Simple Virasoro modules induced from codimension one subalgebras of the positive part

We construct a new five-parameter family of simple modules over the Virasoro Lie algebra.


Introduction and description of the results
We denote by N the set of positive integers and by Z + the set of all non-negative integers. For a Lie algebra a we denote by U(a) the universal enveloping algebra of a.
Let V denote the complex Virasoro algebra, that is the Lie algebra with basis {c, l i : i ∈ Z} and the Lie bracket defined (for i, j ∈ Z) as follows: This algebra plays an important role in various questions of mathematical physics, see [KR]. Classical classes of simple weight V-modules are simple highest weight modules, see [FF], and intermediate series modules. Put together, these two classes exhaust all simple weight V-modules with finite dimensional weight spaces, see [Mt], and even those containing a finite dimensional weight space, see [MZ1]. We also refer the reader to the recent monograph [IK] for a detailed survey of the classical part of the representation theory of V. There are a number of other examples of simple V-modules constructed in [Zh,OW,GWZ,LGZ,MZ2] using various tricks. The present note contributes with a new trick leading to a new family of examples depending on five parameters.
For a nonzero z ∈ C, denote by a z the linear span of l k − z k−1 l 1 , where k ≥ 2, in V. It is easy to check that a z is in fact a Lie subalgebra in V. For a fixed m := (m 2 , m 3 , m 4 ) ∈ C 3 , define an a z -action on C by • (l i − z i−1 l 1 ) · 1 = m i for i = 2, 3, 4; • (l i − z i−1 l 1 ) · 1 = −(i − 4)m 3 z i−3 + (i − 3)m 4 z i−4 for i > 4.
It's straightforward to verify that this gives an a z -module. We denote it by C m .
For a fixed θ ∈ C consider the V-module Our main result is the following claim.
Then for any θ ∈ C the V-module Ind z,θ (C m ) is simple.
We start the paper by characterizing a z as a special family of codimension one subalgebras in the positive part n of V in Section 2. Our proof of Theorem 1 consists of three steps: we first induce C m up to n in Section 3, then we study the induction of C m to the Borel subalgebra of V in Section 4 and, finally, we complete the proof in Section 5.
Acknowledgement. The major part of this work was done during the visit of the second author to Uppsala in May 2012. The hospitality and financial support of Uppsala University are gratefully acknowledged. The first author is partially supported by the Royal Swedish Academy of Sciences and the Swedish Research Council.

Subalgebras in n of codimension one
Denote by n the "positive part" of V, that is the Lie subalgebra spanned by l i , where i ≥ 1. In this section we characterize a z as a special family of codimension one subalgebras in n. We start with the following lemma.
Lemma 2. Let x ⊆ n be a Lie subalgebra of codimension one. (a) If l 1 ∈ x, then l k , l k+1 , . . . ∈ x for some k ∈ N.
. ∈ x for some s, i ∈ N, then l k , l k+1 , . . . ∈ x for some k.
Proof. If l k ∈ a for some k > 1, then claim (a) follows immediately. Therefore, assume that this is not the case and in particular that claim (a) is false. Since x has codimension 1 and l k ∈ x for any k > 1, we must have that l k − a k l k+1 ∈ x for all k ≥ 2 and some 0 = a k ∈ C. Also, [l 1 , l k − a k l k+1 ] ∈ x, which implies a k+1 = k k−1 a k = ka 2 . On the other hand, [l 2 − a 2 l 3 , l 3 − a 3 l 4 ] = l 5 − 2a 3 l 6 + a 2 a 3 l 7 = l 5 − 4a 2 l 6 + 2a 2 2 l 7 . Since l 5 − a 5 l 6 = l 5 − 4a 2 l 6 , this implies that l 7 ∈ x, a contradiction. This proves claim (a). Claim (b) follows from claim (a) as the span of all l js , j ∈ N, is isomorphic to n as a Lie algebra.
