Unitary representations of the $\mathcal{W}_3$-algebra with $c\geq 2$

We prove unitarity of the vacuum representation of the $\mathcal{W}_3$-algebra for all values of the central charge $c\geq 2$. We do it by modifying the free field realization of Fateev and Zamolodchikov resulting in a representation which, by a nontrivial argument, can be shown to be unitary on a certain invariant subspace, although it is not unitary on the full space of the two currents needed for the construction. These vacuum representations give rise to simple unitary vertex operator algebras. We also construct explicitly unitary representations for many positive lowest weight values. Taking into account the known form of the Kac determinants, we then completely clarify the question of unitarity of the irreducible lowest weight representations of the $\mathcal{W}_3$-algebra in the $2\leq c\leq 98$ region.


Introduction
The W N (N = 2, 3, • • • ) algebras are "higher spin" extensions of the Virasoro algebra [Zam85,FZ87,FL88], with W 2 being the Virasoro algebra itself and W 3 in some sense the simplest one without a Lie algebra structure.For general N, the W N -algebra is generated by N − 1 fields, the first one of which is the Virasoro field.For some discrete values of the central charge c < N − 1, they have been recently realized as a certain coset, showing unitarity of their vacuum representations (i.e. the irreducible representations with zero lowest weights) as well as many other representations [ACL19].In the Virasoro case (N = 2), this is the famous construction of Goddard, Kent and Olive [GKO86] and the corresponding central charge values are c = 1 − 6 m(m + 1) m = 3, 4, 5 . . .whereas for the W 3 -algebra, these values are [Miz89,Miz91] c = 2 1 − 12 m(m + 1) m = 4, 5, 6 . . .and in both cases N = 2, 3 it is known that there are no other unitary representations in the c < N − 1 region than the ones obtained in this manner.Though this coset realization is recently generalized [ACL19] to an even wider class of W-algebras, it is not expected to take us above the central charge value c = N − 1, where rationality cannot hold.Indeed, as far as we know, unitarity has never been shown for any central charge value c > N − 1 ≥ 2. Note that unlike in the Virasoro (or in the affine Kac-Moody) case, when N ≥ 3 -because of the lack of a Lie algebra structure -one cannot simply produce representations of W N by e.g.taking tensor products of known ones.Because of the difficulty of finding explicitly unitary constructions, some even expected the W N -algebras to not to have unitary vacuum representations for c > N − 1 ≥ 2 (see e.g.[AJCH + 18]).In this paper, we prove in fact that the vacuum representation of the W 3 -algebra is unitary for any value of the central charge c ≥ 2. In the Virasoro case, unitarity for c > 1 can be settled using Kac determinants see e.g.[KR87,Section 8.4].At any "energy level" (i.e.eigenspace of the conformal Hamiltonian), the Kac determinant is a polynomial of the central charge c and lowest weight h.Since all Kac determinants are strictly positive in the region {(c, h) : c > 1, h > 0}, by a continuity argument, unitarity in a single case inside that region (which can be easily obtained e.g. by taking tensor products) implies unitarity for the whole closure {(c, h) : c ≥ 1, h ≥ 0}.In case of the W 3 -algebra, the difficulty is twofold.First, one cannot obtain unitary representations with c > 2 by tensor product.Second, the Kac determinants -which are this time rational functions of the central charge c and lowest weights h, w and are explicitly worked out in [Miz89] by Mizoguchi -show that when c > 2, no irreducible lowest weight representation can be unitary in a neighbourhood of h = w = 0 (apart from the vacuum itself).Hence the physically most important representation, the vacuum one, cannot be accessed in this manner from the (h, w) = (0, 0) region.With the usual indirect method ruled out, we are lead to consider unitarity in a more constructive approach.
The explicit construction of unitary vacuum representations in the c > N − 1 region is not trivial even in the Virasoro (N = 2) case.Buchholz and Schulz-Mirbach [BSM90] provided an interesting construction in this regard.They first realized the Virasoro algebra with central charge c > 1 with the help of the U(1)-current (a field whose Fourier modes form a representation of the Heisenberg algebra) in a -strictly speaking -non-unitary way.These representations (which we simply call the BS-M construction) turn out to be "almost unitary": the only problem is a singularity at just one point (indeed, they only needed their construction to be defined on the punctured circle).As observed in [Wei08], the BS-M construction may be viewed as a non-unitary representation of the Virasoro algebra admitting an invariant subspace containing the vacuum vector Ω, on which it is unitary.Inspired by the BS-M construction and the mentioned observation, we start with a pair of commuting U(1)-currents in their unitary vacuum representation and modify them so that the Fateev-Zamolodchikov free field realization of the W 3 -algebra [FZ87] associated with this modified representation of the Heisenberg algebra gives a stress-energy field corresponding to the BS-M one.Similarly to the BS-M case, the obtained new stress-energy and W (z) fields will not give a unitary representation of the W 3 -algebra on the full space but they become so on a subspace generated by Ω.However, the proof of this relies on a rather involved argument exploiting the degeneracy of the vacuum representation: the same construction with nonzero lowest weights does not have unitarity on the minimal invariant subspace containing the lowest weight vector.
Whereas unitarity of the vacuum is difficult to treat, it turns out that some non-vacuum representations can be shown to be unitary in a relatively simple, constructive manner.Making another suitable use of the realization of Fateev and Zamolodchikov, we obtain a manifestly unitary representation of the W 3 -algebra on a full unitary representation space of two U(1)-currents.In this way, we produce unitary representations with h ≥ c−2 24 ≥ 0 and w limited in a certain interval depending on c and h.This is similar to the Virasoro case, where an oscillator representation with a modified Sugawara construction gives manifestly unitary representations for all h ≥ c−1 12 ≥ 0; see e.g.[KR87,Section 3.4].Having already found some unitary representations, one can use the known form of the Kac determinant to arrive at even further values of c, h and w.In this way, for 2 ≤ c ≤ 98 we completely clear the question of unitarity.When c > 98, determining the sign of the Kac determinant becomes harder; our results there remain partial.
This paper is organized as follows.In Section 2 we give a summary of formal series with operator coefficients on Hermitian vector spaces and on the W 3 -algebra, the current algebras and their representations.Apart from self-containment, we use the occasion to fix notations and conventions.An important tool for unitarity, the Kac determinant, is also introduced.Our main results are in Section 3, where we prove the unitarity of various representations of the W 3 -algebra and completely classify unitary lowest weight representations with central charge c ∈ [2, 98].We also briefly explain in a remark how each unitary vacuum representation gives rise to a simple unitary vertex operator algebra.Finally, in Section 4 we collect possible future directions and open problems.
The non-constructive part of our work (where we exploit Kac determinants) is based on the existence of lowest weight representations with invariant forms.Yet, as the W 3 -algebra is not a Lie algebra, the existence of lowest weight representations with invariant forms for all values of lowest weights is not straightforward.Though implicit in the literature, we could not find a reference suitable for our needs, so we added an Appendix A to our work where we clarify this issue by a novel, analytic method.

