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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!

-----Partial Answer-----

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?

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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.

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    $\begingroup$ Hi Ted, thanks for your answer! So it's all because the eigenvectors maximise $\langle v, Av \rangle$ ? Thanks again :D $\endgroup$ – valcofadden Apr 4 at 18:21
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    $\begingroup$ 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. $\endgroup$ – Ted Shifrin Apr 4 at 18:23

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