What does $δF=0$ imply?

Please have a look here section 12.6. It says here that

Given the action of a supergravity theory, it is generally useful to search for solutions of the classical equations of motion. It is most useful to obtain solutions that can be interpreted as backgrounds or vacua. Fluctuations above the background are then treated quantum mechanically. The backgrounds that are considered have vanishing values of fermions, and are thus determined by a value of the metric, the vector fields (or higher forms) and scalar fields. One common background is Minkowski space, but there are others such as anti-de Sitter space, certain black holes, cosmic strings, branes or pp-waves, which are all supersymmetric, i.e. they ‘preserve some supersymmetry’. This means that the background is invariant under a subset of the local supersymmetries of the supergravity theory. For a preserved supersymmetry, the local SUSY variations of all fields must vanish when the background solution is substituted. This leads to conditions of the generic form $$δ(\epsilon) \text{boson} = \text{fermion} = 0, \hspace{.5cm} δ(\epsilon)\text{fermion} = \text{boson} = 0. \tag{12.15}$$

I discussed this a little with ACuriousMind in this thread. My question was if supersymmetric background means that it preserves some supersymmetry, then as the text above says, the SUSY variations (12.15) must vanish. This results in for example that $$\delta \text{fermion}=0.$$ So, what does this imply? Does it imply that for a background to be supersymmetric, then fermions must not transform into bosons but rather stay fermions? I must be confused about this because this doesn't make much sense now does it?

• physics.stackexchange.com/q/226231 Dec 28, 2015 at 18:51
• sorry @Rexcirus? can you elaborate what you meant by that posting? Dec 28, 2015 at 18:54

If a particular background is supersymmetric then we have smth like $$\delta\lambda=B_1,\quad \delta \psi_M=B_2,\quad ...$$ where $B_{1,2}$ are some combinations of the bosonic fields, and left hand sides are variations of the fermionic fields.

Now if we remove all the fermions, the background still exists, but we get some restrictions on the combinations of the bosonic fields: $$B_1=0,\quad B_2=0,\quad ...$$ These are SUSY equatons.

• Thanks for your answer. I just have to say two things here 1) My question was: "Does it imply that for a background to be supersymmetric, then fermions must not transform into bosons but rather stay fermions?" So if you could please elaborate on this. Dec 29, 2015 at 2:27
• 2) If you check ACuriousMind's answer on the other thread that I have linked above in my question, he says in an answer to my question, "You're completely misunderstanding what a "preserved supersymmetry" is. That's not a requirement on the theory, it is just a requirement on the classical solutions, that they be mapped onto themselves under the preserved supersymmetric transformation (i.e. $δX=0$ for all fields), and not onto another solution (this is just a weird way of stating that the solution doesn't break the supersymmetry)." What does he mean, this got me very lost. Dec 29, 2015 at 2:30