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Hyperbolic geometry was discovered independently by Nikolai Lobachevsky, János Bolyai, and Carl Friedrich Gauss in the early 19th century. It has important applications in fields such as art, architecture, and physics.

In hyperbolic geometry, the sum of the angles in a triangle is always less than 180 degrees, and the area of a triangle is proportional to its excess angle. This leads to some unusual and counterintuitive properties, such as the fact that circles and lines behave differently than in Euclidean geometry. For example, in hyperbolic geometry, circles have negative curvature and get smaller as you move away from the center, while lines have positive curvature and get farther apart as you move away from them.

Hyperbolic geometry has also been used to model certain types of curved spaces, such as the surface of a saddle or a hyperbolic plane. These models have been used in computer graphics, virtual reality, and other applications.

Consider the parametrized surface $$ \mathbf{r}(u, v)=\left(u-\frac{u^3}{3}+u v^2, v-\frac{v^3}{3}+v u^2, u^2-v^2\right) . $$ Show that (a) The coefficients of the first fundamental form are $$ E=G=\left(1+u^2+v^2\right)^2, F=0 . $$

To find the coefficients of the first fundamental form, we need to compute the dot products of the partial derivatives of the surface with respect to the parameters $u$ and $v$:

\begin{align*} \mathbf{r}_u &= \left(1-u^2+v^2, 2uv, 2u\right) \ \mathbf{r}_v &= \left(2uv, 1-v^2+u^2, -2v\right) \end{align*}

Then, we can compute the coefficients using the formulas:

\begin{align*} E &= \mathbf{r}_u \cdot \mathbf{r}_u = \left(1-u^2+v^2\right)^2 + 4u^2v^2 + 4u^2 \ F &= \mathbf{r}_u \cdot \mathbf{r}_v = 2u^2v – 2uv^3 + 2uv \ G &= \mathbf{r}_v \cdot \mathbf{r}_v = 4u^2v^2 + \left(1-v^2+u^2\right)^2 + 4v^2 \end{align*}

To simplify these expressions, we can use the identity $a^2-b^2 = (a+b)(a-b)$:

\begin{align*} E &= \left(1+u^2+v^2\right)^2 \ F &= 2uv\left(u^2-v^2+1\right) = 0 \ G &= \left(1+u^2+v^2\right)^2 \end{align*}

Therefore, we have $E=G=\left(1+u^2+v^2\right)^2$ and $F=0$, which confirms the result.

Consider the parametrized surface $$ \mathbf{r}(u, v)=\left(u-\frac{u^3}{3}+u v^2, v-\frac{v^3}{3}+v u^2, u^2-v^2\right) $$ (b) The coefficients of the second fundamental form are $$ L=2, M=-2, N=0 . $$

To find the coefficients of the second fundamental form, we first need to compute the first fundamental form and the normal vector.

The first fundamental form is given by:

$\begin{aligned} & E=\mathbf{r}_u \cdot \mathbf{r}_u=\left(1-u^2+v^2\right)^2+4 u^2 v^2 \ & F=\mathbf{r}_u \cdot \mathbf{r}_v=\mathbf{r}_v \cdot \mathbf{r}_u=2 u v\left(1-u^2-v^2\right) \ & G=\mathbf{r}_v \cdot \mathbf{r}_v=\left(1+u^2-v^2\right)^2+4 u^2 v^2\end{aligned}$

The normal vector is given by:

$\mathbf{N}=\frac{\mathbf{r}_u \times \mathbf{r}_v}{\left|\mathbf{r}_u \times \mathbf{r}_v\right|}=\frac{1}{\sqrt{2\left(1+u^2+v^2\right)}}\left(-2 u,-2 v, 1-u^2-v^2\right)$

Now, we can compute the coefficients of the second fundamental form:

$\begin{aligned} & L=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial u^2}=\frac{4 u^2 v^2-2\left(1-u^2+v^2\right)^2}{2\left(1+u^2+v^2\right)^{3 / 2}}=2 \ & M=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial u \partial v}=\frac{2 u v\left(1-u^2-v^2\right)}{2\left(1+u^2+v^2\right)^{3 / 2}}=-2 \ & N=\mathbf{N} \cdot \frac{\partial^2 \mathbf{r}}{\partial v^2}=\frac{4 u^2 v^2-2\left(1+u^2-v^2\right)^2}{2\left(1+u^2+v^2\right)^{3 / 2}}=0\end{aligned}$

Therefore, the coefficients of the second fundamental form are $L=2$, $M=-2$, and $N=0$.

(c) The principal curvatures are $$ \kappa_1=\frac{2}{\left(1+u^2+v^2\right)^2}, \quad \kappa_2=-\frac{2}{\left(1+u^2+v^2\right)^2} . $$

(c) The principal curvatures are $$ \kappa_1=\frac{2}{\left(1+u^2+v^2\right)^2}, \quad \kappa_2=-\frac{2}{\left(1+u^2+v^2\right)^2} . $$

$\mathbf{n}=\frac{\mathbf{r}_u \times \mathbf{r}_v}{\left|\mathbf{r}_u \times \mathbf{r}_v\right|}=\frac{\left(2 u-u^3+v^2,-2 v+v^3-u^2, 2 u v\right)}{\sqrt{4 u^2 v^2\left(1+u^2+v^2\right)}}$

The Gaussian curvature $K$ and mean curvature $H$ are given by

\begin{align*} K &= \frac{LN – M^2}{EG – F^2} = -\frac{2}{(1+u^2+v^2)^4} \ H &= \frac{1}{2}(L + N) = \frac{2u(2-3u^2-v^2) + 2v(2-3v^2-u^2)}{2(1+u^2+v^2)^{5/2}}. \end{align*}

The principal curvatures $\kappa_1$ and $\kappa_2$ are the solutions to the characteristic equation

$\operatorname{det}\left(\begin{array}{cc}E-\kappa & F \ F & G-\kappa\end{array}\right)=0$.

Substituting in the coefficients above, we get

$\operatorname{det}\left(\begin{array}{cc}1-(\kappa-E) & 2 u v\left(1-u^2-v^2\right) \ 2 u v\left(1-u^2-v^2\right) & 1-(\kappa-G)\end{array}\right)=0$

Simplifying, we get

$(1-\kappa)^2-4 u^2 v^2\left(1+u^2+v^2\right)^{-4}=0$.