January 01, 1970
We describe finite simple Novikov conformal algebras over an algebraically closed field of characteristic zero. The classification includes the current conformal algebra over the base field and a one-parameter list of Virasoro-like conformal algebras. A semisimple finite Novikov conformal algebra is proved to be a direct sum of simple ones. We also describe deformations and central extensions of finite simple Novikov conformal algebras.
The notion of a conformal algebra [1] emerged as a formal language to describe the algebraic properties of coefficients in the singular part of the operator product expansion (OPE) of quantum fields in 2-dimensional conformal field theory. Namely, vertex algebras may be considered from the algebraic point of view as “breeds” of differential pre-Lie algebras and Lie conformal algebras [2] in a similar way as Poisson algebras are “breeds” of commutative and Lie algebras.
Every Lie conformal algebra may be presented by formal distributions over an “ordinary” Lie algebra. For example, the Virasoro conformal algebra \(\mathrm{Vir}\) over a field \(\Bbbk\) of characteristic zero is generated by the single formal distribution \(v(z) = \sum\limits_{n\in \mathbb{Z}} t^n\partial_t z^{-n-1}\) over the Witt Lie algebra \(\mathrm{Der}\,\Bbbk [t,t^{-1}]\).
Given an arbitrary non-associative algebra \((A,\cdot )\) over a field \(\Bbbk\), \(\mathrm{char}\,\Bbbk =0\), a subspace \(C\subseteq A[[z,z^{-1}]]\) of the space of formal distributions with coefficients in \(A\) is called a conformal algebra of formal distributions if the following three conditions hold:
for every \(a(z),b(z)\in C\) there exists an integer \(N\ge 0\) such that \[a(w)b(z)(w-z)^N = 0\] in \(A[[w,w^{-1},z,z^{-1}]]\);
if \(a(z)\in C\) then \(\partial_z a(z)\in C\);
if \(a,b\in C\) then the formal distributions—coefficients of the polynomial \[(a\mathbin{{}_{(\lambda )}}b)(z) =\mathop{\fam 0 Res}\limits_{w=0} a(w)b(z)\exp (\lambda (w-z))\] (at \(\lambda^n\), \(n\ge 0\)) belong to \(C\).
From the abstract point of view, a conformal algebra \(C\) may be considered as a linear space equipped with a linear map \(\partial\) and with a polynomial-valued bilinear operation \((\cdot \mathbin{{}_{(\lambda )}}\cdot )\). The product rule for the derivation implies \[\label{eq:sesqui-lin} (\partial a\mathbin{{}_{(\lambda )}}b) = -\lambda (a\mathbin{{}_{(\lambda )}}b), \quad (a\mathbin{{}_{(\lambda )}}\partial b) = (\partial+\lambda) (a\mathbin{{}_{(\lambda )}}b),\tag{1}\] known as the sesqui-linearity condition.
For every conformal algebra \(C\) one may construct an “ordinary” algebra \(A=\mathcal{A}(C)\) such that \(C\) embeds into \(A[[z,z^{-1}]]\) as a conformal algebra of formal distributions. If \(\mathcal{A}(C)\) is associative (Lie, commutative, etc.) then \(C\) is said to be an associative (resp., Lie, commutative, etc.) conformal algebra. In fact, for a variety \(\mathfrak V\) of ordinary algebras a conformal algebra \(C\) is a \(\mathfrak V\)-conformal algebra if and only if there exists an algebra \(A\) from \(\mathfrak V\) such that \(C\) is isomorphic to a conformal algebra of formal distributions over \(A\).
Example 1. Given an “ordinary” algebra \((P,\cdot)\), construct the free module \(C=\Bbbk [\partial ]\otimes P\) and define \[(x\mathbin{{}_{(\lambda )}}y) = xy,\quad x,y\in P,\] then extend to all \(C\) by 1 . Then \(C\) is a conformal algebra called current conformal algebra over \(P\) denoted \(\mathop{\fam 0 Cur}\nolimits P\). In this case, \(\mathcal{A}(C) = P[t,t^{-1}]\), so \(\mathop{\fam 0 Cur}\nolimits P\) is a \(\mathfrak V\)-conformal algebra if and only if \(P\) belongs to \(\mathfrak V\).
An alternative but equivalent approach to the definition of what is a \(\mathfrak V\)-conformal algebra was proposed in [3]. A conformal algebra may be considered as an algebra in the pseudo-tensor category \(H\)-mod for \(H=\Bbbk [\partial ]\). If we denote by \(\mathcal{O}_{\mathfrak V}\) the operad governing the variety \(\mathfrak V\) of algebras [4] then the class of \(\mathfrak V\)-conformal algebras coincides with the class of all morphisms from \(\mathcal{O}_{\mathfrak V}\) to \(H\)-mod.
A conformal algebra is said to be finite if it is finitely generated as a module over \(H=\Bbbk [\partial ]\). An ideal \(I\) of a conformal algebra \(C\) is an \(H\)-submodule which is closed with respect to \((\cdot\mathbin{{}_{(\lambda )}}\cdot)\), i.e., \((I\mathbin{{}_{(\lambda )}}C) + (C\mathbin{{}_{(\lambda )}}I)\subseteq I[\lambda ]\). A conformal algebra \(C\) is said to be simple if \((C\mathbin{{}_{(\lambda )}}C)\ne 0\) and there are no non-zero proper ideals of \(C\).
If \(\mathfrak V\) is a 2-variety (i.e., square of an ideal is again an ideal; so are the varieties of Lie, associative, and Novikov algebras) then, given an ideal \(I\) in a \(\mathfrak V\)-algebra, it is natural to consider the descending series of ideals \(I^{(1)}=I\), \(I^{(n+1)} = I^{(n)}I^{(n)}\). An ideal \(I\) is solvable if \(I^{(n)}=0\) for some \(n\ge 1\). An algebra without nonzero solvable ideals is called semisimple. For \(\mathfrak V\)-conformal algebras, the definition of semisimplicity is completely similar.
Simple finite Lie and associative conformal algebras were described in [5]. The case of Jordan algebras was considered in [6]. A more complicated description of simple finite Lie conformal superalgebras was found in [7], [8], the Jordan case was considered in [9].
The purpose of this paper is to describe simple finite conformal algebras corresponding to the variety of Novikov algebras. A Novikov algebra is a linear space \(V\) equipped with a bilinear operation \(\circ\) satisfying the following identities: \[(a\circ b)\circ c - a\circ (b\circ c) = (b\circ a)\circ c -b\circ (a\circ c),\] \[(a\circ b)\circ c = (a\circ c)\circ b,\] for all \(a,b,c \in V\). These identities first appeared in the paper [10] as a description of polynomial relations on the coefficients of a Hamiltonian differential operator. In [11], the same identities were used to describe the conditions on the coefficients of a generalized Poisson bracket in the study of partial differential equations of hydrodynamic type.
As a standard example, one may construct a Novikov algebra from an associative and commutative algebra \((A,\cdot )\) equipped with a derivation \(d\): then for every \(a\in A\) the new operation \[\label{eq:GD-construction} x\circ y = xd(y) + axy,\quad x,y\in A,\tag{2}\] turns the same space \(A\) into a Novikov algebra denoted \(A^{(d,a)}\). This series of examples appeared in [10] so it is called a Gelfand–Dorfman construction.
In some sense, the generalized Poisson bracket in [11] is a particular case of the commutator OPE formula in [1], so it is not surprising that Novikov algebras have tight relations to Lie conformal algebras (see also [3]).
The first structure result on Novikov algebras was obtained in [12]: a finite-dimensional simple Novikov algebra over a field of characteristic zero is a field; for a semisimple Novikov algebra (without nonzero solvable ideals) it was proved that it is a direct sum of simple ones. Further results on the structure of Novikov algebras and their representations were obtained in [13], [14], [15]. In particular, in the last paper it was shown that every simple finite-dimensional Novikov algebra (in arbitrary characteristic) is a Gelfand–Dorfman construction of a differentiably simple finite-dimensional commutative algebra.
We will describe simple finite Novikov conformal algebras over an algebraically closed field \(\Bbbk\) of characteristic zero. The methods of our study are based on the structure theory of associative conformal algebras with a finite faithful representation developed in [16]. In some sense, the classification obtained in this way agrees with the results of [15]: every simple finite Novikov conformal algebra is a (conformal) Gelfand–Dorfman construction over the only simple finite commutative conformal algebra \(\mathop{\fam 0 Cur}\nolimits\Bbbk\).
Let \(\mathrm{Nov}\) be the class of all Novikov algebras over the field \(\Bbbk\). Denote by \(\mathcal{O}_{\mathrm{Nov}}\) the corresponding operad. Then a Novikov conformal algebra is a morphism from \(\mathcal{O}_{\mathrm{Nov}}\) to the pseudo-tensor category \(\Bbbk [\partial ]\)-mod. Being translated to the language of \(\lambda\)-products, this definition turns into the following one.
Definition 1 ([17]). A Novikov conformal algebra is a left (unital) module \(V\) over \(H=\Bbbk [\partial ]\) equipped with a sesqui-linear operation \[(\cdot \mathbin{{\circ}_{(\lambda)}} \cdot ): V\otimes V \to \Bbbk [\partial,\lambda ]\otimes _H V\cong V[\lambda ]\] such that the following identities hold for all \(a,b,c\in V\): \[\label{eq:LSym} (a\mathbin{{\circ}_{(\lambda)}} b)\mathbin{{\circ}_{(\lambda +\mu)}} c - a\mathbin{{\circ}_{(\lambda)}} (b\mathbin{{\circ}_{(\mu)}} c) = (b\mathbin{{\circ}_{(\mu)}} a)\mathbin{{\circ}_{(\lambda +\mu)}} c - b\mathbin{{\circ}_{(\mu)}} (a\mathbin{{\circ}_{(\lambda)}} c),\tag{3}\] \[\label{eq:RCom} (a\mathbin{{\circ}_{(\lambda)}} b)\mathbin{{\circ}_{\mu }}c = \{(a\mathbin{{\circ}_{(\lambda)}} c)\mathbin{{\circ}_{(\mu-\lambda )}} b\}.\tag{4}\] Hereinafter, we use the notation \[\{x\mathbin{{\circ}_{(\lambda )}} y\} = (x\mathbin{{\circ}_{(-\partial-\lambda)}} y).\]
Example 2. If \(A\) is a Novikov algebra with a binary product \(\circ : A\otimes A\to A\) then the current conformal algebra \(\mathop{\fam 0 Cur}\nolimits A\) (see Example 1) is a Novikov conformal algebra.
Example 3. Let \(A=\Bbbk[t,t^{-1}]\) be the (commutative) algebra of Laurent polynomials. Let us fix a scalar \(\alpha \in \Bbbk\) and define the new operation \(\circ\) on \(A\) by the rule \[f(t)\circ g(t) = f(t)g'(t) +\alpha f(t)g(t),\] where \(g' = \partial_t g\) is the ordinary derivative. Then \((A,\circ )\) is a Novikov algebra. Consider the formal distribution \[v(z) = \sum\limits_{n\in \mathbb{Z}} t^nz^{-n-1} \in A[[z,z^{-1}]].\] Then \(v(w)\circ v(z) (w-z)^2 =0\) and \[(v\mathbin{{\circ}_{(\lambda)}} v)(z) = \alpha v(z) -\partial_z v(z) -\lambda v(z),\] so the distribution \(v(z)\) together with all its formal derivatives span a conformal algebra of formal distributions over the Novikov algebra \((A,\circ )\).
