5
$\begingroup$

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 projectively equivalent.

$\endgroup$
3
  • 3
    $\begingroup$ Your last sentence is inconsistent with the title of the question; I guess "isomorphic" should be replaced by "projectively equivalent". $\endgroup$
    – user5117
    Commented Jul 8, 2013 at 16:53
  • $\begingroup$ 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. $\endgroup$
    – Ben McKay
    Commented Apr 20, 2015 at 11:03
  • $\begingroup$ I guess that "variety" is a mistranslation of French "variété" (which both translates in the language used on this site as "variety" and "manifold"), or from another language with the same terminology collapse, and that the OP means "projective manifold". This would be more in keeping with @BenMcKay's new answer. Anyway, the question is way too vague and the OP, at least under his/her posting name, has vanished from MO from the very minute he/she asked this question, so we won't expect any clarification. $\endgroup$
    – YCor
    Commented Mar 1, 2018 at 13:20

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Projective invariants, especially local ones, are often discussed in depth in the work of J.M. Landsberg. A particularly nice introduction constitutes chapter 3 of the book T. Ivey and J.M. Landsberg, Cartan for Beginners. You might also look at Wilcysnki, Projective Differential Geometry of Curves and Ruled Surfaces to see the local projective differential invariants of curves.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .