Affine arc length

Question

I was looking for an analog of arc length for plane curves in affine geometry, but I have only found the equi-affine arc length $d\sigma ={ || \gamma '(t)\wedge \gamma ''(t) ||}^{1 \over 3}dt$. On Wikipedia there is something, but it is not invariant under the action of the affine group. Any suggestion? And why everyone cares only about equi-affine geometry?

Answer 1

Marco,

I'm not aware of any notion of (non-equi-)affine arc length defined in the the literature - but the one you've constructed above, i.e., $$d\sigma = \sqrt{\alpha + \tfrac{2}{9} \beta^2 - \tfrac{1}{3} \beta'}\, dt, $$ looks pretty natural to me. I did a quick Maple calculation and verified that it is, in fact, invariant under arbitrary reparametrizations of the curve, so it seems as good a definition as any.

As to your point about the sign - what's really invariant is the quadratic form $$ Q = \left(\alpha + \tfrac{2}{9} \beta^2 - \tfrac{1}{3} \beta' \right) dt^2. $$ Of course, one could just as well have chosen the invariant quadratic form $-Q$, and then it would seem equally natural to define $$d\sigma = \sqrt{-\left(\alpha + \tfrac{2}{9} \beta^2 - \tfrac{1}{3} \beta'\right)}\, dt. $$ Personally, my inclination would be to call the curve "nondegenerate" (or some other word meaning "nice") if the quadratic form $Q$ never vanishes, and then define the affine arc length element to be $$ d\sigma = \sqrt{|\alpha + \tfrac{2}{9} \beta^2 - \tfrac{1}{3} \beta'|}\, dt. $$

Answer 2

I write two words about what I've found.

I followed the method exposed in La théorie des groupes finis et continus et la géométrie différentielle traitées par la méthode du repère mobile by E. Cartan. I found this book in my university's library.

I started with two frames, both with $(e_1, e_2)$ linearly indipendent and $e_1$ parallel to the tangent line. The change of frame is given by

$ e_1 = \lambda\bar {e}_1 \\ e_2 = \mu \bar {e}_1 + \eta \bar e_2$

which depends on 3 secondary parameters $(\lambda, \mu, \eta)$, with $\lambda \eta \neq 0$. I got

$ \omega ^1 = {1 \over \lambda} \bar \omega^1 \\ \omega^2 = \bar \omega ^2 = 0$

These are the two principal components of order zero. Then I got

$\omega_{1}^2 = {\lambda \over \eta} \bar \omega_1^2$

This is the principal component of first order. I chose $\omega_1^2 = \omega^1$ and I got

$\omega^1 = {\lambda \over \eta} \bar \omega^1 = {\lambda^2 \over \eta} \omega^1$

So $\eta = \lambda ^ 2$. Then I got

$\omega_1^1 = \bar \omega_1^1 - {\mu \over \lambda^2}\bar \omega_1^2 + {1 \over \lambda}d\lambda \\ \omega_2^2 = \bar \omega_2^2 + {\mu \over \lambda^2}\bar \omega_1^2 + {2 \over \lambda}d\lambda$

So the principal component of second order is $\omega_2^2 - 2\omega_1^1$. I chose $\omega_2^2 - 2\omega_1^1 = 0$. I got

$ 0 = 3{\mu \over \lambda^2}\bar \omega_1^2$

So $\mu = 0$. Then I got

$\omega_2^1 = \lambda \omega_2^1$

which is the principal component of third order. I chose $\omega_2^1 = \omega^1$. I got

$\omega^1 = \lambda \bar \omega^1 = \lambda^2 \omega^1$

So $\lambda^2 = 1$ and I chose $\lambda = 1$.

I have no more secondary parameters, so this is the Frenet frame. I have two invariants: $\omega^1$ and $\omega_1^1$. I defined $d\sigma$ and $k$ by $d\sigma = \omega^1$ and $\omega_1^1 = k\omega^1$. The Frenet formulas are

$dP = e_1d\sigma \\ de_1 = (ke_1 + e_2)d\sigma \\ de_2 = (e_1 + 2ke_2)d\sigma$

To obtain explicit formulas I considered $\bar e_1 = P'$ and $\bar e_2 = P''$. I got

$\bar \omega_1 = dt \\ \bar \omega_2 = 0 \\ \bar \omega_1^1 = 0 \\ \bar \omega_1^2 = dt \\ \bar \omega_2^1 = \alpha dt\\ \bar \omega_2^2 = \beta dt\\$

where $P''' = \alpha P' + \beta P''$. Considering $\omega_1^2 = \omega_2^1 = \omega^1$ and $\omega_2^2 = 2\omega_1^1$ I got

$e_1 = \lambda P' \\ e_2 = -{1 \over 3}\beta \lambda^2 P' + \lambda^2 P''$

where ${1 \over \lambda} = \sqrt{\alpha + {2 \over 9}\beta^2 - {1 \over 3}\beta '}$. Eventually I got

$d\sigma = {1 \over \lambda} dt \\ k = \lambda ' + {1 \over 3} \beta \lambda$

I hope there are not too many errors.