Invariant polynomial functions on the Poisson algebra in characteristic p

Serge Skryabin

Classical results of Chevalley and Harish-Chandra describe the ring of invariant polynomial functions k[𝔤]^{𝔤} on a complex semisimple Lie algebra 𝔤 and the center Z of the universal enveloping algebra U(𝔤). If now 𝔤 is a finite dimensional Lie algebra over an algebraically closed field k of characteristic p > 0 then both k[𝔤]^{𝔤} and Z are essentially bigger than in the complex case. Indeed, k[𝔤]^{𝔤} always contains the subalgebra k[𝔤]^{(p)} consisting of the powers φ^{p} of all functions φ ∈ k[𝔤] and, similarly, Z contains the so-called p-center Z_{p} over which U(𝔤) is a finite module (see [25]). However, if 𝔤 is the Lie algebra of a semisimple algebraic group G then, under some restrictions on p, there are precise analogs of classical results for the subrings of G-invariants k[𝔤]^{G} and Z^{G}, as was shown by Veldkamp [21], and Kac and Weisfeiler [5]. Furthermore, k[𝔤]^{𝔤} = k[𝔤]^{(p)} · k[𝔤]^{G} and Z = Z_{p} · Z^{G} (see also [3]).

There is another big class of simple finite dimensional Lie algebras over k called the Lie algebras of Cartan type, for which the situation with the invariants is very little understood until now. A significant progress was earlier achieved only in one case by Premet [15] who completely described the ring of invariants k[𝔤]^{G} where 𝔤 = W_{n} is the Jacobson-Witt algebra and G its automorphism group. Premet established analogs of many classical results, although G has a big unipotent radical and the Lie algebra of G is a proper subalgebra of 𝔤. One no longer has an invariant bilinear form on 𝔤, and so one cannot pass to invariants in the symmetric algebra S(𝔤), all the more to the central elements in U(𝔤). From the viewpoint of invariant theory the coadjoint 𝔤-module looks very much different from the adjoint module. As follows from computations of Krylyuk [8], the stabilizers of generic linear functions on 𝔤 have big [p]-nilpotent parts whose dimension grows exponentially when n increases.

In the present paper I consider another class of Lie algebras over k. Denote by B_{2n} = k[x_{1}, …, x_{2n}], x_{i}^{p} = 0, the commutative associative algebra of truncated polynomials in 2n variables. The Poisson bracket defined by the formula

[f, g] = ∑_{i=1}^{n} (∂_{i}(f)∂_{n+i}(g) − ∂_{n+i}(f)∂_{i}(g)), f, g ∈ B_{2n},

where ∂_{1}, …, ∂_{2n} denote the partial derivatives in x_{1}, …, x_{2n} gives a Lie algebra structure on B_{2n}. The resulting Lie algebra has a center coinciding with the scalars k, and the factor algebra B_{2n}/k is isomorphic to a certain Hamiltonian Lie algebra L whose commutator subalgebra [L, L] is simple (with an exception for p = 2, n = 1) and nonclassical (with an exception for p = 2, n = 2 or p = 3, n = 1). The Lie algebras appearing here are rather special in the whole class of Hamiltonian algebras (see [4], [11], [23]). In fact these algebras admit a [p]-map which makes them into p-Lie algebras or restricted Lie algebras, as defined by Jacobson [1]. The [p]-structure turns out to be essential for the construction of invariants. The reason that I work with the Poisson algebra B_{2n} rather than with L is that the results are stated nicer in this case.

The main results are as follows. Denote by G the group of Poisson automorphisms of B_{2n}, i.e., invertible transformations preserving both the associative and the Lie algebra structures. Let G_{[p]} ⊂ G be the subgroup whose elements respect the [p]-map as well. Consider also the action of L on B_{2n} induced by the adjoint action of B_{2n}. Theorem 3.4 states that k[B_{2n}]^{L} is generated over k[B_{2n}]^{(p)} by certain G_{[p]}-invariant functions φ_{1}, …, φ_{p^{n}}. Moreover, k[B_{2n}]^{L} is free over k[B_{2n}]^{(p)} and is a locally complete intersection ring. In Theorem 5.2 I prove that k[B_{2n}]^{G_{[p]}} and k[B_{2n}]^{G} are polynomial rings generated by p^{n} + n and p^{n} elements, respectively. Moreover, there are affine subspaces S, S_{0} ⊂ B_{2n} such that the restriction of functions gives isomorphisms k[B_{2n}]^{G} ≅ k[S_{0}] and k[B_{2n}]^{G_{[p]}} ≅ k[S]. Clearly, S, S_{0} are analogs of Kostant’s transversal plane to a principal nilpotent element in the classical settings [10]. The subsets S, S_{0} give a full system of representatives for the G_{[p]}- (respectively G-) conjugacy classes on a Zariski open subset U ⊂ B_{2n} whose elements are “regular” for the results of the present paper. In particular, the Lie centralizers 𝔷(f) of elements f ∈ U all have dimension p^{n} (see Proposition 2.2). To prove that the invariants separate the orbits on U I need Theorem 4.3 which ensures that every [p]-compatible isomorphism of Lagrangian subalgebras of B_{2n} (see section 1 for the definition) admits an extension to an element of G.

In Theorem 6.4 I prove that the cone N of [p]-nilpotent elements in B_{2n} is an irreducible normal complete intersection of codimension n. This is again an analog of Kostant’s results. It should be mentioned for comparison that in case of W_{n} the singular locus of the nilpotent cone has codimension 1 in N, and therefore N is not normal, as was proved by Premet [15]. In [14] Premet conjectured that the nilpotent cone N is irreducible for any p-Lie algebra, and proved under this assumption that N is a set-theoretic complete intersection (apparently the irreducibility assumption can be removed here). The available information suggests that N is always a strict complete intersection. There are still certain classical facts which do not carry over to the Poisson algebra. The nilpotent cone does not contain a dense orbit, and k[B_{2n}] is a free module neither over k[B_{2n}]^{G_{[p]}} nor over k[B_{2n}]^{G}, at least when either p or n is big enough. It seems that these deviations are related somehow to the fact that the generic centralizers 𝔷(f) are not [p]-semisimple, and the [p]-semisimple elements are not dense in B_{2n}.

Luckily enough, an invariant bilinear form on B_{2n} does exist, yielding a G- and L-equivariant isomorphism S(B_{2n}) ≅ k[B_{2n}]. Denote by φ_{1}^{∨}, …, φ_{p^{n}}^{∨} ∈ S(B_{2n}) the images of the invariant functions φ_{1}, …, φ_{p^{n}} mentioned before. Then φ_{a}^{∨} is homogeneous of degree a. Consider the canonical increasing filtration U_{m}(B_{2n}), m ≥ 0, in the universal enveloping algebra. If 0 < a < p^{n} and a ≢ 0 (mod p), Theorem 7.2 ensures the existence of a G_{[p]}-invariant element z_{a} ∈ Z ∩ U_{a}(B_{2n}) whose image in S^{a}(B_{2n}) coincides with φ_{a}^{∨}. The major unsolved problem is to verify this property for the remaining invariants in the symmetric algebra. I will explain briefly the construction of z_{a}. If a ≢ 0 (mod p) then φ_{a}^{∨} is an element of an L-submodule, call it V ⊂ S^{a}(B_{2n}), which is induced from a onedimensional L_{0}-submodule, call it V_{0} ⊂ S^{a}(B_{2n}), where L_{0} is a subalgebra of codimension 2n in L. It turns out that V_{0} ⊂ S(I) where I ⊂ B_{2n} is an abelian Lie subalgebra. As U(I) ≅ S(I), we can generate by V_{0} an L-submodule V′ ⊂ U_{a}(B_{2n}). The latter is mapped isomorphically onto V under the projection U_{a}(B_{2n}) → S^{a}(B_{2n}). So we can find z_{a} inside V′.

1. Preliminaries

Denote by B_{n} = k[x_{1}, …, x_{n}], x_{i}^{p} = 0, the truncated polynomial algebra in n variables where k is an algebraically closed field of characteristic p > 0. One has f^{p} ∈ k for all f ∈ B_{n}, and so B_{n} is local with the maximal ideal 𝔫 consisting of those f ∈ B_{n} for which f^{p} = 0. Any minimal system of generators of B_{n} consists of n elements y_{1}, …, y_{n} which are linearly independent modulo k + 𝔫^{2}. One has then y_{i}^{p} = α_{i} for some α_{1}, …, α_{n} ∈ k and the monomials y_{1}^{r_{1}} ⋯ y_{2n}^{r_{n}} with 0 ≤ r_{i} < p constitute a basis for B_{n} over k. If z_{1}, …, z_{n} is another minimal system of generators with z_{i}^{p} = β_{i} then there exists an automorphism θ ∈ Aut B_{n} sending each y_{i} to z_{i} if and only if α_{i} = β_{i} for all i. The derivation algebra W_{n} = Der B_{n} is a Lie algebra called the Jacobson-Witt algebra. Note that W_{n} is a free B_{n}-module with a basis consisting of the partial derivatives ∂_{1}, …, ∂_{n} in x_{1}, …, x_{n}.

If C is a commutative, associative and unital algebra, then a Poisson bracket on C is a Lie algebra structure satisfying the identity [f, gh] = [f, g]h + [f, h]g where f, g, h ∈ C. This identity means that the linear transformations D_{f} : C → C given by D_{f}(g) = [f, g] are derivations of C. The algebra C together with a fixed Poisson bracket on it is called a Poisson algebra. We will be considering the truncated polynomial algebra B_{2n} in 2n variables x_{1}, …, x_{2n} as a Poisson algebra using the Poisson bracket explicitly written out in the introduction. Thus [f, g] = D_{f}(g) where

D_{f} = ∑_{i=1}^{n} (∂_{i}(f)∂_{n+i} − ∂_{n+i}(f)∂_{i}) ∈ W_{2n}.

