# Can a symmetry of a topological field theory be spontaneously broken?

There are examples of topological "terms" causing spontaneous symmetry breaking. One that comes to mind is the $$\theta$$ term in $$4d$$ $$SU(N)$$ Yang-Mills, which at $$\theta=\pi$$ spontaneously breaks time reversal symmetry.

I am curious of a purely topological field theory's ability to spontaneously break a symmetry. My understanding is that spontaneous symmetry breaking requires non trivial dynamics to be present, if only because I am unaware of any examples to the contrary.

Topological field theories have no propagating degrees of freedom, and so this would lead me to believe that it is impossible. But I have not much else in terms of reasoning. Is there anything that can be said about this?

• Prerequisite: How would you propose to define spontaneous symmetry breaking for a TQFT? May 29, 2020 at 2:29
• Presumably the same way it's usually done. The ground state is degenerate and is not invariant under said symmetry. As far as finding an operator with a non-zero VEV that implies this, I have no clue! May 29, 2020 at 2:32
• Does SSB of a $p$-form symmetry count? "Nontrivial $d−1$-form symmetries of TQFTs in $d+1$ dimensions are always spontaneously broken... [and] $0$-form symmetries of TQFTs are always unbroken" (from page 3 in arxiv.org/abs/2001.11938 by Ryan Thorngren). The author's definition of SSB is on page 2. May 29, 2020 at 3:11
• This is exactly what I was looking for. It seems like the underlying assumption about the unbroken $0$-form symmetry is that a TQFT necessarily has some gauge invariance which removes local operators. Is this a justified assumption? I can't think of any TQFT that doesn't have some sort of "gauge" invariance. May 29, 2020 at 3:50

TQFT is most interesting when we're allowed to vary the topology of the manifold, but in ordinary QFT we normally consider the QFT on a fixed spacetime manifold. For a fair comparison, I'll consider a TQFT on a fixed manifold (without metric, of course).

Disclaimer: I'm a relative newcomer to the subjects of TQFT and higher-form symmetries. This answer reflects my current understanding, but I might be missing something.$$^\dagger$$

$$^\dagger$$ Edit: Turns out that I was missing something, but maybe it doesn't completely invalidate the conclusion. See the comment by Ryan Thorngren for details.

Let $$S$$ be the set of observables of this TQFT, represented as operators on a Hilbert space $${\cal H}$$. To define spontaneous symmetry breaking (SSB), first we need to define symmetry. Suppose we define a "symmetry" to be any unitary transformation $$U$$ that preserves the set $$S$$ but that has a non-trivial effect on at least one observable in $$S$$. Then we can define SSB to be the condition that at least one ground state is not invariant under $$U$$.

With this definition of symmetry, every non-trivial symmetry of a TQFT is spontaneously broken, simply because every state in $${\cal H}$$ is a ground state, so if any state in $${\cal H}$$ is not invariant under $$U$$, we could call it SSB. If $$U$$ is a non-trivial symmetry (not the identity operator), then $${\cal H}$$ must have at least one state that is not invariant under $$U$$.

Does this definition of SSB agree with the one we normally use in ordinary QFT? I think it does, because of the phrase "at least one ground state." Even if the symmetry group in question is $$\mathbb{Z}_2$$, so that we can take the direct sum of the two SSB Hilbert spaces and construct a ground state that is invariant under the symmetry (disregarding the cluster property), the theory still admits at least one ground state (in some representation) that is not invariant under $$U$$. So I the definition I described above is consistent with the usual one.

## ... or is the answer no (for conventional symmetries)?

On the other hand, page 3 in Ryan Thorngren's paper https://arxiv.org/abs/2001.11938 says

Nontrivial $$d-1$$-form symmetries of TQFTs in $$d+1$$ dimensions are always spontaneously broken... [and] $$0$$-form symmetries of TQFTs are always unbroken...

A $$0$$-form symmetry is a symmetry in the conventional sense. On page 2, the cited paper defines SSB in terms of long range order. For a $$0$$-form symmetry, the definition of long-range order relies on local observables, and since local observables don't exist in a TQFT, we immediately conclude that $$0$$-form symmetries are never spontaneously broken in TQFT, as stated in the excerpt.

## Reconciling the two conclusions

The two different definitions of SSB shown above might seem to lead to opposite conclusions: one says that non-trivial symmetries in TQFT are always spontaneously broken, and the other says that conventional ($$0$$-form) symmetries in TQFT are never spontaneously broken. And yet, if I'm not mistaken, both definitions agree with the one we would normally use for conventional ($$0$$-form) symmetries in ordinary QFT.

How is this possible? If both definitions agree with the one we normally use in ordinary QFT, then how can they give different answers in a TQFT? After all, we can get a TQFT by taking the extreme low-energy limit of an ordinary gapped QFT. What's going on here?

I think$$^\dagger$$ this is resolved by recognizing that a "symmetry" according to the first definition is never a $$0$$-form symmetry. It can't be, because a non-trivial $$0$$-form symmetry must (by definition) have a non-trivial effect on local observables (observables localized in a contractible region of the spacetime), but a TQFT doesn't have any local observables for the symmetry to affect. The first definition implicitly catches all of the theory's symmetries, including $$k$$-form symmetries for $$k\geq 1$$, so it catches the fact that non-trivial $$k$$-form symmetries in TQFT can be spontaneously broken. When we take the extreme low-energy limit of an ordinary gapped QFT, we lose all of the local observables, so whatever $$0$$-form symmetries the theory had become trivial, whether or not they were spontaneously broken before the limit.

Altogether, the answer is yes: a TQFT can have SSB, if we consider $$k$$-form symmetries for $$k\geq 1$$. If we only consider conventional ($$0$$-form) symmetries, then the answer is no: a TQFT can't have SSB for a $$0$$-form symmetry simply because it can't have any non-trivial $$0$$-form symmetries (broken or not).

$$^\dagger$$ Edit: The reasoning in these last two paragraphs is incorrect, as clarified by Ryan Thorngren's comment. TQFTs can have non-trivial $$0$$-form symmetries. That makes today a good day — I learned something new!

• Thanks for the shoutout! I would just clarify that TQFTs do have non-trivial 0-form symmetries, such as anyon permutation symmetries, but because there are no local operators, the ground state is nondegenerate on a sphere. However, these symmetries would look "spontaneously broken" on a torus, for instance, as they would permute the ground states. I don't consider this to be 0-form SSB in the usual sense because there is no local order parameter. May 29, 2020 at 18:57
• @RyanThorngren Thank you for the clarification! I edited the answer to acknowledge my error and to direct readers to your comment. May 29, 2020 at 20:40
• Do any of these arguments from Thorngren's paper apply to spacetime symmetries? These transform local as well as nonlocal operators, so it can't quite be categorized as a $p$-form symmetry. May 29, 2020 at 20:54
• @LucashWindowWasher I'm sure the same claims hold for spacetime symmetries. Spatial symmetries act as internal symmetries in the effective field theory, and the theory of anti-unitary (0-form) symmetries of TQFTs is basically just like unitary symmetries but with complex conjugates in the correct places. These general arguments should still apply. May 30, 2020 at 4:36
• For any readers who had the same misconception I had regarding $0$-form symmetries in TQFT, here's a paper on the subject, which I'm studying now to help cure my misconception: Symmetries of Abelian Chern-Simons Theories and Arithmetic. The focus of the whole paper is $0$-form symmetries in (2+1 dimensional abelian) TQFTs. May 30, 2020 at 13:26