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Given $k_o,\tau_o: I \to \mathbb R, \mathcal C^\infty, k_o>0,$ there exists a curve $\alpha:I\to \mathbb R^3$ parameterized by arc length, such that $k(s)=k_o(s)$ and $\tau(s)=\tau_o(s)$ that is unique up to direct isometries of $\mathbb R^3.$

Consider $$x'(s) = A_o(s) x(s)\tag{*}$$ with

$$A_o=\begin{bmatrix} \mathcal O_3 & k_o (s)\bf I_3 & \mathcal O_3\\ -k_o(s)\bf I_3 & \mathcal O_3 & -\tau_o(s)\bf I_3\\ \mathcal O_3 & \tau_o (s)\bf I_3 & \mathcal O_3 \end{bmatrix}$$

corresponding to a $9\times 9$ block matrix. We choose $\bf a\in \mathbb R^9,$ such that

$$\bf a=\begin{bmatrix}\bf t_o,\bf m_o,\bf b_o\end{bmatrix}^\top=\begin{bmatrix}(a_1,\dots,a_3),(a_4,\dots,a_6),(a_7,\dots,a_9)\end{bmatrix}^\top$$ form an oriented orthonormal basis of $\mathbb R^3.$ We define $f:I\to \mathbb R^9$ to be a solution of $(*)$ as ${\bf f}=[f_1,f_2,f_3]^\top,$ ${\bf m}=[f_4,f_5,f_6]^\top$ and ${\bf b}=[f_7,f_8,f_9]^\top,$ functions of $s,$ with initial values $\bf a.$

The proof proceeds by proving that $\bf f, m$ and $\bf b$ forms a positive orthonomal basis of $\mathbb R^3$ for any $s$ by looking at the matrix

$$M(s)= \begin{bmatrix} \langle \bf t, t \rangle & \langle \bf t, n \rangle & \langle \bf t, b \rangle \\ \langle \bf t, n \rangle & \langle \bf n, n \rangle & \langle \bf n, b \rangle \\ \langle \bf t, b \rangle & \langle \bf n, b \rangle & \langle \bf b, b \rangle \end{bmatrix}= \bf I[3\times 3]$$

And after all this "it is a simple computation"...

$$2M'(s) = A(s) M(s) - M(s)A(s),$$ where

$$A=\begin{bmatrix} 0 & k_o (s) & O\\ -k_o(s) & O & -\tau_o(s)\\ O & \tau_o (s) & O \end{bmatrix}$$

Can I get a hint as how this simple computation would take shape?

amWhy
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Antoni Parellada
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1 Answers1

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First note that $$ M'(s) = \begin{bmatrix} 2\langle \mathbb{t}',\mathbb{t}\rangle & \langle \mathbb{t}',\mathbb{n}\rangle + \langle \mathbb{t},\mathbb{n}'\rangle & \langle \mathbb{t}',\mathbb{b}\rangle+\langle \mathbb{t},\mathbb{b}'\rangle\\ \langle \mathbb{t}',\mathbb{n}\rangle + \langle \mathbb{t},\mathbb{n}'\rangle & 2\langle \mathbb{n}',\mathbb{n}\rangle & \langle \mathbb{n}',\mathbb{b}\rangle + \langle \mathbb{n},\mathbb{b}'\rangle\\ \langle \mathbb{t}',\mathbb{b}\rangle + \langle \mathbb{t},\mathbb{b}'\rangle & \langle \mathbb{n}',\mathbb{b}\rangle + \langle \mathbb{n},\mathbb{b}'\rangle & 2\langle \mathbb{b}',\mathbb{b}\rangle \end{bmatrix}, $$ then use the Frenet formulas \begin{align*} \mathbb{t}' &= k_0(s)\mathbb{n},\\ \mathbb{n}' &= -k_0(s)\mathbb{t} + \tau_0(s)\mathbb{b},\\ \mathbb{b}' &= -\tau_0(s)\mathbb{n} \end{align*} to write $M'(s)$ in terms of $\mathbb{t}$, $\mathbb{n}$ and $\mathbb{b}$ (and not of their derivatives). By the other hand, compute explicitely $A(s)M(s)-M(s)A(s)$ and compare. It is a little bit tedious, but it is a simple calculation.

Albert
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