We read at 0912.4261[end of page 9 and beginning of page 10] that

any susy field configuration of a theory must obey the conditions implied by setting the supersymmetric variations of fermions to zero. In a (untwisted) theory without any (defects) surface operators, this implies that any susy configuration must obey $F=0$ $\ldots$ .

Now, I do not get this. In, e.g. $\mathcal{N}=1$ supersymmetry with one chiral superfield $\{\phi, \psi \}$ we do not require that $\delta_{\epsilon}\psi = 0$, rather the susy transformation must be proportional to the derivative of $\phi$.

Furthermore, how is the flat connection related to the susy transformations of fermions? In a $\mathcal{N}=2$ with a vector superfield $\{ \lambda, A \}$, with $A$ a gauge field, why should it be a flat connection? Can we not have non-trivial instanton sectors etc with non-trivial fluxes etc?

  • $\begingroup$ Regarding the first point: Here, "susy configuration" means "state invariant under susy transformations". Since we have $\left<\psi\right>=0$ in a Lorentz-invariant state (implictly assumed), we also need $\delta_\epsilon\psi=0$. $\endgroup$ – Toffomat Aug 8 '17 at 11:24
  • $\begingroup$ Sorry, why $\langle \psi \rangle = 0$? $\endgroup$ – Marion Aug 8 '17 at 11:30
  • $\begingroup$ Because a VEV for fermions would break Lorentz invariance (unlike scalars, so you can have a nonzero $\langle\phi\rangle$). $\endgroup$ – Toffomat Aug 8 '17 at 11:41
  • $\begingroup$ $\langle \phi \rangle $ is not gauge invariant though. In any case, it is not clear to me the thing about the SUSY transformation. In any textbook the SUSY variation of fermion is not zero. What you say is correct but I can easily write it as $\langle \delta_{\epsilon}\psi \rangle =\delta_{\epsilon} \langle \psi \rangle =0$ due to Lorentz invariance but not due to SUSY. $\endgroup$ – Marion Aug 8 '17 at 11:52
  • $\begingroup$ Well, if $\phi$ is charged, then a VEV breaks the gauge symmetry, but that is not necessarily bad. And your last equation seems wrong. I'll try to write an answer in a few minutes. $\endgroup$ – Toffomat Aug 8 '17 at 12:08

The question really is about the difference between the susy variations of fields and their values in a given state.

Let's consider the transfomations in a $\mathcal N=1$ theory (very schematically). We have a chiral multiplet $\left\{\phi,\psi\right\}$ with variations $$\delta_\epsilon\phi\sim \bar\epsilon\psi \,, \qquad\qquad \delta_\epsilon\psi\sim \partial_\mu \phi \gamma^\mu\epsilon\,,\tag{$*$}$$ and the vector multiplet $\left\{\lambda,A_\mu\right\}$ with variations $$\delta_\epsilon \lambda\sim \partial_\mu A_\nu\gamma^{\mu\nu}\epsilon\,,\qquad\qquad \delta_\epsilon A_\mu \sim \bar\epsilon\gamma_\mu\lambda\,.\tag{$\dagger$}$$ In particular, in both cases we have $$\delta\left(\text{boson}\right)\sim\text{parameter}\cdot\text{fermion} $$ and $$ \delta\left(\text{fermion}\right)\sim\text{parameter}\cdot \partial\left(\text{boson}\right)\,.$$ (This is basically required by the susy algebra which implies $\left[\delta,\delta\right]\sim\partial$)

Now, when we look for a suspersymmetric configuration, we mean a set of field values such the state is invariant under susy variations. What fields can get VEVs?

  • $\phi$ is a scalar, so its VEV respects Lorentz symmetry. It will break gauge symmetry if it's charged.
    • $\psi$ and $\lambda$ are fermions, so a VEV would break Lorentz symmetry, and possibly gauge symmetry.
    • $A_\mu$ is a vector, so it again breaks Lorentz symmetry and possibly gauge symmetry, but this is more subtle.

Phenomenologically, we know that Lorentz symmetry is unbroken, so usually, states with VEVs for fermions or vectors are not considered. Gauge symmetry may or may not be broken, as in QED vs. weak interactions. Hence, we are looking for a state with $\langle\psi\rangle=\langle\lambda\rangle=0$.
Finally, we explicitly want supersymmetric states, i.e. states that do not change under a susy transformation. In particular, since $$0=\langle\psi\rangle\longrightarrow\langle\psi\rangle+\langle\delta_\epsilon\psi\rangle=\langle\delta_\epsilon\psi\rangle\,,$$ we require that the susy variations of $\psi$ and $\lambda$ vanish (in this state, i.e. with the chosen values for the other fields). Inspecting the variations $(*)$ and $(\dagger)$ above, this translates to $$\langle\partial_\mu\phi\rangle=0 \qquad \text{and} \qquad \langle F_{\mu\nu}\rangle=0\,.$$ This is what the paper means with

... any susy configuration must obey $F=0$ ...

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  • $\begingroup$ Thanks for your excellent answer. Can you also comment on which situations the broken gauge invariance is not that bad as you said? $\endgroup$ – Marion Aug 8 '17 at 12:44
  • 1
    $\begingroup$ @Marion The gauge boson(s) will become massive, and as long as susy is unbroken, the gauginos will have the same mass. However, the situation becomes more subtle and depends on the fields and the Lagrangean (e.g. the superpotential that generates the VEV). I seem to remember that susy breaking may ensue, but I don't have the references now to give a good answer. $\endgroup$ – Toffomat Aug 8 '17 at 13:01

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