The assignment f ↦ D_{f} gives a Lie algebra homomorphism B_{2n} → W_{2n}. Its image L = {D_{f} | f ∈ B_{2n}} is therefore a Lie subalgebra in W_{2n} (it is an ideal in the Hamiltonian Lie algebra H_{2n} consisting of all D ∈ W_{2n} such that Dω = 0 where ω = dx_{1} ∧ dx_{n+1} + … + dx_{n} ∧ dx_{2n} is a Hamiltonian differential form). Taking f = x_{1}, …, x_{2n} in the formula for D_{f}, one sees that ∂_{1}, …, ∂_{2n} ∈ L. It is immediate thereof that the Lie center {f ∈ B_{2n} | [f, B_{2n}] = 0} coincides with k, and L ≅ B_{2n}/k. Denote by 𝔪 the maximal associative ideal of B_{2n}, and put L_{0} = {D_{f} | f ∈ 𝔪^{2}}, which is a subalgebra of L.

It is well known that D^{p} ∈ L for all D ∈ L. Hence one can define on B_{2n} a [p]-map f ↦ f^{[p]} satisfying the identities

([p]_{1}) D_{f}^{p} = D_{f^{[p]}},

([p]_{2}) (λf)^{[p]} = λ^{p}f^{[p]},

([p]_{3}) (f + g)^{[p]} = f^{[p]} + ∑_{l=1}^{p−1} s_{l}(f, g) + g^{[p]}

where f, g ∈ B_{2n}, λ ∈ k and ls_{l}(f, g) is equal to the coefficient of t^{l−1} in the expansion of the expression (D_{f} + tD_{g})^{p−1}(g) with t a scalar indeterminate (see [2], [19]). Because of a nontrivial center the [p]-map on B_{2n} is not determined uniquely. Since L_{0} coincides with the stabilizer of 𝔪 in L, we have D^{p} ∈ L_{0} for all D ∈ L_{0}. We can therefore specify the [p]-map on k + 𝔪^{2} by the requirement that 1^{[p]} = 0 and f^{[p]} ∈ 𝔪^{2} for all f ∈ 𝔪^{2}. Any [p]-structure with this property will be called normalized. Denote by G the group of automorphisms of both the associative and the Lie algebra structures on B_{2n}.

Lemma 1.1. Any two normalized [p]-structures on B_{2n} are transformed to one another by an element of G.

Proof. Since the derivations ∂_{1}, …, ∂_{2n} have zero [p]-power, x_{i}^{[p]} = α_{i} ∈ k for all i. Given 1 ≤ j ≤ 2n, we put j′ = j + n if j ≤ n and j′ = j − n if j > n. Let y_{i} = x_{i} for i ≠ j and y_{j} = x_{j} − λx_{j′}^{p−1} where λ ∈ k. There exists θ ∈ Aut B_{2n} sending each x_{i} to y_{i}. In fact θ ∈ G since [y_{i}, y_{j}] = [x_{i}, x_{j}] for all i ≠ j. If p > 2 then (x_{j′}^{p−1})^{[p]} = 0 and the terms s_{l}(x_{j}, x_{j′}^{p−1}) in the formula ([p]_{3}) likewise vanish for 1 < l < p. We

get then y_{j}^{[p]} = x_{j}^{[p]} − D_{x_{j}}^{p−1}(λx_{j′}^{p−1}) = α_{j} + λ. If p = 2 then y_{j}^{[2]} = α_{j} + λ + λ^{2}α_{j′}.

In any case we can find λ such that y_{j}^{[p]} takes any given value in k. Repeating this procedure successively for j = 1, …, 2n, we get the required element in G. □

From now on we fix a normalized [p]-structure on B_{2n} and denote by G_{[p]} ⊂ G the subgroup of automorphisms commuting with the [p]-map.

Remark. In this paper we don’t use the structural properties of G. It can be shown that G coincides with the group of those θ ∈ Aut B_{2n} which stabilize the Hamiltonian form ω, and this group was studied in [22]. In particular, G is a semidirect product of Sp_{2n} and the unipotent radical of G. The Lie algebra of G coincides with the stabilizer of 𝔪 in H_{2n}, and it can be shown that the Lie algebra of G_{[p]} is L_{0} for any normalized [p]-map.

We call an associative subalgebra B ⊂ B_{2n} (respectively an associative ideal I ⊂ B_{2n}) Lagrangian if it is generated by n elements y_{1}, …, y_{n} ∈ 𝔪 which are linearly independent modulo 𝔪^{2} and [B, B] = 0 (respectively [I, I] ⊂ I). Since y_{1}, …, y_{n} can be included in a minimal system of generators for B_{2n}, it is clear that B ≅ B_{n} and B_{2n} is free over B (respectively B_{2n}/I ≅ B_{n}) in this case. We state below a few simple properties of Lagrangian subalgebras and ideals.

Lemma 1.2. Suppose that B ⊂ B_{2n} is a Lagrangian subalgebra generated by y_{1}, …, y_{n} ∈ 𝔪. Then:

(1) D_{y_{1}}, …, D_{y_{n}} are a basis for Der_{B} B_{2n} ⊂ W_{2n}, the subalgebra of B-linear derivations, as a B_{2n}-module.

(2) The Lie centralizer 𝔷(B) = {g ∈ B_{2n} | [g, B] = 0} coincides with B. In particular, B^{[p]} ⊂ B. If f ∈ 𝔪_{B}^{2} where 𝔪_{B} = 𝔪 ∩ B then f^{[p]} = 0.

(3) There exists a Lagrangian ideal I ⊂ B_{2n} such that I^{[p]} ⊂ I and B_{2n} = B ⊕ I.

Proof. Define κ : B_{2n} → k by κ(λ + f) = λ for λ ∈ k and f ∈ 𝔪. The formula (g, h) = κ([g, h]) where g, h ∈ B_{2n} gives an alternating bilinear form on B_{2n}. Since [𝔪^{2}, B_{2n}] ⊂ 𝔪 = ker κ, the subspace 𝔪^{2} is contained in the kernel of this bilinear form. On the other hand, (x_{i}, x_{j}) ≠ 0 precisely when j − i = ±n. It follows that the induced bilinear form on the factor space 𝔪/𝔪^{2} is nondegenerate.

Since y_{1}, …, y_{n} ∈ 𝔪 are linearly independent modulo 𝔪^{2}, we can find elements z_{1}, …, z_{n} ∈ 𝔪 such that [y_{i}, z_{j}] ≡ δ_{ij} (mod 𝔪) for 1 ≤ i, j ≤ n. Then y_{1}, …, y_{n}, z_{1}, …, z_{n} form a minimal system of generators for B_{2n}. It follows that the monomials z_{1}^{r_{1}} ⋯ z_{n}^{r_{n}} with 0 ≤ r_{j} < p give a basis for B_{2n} as a module over B. It is now clear that the B_{2n}-module Der_{B} B_{2n} is free with a basis consisting of the derivations D_{1}, …, D_{n} ∈ W_{2n} such that D_{i}(y_{j}) = 0 and D_{i}(z_{j}) = δ_{ij}. The derivations D_{y_{i}} are B-linear as [B, B] = 0. Write D_{y_{i}} = ∑_{j=1}^{n} h_{ij}D_{j} with h_{ij} ∈ B_{2n}. Then h_{ij} = D_{y_{i}}(z_{j}) ≡ δ_{ij} (mod 𝔪). Hence the n × n matrix with entries h_{ij} is invertible, yielding (1).

Suppose g ∈ 𝔷(B). Then D_{y_{i}}(g) = 0, and by (1) D_{i}(g) = 0, for all i. Expressing g as a B-linear combination of monomials in z’s, we deduce that the coefficients of monomials of positive degree have to be zero. Thus 𝔷(B) = B. Suppose that f = gh where g, h ∈ 𝔪_{B}. Then D_{f}^{p} = (gD_{h} + hD_{g})^{p} = g^{p}D_{h}^{p} + h^{p}D_{g}^{p} = 0 as g^{p} = h^{p} = 0. It follows that D_{f}^{p} = 0 for any f ∈ 𝔪_{B}^{2}. Since f^{[p]} ∈ 𝔪^{2} is a unique element such that ([p]_{1}) holds, we get f^{[p]} = 0.

Let J ⊂ {1, …, 2n} be a maximal subset subject to the two conditions:

(a) for each i ∈ {1, …, n} the indices i and i + n are not both in J,

(b) the elements {y_{1}, …, y_{n}} ∪ {x_{i} | i ∈ J} are linearly independent modulo 𝔪^{2}.

Note that (a) means that [x_{i}, x_{j}] = 0 for all i, j ∈ J. We claim that J has cardinality n. Suppose the contrary. Then there exists j ∈ {1, …, n} such that j ∉ J and j + n ∉ J. Since neither J ∪ {j} nor J ∪ {j + n} is an admissible subset, we must have x_{j} ≡ u_{1} + v_{1} and x_{j+n} ≡ u_{2} + v_{2} (mod 𝔪^{2}) where u_{1}, u_{2} are linear combinations of elements y_{1}, …, y_{n} and v_{1}, v_{2} are linear combinations of elements {x_{i} | i ∈ J}. Since both J ∪ {j} and J ∪ {j + n} satisfy (1), we see that the three commutators [v_{1}, v_{2}], [x_{j}, v_{2}], [v_{1}, x_{j+n}] all vanish. Then [u_{1}, v_{2}] = [v_{1}, u_{2}] = 0 as well. We have also [u_{1}, u_{2}] = 0 since u_{1}, u_{2} ∈ B. It follows [x_{j}, x_{j+n}] = 0, a contradiction.

