quantum field theory - Does de Sitter space admit an asymptotic S-matrix?

From the Penrose diagram of de Sitter space, we see it has a future and past conformal boundary, and they are both spacelike. So, does de Sitter space admit an asymptotic S-matrix? Sure, in the usual coordinates which only cover half of the full space, we don't have an S-matrix because it's not causally complete, but in a coordinate system which covers the entire space, why not? Just because we don't have a globally timelike Killing vector field doesn't mean we can't have an S-matrix.