Example 4. Let \(P\) be a Novikov–Poisson algebra, i.e., a linear space with two operations \(\circ, *:P\otimes P\to P\) such that \((P,\circ)\) is a Novikov algebra, \((P,*)\) is an associative and commutative algebra, and the following identities hold [18]: \[(x\circ y)*z = (x*z)\circ y, \quad x*(y\circ z) - x*(z\circ y) = y\circ (x*z) - z\circ (x*y).\] Then the free \(H\)-module \(H\otimes P\) equipped with the \(\lambda\)-product \[(1\otimes x)\mathbin{{\circ}_{(\lambda)}} (1\otimes y ) = 1\otimes (x\circ y) +\lambda (1\otimes (x*y)) + \partial \otimes (x*y), \quad x,y\in P,\] is a Novikov conformal algebra.
In particular, for a fixed scalar \(\alpha \in \Bbbk\), the 1-dimensional space \(P=\Bbbk v\) is a Novikov–Poisson algebra relative to the operations \[v\ast v= v,\quad v\circ v=\alpha v.\] The Novikov conformal algebra constructed according to Example 4 is denoted \(\mathcal{V}_\alpha\), it is isomorphic to the algebra from Example 3.
Example 5. Let \(C\) be an associative and commutative conformal algebra with operations \(\partial\) and \((\cdot\mathbin{{}_{(\lambda )}}\cdot)\). Suppose \(D:C\to C\) is a derivation of \(C\), i.e., a \(\partial\)-invariant linear map such that \(D(a\mathbin{{}_{(\lambda )}}b ) = D(a)\mathbin{{}_{(\lambda )}}b + a\mathbin{{}_{(\lambda )}}D(b)\) for all \(a,b\in C\). Fix a scalar \(\alpha\in \Bbbk\) and define the new \(\lambda\)-product on the same \(H\)-module \(C\): \[a\mathbin{{\circ}_{(\lambda)}} b = a\mathbin{{}_{(\lambda )}}D(b) +\alpha (a\mathbin{{}_{(\lambda )}}b) , \quad a,b\in C,\] is a Novikov conformal algebra denoted \(C^{(D,\alpha)}\).
For example, \(D=\partial\) is a derivation in the above-mentioned sense, so every commutative conformal algebra \(C\) is a Novikov conformal algebra relative to the new operation \[a\mathbin{{\circ}_{(\lambda)}} b = (\partial +\lambda +\alpha)(a\mathbin{{}_{(\lambda )}}b), \quad a,b\in C.\]
Remark 1. The notion of derivation on conformal algebra mentioned above should not be confused with conformal derivation, which will be discussed later.
In contrast to ordinary Novikov algebras over a field, there exist Novikov conformal algebras that cannot be embedded into a commutative conformal algebra with a derivation. However, every finitely generated Novikov conformal algebra may be presented as a subalgebra of \(C^{(D,0)}\) for an appropriate commutative conformal algebra \(C\) with a derivation \(D\) [19].
Let \(H=\Bbbk[\partial ]\) be the polynomial algebra as above, and let \(V\) be an \(H\)-module. A conformal endomorphism \(f\) of \(V\) [1] is a linear map \[f_\lambda : V\to \Bbbk[\partial,\lambda ]\otimes_H V\cong V[\lambda ], \quad\] such that \[f_\lambda (\partial v) = (\partial+\lambda) f_\lambda (v), \quad v\in V.\] If \(V\) is a finitely generated \(H\)-module then the space \(\mathop{\fam 0 Cend}\nolimits V\) of all conformal endomorphisms of \(V\) is an associative conformal algebra relative to the operations \[(\partial f)_\lambda (v) = -\lambda f_\lambda (v),\] \[(f\mathbin{{}_{(\lambda )}}g)_\mu (v) = f_\lambda (g_{\mu-\lambda }(v)),\] for \(f,g\in \mathop{\fam 0 Cend}\nolimits V\), \(v\in V\).
If \(V\) is the free \(H\)-module of rank \(n\ge 1\) then \(\mathop{\fam 0 Cend}\nolimits V\) is often denoted \(\mathop{\fam 0 Cend}\nolimits_n\). The structure of this associative conformal algebra was systematically studied in [20]. Let us state here an isomorphic presentation following [16].
An element of \(V\) may be presented by a column of polynomials, i.e., \(V\cong H\otimes \Bbbk ^n\). A conformal endomorphism is presented by a matrix \(a(\partial, x) \in M_n(\Bbbk [\partial , x])\), so that \[a(\partial, x)_\lambda (h(\partial)\otimes u) = h(\partial+\lambda )a(-\lambda , \partial) u,\] for \(h\in H\), \(u\in \Bbbk ^n\). Hence, the associative conformal algebra structure on \(\mathop{\fam 0 Cend}\nolimits_n = M_n(\Bbbk [\partial, x])\) is given by \[a(\partial, x)\mathbin{{}_{(\lambda )}}b(\partial, x) = a(-\lambda , x)b(\partial+\lambda , x+\lambda ).\]
Suppose \(\mathcal{S}\) is a conformal subalgebra of the associative conformal algebra \(\mathop{\fam 0 Cend}\nolimits_n\), and let \(U\) be an \(H\)-submodule of \(V=H\otimes \Bbbk^n\). Then \(\mathcal{S}(U)\subseteq V\) stands for the linear span of all coefficients (at \(\lambda ^s\), \(s\ge 0\)) of all \(f_\lambda (u)\), \(f\in \mathcal{S}\), \(u\in U\). Since the base field \(\Bbbk\) is infinite, we may replace the collection of coefficients with the collection of values, i.e., \[\label{eq:Cend40U4195defn} \mathcal{S}(U) = \mathrm{span}_{\Bbbk} \{ f_\alpha(u) \mid f\in \mathcal{S}, u\in U, \alpha\in \Bbbk \}.\tag{5}\] This is obviously an \(H\)-submodule of \(V\).
A submodule \(U\) of \(V\) is said to be \(\mathcal{S}\)-invariant if \(\mathcal{S}(U)\subseteq U\). If there are no nontrivial \(\mathcal{S}\)-invariant submodules in \(V\) then the conformal subalgebra \(\mathcal{S} \subseteq \mathop{\fam 0 Cend}\nolimits_n\) is said to be irreducible [20].
As an example, consider \(\mathop{\fam 0 Cur}\nolimits_n = M_n(\Bbbk [\partial ])\subset \mathop{\fam 0 Cend}\nolimits_n\), this is an irreducible conformal subalgebra. Another series of examples may be constructed as follows: choose a matrix \(Q(x)\in M_n(\Bbbk [x])\), \(\det Q\ne 0\), and consider \(\mathop{\fam 0 Cend}\nolimits_{n,Q} = M_n(\Bbbk [\partial, x])Q(x-\partial )\). This is also an irreducible conformal subalgebra of \(\mathop{\fam 0 Cend}\nolimits_n\).
In [20] it was conjectured that the images of \(\mathop{\fam 0 Cur}\nolimits_n\) under automorphisms of \(\mathop{\fam 0 Cend}\nolimits_n\) and conformal subalgebras of the form \(\mathop{\fam 0 Cend}\nolimits_{n,Q}\), \(\det Q\ne 0\), exhaust all irreducible conformal subalgebras of \(\mathop{\fam 0 Cend}\nolimits_n\). This conjecture was proved in [16], and this result led to the structure theory of associative conformal algebras with finite faithful representation.
Namely, we say that an associative conformal algebra \(\mathcal{S}\) has a finite faithful representation (FFR) if \(\mathcal{S}\) is isomorphic to a subalgebra of \(\mathop{\fam 0 Cend}\nolimits V\) for some finitely generated \(H\)-module \(V\).
Theorem 1 ([16]). Let \(C\) be an associative conformal algebra with an FFR.
If \(C\) is simple then \(C\) is isomorphic either to \(\mathop{\fam 0 Cur}\nolimits_n\) or to \(\mathop{\fam 0 Cend}\nolimits_{n,Q}\), for some \(n\ge 1\) and \(Q\in M_n(\Bbbk[x])\), \(\det Q\ne 0\).
If \(C\) is semisimple then \(C\) is a finite direct sum of simple ones.
If \(C\) is not semisimple then there exists a maximal nilpotent ideal (or nilpotent radical) \(N\) of \(C\) such that \(C/N\) is a semisimple conformal algebra with an FFR.
Suppose \(V\) is a finite nonassociative conformal algebra with a \(\lambda\)-product \((\cdot \mathbin{{\circ}_{(\lambda)}} \cdot ): V\otimes V\to V[\lambda ]\). Consider the following conformal endomorphisms of \(V\) as of an \(H\)-module: \[L^a_\lambda : v\mapsto a\mathbin{{\circ}_{(\lambda )}} v, \quad R^a_\lambda : v\mapsto \{v\mathbin{{\circ}_{(\lambda)}} a\},\] for all \(a,v\in V\). Denote by \(\mathcal{L}\) the conformal subalgebra of \(\mathop{\fam 0 Cend}\nolimits V\) generated by all \(L^a\), and let \(\mathcal{R}\) stand for the subalgebra of \(\mathop{\fam 0 Cend}\nolimits V\) generated by all \(R^a\), \(a\in V\). Similarly, denote by \(\mathcal{M}\) the subalgebra generated jointly by \(\mathcal{L}\) and \(\mathcal{R}\) in \(\mathop{\fam 0 Cend}\nolimits V\).
Note that a conformal \(\mathcal{M}\)-submodule of \(V\) is exactly an ideal of \(V\). Hence, if \(V\) is simple then \(\mathcal{M}\) is an irreducible conformal subalgebra of \(\mathop{\fam 0 Cend}\nolimits V\), hence, \(\mathcal{M}\) is a simple associative conformal algebra with an FFR.
A conformal endomorphism \(d\in \mathop{\fam 0 Cend}\nolimits V\) is called a conformal derivation of the conformal algebra \(V\) if \[\label{eq:CDer-definition} d_\lambda (a\mathbin{{\circ}_{(\mu)}} b) = (d_\lambda (a) \mathbin{{\circ}_{(\lambda +\mu )}} b) + (a\mathbin{{\circ}_{(\mu )}} d_\lambda (b)),\tag{6}\] for all \(a,b\in V\).
For example, if \(V\) is an associative conformal algebra then for every \(a\in V\) the operation of commutator \(d_\lambda(x) = [a\mathbin{{\circ}_{(\lambda)}} x] = (a\mathbin{{\circ}_{(\lambda )}}x) - \{x\mathbin{{\circ}_{(\lambda)}} a\}\), \(x\in V\), is a conformal derivation.
In general, the space \(\mathop{\fam 0 CDer}\nolimits V\) of all conformal derivations of a (not necessarily associative) conformal algebra \(V\) is a Lie conformal subalgebra of \(\mathop{\fam 0 Cend}\nolimits V\).
