No-Go Theorem for Boson Condensation in Topologically Ordered Quantum Liquids

Certain phase transitions between topological quantum field theories (TQFT) are driven by the condensation of bosonic anyons. However, as bosons in a TQFT are themselves nontrivial collective excitations, there can be topological obstructions that prevent them from condensing. Here we formulate such an obstruction in the form of a no-go theorem. We use it to show that no condensation is possible in SO(3)$_k$ TQFTs with odd $k$. We further show that a layered theory obtained by tensoring SO(3)$_k$ TQFT with itself any integer number of times does not admit condensation transitions either. This includes (as the case $k=3$) the noncondensability of any number of layers of the Fibonacci TQFT.

Certain phase transitions between topological quantum field theories (TQFT) are driven by the condensation of bosonic anyons. However, as bosons in a TQFT are themselves nontrivial collective excitations, there can be topological obstructions that prevent them from condensing. Here we formulate such an obstruction in the form of a no-go theorem. We use it to show that no condensation is possible in SO(3) k TQFTs with odd k. We further show that a "layered" theory obtained by tensoring SO(3) k TQFT with itself any integer number of times does not admit condensation transitions either. This includes (as the case k = 3) the noncondensability of any number of layers of the Fibonacci TQFT.
The phase transitions between topological phases are, most of the times, driven by the condensation of bosons [11,[15][16][17][18][19][20][21][22][23][24][25]. In the context of TQFTs, a boson is an emergent quasiparticle in the topologically ordered phase with bosonic selfstatistics, but which could have nontrivial fusion and braiding relations with the other anyons. Such a quasiparticle can potentially undergo Bose-Einstein condensation, causing a phase transition to another topologically ordered phase. The topological data of the new phase can be inferred from those of the initial topological order [25].
One motivation to study condensation transitions is to classify topological order. An important example are the 16 types of gauged chiral superconductors introduced by Kitaev [3]. Kiteav showed that while two-dimensional superconductors are classified by an integer Z, only 16 bulk phases are topologically distinct. This construction can be understood by considering several layers of the elementary (Ising) TQFT. Coupling the layers by condensing inter-layer cooper pairs, one obtains exactly 16 distinct TQFTs including Ising, the toric code and the double semion model. They determine the nature of the topologically protected excitations in the vortices of each superconductor, including their braiding statistics. In essence, Z 16 classification is a property of the Ising TQFT.
It is imperative to ask whether multi-layer systems of other TQFTs show a similar collapse of the classification from Z to Z N for some integer N . In this paper, we derive a criterion for when this is not the case, i.e., when the Z classification generated by a given TQFT is stable. This criterion is based on the fact that there exist bosonic anyons that cannot be condensed. An example are the bosons in multi-layered Fibonacci topological order [17,25,26]. In this work, we gen-eralize this observation by formulating a no-go theorem that constitutes a sufficient obstruction against the condensation of a boson. Our criterion and its proof are given using the tensor category formulation of topological order [3,[27][28][29][30][31][32][33][34], which we can use to describe the condensation transition axiomatically [16,17,25]. We apply our no-go theorem to several examples, including the forementioned multi-layer Fibonacci TQFTs.
Formalism -We use the algebraic formulation of anyon condensation discussed in Ref. 25. Here we simply restate the important relations and refer the reader to Ref. 25 for details. A fusion category is characterized by a set of anyons a, b, c, . . . and fusion rules a × b = c N c ab c between them. The quantum dimension d a gives the size of the nonlocal internal Hilbert space associated with anyon a, and is equal to largest eigenvalue of the matrix N a with elements (N a ) bc ≡ N c ab . A braided tensor category has additional structure, of which we will use the topological spin θ a of a, a complex number with |θ a | = 1. Bosons are defined by θ a = 1. A special role is played by the vacuum anyon as the unique identity element of fusion. It is a boson with quantum dimension 1.
Condensation is based on a mapping, called restriction, between the anyons a in the original TQFT A and the anyons t in the condensed fusion category T characterized by integers n t a ∈ Z ≥0 : If more than one particle appears on the right-hand side of Eq. (1), we say that the a particle splits. If n t a = 0, we say t is in the restriction of a or t ∈ a ↓ . We require that n ϕ 1 = 1, where ϕ and 1 are the vacua in T and A, respectively. Imposing that condensation commutes with fusion implies the fundamental relation [25] r,s∈T n r a n s bÑ t The restriction is compatible with conjugation to antiparticles, i.e., n t a = ntā, where bar denotes the (unique) antiparticle of an anyon. We say particle a condenses if ϕ ∈ a ↓ , i.e., n ϕ a = 0. Common knowledge in condensed matter physics says that any bosons can condense. However, it may also occur that a specific boson a cannot condense, i.e., there is no solution to the above equations with n ϕ a = 0. This is the situation we shall analyze in this paper.