Formal series and fields
Let V be a vector space and A n : V → V (n ∈ Z) be a sequence of linear operators acting on V .We say that the formal series A(z) = n∈Z A n z −n is a field on V if for every v ∈ V , there is n v such that A n v = 0 whenever n ≥ n v .We shall refer to the operators {A n } n∈Z as the Fourier modes of A(z).
The (formal) derivative of is a formal series in two variables z, ζ and we shall use the notations ∂ ζ , ∂ z in the obvious way.Moreover, we shall also use the notation which we call the "derivative along the circle".
Although the product of two formal series of the same variables does not make sense in general, there are some pairs of formal series that can be multiplied.For example, the product of a formal series in variables z and ζ of the form B(z/ζ) with any other formal series in either z or ζ (but not in both!) makes sense.In particular, the product (where N, k ∈ Z) becomes finite on every vector and hence gives rise to a well-defined linear map.In particular, every field can be multiplied with a formal series of the form n≤N c n z −n (where the coefficients c n may be scalars or themselves linear maps).It then turns out that if is well-defined even at z = ζ (i.e. after replacing ζ by z) and the obtained formal series : Note that in general the normal product of fields is neither commutative nor associative; in particular, to have an unambiguous meaning, we need to specify what we mean by the normal power : F (z) n : .Following the standard conventions, we define the n-th power in a recursive manner by the formula : F (z) n : = : F (z)( : F (z) n−1 : ) : , and more in general, :

Formal adjoints of formal series and fields
Let V be a C-linear space equipped with a Hermitian form •, • (i.e. a self-adjoint sesquilinear form) and A, B : then we say that A and B are adjoints of each other and with some abuse of notation we write B = A † .Note however, the following: 1) such an A † might not exist, 2) when •, • is degenerate, A † may not even be unique.Nevertheless, for any two operators A, B the statement B = A † is unambiguous: it simply means that they satisfy equation (2).We also say that A is symmetric1 when A = A † .We define the adjoint of the formal series A(z) = n∈Z A n z −n to be the formal series i.e. we treat the variable z as if it were a complex number in is symmetric, then so is its circle derivative A ′ (z) of (1): this is exactly why we shall prefer it to ∂ z A(z).Note that this is also the convention found in the paper [BSM90] of Buchholz and Schulz-Mirbach.
If f (z) is a trigonometric polynomial, i.e. a finite series f (z) = |n|<N c n z −n , and A(z) is a symmetric field, then one finds that In particular, if c n = c −n for all n -or equivalently: if f takes only real values on S 1then f (z)A(z) is symmetric.This is not surprising at all; in fact, more in general, one has that if A(z) and B(z) are commuting symmetric fields, then their product A(z)B(z) is also a symmetric field.However, in this paper we shall often consider expressions of the type ρ(z)A(z), ρ ′ (z)A(z) where ρ(z) = −i z−1 z+1 .In order to give an unambiguous meaning2 to the expression ρ(z)A(z), we take the expansion around z = 0, where it holds that Accordingly we regard ρ(z) as a field (note that ρ n = 0 for n > 0), and since it is scalar valued, it commutes with anything and its product with another field A(z) is meaningful without need of normal ordering: Similarly, the product ρ ′ (z)A(z), with ρ ′ (z) given by (1), is defined as a field.Although ρ(z) is not defined at z = −1 as a function (it has a singularity there), it takes only real values on the punctured circle S 1 \ {−1} and hence so does its circle derivative ρ ′ (z).So one might wonder whether ρ(z)A(z) and ρ ′ (z)A(z) are still symmetric if A(z) is a symmetric field.A quick check reveals that the answer in general is negative: the problem is caused by the non-symmetric expansion (3).But if r(z) is a trigonometric polynomial and r(−1) = 0, then the singularity of r(z)ρ(z) at z = −1 is removable.Actually, it is clear that in this case r(z) = (z + 1)t(z) where t is another trigonometric polynomial and hence s(z) = r(z)ρ(z) = −i z−1 z+1 (z + 1)t(z) = −i(z − 1)t(z) is also a trigonometric polynomial for which s(z) = r(z)ρ(z).Hence in this case (r(z)ρ(z)A(z)) † = r(z)ρ(z)A(z), as if ρ(z)A(z) were symmetric.If further r ′ (−1) = 0, then also the singularity of r(z)ρ ′ (z) will be removable, resulting in (r(z)ρ ′ (z)A(z)) † = r(z)ρ ′ (z)A(z).These observation will become important in the proof of unitarity of vacuum representations.

The W 3 -algebra
For our purposes the W 3 -algebra (see [BS93,Art16] for reviews) at central charge c ∈ C, c = − 22 5 , consists of two fields where b 2 = 16 22+5c and Λ(z) = : L(z) 2 : − 3 10 ∂ 2 z L(z).Equivalently, in terms of Fourier modes the requirements read where again b 2 = 16 22+5c and The first of these commutation relations says that the operators {L n } n∈Z form a representation of the Virasoro algebra and consequently, we shall say that L(z) is a Virasoro (or alternatively: a stress-energy) field.
Note that one cannot consider (5) (together with the definitions of b and Λ n ) as the defining relations of an associative algebra (as it is sometimes loosely stated in the literature), since the infinite sum appearing in Λ n does not have an a priori meaning: it makes sense if {L n } form a field on V .Under the term "W 3 -algebra", one studies general properties that hold for operators {L n , W n } n∈Z satisfying the above relations.On the other hand, a concrete realization on a linear space is referred to as a representation, although we do not define here an associative algebra called the W 3 -algebra.A universal object with these relations can be defined in the context of vertex operator algebras [DSK05,DSK06]; however, here we do not wish to follow that way.
We shall say that a Hermitian form •, • is invariant for a representation of the W 3algebra, if it makes the fields Equivalently, in terms of Fourier modes, the requirement of invariance is that A representation together with an inner product -or as is also called: scalar product -(i.e. a positive definite Hermitian form) is said to be unitary.
Note that while in papers concerned with vertex operator algebras, the Virasoro field is typically denoted by L(z) (as in our work), physicists often use T (z) for the same object.
Here we chose to reserve this symbol for the "shifted" field T (z) = z 2 L(z) in part to follow the notations of [BSM90] used by Buchholz and Schulz-Mirbach, and in part simply because being interested by unitarity, we will actually use more the combination z 2 L(z) than L(z) on its own.

