Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A projective transformation is a morphism of $P^n$ to $P^n$, for some $n$, determined by an $(n + 1) \times (n + 1)$ invertible matrix $A$ in the obvious way. The sets $Q$, $R$ are projectively equivalent if and only if there exists a projective transformation $f$ such that $f(Q) = R$.

I would like to know useful criteria to determine if two projective varieties are isomorphic or not.

share|improve this question
1  
Your last sentence is inconsistent with the title of the question; I guess "isomorphic" should be replaced by "projectively equivalent". –  Artie Prendergast-Smith Jul 8 '13 at 16:53
    
You might look at Wilcysnki's book Projective Differential Geometry of Curves and Ruled Surfaces to get a sense of how complicated the projective invariants of curves can get. But Wilcynski is only thinking about local geometry of $C^{\infty}$ immersed curves in real projective space, so a very different story globally. –  Ben McKay Apr 20 at 11:03

1 Answer 1

Equivalence of projective varieties does imply that the varieties are isomorphic. But the converse does not hold. For example $\mathbb{P}^1$ is isomorphic to the conic defined by $(xy - z^2)$ in $\mathbb{P}^2$ but there is clearly no way that these are projectively equivalent because the degrees do not match. I'm not sure if that answers your question completely but I think it might be helpful to people who stumble onto this page.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.