Consider the following two smooth projective, complex plane curves
C: x^d+y^d+z^d+(ax+by+cz)^d=0 and C': x^d+y^d+z^d+(a'x+b'y)^d=0
where a,b,c,a',b' are small nonzero complex numbers. Can you see a geometric reason why C and C' are not projectively equivalent ?
Not sure if it will help since I believe you may be aware of the following. However it makes a point on geometrical invariants that are different for both curves.
For two projectively equivalent curves we will have that their associated covariants will be projective equivalent as well. However, in the case of C and C' we have that their Hessian curves have distinct geometric behaviour. Indeed the Hessian curve of C' is always completely reducible but the Hessian curve of C may be irreducible (e.g. a=b=c) .
Dear Alan, thank you very much for your answer !
A more difficult question would be, assuming that C and C' are both smooth, to decide when they are isomorphic.
For sure! I started thinking about the family x^d+y^d+z^d +L^d parameterized by linear forms L, but I did not find any clue. I think such a isomorphism would come from the automorphism group of the Fermat curve and this is finite, order 6d^2, hence only very special pairs L and L' would give isomorphic curves.
Have a look at http://www.ag.jku.at/pubs/2018hj.pdf
https://math.stackexchange.com/questions/3282385/equivalence-between-function-fields-and-curves
- Badges
- Science topic
- Similar topics
- Geoscience
- Asia
- India