Is there a geometric explanation for why principal curvature directions are orthogonal?

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!

Solution 1:

I only know that operator $df(X) \cdot dN(Y)$ is symmetric because the partial derivatives commute. The formal proof of that is just one line in W.P.A. Klingenberg "A Course in Differential Geometry", p.38. Can be this treated as a geometric intuition?

Informally, I would think of $df(X) \cdot dN(Y)$ as being defined completely just by by $df$ (and the dot product in the ambient space, which is $\mathbb{R}^3$ in the question), so the changes in direction $X$ are reflected automatically (via $N=\frac{df}{\left\Vert df\right\Vert }$) in direction $Y$. Maybe, this observation can be made more rigorous.

Edit. I've just noticed an error in the above. Actually, $N=\pm \frac{ds}{\left\Vert ds\right\Vert }$ for a defining function $s$ of the hypersurface.

Solution 2:

My answer is really just an explanation of Qiaochu Yuan's comments.

Let's first take a look at how curvature is defined on 1-dimensional curve: suppose that $C$ is a twice continuously differentiable immersed plane curve represented by $\gamma (s)=(x(s),y(s))$, then curvature is defined as $\kappa (s)= \|T'(s)\|=\|\gamma{''}(s)\|$ ($T(s)$ the unit tangent vector and $s$ the arc length). We can see $\kappa(s)$ only depends up to second derivative.

Have this in mind, we are safe to replace any smooth surface with its quadratic approximation locally to find corresponding principal curvatures/directions. And you can easily check for any quadratic surface the principal directions are orthogonal.

Solution 3:

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 $W(v)$.