Finally, the following definition is useful for formulating our no-go theorem: For a given anyon b, a subset I b = {a 1 , . . . , a m } of anyons is called a set of zero modes localized by b [35] if for all i, j = 1, . . . , m: 1. The fusion products a i × a j do not contain condensable bosons, except the identity if a i =ā j , [36] 2. all a i are zero modes of b, by which we mean a i × b = b + . . ., (i.e. N b aib > 0) 3. if a particle a i is in I b then so is its antiparticle.
Note that the choice of I b for a given boson b is not unique and that I b may or may not contain the identity. (The above conditions are satisfied in both cases.) Typically, we will be interest to find a set I b that is as large as possible. To motivate the terminology of the set I b , observe that N b ab > 0 implies that a anyons can always be emitted or absorbed by b. Therefore, b must carry a zero-mode excitation of a. We can now state our first main result, a general condition under which a boson B cannot condense. It is an obstruction that is sufficient to show that condensation of B cannot occur.
No-go theorem -A boson B cannot condense if there exists a set I B , such that the sum of the quantum dimensions of all anyons in I B exceeds the quantum dimension of B, i.e., if Proof. We start by showing that all particles in I B do not split, and have distinct restrictions. This follows from inspection of Eq. (2) for t = ϕ, a = a i , b =ā j , r∈T n r ai n r where we used n ϕ c = n φ c . By assumption, there are no condensable bosons in a i ×ā j , hence N c aiāj and n ϕ c cannot be both nonzero for any c = 1. Thus r n r ai n r ai = 1, implying a single restriction a ↓ i of a i , with d a ↓ i = d ai using Eq. (3).
Moreover, r n r ai n r aj = 0 if i = j, implying that the restrictions of a i = a j are distinct particles.
With this knowledge about the restrictions of the a i , Eq. (2) for t = ϕ, a = a i , b =B evaluates to where we used NB aiB = N B aiB Inserting this inequality in Eq. (3) for a = B, and using It follows that in a situation where Eq. (4) holds, Eq. (7) im- To follow up with a pictorial representation of these equations, consider the tunneling of anyons across the domain wall as shown in Fig. 1, where each particle a in the uncondensed theory is converted into its restriction a ↓ in the gray region. Figure 1 (a) shows a vertex allowed by the fusion rule a i × B → B in the uncondensed phase. The boson B enters the condensed phase, where it can disappear as it is part of the condensate (one of its restrictions is the vacuum ϕ, the world lines of which can be removed at will). By the fundamental assumption that fusion and condensation commute [which is at the heart of Eq. (2)], Fig. 1 (a) is equivalent to Fig. 1 (b). The latter represents a coherent tunneling process that is mediated by the condensate and converts B into any of the a i . The existence of this process implies that the distinct restriction a ↓ i of any a i must be in the restriction of B. Hence, by Eq. (3), the quantum dimension of B must be large enough to accommodate all the distinct restrictions of the a i , if B condenses. Therefore if we find sufficiently many a i such that Eq. (4) holds, B cannot condense.
Note that the no-go theorem does not a priori require knowing the braiding data of A -although the modular tensor category structure fixes that data to some extend. The theorem involves only data obtainable from N c ab . We remark that the no-go theorem can only ever yield an obstruction against the condensation of non-Abelian bosons. For Abelian bosons, the theory after condensation can be constructed explicitly, which is a constructive proof that there is no obstruction. [25] We now demonstrate that the no-go theorem is practically useful by considering three examples: (i) multiple layers of the Fibonacci TQFT, (ii) single layers of the SO(3) k TQFT for k odd, and (iii) multiple layers of the latter. We will show that all these theories, while containing bosons, do not admit condensation transitions. All the bosons are noncondensable. Additional general results, concerning for instance TQFTs with a condensing Abelian sector and with only a single boson, are given in the Supplemental Information. [41] Example (i): Multiple layers of Fibonacci -The Fibonacci category A Fib is a non-Abelian TQFT containing just one nontrivial particle τ with a fusion rule τ × τ = 1 + τ , a topological spin θ τ = e i4π/5 , and a quantum dimension d τ = φ given by the golden ratio φ = (1 + √ 5)/2. As A Fib does not contain any nontrivial boson, it cannot undergo a condensation transition. We are interested whether the TQFT formed by N identical layers of A Fib i.e., the TQFT A ⊗N Fib , admits a condensation transition. The TQFT A ⊗N Fib contains 2 N particles corresponding to all possible distributions of τ -particles over the N layers. For each r = 0, . . . , N there are N r socalled (rτ ) particles with τ 's in exactly r layers, each with spin θ (rτ ) = e i4πr/5 and quantum dimension d (rτ ) = φ r . The unique r = 0 particle is the identity of A ⊗N Fib . From the topological spin, the bosons in A ⊗N Fib are (rτ ) particles with r = 5n, n ∈ Z. Using the no-go theorem, we show that none of these bosons can condense.
