TITLE: Is there a geometric explanation for why principal curvature directions are orthogonal?
QUESTION [26 upvotes]: Let $f: \mathbb{R}^2 \supset M \rightarrow \mathbb{R}^3$ be a smooth immersion and let $N: M \rightarrow S^2 \subset \mathbb{R}^3$ be the corresponding Gauss map.  The normal curvature along a unit tangent direction $X \in TM$ can be expressed as
$$ df(X) \cdot dN(X), $$
where $\cdot$ is the usual Euclidean inner product on $\mathbb{R}^3$.  The principal curvature directions $X_1, X_2 \in TM$ are the unit directions along which normal curvature is minimized and maximized, respectively.  It is a well-established fact that (away from umbilic points) these directions are orthogonal in the sense that
$$ df(X_1) \cdot df(X_2) = 0. $$
Question 1 Is there any purely geometric intuition for why principal curvature directions are orthogonal?
Question 2 Can you argue that the bilinear form $df(X) \cdot dN(Y)$ is symmetric w.r.t. $X$ and $Y$ without resorting to the use of coordinates?
In fact, I'd be satisfied with an answer to Question 2 alone, since a self-adjoint operator has orthogonal eigenvalues.  I'm really looking for intuition here, not just a formal proof -- if all I get is a bunch of $E,F,G$ and $e,f,g$ or (god forbid) $\Gamma_{ij}^k$ I just might scream! ;-)
Thanks!

REPLY [3 votes]: The principal directions are orthogonal because they are the eigenvectors of a selfadjoint operator acting on the tangent plane, namely the Weingarten operator (shape operator) $W(v)$. This is defined by differentiating the normal vector in the direction of $v$, obtaining $W(v)$.