The U(1)-current (or Heisenberg) algebra
The U(1)-current (or Heisenberg) algebra is an infinite-dimensional Lie algebra spanned freely by the elements {a n } n∈Z and a central element Z with commutation relations [a m , a n ] = mδ m+n,0 Z. (6) We shall be only interested in representations of this algebra where Z acts as the identity and the formal series (where, by the usual abuse of notations, we denote the representing operators with the same symbol as the abstract Lie algebra elements) is a field.Note that in many relevant works regarding the W 3 -algebra and published in physics journals, this field appears as "the derivative of the massless free field" and is denoted by ∂ z ϕ(z) (e.g. in [FZ87] and in [Miz89]), although in our sense, in general 3 there is no field ϕ(z) whose derivative is a(z).Note also that the commutation relation (6 Suppose now that we are also given a Hermitian form •, • on our representation space.We say that it is invariant for our representation, if it makes 3 Unless we are in a representation where a 0 = 0 symmetric; this is equivalent to the condition a † n = a −n for all n ∈ Z.A representation together with an invariant inner product, i.e. an invariant positive definite Hermitian form, is said to be unitary. Similarly to what we did for J(z) and a(z), we also introduce in general the "shifted" normal powers : J n : (z) = z n : a(z) n :.Again, the reason for working with them (rather than with the usual powers4 ) is symmetry: given an invariant Hermitian form, it is this combination which becomes symmetric.For example, for n = 2 we have : Moreover, as a † 0 = a 0 commutes with all a n , putting all together we have that For higher powers, symmetry of : J n : (z) is justified in a similar manner.
If a(z) = n∈Z a n z −n−1 is a field satisfying the commutation relation ( 7), then its associated (or canonical) stress-energy field is ) form a representation of the Virasoro algebra with central charge c = 1.By elementary computations, [L n , a m ] = −ma n+m and it then follows that for any η, κ ∈ C, the operators also form a representation of the Virasoro algebra with central charge c(η, κ) = 1 + 12κ 2 ; see e.g.[KR87, Section 3.4].Using circle derivatives, the corresponding "shifted" stress-energy field can be written as 1 2 : For the formal series J(z) = za(z) = n∈Z a n z −n where a(z) satisfies (7), a nonzero vector Ω q is said to be a lowest weight vector with lowest weight q ∈ C if for all m > 0 : a m Ω q = 0, a 0 Ω q = qΩ q .
If Ω q is also cyclic, then the whole representation is said to be a lowest weight representation.It turns out that for every q ∈ C, such a representation exists (up to equivalence) uniquely; this is the Verma module V U(1) q .In this representation one has that vectors of the form where 1 ≤ n 1 ≤ . . .≤ n k , form a basis, the formal series a(z) is a field and further that a 0 is the (multiplication by the) scalar q.Moreover, when q ∈ R, there exists a unique Hermitian form •, • on V U(1) q with normalization Ω q , Ω q = 1, which is invariant for the representation (the "canonical Hermitian form").This form is automatically positive definite, making the representation unitary.For proofs of these statements see e.g.[KR87].

Lowest weight representations of the W 3 -algebra
Given a representation of the W 3 -algebra {L n , W n } n∈Z with central charge c, a nonzero vector Ω c,h,w =: Ω is said to be a lowest weight vector with lowest weight (h, w) ∈ C 2 , if for all n > 0 : In case h = w = 0, Ω is said to be a vacuum vector.In case the lowest weight vector is cyclic, the whole representation is said to be a lowest weight representation.
Using the W 3 -algebra relations, it is rather easy (however, the induction should go with respect to g in Appendix A instead of the number of operators, see e.g.[BMP96]) to show that for any lowest weight representation, the vectors of the form and lowest weight (h, w) ∈ C 2 there is indeed a representation, the Verma module V W 3 c,h,w , where these vectors form a basis.It is rather clear that such a representation is essentially unique; what is less evident, is its existence.For a Lie algebra, Verma modules are constructed as a quotient of the universal covering algebra, see e.g.[Jac79].As the W 3 -algebra is not a Lie algebra and the commutator [W m , W n ] contains an infinite sum in L's, it is actually nontrivial that Verma modules exist.We show this in a novel, analytic manner in Appendix A.
Using the W 3 -algebra relations, it is not difficult to see that the Verma module can admit at most one invariant Hermitian form •, • with normalization Ω c,h,w , Ω c,h,w = 1.We will call this the "canonical" form.It is also rather trivial that if c, h, w are not all real, then such a Hermitian form cannot exists.Again, what is less evident is the existence for c, h, w ∈ R. We give a proof of this fact in Appendix A. Since the goal of this paper is to deal with unitarity, we will focus on the case when c, h, w ∈ R.
Let us now take some c, h, w ∈ R, c = − 22 5 .Any nontrivial subrepresentation in the Verma module is included in the kernel ker It then turns out that with the given values of c, h, w, there is (an up-to-isomorphism) unique irreducible lowest weight representation V W 3 c,h,w : namely, the one obtained by taking the quotient of the Verma module with respect to ker •, • .The canonical form on a Verma module is positive semidefinite if and only if the corresponding irreducible representation admits a invariant inner product, making it unitary.
Actually, standard arguments show that (for given (c, h, w)) any lowest weight representation with a non-degenerate, invariant Hermitian form •, • is isomorphic to the unique irreducible representation.This is due to the fact that the value of Ψ, Ψ ′ , where Ψ, Ψ ′ are vectors of the form (10), is "universal": it depends on c, h, w but not on the actual representation; see Proposition A.1.In particular, for each triplet (c, h, w), there is (up to isomorphism) at most one lowest weight representation with an invariant inner product; namely, V W 3 c,h,w .

The Kac determinant
The question of when the canonical form •, • on the Verma module V W 3 c,h,w is degenerate or positive semidefinite can be studied through the Kac determinant.See [KR87, Chapter 8] for an overview of the methods used here, which are written for infinite-dimensional Lie algebras, but apply to the W 3 -algebra as well.
The Hermitian form •, • vanishes on pairs of vectors of the form (10) when the eigenvalue N = j m j + j n j of L 0 are different, hence the question can be studied for each N ≥ 0 separately.There are finite many vectors Ψ .Note that these values are real polynomials of c, 1 22+5c , h, w (see Appendix A).Evidently, we have the following.
• V W 3 c,h,w is irreducible if and only if all of these matrices are nondegenerate.
• The canonical form on V W 3 c,h,w is positive (semi)definite if and only if these matrices are all positive (semi)definite.
However, it is difficult to determine the rank and positive (semi)definiteness of all these matrices at once.Nevertheless, a rather compact formula can be given for the determinant det(M N,c,h,w ) at level N -called the Kac determinant -of these matrices.We can use it in the following ways.
• If V W 3 c,h,w is reducible, then det(M N,c,h,w ) = 0 for some N.
• If the canonical form on V W 3 c,h,w is positive-definite, then det(M N,c,h,w ) > 0 for all N.
