# Showing directly that the principal directions are going to be orthogonal

So first of all, I know that the Weingarten map (which from now on I shall denote by $L$) is a symmetric linear operator, so there is an orthonormal basis of eigenvalues (Spectral Theorem).

I have been trying this concrete example for a while but I am just stuck and I would appreciate an extra pairs of eyes.

I will use the result that claims $L=G^{-1}H$ where $G$ is the matrix for the first fundamental form and $H$ is the matrix for the second fundamental form.

Our surface is given by $f(u,v)=(u\cos v, u\sin v , v)$. Then:

$f_u=(\cos v, \sin v, 0)$, $f_v=(-u\sin v, u \cos v, 1)$, $f_{uu}=(0,0,0)$, $f_{uv}=f_{vu}=(-\sin v, \cos v, 0)$, and $f_{vv}=(-u\cos v, -u \sin v, 0)$.

Thus, $g_{11}=f_u\cdot f_u=1$, $g_{12}=g_{21}=f_u\cdot f_v=0$, and $g_{22}=f_v\cdot f_v =u^2+1$.

Then, $f_u\times f_v=(\sin v, -\cos v, u)$, so $n=\frac{1}{\sqrt{1+u^2}}(\sin v, -\cos v, u)$.

Then, $h_{11}=n\cdot f_{uu}=0$, $h_{12}=h_{21}=n\cdot f_{uv}=\frac{-1}{\sqrt{u^2+1}}$, and $h_{22}=0$.

Then as $G^{_1}=Diag[1,\frac{1}{1+u^2}]$, we have that $L=G^{-1}H$, is: $L_{11}=0$, $L_{12}=\frac{-1}{\sqrt{u^2+1}}$, $L_{21}=\frac{-1}{(u^2+1)^{3/2}}$, and $L_{22}=0$ (here is where I start to doubt because I thought I would always wind up with a symmetric matrix).

I get that the eigenvalues of this matrix are given by $\det(\lambda I-L)=\lambda^2-\frac{1}{(1+u^2)^2}$, so $k_1=\frac{1}{u^2+1}$, and $k_2=-k_1$. Then to find the first principal direction, we have to find the eigenvector correspoding to $k_1$, which turned out to be $(\frac{-1}{\sqrt{u^2+1}},1)$, and the other principal direction turned out to be $(\frac{1}{\sqrt{u^2+1}},1)$, which are not always orthogonal, so I dont know where I went wrong.

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I agree with almost all of your computations. That is, I agree with everything until you compute eigenvectors.

The eigenvectors for $k_1$ and $k_2$ are (according to Maple and my own work) respectively, $$\begin{bmatrix} 1 \\ \frac{1}{\sqrt{1+u^2}}\end{bmatrix} \text{ and } \begin{bmatrix} 1 \\ -\frac{1}{\sqrt{1+u^2}}\end{bmatrix}.$$

Unfortunately, these seem to keep the same issue your calculation gives, right? They seem to fail to be orthogonal (except when $u=0$ of course). What's going on?

I claim that, in fact, these two vectors are everywhere orthgonal as measured by someone living on your coordinate patch. The key point is that all of this has been expressed in the basis $\{f_u, f_v\}$, but this is not an orthonormal basis. If it was, you'd have $G = I$ everywhere.

So, the dot product $\left\langle \begin{bmatrix} 1 \\ \frac{1}{\sqrt{1+u^2}}\end{bmatrix}, \begin{bmatrix} 1 \\ -\frac{1}{\sqrt{1+u^2}} \end{bmatrix} \right\rangle$ is not $1-\frac{1}{1+u^2}$, but it must take the metric $G$ into account.

The dot product is actually given by \begin{align*} \left\langle \begin{bmatrix} 1 \\ \frac{1}{\sqrt{1+u^2}}\end{bmatrix}, \begin{bmatrix} 1 \\ -\frac{1}{\sqrt{1+u^2}} \end{bmatrix} \right\rangle &= G_{11}\cdot 1\cdot 1 + G_{22} \frac{1}{\sqrt{1+u^2}}\cdot \frac{-1}{\sqrt{1+u^2}} \\ &= 1+(1+u^2)\left(\frac{-1}{1+u^2} \right) \\ &= 1-1 \\ &= 0.\end{align*}

In other words, your two eigenvectors are orthogonal with respect to the metric on your coordinate patch.

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If any latex guru reads this, I used the \left command to size the \langle commands. This worked great in the usual (single dollar sign) environment, but the \langle in the align environment is too small. What's the best way to adjust this? –  Jason DeVito Dec 17 '12 at 4:04