To prove claim (c) we show that for any 1 ≤ p < s there is j ∈ Z + such that l p+js ∈ x. Assume that this is not the case. Then l p+js − a j l p+(j+1)s ∈ x for all j ∈ Z + and some 0 = a j ∈ C. Also, On the other hand, [l 2is , l p+js − a j l p+(j+1)s ] ∈ x, which implies a 2i+j = p+(j+1−2i)s p+(j−2i)s a j . This gives for all j ∈ Z + . This means that the following multisets (the negatives of the roots with respect to the variable js) coincide: Since both i, s ∈ N, the only way to have a zero on the left hand side is to have i = 1, which implies Now the left hand side contains a positive integer, while the right hand side does not. This is a contradiction which proves claim (c).
Proposition 3. Suppose x ⊆ n is a subalgebra of codimension one and there is no k such that l k , l k+1 , . . . ∈ x. Then x has a basis of the form Proof. By Lemma 2 we have l s ∈ x for any s ∈ N. As x has codimension one in n, it must have a basis of the form {l k − a k l 1 | k ≥ 2} for some nonzero a k ∈ C. For 2 ≤ i < j consider Taking i = 2 and j = 3, 4, . . . we get a recursive formula which uniquely determines a 5 , a 6 , . . . in terms of a 2 , a 3 , a 4 .
Let I be the ideal in C[a 2 , a 3 , a 4 , a 5 , a 6 , a 7 , a 8 , a 9 ] generated by D 2,3 , D 2,4 , D 2,5 , D 2,6 , D 2,7 , D 3,4 , D 3,5 , D 3,6 and D 4,5 . Computing the Gröbner basis of I with respect to the lexicographic order for which a 9 > a 8 > a 7 > a 6 > a 5 > a 4 > a 3 > a 2 , we get that I contains a 6 3 − a 5 3 a 2 2 , which implies a 3 = a 2 2 since a 3 = 0. Computing the Gröbner basis of I with respect to the lexicographic order for which a 9 > a 8 > a 7 > a 6 > a 5 > a 2 > a 3 > a 4 , we get that I contains a 3 4 a 2 − a 2 4 a 2 3 . Using a 3 = a 2 2 and a 4 , a 2 = 0, we get a 4 = a 3 2 . This means that all a k are uniquely determined by the value of a 2 .
At the same time, it is easy to check that a k = a k−1 2 satisfies (2.1) and hence defines a subalgebra.
For a nonzero z ∈ C we denote by a z the subalgebra constructed in Proposition 3. This one-parameter family of subalgebras exhausts all codimension one subalgebras of n which do not contain l k , l k+1 , . . . for some k. Modules induced from subalgebras of n containing l k , l k+1 , . . . for some k were studied in [MZ2]. All such modules fit into the general Whittaker setup for V defined in [BM]. Simple modules which are induced from simple 1-dimensional a z -modules do not fit into this general Whittaker setup. They are the objects of our study in the present paper.

Induction to n
For m := (m 1 , m 2 , m 3 ) ∈ C 3 define V m = Ind n az (C m ). Since n has a basis l 1 , l 2 − zl 1 , l 3 − z 2 l 1 , . . ., the PBW Theorem implies that V m has a basis {l k 1 ⊗ 1 | k ≥ 0}. Proposition 4. The n-module V m is simple if and only if am 3 = m 4 .
Proof. Suppose 0 = M ⊆ V m . Since l 1 acts freely on M, it follows that V m /M is finite-dimensional. In particular, we may assume that M is chosen so that V m /M is simple.
Let I ⊆ n be the annihilator of this quotient module. In [MZ2,Subsection 3.3] it is shown that any ideal in n of finite codimension contains l k , l k+1 , . . . for some k > 0. This implies that n/I is nilpotent, V m /M is one-dimensional, and [n, n] ⊆ I. Therefore, l 3 and l 4 act on V m /M as zero and thus z(l 3 − z 2 l 1 ) and l 4 − z 3 l 1 act equally, which yields zm 3 = m 4 .
