Upper and Lower Bounds and Modulus of Continuity of Decomposed Möbius Energies

Abstract

The Möbius energy is one of the knot energies, and is named after its Möbius invariant property. It is known to have several different expressions. One is in terms of the cosine of conformal angle, and is called the cosine formula. Another is the decomposition into Möbius invariant parts, called the decomposed Möbius energies. Hence the cosine formula is the sum of the decomposed energies. This raises a question. Can each of the decomposed energies be estimated by the cosine formula? Here we give an affirmative answer: the upper and lower bounds, and modulus of continuity of decomposed parts can be evaluated in terms of the cosine formula. In addition, we provide estimates of the difference in decomposed energies between the two curves in terms of Möbius invariant quantities.

Appendix

We derive the expression

$$\begin{aligned} \cos \varphi ( s_1 , s_2 ) = \frac{1}{2} \Vert \Delta \mathbf{f} \Vert ^2 \frac{ \partial ^2 }{ \partial s_1 \partial s_2 } \log \Vert \Delta \mathbf{f} \Vert ^2 \end{aligned}$$

for the conformal angle $\varphi ( s_1 , s_2 )$. Let $C_{12}$ be the circle passing through two points $\mathbf {f} (s_1)$ and $\mathbf {f} (s_2)$ whose tangent line at $\mathbf {f} (s_1)$ coincides with that of $\mathrm {Im} \mathbf {f}$. And, let r and $\mathbf{c}$ be the radius and center of $C_{12}$, respectively. When $\{ {\varvec{\tau }}( s_1 ) , \Delta \mathbf{f} \}$ is linearly dependent, we interpret $C_{12}$ as the line passing through $\mathbf{f} ( s_1 )$ and $\mathbf{f} ( s_2 )$.

Case 1 $\dim \mathrm {span} \{ {\varvec{\tau }}( s_1 ) , \Delta \mathbf{f} \} = 2$. Since $\Delta \mathbf{f}$ is not parallel to ${\varvec{\tau }}( s_1 )$ in the case,

$$\begin{aligned} \Vert \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) \Vert \ne 0 . \end{aligned}$$

We set

$$\begin{aligned} \mathbf{e} = - \frac{ ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) }{ \Vert \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) \Vert } . \end{aligned}$$

Then

$$\begin{aligned} \mathbf{c} = \mathbf{f} ( s_1 ) + r \mathbf{e} , \end{aligned}$$

and $C_{12}$ is represented as

$$\begin{aligned} \mathbf{x} ( \theta ) - \mathbf{c} = \left( \cos \frac{\theta }{r} \right) ( \mathbf{f} ( s_1 ) - \mathbf{c} ) + r \left( \sin \frac{\theta }{r} \right) {\varvec{\tau }}( s_1 ) , \end{aligned}$$

where $\theta \in {\mathbb {R}} / 2 \pi {\mathbb {Z}}$ is the arc-length parameter of $C_{12}$. Since $\mathbf{c} = \mathbf{f} ( s_1 ) + r \mathbf{e}$,

$$\begin{aligned} \mathbf{x} ( \theta ) =&\ \mathbf{f} ( s_1 ) + r \mathbf{e} - r \left( \cos \frac{\theta }{r} \right) \mathbf{e} + r \left( \sin \frac{\theta }{r} \right) {\varvec{\tau }}( s_1 ) , \\ {\dot{\mathbf{x }}} ( \theta ) =&\ \left( \sin \frac{\theta }{r} \right) \mathbf{e} + \left( \cos \frac{\theta }{r} \right) {\varvec{\tau }}( s_1 ) . \end{aligned}$$

Since $C_{12}$ passes through $\mathbf{f} ( s_2 )$, there exists $\theta _*\in {\mathbb {R}} / 2 \pi {\mathbb {Z}}$ such that $\mathbf{x} ( \theta _*) = \mathbf{f} ( s_2 )$. Hence

$$\begin{aligned} \mathbf{f} ( s_2 ) = \mathbf{f} ( s_1 ) + r \mathbf{e} - r \left( \cos \frac{ \theta _*}{r} \right) \mathbf{e} + r \left( \sin \frac{ \theta _*}{r} \right) {\varvec{\tau }}( s_1 ) , \end{aligned}$$

and therefore

$$\begin{aligned} r \mathbf{e} = - \Delta \mathbf{f} + r \left( \cos \frac{ \theta _*}{r} \right) \mathbf{e} - r \left( \sin \frac{ \theta _*}{r} \right) {\varvec{\tau }}( s_1 ) . \end{aligned}$$

Taking the inner product with $\mathbf{e}$ and with ${\varvec{\tau }}( s_1 )$, we have

$$\begin{aligned} r =&\ - \Delta \mathbf{f} \cdot \mathbf{e} + r \cos \frac{ \theta _*}{r} , \\ 0 =&\ - \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) - r \sin \frac{ \theta _*}{r} . \end{aligned}$$

It follows from these that

$$\begin{aligned} r^2 =&\ r^2 \left( \cos ^2 \frac{ \theta _*}{r} + \sin ^2 \frac{ \theta _*}{r} \right) \\ =&\ \left( r + \Delta \mathbf{f} \cdot \mathbf{e} \right) ^2 + \left( - \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) \right) ^2 \\ =&\ \left( \Delta \mathbf{f} \cdot \mathbf{e} \right) ^2 + 2r \Delta \mathbf{f} \cdot \mathbf{e} + r^2 + \left( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) \right) ^2 \\ =&\ \Vert \Delta \mathbf{f} \Vert ^2 + 2r \Delta \mathbf{f} \cdot \mathbf{e} + r^2 . \end{aligned}$$

