Suppose I have the curve $E/F_{59}: y^2 = x^3+1$ -i.e. it is supersingular. Hence number of points is $\#E(F_q) = 59+1$ and for $r=3$ (i.e $3\ |\ 60$ but $3^2\not|\ 60$), the embeding degree is $k=2$ and construct the extension $F_q /\langle i^2+1\rangle$ (see the picture).
Now since $59 \equiv 2 \pmod{3}$ I choose the following distortion map $\phi(x,y) = (\zeta_3 x,y)$, where $\zeta_3 = 24i+29$ and $\zeta_3^3 = 1$.
The $r$-torsion and the action of the map $\phi$ over it are painted below:
The other maps are the trace map(Tr) and anti-trace map(aTr). Now I have this definition
A distortion map on E is an endomorphism $\phi$ of E such that $\phi(P) \notin \langle P\rangle$.
My question. Why $\phi$ fails to map the red elements out of their subgroup ? Since it is a distortion map it should not do that, right?
It seems that for any $P$ in red subgroup $\phi(P) = P$. Hence the endomorphism (on that subgroup) has an eigenvalue $\lambda = 1$ with the corresponding eigenvectors in the red subgroup.
It seems that distortion maps fail to move out the elements from their subgroup when are defined over their eigenvectors.