A Casimir element inexpressible as a Lie polynomial

Let $q$ be a scalar that is not a root of unity. We show that any polynomial in the Casimir element of the Fairlie-Odesskii algebra $U_q'(\mathfrak{so}_3)$ cannot be expressed in terms of only Lie algebra operations performed on the generators $I_1,I_2,I_3$ in the usual presentation of $U_q'(\mathfrak{so}_3)$. Hence, the vector space sum of the center of $U_q'(\mathfrak{so}_3)$ and the Lie subalgebra of $U_q'(\mathfrak{so}_3)$ generated by $I_1,I_2,I_3$ is direct.


Introduction
Important quantum groups and associative algebras have presentations with deformed commutation relations, by which we mean defining relations that involve expressions of the form c 1 U V − c 2 V U where U, V are elements of the associative algebra, and the scalar parameters c 1 , c 2 are not necessarily both 1. These expressions resemble the usual Lie bracket in the associative algebra, which is [U, V ] := U V − V U . Some specific examples are the quantum group U q (sl 2 ) in its "equitable presentation" [13], the Fairlie-Odesskii algebra U ′ q (so 3 ) in its usual presentation [7,8,9,12], the parametric family of Askey-Wilson algebras [15,16], and the parametric family of q-deformed Heisenberg algebras [10,11] whose defining relations are q-deformations of the Heisenberg-Weyl relation.
However, the existence of an associative algebra structure in some vector space implies the existence of a Lie algebra structure in the same vector space. Thus, even if the defining relations of an associative algebra involve deformed commutation relations, it is still possible to compute Lie polynomials in the generators of the algebra. This notion first appeared in [15,Problem 12.14] for the universal Askey-Wilson algebra. It was found that the defining relations in this algebra and the related family of Askey-Wilson algebras are not Lie polynomials in the generators [2], and so the usual algebraic machinery for finitely generated and finitely presented Lie algebras is not directly applicable, but it was further shown that the Lie subalgebra generated by the generators of the algebra is not free [2]. Similar studies were done for q-deformed Heisenberg algebras [3,4]. In all such studies [2,3,4] the focus was on the consequences of the non-Lie polynomial, deformed commutation relations on the Lie polynomials in the same algebra.
For this paper, we study the Fairlie-Odesskii algebra U ′ q (so 3 ), and the consequences of its deformed commutation relation on the Lie polynomials in relation to the Casimir element, an element which is significant in studying representations and central elements of the algebra [7,8,9]. More specifically, we show that any nonzero polynomial in the Casimir element of the Fairlie-Odesskii algebra is not a Lie polynomial in the generators of the algebra.

Preliminaries
Denote the set of all nonnegative integers by N, and the set of all positive integers by Z + . Let F denote a fixed but arbitrary field. Throughout, by an algebra we mean a unital associative algebra A over F with unity element I A . We use the convention that U 0 = I A for any U ∈ A, and we denote I A simply by I if no confusion shall arise. Any subalgebra is assumed to contain the unity element. We also note that every algebra A has a Lie algebra structure induced by [U, V ] := U V −V U for all U, V ∈ A. Throughout, whenever we refer to a Lie algebra structure on an algebra A, we shall always mean that which is induced by the Lie bracket operation [·, ·] just mentioned. Let X 1 , X 2 , . . . , X n ∈ A. If K is the Lie subalgebra of A generated by X 1 , X 2 , . . . , X n , then we call the elements of K the Lie polynomials in X 1 , X 2 , . . . , X n . Given U ∈ A, the linear map ad U : A → A is defined by the rule V → [U, V ]. Also, for any linear map ϕ whose domain and codomain are equal, and for any n ∈ N, by ϕ n we mean composition of ϕ with itself n times, where ϕ 0 is interpreted as the identity linear map.
The Fairlie-Odeskii algebra. Fix a nonzero q ∈ F. The Fairlie-Odeskii algebra is the algebra U ′ q (so 3 ) that has a presentation by generators I 1 , I 2 , I 3 and relations A torus algebra related to U ′ q (so 3 ). The algebra U ′ q (so 3 ) can be interpreted as an algebra of quantum geodesics or an algebra of quantized geodesic functions on a coordinate algebra of a torus, which is related to quantum gravity [5,6,12,14]. More precisely, U ′ q (so 3 ) is a subalgebra of some other algebra related to a torus, which we describe in the following. Denote by A q the algebra with a presentation given by six Thus, (7) holds for the cases covered in (10), (11), (12). We now consider arbitrary m, n ∈ Z. If one of m, n is zero, then we are left with the trivial equation u = q 0 u where u is either I or a power of one generator. In this case, (7) still holds. Consider the case that both m, n are nonzero. Given h ∈ {m, n} such that h is negative, there exists k ∈ Z + such that h = −k. This implies that one of the three cases (10), (11), (12) is applicable, and so (7) holds for any m, n ∈ Z. Apply Φ to both sides of (7) to get (8). To get (9), observe that since m, n are arbitrary, (8) can be written as z n 2 z m 3 = q mn z m 3 z n 2 . Apply Φ to both sides of this equation, and then solve for z m 1 z n 3 . From this, we get (9). The reordering formula from Proposition 3.1 may be easily used to rewrite the product or Lie bracket of any two basis elements from (3) into a linear combination of (3).
Corollary 3.2. Given any integers h, m, n, u, v, w, the relations hold in A q .
Proof. Reorder z h 3 z m 2 z n 1 · z u 3 z v 2 z w 1 using (7), (8), (9). From this, we get (13). The relation (14) is a routine application of (13). Corollary 3.3. Given any integers h, m, n, u, v, w, the relation where Proof. Use (13) and (14).  (ii) If q is not a root of unity and if H = 0, then the identity (15) implies that scalar multiplication can be used to "convert" the product of two basis vectors from (3) into the Lie bracket of the same two vectors.
which implies that which proves (17) for the case T = −1. Suppose that (17) holds for some T ∈ Z\N. That is, We emphasize here that in (19), the exponent of ad z −1 1 is −T > 0, and so ad z −1 1 −T is a valid composition of mappings. We then apply the map ad z −1 1 on both sides of (19) and use the reordering formula (7), (8), (9). From this, we get , which further implies that By induction (17) holds for all T ∈ Z\N, and this completes the proof. The relevance of (16) and (17) will now be apparent in the proof of the following.