Example 6. The space \(\mathop{\fam 0 CDer}\nolimits\mathop{\fam 0 Cur}\nolimits_1\) is an \(H\)-module of rank one.
Indeed, suppose \(d\in \mathop{\fam 0 CDer}\nolimits\mathop{\fam 0 Cur}\nolimits_1\), \(\mathop{\fam 0 Cur}\nolimits_1=He\), \(e\mathbin{{}_{(\lambda )}}e =e\), \(d_\lambda (e) = f(\partial, \lambda )e\). Then \(f(\partial,\lambda )e = d_\lambda (e\mathbin{{}_{(\mu )}}e) = (f(\partial,\lambda)e\mathbin{{}_{(\mu+\lambda )}} e) + (e\mathbin{{}_{(\mu )}}f(\partial,\lambda )e) = f(-\mu-\lambda, \lambda )e + f(\partial+\mu, \lambda )e.\) The equation \[f(\partial,\lambda ) - f(\partial+\mu, \lambda ) = f(-\mu-\lambda, \lambda)\] holds only for \(f(\partial,\lambda ) = (\partial+\lambda )h(\lambda )\), so it depends only on one polynomial \(h\in H\). Therefore, the polynomial presentation of \(d\in \mathop{\fam 0 Cend}\nolimits_1\) is \[d = h(-\partial )(x-\partial), \quad h\in H,\] so the single generator of \(\mathop{\fam 0 CDer}\nolimits\mathop{\fam 0 Cur}\nolimits_1\) over \(H\) is \(v = x-\partial\). As a Lie conformal algebra, \(\mathop{\fam 0 CDer}\nolimits\mathop{\fam 0 Cur}\nolimits_1\) is isomorphic to the Virasoro conformal algebra: \[[v\mathbin{{}_{(\lambda )}}v] = (v\mathbin{{}_{(\lambda )}}v) - \{v\mathbin{{}_{(\lambda )}}v\} = (2\lambda +\partial )v.\]
Recall that an algebra (or a conformal algebra) is called weakly Noetherian if it satisfies the ascending chain condition for two-sided ideals.
Lemma 1. Let \(C\) be a weakly Noetherian associative conformal algebra and let \(I\) be a nilpotent ideal of \(C\). Then for every \(\varphi \in \mathop{\fam 0 CDer}\nolimits(C)\) the \(H\)-submodule \(I+\varphi(I)\) is a nilpotent ideal of \(C\).
Proof. Let \(a\in I\), \(\alpha \in \Bbbk\), then \(\varphi_\alpha (a) = \varphi_\lambda (a)|_{\lambda=\alpha }\in C\). The \(H\)-submodule \(\varphi(I)\) is the linear span of all such elements. For every \(x\in C\) we have \[x\mathbin{{}_{(\lambda )}}\varphi_\alpha (a) =\varphi_\alpha (x\mathbin{{}_{(\lambda )}}a) - \varphi_\alpha(x)\mathbin{{}_{(\lambda +\alpha )}} a \in (\varphi_\alpha(I)+ I)[\lambda ].\] In a similar way, \(\varphi_\alpha(a)\mathbin{{}_{(\lambda )}}x \in (I+\varphi_\alpha (I))[\lambda ]\), so \(I+\varphi_\alpha (I)\) is an ideal of \(C\).
Now suppose \(I\) is nilpotent, i.e., there exists \(n\ge 1\) such that \[I^n_{\lambda_1,\dots, \lambda_{n-1}} = I\mathbin{{}_{(\lambda_1)}}(I\mathbin{{}_{(\lambda_2)}} ( \dots (I\mathbin{{}_{(\lambda_{n-1})}} I)\dots )) = 0\] in \(C[\lambda_1,\dots , \lambda _{n-1}]\). Then for every fixed \(\alpha \in \Bbbk\) the expression for \(\varphi_\alpha ^n(I^n_{\lambda_1,\dots, \lambda_{n-1}}) = 0\) leads us to \[(\varphi_\alpha(I) )^n_{\lambda_1+\alpha,\dots ,\lambda_2+\alpha } \subseteq I.\] Hence, \(I+\varphi_\alpha (I)\) is a nilpotent ideal of index \(\le n^2\).
If \(I_0=I\) is not \(\varphi\)-invariant then there exists \(\alpha _1\in \Bbbk\) such that \(I\subsetneq I_1 = I+\varphi_{\alpha_1}(I)\). If the nilpotent ideal \(I_1\) is not \(\varphi\)-invariant then there exists \(\alpha _2\in \Bbbk\) such that \(I_1\subsetneq I_2=I_1+\varphi_{\alpha_2}\), and so on. We obtain an ascending chain of nilpotent ideals \(I_k\), \(k\ge 0\). Since \(C\) is Noetherian, there exists a number \(N\ge 0\) such that \(I_N\) is \(\varphi\)-invariant. In particular, \(I+\varphi(I)\subseteq I_N\), thus \(I + \varphi(I)\) is nilpotent. ◻
Proposition 1. For a finite Novikov conformal algebra \(V\), the generators \(L^a\), \(R^a\) (\(a,b\in V\)) of the corresponding conformal subalgebra \(\mathcal{M}\subseteq \mathop{\fam 0 Cend}\nolimits V\) satisfy the following relations: \[\begin{gather} (R^a\mathbin{{}_{(\lambda )}}R^b) = \{R^b\mathbin{{}_{(\lambda )}}R^a\}, \label{eq:R-R95commute}\\ (R^a\mathbin{{}_{(\lambda )}}L^b) = L^{\{b\mathbin{{\circ}_{\lambda }}a\}}, \label{eq:R-L95product} \\ [L^a\mathbin{{}_{(\lambda )}}R^b] = R^{(a\mathbin{{\circ}_{\lambda }}b )} - \{R^b\mathbin{{}_{(\lambda )}}R^a\}. \label{eq:L-R95commute} \end{gather}\] {#eq: sublabel=eq:eq:R-R95commute,eq:eq:R-L95product,eq:eq:L-R95commute}
Proof. On the one hand, the identities 3 and 4 imply, respectively, \[\begin{gather} a\mathbin{{\circ}_{(\lambda)}} \{ c\mathbin{{\circ}_{(\mu)}} b\} - \{(a\mathbin{{\circ}_{(\lambda)}} c)\mathbin{{\circ}_{(\mu )}}b\} = \{c\mathbin{{\circ}_{(\lambda +\mu)}} \{ a\mathbin{{\circ}_{(\mu)}} b \}\} - \{ \{ c\mathbin{{\circ}_{(\lambda)}} a \} \mathbin{{\circ}_{(\mu)}} b \}, \tag{7}\\ \{\{ c\mathbin{{\circ}_{(\mu)}} b\}\mathbin{{\circ}_{(\lambda)}} a \} = \{\{ c\mathbin{{\circ}_{(\lambda)}} a\}\mathbin{{\circ}_{(\mu)}} b \} \tag{8} \end{gather}\] in the same way as in [21]. These identities are equivalent to ?? and ?? . The identity ?? is a different form of 4 .
On the other hand, a finite Novikov algebra can always be embedded into an associative and commutative conformal algebra with an ordinary derivation \(D\) in such a way that \(a\mathbin{{\circ}_{(\lambda)}} b = a\mathbin{{}_{(\lambda )}}Db\). In this way, for example, the equation 7 is easy to prove: \[\begin{gather} a\mathbin{{\circ}_{(\lambda )}}\{ c\mathbin{{\circ}_{(\mu)}} b\} - \{(a\mathbin{{\circ}_{(\lambda)}} c)\mathbin{{\circ}_{(\mu)}} b\} = a\mathbin{{}_{(\lambda)}} D\{c\mathbin{{}_{(\mu )}}Db \} - \{(a\mathbin{{}_{(\lambda )}}Dc)\mathbin{{}_{(\mu )}}Db\} \\ = a\mathbin{{}_{(\lambda )}}\{ c \mathbin{{}_{(\mu )}}D^2 b\} = a\mathbin{{}_{(\lambda )}}(D^2b \mathbin{{}_{(\mu )}}c), \end{gather}\] \[\begin{gather} \{c\mathbin{{\circ}_{(\lambda +\mu)}} \{ a\mathbin{{\circ}_{(\mu )}}b \}\} - \{ \{ c\mathbin{{\circ}_{(\lambda)}} a \} \mathbin{{\circ}_{(\mu)}} b \} = \{c\mathbin{{}_{(\lambda +\mu)}} D\{ a\mathbin{{}_{(\mu )}}Db \}\} - \{ \{ c\mathbin{{}_{(\lambda )}}Da \} \mathbin{{}_{(\mu )}}Db \} \\ = (D(Db\mathbin{{}_{(\mu )}}a)\mathbin{{}_{(\lambda +\mu)}} c) - (Db\mathbin{{}_{(\mu)}}(Da\mathbin{{}_{(\lambda )}}c)) = ((D^2b\mathbin{{}_{(\mu )}}a)\mathbin{{}_{(\lambda +\mu )}} c) = D^2b\mathbin{{}_{(\mu )}}(a\mathbin{{}_{(\lambda )}}c). \end{gather}\] The right-hand sides are equal since \(x\mathbin{{}_{(\lambda )}}(y\mathbin{{}_{(\mu )}}z) = y\mathbin{{}_{(\mu )}}(x\mathbin{{}_{(\lambda )}}z)\) in a commutative conformal algebra. The equation 8 can be checked similarly. ◻
In particular, the elements \(L^a\in \mathop{\fam 0 Cend}\nolimits V\) determine derivations \([L^a\mathbin{{}_{(\lambda )}}\cdot ]\) of \(\mathop{\fam 0 Cend}\nolimits V\), and it follows from ?? that the subalgebra \(\mathcal{R}\) is invariant under these derivations. Suppose \(\mathcal{J}\) is an ideal of \(\mathcal{R}\). If \([L^v\mathbin{{}_{(\lambda)}} \mathcal{J}] \subseteq \mathcal{J}[\lambda ]\) for all \(v\in V\) then we say \(\mathcal{J}\) is ad-invariant.
Lemma 2. Let \(V\) be a finite Novikov conformal algebra. Suppose \(I\) is a left ideal of \(V\) and \(\mathcal{J}\) is an ad-invariant ideal of the commutative conformal algebra \(\mathcal{R}\). Then \(\mathcal{J}(I)\) is a two-sided ideal of \(V\).
Proof. We just need to show that \(L^v_\lambda (\mathcal{J}(I)) \subseteq \mathcal{J}(I)[\lambda ]\) since \(\mathcal{J}(I)\) is already closed with respect to right multiplications.