Using proof by induction on n ≥ 1, we show that for any (5nτ ) boson B, there exists a set I (5nτ ) such that Eq. (4) holds. We first consider the case n = 1. Given a (5τ ) boson, we must construct a set I (5τ ) for this boson. Consider the set formed by all (2τ ) particles obtained by replacing any 3 τ 's in the boson with a 1. There are 5 2 = 10 such (2τ ) particles for a given (5τ ) boson. They form a set I (5τ ) that obeys point 1-3 from the definition: point 1 holds as any product of two of these particles has at most 4 τ s and is therefore not a (potentially condensable) boson. Points 2 and 3 can be checked by using the Fibonacci fusion rules in each layer. Finally, Eq. (4) holds because evaluates to about 11.1 < 26.2. We conclude that none of the (5τ ) bosons condense for any number N of layers of Fibonacci TQFT.
We note that for k odd, all particles have distinct quantum dimensions. The fusion rules are The smallest odd k for which SO(3) k contains a boson is k = 13, in which j = 5 is a boson -an uncondensable one, as we shall see.
The topological spins θ j yield the condition j(j+1) = k+2 for the lowest j that may correspond to a boson (aside from the vacuum j = 0). (Frequently, this condition cannot be met with integer j, as in the k = 13 example, and the lowest boson appears at even higher j.) We conclude that the first boson after j = 0 cannot occur for j lower than We will now discuss separately bosons j in the three ranges (see Fig. 2 for two examples) Due to Eq. (12), bosons j B in range III have no bosons in their fusion product j B × j B , other than the identity. Thus, from Eq. (2) for t = ϕ, and the fact that B are their own antiparticles, we conclude that they cannot split. Using Eq. (3) and the fact that they have d j B > 1, we conclude that they cannot restrict to the vacuum i.e., they cannot condense.
We now use our no-go theorem to show that bosons j B in range I are non-condensable. Specifically, we show that the particles 0 < j < j B /2 form a set I j B of j B obeying Eq. (4). Before establishing that they satisfy the conditions for a set I j B , let us show that Eq. (4) holds for I j B . For large k, we can rely on the following asymptotic estimate. Using that the sine function in Eq. (10) is monotonously increasing with negative second derivative for j ≤ k/4 , the estimate implies Eq. (4) for j B in range I. This inequality holds for all j B ≥ 10. Using Eq. (12) we conclude that it applies to all bosons in range I for k ≥ 109. We verified explicitly that inequality (4) holds (using the exact values of the quantum dimensions) for all bosons in range I for k < 109. Finally, it is readily verified using Eq. (11) that I j B form a set of zero modes localized by j B provided that all bosons with j < j B cannot condense. The proof then proceeds straightforwardly by induction. We apply our no-go theorem successively to bosons j B in range II in order of increasing j B . Using the result that all bosons in range I are uncondensable, one verifies that the particles j with 1 ≤ j ≤ min{k − 2j B , j B /2 − 1} form a set I j B . As for range I, we can estimate the quantum dimensions. From the relation sin[π(2j B + 1)/(k + 2)] = sin[π(k − 2j B + 1)/(k + 2)] we can estimate the quantum dimension of j B using sin[π(2j B + 1)/(k + 2)] < π(k − 2j B + 1)/(k + 2). The quantum dimensions of the anyons in I j B are estimated as for range I with sin[π(2j + 1)/(k + 2)] < π(2j + 1)/(k + 2). Using these estimates we find that if holds, Eq. (4) follows. In the case k − 2j B < j B /2 − 1, Eq. (15) reduces to 1 < (k − 2j B ) 2 + (k − 2j B ), which is true for all j B in range II for all k. In the case k − 2j B > j B /2 − 1, Eq. (15) simplifies to k + 2 < 2j B + ( j B /2 ) 2 , which holds for all j B in range II if k ≥ 37. We verified explicitly that Eq. (4) holds for all bosons in range II if k < 37 (they appear in k = 13,19,31). This concludes our proof that no condensation transition is possible in the SO(3) k TQFT for any odd k. We note that this result can be readily extended to SU(2) k with k odd, since SO(3) k is the projection of SU(2) k to anyons with integer j. One simply includes the half-integer j anyons in the theory (none of which are bosons). The sets I b as defined above remain the same and so do all the quantum dimensions. Hence, we also showed the noncondensability of SU(2) k , with k odd. This is consistent with the ADE classification of SU(2) k [37]: There are no off-diagonal modular invariant partition functions for odd k in SU(2) k [38]. Thus, the no-go theorem provides a proof of this fact that is complementary to the ADE classification.
