# Why do the principle curvatures provide the maximum and minimum values of the normal curvature of a curve?

I am having difficulty showing that the principle curvatures maximise and minimise the normal curvature of a given curve, say $$\alpha(t) \in S$$ where $$S$$ is a regular surface.

Given $$T$$ is the unit tangent of $$\alpha$$ and $$N$$ being the Guass map on $$S$$, the normal curvature is defined as $$k_n:=\langle \dot{T}, N \rangle$$ I want to understand why the eigenvalues of $$-dN$$ (i.e. the Wingarten map) maximise $$k_n$$ above.

Furthermore I'm really curious to see if this is a consequence of some theorem /more general result describing how the eigenvalues/vectors of a matrix optimise various quantities associated with the matrix in question.

If anyone has a spare moment to comment on either of these two questions I'd be very grateful!

Ok after looking at the problem for a little while I realised you can express $$k_n$$ as $$k_n=-\langle T, \dot{N} \rangle=\langle T, -dN(\alpha') \rangle=\left\langle \frac{\alpha'}{|\alpha'|}, -dN(\alpha') \right\rangle$$

If we let $$\alpha$$ have unit speed to simplify the algebra, and express $$-dN$$ in some eigenbasis say $$\{v_1,v_2\}$$ so it is diagonal, we have: $$\alpha'=(\lambda_1,\lambda_2)$$ as written in the $$\{v_1,v_2\}$$ basis.

The expression for normal curvature becomes

$$k_n=\left \langle \begin{pmatrix}\lambda_1 \\ \lambda_2\end{pmatrix}, \begin{pmatrix}k_1 & 0\\ 0 & k_2\end{pmatrix} \begin{pmatrix}\lambda_1 \\ \lambda_2\end{pmatrix}\right \rangle=\begin{pmatrix}\lambda_1 & \lambda_2\end{pmatrix} \begin{pmatrix}k_1 & 0\\ 0 & k_2\end{pmatrix} \begin{pmatrix}\lambda_1 \\ \lambda_2\end{pmatrix}=k_1\lambda_1^2+k_2\lambda_2^2$$ You can check that the above expression has maxima $$k_1$$ and minima $$k_2$$ when $$\lambda_1^2+\lambda_2^2$$ is bounded (as we have assumed because of $$\alpha$$ having unit speed). QED?

Yes, precisely, this is a fact about the eigenvalues (eigenvectors) of a symmetric matrix $$A$$. In particular, say a $$2\times 2$$ symmetric matrix has orthonormal eigenvectors $$e_1,e_2$$ with corresponding eigenvalues $$\lambda_1,\lambda_2$$. Then for a general unit vector $$v=\cos\theta e_1+\sin\theta e_2$$, you will have \begin{align*} Av\cdot v &= \big(A(\cos\theta e_1+\sin\theta e_2)\big)\cdot(\cos\theta e_1+\sin\theta e_2) \\&= (\lambda_1\cos\theta e_1+\lambda_2\sin\theta e_2)\cdot(\cos\theta e_1+\sin\theta e_2) \\ &= \lambda_1\cos^2\theta + \lambda_2\sin^2\theta \end{align*} and you can easily check (using trigonometry or calculus) that $$\lambda_1$$ and $$\lambda_2$$ are the max/min values.
• Hi Ted, thanks for your answer! So it's all because the eigenvectors maximise $\langle v, Av \rangle$ ? Thanks again :D – valcofadden Apr 4 at 18:21
• Maximize/minimize in the 2-D case ... in general, they are the critical points of the function $Av\cdot v$ on the unit sphere. P.S. You might find my differential geometry text, linked in my profile, useful. – Ted Shifrin Apr 4 at 18:23