Suppose \(f \in \mathcal{J}\), \(a\in I\), \(v\in V\). Then \[L^v_\lambda (f_\mu(a)) = (L^v \mathbin{{}_{(\lambda )}}f)_{\mu+\lambda }(a) = \{f\mathbin{{}_{(\lambda )}}L^v \}_{\mu+\lambda }(a) + [L^v\mathbin{{}_{(\lambda )}}f]_{\lambda +\mu }(a).\] The coefficients of the second summand belong to \(\mathcal{J}(I)\). The first summand \(\{f\mathbin{{}_{(\lambda )}}L^v \}_{\mu+\lambda }(a)\) belongs to \(\mathcal{J}(I)[\lambda,\mu ]\) since \[\{f\mathbin{{}_{(\lambda )}}L^v \}_{\mu+\lambda }(a) = (f\mathbin{{}_{(-\partial -\lambda)}} L^v )_{\mu+\lambda }(a) = (f\mathbin{{}_{(\mu)}} L^v )_{\mu+\lambda }(a) = f_\mu (L^v_\lambda (a))\] and \(L^v_\lambda (a)\in I[\lambda ]\). ◻
Corollary 1. If \(\mathcal{J} = \mathcal{R}\) and \(I\ne 0\) then either \(I\) is an ideal of \(V\) such that \(I\mathbin{{\circ}_{(\lambda )}}V=0\), or \(\mathcal{R}(I)\) is a nonzero ideal of \(V\).
Lemma 3. Let \(V\) be a finite Novikov conformal algebra and let \(\mathcal{J}\) be an ad-invariant ideal of \(\mathcal{R}\). Then \(\mathcal{J}(V)\mathbin{{\circ}_{(\lambda )}} \mathcal{J}(V) \subseteq \mathcal{J}^2(V)[\lambda ]\).
Note that if \(\mathcal{J}\ne 0\) then \(\mathcal{J}(V)\) is a nonzero ideal of \(V\).
Proof. For every \(f,g\in \mathcal{J}\), \(u,v\in V\), \(\alpha,\beta,\gamma\in \Bbbk\) we have to show \[f_\alpha(u)\mathbin{{\circ}_{(\gamma )}} g_\beta (v) = L^{f_\alpha(u)}_\gamma g_\beta (v)\in \mathcal{J}^2(V).\] It follows from ?? that \[\label{eq:R-L95Cend} L^{f_\alpha(u)}_\gamma = f_\alpha L^u_{\gamma-\alpha } \in \mathop{\fam 0 Cend}\nolimits V.\tag{9}\] Indeed, a generic element of \(\mathcal{R}\) may be presented as \[f = \sum \partial^s(R^{a_1}\mathbin{{}_{(\alpha_1)}}(R^{a_2}\mathbin{{}_{(\alpha_2)}}\dots (R^{a_n}\mathbin{{}_{(\alpha_n)}} R^{a_{n+1}})\dots )).\] Then \[f_\alpha(u) = \sum (-\alpha)^s R^{a_1}_{\alpha_1}R^{a_2}_{\alpha_2}\dots R^{a_n}_{\alpha_n}R^{a_{n+1}}_{\alpha-\alpha_1-\dots -\alpha_n}(u),\] and consecutive application of ?? leads us to 9 .
Hence, \[\begin{gather} f_\alpha(u)\mathbin{{\circ}_{(\gamma )}} g_\beta (v) =f_\alpha L^u_{\gamma-\alpha}g_\beta (v) = f_\alpha g_\beta L^u_{\gamma-\alpha }(v) +f_\alpha [L^u_{\gamma-\alpha }, g_\beta](v) \\ = (f\mathbin{{}_{(\alpha )}}g)_{\beta+\alpha}(u\mathbin{{\circ}_{(\gamma-\alpha)}} v) + (f\mathbin{{}_{(\alpha)}} [L^u\mathbin{{}_{(\gamma-\alpha)}}g])_{\beta+\gamma}(v). \end{gather}\] Both summands in the right-hand side of this expression belong to \(\mathcal{J}^2(V)\). ◻
Remark 2. Note that if \(\mathcal{J}\) is an ad-invariant ideal of \(\mathcal{R}\) then so is \(\mathcal{J}^2\). Therefore, if \(\mathcal{J}\) is nilpotent then \(\mathcal{J}(V)\) is solvable.
Proposition 2. If \(V\) is a finite (semi)simple Novikov conformal algebra then \(\mathcal{R}\) is a semisimple subalgebra of \(\mathop{\fam 0 Cend}\nolimits V\).
Proof. Assume the converse: there exists a nonzero nilpotent ideal in \(\mathcal{R}\). Then the nilpotent radical \(\mathcal{N}\) (i.e., the maximal nilpotent ideal) of \(\mathcal{R}\) is nonzero.
Since \(\mathcal{R}\) is generated by the operators of right multiplication in \(V\) and the latter is a finite conformal algebra, we may conclude \(\mathcal{R}\) is a finitely generated commutative conformal algebra. Hence, \(\mathcal{R}\) is Noetherian [22]. By Lemma 1, the maximal nilpotent ideal \(\mathcal{N}\) of \(\mathcal{R}\) is invariant with respect to all conformal derivations of \(\mathcal{R}\).
In particular, it works for \(\varphi_\lambda = [L^u\mathbin{{}_{(\lambda )}}\cdot ]\), \(u\in V\). By Lemma 2 \(\mathcal{N}(V)\) is an ideal of \(V\) which is solvable by Lemma 3.
Therefore, \(\mathcal{N}=0\) as desired. ◻
Throughout the section, let \(V\) be a simple finite Novikov conformal algebra. Then \(V\) is a torsion-free \(H\)-module since the \(H\)-torsion is always an annihilator ideal in a conformal algebra (see, e.g., [1]).
Suppose \(\rho : V\to \mathcal{R}\subseteq \mathop{\fam 0 Cend}\nolimits V\) is the right regular representation of \(V\), i.e., \[\rho(a) = R^a, \quad a\in V.\] This is an \(H\)-linear map. Denote by \(A\) the kernel of \(\rho\), i.e., the space of all \(a\in V\) such that \(V\mathbin{{\circ}_{(\lambda)}} a =0\).
Lemma 4. If \(a\in A\), \(a\ne 0\) then \(\mathcal{R}(Ha)=V\).
Proof. If \(a\in A\) then \(Ha\) is a left ideal of \(V\): \(V\mathbin{{\circ}_{(\lambda)}} a = 0\). By Lemma 2, \(\mathcal{R}(Ha)\) is an ideal of \(V\), so either \(\mathcal{R}(Ha)=0\) or \(\mathcal{R}(Ha)=V\). If the first option works for at least one \(A\ni a\ne 0\) then the space of all such \(a\in A\) is an annihilator ideal of \(V\). Therefore, \(\mathcal{R}(Ha)=V\) for all \(A\ni a\ne 0\). ◻
Corollary 2. The conformal algebra \(\mathcal{R}\) is isomorphic to \(\mathop{\fam 0 Cur}\nolimits_1 = \mathop{\fam 0 Cur}\nolimits\Bbbk\).
Proof. As a semisimple associative conformal algebra with FFR, \(\mathcal{R}\) has to be isomorphic to a direct sum of simple ones according to Theorem 1. Since \(\mathcal{R}\) is commutative by Proposition 1, so are all the simple summands, but the only commutative simple conformal algebra with FFR is \(\mathop{\fam 0 Cur}\nolimits_1\).
Hence, \(\mathcal{R}\) is a direct sum of several copies of \(\mathop{\fam 0 Cur}\nolimits_1\): \(\mathcal{R} = \mathcal{R}_1\oplus \dots \oplus \mathcal{R}_m\), \(\mathcal{R}_i\cong \mathop{\fam 0 Cur}\nolimits_1\). Each \(\mathcal{R}_i\) is an ideal of \(\mathcal{R}\) which is closed under all conformal derivations of \(\mathcal{R}\), in particular, under the derivations \([L^v\mathbin{{}_{(\lambda )}}\cdot ]\), \(v\in V\). Hence, the \(H\)-submodule \(\mathcal{R}_i(V)\) is \(\mathcal{M}\)-invariant, i.e., \(V=\mathcal{R}_i(V)\) for every \(i=1,\dots, m\).
Assume \(m>1\). Then \(V=\mathcal{R}_1(V)=\mathcal{R}_2(V)\) implies \(V=\mathcal{R}_1(\mathcal{R}_2(V))=(\mathcal{R}_1\mathcal{R}_2)(V)=0\), a contradiction. Hence, \(\mathcal{R}\cong \mathop{\fam 0 Cur}\nolimits_1\). ◻
Theorem 2. Let \(V\) be a simple Novikov conformal algebra over an algebraically closed field \(\Bbbk\) of characteristic zero. Then \(V\) is an \(H\)-module of rank 1.
Proof. As above, \(\mathcal{M}\) stands for the conformal subalgebra of \(\mathop{\fam 0 Cend}\nolimits V\) generated by all operators of left (\(L^v\)) and right (\(R^v\)) multiplication. Recall that if \(A=\{a\in V \mid R^a = 0\}\) then for every \(0\ne a\in A\) we have \(\mathcal{R}(Ha)=V\) by Lemma 4.
Consider the map \[\ell : V\to \mathop{\fam 0 CDer}\nolimits\mathcal{R}, \quad v\mapsto [L^v\mathbin{{}_{(\lambda )}}\cdot ].\] By Example 6, the rank of the \(H\)-module \(\mathop{\fam 0 CDer}\nolimits\mathcal{R}\) is equal to 1.
Therefore, we have two \(H\)-linear maps \(\rho : V\to \mathcal{R}\cong \mathop{\fam 0 Cur}\nolimits_1\) and \(\ell: V\to \mathop{\fam 0 CDer}\nolimits\mathcal{R}\cong \mathrm{Vir}\). If the rank of \(V\) is greater than 2 then the intersection of two kernels \(A =\ker \rho\) and \(\ker\ell\) is nonzero since each of them is of corank 1.
Assume \(0\ne a\in A\cap \ker\ell\). Then \(\mathcal{R}(Ha)=V\), so \[L^a_\lambda (V) = L^a_\lambda (\mathcal{R}(Ha)) =\mathcal{R}(L_\lambda^a Ha) = 0\] since \(a\mathbin{{\circ}_{(\lambda )}} a =0\). In this case, \(a\) belongs to the two-sided annihilator of \(V\), a contradiction to simplicity of \(V\).
Hence, the rank of \(V\) does not exceed 2. Assume the rank of \(V\) is equal to 2. Then there exists a basis \(e_1,e_2\) of \(V\) over \(H\) such that \(\rho (e_2)=0\), i.e., \(R^{e_2} =0\). The conformal algebra structure on \(V\) is then completely defined by \(R^{e_1}\in \mathop{\fam 0 Cend}\nolimits_2\). Suppose \[R^{e_1} = \begin{pmatrix} a(x,x-\partial) & f(x,x-\partial ) \\ b(x,x-\partial ) & g(x,x-\partial ) \end{pmatrix},\] where \(a,b,f,g\in \Bbbk [x,\partial ]\). In particular, \(\{e_2\mathbin{{\circ}_{(\lambda)}} e_1\} = R^{e_1}_\lambda (e_2) = f(\partial,\partial+\lambda )e_1 + g(\partial, \lambda +\partial)e_2\), so \(e_2\mathbin{{\circ}_{(\lambda )}} e_1 = f(\partial,-\lambda )e_1+g(\partial, -\lambda )e_2\). Hence, \[L^{e_1} = \begin{pmatrix} a(x,\partial ) & 0 \\ b(x,\partial) & 0 \end{pmatrix}, \quad L^{e_2} = \begin{pmatrix} f(x,\partial) & 0 \\ g(x,\partial) & 0 \end{pmatrix}.\] Note that \(f\ne 0\): otherwise, \(He_2\) is a non-trivial ideal of \(V\).