Assume now that zm 3 = m 4 and consider the n-module C m 2 ,m 3 on which l 1 acts via − 1 z 2 m 3 and l 2 acts via m 2 − 1 z m 3 . Computing its restriction to a z one gets that the restriction is isomorphic to C m . From the universal property of induced modules it follows that V m surjects onto this n-module C m 2 ,m 3 and hence is reducible.
From the above proof it follows that in the case zm 3 = m 4 the module V m has a unique simple top isomorphic to the n-module C m 2 ,m 3 on which l 1 acts via − 1 z 2 m 3 and l 2 acts via m 2 − 1 z m 3 . The induced module Ind z,θ (C m 2 ,m 3 ) is completely described in [OW,MZ2].
For k > 4 set To simplify our notation, for k ≥ 2 we will denote byl k the element l k − z k−1 l 1 ∈ a z . We will need the following property of V m : Lemma 5. Let k ≥ 2. Then the module V m has a basis {v Proof. Let v be the canonical generator of V m . For n ∈ N 0 denote by V (n) the linear span of v, l 1 v, . . . , l n 1 v. We claim that V (n) is invariant under the action ofl k and, moreover, . We show this by induction on n. If n = 0, the claim is clear. To show the induction step, we compute (by movingl k through one l 1 and using the inductive assumption in the first equality): This implies (3.2) and the statement of the lemma follows.

Induction to the Borel
Denote by b the standard Borel subalgebra of V, that is the subalgebra generated by n and l 0 . Define W m = Ind b n (V m ). Proposition 6. If zm 3 = m 4 and m = (m 2 , 2zm 2 , 3z 2 m 2 ), then the b-module W m is simple.
Proof. Consider V = V m and let W be the induced b-module. Then every element in W can be written as i≥0 l i 0 v i where v i ∈ V and only finitely many of them are nonzero. Let X be a nonzero submodule of W and x ∈ X be such that, when written in the above form, the maximal i such that v i = 0 is minimal possible, let it be N.
Since V is a simple n-module, using the action of n we may assume that v N is the canonical generator v of V . We havel k · v = m k v.
We claim that y = (l k − m k )x is nonzero for some k (which reduces N giving a contradiction). For this we show that when we write y in the above form, then the coefficient at l N −1 0 will be nonzero. Clearly, y will not contain any coefficient at l N 0 .
Look at l N −1 0 V (modulo smaller powers of l 0 ) and choose there an eigenbasis forl k as given by Lemma 5. There is one eigenvector l N −1 0 v with eigenvalue m k and the spectrum ofl k is simple, which means that this eigenvector is not in the image ofl k − m k . Hence it is enough to show that y contains a nonzero component at this eigenvector coming from level N (and hence this component cannot cancel with anything from level N − 1 as it is not in the image).
Without loss of generality (namely, by factoring out smaller powers of l 0 ) we may assume N = 1. In this case we get that the eigenvector at level 0 with eigenvalue m k + (1 − k)z k is m k+1 v + z k l 1 v while the contribution from level 1 is km k v + (k − 1)z k−1 l 1 v (up to a nonzero constant). For these two vectors to be linearly dependent we get the equality (k − 1)m k+1 z k−1 = km k z k . This reduces, by recursion, to the equations m k = (k − 1)m 2 z k−1 for k > 2. The claim follows.
Consider the filtration of W m by n-submodules given by the degree of l 0 . Every layer of this filtration is isomorphic to V m and from Lemma 5 it follows that the kernel ofl 2 − m 2 on every layer of this filtration is 1-dimensional. The computation from the previous paragraph implies that each nonzero kernel element at the layer k is sent to a nonzero kernel element at the layer k − 1. This implies that the kernel of l 2 − m 2 on W m coincides with the kernel ofl 2 − m 2 on the first layer (corresponding to l 0 0 ). The latter one is exactly v by Lemma 5. This proves claim (a) forl 2 − m 2 and forl 3 − m 3 the arguments are similar.
From claim (a) and the computation above it follows that the equation (l 2 − m 2 ) · x = v has solutions y 2 + αv, α ∈ C. None of these equals y 3 which implies claim (b), completing the proof.

