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The curve C(𝜃) drawn on a smoothly embedded surface 𝜮 in 3-space — where C(𝜃) is defined as the intersection of 𝜮 with a 2-plane perpendicular to 𝜮 at P — leaving the point P at angle 𝜃 will have the well-known formula for curvature:

k(𝜃) = k1 cos2(𝜃) + k2 sin2(𝜃)

at P, where k1 and k2 are the principal curvatures of 𝜮 at P.

Likewise, we can look at a point P of a smoothly embedded surface 𝜮 in Euclidean n-space, and consider the curves C(𝜃) that are each the intersection of 𝜮 with a hyperplane that is perpendicular to 𝜮 at P and ask what their curvatures are as a function of angle 𝜃.

Does there exist a formula (like k(𝜃) = k1 cos2(𝜃) + k2 sin2(𝜃), or different) for the curvatures k(𝜃) of the curves C(𝜃) in this case?

And what about the special case where 𝜮 is a Riemann surface holomorphically embedded in Cm = R2m?

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  • $\begingroup$ You might want to look at the second fundamental form, which becomes vector valued in codimension more than 1 $\endgroup$
    – alesia
    Commented Feb 22, 2023 at 0:59
  • $\begingroup$ I certainly expect the answer is essentially some symmetric combination of second derivatives. $\endgroup$ Commented Feb 22, 2023 at 1:19
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    $\begingroup$ Since hyperplanes are totally geodesic: if you denote by $II_p(v,w)$ the vector valued second fundamental form of $\Sigma$ at the point $p$, and $v$ the tangent vector to $C$, then the curvature vector (the acceleration of $C$ with unit-speed parametrization) is exactly $II_p(v,v)$. You don't always as nice a formula as $k_1\cos^2(\theta) + k_2\sin^2(\theta)$, since the different (vector) components of $k$ may not be simultaneously diagonalizable (so what you call $\theta$ may be not well-defined). $\endgroup$ Commented Feb 22, 2023 at 5:12
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    $\begingroup$ For that matter, the principal extrinsic curvatures are also not always well-defined. The same problem should still persist in the holomorphic setting. $\endgroup$ Commented Feb 22, 2023 at 5:19
  • $\begingroup$ At any given point P of the surface 𝜮: What I call 𝜃 is always well-defined (relative to a choice of reference angle). After all, the surface is embedded, so it has a well-defined 2-plane T in n-space as its tangent plane, as well as a well-defined normal (n-2)-space N. So the sum of N and any tangent line to 𝜮 at P is a hyperplane perpendicular to T, intersecting it in a line. The angle between any two such lines is well-defined, and there are 2π worth of tangent rays. $\endgroup$ Commented Feb 23, 2023 at 5:23

1 Answer 1

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Let $u$ be a unit tangent vector at $p$. The curvature $k$ in the direction of $u$ is $|\mathrm{I\!I}_p(u,u)|$, where $\mathrm{I\!I}_p$ is the second fundamental form at $p$; it is a quadratic form on tangent plane $\mathrm{T}_p$ with values in the normal space $\mathrm{N}_p$.

We may write $\mathrm{I\!I}_p$ in a basis $e_1,\dots,e_{n-2}$ of $\mathrm{N}_p$, $$\mathrm{I\!I}_p=s_1\cdot e_1+\dots+s_{n-2}\cdot e_{n-2};$$ so, $$k=\sqrt{s_1(u,u)^2+\dots+s_{n-2}(u,u)^2}.$$

Each form $s_i$ is diagonalizable, denote by $a_i$ and $b_i$ its eigenvalues; let $\phi_i$ be the angle to the first eigenvector. Then we get $$k=\sqrt{\sum_i(a_i\cdot \cos(\theta-\phi_i)^2+b_i\cdot \sin(\theta-\phi_i)^2)^2}.$$ (I do not think you need it, but that is what you asked for.)

By the way, the image of $\mathrm{I\!I}_p$ is at most 3-dimensional subspace of $\mathrm{N}_p$. So you may reduce number of terms to three.

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