By ?? , \([L^{e_2}\mathbin{{}_{(\lambda )}}R^{e_1}] = R^{(e_2\mathbin{{\circ}_{(\lambda)}} e_1)} = f(\partial, -\lambda )R^{e_1}\). The latter leads us to the \(2\times 2\) matrix equation \[\label{eq:LR-comm-V2} L^{e_2}\mathbin{{}_{(\lambda )}}R^{e_1} - R^{e_1}\mathbin{{}_{(-\partial-\lambda )}} L^{e_2} = f(\partial, -\lambda ) R^{e_1},\tag{10}\] i.e., \[\begin{gather} \begin{pmatrix} f(x,-\lambda) & 0 \\ g(x,-\lambda) & 0 \end{pmatrix} \begin{pmatrix} a(x+\lambda,x-\partial) & f(x+\lambda,x-\partial ) \\ b(x+\lambda,x-\partial ) & g(x+\lambda,x-\partial ) \end{pmatrix} \\ - \begin{pmatrix} a(x,x-\partial-\lambda) & f(x,x-\partial -\lambda) \\ b(x,x-\partial-\lambda ) & g(x,x-\partial-\lambda ) \end{pmatrix} \begin{pmatrix} f(x-\partial-\lambda ,-\lambda) & 0 \\ g(x-\partial-\lambda ,-\lambda) & 0 \end{pmatrix} \\ = f(\partial,-\lambda ) \begin{pmatrix} a(x,x-\partial) & f(x,x-\partial ) \\ b(x,x-\partial ) & g(x,x-\partial ) \end{pmatrix}. \end{gather}\] Consider the (1,2)-component of this matrix equation: \[f(x,-\lambda ) f(x+\lambda,x-\partial ) = f(\partial,-\lambda ) f(x,x-\partial).\] This is a quadratic equation for a polynomial \(f(x,y)\) in two formal variables \(x\), \(y\). Let us make a substitution \(\partial \to x-\partial\) to get an equivalent equation \[\label{eq:12-f-equation} f(x,-\lambda ) f(x+\lambda,\partial ) = f(x-\partial,-\lambda ) f(x,\partial).\tag{11}\] It is obvious that \(f(x,y)=q(y)\) is a solution of this equation for every \(q\in \Bbbk [y]\). Let us show that there are no other solutions.
Assume \(f(x,y)\ne 0\) is a solution of 11 . Then put \(\lambda =0\) to obtain \(f(x,0)=f(x-\partial, 0)\). The latter means \(f(x,0)=c\in \Bbbk\). Expand left- and right-hand sides of 11 by the Taylor’s formula and compare the terms at \(\lambda\): \[\big ( c - \lambda f'_y(x,0) +\lambda^2(\dots ) \big ) \big ( f(x,\partial ) + \lambda f'_x(x,\partial ) + \lambda ^2(\dots ) \big ) \\ = f(x,\partial ) \big ( c -\lambda f'_y(x-\partial, 0) +\lambda^2(\dots ) \big ),\] \[c f'_x(x,\partial ) = f(x,\partial )f'_y(x,0) -f(x,\partial)f'_y(x-\partial, 0).\] If \(c\ne 0\) then we obtain that \(f(x,y)\) divides its derivative \(f'_x(x,y)\) which is only possible for \(f'_x(x,y)=0\), i.e., \(f(x,y)=q(y)\) as desired. If \(c=0\) then \(f(x,y)=y \hat{f}(x,y)\) for some \(\hat{f}\in \Bbbk [x,y]\) of smaller degree. It is obvious that \(\hat{f}\) satisfies the same equation 11 , so \(\hat{f}(x,y)\) depends only on \(y\) by induction reasoning. Therefore, \(f(x,y)=q(y)\in \Bbbk[y]\), so \[R^{e_2} = \begin{pmatrix} * & q(x-\partial ) \\ * & * \end{pmatrix}, \quad q\ne 0.\]
Recall that \(R^{e_1}\) is a generator of the conformal subalgebra \(\mathcal{R}\subset \mathop{\fam 0 Cend}\nolimits_2\) isomorphic to \(\mathop{\fam 0 Cur}\nolimits_1\). Hence, \[R^{e_1} = h(\partial )e,\] where \(h(\partial )\in H\), \(e=e(x,\partial )\in \mathop{\fam 0 Cend}\nolimits_2\) is an idempotent: \(e\mathbin{{}_{(\lambda )}}e = e\). Moreover, we know from Lemma 4 that \(\mathcal{R}(He_2)=V\). Since \((e\mathbin{{}_{(0)}} \cdot )=\mathrm{id}_{\mathcal{R}}\), we conclude \(e(x,0)=I\), the identity \(2\times 2\)-matrix, \(e_{12}(x,0)=0\). Therefore, \[R^{e_2}_0 = h(0) e(\partial, 0) \in M_2(H),\] but \(h(\partial) e_{12}(x,\partial ) = f(x,x-\partial) = q(x-\partial )\) implies \(h = 0\), a contradiction.
As a result, the hypothesis \(\mathrm{rank}\,(V) \ge 2\) contradicts to the simplicity of \(V\), so the only option is \(\mathrm{rank}\,(V)=1\). ◻
Proposition 3. Let \(V = Hv\) be the free \(H\)-module of rank one equipped with a sesqui-linear \(\lambda\)-bracket \[v\mathbin{{\circ}_{(\lambda)}} v = f(\partial,\lambda ) v.\] Then \(V\) is a Novikov algebra if and only if \(f=\beta(\partial+\lambda ) + \alpha\), \(\alpha,\beta \in \Bbbk\).
Proof. For the “if” part, it is enough to check 3 and 4 for \(a=b=c=v\) which is straightforward (see also [17]).
For the “only if” part, suppose \[v\mathbin{{\circ}_{(\lambda)}} v = f(\partial,\lambda )v\] for some nonzero polynomial \(f(x,y)\in \Bbbk [x,y]\). Then the right commutativity 4 is equivalent to \[\label{eq:RCom-rank1} f(-\mu, \lambda ) f(\partial , \mu) = f(\partial+\mu-\lambda, \lambda )f(\partial, -\partial-\mu+\lambda ).\tag{12}\] Let us prove that 12 holds if and only if \(f(x,y)\) is a polynomial in \(x+y\). The “if” part is obvious. To show “only if”, put \(f(x,y)=q(x,x+y)\) for some polynomial \(q(x,y)\). Then \[q(-\mu, \lambda -\mu)q(\partial, \partial+\mu)=q(\partial+\mu-\lambda ,\partial+\mu)q(\partial, \lambda-\mu).\] Make a substitution \(\partial=z\), \(\partial+\mu =x\), \(\lambda -\mu=y\) to obtain an equation for \(q\): \[\label{eq:RCom-Q} q(z-x,y)q(z,x)=q(z-y,x)q(z,y).\tag{13}\] Put \(y=0\) into 13 to get \(q(z-x,0)q(z,x) = q(z,x)q(z,0)\). If \(q\ne 0\) then \(q(x,0)=c_0\in \Bbbk\).
Next, evaluate the derivative \(\partial/\partial y\) of 13 at \(y=0\): \[q'_2(z-x,0)q(z,x)=-q'_1(z,x)q(z,0) + q(z,x)q'_2(z,0),\] where \(q'_1\) and \(q'_2\) stand for the partial derivatives of \(q(x_1,x_2)\) relative to \(x_1\) and \(x_2\), respectively.
As a result, \[(q'_2(z,0)-q'_2(z-x,0))q(z,x)= c_0 q'_1(z,x) ,\] i.e., either \(c_0=0\) or \(q\) divides its derivative \(q'_1\). The second option means that \(q(x,y)\) depends only on \(y\), as desired. If \(c_0=0\) then \[q(x,y) = y \hat{q}(x,y),\] where \(\hat{q}\) satisfies the same equation 13 . Hence, \(q(x,y)\) depends only on \(y\), and \(f(x,y) = q(x+y)\).
The left symmetry 3 turns into the following equation on \(f(\partial,\lambda )\): \[f(-\lambda -\mu,\lambda )f(\partial, \lambda+\mu) - f(\partial+\lambda, \mu)f(\partial,\lambda ) = f(-\mu -\lambda,\mu )f(\partial, \lambda+\mu) - f(\partial+\mu, \lambda )f(\partial,\mu ).\] For \(f(x,y)=q(x+y)\), we have \[q(-\mu )q(\partial+\lambda+\mu) - q(\partial+\lambda+\mu)q(\partial+\lambda ) = q( -\lambda )q(\partial+\lambda+\mu) - q(\partial+\mu+\lambda )q(\partial+\mu ).\] Since \(q\ne 0\), we have \[q(\partial+\mu ) + q(-\mu ) = q(\partial+\lambda ) + q( -\lambda ) .\] The latter means that \(q(y) = \alpha + \beta y\), \(\alpha,\beta \in \Bbbk\), as desired. ◻
Denote by \(\mathcal{V}_\alpha\) the rank one Novikov conformal algebra \(Hv\) with \[(v\mathbin{{\circ}_{(\lambda )}} v) = (\partial+\lambda +\alpha) v,\] for \(\alpha \in \Bbbk\).
Corollary 3. A simple finite Novikov conformal algebra \(V\) over an algebraically closed field of characteristic zero is isomorphic either to \(\mathop{\fam 0 Cur}\nolimits_1\) or to \(\mathcal{V}_\alpha\).
As it was shown in [5], it may happen that a semisimple finite Lie conformal algebra is not isomorphic to the direct sum of simple ones. Therefore, in contrast to the case of “ordinary” algebras, a semisimple Lie conformal algebra may be not classically semisimple. The purpose of this section is to show that for finite Novikov conformal algebras every semisimple object is a direct sum of ideals each isomorphic to a simple Novikov conformal algebra from Corollary 3.
Throughout this section, \(V\) is a finite semisimple Novikov conformal algebra, \(\mathcal{R}\) is the conformal subalgebra of \(\mathop{\fam 0 Cend}\nolimits V\) generated by all right multiplications \(R^a\), \(a\in V\). By Proposition 2, \(\mathcal{R}\) is semisimple, so we may conclude \(\mathcal{R}\cong \mathop{\fam 0 Cur}\nolimits_1 \oplus \dots \oplus \mathop{\fam 0 Cur}\nolimits_1\), i.e., \[\mathcal{R} = H e^1 \oplus \dots \oplus He^k, \quad e^i\in \mathcal{R},\;(e^i\mathbin{{}_{(\lambda )}}e^j) = \delta_{ij}e^j.\]
Lemma 5. The conformal algebra \(V\) is perfect, i.e., \(V^2=V\).
Proof. By definition, \(V^2=\mathcal{R}(V)\). Assume there exists \(a\in V\) but \(a\notin V^2\). Then \[\bar a = a -e^1_0(a)-\dots - e^k_0(a) \notin V^2\] since \(e^i_\alpha (a)\in \mathcal{R}(V) = V^2\). In particular, \(\bar a\ne 0\).