Proof of Theorem 1
To prove Theorem 1 we will use a variation of the argument from the proof of Proposition 6. Assume that m ∈ C 3 satisfies (1.1). For θ ∈ C, consider the module Ind z,θ (C m ).
Denote by M the set of all infinite vectors i = (. . . , i 2 , i 1 ) with nonnegative integer coefficients in which only finitely many coordinates are nonzero. For i ∈ M set l i := . . . l i 2 −2 l i 1 −1 ∈ U(V). For i ∈ M, the degree d(i) is defined as s≥0 i s , and the weight w(i) is defined as s≥0 si s . Let 0 = (. . . , 0, 0, 0) and for s ∈ Z 0 let ε s be the element (. . . , i 2 , i 1 ) such that i s = 1 and i t = 0 for all t = s. Define a total order ≺ on M recursively as follows: i ≺ j if and only if • w(i) < w(j); or • w(i) = w(j) and d(i) < d(j); or • w(i) = w(j) and d(i) = d(j) and min{s|i s = 0} > min{s|j s = 0}; or • w(i) = w(j) and d(i) = d(j) and min{s|i s = 0} = min{s|j s = 0} = p and i − ε p ≺ j − ε p . Clearly, the element 0 is the minimum element with respect to this order.
Every element in Ind z,θ (C m ) can be uniquely written as a sum of l i v i , where i ∈ M and v i ∈ W m (where only finitely many of the v i 's are non-zero). If x is a non-zero element written in this form, then the support of x is defined as the following finite set: supp ( Proof. By linearity, it is enough to prove the claim for u = l k and x = l i v i . We have l k · l i v i = l i l k · v i + [l k , l i ]v i . From the definition of ≺ it follows that, moving l k to the right in [l k , l i ], we get a linear combination of some l j (with possible element from b on the right) where j i. The claim follows.
Assume that i = 0 and let x ∈ M be some non-zero element with maximal term i. Since W m is a simple module by Proposition 6, without loss of generality we may assume v i = v (the canonical generator of V m ). For k ≥ 2 consider the element y k := (l k − m k ) · x and we have either y k = 0 or i ∈ supp(y k ). Therefore y k = 0 implies t(y k ) ≺ i by Lemma 8, which gives a contradiction completing the proof of Theorem 1. In what follows we show that either y 2 = 0 or y 3 = 0.
Let p := min{s : i s = 0} and consider first the case p > 1. We claim that j := i − ε p + ε p−1 belongs to either supp(y 2 ) or supp(y 3 ), which implies that at least one of these elements is nonzero. Assume that this is not the case and let k ∈ {2, 3}. Let i ′ ∈ supp(x) be different from i. Using the definition of ≺, it is easy to see that, writing [l k − m k , l i ′ ] as a linear combination of l s u s , where u s ∈ b, all s which will appear with nonzero u s satisfy s ≺ j.
At the same time, [l k − m k , l i ] contributes with −i p (p + 1)z k−1 l j . If y k = 0, then to cancel this contribution, we thus must have j ∈ supp(x) and v j ∈ W m should satisfy (l k − m k ) · v j = i p (p + 1)z k−1 v. However, such v j does not exist by Lemma 7(b), which completes the proof in the case p > 1.
Assume now that p = 1. Our argument will be similar to the one we used above, however, it will require a computationally more complicated analogue of Lemma 7. We claim that j := i − ε 1 belongs to either supp(y 2 ) or supp(y 3 ). Assume that this is not the case. Consider the coefficient at l j in both y 2 and y 3 . Similarly to the above, the only contribution to this coefficient comes from [l k − m k , l i ] and from (l k − m k ) · v j . So, if this coefficient is zero, these two contributions should cancel each other for both k = 2 and k = 3.
Both solutions in the proof of Lemma 9 were found and also Gröbner basis computations in the proof of Proposition 3 were performed using MAPLE, but it is straightforward to check that they are correct. Note that from the definition of our modules it follows immediately that different values of parameters give rise to non-isomorphic simple modules. It is also easy to check (looking at the action of a z ) that our modules are not isomorphic to any of the previously known simple Virasoro modules.