈ F_2(U)$ such that the local first fundamental forms agree $A_1=A_2$.

The catenoid $C$ is the surface of revolution obtained from the catenary $X=\cosh y, Z=y$ by rotating about the $Z$-axis, where $(X, Y, Z)$ are the standard coordinates on $ℝ^3$. Write a formula for a local parametrization $F(x, y)$ for $C$, with $x ∈(-π, π)$ the angle of rotation. Prove that the surfaces $H$ and $C$ are locally isometric.

\[ F(x, y)=(\cosh y \cos x, \cosh y \sin x, y) \] For that parametrization of $C$, \begin{aligned} ∂_x F&=(-\cosh y \sin x, \cosh y \cos x, 0) \\ ∂_y F&=(\sinh y \cos x, \sinh y \sin x, 1) \end{aligned} So the first fundamental form is $I=\cosh^2 y d x^2+\cosh^2 y d y^2$ (the second term arises from $\sinh^2 y+1=\cosh^2 y)$. This is the same as the $I$ in part (b), so the surfaces are locally isometric.