Suppose \(f\in \mathcal{R}\), \[f = h_1(\partial )e^1+\dots +h_k(\partial )e^k, \quad h_i\in H.\] Then \(f_\alpha (a) =\sum\limits_{i=1}^k h_i(-\alpha )e^i_\alpha (a)\), \(f_\alpha (e^j_0 (a)) = (f\mathbin{{}_{(\alpha )}}e^j)_\alpha (a) = h_j(-\alpha )e^j_\alpha (a)\). Hence, \(f_\alpha(\bar a)=0\) and, in particular, \(R^v_\lambda (\bar a)=0\), so \(\bar a\mathbin{{\circ}_{(\lambda )}}V=0\). The left annihilator of \(V\) is an Abelian ideal of \(V\), which cannot be nonzero. ◻
Lemma 6. Let \(I\) be a proper ideal of \(V\). Then \(\mathrm{Ann}_l (I) = \{ v\in V \mid v\mathbin{{\circ}_{(\lambda )}}I = 0\}\) is nonzero ideal of \(V\).
Proof. It is clear that \(\mathrm{Ann}_l(I)\) is an ideal of \(V\). Indeed, \[\mathrm{Ann}_l(I) = \{u\in V \mid R^a_\lambda (u) = 0, a\in I\}.\] It follows from ?? and ?? that if \(u\in \mathrm{Ann}_l(I)\) then so are \(R^v_\alpha (u)\) and \(L^v_\alpha (u)\) for all \(v\in V\), \(\alpha\in \Bbbk\).
Denote by \(\mathcal{R}^I\) the ideal of \(\mathcal{R}\) generated by all \(R^a\), \(a\in I\). Then \(\mathcal{R}^I(V)\subseteq I\ne V\), in particular, \(\mathcal{R}^I\ne \mathcal{R}\) by Lemma 5.
Every proper ideal in \(\mathcal{R}\cong \mathop{\fam 0 Cur}\nolimits_1\oplus \dots \oplus \mathop{\fam 0 Cur}\nolimits_1\) has a nonzero annihilator (at least, one of \(e^i\)s). Suppose \(\mathcal{R}^I\mathbin{{}_{(\lambda )}}e = 0\) for some \(0\ne e\in \mathcal{R}\). Then for every \(\alpha,\beta \in \Bbbk\) and for every \(a\in I\), \(v\in V\) we have \[0=(R^a\mathbin{{}_{(\alpha )}}e)_{\beta+\alpha }(v) = R^a_\alpha (e_\beta (v)).\] Since \(e\ne 0\), there exist \(\beta\) and \(v\) such that \(u = e_\beta(v)\ne 0\), this is an element of the left annihilator of \(I\). ◻
Theorem 3. A finite semisimple Novikov conformal algebra \(V\) is isomorphic to a direct sum of simple ones: \[V = V_1\oplus \dots \oplus V_k,\] \(V_i\cong \mathop{\fam 0 Cur}\nolimits_1\) or \(V_i\cong \mathcal{V}_{\alpha_i}\), \(\alpha_i\in \Bbbk\), \(i=1,\dots, k\).
Proof. Proceed by the rank of \(V\) as \(H\)-module. If the rank is zero then \(V=0\) (since torsion elements annihilate \(V\)). For \(V\ne 0\), find a maximal ideal \(I\) of \(V\). By Lemma 6 there exists \(J=\mathrm{Ann}_l(I)\ne 0\). Since \(I\cap J\) is an Abelian ideal of \(V\) and \(I\) is maximal, we conclude \(V = I\oplus J\), \(J\cong V/I\). If the rank of \(J\) is zero then it belongs to the torsion of \(V\) as \(H\)-module, which is impossible. Also, an ideal of \(I\) is an ideal of \(V\).
Hence, the rank of \(J\) is positive and \(I\) is a semisimple Novikov conformal algebra, the rank of \(I\) is smaller than that of \(V\). Therefore, \(I = V_2\oplus \dots \oplus V_k\), each \(V_i\) is simple, and the claim follows for \(V_1=J\cong V/I\). ◻
An infinitesimal conformal deformation of a Novikov conformal algebra \(V\) with a \(\lambda\)-product \((\cdot \mathbin{{\circ}_{(\lambda )}}\cdot )\) may be defined in a similar way to that of Lie, associative [23], or Poisson conformal algebras [24]. Denote by \(\hbar\) a formal variable. Then \(\Bbbk \subset \bar \Bbbk_\hbar :=\Bbbk +\Bbbk \hbar \cong \Bbbk[[\hbar ]]/\hbar^2 \Bbbk [[\hbar ]]\). The \(H\)-module \(V\) may be considered as a subspace of \(\bar V_\hbar := V[[\hbar ]]/\hbar^2 V[[\hbar ]]\).
An infinitesimal conformal deformation of \(V\) is an \(\bar\Bbbk_\hbar\)-linear operation \[u\mathbin{\circ_{(\lambda )}^\hbar } v = u\mathbin{{\circ}_{(\lambda )}} v + \hbar f_\lambda (u,v),\] which turns \(\bar V_\hbar\) into a Novikov conformal algebra over \(\bar \Bbbk_\hbar\). The latter condition can be written as a collection of conditions on \(f_\lambda : V\otimes V \to V[\lambda ]\): \[f_\lambda (\partial u, v) = -\lambda f_\lambda(u,v),\quad f_\lambda (u, \partial v) = (\lambda +\partial) f_\lambda(u,v); \label{eq:sesqui-cochain}\tag{14}\] \[\begin{gather} u\mathbin{{\circ}_{(\lambda )}} f_\mu(v,w) - f_{\lambda+\mu}(u\mathbin{{\circ}_{(\lambda)}}v, w) + f_\lambda (u, v\mathbin{{\circ}_{(\mu)}} w) - f_\lambda(u,v)\mathbin{{\circ}_{(\lambda+\mu)}} w \\ {} = v\mathbin{{\circ}_{(\mu)}} f_\lambda(u,w) - f_{\lambda+\mu}(v\mathbin{{\circ}_{(\mu )}}u, w) + f_\mu (v, u\mathbin{{\circ}_{(\lambda)}} w) - f_\mu(v,u)\mathbin{{\circ}_{(\lambda+\mu)}} w ;\label{eq:LSym-cochain} \end{gather}\tag{15}\] \[f_{\lambda+\mu} (u\mathbin{{\circ}_{(\lambda)}} v, w) + f_\lambda(u,v)\mathbin{{\circ}_{(\lambda+\mu)}}w = \{ f_\lambda(u,w)\mathbin{{\circ}_{(\mu)}} v \} + f_{-\partial-\mu}(u\mathbin{{\circ}_{(\lambda)}} w, v). \label{eq:RCom-cochain}\tag{16}\] Denote by \(Z^2(V,V)\) the space of all such \(f\) satisfying 14 –16 .
Two infinitesimal conformal deformations \(\bar V_\hbar\) and \(\bar V_\hbar'\) of a Novikov conformal algebra \(V\) are equivalent if there exists a \(\bar \Bbbk _\hbar\)-linear map \(\varphi : \bar V_\hbar = V+\hbar V \to \bar V_\hbar' = V+\hbar V\) preserving the deformed operations such that \[\varphi (u+\hbar v) = u +\hbar (v+\tau(u)), \quad u,v\in V,\] where \(\tau : V\to V\) is an \(H\)-linear map. In other words, \(\varphi = \mathrm {id}+\hbar \tau \pmod {\hbar ^2}\).
For example, if the deformations \(\bar V_\hbar\) and \(\bar V_\hbar'\) are respectively defined by cocycles \(f,f'\in Z^2(V,V)\) then these deformations are equivalent if and only if \[f_\lambda(u,v) - f'_\lambda(u,v) = u\mathbin{{\circ}_{(\lambda )}}\tau (v) - \tau(u\mathbin{{\circ}_{(\lambda )}} v) + \tau(u)\mathbin{{\circ}_{(\lambda )}} v = : (d^1\tau)_\lambda (u,v),\] for all \(u,v\in V\) (similar to [24]).
Denote by \(B^{2}(V,V)\) the space of all those \(f\in Z^2(V,V)\) that are of the form \(d^1\tau\) for some \(H\)-linear \(\tau : V\to V\). Therefore, the deformations of \(V\) up to equivalence are described by the elements of the space \(Z^2(V,V)/B^2(V,V)\).
This “naive” approach to conformal cohomology of Novikov algebras has a natural explanation in terms of operads.
Let us define the cohomology of a Novikov conformal algebra \(V\) in the same way as it was done for Lie conformal algebras in [23] (BKV-cohomology, for short).
The structure of a Novikov conformal algebra on an \(H\)-module \(V\) is defined by a morphism \(\nu\) from the operad \(\mathcal{O}_{\mathrm{Nov}}\) to the operad \(\mathrm{Chom}_V\) of conformal endomorphisms of \(V\) (see [25]), a full subcategory of the pseudo-tensor category \(H\)-mod: \[\nu : \mathcal{O}_{\mathrm{Nov}} \to \mathrm{Chom}_V, \quad \nu(x_1\circ x_2) \mapsto (\cdot \mathbin{{\circ}_{(\lambda )}} \cdot ).\]
Recall that the Koszul dual operad \(\mathcal{O}_{\mathrm{Nov}}^!\) represents the class of opposite Novikov algebras (right-symmetric and left-commutative). Hence, \(\mathcal{O}_{\mathrm{Nov}}^!\) can be considered as a sub-operad of the operad \(\Omega\mathrm{Com}\) of differential associative and commutative algebras generated by \(x_1\bullet x_2 = x_1' x_2\), where \(x'=d(x)\), \(d\in \Omega\mathrm{Com}(1)\) is the derivation. In particular, \(\mathcal{O}_{\mathrm{Nov}}^!(3)\) is spanned by linearly independent monomials \[x_1''x_2x_3,\;x_1x_2''x_3,\;x_1x_2x_3'',\; x_1'x_2'x_3,\; x_1'x_2x_3',\;x_1x_2'x_3'.\]
Consider the canonical morphism of operads [4] \[\iota : \mathcal{O}_{\mathrm{Lie}} \to \mathcal{O}_{\mathrm{Nov}}^!\otimes \mathcal{O}_{\mathrm{Nov}}\] sending the generator \([x_1,x_2]\in \mathcal{O}_{\mathrm{Lie}}(2)\) to \[x_1'x_2\otimes x_1\circ x_2 - x_1x_2'\otimes x_2\circ x_1.\] The composition of \(\iota\) with \(\mathrm{id}\otimes \nu\) defines a morphism \[\mathcal{O}_{\mathrm{Lie}} \to \mathcal{P}_V:=\mathcal{O}_{\mathrm{Nov}}^!\otimes \mathrm{Chom}_V.\] Define a complex \(C^*(V,V)\) as the reduced BKV-cohomology complex for the operad \(\mathcal{P}_V\). Namely, \[C^n(V,V) = \{f\in \mathcal{P}_V(n) \mid f^{\sigma }= (-1)^\sigma f, \;\sigma\in S_n\}, \quad n\ge 1,\] and the differential \(d^n: C^n(V,V)\to C^{n+1}(V,V)\) is given by [23]. In particular, for \(n=1\) the space \(C^1(V,V)\) coincides with \(\mathrm{Chom}_V(1)=\mathrm{End}_H(V)\). For \(n=2\), an element \(x_1'x_2\otimes f - x_1x_2'\otimes g \in \mathcal{P}_V(2)\) belongs to \(C^2(V,V)\) if and only if \(g = f^{(12)} \in \mathrm{Chom}_V(2)\). Therefore, \(C^2(V,V) \cong \mathop{\fam 0 Cend}\nolimits(V)\). For \(n=3\), the skew-symmetry of \[x_1''x_2x_3\otimes f_1 +x_1x_2''x_3\otimes f_2 + x_1x_2x_3''\otimes f_3 + x_1x_2'x_3' \otimes g_1 + x_1'x_2x_3' \otimes g_2 + x_1'x_2'x_3 \otimes g_3 \in \mathcal{P}_V(3),\] \(f_i,g_i\in \mathrm{Chom}_V(3)\), means that \(f_2 = - f_3^{(23)} = -f_1^{(12)}\) and \(g_2 = -g_3^{(23)} = -g_1^{(12)}\), so a 3-cochain is defined by a pair \((f_1,g_3)\in \mathrm{Chom}_V(3)^2\).
