Affine arc length 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?
 A: 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
$$ d\sigma = \sqrt{|\alpha + \tfrac{2}{9} \beta^2 - \tfrac{1}{3} \beta'|}\, dt. $$
A: 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.