Lemma 4.2.
For any h ∈ Z\{0}, we have Proof. By (16) and (17), both z 3 z 2 2 z 1 and z −1 are elements of L q . But by (14), we have where the existence of (1 − q h ) −1 follows from the assumption that q is not a root of unity. Thus, z h 1 ∈ L q . We now show z h 2 ∈ L q . Apply Φ on both sides of (20), and we obtain where the left-hand side was obtained by using the fact that Φ is an algebra homomorphism, while the right-hand side was obtained using the fact that Φ, being an algebra homomorphism, is necessarily a Lie algebra homomorphism. To evaluate Φ z 3 z 2 2 z 1 and Φ z −1 3 z −2 2 z h−1 1 , we use (16) and (17), and we use the property of Φ that it is a Lie algebra homomorphism to evaluate the resulting right-hand sides. This gives us We note that the scalar coefficients in (21), (22), (23) are all defined in the field because q is assumed to be nonzero and not a root of unity. By inspecting the right-hand sides in (22), (23), we find that both Φ z 3 z 2 2 z 1 and Φ z −1 are in L q , and so by (21), we have z h 2 ∈ L q . A similar argument can be made to prove that z h 3 ∈ L q , and this should start by applying Φ 2 on both sides of (20). We now look for more basis elements from (3) that are in L q . But first, we need the following.
hold in A q .
We immediately find that {Λ N : N ∈ Z} is a Z-gradation of A q .
It is routine to show that where (53) is obtained from (52) by using the reordering formula from Section 3 on the first term of (52). Now, we use (47) and (53) to compute for −G 1 G 2 G 3 , and we obtain By (44) and (48) to (51), we find that ν j ∈ Λ j for all j ∈ {−4, −2, 0, 2}. Then 1 i=−2 ν 2i ∈ 1 i=−2 Λ 2i . This further implies that the terms (55) to (58) are elements of respectively. By the properties of the gradation subspaces, the second and fourth vector spaces in (59) can be simplified as and so we deduce that the terms (55) to (58) are elements of respectively. Denote by R the sum of (55) to (58). By inspection of the limits of the indices of the direct sums in (60), we find that R ∈ 2 i=−3 Λ 2i . We now rewrite (54) to (58) as −G 1 G 2 G 3 + q 1 2 z 2 3 (z 2 z 1 ) 2 = R, in which we further rewrite the term q 1 2 z 2 3 (z 2 z 1 ) 2 using the reordering formula for A q , and by doing this we get Lemma 5.2. For any n ∈ Z + , the Casimir element C satisfies the property C n − (−1) n q 2(n 2 −n+2) (q 2 − 1) −2n z 2n 3 z 2n 2 z 2n