At each level N, det(M N,c,h,w ) is a polynomial of c, 1 22+5c , h, w.Therefore, if one finds a vector in ker •, • in a Verma module V W 3 c,h,w , one can extract a factor from det(M N,c,h,w ) for some N.With sufficiently many such vectors in ker •, • , one can determine det(M N,c,h,w ) up to a multiplicative positive constant.According to [Miz89, AJCH + 18], the Kac determinant at level N is where "∼" means equality up to a positive multiplicative constant that can depend on N (but not on c, h, w) and We shall exploit the knowledge of the signs of the Kac determinant (given by these explicit formulas) in two ways: • Let H ⊂ R 3 be a connected set where for any (c, h, w) ∈ H and any N ∈ N it holds that det(M N,c,h,w ) > 0. In this situation, if then it is so for all triples in the closure H.
c,h,w is not unitary.By the observation of [AJCH + 18, (A.10)], if 2 < c < 98, the contributions from f mn with m = n are non-zero positive because α ± in (11) have non-zero imaginary parts, and since is increasing with respect to m, hence all Kac determinants are positive if Note that regardless of the value of the central charge, f 11 (h, c) − w 2 ≥ 0 is a necessary condition for unitarity since f 11 (h, c) − w 2 is the first Kac determinant up to a positive constant.
The case h = 0 is of particular importance, as this is when the lowest weight vector is a "vacuum vector" for the Virasoro subalgebra.From the observation above, unitarity together with h = 0 implies w = 0.
3 Unitarity of lowest weight representations

The free field realization of Fateev and Zamolodchikov
Given a pair of commuting fields a , both satisfying the U(1)-current relation (7), one can construct a family of representations of the W 3 -algebra depending on a complex parameter α 0 .Following Fateev and Zamolodchikov [FZ87], we set . Then the above defined L(z; α 0 ), W (z; α 0 ) fields satisfy the W 3 -algebra relations (4) with central charge c(α 0 ).Remark 3.2.We think it useful to make some comments on the computations justifying the above theorem.First of all, instead of commutation relations, it is more common to work in terms of operator product expansions (OPEs).The OPE of two fields F 1 (z), F 2 (z) is usually written in the form where G j (z), j = 1, • • • , N are some other fields.As formal series, this relation should be interpreted as (see . It is possible to write the OPE between a field F (z) and a normal product : G(z)H(z) : in terms of the OPE between F, G, H and the fields appearing in their OPE; again, for details we refer to [Kac98].Thus, if the OPE algebra of the basic fields is closed -like in our case: then in principle the OPE of any pair of normal products can be determined in terms of the basic fields.Therefore, Theorem 3.1 can be indeed proved only in terms of the commutation relation (7).Although actual computations of OPE of composite fields can be tedious and painful, these computations are fortunately very established procedures and can be carried out by computers, too.The most widely used software for this the Mathematica package5 OPEdefs [Thi91] by Thielemans (although there are also other packages, e.g.[Eks11]).As is indicated in the text, the authors of [RSW18] also used this package to make computations with OPEs related to the free-field realizations of the W-algebras, and this is what we also used6 in part to have an independent verification and in part to check that our constants (which, due to differing conventions, slightly differ from the one appearing in [FZ87]) are indeed rightly set.
Since we are interested by unitarity, it is worth rewriting our fields using the circle derivative F ′ (z) = iz∂ z F (z) and performing computations with the "shifted fields" we introduced above.Also, we prefer to make some different choices of variables -e.g.instead of α 0 as in the previous theorem, we will use κ := −i √ 2α 0 -so that in the unitary case we will need to deal with real constants, only.We thought it useful for the reader to summarize our conventions in a table (which are actually mainly the ones used by Buchholz and Schulz-Mirbach in [BSM90] and hence will be referred as the "B-SM conventions") and put it in contrast with the one used by the physicist and the one used by the VOA community.
the following way: Assume that J [1] (z), J [2] (z) have a common lowest weight vector Ω q 1 ,q 2 with lowest weights q 1 , q 2 .It is straightforward to check that Ω q 1 ,q 2 is annihilated by all positive Fourier modes of fields like : J 3 [2] : (z) or J ′ [1] (z)J [2] (z) and hence also by those of T (z; κ) = n∈Z Lκ,n z −n and M(z; κ) = n∈Z Wκ,n z −n .One also computes that Hence we have the following.
Proposition 3.3.If Ω q 1 ,q 2 is a lowest weight vector for the two commuting U(1)-currents (z) with corresponding lowest weights q 1 and q 2 , respectively, then it is also a lowest weight vector for the representation of the W 3 -algebra given by the fields (13) and (14) with central charge c = 2 − 24α 2 0 = 2 + 12κ 2 and lowest weight (h, w) where Now suppose we have an inner product on our representation space making the currents all symmetric, but the linear combination giving T (z; κ) is only symmetric for κ = 0; i.e. for the central charge c = 2 case (and we have the same situation regarding M(z)).
One possible remedy would be a modification of our inner product; instead of the invariant form for our currents, we should try to use a "strange" one that does not make J [1] (z), J [2] (z) symmetric.Here we will follow a -in some sense -dual approach.Namely, we retain our original inner product, but instead modify our currents by applying an automorphisms of the algebra (7).

New representations by automorphisms of the U(1)-current
Suppose the field J(z) = n∈Z a n z −n is a U(1)-current and f (z) = n∈Z c n z −n is a scalar valued field (i.e.c n = 0 for n large enough).Then, because scalars commute with everything, the sum J(z) + f (z) satisfies the same commutation relation of the U(1)-current field.In terms of Fourier modes, the transformation is a n → a n + c n .If further c n = 0 for all n > 0 and Ψ is a lowest weight vector for J(z) with weight q (i.e.we have a n Ψ = 0 for all n > 0 and a 0 Ψ = qΨ), then Ψ is a lowest weight vector for J(z) + f (z) with lowest weight q + c 0 .Representations of this kind play a central role in [BMT88].
Evidently, the map a n → a n + c n can be interpreted as a composition of a representation with an automorphism of our Lie algebra.Thus, if we further used our current to construct something -say a stress-energy field -then by composition with such an automorphism, we get a "transformed" stress-energy field.As an expression involving only normal powers and derivatives of J(z) + f (z), it still satisfies the same commutation relations with the same central charge, because the latter relations are determined by the U(1) commutation relation.
Following the ideas of Buchholz and Schulz-Mirbach [BSM90, (4.6)], we consider the above transformation with f (z) = κρ(z) + η, where κ, η are scalar constants and ρ(z) = −i z−1 z+1 .As was explained in Section 2.1, here we interpret ρ(z) as the formal series (3), rather than a function.Accordingly, ρ n = 0 for all n > 0 and in terms of Fourier modes, our transformation is where ρ ′ (z) denotes the derivative along the circle (1).The transformed U(1)-current field gives rise to a new associated stress-energy field.By an abuse of notation, we denote (the shifted version of) this by ϕ κ,η (T (z)), even though ϕ κ,η does not formally act on T (z).After a straightforward computation, we find that where T (z) = 1 2 : J 2 : (z) is the canonical stress-energy field of the original representation.