Example (iii): Multiple layers of SO(3) k -We can show that any number of layers of SO(3) k , with k odd, does not contain condensable bosons. Fixing k, the proof proceeds again by induction. As induction base, we proof that all multi-layer anyons with a nontrivial particle in only a single layer (and the identity anyon in the other k − 1 layers) cannot condense nor split. To show that, we can use the single-layer result from Example (ii). For the induction step, we assume that for a fixed k 0 < k all multi-layer anyons with nontrivial particles in l layers, 1 ≤ l ≤ k 0 , cannot condense and do not split. We can then show that the same holds for multilayer anyons with nontrivial particles in k 0 + 1 layers, completing the induction. The details of this proof are given in [41].
Summary -We have presented a generally applicable nogo theorem against the condensation of a topological boson and illustrated it with several examples. The proof of our theorem uses mostly the fusion (as compared to the braiding) information of the TQFT. We showed a connection between our results and the ADE classification of SU(2) k theories, indicating that the no-go theorem might be useful for the classification of modular invariant partition functions of conformal field theories more broadly. [25] It would be interesting to study, whether other obstructions against boson condensation exist or whether our no-go theorem actually constitutes a necessary condition. In all examples we know, noncondensability is captured by the no-go theorem.
The no-go theorem can not only be used to directly classify (symmetry enriched) TQFTs, but also symmetry protected topological phases without intrinsic topological order. The classification of the latter is often related to the former upon gauging the protecting symmetry. [39,40] Finally, we note that a particle physics analogy of the no-go theorem is that a particle cannot decay if its mass is smaller than the total mass of its potential products. In this picture desintegration amounts to condensation and mass to quantum dimension.
The authors thank Parsa Bonderson for useful discussions. with a shorthand notation. An anyon that has the identity particle from SO(3) k in all layers, except for the k 0 layers i 1 , i 2 , · · · , i k0 , is denoted by {j i1 j i2 · · · j i k 0 }. Here 1 ≤ j i l ≤ (k − 1)/2 can stand for any anyon from SO(3) k (except the identity 0), for all l = 1, · · · , k 0 .

Induction base
First, consider particles {j i } with just one nontrivial anyon in some layer i. This will serve as the induction base. By the no-go theorem and our proof in Example (ii), we know that no bosons of form {j i } can condense. [Use the particles with only one nontrivial in that same layer i to build the set I ji as elaborated for Example (ii).] As a corollary, the anyons {j i } do not split: when fused with themselves no condensable boson appears in the fusion product, which prevents splitting by Eq. (2) from the main text for t = ϕ.

Induction step
We assume that for any 1 ≤ l ≤ k 0 all {j i1 j i2 · · · j i l } 1. do not condense and 2. do not split.
We now show the induction step, namely that all particles with nontrivial anyons in (k 0 + 1) layers {j i1 j i2 · · · j i k 0 +1 } neither condense nor split.
