Relation between projective equivalence and linear equivalence of divisors

Question

For the whole question I'll be working in $\mathbb{P}^n_{\mathbb{C}}$ and assume that everything is smooth.

We say that two sets $U,V\subseteq \mathbb{P}^n$ are projectively equivalent if there exists an element $A\in PGL(n)$ such that $A\cdot U=V$.

With these concepts I think it's reasonable to say (define) that two Weyl divisors $D,D'$ are projectively equivalent if there is some automorphism $A$ such that $A\cdot D=D'$, where we define a linear action of $PGL(n)$ in the divisors in each summand, i.e., if $D=\sum a_iU_i$, then $A\cdot D :=\sum a_i (A\cdot U_i)$.

First of all I'd like to know if this makes sense, since I can't find this definition anywhere. In case it does, I'd like to know how is being projectively equivalent related to the divisors being linearly equivalent.

The question can be translated to showing that for any given divisor $D$ there exists and automorphism $A$ such that $(1-A)\cdot D$ is principal.

The motivation for this question is that all hyperplanes in $\mathbb{P}^n$ are linearly equivalent and projectively equivalent under the previous definition, yet both facts seem to be unrelated at a first glance.

Answer 1

As requested, here is a summary of my comments on this "projective equivalence".

I see several problems for this notion to be interesting :

• It is not compatible with the sum of divisors, though you can easily change the definition to have a better behaviour.
• On an arbitrary variety, you need to replace $PGL$ by the automorphism group, and this group can be very small, finite or even trivial. In that case, this notion is not really interesting.
• An irreducible divisor will only be equivalent to an isomorphic subset. This is very restrictive. In particular, a conic will not be equivalent to the union of two lines. But also two cubics can't be equivalent if they are not isomorphic.

Now for the comparison question : this equivalence is not finer nor coarser than the linear equivalence.

• As noted before, a conic and the union of two lines in $\mathbb{P}^2$ are linearly equivalent but they are not "projectively equivalent".
• On an elliptic curve, two distinct points are "automorphically equivalent" but not linearly equivalent.