"Almost" symmetric stress-energy tensor with c > 1.Following the work of Buchholz and Schulz-Mirbach, given a U(1)-current field J(z), apart from the canonical (shifted) stressenergy field T (z) = 1 2 : J(z) 2 :, we shall also consider T κ (z) = n∈Z L κ,n z −n where and of course the product ρ(z)J(z) is understood in the sense of fields; i.e. its coefficient of z −n is m iκ(δ m,0 + 2(−1) m χ (−∞,0) (m))J n−m .Note that T 0 (z) = T (z); i.e. for κ = 0 the construction reduces to the canonical one.One can show that the operators {L κ,n } {n∈Z} form a representation of the Virasoro algebra with central charge c = 1+12κ 2 by a straightforward computation.However, we will not need that since we see this below in another way.The representation ( 16) is different from (8): the construction (8) does not yield a manifestly unitary vacuum representation with central charge c > 1.On the other hand, if 0 = κ ∈ R then c > 1 and if J(z) is symmetric and Ω is a lowest weight vector for J(z) with zero lowest weight q = 0 (i.e. if Ω was a vacuum vector for J(z)), then -as is easily checked -Ω is still a vacuum vector for the representation {L κ,n } {n∈Z} (Ω is not necessarily cyclic for {L κ,n } {n∈Z} , even if it was so for J(z)).Moreover, even if it is not properly symmetric, T κ (z) has a certain weakened symmetry property.Since the fields T (z), J(z), J ′ (z) appearing in our formula are symmetric, κ ∈ R and ρ is also real on the unit circle -as was explained at the end of Section 2 -we have that for any (scalar valued) trigonometric polynomial p(z) = |n|<N c n z −n satisfying the additional property p(−1) = 0.
Although different, this construction is closely related to (8).Indeed, if we apply the construction (16) to the current ϕ κ,η (J(z)) instead of J(z) (i.e.we apply the transformation ϕ κ,η with the same κ) then we obtain the stress-energy field of (8): where we used that ρ(z) satisfies the differential equation ( 15).This also shows that the operators {L κ,n } {n∈Z} indeed satisfy the Virasoro relations with central charge c = 1 + 12κ 2 , since the last expression coincides with (8).
Restoring unitarity to the Fateev-Zamolodchikov realization The transformation ϕ −κ,iκ will be of special interest.Since ρ 0 = i, it changes the lowest weight value for J(z) by −iκ + iκ = 0; i.e. it preserves the lowest weight.Moreover, by substituting η = iκ in (17) and taking account of the fact that ϕ −κ,iκ = ϕ −1 κ,−iκ , we see that suggesting that by applying ϕ −κ,iκ to the first of our commuting currents appearing in the Fateev-Zamolodchikov construction, we could turn our "very much non symmetric" fields into ones that have a discussed weak form of symmetry without changing lowest weight values.So let us take again two commuting U(1)-current fields J [1] (z), J [2] (z) and consider them as a representation of the direct sum of the Heisenberg algebra with itself.Then letting ϕ −κ,iκ act on the first one while not doing anything with the second one, i.e. the transformation φ−κ,iκ defined by can be viewed as a composition of our representation with an automorphism.Accordingly, we can apply the Fateev-Zamolodchikov realization (13)(14) to these representations φ−κ,iκ (J [1] (z)), φ−κ,iκ (J [2] (z)) and obtain a shifted pair of fields, which we denote by T (z; κ) and M (z; κ).Setting T 16) for j = 1, 2, by a straightforward computation we find that φ−κ,iκ ( T (z; κ) Since we obtained them by a transformation which is in fact a composition with an automorphism of a pair of U(1)-currents, the fields z 2 φ−κ,iκ ( T (z; κ)), z 3 φ−κ,iκ ( M(z; κ)) must still result in a representation of the W 3 -algebra.Moreover, since φ−κ,iκ transforms our currents in a manner that leaves every lowest weight vector a lowest weight vector with the same weight, by Proposition 3.3, we have that if Ω q 1 ,q 2 was a common lowest weight vector for J [1] (z) and J [2] (z) with lowest weights q 1 and q 2 respectively, then it will be also a lowest weight vector for the representation of the W 3 -algebra given by ( 19) with lowest weight value (h, w) given by Proposition 3.3.
Corollary 3.4.Let κ, q 1 , q 2 , b ∈ R be such that b 2 = 16 22+5c where c = 2 + 12κ 2 .Then there exists a lowest weight representation {(L n , W n )} n∈Z of the W 3 -algebra with central charge c = 2 + 12κ 2 and lowest weight (h, w) = ( on an inner product space such that the fields T (z) = n∈Z L n z −n and M(z) = n∈Z W n z −n satisfy the weak symmetry condition for all trigonometric polynomials p, r with p(−1) = r(−1) = r ′ (−1) = 0.
Proof.By taking a tensor product of two lowest weight representations, it is clear that we can construct two commuting symmetric U(1)-current fields J [1] (z), J [2] (z) on an inner product space having a common lowest weight vector Ω q 1 ,q 2 of lowest weight q 1 and q 2 , respectively.(Note: this is the point where we use that q 1 , q 2 are real: with a nonzero imaginary part, we could not have an invariant inner product for our currents).Now consider the representation z 2 T (z), z 3 M(z) of the W 3 -algebra constructed through (19) with the help of the fields J [1] (z) and J [2] (z).Taking account of the symmetry of our currents, the fact that κ, b ∈ R and the comments at the end of Section 2, we see that T (z) and M(z) indeed satisfy the required symmetry condition.Moreover, by Proposition 3.3 and the observation above the current corollary, Ω q 1 ,q 2 is a lowest weight vector for this representation with the claimed lowest weight value.Thus, restricting our representation of the W 3 -algebra to the cyclic subspace of Ω q 1 ,q 2 gives a lowest weight representation with all the desired properties.
Remark 3.5.One might wonder whether our "weak" symmetry condition in the above corollary actually implies "true" symmetry.It turns out that in the vacuum case this is exactly what happens -we shall see this in the next section.However, note that in general, the answer is: "no".In fact, if q 1 = 0, then h is not real, so we cannot even have an invariant Hermitian form (let alone an inner product).Actually, by (12), even if we set q 1 = 0 (and hence have real h and w), in general we cannot have unitarity (see Theorem 3.8 for some values of h, w for which unitarity fails).Indeed, our argument in the next section will use in a crucial way that h = w = 0.In contrast, in the Virasoro case, the "weak" symmetry can indeed be turned into "true" one; see Proposition 3.11.

Proof of unitarity for h = w = 0
In this section we will work in an abstract setting: we suppose that {(L n , W n )} {n∈Z} form a representation of the W 3 -algebra with central charge c ≥ 2 and that we are also given a nonzero vector Ω as well as an inner product •, • satisfying the following requirements: (i) Ω is a cyclic lowest weight vector for our representation and for all trigonometric polynomials p, r with p(−1) = r(−1) = r ′ (−1) = 0 (where the adjoint is considered w.r.t. the given inner product •, • ).
