# 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!

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The principal curvature directions are orthogonal for the same reason that the semimajor and semiminor axes of an ellipse are orthogonal. –  Qiaochu Yuan Oct 12 '11 at 1:36
Which is? :-) Usually I think of an ellipse as the image of a circle w.r.t. a self-adjoint linear operator, which doesn't help me much here! –  PolyKnowMeAll Oct 12 '11 at 1:42
fuzzytron, that's funny, because usually I do the reverse of that. –  anon Oct 12 '11 at 3:10
Ah, I see -- you're saying: consider the local quadratic approximation, whose level sets look like ellipses (or hyperbola). –  PolyKnowMeAll Oct 12 '11 at 16:57
What a pity you already accepted an answer! I would really like to see some intuition as to why principal curvature directions must be orthogonal (away from umbilic points)! I can see some counterexamples in my mind, but I must be violating the smoothness condition somewhere, I guess... –  Bruno Stonek Jun 7 '13 at 1:23
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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.