The differential \(d^1\) maps \(f\in \mathrm{End}_H(V)\) into \(d^1f\in \mathop{\fam 0 Cend}\nolimits(V)\), where \[(d^1f)_\lambda (u,v) = u\mathbin{{\circ}_{(\lambda )}} f(v) - f(u\mathbin{{\circ}_{(\lambda )}} v) + f(u)\mathbin{{\circ}_{(\lambda )}} v\] for \(u,v\in V\). Similarly, we may evaluate \(d^2(f)\) for \(f\in \mathop{\fam 0 Cend}\nolimits(V)\). Consider \[p = x'_1x_2\otimes f - x_1x_2'\otimes f^{(12)} \in \mathcal{P}_V(2)\] and apply the differential formula of the BKV-cohomology theory to the morphism \[(\mathrm{id}\otimes \nu )\iota: [x_1,x_2]\mapsto \theta : = x_1'x_2\otimes (\cdot\mathbin{{\circ}_{(\lambda )}}\cdot) - x_1x_2'\otimes \{\cdot\mathbin{{\circ}_{(\lambda )}} \cdot\}^{(12)}\] to obtain \[\label{eq:Diff-2} d^2(p) = \theta (1,p) -\theta (1,p)^{(12)} + \theta (1,p)^{(123)} - p(\theta ,1) + p(\theta ,1) ^{(23)} - p(\theta ,1) ^{(132)},\tag{17}\] where \(1\) stands for the identity in \(\mathcal{P}_V(1)\).
Calculate the composition \(\theta (1,p) \in \mathcal{P}_V(3)\) following the rules of [25] in the second tensor factor. In the non-reduced form we have \[p_{\lambda_1,\lambda_2}(u,v) = x_1'x_2 \otimes f_{\lambda _1}(u,v) - x_1x_2'\otimes f_{\lambda _2}(v,u),\] \[\theta _{\lambda_1,\lambda_2}(u,v) = x_1'x_2\otimes (u\mathbin{{\circ}_{(\lambda _1)}} v) -x_1x_2'\otimes (v\mathbin{{\circ}_{(\lambda_2)}} u),\] where \(\lambda_1+\lambda_2 = -\partial\). Then \[\begin{gather} \label{eq:Cohom-comp-1} \theta (1,p)_{\lambda_1,\lambda_2,\lambda_3}(u,v,w) =\theta _{\lambda_1,\lambda_2+\lambda_3}(u,p_{\lambda_2,\lambda_3}(v,w)) \\ = x_1'x_2'x_3\otimes (u \mathbin{{\circ}_{(\lambda_1)}} f_{\lambda_2}(v,w)) -x_1'x_2x_3' \otimes (u \mathbin{{\circ}_{(\lambda_1)}} f_{\lambda_3}(w,v)) \\ - x_1(x_2'x_3)'\otimes (f_{\lambda_2}(v,w)\mathbin{{\circ}_{(\lambda _2+\lambda_3)}} u ) + x_1(x_2x_3')'\otimes (f_{\lambda_3} (w,v) \mathbin{{\circ}_{(\lambda_2+\lambda_3 )}} u), \end{gather}\tag{18}\] where \(\lambda_1+\lambda_2+\lambda_3=-\partial\). Similarly, \[\begin{gather} \label{eq:Cohom-comp-2} p(\theta ,1)_{\lambda_1,\lambda_2,\lambda_3 }(u,v,w) =p_{\lambda_1+\lambda_2,\lambda_3} (\theta _{\lambda_1,\lambda_2}(u,v),w) \\ = (x_1'x_2)'x_3\otimes f_{\lambda_1+\lambda_2}(u\mathbin{{\circ}_{(\lambda_1)}} v, w) - (x_1x_2')'x_3 \otimes f_{\lambda_1+\lambda_2}(v\mathbin{{\circ}_{\lambda_2}} u, w) \\ -x_1'x_2x_3'\otimes f_{\lambda_3}(w, u\mathbin{{\circ}_{(\lambda_1)}} v) +x_1x_2'x_3'\otimes f_{\lambda_3}(w, v\mathbin{{\circ}_{(\lambda_2)}} u). \end{gather}\tag{19}\] All terms of 17 may be evaluated from 18 and 19 by permutations, for example, \[\theta (1,p)^{(123)}_{\lambda_1,\lambda_2,\lambda_3}(u,v,w) =\theta (1,p)_{\lambda_3,\lambda_1,\lambda_2}(w,u,v),\] etc. For example, the first tensor factor \(x_1''x_2x_3\) appears in \(p(\theta ,1)\), \(p(\theta ,1)^{(23)}\), \(\theta (1,p)^{(12)}\), and \(\theta (1,p)^{(123)}\). Collecting the respective summands of 17 , we obtain that \(d^2(p)=0\) implies 16 for \(\lambda_1=\lambda\), \(\lambda_2=\mu\), \(\lambda_3 =-\partial - \lambda -\mu\). Similarly, comparing the coefficients at \(x_1'x_2'x_3\) leads us to 15 . Conversely, for every \(f\in Z^2(V,V)\), other summands of 17 turn into zero due to skew symmetry.
Therefore, the second cohomology of \(C^*(V,V)\) coincides with \(H^2(V,V) = Z^2(V,V)/B^2(V,V)\) defined above. Let us calculate \(H^2(V,V)\) for simple finite Novikov algebras \(V=\mathop{\fam 0 Cur}\nolimits_1\) and \(V=\mathcal{V}_\alpha\), \(\alpha\in \Bbbk\). As a result, we describe infinitesimal conformal deformations of these Novikov conformal algebras.
In the following statements, \(\bar \varphi = \varphi + B^2(V,V)\in H^2(V,V)\) for \(\varphi \in Z^2(V,V)\).
Theorem 4. For \(V=\mathop{\fam 0 Cur}\nolimits_1=Hv\) with \((v\mathbin{{\circ}_{(\lambda)}} v)=v\) we have \(H^2(V,V)=\Bbbk \bar\omega\), where \(\omega_\lambda(v,v) = (\partial+\lambda)v\).
Proof. Suppose a cocycle \(\varphi \in Z^2(V,V)\) acts on the generator of \(V\) as \[\varphi_\lambda (v,v) = f(\lambda,\partial ) v\] for some polynomial \(f(x_1,x_2)\in \Bbbk [x_1,x_2]\). Due to 14 the polynomial \(f\) completely determines the values of \(\varphi_\lambda\) on the \(H\)-module generated by \(v\).
Relations 15 and 16 for \(\varphi\) turn into the following polynomial equations on \(f\): \[\label{eq:Z240141Cur} f(\mu, \partial+\lambda ) + f(\lambda , \partial ) -f(\lambda , -\lambda -\mu) = f(\lambda , \partial+\mu) + f(\mu, \partial )- f(\mu, -\lambda -\mu),\tag{20}\] \[\label{eq:Z240241Cur} f(\lambda+\mu, \partial ) + f(\lambda, -\lambda -\mu) = f(\lambda , \partial+\mu) + f(-\partial-\mu, \partial).\tag{21}\]
A cocycle \(\varphi\) defined by a polynomial \(f\) as above is a coboundary if and only if \[\varphi_\lambda (v,v) = v\mathbin{{\circ}_{(\lambda )}} \tau(v) - \tau(v\mathbin{{\circ}_{(\lambda )}} v) + \tau(v) \mathbin{{\circ}_{(\lambda )}} v\] for some \(H\)-linear map \(\tau : V\to V\). As \(\tau (v) = g(\partial )v\) for some polynomial \(g\in \Bbbk [x]\), we have \[\label{eq:B295Cur} f(\lambda , \partial ) = (d^1g)(\lambda , \partial ):= g(\partial+\lambda) -g(-\lambda ) - g(\partial ).\tag{22}\]
Since the equations 20 and 21 are homogeneous, we may assume the polynomial \(f(x_1,x_2)\) is homogeneous as well, \(m=\deg f\ge 0\).
For \(m=0\), a constant polynomial \(f(\lambda ,\partial )=c\in \Bbbk\) is a coboundary (\(g=c\) works), so there are no non-trivial cocycles in degree zero.
Suppose \(m>0\). Put \(\mu=0\) in 21 to get \[f(\lambda ,-\lambda ) = f(-\partial, \partial ),\] i.e., \(\lambda +\partial\) divides \(f(\lambda , \partial )\).
Put \(\mu=0\) in 20 to get \[\label{eq:Z240lin41Cur} f(0,\partial+\lambda ) = f(0,\partial ) - f(0,-\lambda ).\tag{23}\] If \(m=1\) then \(f(\lambda ,\partial ) = c_1(\lambda +\partial )\), \(c_1\in \Bbbk\). This polynomial defines a cocycle which is not a coboundary for \(c_1\ne 0\) since no terms of degree one may appear in \(d^1g\). Therefore, there is a 1-dimensional space of cohomologies spanned by the cocycle \(\omega_\lambda (v,v) = (\partial +\lambda ) v\).
Suppose \(m>1\). Then \[f(\lambda , \partial ) = a_0\lambda ^m + \dots + a_{m-1}\lambda \partial^{m-1} + a_m\partial ^m.\] It follows from 23 that \(a_m=0\). We may also eliminate the summand with \(a_{m-1}\) modulo an appropriate coboundary: choose \(g(x) = \frac{1}{m} a_{m-1}x^m\), then \[d^1g(\lambda,\partial ) = \lambda (\lambda (\dots )+a_{m-1}\partial^{m-1}).\] Hence, modulo \(B^2(V,V)\), we may assume that \(\lambda^2\) divides \(f(\lambda , \partial )\).
Apply the partial derivative \(\partial /\partial \lambda\) to 21 : \[f_1'(\lambda+\mu,\partial ) + f_1'(\lambda , -\lambda -\mu) - f'_2(\lambda , -\lambda -\mu) = f'_1(\lambda , \partial+\mu),\] where \(f'_i(x_1,x_2)\) stand for \(\partial f/\partial x_i\), \(i=1,2\). Put \(\lambda =0\) into the last expression. The summands \(f'_1(0,x)\) are zero since \(\lambda^2\mid f\), so we obtain \[\label{eq:DiffEq95Cur} f'_1(\mu,\partial )- f'_2(0,-\mu) = 0.\tag{24}\] Finding all homogeneous solutions of 24 is a routine problem on the method of indeterminate coefficients, the only solution is \[f(x_1,x_2) = c_0x_1^m.\] Recall that \(x_1+x_2\) divides \(f\), so we have \(c_0=0\) as required.