Such a representation and inner product indeed exists; this is clear by considering Corollary 3.4 with q 1 = q 2 = 0 and κ = c−2 12 .From now on we shall not be interested how these objects were explicitly constructed; we will only use to above listed properties.Our aim will be to conclude that •, • is in fact an invariant inner product for our representation, making it unitary.Since we work with Fourier modes rather than fields, we begin with reformulating property (ii).
Lemma 3.6.Let p(z) = |n|<N c n z −n and r(z) = |n|<N d n z −n be a trigonometric polynomials satisfying p(−1) = r(−1) = r ′ (−1) = 0. Then Proof.This is evident by considering the zero mode of the products appearing in the equalities of property (ii).
This implies in particular that (L n 1 − (−1) The next Lemma follows from Assumption (ii), and the form •, • is not necessarily the canonical one for (c, h, w).

Consequently, the representation is unitary.
Proof.We fist show that L † 0 = L 0 .Each vector of the form (10) is an eigenvector of L 0 with a real eigenvalue and since we are in a lowest weight representation, these vectors -and hence also the eigenspaces of L 0 -span the full space.So to prove that L † 0 = L 0 , it is enough to check that these vectors are orthogonal to each other whenever the associated eigenvalues of L 0 are not equal.We will do this by performing an induction.
Induction on gr := 2×(number of L's) + 3×(number of W 's). Assume that for some j ∈ {0, 1, . ..} it holds that whenever ℓ, ℓ ′ , k, k ′ are nonnegative integers of "total grade value" (see [BMP96] for a similar grading) e. unless they correspond to the same eigenvalue of L 0 ).Note that for gr = 0, our assumption is trivially true as in that case we have a single possible pair of vectors only: Ψ = Ψ ′ = Ω.We have to show that this remains true for gr = j + 1.We will do this by considering all possible pairs of vectors Ψ, Ψ ′ of the form (21) with gr = 2(ℓ there is nothing to prove, so assume λ = λ ′ .Since now we treat the case when the sum of ℓ and ℓ ′ is positive, at least one of them must be nonzero; so say ℓ ≥ 1, meaning that Ψ must contain at least one L operator in its defining expression.Let then ξ be the vector obtained by removing the first L operator from the expression of Ψ, namely, Ψ = L −m 1 ξ.The vector ξ is still given by an expression of the same form than Ψ or Ψ ′ , but the corresponding eigenvalue of L 0 is (λ − m 1 ) and hence . Putting all this together, we have We will argue that both terms in the above sum are separately zero and we begin with the second term.The total number of L's in the expression giving ξ and Ψ ′ is (ℓ − 1) + ℓ ′ and the total number of W 's is k + k ′ .Thus, by the inductive hypothesis, their inner product is zero unless they correspond to the same eigenvalue of L 0 , in which case we must have λ −m 1 = λ ′ .In either case, the product (λ where we just symbolically wrote "L . . .LW . . .W " without detailing the indices.Using the W 3 -algebra commutation relations, the above vector can be rewritten as a linear combinations of vectors of the form (21) with the same associated eigenvalue of L 0 -i.e. with eigenvalue λ ′ − m 1 -but with strictly smaller values of the quantity "2 × the number of L's + 3 × the number of W 's ". (E.g.note that when exchanging the two W operators, then, due to the commutation relations, two "new" L operators can appear -but only on the "cost" of having two W operators less.This is why we gave more weight to a W operator than an L operator.)Therefore, again by the inductive hypothesis and λ = λ ′ , we have ξ, L m 1 Ψ ′ = 0 and thus Ψ, Ψ ′ = 0.
Case 2: ℓ = ℓ ′ = 0.In this second case we have no L operators at all in the defining expressions of our two vectors: Again we may assume that λ = λ ′ , and so in particular we must have at least one W operator in our expressions (otherwise Ψ = Ψ ′ = Ω).So say k ≥ 1 and let ξ be the vector obtain by removing the last W from the expression of Ψ.Then W −n 1 ξ = Ψ and L 0 ξ = (λ − n 1 )ξ.
By Lemma 3.7, W s Ω = W −s Ω = 0 for s ∈ {0, 1, 2}.Since the index set {0, 1, 2} has three elements, there must exists at least two different r, s ∈ {0, 1, 2} such that neither W −r ξ nor W −s ξ does not correspond to the same eigenvalue of L 0 as Ψ ′ ; i.e. that λ ′ = (λ − n 1 + s), (λ − n 1 + r).Then by (20), we have some real numbers u, d such that we have the adjoint relation Since both A = W n 1 + uW r + dW s and B = (uW −r + dW −s ) annihilate Ω, one can rewrite the above expressions using commutators as a linear combination of terms with strictly smaller total value of the quantity "2 × the number of L's + 3 × the number of W 's" than the original value gr.Moreover, by our choice of s and r, the corresponding eigenvalues of L 0 of the terms on the two sides of the inner product never coincide.So again by the inductive hypothesis, each of those inner product values are zero and hence Ψ and Ψ ′ are orthogonal.Now we know that We then have Corollary 3.9.The irreducible lowest weight representation of the W 3 -algebra with central charge c ≥ 2 and lowest weights h = w = 0 is unitary.
Remark 3.10.By the existence theorem [Kac98, Theorem 4.5], any lowest weight representation where the lowest weight vector Ω satisfies the extra condition L −1 Ω = 0, generates a vertex algebra with translation operator T = L −1 .(This condition implies that the lowest weight must be (h, w) = (0, 0) but not the other way around.Note however that in the unitary case, h = 0 alone implies L −1 Ω = 0.) This vertex algebra evidently has a Virasoro element ν = L −2 Ω whose corresponding field has T as a component, and since the representation space is the direct sum of eigenspaces of L 0 with non-negative integral eigenvalues and each eigenspace is finite dimensional as it is spanned by finite many vectors of the form (10), the resulting structure is actually a vertex operator algebra (VOA).Moreover, if the representation we started with was unitary, then the obtained VOA is also unitary in the sense of [ It is worth noting that with the same induction technique we used in this section, we can show that if a lowest weight representation of the Virasoro algebra {L n } n∈Z on an inner product space satisfies (ii) in the sense that L 0 − (−1) n L n = (L 0 − (−1) n L −n ) † for all n ∈ Z, then in fact our inner product is an invariant form for the representation; in this case we do not need to assume that h = 0. Proposition 3.11.Let {L n } {n∈Z} be a lowest weight representation of the Virasoro algebra with lowest weight h ∈ R and lowest weight vector Ω, and suppose that (L 0 − (−1) n L −n ) † = L 0 − (−1) n L n for all n ∈ Z with respect to a given Hermitian form •, • (not necessarily the canonical one).Then L n = L † −n for all n ∈ Z.