We begin by showing that {j i1 j i2 · · · j i k 0 +1 } cannot condense. The particles {j i1 j i2 · · · j i k 0 +1 } can be obtained by fusing a {j i1 j i2 · · · j i k 0 } with a {j i k 0 +1 }, where i k0+1 / ∈ {i 1 , · · · , i k0 }. In this case, Eq. (2) from the main text reads for t = φ N ϕ Now, because of the uniqueness of the antiparticle, N ϕ {ji 1 ···ji k 0 } ↓ ,{ji k 0 +1 } ↓ can be either 0 or 1. If it was 1, {j i1 j i2 · · · j i k 0 } ↓ would be the antiparticle of {j i k 0 +1 } ↓ . Because all particles are their own antiparticles, this would imply {j i1 j i2 · · · j i k 0 } ↓ = {j i k 0 +1 } ↓ . However, this is not possible for k 0 > 1, because the associativity of fusion would then also imply that {j i1 } ↓ is the antiparticle (and coinciding with) {j i2 · · · j i k 0 j i k 0 +1 } ↓ , i.e., {j i1 } ↓ = {j i2 · · · j i k 0 j i k 0 +1 } ↓ . Remembering that {j i1 j i2 · · · j i k 0 }, {j i2 j i2 · · · j i k 0 +1 } do not split, and equating the quantum dimensions of the particles for these two identifications we have For k 0 > 1, this contradicts the fact that all nontrivial particles in this theory have quantum dimensions d > 1. This rules out the possibilityÑ ϕ The case k 0 = 1 needs to be considered separately, as both lines in Eq. (17) are identical in this case, and therefore do not lead to a contradiction. Assume thatÑ ϕ {ji 1 } ↓ ,{ji 2 } ↓ = 1. In the case j i1 = j i2 , we can rely on the following argument to disprove this assumption: As all anyons in SO(3) k with k odd have distinct quantum dimension, it follows that the two anyons {j i1 } and {j i2 } restrict to distinct particles and in par- (2) from the main text this implies that {j i1 j i2 } neither splits nor condenses. In the case j i1 = j i2 ≡ j, defineĵ ≡ {j i1 j i2 }. We want to show thatĵ does not condense. As there are no fermions in SO(3) k with k odd,ĵ can only be a boson if θ j = 1, i.e., if {j i1 } and {j i2 } are bosons. Our no-go theorem applies to all bosons {j i1 } and {j i2 } with zero mode sets I {ji 1 } and I {ji 2 } . We can then use the set Iĵ = I {ji 1 } × I {ji 2 } , containing the fusion product of any particle in I {ji 1 } with any particle in I {ji 2 } , to prove that j cannot condense. To show that Iĵ is a set of zero modes of j, the main challenge is to show that the product of any two elements from Iĵ cannot condense. The product of any two elements from Iĵ is always of the form {j i1 j i2 }. We have shown that when j i1 = j i2 such particles cannot condense. We therefore need only show that non-trivial particles of form {j i1 j i2 } with j i1 = j i2 both bosons cannot condense. In order to show they are not condensable, we can use the proof given for Example (ii). For that, observe that the anyonsĵ have the same fusion coefficients among themselves as the j anyons in SO(3) k in Example (ii) have, i.e., Nĵ ĵ ,ĵ = N j j,j , where j, j , j ∈ SO(3) k . Recall that conditions 1-3 from the definition of a set of zero modes only depend on the fusion coefficients N j j,j and the information, which particles are bosons. Hence, conditions 1-3 are satisfied for Iĵ whenever they are satisfied for I j in Example (ii). It remains to show that Iĵ is of large enough quantum dimension to satisfy the fundamental inequality Eq. (4) from the main text. Forĵ, Eq. (4) from the main text takes the form Upon taking the square root, this is equivalent to Eqs. (14) and (15) from the main text, which were shown to hold in Example (ii). Therefore theĵ = {j i1 j i2 } anyons do not condense and all {j i } have distinct restrictions. We conclude that for any k 0 ≥ 1 onlỹ N ϕ {ji 1 ···ji k 0 } ↓ ,{ji k 0 +1 } ↓ = 0 is permitted and hence Eq. (16) implies that {j i1 j i2 · · · j i k 0 +1 } does not restrict to the identity ϕ, i.e., it does not condense. This proves the assumption 1 of the induction step for k 0 + 1.
To complete the induction step, we need to show that {j i1 j i2 · · · j i k 0 +1 } does not split. For that, consider Eq. (2) from the main text for {j i1 · · · j i k 0 +1 } with itself and t = ϕ r n r We have used that none of the {j i1 j i2 · · · j i l } with 1 ≤ l ≤ k 0 +1 can restrict to the identity ϕ since they cannot condense. This implies none of {j i1 · · · j i k 0 +1 } splits, which proves the assumption 2 of the induction step for k 0 + 1.