Denote by I ⊂ B_{2n} the associative ideal generated by n elements {x_{i} | i ∈ J}. Then (a) ensures that [I, I] ⊂ I. As {y_{1}, …, y_{n}}∪{x_{i} | i ∈ J} is a minimal system of generators for B_{2n}, we get also B_{2n} = B ⊕ I. In view of lemma 1.1 we may assume, adjusting the generators x_{1}, …, x_{2n} if necessary, that x_{i}^{[p]} = 0 for all 1 ≤ i ≤ 2n. It will follow then from the next lemma that I^{[p]} ⊂ I. □

Lemma 1.3. Suppose that I ⊂ B_{2n} is a Lagrangian ideal generated by elements y_{1}, …, y_{n} ∈ 𝔪. For each f ∈ I denote by \overline{D}_{f} the derivation of the factor algebra \overline{B} = B_{2n}/I induced by D_{f}. Then:

(1) \overline{D}_{y_{1}}, …, \overline{D}_{y_{n}} are a basis for Der \overline{B} as a \overline{B}-module.

(2) The Lie normalizer 𝔫(I) = {g ∈ B_{2n} | [g, I] ⊂ I} coincides with k + I. In particular, I^{[p]} ⊂ k + I. If y_{i}^{[p]} ∈ I for all i then I^{[p]} ⊂ I.

(3) There exists a Lagrangian subalgebra B ⊂ B_{2n} such that B_{2n} = B ⊕ I.

Proof. Let y_{1}, …, y_{n}, z_{1}, …, z_{n} ∈ 𝔪 be a minimal system of generators for B_{2n} such that [y_{i}, z_{j}] ≡ δ_{ij} (mod 𝔪). Then \overline{B} has a basis consisting of the monomials \overline{z}_{1}^{r_{1}} ⋯ \overline{z}_{n}^{r_{n}} with 0 ≤ r_{i} < p where \overline{z}_{i} denotes the image of z_{i} in \overline{B}. Furthermore, \overline{z}_{i}^{p} = 0 for all i. The \overline{B}-module Der \overline{B} has a basis consisting of the derivations D_{1}, …, D_{n} such that D_{i}(\overline{z}_{j}) = δ_{ij}. Write \overline{D}_{y_{i}} = ∑_{j=1}^{n} h_{ij}D_{j} with h_{ij} ∈ \overline{B}. Then h_{ij} = \overline{D}_{y_{i}}(\overline{z}_{j}) ≡ δ_{ij} modulo the maximal ideal 𝔪/I of \overline{B}. Hence the n × n matrix with entries h_{ij} is invertible, and (1) is immediate.

Suppose that g ∈ 𝔫(I). Then D_{y_{i}}(g) ∈ I, and therefore \overline{D}_{y_{i}}(\overline{g}) = 0 for all i, where \overline{g} denotes the image of g in \overline{B}. It follows from (1) that \overline{g} ∈ k, i.e., g ∈ k + I. For every f ∈ I∩𝔪^{2} we have f^{[p]} ∈ (k+I)∩𝔪^{2} ⊂ I. Now I = ky_{1}+…+ky_{n}+I∩𝔪^{2}. If y_{i}^{[p]} ∈ I for all i, we get f^{[p]} ∈ I for every f ∈ I applying the formula ([p]_{3}).

Let J ⊂ {1, …, 2n} be a subset constructed as in the proof of lemma 1.2 with respect to y_{1}, …, y_{n}. Then the associative subalgebra B ⊂ B_{2n} generated by elements {x_{i} | i ∈ J} is Lagrangian and satisfies (3). □

Given two vector spaces V, W over k, denote by Pol(V, W) the vector space of all polynomial maps V → W. If φ ∈ Pol(V, W) and u, v ∈ V then there is an expansion as a power series in t ∈ k,

φ(v + tu) = φ(v) + tφ′(v, u) + t^{m}-terms with m ≥ 2.

Here φ′(v, u) is polynomial in v and linear in u. The differential of φ at v is the linear map (dφ)_{v} : V → W given by u ↦ φ′(v, u). If V, W are modules over a Lie algebra 𝔤 or rational modules over an algebraic group then Pol(V, W) inherits the same structure, which comes in fact from a linear isomorphism Pol(V, W) ≅ S(V^{∗}) ⊗ W where S(V^{∗}) is the symmetric algebra of the dual space V^{∗}. If D ∈ 𝔤 then (Dφ)(v) = D(φ(v)) − φ′(v, Dv). In the special case where W = k we denote by k[V] = Pol(V, k) the algebra of polynomial functions on V.

Suppose that 𝔤 is a p-Lie algebra operating on V via a p-representation. Put 𝔤_{v} = {D ∈ 𝔤 | Dv = 0} for v ∈ V, and

c_{𝔤}(V) = max_{v∈V} codim_{𝔤} 𝔤_{v}.

The maximal ideal 𝔪_{v} = {φ ∈ k[V] | φ(v) = 0} of the function algebra is generated by functions ξ − ξ(v)1 with ξ ∈ V^{∗}. Hence 𝔪_{v} is stable under D ∈ 𝔤 if and only if (Dξ)(v) = −ξ(Dv) = 0 for all ξ, if and only if D ∈ 𝔤_{v}. Denote by k[V]^{(p)} ⊂ k[V] the subalgebra consisting of pth powers φ^{p} of functions φ ∈ k[V]. We can now state a special case of [18], theorem 5.4:

Theorem 1.4. Suppose that φ_{1}, …, φ_{m} ∈ k[V]^{𝔤} are 𝔤-invariant polynomial functions where m = dim V − c_{𝔤}(V). Suppose also that their differentials are linearly independent at all points of an open subset U ⊂ V whose complement has codimension at least 2 in V. Then k[V]^{𝔤} = k[V]^{(p)}[φ_{1}, …, φ_{m}]. Moreover, k[V]^{𝔤} is free of rank p^{m} over k[V]^{(p)} and is a locally complete intersection ring.

The arguments used by Kostant to prove the normality of nilpotent cones in complex semisimple Lie algebras work equally well in our case. For convenience we state below a lemma which incorporates several standard facts from algebraic geometry.

Lemma 1.5. Let N ⊂ V be the zero set of homogeneous polynomial functions φ_{1}, …, φ_{m} ∈ k[V]. Suppose that U ⊂ V is an open subset and E ⊂ V a vector subspace such that E∩N ⊂ {0}∪U and (dφ_{1})_{u}, …, (dφ_{m})_{u} are linearly independent at all points u ∈ U.

(1) If dim E ≥ m + 1 then N is a complete intersection of codimension m in V and the ideal I_{N} = {φ ∈ k[V] | φ(N) = 0} is generated by φ_{1}, …, φ_{m}.

(2) If dim E ≥ m + 2 then N is normal and irreducible.

Proof. Denote by X the closed subset of those u ∈ N for which (dφ_{1})_{u}, …, (dφ_{m})_{u} are linearly dependent. Since φ_{1}, …, φ_{m} are homogeneous, X is conical, and by the hypotheses E ∩ X = {0}. Since every irreducible component of X meets E at 0, the theorem on dimensions of intersections in an affine space ensures that codim_{V} X ≥ dim E ≥ m + 1. Now N is a fiber of the morphism π : V → 𝔸^{m} given by functions φ_{1}, …, φ_{m}. By the theorem on dimension of fibers codim_{V} Z ≤ m, and so Z ⊄ X, for each irreducible component Z of N. If u ∈ Z ∖ X, then (dπ)_{u} is surjective, whence π is smooth at u. This shows, in particular, that codim_{V} Z = m. Furthermore, each u ∈ N ∖ X is a nonsingular point of N, and the kernel of the canonical homomorphism of local rings O_{V,u} → O_{N,u} is generated by φ_{1}, …, φ_{m}. Let I be the ideal of k[V] generated by φ_{1}, …, φ_{m}. The ring A = k[V]/I is Cohen-Macaulay by [13], (16.A), (16.B). Each minimal prime ideal 𝔭 ⊂ A corresponds to some Z above, and if 𝔪 ⊂ A is the maximal ideal corresponding to u ∈ Z ∖ X, then the local ring A_{𝔪} ≅ O_{N,u} is regular and, in particular, integral. It follows that A_{𝔭} is integral and artinian, i.e., a field. If now a ∈ A is nilpotent, then its annihilator ideal ann(a) is not contained in any such 𝔭. As the set of zero divisors in a Cohen-Macaulay ring coincides with the union of minimal primes [13], (16.C), ann(a) contains a regular element, which means a = 0. Thus A is reduced, and (1) is proved. Under assumptions of (2), codim_{Z} Z ∩ X ≥ 2 for each irreducible component Z of N. Then N is smooth in codimension 1, whence N is normal by Serre’s criterion [13], (17.I). As N is a disjoint union of its irreducible components, each of which has to contain 0, the variety N is irreducible. □

2. The open set of regular elements

Given f ∈ B_{2n}, we define f^{[1]} = f and, inductively, f^{[p^{i}]} = (f^{[p^{i−1}]})^{[p]} for i > 0. For each integer a ≥ 0 denote by a_{0}, a_{1}, a_{2}, … the coefficients in the p-adic expansion a = ∑_{i≥0} a_{i}p^{i} with 0 ≤ a_{i} < p, and put

f^{[a]} = ∏_{i≥0} (f^{[p^{i}]})^{a_{i}} ∈ B_{2n}.