Proof.As in Theorem 3.8, it is enough to prove that L † 0 = L 0 .Let V h+n be the eigenspaces of L 0 .Assume that V h , • • • , V n+h are pairwise orthogonal.(For n = 0 this is trivial.)This implies that L 0 is symmetric when restricted to We have to show that ξ, η = 0. We may assume that ξ = L −j ζ, where ζ ∈ V h+n−j+1 , as the general case is a linear combination.We have (L −j − (−1) where the 3rd equality holds since L 0 is symmetric on V 0 ⊕ • • • ⊕ V h+n , and the last equality follows from L j η ∈ V h ⊕ • • • ⊕ V h+n−j and the hypothesis of induction.

More unitary representations
It is also possible to construct unitary representations on the full space of the two commuting currents we used.Suppose again that we have two commuting U(1)-current fields J [j] (z) = n∈Z J n z −n = za [j] (z) (j = 1, 2) having a common lowest weight vector Ω q 1 ,q 2 with lowest weights q 1 and q 2 , respectively and that we have a fixed inner product on our representation space making our currents J [1] (z), J [2] (z) symmetric.Such currents on an inner product space indeed exist if q 1 , q 2 ∈ R (e.g.consider the tensor product of two lowest weight representations).We now perform transformation φ0,iκ ; i.e. while remaining on the same inner product space, we consider the currents φ0,iκ (J instead of the original ones J [1] (z), J [2] (z).The vector Ω q 1 ,q 2 is still a common lowest weight vector for these currents, but this time with lowest weights q1 = q 1 + iκ and q2 = q 2 .Recall that the transformation φ0,iκ can be viewed as a composition of a representation with a Lie algebra automorphism, and can be further composed with the Fateev-Zamolodchikov realization of the W 3 -algebra.By the usual abuse of notation, we shall denote the fields constructed from φ0,iκ (J [1] (z)) and φ0,iκ (J [2] (z)) using the formulas (13) and ( 14) by φ0,iκ ( T (z; κ)) and φ0,iκ ( M (z; κ)).Note that κ appears twice in these expressions: its value effects both the transformation we perform on the currents and the Fateev-Zamolodchikov construction.In For the convenience of the algebra-oriented reader, we show the "unshifted" fields: φ0,iκ ( L(z; α 0 )) This results allows us to completely characterize unitarity in the region 2 ≤ c ≤ 98.
Corollary 3.13.Let 2 ≤ c ≤ 98.Then the irreducible lowest weight representation of the Proof.As we already mentioned at (12), the condition f 11 (h, c) − w 2 ≥ 0 is necessary for unitarity, so we only need to show the "if" part.Consider the open region H and the closed region R defined by Our aim is to prove unitarity in the region R. Now one that R = H ∪ {(c, 0, 0)|2 ≤ c ≤ 98} and Corollary 3.9 tells us that we indeed have unitarity on the line {(c, 0, 0)|2 ≤ c ≤ 98}; so let us turn our attention to the region H.
It is clear that f 11 (h, c) is monotonically increasing with respect to h and hence that (c, h, w) ∈ H if and only if 2 < c < 98, h > c−2 32 , |w| < f 11 (c, h).In particular, H is connected.As we already mentioned at (12), in this region all Kac determinants are positive and hence, as was explained in Section 2.6, unitarity at a single point of H implies unitarity for the entire closure H. Since e.g.(3, 1 24 , 0) ∈ H, and at c = 3, h = 1 24 , w = 0 unitarity holds by the previous theorem, therefore, we have unitarity on H.

Outlook
The existence of unitary vacuum representations urges us to investigate the conformal field theories (conformal nets and vertex operator algebras, see e.g.[CKLW18]) related with these representations.Specifically, we are interested in the following questions.
• Can one always construct a conformal net using the unitary vacuum representations?
• Are all other unitary representations associated with DHR sectors of these conformal nets?(C.f.[Car04,Wei17] for the similar question regarding the Virasoro algebra.) • How does the present result generalize to other W-algebras?
and and ν where d(L) = 2 and d(W ) = 3 (see [BMP96] for a similar grading).Note that both g and λ are completely symmetric in their arguments.Let {L n , W n } n∈Z form a lowest weight representation of the W 3 -algebra with central charge c = − 22 5 , lowest weight (h, w) ∈ C 2 and lowest weight vector Ψ.Then, using the W 3 -algebra relations (5) and that Ψ is a lowest weight vector, it is straightforward to show that for any permutation σ, the difference can be written as a linear combination of terms of the form . ., ν ′ s ) strictly smaller9 than g(ν 1 , . . ., ν s ) and coefficients which are real polynomials of c, 1 22+5c , h and w.In particular, it follows that the cyclic space obtained from Ψ is spanned by vectors of the form and m < n).However, this is not the only important conclusion one can draw.
there exists a real polynomial p such that whenever {L n , W n } is a representation of the W 3 -algebra with central charge c = − 22 5 on a space V with a twisted-invariant bilinear form (•, •) and lowest weight vector Ψ ∈ V with lowest weights (h, w) and (Ψ, Ψ) = 1, then 22+5c , h, w).Proof.We shall inductively construct such polynomials without any particular knowledge about the actual representation.It is enough to deal with the case s = 0, since by the invariance of the form, we can put everything on one side: If further r = 0, then the claim is trivially true, while for r = 1, we have the expression (K ν 1 Ψ, Ψ) = (Ψ, K −ν 1 Ψ), showing that it is zero unless λ(ν 1 ) = 0, in which vase it is h when ν 1 = (L, 0) and w when ν 1 = (W, 0).Thus the claim is true for g ), depending on whether ν 1 = (L, 0) or (W, 0).In both cases we are done, as by the inductive hypothesis, we already have a polynomial giving the value of (K ν 2 • • • K νr Ψ, Ψ).If finally λ(ν 1 ) > 0, then K ν 1 annihilates Ψ and which, as was mentioned, can be rewritten as a linear combination of terms of the form . ., ν ′ s ) strictly smaller than g(ν 1 , µ 1 , . . ., ν r ) and coefficients which are real polynomials of c, 1 22+5c , h and w.This concludes the induction.Corollary A.2.The W 3 -algebra admits a lowest weight representation with a symmetric, non-degenerate twisted-invariant bilinear form form for every value of the central charge c = − 22 5 and lowest weight (h, w) ∈ C 2 .If further c, h, w ∈ R, then the same remains true even if we replace the words "symmetric bilinear" by "Hermitian".
Proof.Consider a lowest weight representation with either a non-degenerate, symmetric twisted-invariant bilinear form (•, •) or a non-degenerate Hermitian invariant sesquilinear form •, • .If c, h, w ∈ R, then the arguments used in our previous proof remain valid regardless whether we apply them for (•, •) or •, • and show that the product of elements from the real subspace M spanned by vectors of the form K ν 1 • • • K νr Ψ is real and hence -because of the non-degeneracy of the form -that M ∩ iM = {0}.It then follows that starting from either (•, •) or from •, • , the equation defines unambiguously the other object with all the desired properties.
