I've been looking at a problem in Elementary Differential Geometry by Barrett O'Neill (Revised 2nd Edition), and Im having trouble confirming what the question asks is true.
The problem goes as follows (5.4.10):
"Show that the saddle surface z = xy the two vector fields
$$ ( \sqrt{1+x^2} \pm \sqrt{1+y^2}, y\sqrt{1+x^2} \pm x\sqrt{1+y^2}) $$
are principal at each point. Check that they are orthogonal and tangent to M."
The first issue I have is making sure that the 2 vector fields Im using are correct. I assume the author means the following 2 vector fields in $R^2$:
$$ ( \sqrt{1+x^2} + \sqrt{1+y^2}, y\sqrt{1+x^2} + x\sqrt{1+y^2}) \hspace{1cm} (1) $$ and $$ ( \sqrt{1+x^2} - \sqrt{1+y^2}, y\sqrt{1+x^2} - x\sqrt{1+y^2}) \hspace{1cm} (2)$$
If I assume that the Monge Patch for this problem is $A: R^2 \mapsto M$ where $A(x,y) = (x,y,xy)$ then the tangent map of A $(A_{*})$ is:
\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ y & x \\ \end{pmatrix}
and the Shaper operator, S, of M (where $S: R^2 \mapsto T_{p}M$) is: $$ \frac{1}{(1+x^2+y^2)^{\frac{3}{2}}} \begin{pmatrix} -xy & 1+x^2 \\ 1+y^2 & -xy \\ \end{pmatrix} $$
I initially tried solving this out for the general case by showing that the vectors from (1) and (2) (which I will call $V_{1}$ and $V_{2}$ respectively) work, but wasnt able to show that $S(V_{1}) = k_{1}V_{1}$.
Instead I tried to see if any of this worked out at the origin, $(0,0)$.
This means that $V_{1} = (2,0)$, $V_2 = (0,0)$, the tangent map (in this case $R^2 \mapsto T_{(0,0)}(M)$) equals:
\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}
And the Shape operator is:
\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}
However, with the above Shape operator, the eigen values are $\pm 1$ and the eigenvectors are $(1,1)$ and $(1,-1)$ which are definitely not parallel to $V_1$ and $V_2$ as show above.
Is there a typo in the question? Or did I make a mistake somewhere?