Euler's Theorem in Differential Geometry

Let $w_1$ and $w_2$ be the preimages of $T_1$ and $T_2$, respectively, in the parametric domain. By definition, $T$ is a linear combination of the vectors $T_1$ and $T_2.$ Then, $w_1$ and $w_2$ span the parametric domain, so we are able to define the preimage $w$ of the vector $T$ as a linear combination of vectors $w_1$ and $w_2$ as such

$$\begin{align*} T &= \cos(\theta)\,T_1 + \sin(\theta)\,T_2\\ &= \cos(\theta)\,\langle X_u, X_v\rangle w_1 + \sin(\theta)\langle X_u, X_v\rangle w_2\\ &= \langle X_u, X_v\rangle \underbrace{(w_1 \cos(\theta) + w_2 \sin(\theta))}_{w}. \end{align*}$$

Now we have $T$ in terms of $w_1$ and $w_2$, we can plug it into the formula the normal surface curvature.

$$\begin{align*} \kappa_n(w) &= \frac{\mathrm{II}}{\mathrm{I}}\\ &= \frac{e\mathrm{d}u^2 + 2f\mathrm{d}u^2\mathrm{d}v^2+g\mathrm{d}v^2}{E\mathrm{d}u^2+2F\mathrm{d}u^2\mathrm{d}v^2+G\mathrm{d}v^2}\\ &= \frac{w^T H_{\mathrm{II}}w}{w^TH_{I}w} \end{align*}$$

where $H_I$ and $H_{\mathrm{II}}$ are given by

$$H_I = \begin{pmatrix}E & F \\ F & G\end{pmatrix},\quad H_{\mathrm{II}} = \begin{pmatrix}e & f \\ f & g\end{pmatrix}.$$

Plugging in $w = w_1 \cos(\theta) + w_2 \sin(\theta),$

$$\begin{align*} \kappa_n(w) &= \frac{(w_1\cos\theta + w_2\sin\theta)^{\intercal}H_{\mathrm{II}}(w_1\cos\theta + w_2\sin\theta)}{(w_1\cos\theta + w_2\sin\theta)^{\intercal}H_{\mathrm{I}}(w_1\cos\theta + w_2\sin\theta)}\\ &= \frac{\cos^2(\theta)w_1^{\intercal} H_{\mathrm{II}}w_1 + 2\cos(\theta)\sin(\theta)w_1^{\intercal} H_{\mathrm{II}}w_2 + \sin^2(\theta)w_2^{\intercal} H_{\mathrm{II}}w_2}{\cos^2(\theta)w_1^{\intercal} H_{\mathrm{I}}w_1 + 2\cos(\theta)\sin(\theta)w_1^{\intercal} H_{\mathrm{I}}w_2 + \sin^2(\theta)w_2^{\intercal} H_{\mathrm{I}}w_2}. \end{align*}$$

Using some of the properties of $T_1$ and $T_2$, we can simplify the above expresion. First, $T_1$ and $T_2$ are unit vectors,

$$\begin{align*} 1 = T_1^{\intercal} \cdot T_1 &= w_1^{\intercal}\cdot\langle X_u^{\intercal}, X_v^{\intercal}\rangle\langle X_u, X_v\rangle^{\intercal} \cdot w_1\\ &= w_1^{\intercal} \begin{pmatrix}X_u^{\intercal}X_u & X_u^{\intercal}X_v \\ X_u^{\intercal}X_v & X_v^{\intercal}X_v\end{pmatrix}\\ &= w_1^{\intercal} H_{\mathrm{I}} w_1. \end{align*}$$

Doing the same for $w_2,$ we get that $w_2^{\intercal}H_1w_2 = 1.$

Next, using the additional fact that $T_1$ and $T_2$ are orthogonal gives

$$\begin{align*} 0 = T_1^{\intercal}\cdot T_2 &= w_1^{\intercal} \cdot \langle X_u^{\intercal}, X_v^{\intercal}\rangle\langle X_u, X_v\rangle^{\intercal} w_2\\ &= w_1^{\intercal} H_{\mathrm{I}} w_2. \end{align*}$$

These results allow us to simplify the denominator of $\kappa_n(w),$

$$\begin{align*} &\cos^2(\theta)\underbrace{\cancel{w_1^\intercal H_{\mathrm{I}} w_1}}_{1} + 2\cos(\theta)\sin(\theta)\underbrace{\cancel{w_1^{\intercal} H_{\mathrm{I}} w_2}}_{0} + \sin^2(\theta)\underbrace{\cancel{w_2^{\intercal}H_{\mathrm{I}}w_2}}_1\\ =&\ \cos^2\theta + \sin^2 \theta\\ =&\ 1. \end{align*}$$

Then, the curvature is expressed as

$$\kappa_n(w) = \cos^2(\theta)\left(w_1^{\intercal} H_{\mathrm{II}} w_1\right) + 2\cos(\theta)\sin(\theta)\left(w_1^{\intercal} H_{\mathrm{II}} w_2\right) + \sin^2(\theta)\left(w_2^{\intercal} H_{\mathrm{II}} w_2\right).$$

Inspecting $w_1^{\intercal} H_{\mathrm{II}} w_1,$

$$\begin{align*} w_1^{\intercal} H_{\mathrm{II}} w_1 &= \frac{w_1^{\intercal} H_{\mathrm{II}} w_1}{1}\\ &= \frac{w_1^{\intercal} H_{\mathrm{II}} w_1}{w_1^{\intercal} H_{\mathrm{I}} w_1}\\ &= \kappa_n(w_1)\\ &= \kappa_1. \end{align*}$$

Using the same method, we have $\kappa_2 = w_2^{\intercal} H_{\mathrm{II}} w_2.$ Now, we have that the curvature is given by

$$\kappa_n(w) = \kappa_1\cos^2(\theta) + 2\cos(\theta)\sin(\theta)\left(w_1^{\intercal} H_{\mathrm{II}} w_2\right) + \kappa_2\sin^2(\theta).$$

Because $\kappa_1$ and $\kappa_2$ are principal curvatures, we have that $H_{\mathrm{II}}w_2 = \kappa_2 H_I w_2$ which yields the following expression

$$\kappa_n(w) = \kappa_1\cos^2(\theta) + 2\cos(\theta)\sin(\theta)\left(\kappa_2 w_1^{\intercal} H_{\mathrm{I}} w_2\right) + \kappa_2\sin^2(\theta).$$

We've previously shown that $w_1^{\intercal}H_I w_2 = 0,$ therefore cancelling the middle term in the above expression. Finally, the normal surface curvature is given by

$$\kappa_n(w) = \kappa_1\cos^2(\theta) + \kappa_2\sin^2(\theta).$$