By the construction in Section 3.4, there exists a region H ⊂ R 3 with nonempty interior such that for all (c, h, w) ∈ H, there is a lowest weight representation of the W 3 -algebra with central charge c and lowest weight (h, w) having an invariant inner product (see Theorem 3.12 for an actual description of the region H).In particular, for these values of c, h and w we also have the existence of a non-degenerate, symmetric twisted-invariant bilinear form.Now suppose the value of c = − 22 5 , h and w are arbitrary.Let Ṽ be the linear space freely spanned by (at the moment formal) expressions of the form K ν 1 • • • K νr Ψ where r ∈ {0, 1, • • • }.We introduce a bilinear form on Ṽ by setting where for each choice of ν 1 , • • • , ν r and µ 1 , • • • , µ s , p is a (possibly different) polynomial as in Proposition A.1.Note in particular, that the above value given to the form is a rational function of c, h, w, and thus it is completely determined by its values in H.
To check that the introduced form is symmetric, we need to verify that for each choice of ν 1 , • • • , ν r and µ 1 , • • • , µ s .However -though not indicated in notationseach side of the above expression is a rational function of c, h, w, and when (c, h, w) ∈ H, we indeed have an equality.But if an equality of rational functions holds in H, then so does for all of their domain.Let V be the space obtained by factorizing Ṽ with the set of "null-vectors", i.e. by the subspace Ñ := {x ∈ Ṽ : for all y ∈ Ṽ : (x, y) = 0}.On this space, our form is still well-defined, symmetric, bilinear and by its construction, non-degenerate.We have to show that the natural action of the K operators on V is well-defined and gives a lowest weight representation of the W 3 -algebra on the factorized space V .
To show well-definedness, we need to check that if x ∈ Ñ , then K ν x ∈ Ñ; that is, (K ν x, y) = 0 for all (non-commutative) polynomial y in {L n , W n }.We know that the lefthand side is a rational function of (c, h, w) and that its value is indeed zero in H -and hence that it is zero on all of its domain.This proves well-definedness.Lastly, to verify that V gives a lowest weight representations, we only have to repeat the argument: both of the W 3 relations and the lowest weight property are written as equalities between rational functions in c, h, w with only singularity at c = − 22 5 , therefore, their validity in H implies their validity for all (c, h, w), c = − 22 5 .Although we do not need Verma modules for our main results, we think it worth explaining how their existence can be verified using reasoning similar to what we have just employed.In addition, although we will need Kac determinants and in particular the results of Mizoguchi in [Miz89], we note that, for the notion of Kac determinant to be well-defined, there is no need to have a Verma module.Indeed, as was explained, the value of (K is universal : it depends only on the central charge c and lowest weights h, w, but not the particular representation.Indeed, to obtain his result, Mizoguchi never considers Verma modules; he works with some concrete representation to find null-vectors.Therefore, our use in Corollary 3.13 and Proposition A.3 of the Kac determinant computed in [Miz89] does not involve circular arguments and is justified. Proposition A.3.For every value of the central charge c = − 22 5 and lowest weights (h, w) ∈ C 2 , there exists (an up to isomorphism) unique lowest weight representation of the W 3 -algebra with lowest weight vector Ψ in which vectors of the form where n 1 ≤ • • • ≤ n s < 0 and m 1 ≤ • • • ≤ m r < 0, form a basis; i.e. a Verma representation.This representation admits a unique twisted-invariant bilinear form (•, •) with normalization (Ψ, Ψ) = 1, and this form is automatically symmetric.Moreover, if in addition c, h, w ∈ R, then everything remains true even if we replace the words "bilinear" by "sesquilinear" and "symmetric" by "Hermitian".
Proof.By now we know that for every c = − 22 5 and (h, w) ∈ C 2 there is an irreducible lowest weight representation.However, in this representation, when (c, h, w) ∈ H, where H is the set introduced in the proof of Corollary 3.13, the vectors (23) are independent (since in H all Kac determinants are strictly positive) and thus this representation is the Verma one.
For the rest of values, we consider the abstract space V spanned freely by vectors of the form (23).By doing so, seemingly we have linear independence for free.However, we have to check that it carries a corresponding representation!At this point, we use quotation marks and write symbols such as "K ν 1 • • • K νr Ψ", as this is indeed a vector of V by construction, but it is not (yet) the vector Ψ acted on by K. Given a c = − 22 5 and (h, w) ∈ C 2 , our task is then to define, for each ν, an operator K ν acting on V so that they satisfy the following requirements: (i) K ν Ψ = 0 whenever λ(ν) > 0, L 0 Ψ = hΨ, W 0 Ψ = wΨ (ii) if ν, ν 1 • • • , ν r are lexicographically ordered and ℓ(ν), ℓ(ν 1 ), . . .ℓ(ν r ) < 0, then the action of K ν on the (abstract) vector "K ν 1 • • • K νr Ψ" should result in the (abstract) vector "K ν K ν 1 • • • K νr Ψ".
(iii) {K ν } ν∈{L,W }×Z is a representation of the W 3 -algebra with central charge c.
Let us enumerate our basis vectors of the form (23) and denote them by Ψ 0 = Ψ, Ψ 1 , Ψ 2 , . ...An action of K ν can be defined by fixing its matrix-components; i.e. by choosing scalars M ν,j,k (c, h, w) ∈ C and setting K ν Ψ j := k M ν,j,k (c, h, w)Ψ k .When (c, h, w) ∈ H, we know that this can be done in a way so that requirements (i), (ii) and (iii) are met, because for those values we do have Verma representations.However, it is not difficult to see that again, the coefficients M ν,j,k (c, h, w) given by those Verma representations which are already known to exist, are rational expressions of the central charge c and lowest weights (h, w) with real coefficients and possible singularity only at c = − 22 5 .Thus, we can naturally continue them also outside of H.
We use these analytically continued matrix coefficients define the operators K ν .Again, since inside H these coefficients satisfy the properties (i), (ii) and (iii) that are expressed in terms of rational functions of c, h, w with only possible singularity at c = − 22 5 .the same remains true outside.This proves that we obtain a lowest weight representation on V .
span the whole representation space.However, in general, linear independence does not follow.Nevertheless, for each central charge c = − 22 5 d N among (10) for any given N that span a finite dimensional subspace in V W 3 c,h,w and one can consider the Gram matrix M N,c,h,w whose entries are the product values Ψ (N ) j , Ψ (N ) k 0 , where χ (−∞,0) is the characteristic function of the open interval (−∞, 0).The reader might wonder what is the reason behind the choice of the function ρ.As we shall see in the next subsection, what makes [DSK05,DSK06]ition 5.2]; see e.g.[CKLW18, Proposition 5.17], which says that unitarity follows if the VOA is generated by a family of Hermitian 7 quasi-primary fields.The unitary VOAs constructed in this way must coincide with the simple quotients of the freely generated VOAs defined from the Verma modules for the W 3 -algebra in [Lin09, Section 5].The latter can be identified as special cases of the universal VOAs in[DSK05,DSK06], as a consequence of [DSK06, Theorem 3.14], cf. also [DSK05, Proposition 3.11, Example 3.14].