# String Landscape, De Sitter vacua and Broken Supersymmetry

If we assume that the swampland conjectures, etc. regarding de sitter vacuas existence in the string / F-theory landscape turn out to be incorrect (and therefore we can assume the problem is well-posed), would all such solutions have broken supersymmetry?

There are a number of obvious “difficulties” with formulating (at least for the case of generic qft on curved spacetime) particle theories on de Sitter like spaces (ex. non-trivial hamiltonian from vanishing of globally timelike killing vectors, etc.). One feature of de Sitter space, that is probably obvious but likely not trivial (at least from first glance) is the non-existence of positive conserved energy and its requirement for broken supersymmetry.

Have I made an error, or is this a “just so” theorem regarding de Sitter vacua and the necessity for broken supersymmetry?

• My reasoning for believing susy must be broken is as follows: if susy were not broken (with the positive energy condition in mind) the supercharge would be hermitian and pick up non zero value, and its square would be forced non-zero, and would be a conserved positive (?) bosonic value. This doesn’t make sense (unless I’m missing something) forcing susy to be broken by necessity. – alex sharma Jan 10 '20 at 0:38
• If SUSY was unbroken then we would've detected it. – redhood Apr 8 '20 at 19:23
• – Qmechanic Apr 8 '20 at 19:39
• @Qmechanic thank you, i appreciate the link. – alex sharma Apr 9 '20 at 5:22

You're right. Realistic string compactifications require four macroscopic dimensions. For supersymmetry to exist in a given background a non zero globally defined Killing spinor is required. In the d=4 case such a spinor should be (locally) the generator of the $$Spin(4,1)$$ group, the problem is that $$Spin(4,1)$$ has no Majorana representations (condition required to realize supersymmetry in a unitary way). Nevertheless, non unitary realizations of supersymmetry over de-Sitter space are possible.