Getting ovals from hyperovals

Question

It is said that if we are given a hyperoval we can determine possible ovals by finding the stabilizer of the hyperoval. As an example take a translation hyperoval $D(2)=\{(1,t,f(t));t\in GF(4)\}\cup\{(0,1,0),(0,0,1)\}$ , $f(t)=t^2$. Thus the points of hyperovals are $\{(1,0,0), (1,1,1), (0,1,0), (0,0,1), (1,a,a+1), (1,a+1,a)\}$ where $a$ is the root of the polynomial $x^2+x+1=0$. Now, How can I find the stabilizers and then different ovals? I know that by adding intersection of tangents to an oval we can get a hyperoval, but the reverse is complicated. Thanks in advance for responses.

Answer 1

Given a hyperoval, you can obtain an oval by removing any point; it is not complicated at all, removal of any arbitrary point will give you an oval. If you want to look at the possible ovals you can obtain up to isomorphism, then yes, you can look at the stabilizer of the hyperoval; removing two different points from the same orbit under this stabilizer will give isomorphic ovals. Removing points from different orbits will give ovals that might be nonisomorphic.

You would usually compute the stabilizer computationally, using a program like Magma; this would also let you compute the orbits. Then, for each orbit, you remove an arbitrary point from that orbit to obtain an oval. This will give you one oval for each point orbit. They are not guaranteed to be nonisomorphic, so you still have to check this as a last step. In $\mathrm{PG}(2,4)$, the only hyperoval is a hyperconic, and it's stabilizer is transitive on the points. Thus you will get a unique oval up to isomorphism by removing whichever point you want.