# Frenet Frame

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1. Mar 2, 2017

### MxwllsPersuasns

1. The problem statement, all variables and given/known data

The Frenet frame of a curve in R 3 . For a regular plane curve (and more generally for a regular curve on a 2-dimensional surface - e.g. the 2-sphere above) we could construct a unique adapted frame F. This is not the case for curves in higher dimensional spaces. Besides the curve being regular we need more conditions to ensure the existence of a unique adapted frame, which then will give invariants of the curve, which in turn reconstruct the curve up to Euclidean motions. Let γ : I → R 3 be an arclength parametrized curve. Then T = γ' has unit length.

(i) Show that < γ'', T >= 0. Thus, provided that γ'' is nowhere vanishing, we can define N := γ''/||γ''|| and obtain a moving basis {T, N, T × N}. A regular space curve for which γ'' is nowhere vanishing is called a Frenet curve.

(ii) Let γ be an arclength parametrized Frenet curve. Define the curvature function to be κ := ||γ 00|| > 0 and the torsion function τ :=< T ×N, N' >. Show that the adapted frame F = (T, N, T × N): I → SO(3, R) and calculate A = F-1F' in terms of κ and τ .

(iii) If γ is am arclength parametrized plane curve, we can regard it as a space curve. Show that this space curve has τ ≡ 0. Also prove the converse: if a space curve has τ ≡ 0 then it lies in a plane in R3 .

(iv) Show that given κ: I → R, τ (t) > 0 for all t ∈ I, and τ : I → R smooth, there exists a unique (up to Euclidean motion) Frenet curve in R3 whose curvature and torsion are κ and τ respectively.

(v) Classify the Frenet space curves which have curvature and torsion constant

2. Relevant equations

3. The attempt at a solution

Part i) So I defined γ' as (x', y', z')T and γ'' as (x'', y'', z'')T and when I take the dot product i get {x''x' + y''y' + z''z'}. Nothing jumps out at me as to how I can argue this equals 0, though I have a few inclinations...

So I'm thinking that the dot product of two orthogonal vectors is always 0 and thus γ' and γ'' must be orthogonal though I'm not sure how exactly i can show that as it's not a general fact that the derivative of a curve is orthogonal to that curve (its supposed to be tangent at a particular point). The only other thing I can think of is using the fact that gamma is arc length parameterized but I cant immediately see how that leads to 0. Any help is greatly appreciated.

Part ii) So basically this part asks us to construct the 3x3 matrix with column vectors {T, N, T x N} and then show that this matrix is part of the Rotation group (Special Orthogonal in 3D) by i) showing that the matrix times its transpose equal 1 and then that the determinant of the matrix is 1. Correct?

Part iii) For this part I imagine we need to look at the def of torsion < T x N, N'> and notice that either Tx N or N' must be identically zero given that we've flattened down into the plane. Thought I can't quite imagine which one it is.

That's probably enough for now.. Really hoping for some interesting insights and help on this tough problem. Thanks guys/gals!

Last edited: Mar 2, 2017