Note that $\Delta \mathbf{f} \cdot \mathbf{e} \ne 0$ when $\dim \mathrm {span} \{ \Delta \mathbf{f} , {\varvec{\tau }}( s_1 ) \} = 2$. Therefore we obtain

$$\begin{aligned} r = - \frac{ \Vert \Delta \mathbf{f} \Vert ^2 }{ 2 \Delta \mathbf{f} \cdot \mathbf{e} } . \end{aligned}$$

Now we calculate the cosine of the conformal angle:

$$\begin{aligned} \cos \varphi =&\ {\dot{\mathbf{x }}} ( \theta _*) \cdot {\varvec{\tau }}( s_2 ) \\ =&\ \left\{ \left( \sin \frac{ \theta _*}{r} \right) \mathbf{e} + \left( \cos \frac{ \theta _*}{r} \right) {\varvec{\tau }}( s_1 ) \right\} \cdot {\varvec{\tau }}( s_2 ) \\ =&\ - \frac{ \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) }{r} \mathbf{e} \cdot {\varvec{\tau }}( s_1 ) + \left( 1 + \frac{ \Delta \mathbf{f} \cdot \mathbf{e} }{r} \right) {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) \\ =&\frac{ 2 ( \Delta \mathbf{f} \cdot \mathbf{e} ) ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) }{ \Vert \Delta \mathbf{f} \Vert ^2 } ( \mathbf{e} \cdot {\varvec{\tau }}( s_2 ) ) + \left\{ 1 - \frac{ 2 ( \Delta \mathbf{f} \cdot \mathbf{e} )^2 }{ \Vert \Delta \mathbf{f} \Vert ^2 } \right\} {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) . \end{aligned}$$

Here

$$\begin{aligned} \Delta \mathbf{f} \cdot \mathbf{e} =&\ \Delta \mathbf{f} \cdot \left\{ - \frac{ \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) }{ \Vert \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) \Vert } \right\} \\ =&\ - \frac{ \Vert \Delta \mathbf{f} \Vert ^2 - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) )^2 }{ \Vert \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) \Vert } \\ =&\ - \Vert \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) \Vert ,\\ \mathbf{e} \cdot {\varvec{\tau }}( s_2 ) =&\ - \frac{ \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) }{ \Vert \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) \Vert } \cdot {\varvec{\tau }}( s_2 ) \\ =&\ - \frac{ ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_2 ) ) - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) ( {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) ) }{ \Vert \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) \Vert } . \end{aligned}$$

Consequently, we have

$$\begin{aligned} \cos \varphi =&\ \frac{ 2 ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) \{ ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_2 ) ) - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) ( {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) ) \} }{ \Vert \Delta \mathbf{f} \Vert ^2 } \\&\quad + \, \left[ 1 - \frac{ 2 \Vert \Delta \mathbf{f} - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) {\varvec{\tau }}( s_1 ) \Vert ^2 }{ \Vert \Delta \mathbf{f} \Vert ^2 } \right] {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) \\ =&\ \frac{ 2 ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_2 ) ) }{ \Vert \Delta \mathbf{f} \Vert ^2 } - \frac{ 2 ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) )^2 ( {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) ) }{ \Vert \Delta \mathbf{f} \Vert ^2 } \\&\quad + \, \left[ 1 - \frac{ 2 \{ \Vert \Delta \mathbf{f} \Vert ^2 - ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) )^2 \} }{ \Vert \Delta \mathbf{f} \Vert ^2 } \right] ( {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) ) \\ =&\ \frac{ 2 ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_2 ) ) }{ \Vert \Delta \mathbf{f} \Vert ^2 } - ( {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) ) \\ =&\ \frac{1}{2} \Vert \Delta \mathbf{f} \Vert ^2 \frac{ \partial ^2 }{ \partial s_1 \partial s_2 } \log \Vert \Delta \mathbf{f} \Vert ^2 . \end{aligned}$$

Case 2 $\dim \mathrm {span} \{ {\varvec{\tau }}( s_1 ) , \Delta \mathbf{f} \} = 1$. In this case $C_{12}$ can be interpreted as the line passing through $\mathbf{f} ( s_1 )$ and $\mathbf{f} ( s_2 )$. Therefore

$$\begin{aligned} \cos \varphi = {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) . \end{aligned}$$

Also we have

$$\begin{aligned} \Delta \mathbf{f} = \pm \Vert \Delta \mathbf{f} \Vert {\varvec{\tau }}( s_1 ) . \end{aligned}$$

Hence

$$\begin{aligned}&\frac{1}{2} \Vert \Delta \mathbf{f} \Vert ^2 \frac{ \partial ^2 }{ \partial s_1 \partial s_2 } \log \Vert \Delta \mathbf{f} \Vert ^2 \\&\quad = \frac{ 2 ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_1 ) ) ( \Delta \mathbf{f} \cdot {\varvec{\tau }}( s_2 ) ) }{ \Vert \Delta \mathbf{f} \Vert ^2 } - ( {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) ) \\&\quad = \frac{ 2 ( \pm \Vert \Delta \mathbf{f} \Vert ) ( \pm \Vert \Delta \mathbf{f} \Vert {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) ) }{ \Vert \Delta \mathbf{f} \Vert ^2 } - ( {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) ) \\&\quad = {\varvec{\tau }}( s_1 ) \cdot {\varvec{\tau }}( s_2 ) = \cos \varphi . \end{aligned}$$