Normal curvature: does Euler's theorem require an ordering of the principal directions/curvatures?

Question

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1. What the question is asking (in plain language)

A surface (M\subset\mathbb R^{3}) has at each point (p)

two principal directions (\mathbf e_{1},\mathbf e_{2}) (mutually orthogonal unit tangent vectors) and
the corresponding principal curvatures (\kappa_{1},\kappa_{2}) (the normal curvatures in those directions).

Euler’s theorem says that for any unit tangent vector (u) making an angle (\theta) with (\mathbf e_{1}),

[ \kappa(u)=\kappa_{1}\cos^{2}\theta+\kappa_{2}\sin^{2}\theta . ]

The student wonders:

Must we label the direction with the largest curvature as (\mathbf e_{1}) (and thus call its curvature (\kappa_{1}))?
If we do not order them, does the formula still hold?
What does “largest” mean when one curvature is negative?

In short: Is an ordering of the principal directions/curvatures required for Euler’s formula, or is it merely a convenient convention?

2. Detailed answer – step‑by‑step

2.1 Principal directions and the shape operator

At a point (p) the shape operator (Weingarten map) is the linear map

[ S_p:T_{p}M\longrightarrow T_{p}M,\qquad S_p(v)=-\mathrm dN_{p}(v), ]

where (N) is a chosen unit normal field.
(S_p) is symmetric with respect to the first fundamental form, therefore it can be diagonalised:

[ S_p(\mathbf e_{i})=\kappa_{i}\,\mathbf e_{i},\qquad i=1,2, ]

where ({\mathbf e_{1},\mathbf e_{2}}) is an orthonormal basis of eigenvectors (the principal directions) and (\kappa_{1},\kappa_{2}) are the corresponding eigenvalues (the principal curvatures).

The eigenvalues are just numbers; they can be positive, zero or negative. No sign restriction is imposed.

2.2 Normal curvature of an arbitrary direction

Take any unit tangent vector (u\in T_{p}M). Because ({\mathbf e_{1},\mathbf e_{2}}) is a basis we can write

[ u=\cos\theta\,\mathbf e_{1}+\sin\theta\,\mathbf e_{2}, \qquad \text{with }\theta\in[0,2\pi). ]

(The angle (\theta) is precisely the oriented angle from (\mathbf e_{1}) to (u).)

The normal curvature of the surface in the direction (u) is defined by

[ \kappa(u)=\langle S_p(u),\,u\rangle . ]

Insert the decomposition of (u):

[ \begin{aligned} S_p(u) &= S_p\bigl(\cos\theta\,\mathbf e_{1}+\sin\theta\,\mathbf e_{2}\bigr) = \cos\theta\,S_p(\mathbf e_{1})+\sin\theta\,S_p(\mathbf e_{2})
&= \cos\theta\,\kappa_{1}\mathbf e_{1}+\sin\theta\,\kappa_{2}\mathbf e_{2}. \end{aligned} ]

Now take the inner product with (u) (remember that (\mathbf e_{1},\mathbf e_{2}) are orthonormal):

[ \begin{aligned} \kappa(u) &=\big\langle\cos\theta\,\kappa_{1}\mathbf e_{1}+\sin\theta\,\kappa_{2}\mathbf e_{2}, \; \cos\theta\,\mathbf e_{1}+\sin\theta\,\mathbf e_{2}\big\rangle
&=\kappa_{1}\cos^{2}\theta+\kappa_{2}\sin^{2}\theta . \end{aligned} ]

Thus Euler’s formula holds for every orthonormal pair of principal directions, no matter how we label them.

2.3 What happens if we interchange the labels?

Suppose we decide to call the eigenvector with curvature (\kappa_{2}) “(\mathbf e_{1})” and the other “(\mathbf e_{2})”.
Then the decomposition of the same vector (u) becomes

[ u = \cos\theta’\,\mathbf e_{1}^{\text{new}} + \sin\theta’\,\mathbf e_{2}^{\text{new}}, ]

where (\theta’ = \theta - \frac{\pi}{2}) (the angle measured from the new (\mathbf e_{1})).

