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Notes on Menger Curvature

F. Leymarie

January 2000 - updated September 2003


Update (Sept. 2003):

The number K(x_1 , x_2 , x_3) = 1/ R , where R is the radius of the circumcircle through the three points x_1 , x_2 , x_3 , is called the Menger curvature of the triple [3].

The derivation of the formula below, then relies on using the well known relation between the area of the triangle, A, through these three points and R:

K=1/R = 4 A / abc , where a, b and c are the length of the sides of the triangle.

Note that Heron's formula gives us:

16 A^2 = (a+b+c) (b+c-a) (a+c-b) (a+b-c)

and this directly give the formula below. An alternative formula is provided by the Cayley-Menger determinant [1], which avoids directly computing the side lengths (as square roots): 16A^2=| a^2(a^2-b^2-c^2)+b^2(b^2-a^2-c^2)+c^2(c^2-a^2-b^2)|.


Let f be a regular curve of class C2 in a Euclidean space, En. Let x1,x2,x3 be distinct points of f. Then, define [1, vol.1, p.273]:

\begin{displaymath} K(x_{1},x_{2},x_{3})=\frac{\sqrt{(x_{1}x_{2}+x_{2}x_{3}+x_{3...  ...}x_{1})}}{x_{1}x_{2}\cdot x_{2}x_{3}\cdot x_{3}x_{1}}\, \, \, ,\end{displaymath}

where the (non-negative) number K is called Menger's curvature. As x2 and x3 approach x1 on f, K(x1,x2,x3) tends towards the curvature of f at x1. Also, K=0 if and only if x1, x2 and x3 are collinear .

In the complex domain, for \( z_{1},z_{2},z_{3}\in C \), this notion is called the Menger-Melnikov curvature [2]:

\begin{displaymath} c(z_{1},z_{2},z_{3})^{2}=\sum _{\sigma }\frac{1}{(z_{\sigma ...  ...igma (3)})\overline{(z_{\sigma (2)}-z_{\sigma (3)})}}\, \, \, ,\end{displaymath}

where the sum is taken over all permutations of \( \sigma \) of \( \{1,2,3\} \).This identity is transformed for 1-sets in En to:

\begin{displaymath} c(x_{1},x_{2},x_{3})^{2}=\sum _{\sigma }\frac{(x_{\sigma (1)...  ...t\vert x_{\sigma (2)}-x_{\sigma (3)}\right\vert ^{2}}\, \, \, ,\end{displaymath}

or equivalently, after some manipulations:

\begin{displaymath} c(x_{1},x_{2},x_{3})^{2}=4\, \left\{ \frac{\left\vert x_{1}-...  ...2}\, \left\vert x_{1}-x_{2}\right\vert ^{2}}\right\} \, \, \, ,\end{displaymath}

which, by Schwartz inequality, can be shown to always be non-negative.



 



Frederic Leymarie
1/26/2000 - 9/5/2003