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! ;-)


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    $\begingroup$ The principal curvature directions are orthogonal for the same reason that the semimajor and semiminor axes of an ellipse are orthogonal. $\endgroup$ Oct 12, 2011 at 1:36
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    $\begingroup$ 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! $\endgroup$ Oct 12, 2011 at 1:42
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    $\begingroup$ fuzzytron, that's funny, because usually I do the reverse of that. $\endgroup$
    – anon
    Oct 12, 2011 at 3:10
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    $\begingroup$ Ah, I see -- you're saying: consider the local quadratic approximation, whose level sets look like ellipses (or hyperbola). $\endgroup$ Oct 12, 2011 at 16:57
  • $\begingroup$ 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... $\endgroup$ Jun 7, 2013 at 1:23

3 Answers 3


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.

  • $\begingroup$ For those who are not privileged enough to own the book, here's what's written on p. 38 (rewritten using the notation above). Tangent vectors are all orthogonal to the normal, i.e., Ndf(X) = 0 for any X. Differentiating this expression along any direction Y we get dN(Y)*df(X) + Nd(df(X))(Y) = 0. So by equality of mixed partials we have df(X)*dN(Y) = -Nd(df(X))(Y) = -Nd(df(Y))(X) = df(Y)*dN(X). I.e., the second fundamental form is symmetric, QED. $\endgroup$ Oct 12, 2011 at 23:51
  • $\begingroup$ Thanks, @fuzzytron. This makes sense. $\endgroup$ Oct 13, 2011 at 1:01

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.


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 $W(v)$.


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