Therefore, the only non-trivial cocycle appears in degree one. ◻
Theorem 5. For \(V=\mathcal{V}_\alpha\) with \((v\mathbin{{\circ}_{(\lambda )}} v)=(\partial+\lambda+\alpha )v\) we have \(H^2(V,V) = \Bbbk \bar \varepsilon\), where \(\varepsilon_\lambda (v,v)=v\).
Proof. Suppose \(\varphi \in Z^2(V,V)\). Then it is completely determined by a polynomial \(f(x_1,x_2)\in \Bbbk [x_1,x_2]\) such that \(\varphi_\lambda (v,v) = f(\lambda , \partial )v\) for the generator \(v\) of \(\mathcal{V}_\alpha\). Relations 15 and 16 turn into \[\begin{gather} \label{eq:Z2V4014195a} (\partial+\lambda+\alpha)f(\mu, \partial+\lambda )- (\alpha-\mu)f(\lambda+\mu,\partial ) + (\partial+\lambda+\mu+\alpha) (f(\lambda,\partial ) - f(\lambda , -\lambda -\mu)) \\ = (\partial+\mu+\alpha)f(\lambda , \partial+\mu ) - (\alpha -\lambda )f(\lambda+\mu, \partial ) + (\partial+\lambda+\mu+\alpha) (f(\mu,\partial ) - f(\mu, -\lambda -\mu)) \end{gather}\tag{25}\] and \[\begin{gather} \label{eq:Z2V4024195a} (\mu-\alpha )f(\lambda+\mu,\partial )- (\partial+\lambda+\mu+\alpha)f(\lambda, -\lambda-\mu) \\ = (\mu-\alpha )f(\lambda , \partial+\mu) - (\partial+\lambda+\mu+\alpha)f(-\partial-\mu, \partial ). \end{gather}\tag{26}\] A cocycle \(\varphi\) defined by a polynomial \(f\) is a coboundary if and only if \[f(\lambda, \partial ) = (d^1g)(\lambda ,\partial ) := (\partial+\lambda+\alpha) (g(\partial+\lambda )+g(-\lambda ) -g (\partial ))\] for some polynomial \(g\in \Bbbk [x]\).
For example, \(f(x_1,x_2)=c \in \Bbbk\) defines a cocycle which is not a coboundary (for \(c\ne 0\)) since it is not divisible by \((\partial+\lambda+\alpha)\).
Let us show that every polynomial \(f\) satisfying 15 , 16 is constant modulo \(d^1g\), \(g\in \Bbbk [x]\).
Put \(\mu=0\) in 26 to obtain \[f(\lambda , -\lambda ) = f(-\partial, \partial ).\] Hence, \(f(x,-x) = c\); we may assume \(c=0\) modulo the constant cocycle. Therefore, \(\partial+\lambda\) divides \(f(\lambda, \partial)\).
Then put \(\mu=0\) in 25 to get \[(\partial+\lambda+\alpha)(-f(0,\partial+\lambda ) + f(0,\partial)- f(0,-\lambda )) = 0.\] The latter means \(f(0,\partial ) = c_1\partial\), \(c_1\in \Bbbk\), so \[\label{eq:Z2V240h41} f(\lambda , \partial ) = c_1(\partial+\lambda ) + \lambda(\partial+\lambda )h(\lambda ,\partial)\tag{27}\] for some polynomial \(h\). Let us decompose \(h\) into homogeneous summands to present \[\label{eq:Z2V4034195a} f(\lambda , \partial ) = c_1(\lambda +\partial ) + f_2(\lambda,\partial) + \dots + f_m(\lambda, \partial ),\tag{28}\] where \(\deg f_k = k\). Note that \(\lambda (\lambda+\partial )\) divides every \(f_k(\lambda , \partial )\), \(k=2,\dots, m\).
Suppose \[f_k(\lambda, \partial ) = (c_k\lambda \partial^{k-2} + \lambda^2(\dots ))(\lambda +\partial ).\] For \(k>1\), there exists a polynomial \(g_k(x)\) such that \(d^1g_k(\lambda ,\partial ) = (c_k\lambda \partial^{k-2} + \lambda^2(\dots ))(\lambda +\partial +\alpha )\). The highest homogeneous component of \(d^1g_k\) is of the same form as \(f_k\): the coefficient at \(\lambda\partial^{k-1}\) is \(c_k\). Hence, modulo \(B^2(V,V)\), we may assume that \(\lambda^2\) divides the highest homogeneous component of \(f\): \[x_1^2 \mid G(x_1,x_2) := f_m(x_1, x_2 ).\] The latter means \(G'_i(0,x_2) = 0\), where \(G'_i\) stand for \(\partial G/\partial x_i\), \(i=1,2\).
Now put the presentation 28 into 26 and consider the component of highest degree \(m+1\): \[\label{eq:Z2V40pr41} \mu G(\lambda+\mu , \partial ) - (\partial+\lambda+\mu) G(\lambda, -\lambda -\mu) = \mu G(\lambda , \partial +\mu ) - (\partial+\lambda +\mu)G(-\partial-\mu,\partial ).\tag{29}\] Apply partial derivative \(\partial/\partial\lambda\) to 29 : \[\begin{gather} \mu G'_1(\lambda +\mu,\partial ) - G(\lambda , -\lambda -\mu) -(\partial +\lambda +\mu)(G'_1(\lambda , -\lambda -\mu ) - G'_2(\lambda, -\lambda -\mu )) \\ = \mu G'_1(\lambda , \partial+\mu) - G(-\partial-\mu, \partial ). \end{gather}\] Put \(\lambda =0\) into the last expression to get \[\label{eq:Z2V40fi41} \mu G'_1(\mu,\partial ) = - G(-\partial-\mu, \partial ).\tag{30}\] It is not hard to note that the only homogeneous solution \(G(x_1,x_2)\) of 30 is proportional to \(x_1x_2^{m-1}+x_2^m\) which is not divisible by \(x_1^2\) apart from \(G=0\).
Therefore, the polynomial \(h\) in 27 has no nonzero homogeneous components and \(f(\lambda , \partial ) = c_1(\lambda +\partial ) \equiv -\alpha c_1 \pmod {B^2(V,V)}\), as required. ◻
Let \(V\) be a Novikov conformal algebra, and let \(M\) be a Novikov conformal bimodule over \(V\). The second cohomology space of \(V\) with coefficients in \(M\) describes the equivalence classes of extensions \[0\to M\to E\to V\to 0\] with \(M\mathbin{{\circ}_{(\lambda )}}M = 0\) in \(E\). The details are completely similar to what is known in Lie or associative case [23].
It is well known that the Virasoro conformal algebra \(Hv\) with \([v\mathbin{{}_{(\lambda )}} v] = (2\lambda+\partial ) v\) has a non-trivial central extension by means of the scalar module \(M=\Bbbk e\), \(\partial e = 0\). Let us consider a similar question for simple Novikov conformal algebras.
Let \(V\) be a Novikov conformal algebra and let \(M=\Bbbk e\) be a 1-dimensional \(H\)-module such that \(\partial e=0\). Then \(M\) is a conformal bimodule over \(V\) so that \[u\mathbin{{\circ}_{(\lambda )}} e = e\mathbin{{\circ}_{(\lambda )}} u = 0\] for all \(u\in V\). Consider an extension \(E\) of \(V\) by means of \(M\). The structure of \(E\) is completely determined by a mapping \[f_\lambda : V\otimes V \to M[\lambda ]\cong \Bbbk [\lambda ]e\] such that \[\begin{gather} f_\lambda (\partial u,v) = -\lambda f_\lambda (u,v), \quad f_\lambda (u,\partial v) = \lambda f_\lambda (u,v), \tag{31} \\ f_{\lambda +\mu} (u\mathbin{{\circ}_{(\lambda )}}v, w) - f_\lambda (u, v\mathbin{{\circ}_{(\mu )}} w) = f_{\lambda+\mu} (v\mathbin{{\circ}_{(\mu )}} u, w) - f_\mu (v, u\mathbin{{\circ}_{(\lambda )}}w), \tag{32} \\ f_{\lambda+\mu} (u\mathbin{{\circ}_{(\lambda )}} v, w) = f_{-\partial-\mu } (u\mathbin{{\circ}_{(\lambda )}} w, v). \tag{33} \end{gather}\] The last two equations are obtained from 15 and 16 assuming the action of \(V\) on \(M\) is trivial.
Two extensions \(E\) and \(E'\) defined respectively by maps \(f\) and \(f'\) are said to be equivalent if there exists an isomorphism of conformal algebras \(E\to E'\) which is an identity map on \(M\). As in the case of Lie or associative conformal algebras [23], \(E\) and \(E'\) are equivalent if and only if \[f_\lambda (u,v) - f'_\lambda(u,v) = (d^1g)_\lambda (u,v):= g(u\mathbin{{\circ}_{(\lambda )}} v), \quad u,v\in V,\] for some \(H\)-linear map \(g:V\to M\).
Theorem 6. Let \(V=\mathop{\fam 0 Cur}\nolimits_1\) or \(V=\mathcal{V}_\alpha\), \(\alpha \in \Bbbk\). Then \(V\) has no nontrivial extensions by means of the scalar module \(M=\Bbbk e\).
Proof. Suppose an extension \(E\) of \(V=Hv\) by means of the scalar module \(M\) is defined by a cocycle \(\varphi\) such that \[\varphi_\lambda (v,v) = f(\lambda )e\] for some polynomial \(f(t)\in \Bbbk [t]\).
Suppose \(V=\mathop{\fam 0 Cur}\nolimits_1\). Then 32 turns into \(f(\lambda ) = f(\mu )\), so \(f(t)=c_0\in \Bbbk\). Obviously, the cocycle \(\varphi_\lambda (v,v)=c_0e\) is a coboundary: \(\varphi = d^1g\), where \(g(v)=c_0e\).
Suppose \(V=\mathcal{V}_\alpha\). Then 32 and 33 turn into \[\begin{gather} (\lambda -\mu)f(\lambda +\mu ) = (\lambda+\mu+\alpha )(f(\lambda ) - f(\mu )), \\ (\alpha -\mu)f(\lambda +\mu ) = (\lambda+\mu+\alpha )f(-\mu). \end{gather}\] Since \(\lambda -\mu\) and \(\lambda+\mu+\alpha\) are mutually prime, we derive that \(t+\alpha\) divides \(f(t)\). Substitute \(f(t)=(t+\alpha )h(t)\) into the second equation to get \(h(\lambda+\mu)=h(-\mu)\), i.e., \(f(t) = (t+\alpha )c_1\), \(c_1\in \Bbbk\). The corresponding cocycle \(\varphi\) is a coboundary: for \(g(v)=c_1e\) we have \((d^1g)_\lambda (v,v) = g((\partial+\lambda+\alpha)v)= (\lambda+\alpha)c_1 e\). ◻