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Suppose that $v$ and $w$ are orthogonal unit vectors in $T_p\Sigma.$ Show that $\kappa_p(v)+\kappa_p(w)$ is independent of the specific choice of $v$ and $w$ as long as they are orthogonal.

$T_p\Sigma$ is the tangent plane of a surface $\Sigma$ at a point $p$.

How can I prove this? Do I have to use the second fundamental form to show this or is there another straight forward way?

If $v$ and $w$ are orthogonal unit vectors of $T_p\Sigma$ then wouldn't $\kappa_p(v)$ just be the normal curvature? If it is the normal curvature then we can define it by taking an arc-length parametrized curve $\alpha$ in $\Sigma$ such that $\alpha(0)=p$ and $\alpha'(0)=v$ which passes through the point $p$ and has $v$ as its tangent vector at the point $p$. But then wouldn't the value of the normal curvature of $\alpha$ depend only on the unit tangent vector $v$?

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Hint: Look at Euler's formula relating the normal curvature in direction $v$ to the principal curvatures.

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  • $\begingroup$ Ted, in order to show this do I have to use the second fundamental form? $\endgroup$ – Lays Oct 15 '13 at 6:28
  • $\begingroup$ Well, since the second fundamental form is what gives you normal curvature in the direction of $v\in T_pM$, yes. Or, as @John suggested, you need the associated linear map often called the Weingarten map from $T_pM$ to itself. $\endgroup$ – Ted Shifrin Oct 15 '13 at 11:07
  • $\begingroup$ Alright thanks. I think I know what to do now. $\endgroup$ – Lays Oct 15 '13 at 23:51
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Hint: By some orthogonal transformation you may assume that $\Sigma$ is the surface given by a function $f$ (that is, $\Sigma = (x, y, f(x, y))$), $p = 0$ and $T_p\Sigma$ is the $(x, y)$ plane. You can check directly that $k_1 +k_2$ is independent of unit vectors chosen. Indeed $k_1 + k_2$ can be considered as trace of a linear map, so does not depends on the choices of the vectors.

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Here is a solution for future viewers:

If the principal curvatures are the same at $p$ then all the normal curvatures at $p$ are the same. If the principal curvatures are different then we know there are two unit orthonormal eigenvectors $e_1$ and $e_2$ for the shape operators $S_p$ corresponding to the principal curvatures (which are also eigenvalues of $S_p$) $\kappa_1$ and $\kappa_2.$

Let $v$ and $w$ be two orthogonal unit vectors in $T_p\Sigma.$ We know that there is some $\theta$ such that $v = (\cos\theta)e_1 + (\sin\theta)e_2.$ Since $w$ is orthogonal to $v$ then $$w = \left(cos(\theta\pm \frac{\pi}{2})\right) e_1 + \left(sin(\theta\pm\frac{\pi}{2})\right)e_2.$$ Now we see that $\kappa_p(v)=(\cos^2\theta)\kappa_1 +(\sin^2\theta)\kappa_2.$ Thus $\kappa_p(w)=(\sin^2\theta)\kappa_1 +(\cos^2\theta)\kappa_2.$ Therefore $\kappa_p(v)+\kappa_p(w) = (\sin^2\theta +\cos^2\theta)\kappa_1 +(\sin^2\theta +\cos^2\theta)\kappa_2 = \kappa_1 +\kappa_2$ so we see that $\kappa_p(v)+\kappa_p(w) = \kappa_1 +\kappa_2$ independent of which the two orthonormal vectors we used.

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