Denote also

U = {f ∈ B_{2n} | f, f^{[p]}, …, f^{[p^{n−1}]} are linearly independent modulo k + 𝔪^{2}}

where 𝔪 is the maximal ideal of B_{2n}. The elements in U are regular in as much as our further results are concerned. Because the term “regular” has so many different meanings, the reader is warned that its usage here is limited to the specific context of the present paper.

Lemma 2.1. Let T_{𝔽_{p}} ⊂ B_{2n} be the 𝔽_{p}-linear span of elements t_{1}, …, t_{n} where t_{i} = (1 + x_{i})x_{n+i} and 𝔽_{p} ⊂ k the prime subfield. Then every 𝔽_{p}-linear invertible transformation τ of T_{𝔽_{p}} extends to an automorphism θ ∈ G such that θ(V) = V where V = k + kx_{1} + … + kx_{n} + 𝔪^{2}.

Proof. Put y_{i} = 1 + x_{i} for i = 1, …, n. The elements t_{1}, …, t_{n}, y_{1}, …, y_{n} form a minimal system of generators for B_{2n} and satisfy the relations t_{i}^{p} = 0, y_{i}^{p} = 1, [t_{i}, t_{j}] = [y_{i}, y_{j}] = 0, [y_{j}, t_{i}] = δ_{ij}y_{j}.

Denote X = Hom_{𝔽_{p}}(T_{𝔽_{p}}, 𝔽_{p}), and let B ⊂ B_{2n} be the Lagrangian subalgebra generated by x_{1}, …, x_{n}. For each α ∈ X the algebra B contains a unique element y_{α} such that y_{α}^{p} = 1 and [y_{α}, t] = α(t)y_{α} for all t ∈ T_{𝔽_{p}}. In fact y_{α} = y_{1}^{r_{1}} ⋯ y_{n}^{r_{n}} where r_{i} ∈ ℤ are such that α(t_{i}) = r_{i}1. We have y_{α}y_{β} = y_{α+β} for α, β ∈ X so that B is isomorphic with the group algebra of the additive group X.

Put t′_{i} = τ(t_{i}). Then t′_{1}, …, t′_{n} are a basis for T_{𝔽_{p}} over 𝔽_{p}. Denote by ε′_{1}, …, ε′_{n} the elements of the dual basis of X, and put y′_{i} = y_{ε′_{i}}. Then y′_{1}, …, y′_{n} generate the algebra B, whence t′_{1}, …, t′_{n}, y′_{1}, …, y′_{n} is a minimal system of generators for B_{2n}. We have the relations t′_{i}^{p} = 0, y′_{i}^{p} = 1, [t′_{i}, t′_{j}] = [y′_{i}, y′_{j}] = 0, [y′_{j}, t′_{i}] = δ_{ij}y′_{j}. Now the assignments t_{i} ↦ t′_{i}, y_{i} ↦ y′_{i} define an automorphism θ ∈ Aut B_{2n}. Furthermore, θ([f, g]) = [θ(f), θ(g)] for all f, g ∈ B_{2n} since it suffices to verify this identity on a set of generators for B_{2n}. Thus θ ∈ G. By construction θ(B) = B. Since V = B + 𝔪^{2} and 𝔪^{2} is stable under automorphisms, we get θ(V) = V. □

Proposition 2.2. (1) The subset U is nonempty, Zariski open in B_{2n} and stable under G. Its complement B_{2n} ∖ U has codimension at least 2 in B_{2n}.

(2) For each f ∈ U its centralizer 𝔷(f) = {g ∈ B_{2n} | [f, g] = 0} is a Lagrangian subalgebra with basis elements f^{[a]}, 0 ≤ a < p^{n}. In particular, dim 𝔷(f) = p^{n}.

(3) The subset U_{s} = {f ∈ B_{2n} | f_{s} ∈ U}, where f_{s} denotes the [p]-semisimple component of f, is nonempty, Zariski open in B_{2n} and G-stable. Furthermore, U_{s} ⊂ U and 𝔷(f) = 𝔷(f_{s}) for every f ∈ U_{s}.

(4) f^{[p]} ∈ U_{s} if and only if f ∈ U_{s}.

Proof. (1) If θ ∈ G then θ(f^{[p]}) ≡ θ(f)^{[p]} (mod k), whence θ(f^{[p^{i}]}) ≡ θ(f)^{[p^{i}]} (mod k) for all f ∈ B_{2n} and i ≥ 0 by induction on i. Since k + 𝔪^{2} is stable under θ, it is immediate that so is U too.

Suppose that V is a vector subspace of codimension n in B_{2n} such that V ⊃ k + 𝔪^{2}. Let e_{1}, …, e_{n} be a basis for a complement of V in B_{2n}. For each i we can write f^{[p^{i}]} ≡ ∑_{j=1}^{n} λ_{ij}(f)e_{j} (mod V) where λ_{ij} are homogeneous polynomial functions of degree p^{i} on B_{2n}. Denote by Δ(f) the determinant of the square matrix [λ_{ij}(f)]_{j=1,...,n}^{i=0,...,n−1}. Then Δ is a homogeneous polynomial function of degree (p^{n} − 1)/(p − 1) on B_{2n}. Note that Δ(f) ≠ 0 if and only if f, f^{[p]}, …, f^{[p^{n−1}]} are linearly independent modulo V. It is clear that f ∈ U if and only if f lies in the open subset {f ∈ B_{2n} | Δ(f) ≠ 0} defined with respect to a suitable subspace V as above. Hence U is open.

We take now V as in lemma 2.1 and put Z = {f ∈ B_{2n} | Δ(f) = 0}, so that B_{2n} ∖ U ⊂ Z. Let t_{i}, y_{i}, T_{𝔽_{p}}, X have the same meaning as in lemma 2.1. Extend each α ∈ X to a linear function on T = kT_{𝔽_{p}}. Note that the monomial y_{1}^{r_{1}} ⋯ y_{n}^{r_{n}}t_{1}^{s_{1}} ⋯ t_{n}^{s_{n}} is an eigenvector for D_{t_{i}} with eigenvalue −r_{i}1. Thus D_{t_{i}} is a diagonalizable linear transformation of B_{2n} with all eigenvalues in 𝔽_{p}, that is, D_{t_{i}}^{p} = D_{t_{i}}. We see that D_{t_{1}}, …, D_{t_{n}} span an n-dimensional torus. If t ∈ T then t ∈ Z if and only if D_{t} generates a torus of dimension less than n, which means that t ∈ T_{α} = ker α for some 0 ≠ α ∈ X. There are p^{n} − 1 nonzero elements in X, hence (p^{n} − 1)/(p − 1) distinct hyperplanes T_{α} ⊂ T. It follows that the restriction of Δ to T coincides up to a scalar multiple with the product of (p^{n} − 1)/(p − 1) nonproportional elements of X.

Put G_{V} = {θ ∈ G | θ(V) = V }. If θ ∈ G_{V} then clearly Δ(θf) = det(θ_{V})Δ(f) where θ_{V} is the invertible linear transformation of the vector space B_{2n}/V induced by θ. Hence Z is stable under G_{V}, and G_{V} permutes the irreducible components of Z. Let Z_{1}, …, Z_{m} be all irreducible components of Z which are G_{V} -conjugate to one chosen irreducible component. Then Z_{1} ∪ … ∪ Z_{m} is the zero locus of a polynomial function Δ′ on B_{2n} which divides Δ. Now Δ′|_{T} is up to a scalar multiple a product of some α ∈ X. Suppose that 0 ≠ α ∈ X occurs as a factor of Δ′|_{T}, and let 0 ≠ β ∈ X be any element. Both T_{α} and T_{β} are spanned by elements from T_{𝔽_{p}}. Hence there exists an invertible linear transformation τ : T → T such that τ(T_{𝔽_{p}}) = T_{𝔽_{p}} and τ(T_{α}) = T_{β}. By lemma 2.1 τ extends to an automorphism θ ∈ G_{V}. Since T_{α} is contained in a G_{V} -stable set Z_{1} ∪ … ∪ Z_{m}, the same is valid for T_{β} as well. In other words, β divides Δ′|_{T}. It follows that deg Δ′ = deg Δ, and Δ′ = Δ up to a scalar multiple. Thus all irreducible components of Z are conjugate to each other with respect to G_{V}. If one of the components were contained in B_{2n} ∖U, we would get B_{2n} ∖U = Z. This is, however, impossible. Indeed, consider the linear span T′ of t′_{1}, …, t′_{n} where t′_{i} = x_{i}(1 + x_{n+i}). As T′ ∩ (k + 𝔪^{2}) = 0 and D_{t′_{1}}, …, D_{t′_{n}} span an n-dimensional torus, we have T′ ∩ U ≠ ∅. On the other hand T′ ⊂ V ⊂ Z by construction. Thus no irreducible component of B_{2n} ∖ U has codimension 1 in B_{2n}.

(2) Assume f ∈ U. Denote by B ⊂ B_{2n} the associative subalgebra generated by f, f^{[p]}, …, f^{[p^{n−1}]}. Then B is Lagrangian by the definition of U, and so the elements f^{[a]} with 0 ≤ a < p^{n} form a basis for B. Note that 𝔷(f) = 𝔷(B) where 𝔷(B) = B by lemma 1.2.