Applying the same computation gives

[ \kappa(u)=\kappa_{2}\cos^{2}\theta’ + \kappa_{1}\sin^{2}\theta’, ]

which is exactly the same number, because (\cos^{2}\theta’ = \sin^{2}\theta) and (\sin^{2}\theta’ = \cos^{2}\theta).
Hence the formula is invariant under swapping the labels; the only thing that changes is the name of the angle (\theta).

2.4 Why many texts say “largest” and “smallest”

Because the shape operator has two eigenvalues, it is customary to order them for ease of reference:

[ \kappa_{1}\ge \kappa_{2}. ]

Here “largest” and “smallest” refer to the numerical ordering on the real line, including the sign.
Examples:

If (\kappa_{1}=5,\;\kappa_{2}=-2), then (\kappa_{1}) is the largest (5 > −2).
If (\kappa_{1}=-1,\;\kappa_{2}=-4), then the “largest” is (-1) (because (-1>-4)).

Ordering is purely conventional; the geometry does not depend on which eigenvalue we call (\kappa_{1}). One may also order by absolute value ((

\kappa_{1}

\ge

\kappa_{2}

)) if a problem specifically needs “principal curvature of greatest magnitude”, but that is a different convention that must be stated explicitly.

2.5 Summary of the answer to the original question

Euler’s theorem does not require us to label the principal direction with the numerically largest curvature as (\mathbf e_{1}).
Any orthonormal eigenbasis ({\mathbf e_{1},\mathbf e_{2}}) together with its eigenvalues (\kappa_{1},\kappa_{2}) satisfies

[ \boxed{\;\kappa(u)=\kappa_{1}\cos^{2}\theta+\kappa_{2}\sin^{2}\theta\;}, ]

where (\theta) is the angle from the chosen (\mathbf e_{1}) to (u).

If one wishes to adopt a standard naming, the usual convention is to order the eigenvalues so that (\kappa_{1}\ge\kappa_{2}). “Largest” means greater in the usual order of real numbers, sign included. This convention is only for bookkeeping; the formula itself is independent of it.

3. Final answer

Euler’s formula holds for any orthonormal pair of principal directions, regardless of which one is called (\mathbf e_{1}) or (\mathbf e_{2}).
The only requirement is that the curvature (\kappa_{i}) attached to (\mathbf e_{i}) is the corresponding eigenvalue of the shape operator.

The common practice of naming (\kappa_{1}) the “largest” curvature simply means that we order the two eigenvalues so that (\kappa_{1}\ge\kappa_{2}) (numerically, sign included). This ordering is a convenient convention, not a mathematical necessity for Euler’s theorem.

4. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Assuming Euler’s formula fails if (\kappa_{1}<\kappa_{2}).	The derivation uses only the eigen‑pair relationship; the inequality is irrelevant.	Remember the proof: (\kappa(u)=\langle S(u),u\rangle). The result is symmetric in the two eigenvalues.
Swapping (\mathbf e_{1},\mathbf e_{2}) but keeping the same (\theta).	After swapping, the angle measured from the new (\mathbf e_{1}) changes ((\theta’=\theta-\pi/2)).	When you relabel the basis, also relabel the angle accordingly.
Confusing “largest” with “largest in absolute value”.	“Largest” in the usual statement means greater as a real number, not larger magnitude.	State explicitly which ordering you are using; if you need magnitude, write (	\kappa_{1}	\ge	\kappa_{2}	).
Using a non‑orthonormal eigenbasis.	The derivation uses (\langle\mathbf e_{i},\mathbf e_{j}\rangle=\delta_{ij}).	Always normalise eigenvectors to unit length before applying the formula.
Ignoring sign of curvature.	Principal curvatures can be negative; the formula works unchanged.	Keep the signs of (\kappa_{1},\kappa_{2}) in the computation; do not replace them by absolute values unless the problem explicitly asks for it.

By keeping these points in mind, the correct application of Euler’s theorem becomes straightforward.

Original question: Normal curvature: does Euler's theorem require an ordering of the principal directions/curvatures? on Mathematics Stack Exchange, licensed CC BY-SA.