We have thus shown inductively that none of the particles (except for the vacuum) restricts to the vacuum in the N -layer theory SO(3) ⊗N k . Thus, there is no condensate and with it no condensation in any number N of layers of SO(3) k with k odd.

No-go theorem with Abelian sector
We have seen from the examples discussed in the main text, that the no-go theorem can often be used to not only show that individual bosons in a TQFT cannot condense, but that an entire TQFT is not condensable. Here, we extend this discussion to examples of TQFTs that have noncondensable substructures. This problem is motived by physical examples: in the fractional quantum Hall effect, for instance, one frequently discusses phases that are described by a direct (or semi-direct) product of an Abelian and a non-Abelian TQFT. A simple example is the Z 3 Read-Rezayi state of bosons, which is described by the TQFT A Fib × Z 2 . While such a theory admits condensations, already in the Z ⊗N 2 sector, when enough layers are considered, one has the intuition that the noncondensability of Fibonacci should still constrain the possible condensations.

Lemma 1. Consider a TQFT
where X is an Abelian TQFT (i.e., all its anyons have quantum dimension 1). Further, for all particles b ∈ A (not only for the bosons), except for the vacuum, let there exist a set I b = {a 1 , . . . , a m } of zero modes of b, containing anyons from A, such that the quantum dimensions satisfy Then, any possible condensation transition will lead to a theory of the form where the Abelian TQFT Y can be obtained from X through a condensation.
Proof. This Lemma follows almost directly from the no-go theorem. Let us denote a particle from A × X by the pair (b, x) where b ∈ A and x ∈ X . If (b, x) is boson, we can show that it has to be an uncondensable one, except if b = 1.
The set (where a 1 , · · · , a m form a set I b of zero modes of b whose existence is guaranteed by assumption) satisfies all the conditions 1-3 form the definition of a set of (b, x) zero modes. Since x is an Abelian particle, d x = 1 and Eq. (21) directly implies that the sum of the quantum dimensions of the particles in I (b,x) satisfies the inequality (4) from the main text. Hence, (b, x) cannot condense. In turn, this implies any condensable boson in A×X is of the form (1, x). A condensate of this form is transparent to the anyons in A and will thus leave this sub-TQFT unaffected. It will only induce a condensation X → Y, so that the final theory is of the from (22).
We return to the example of A Fib × Z 2 . Consider N layers of this theory, i.e., A ⊗N Fib × Z ⊗N 2 . This multi-layer TQFT satisfies all assumptions of Lemma 1: for each anyon b ∈ A ⊗N Fib , a choice for the set I b is given by I b = {1, b}. This is so because all possible bosons appearing in the fusion product of b × b are uncondensable by the no-go theorem and the sum of the quantum dimensions of I b , given by 1 + d b is larger than d b . We conclude that the A ⊗N Fib structure is preserved under any condensation transition in such a theory.

General constraints on boson condensation
In this section, we list lemmas that pose other general constraints on condensation transitions in TQFTs.
Lemma 2. Suppose S = {a 1 , · · · , a m } is a collection of particles in a TQFT A with a i ×ā i containing no bosons other than the identity -i.e., n t ai = δ t a ↓ i and a i does not split. Moreover assume a ↓ i = a ↓ j for i = j. Then if a boson B appears in the fusion of a i andā j , a i ×ā j = B + · · · for any i = j, then B is not condensable.
Proof. Using Eq. (2) from the main text for a = a i , b =ā j and t = ϕ, we have δ ij = t n t ai n t aj = c n ϕ c N c aiāj . For i = j we get B n ϕ B N B aiāj = 0. So if boson B appears in a i × a j , we must have n ϕ B = 0, so that B is not condensable. Lemma 3. Consider a TQFT A with no fusion multiplicity and just one boson B. If B is condensed then either B is abelian or B ↓ = ϕ + r where r is a single anyon.
Proof. As there is just a single boson, B = B. Equation (2) from the main text implies Notice, however, that the left-hand side is greater or equal to n ϕ B n ϕ B . For condensation, this implies n ϕ B = 1, and tells us that t n t B n t B = 1 or 2. In the former case, B ↓ = ϕ. This implies d B = 1, and so B is a quantum dimension 1 boson hence must have N a BB = δ a,1 . In the latter case, B restricts to just two particles with multiplicity 1 each, so that B ↓ = ϕ + r.
Consider now Eq. (2) from the main text for b = 1, B and for a = b, which gives t n t a n t b = N B ab ≥ n r a n r b Combining the a, B and b, B and a, b equations gives the inequality