(3) and (4) If θ ∈ G then θ ∘ D_{f} ∘ θ^{−1} = D_{θf} for all f ∈ B_{2n}. If f = f_{s} + f_{n} is the Jordan-Chevalley decomposition then we deduce that D_{θf} = D_{θf_{s}} +D_{θf_{n}} where D_{θf_{s}} is semisimple, D_{θf_{n}} nilpotent, and [θf_{s}, θf_{n}] = 0. It follows that D_{(θf)_{s}} = D_{θf_{s}}, i.e., (θf)_{s} ≡ θf_{s} (mod k). Since U is G-stable, it is immediate now that so is U_{s} too. Let T be as in the proof of (1). As we have seen, T ∩ U ≠ ∅. Take f ∈ T ∩ U. Since D_{f} is diagonalizable, we have f ≡ f_{s} modulo k. It follows f_{s} ∈ U, and so U_{s} ≠ ∅.

Given f ∈ B_{2n}, let m ≥ 0 be the largest integer such that f, f^{[p]}, …, f^{[p^{m−1}]} are linearly independent modulo k + 𝔪^{2}. Since k + 𝔪^{2} is a p-Lie subalgebra in B_{2n}, we deduce that f^{[p^{i}]} is a linear combination of f, f^{[p]}, …, f^{[p^{m−1}]} modulo k + 𝔪^{2} for all i ≥ m. Hence f ∈ U if and only if dim 𝔤(f)/𝔤(f) ∩ (k + 𝔪^{2}) ≥ n where 𝔤(f) ⊂ B_{2n} denotes the linear span of all elements f^{[p^{i}]} with i ≥ 0. Since f_{s} ∈ 𝔤(f), we have 𝔤(f_{s}) ⊂ 𝔤(f), and so f_{s} ∈ U implies f ∈ U. Thus U_{s} ⊂ U.

Since 𝔤(f_{s}) = 𝔤(f_{s}^{[p]}), our previous observations show that f_{s} ∈ U if and only if f_{s}^{[p]} ∈ U. This gives (4) since f_{s}^{[p]} is the semisimple component of f^{[p]}. There exists r > 0 such that f^{[p^{r}]} is semisimple for every f ∈ B_{2n}. Then f ∈ U_{s} if and only if f^{[p^{r}]} ∈ U. In other words, U_{s} = q^{−1}(U) where q : B_{2n} → B_{2n}, f ↦ f^{[p^{r}]}, is a polynomial map. Since U is open, so is U_{s} too.

Obviously, 𝔷(f) ⊂ 𝔷(f_{s}). If f ∈ U_{s} then dim 𝔷(f) = dim 𝔷(f_{s}) in view of (2), whence 𝔷(f) = 𝔷(f_{s}). □

Remark. Krylyuk [7] computed the dimensions of generic centralizers and stabilizers of generic linear functions in the Hamiltonian algebras related to the Poisson algebra B_{2n} using essentially the open subset U_{s}.

3. Construction of invariants

Lemma 3.1. The assignment f ↦ f^{[a]} defines a homogeneous of degree a polynomial map B_{2n} → B_{2n} which is G_{[p]}-invariant and L-invariant. If t is a scalar indeterminate and f, g ∈ B_{2n} then

(f + tg)^{[a]} = f^{[a]} + t ∑_{{i≥0|p^{i}≤a}} a_{i}f^{[a−p^{i}]}D_{f}^{p^{i}−1}(g) + terms divisible by t^{2}.

Proof. The homogeneity and invariance are immediately derived from the special case of a = p. The formula ([p]_{3}) from section 1 gives (f + tg)^{[p]} ≡ f^{[p]} + tD_{f}^{p−1}(g),

and by iteration (f + tg)^{[p^{i}]} ≡ f^{[p^{i}]} + tD_{f}^{p^{i}−1}(g) (mod t^{2}) for all i ≥ 0. The formula for (f + tg)^{[a]} is also immediate now. □

Proposition 3.2. (1) There are uniquely determined polynomial functions φ_{1}, …, φ_{p^{n}} on B_{2n} such that the following relation holds true for all f ∈ B_{2n}:

f^{[p^{n}]} + ∑_{a=1}^{p^{n}} φ_{a}(f)f^{[p^{n}−a]} = 0.  (∗)

(2) The functions φ_{a} are G_{[p]}-invariant and L-invariant. Furthermore, φ_{a} is homogeneous of degree a.

(3) aφ_{a}(f) = φ_{1}(f^{[a]}) for 0 ≤ a < p^{n} with the convention that φ_{0}(f) = 1.

(4) φ′_{a}(f, g) = φ_{1}(f^{[a−1]}g) for 0 < a ≤ p^{n}.

(5) φ_{1} is a nonzero linear function such that ker φ_{1} = [B_{2n}, B_{2n}].

(6) The symmetric bilinear form β(f, g) = φ_{1}(fg) defined on B_{2n} is nondegenerate. If f ∈ U_{s} then the restriction of β to 𝔷(f) is nondegenerate.

Proof. Clearly, f^{[p^{n}]} ∈ 𝔷(f). If f ∈ U then it follows from the description of 𝔷(f) in proposition 2.2 that f^{[p^{n}]} is a uniquely determined linear combination of the elements f^{[a]} with 0 ≤ a < p^{n}. Hence we can define the functions φ_{a} on U so that (∗) is fulfilled for all f ∈ U.

We claim that the functions φ_{a} are regular on U. Suppose that V ⊂ B_{2n} is a subspace of codimension p^{n}. Put U_{V} = {f ∈ U | B_{2n} = 𝔷(f) ⊕ V }, and choose elements e_{b}, 1 ≤ b ≤ p^{n}, such that B_{2n} = ke_{1} ⊕ … ⊕ ke_{p^{n}} ⊕ V. We can write f^{[a]} ≡ ∑_{b=1}^{p^{n}} λ_{ab}(f)e_{b} (mod V) where λ_{ab} are polynomial functions on B_{2n}. Then U_{V} consists precisely of those f ∈ U for which the square matrix [λ_{ab}(f)]_{b=1,...,p^{n}}^{a=0,...,p^{n}−1} is invertible. In particular, U_{V} is an open subset of U. For f ∈ U_{V} we have an expression e_{b} ≡ ∑_{a=0}^{p^{n}−1} μ_{ba}(f)f^{[a]} (mod V) where μ_{ba} are regular functions on U_{V}. Now

f^{[p^{n}]} ≡ ∑_{b=1}^{p^{n}} ∑_{a=0}^{p^{n}−1} λ_{p^{n},b}(f)μ_{ba}(f)f^{[a]} (mod V).

Since 𝔷(f) ∩ V = 0, we get f^{[p^{n}]} = ∑_{b=1}^{p^{n}} ∑_{a=0}^{p^{n}−1} λ_{p^{n},b}(f)μ_{ba}(f)f^{[a]}. It follows that φ_{a}(f) = − ∑_{b=1}^{p^{n}} λ_{p^{n},b}(f)μ_{ba}(f), which is a regular function on U_{V}. Clearly U is the union of its open subsets U_{V} defined for different V. Hence φ_{a} is regular on U.

Now φ_{a} are rational functions on B_{2n}. Since B_{2n} ∖ U has no irreducible components of codimension 1, the φ_{a}’s extend to polynomial functions on the whole B_{2n}. Since (∗) is fulfilled on an open subset of B_{2n}, it is fulfilled everywhere. That completes the proof of (1).

We may always assume that f ∈ U when verifying that two polynomial functions in f coincide. Recall that 1, f, …, f^{[p^{n}−1]} are linearly independent under this assumption. Multiplying (∗) by λ^{p^{n}}, where λ ∈ k, and noting that (λf)^{[c]} = λ^{c}f^{[c]} for all c ≥ 0, we deduce φ_{a}(λf) = λ^{a}φ_{a}(f). Similarly, applying θ ∈ G_{[p]} to (∗), we get φ_{a}(θf) = φ_{a}(f) since θ(f^{[c]}) = (θf)^{[c]}.

We put formally φ_{0}(f) = 1. Then (∗) is rewritten as ∑_{b+c=p^{n}} φ_{b}(f)f^{[c]} = 0. Let f, g, h ∈ B_{2n}. Now substitute f + tg for f in (∗) where t is a scalar indeterminate and compare the coefficients of t. Using lemma 3.1, we get

∑_{b+c=p^{n}} (φ′_{b}(f, g)f^{[c]} + φ_{b}(f) ∑_{{i≥0|p^{i}≤c}} c_{i}f^{[c−p^{i}]}D_{f}^{p^{i}−1}(g)) = 0.  (∗∗)

Substitute here D_{f}(g) for g. We have ∑_{{i≥0|p^{i}≤c}} c_{i}f^{[c−p^{i}]}D_{f}^{p^{i}}(g) = [f^{[c]}, g], and ∑_{b+c=p^{n}} φ_{b}(f)[f^{[c]}, g] = 0 in view of (∗). Hence ∑_{b+c=p^{n}} φ′_{b}(f, [f, g])f^{[c]} = 0. This is a linear combination of 1, f, …, f^{[p^{n}−1]} since φ′_{0} = 0. It follows φ′_{a}(f, [f, g]) = 0, i.e., D_{g}φ_{a} = 0 for all a. That completes the proof of (2).

Suppose now g ∈ 𝔷(f). Then D_{f}^{p^{i}−1}(g) = 0 for all i > 0, and (∗∗) reduces to the equality

\begin{matrix}∑_{b+c=p^{n}} φ′_{b}(f, g)f^{[c]} + ∑_{b+c=p^{n}, c>0} c_{0}φ_{b}(f)f^{[c−1]}g \\ = ∑_{b+c=p^{n}} φ′_{b}(f, g)f^{[c]} − ∑_{b+c=p^{n}−1} bφ_{b}(f)f^{[c]}g = 0 \end{matrix}

since c_{0} ≡ c ≡ −b (mod p) in the left hand sum. Take g = f^{[a]} where 0 ≤ a < p^{n}. Note that f^{[c]}f^{[a]} = λf^{[d]} where λ ∈ k and d is an integer, 0 ≤ d < p^{n}, such that d_{i} = c_{i}+a_{i} or d_{i} = c_{i}+a_{i}−p for each i = 0, …, n−1 depending on whether c_{i}+a_{i} < p or not. We have d_{i} < p−1 in all cases except when c_{i} +a_{i} = p−1. Hence d < p^{n} −1 in all cases except when c + a = p^{n} − 1. Note also that φ′_{1}(f, g) = φ_{1}(g) since φ_{1} is a linear function. Thus the coefficient of f^{[p^{n}−1]} in the displayed equation above equals φ_{1}(f^{[a]}) − aφ_{a}(f). If f ∈ U, then 1, f, …, f^{[p^{n}−1]} are linearly independent, whence φ_{1}(f^{[a]}) = aφ_{a}(f). Since both sides of the last equality are polynomial functions on B_{2n}, it is fulfilled everywhere. We get (3).

We now prove (5). The L-invariance of the linear function φ_{1} means that φ_{1} vanishes on [B_{2n}, B_{2n}]. Suppose that φ_{1} = 0. Then φ_{a} = 0 for all a ≢ 0 (mod p) according to (3). By (∗) f^{[p^{n}]} is a linear combination of elements f^{[c]} with c < p^{n}, c ≡ 0 (mod p). Any such f^{[c]} is, modulo k + 𝔪^{2}, a linear combination of elements f^{[p^{i}]} with 0 < i < n. Thus f^{[p]}, …, f^{[p^{n−1}]}, f^{[p^{n}]} are linearly dependent modulo k + 𝔪^{2}, and so f^{[p]} ∉ U for every f ∈ B_{2n}. But this contradicts proposition 2.2(4). Thus φ_{1} ≠ 0. As is easily checked straightforwardly, [B_{2n}, B_{2n}] is a subspace of codimension 1 in B_{2n} spanned by the monomials x_{1}^{r_{1}} ⋯ x_{2n}^{r_{2n}} with r_{i} < p − 1 for at least one i. It follows ker φ_{1} = [B_{2n}, B_{2n}].

The kernel of β coincides with the largest associative ideal of B_{2n} contained in ker φ_{1}. Any nonzero associative ideal of B_{2n} contains the element y = x_{1}^{p−1} ⋯ x_{2n}^{p−1}. Since y ∉ [B_{2n}, B_{2n}], the bilinear form is nondegenerate. Suppose that f ∈ U_{s} so that 𝔷(f) = 𝔷(f_{s}) by proposition 2.2. We have then B_{2n} = 𝔷(f) ⊕ [f_{s}, B_{2n}] and φ_{1}(u[f_{s}, v]) = φ_{1}([f_{s}, uv]) = 0 for all u ∈ 𝔷(f), v ∈ B_{2n}. As 𝔷(f) is orthogonal to [f_{s}, B_{2n}] with respect to β, the second assertion in (6) is clear.

Now multiply (∗∗) by h ∈ 𝔷(f) and apply φ_{1} to both sides of the equality obtained. Since hf^{[c−p^{i}]}D_{f}^{p^{i}−1}(g) = D_{f}^{p^{i}−1}(ghf^{[c−p^{i}]}) ⊂ [B_{2n}, B_{2n}] = ker φ_{1} for all i > 0, we get

\begin{aligned}φ_{1}( ∑_{b+c=p^{n}} φ′_{b}(f, g)f^{[c]}h) & = −φ_{1}( ∑_{b+c=p^{n}, c>0} c_{0}φ_{b}(f)f^{[c−1]}gh) \\ & = ∑_{b+c=p^{n}−1} φ_{1}(f^{[b]})φ_{1}(f^{[c]}gh) \end{aligned}

where we first used the equality −c_{0}φ_{b}(f) = bφ_{b}(f) = φ_{1}(f^{[b]}) for c > 0 and then substituted c + 1 for c. We claim that

∑_{b+c=p^{n}−1} φ_{1}(f^{[b]})φ_{1}(f^{[c]}gh) = ∑_{b+c=p^{n}−1} φ_{1}(f^{[b]}h)φ_{1}(f^{[c]}g)

when h = f^{[a]} with 0 ≤ a < p^{n}. In fact both left and right sides can be rewritten as

∑_{\begin{matrix}0≤d<p^{n}, 0≤e<p^{n}, \\ d_{i}+e_{i}≡a_{i}−1 (mod p) for i=0,...,n−1 \end{matrix}} λ_{de}φ_{1}(f^{[d]})φ_{1}(f^{[e]}g)

with λ_{de} = ∏_{i∈I_{de}}(f^{[p^{i}]})^{p} ∈ k where I_{de} ⊂ {0, 1, …, n − 1} is the subset consisting of those i for which e_{i} < a_{i} in one case and d_{i} < a_{i} in the other. Note that for any pair of indices d, e occurring in the sum we have either d_{i} + e_{i} = a_{i} − 1 or d_{i} + e_{i} = a_{i} + p − 1. One sees that each of the two conditions e_{i} < a_{i} and d_{i} < a_{i} is equivalent to d_{i} + e_{i} = a_{i} − 1. Thus the subsets I_{de} are the same in both cases. We now arrive at the equality

\begin{aligned}φ_{1}( ∑_{b+c=p^{n}} φ′_{b}(f, g)f^{[c]}h) & = ∑_{b+c=p^{n}−1} φ_{1}(f^{[b]}h)φ_{1}(f^{[c]}g) \\ & = φ_{1}( ∑_{b+c=p^{n}−1} φ_{1}(f^{[c]}g)f^{[b]}h) \end{aligned}

when h = f^{[a]}. Suppose that f ∈ U_{s}. Then 𝔷(f) is spanned by the elements f^{[a]} with 0 ≤ a < p^{n}, so that the equality above holds for all h ∈ 𝔷(f). Applying (6), we get

∑_{b+c=p^{n}} φ′_{b}(f, g)f^{[c]} = ∑_{b+c=p^{n}−1} φ_{1}(f^{[c]}g)f^{[b]}

Extracting the coefficients of f^{[p^{n}−a]}, we deduce (4). □

Remark. Assertions (3) and (5) show that f^{[a]} ∈ [B_{2n}, B_{2n}] whenever 0 ≤ a < p^{n} and p|a.

Denote by κ : B_{2n} → k the homomorphism of associative algebras with kernel 𝔪. Put κ_{i}(f) = κ(f^{[p^{i}]}). Thus κ_{i} is a homogeneous polynomial function of degree p^{i} on B_{2n}. It is G_{[p]}-invariant and L_{0}-invariant. We mention also that κ = κ_{0} is a G-invariant function.

Lemma 3.3. The differentials of p^{n} + n functions φ_{1}, …, φ_{p^{n}}, κ_{0}, …, κ_{n−1} are linearly independent at all f ∈ U.

Proof. For g ∈ B_{2n} we have (dφ_{a})_{f}(g) = φ′_{a}(f, g) = β(f^{[a−1]}, g) by proposition 3.2, and (dκ_{i})_{f}(g) = κ′_{i}(f, g) = κ(D_{f}^{p^{i}−1}(g)) in view of lemma 3.1. Since 1, …, f^{[p^{n}−1]} are linearly independent and β is nondegenerate, (dφ_{1})_{f}, …, (dφ_{p^{n}})_{f} are linearly independent. It remains to show that the linear functions (dκ_{0})_{f}, …, (dκ_{n−1})_{f} have linearly independent restrictions to the subspace V = ⋂_{a=1}^{p^{n}} ker(dφ_{a})_{f}. Note that V is the orthogonal complement of 𝔷(f) with respect to β. Since β is L-invariant, 𝔷(f) coincides with the orthogonal complement of the subspace [f, B_{2n}]. It follows V = [f, B_{2n}]. Now (dκ_{i})_{f}([f, g]) = κ([f^{[p^{i}]}, g]). The elements y_{i} = f^{[p^{i}]} − κ(f^{[p^{i}]}) with 0 ≤ i < n lie in 𝔪 and generate the Lagrangian subalgebra 𝔷(f). As we have seen in the proof of lemma 1.2 there exist elements z_{0}, …, z_{n−1} ∈ 𝔪 such that [f^{[p^{i}]}, z_{j}] = [y_{i}, z_{j}] ≡ δ_{ij} (mod 𝔪) for all 0 ≤ i, j < n. Then [f, z_{j}] ∈ V and (dκ_{i})_{f}([f, z_{j}]) = δ_{ij}. Thus we are done. □

Theorem 3.4. The algebra of invariants k[B_{2n}]^{L} (respectively k[B_{2n}]^{L_{0}}) is generated over k[B_{2n}]^{(p)} by φ_{1}, …, φ_{p^{n}} (respectively by φ_{1}, …, φ_{p^{n}}, κ_{0}, …, κ_{n−1}). Both algebras are free over k[B_{2n}]^{(p)} and are locally complete intersection rings.

Proof. Apply theorem 1.4 for V = B_{2n} and 𝔤 = B_{2n} or 𝔤 = 𝔪^{2} operating on V via the homomorphism 𝔤 → L. Since dim 𝔷(f) = p^{n} for all f in an open subset of B_{2n}, we have c_{𝔤}(V) = p^{2n} − p^{n} in case of 𝔤 = B_{2n}. If f ∈ U then 𝔷(f) is a Lagrangian subalgebra, whence dim 𝔷(f) ∩ 𝔪^{2} = p^{n} − n − 1. Now dim 𝔪^{2} = p^{2n} − 2n − 1, and it follows that c_{𝔤}(V) = codim_{𝔪^{2}} 𝔷(f) ∩ 𝔪^{2} = p^{2n} − p^{n} − n in case of 𝔤 = 𝔪^{2}. Thus we have the correct number of invariant functions m = p^{n} (respectively m = p^{n} + n). Lemma 3.3 and proposition 2.2 show that the hypotheses of theorem 1.4 are fulfilled. □

In order to get the G-invariant functions we need to modify our construction slightly. For a ≥ 0 put

f^{⟨a⟩} = ∏_{i≥0} (f^{[p^{i}]} − κ_{i}(f))^{a_{i}}.

Thus f^{⟨p^{i}⟩} = f^{[p^{i}]} − κ_{i}(f) is the unique element g ∈ 𝔪 such that D_{g} = D_{f}^{p^{i}}. It is clear that the polynomial map B_{2n} → B_{2n} given by assignment f ↦ f^{⟨a⟩} is G-invariant and homogeneous of degree a.

Proposition 3.5. (1) There are uniquely determined polynomial functions φ̃_{1}, …, φ_{p^{n}−1} on B_{2n} such that the following relation holds true for all f ∈ B_{2n}:

f^{⟨p^{n}⟩} + ∑_{a=1}^{p^{n}−1} φ̃_{a}(f)f^{⟨p^{n}−a⟩} = 0.

(2) The functions φ̃_{a} are G-invariant and homogeneous with deg φ̃_{a} = a.

(3) φ̃_{a}(f) = ∑_{b=1}^{a}(−1)^{a−b}\binom{a − 1}{b − 1}κ(f^{[a−b]})φ_{b}(f) for all f ∈ B_{2n} and 0 < a < p^{n}.

(4) φ_{a}(f) = ∑_{b=1}^{a} \binom{a − 1}{b − 1}κ(f^{[a−b]})φ̃_{b}(f) for all f ∈ B_{2n} and 0 < a < p^{n}.

(5) φ_{p^{n}}(f) = −κ_{n}(f) − ∑_{b=1}^{p^{n}−1}(−1)^{b}κ(f^{[p^{n}−b]})φ̃_{b}(f) for all f ∈ B_{2n}.

(6) The differentials of p^{n} + n functions φ̃_{1}, …, φ_{p^{n}−1}, κ_{0}, …, κ_{n} are linearly independent at all f ∈ U.

Proof. If f ∈ U then the elements f^{⟨p^{i}⟩} with 0 ≤ i < n are linearly independent modulo 𝔪^{2}, whence the elements f^{⟨a⟩} with 0 ≤ a < p^{n} form a basis for 𝔷(f). In particular, the equality in (1) characterizes the φ̃_{a}’s uniquely. Now note that

f^{[a]} = ∑_{\begin{matrix}b_{0}+c_{0}=a_{0}, \\ b_{1}+c_{1}=a_{1},... \end{matrix}} ∏_{i≥0} \binom{a_{i}}{c_{i}}κ(f^{[p^{i}]})^{b_{i}}(f^{⟨p^{i}⟩})^{c_{i}} = ∑_{b+c=a} \binom{a}{c}κ(f^{[b]})f^{⟨c⟩}.

as \binom{a}{c} ≡ ∏ \binom{a_{i}}{c_{i}} (mod p). By proposition 3.2

0 = ∑_{b+c=p^{n}} φ_{b}(f)f^{[c]} = ∑_{b+d+e=p^{n}} \binom{d + e}{e}φ_{b}(f)κ(f^{[d]})f^{⟨e⟩} = ∑_{a+e=p^{n}} φ̃_{a}(f)f^{⟨e⟩}

where we put φ̃_{a}(f) = ∑_{b+d=a} \binom{p^{n}−b}{p^{n}−a}φ_{b}(f)κ(f^{[d]}) for a = 0, …, p^{n}. Here φ̃_{0} = 1. If 0 < a < p^{n}, then \binom{p^{n}}{p^{n}−a} ≡ 0 and \binom{p^{n}−b}{p^{n}−a} ≡ (−1)^{a−b}\binom{a−1}{b−1} (mod p) for b > 0, so that the φ̃_{a}’s are given by the formula in (3). Note that f^{⟨0⟩} = 1 and f^{⟨e⟩} ∈ 𝔪 whenever e > 0. It follows from the displayed equality above that φ_{p^{n}} = 0, and we come to (1). Assertion (2) is a consequence of properties of maps f ↦ f^{⟨a⟩}.

Using the relation f^{⟨a⟩} = ∑_{b+c=a}(−1)^{b}\binom{a}{c}κ(f^{[b]})f^{[c]}, we deduce similarly

0 = ∑_{b+c=p^{n}} φ̃_{b}(f)f^{⟨c⟩} = ∑_{b+d+e=p^{n}} (−1)^{d}\binom{d + e}{e}φ̃_{b}(f)κ(f^{[d]})f^{[e]},

which proves (4) and (5). For each a = 1, …, p^{n} the function φ_{a} is expressed as a polynomial in φ̃_{1}, …, φ_{p^{n}−1}, κ_{0}, …, κ_{n}. Hence (dφ_{a})_{f} is a linear combination of (dφ̃_{1})_{f}, …, (dφ_{p^{n}−1})_{f}, (dκ_{0})_{f}, …, (dκ_{n})_{f}. Now (6) follows from lemma 3.3. □

Corollary 3.6. One has φ_{1} = φ̃_{1}. The bilinear form β on B_{2n} is L-invariant and G-invariant.

We will use the notations φ̃_{0} = 1 and φ_{p^{n}} = 0 for consistency reasons. If λ ∈ k then (f +λ)^{⟨a⟩} = f^{⟨a⟩} for all a, whence also φ̃_{a}(f +λ) = φ̃_{a}(f) for all a. This shows that φ̃_{a} induces a G-invariant polynomial function on L ≅ B_{2n}/k.

4. A universal construction and automorphisms

With each Lagrangian subalgebra B ⊂ B_{2n} we will associate a Poisson algebra F (B). The latter is attached to B in a canonical way in the sense that every isomorphism of Lagrangian subalgebras τ : B → B′ compatible with the [p]-map induces an isomorphism of Poisson algebras τ_{∗} : F (B) → F (B′). We will show that B_{2n} ≅ F (B), although this isomorphism depends on additional data. This will enable us to prove that any isomorphism τ above extends to an element of the group G of Poisson automorphisms of B_{2n}.

We first recall one construction (see [20], Theorem 2.2, for a more general treatment). Suppose that 𝔤 is a p-Lie algebra and A an arbitrary algebra over k. Put F (𝔤, A) = Hom(u(𝔤), A) where u(𝔤) is the restricted universal enveloping algebra of 𝔤. With each x ∈ 𝔤 we associate a linear transformation D_{x} of F (𝔤, A) defined by the rule (D_{x}f)(u) = f(ux) for f ∈ F (𝔤, A) and u ∈ u(𝔤). This makes F (𝔤, A) into a restricted 𝔤-module. Define a linear map π : F (𝔤, A) → A by the rule π(f) = f(1). Then for every u(𝔤)-module M and a linear map ξ : M → A there exists a unique 𝔤-module homomorphism η : M → F (𝔤, A) such that ξ = π ∘ η. Explicitly, η(m)(u) = ξ(u · m). In particular, if η satisfies π ∘ η = 0 then necessarily η = 0. This shows also that ker π contains no nonzero 𝔤-invariant subspaces.

Using the universality property above, one sees easily that F (𝔤, A) has a unique 𝔤-invariant multiplication such that π is an algebra homomorphism. The explicit formula involves the comultiplication in u(𝔤) but we will not need it. Now 𝔤 operates on F (𝔤, A) as a p-Lie algebra of derivations. Considering next the trivial 𝔤-module k, we see that for every element a ∈ A there exists a unique 𝔤-module homomorphism η : k → F (𝔤, A) such that π(η(1)) = a. In other words π maps the subalgebra of 𝔤-invariants F (𝔤, A)^{𝔤} isomorphically onto A. We will identify A with F (𝔤, A)^{𝔤} by means of this isomorphism. After this identification we have D_{x}(a) = 0 for all x ∈ 𝔤 and a ∈ A. In our application A will be commutative, associative and unital. Then F (𝔤, A) inherits the same properties.

Suppose that B ⊂ B_{2n} is a Lagrangian subalgebra. By (2) of lemma 1.2 B is a p-Lie subalgebra containing the Lie center k of B_{2n}. We apply the previous construction taking 𝔤 = B/k and A = B considered as an algebra with respect to the associative multiplication. We thus obtain a commutative, associative and unital algebra F (𝔤, B) and a homomorphism of algebras π : F (𝔤, B) → B. With each f ∈ B we have associated a derivation D_{f} of F (𝔤, B) which depends only on the coset of f modulo k, so that D_{1} = 0. Furthermore, B is identified with the subalgebra of elements g ∈ F (𝔤, B) such that D_{f}(g) = 0 for all f ∈ B. We have π(g) = g for all g ∈ B. Put

F (B) = {h ∈ F (𝔤, B) | D_{fg}(h) = fD_{g}(h) + gD_{f}(h) for all f, g ∈ B}.

Clearly F (B) is a subalgebra of F (𝔤, B) and B ⊂ F (B). Since the Lie multiplication in B is abelian, the B-submodule of Der F (𝔤, B) generated by all derivations D_{f} with f ∈ B is an abelian Lie algebra too. It is immediate thereof that F (B) is stable under all derivations D_{f}.

Proposition 4.1. There exists a unique Poisson bracket on F (B) such that the following identities hold for all f ∈ B and g, h ∈ F (B):

(1) D_{f}([g, h]) = [D_{f}(g), h] + [g, D_{f}(h)].

