Symmetries that have non-trivial effects on observables must be preserved by dualities (equivalences between different-looking quantum field theories), because the equivalence relation preserves observables by definition.

Gauge symmetries don't affect observables, so two quantum field theories can be equivalent to each other even if they have different gauge symmetries.

Fermion-odd supersymmetry transformations must also leave observables invariant, because observables cannot be fermion-odd. Does this mean that two quantum field theories can be equivalent to each other (as far as observables are concerned) even if they have different supersymmetry algebras? ...even if one is supersymmetric and the other is not? Are any examples known?



If by duality you mean an exact duality, then yes: dual theories have the exact same amount of supersymmetry. The reason is that two theories are dual if they are actually the same theory, only expressed in different variables. The symmetry of a theory is intrinsic to the theory itself, not to the variables you use to express it (unlike gauge symmetries). In particular, the two theories have the exact same Hilbert space, the same spectrum, same algebra of observables, etc. Whether a theory is supersymmetric can be decided by looking at the Hilbert space, so if the dual theories share the latter, they also share the former.

Nowadays people also use the word duality in a weaker sense, initiated by the work of Seiberg. In this weaker sense, often referred to as an infrared duality, the two theories are actually different, but become identical in some limit, say, at low energies. The two theories describe the same long distance physics, they have "the same vacuum". But the excited states, the short-distance physics, are different.

Weakly dual theories may have different symmetries. The reason is that symmetries may be emergent in the infrared. In this case, one of the theories may have more symmetry than the other, but they both end up with the same symmetry in the limit. Again, see the work of Seiberg for as many examples as one may wish for. His original work was in 4d $\mathcal N=1$ supersymmetry, but the cleanest examples occur in 3d, cf. e.g. https://arxiv.org/abs/1702.07035, where they also comment on SUSY enhancement. See also the nice work of Benini and collaborators, e.g. https://arxiv.org/abs/1803.01784. Here you will find several examples of (conjectured) dualities where the different theories have different amounts of SUSY.

You can find many more papers by googling "infrared supersymmetry 3d" or something like that. Fascinating topic. Have fun!

Sketch of the argument.

Take two theories $T_1,T_2$, which are assumed to be dual in the strict sense. This means, in particular, that these two theories have the same Hilbert space $\mathcal H$ (and the same algebra of observables, such as Poincaré). Assume that $T_1$ is supersymmetric, i.e., there exist some operators $Q_i\in\mathrm{End}(\mathcal H)$ with the usual super-commutation relations. It is trivial to show that $T_2$ also has these operators, i.e., it is also supersymmetric.

The argument is straightforward. In $T_1$ all states are paired up in a supersymmetric fashion: to each boson $|b\rangle\in\mathcal H$ there is a fermion $|f\rangle= Q|b\rangle\in\mathcal H$ (more precisely, there is a whole multiplet whose details depend on the number of spacetime dimensions and amount of supersymmetry $i=1,2,\dots,\mathcal N$).

The states $|b\rangle,|f\rangle$ also exist in $T_2$, by assumption: the two theories have the same Hilbert space and same spectrum of states. This means that, in $T_2$, we also have the supercharges. For example, we can define these intrinsically by specifying its matrix elements, $\langle b|Q|f\rangle:=1$, etc. It may well be the case that, in $T_2$, the supercharges do not act on the fundamental fields (those that appear in the Lagrangian). It would act on non-local operators, or on non-perturbative operators (such as monopoles). In such case, it would be very difficult to guess that $T_2$ is also supersymmetric. But it must be, if $T_1$ and $T_2$ really are dual. If a symmetry exists in $T_1$, it must also exist in $T_2$, and vice-versa. Sometimes theories are more (super)symmetric than one might initially guess.

If $T_1$ and $T_2$ are weakly dual, then they do not share all of $\mathcal H$, but only the vacuum sector. In such case, the argument above breaks down.

  • $\begingroup$ Thanks for the answer! I do mean exact duality... at least I think that's what I mean, but I suppose that "exact" implies that the QFTs need to be well-defined (UV complete, or lattice, etc) in the first place, and something like a lattice regulator might ruin an otherwise perfectly good duality... so maybe I do mean IR duality after all, whether I wanted to or not. I'll have to think about this more carefully. $\endgroup$ Aug 8 '20 at 20:32
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    $\begingroup$ @ChiralAnomaly I updated the answer. The only examples I know of exact dualities are in scale invariant theories (maximally supersymmetric Yang-Mills, topological theories, CFTs, etc.). Well, also in free theories. I believe people only expect this type of dualities in these types of theories. Indeed, exact dualities are extremely rare, and you need a lot of symmetry if you want it to hold. On the other hand, weak dualities are a dime a dozen. $\endgroup$ Aug 9 '20 at 15:16
  • $\begingroup$ Thank you, it seems clear now. A key difference between gauge transfns and fermion-odd SUSY transfns is that physical states must be invariant under the former, but nothing says they must be invariant under the latter. Observables are, but states don't need to be, because we can have superselection sectors: an irreducible repn of the algebra of field operators can be a reducible repn of the algebra of observables. So even if we were only given the algebra of observables, we could find all of its Hilbert-space repns and then realize that those repns are related to each other by SUSY. $\endgroup$ Aug 9 '20 at 17:33
  • $\begingroup$ @ChiralAnomaly Yes, that's exactly right. You put it much more eloquently than myself :-) Cheers! $\endgroup$ Aug 9 '20 at 19:25

Yes. For starters, $T$-duality (is conjectured to) relate string theories with different kinds/numbers of super-symmetries.

  • $\begingroup$ Thank you! I didn't realize that T-duality could change the number of worldsheet supersymmetries -- worldsheet because I'm looking for QFT examples. The paper T-duality and world-sheet supersymmetry (which I found based on your tip) seems to give an example, but the context is string theory rather than just plain CFT. If we think of the CFT just as a 1+1 dimensional QFT, without the string-theory context, do examples like this preserve the CFT's observables? $\endgroup$ Aug 8 '20 at 18:41
  • $\begingroup$ Can you clarify what is the SUSY of these theories? As far as I can see, all the T-dual pairs have the exact same amount of supersymmetry, although I am sure I am missing something. Thanks! $\endgroup$ Aug 8 '20 at 19:18
  • $\begingroup$ @AccidentalFourierTransform see e.g. arxiv.org/abs/hep-th/9410230 and arxiv.org/abs/1712.05750 and refs. therein $\endgroup$
    – Kosm
    Aug 8 '20 at 20:35

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