Question

Burt Totaro has a result that for a certain class of algebraic stacks, having affine diagonal is equivalent to the stabilizers at closed points begin affine. Is there an example of this equivalence failing in general?

Answer 1

The answer is yes since a fiberwise condition (such as affine stabilizers) does not imply a global condition (such as affine diagonal) without extra hypotheses (such as having the resolution property). Think of quasi-finite+proper <=> finite.

There are (non-separated) schemes with non-affine diagonal, for example, two copies of the affine plane glued together outside the origin.

Also see the related question.

Answer 2

Let $X$ be an algebraic stack. Given a point $f:T\to X$, we define $Stab(f)=X\times_{X\times X}T$, where the map $X\to X\times X$ is the diagonal and $T\to X\times X$ is $(f,f)$. We say that the stabilizer if affine if $Stab(f)\to T$ is an affine morphism.

Since affine morphisms are stable under base extension, it is always true that if the diagonal is affine, then stabilizers are affine.

If I've got everything right so far, then I think I have an argument that shows that if stabilizers are affine, then the diagonal is affine. Let $h:U\to X$ be a smooth cover by an affine scheme, then $Stab(h)$ is affine over $U\times U$ by assumption. But affine morphisms are local on the base in the smooth topology, so the diagonal is affine.

Stab(h) --> U×U
  |          | 
  |  cart    | smooth cover
  v          v
  X ------> X×X

But this uses that stabilizers of scheme-theoretic points are affine. Perhaps in the original question, you're only allowed to assume that stabilizers of geometric points are affine (or something like that).

Edit: Somehow I missed that "closed points" part of the question.