(2) [f, g] = D_{f}(g).

(3) π([g, h]) = 0 whenever π(g) = 0 and π(h) = 0.

Proof. Put J = {g ∈ F (B) | π(g) = 0}. Then F (B) = B ⊕ J and J is an ideal of F (B). Define a linear map ξ : F (B)^{⊗2} → B setting

ξ(g ⊗ h) = \begin{cases}0 & when g, h ∈ J, \\ π(D_{g}(h)) & when g ∈ B, \\ −π(D_{h}(g)) & when h ∈ B, \end{cases}

Note that D_{g}(h) = 0 = D_{h}(g) when g, h ∈ B, so that ξ is well defined. If the required Poisson bracket on F (B) exists then the formula η(g ⊗ h) = [g, h] defines a 𝔤-module homomorphism η : F (B)^{⊗2} → F (B) satisfying π ∘ η = ξ. However, by the universality property of F (𝔤, B) there exists a unique 𝔤-module homomorphism η : F (B)^{⊗2} → F (𝔤, B) such that π ∘ η = ξ. We put [g, h] = η(g ⊗ h) ∈ F (𝔤, B) for g, h ∈ F (B). We will have yet to check that η takes values in F (B).

Identities (1) and (3) in the statement of the proposition are immediate from the definition of η. The linear map α : B ⊗ F (B) → F (B) defined by the rule α(f ⊗ h) = D_{f}(h) is clearly a 𝔤-module homomorphism and π ∘ α = ξ. Then α coincides with the restriction of η by the universality property of F (𝔤, B). Hence (2) is also fulfilled.

We will construct several other 𝔤-module homomorphisms to obtain the necessary identities. Consider first β : F (B)^{⊗2} → F (𝔤, B) such that β(g ⊗ h) = [g, h] + [h, g]. Then π ∘ β = 0, whence β = 0. Note that [f, f] = D_{f}(f) = 0 for all f ∈ B. If g ∈ J then π([g, g]) = 0 by (3) and D_{f}([g, g]) = [D_{f}(g), g] + [g, D_{f}(g)] = β(D_{f}(g) ⊗ g) = 0 for all f ∈ B, whence [g, g] ∈ F (𝔤, B)^{𝔤} = B. It follows that [g, g] = 0 for all g ∈ F (B).

Consider next γ : F (B)^{⊗3} → F (𝔤, B), γ(f ⊗ g ⊗ h) = [fg, h] − f[g, h] − g[f, h] for f, g, h ∈ F (B). If h ∈ B then, using (2) and the anticommutativity, we can rewrite the value of γ as −D_{h}(fg) + fD_{h}(g) + gD_{h}(f). As D_{h} is a derivation, we get γ(f ⊗g⊗h) = 0 in this case. The same is true whenever f, g ∈ B as is immediate from (2) and the definition of F (B). If now h ∈ J and either f ∈ J or g ∈ J then π([fg, h] − f[g, h] − g[f, h]) = 0 because J is an ideal and π([J, J]) = 0 by (3). It follows that π ∘ γ = 0, and then γ = 0.

We can now show that [h, h′] ∈ F (B) for all h, h′ ∈ F (B). If f, g ∈ B then

fD_{g}([h, h′]) = f[D_{g}(h), h′] + f[h, D_{g}(h′)]
= [fD_{g}(h), h′] − D_{g}(h)[f, h′] + [h, fD_{g}(h′)] − [h, f]D_{g}(h′)
= [fD_{g}(h), h′] − D_{g}(h)D_{f}(h′) + [h, fD_{g}(h′)] + D_{f}(h)D_{g}(h′).

Setting D_{f,g} = D_{fg} − fD_{g} − gD_{f}, we deduce

D_{f,g}([h, h′]) = [D_{f,g}(h), h′] + [h, D_{f,g}(h′)] = 0,

as desired.

Finally, consider the 𝔤-module homomorphism δ : F (B)^{⊗3} → F (B) such that δ(f ⊗ g ⊗ h) = [f, [g, h]] − [[f, g], h] − [g, [f, h]]. If f ∈ B then δ(f ⊗ g ⊗ h) = 0 as this equality reduces to (1). Since δ is skewsymmetric in its arguments, the same is valid when either g ∈ B or h ∈ B. Now (3) means that [J, J] ⊂ J, and therefore δ(f ⊗ g ⊗ h) ∈ J whenever f, g, h ∈ J. It follows that π ∘ δ = 0, and δ = 0. All necessary identities are now checked. □

Let C be an arbitrary Poisson algebra over k. Suppose that C = C′ ⊕ I where C′ is an associative subalgebra and I an associative ideal of C such that [C′, C′] = 0 and [I, I] ⊂ I. Suppose also that C′ is endowed with a p-semilinear transformation f ↦ f^{[p]} such that 1^{[p]} = 0 and D_{f}^{p} = D_{f^{[p]}} for all f ∈ C′, where D_{f} ∈ Der C denotes the adjoint derivation.

Proposition 4.2. (1) Under the above hypotheses every isomorphism of algebras τ : C′ → B such that τ(f^{[p]}) ≡ τ(f)^{[p]} (mod k) for all f ∈ C′ extends in a unique way to a homomorphism of Poisson algebras σ : C → F (B) such that σ(I) ⊂ ker π.

(2) If C = B_{2n}, C′ = B and τ = id_{B} then the homomorphism σ in (1) is an isomorphism.

Proof. We may regard C′ as an abelian p-Lie algebra which operates on C by means of the representation f ↦ D_{f}. Since [k, C] = 0, there is the induced action of the factor algebra C′/k on C. Now τ induces an isomorphism of p-Lie algebras C′/k → B/k, and so the p-Lie algebra 𝔤 = B/k operates on C via the inverse isomorphism. Let π_{C} : C → C′ be the projection with respect to the decomposition C = C′ ⊕ I. There exists then a unique 𝔤-module homomorphism σ : C → F (𝔤, B) such that π ∘ σ = τ ∘ π_{C}. It is immediate that σ(I) ⊂ ker π and π ∘ σ|_{C′} = τ = π ∘ τ. Hence σ|_{C′} = τ by the uniqueness property for 𝔤-module homomorphisms into F (𝔤, B).

Consider the 𝔤-module homomorphism α : C ⊗ C → F (𝔤, B) defined by the rule α(g ⊗ h) = σ(gh) − σ(g)σ(h) for g, h ∈ C. Since π ∘ σ = τ ∘ π_{C} is a homomorphism of associative algebras, we get π ∘ α = 0, whence α = 0. Thus σ is a homomorphism of associative algebras.

Note that the 𝔤-invariance of σ means that D_{τ(g)}(σ(h)) = σ(D_{g}(h)) for g ∈ C′ and h ∈ C. If f, g ∈ C′ and h ∈ C then

(D_{τ(f)τ(g)} − τ(f)D_{τ(g)} − τ(g)D_{τ(f)})σ(h) = σ((D_{fg} − fD_{g} − gD_{f})(h)) = 0

by the properties of Poisson brackets. Since τ(C′) = B, we deduce σ(h) ∈ F (B).

Consider the 𝔤-module homomorphism β : C ⊗ C → F (B) defined by the rule β(g ⊗ h) = σ([g, h]) − [σ(g), σ(h)] for g, h ∈ C. If g ∈ C′ then β(g ⊗ h) = 0 in view of the 𝔤-invariance of σ. By anticommutativity β(g ⊗ h) = 0 when h ∈ C′ as well. If g, h ∈ I then β(g ⊗ h) ∈ ker π. Thus π ∘ β = 0, and so β = 0. In other words, σ : C → F (B) respects the Poisson brackets.

Clearly ker σ is an associative and a Lie ideal of C such that ker σ ⊂ I. Under the assumptions of (2) ker σ is stable under ∂_{1}, …, ∂_{2n} ∈ L. It is easy to see that B_{2n} contains no nonzero proper associative ideals stable under ∂_{1}, …, ∂_{2n}. It follows ker σ = 0.

Denote by T ⊂ End_{k} F (B) the associative subalgebra generated by all derivations D_{f} with f ∈ B and all operators of associative multiplications by elements f ∈ B. Clearly T is commutative. We may identify B with the subalgebra of associative multiplication operators in T. It is immediate from the identity D_{fg} = fD_{g} + gD_{f} that T is generated as an algebra over B by D_{y_{1}}, …, D_{y_{n}} where y_{1}, …, y_{n} ∈ 𝔪 is any minimal system of generators for B. Since D_{f}^{p} = D_{f^{[p]}} is a B-linear combination of derivations D_{y_{1}}, …, D_{y_{n}} for every f ∈ B, we see that T is generated as a module over B by p^{n} elements D_{y_{1}}^{r_{1}}, …, D_{y_{n}}^{r_{n}} where 0 ≤ r_{i} < p. Now define a linear map λ : F (B) → Hom_{B}(T, B) setting λ(h)(ψ) = π(ψ(h)) for h ∈ F (B) and ψ ∈ T. Then ker λ is the largest T -invariant subspace contained in ker π. As we know, ker π contains no nonzero 𝔤-invariant subspaces, whence ker λ = 0. We conclude

dim F (B) ≤ dim Hom_{B}(T, B) ≤ p^{n} dim B = p^{2n} = dim B_{2n}.

Comparing of dimensions yields σ(B_{2n}) = F (B), i.e., σ is an isomorphism. □

Theorem 4.3. Suppose B, B′ ⊂ B_{2n} are Lagrangian subalgebras and I, I′ ⊂ B_{2n} Lagrangian ideals such that B_{2n} = B ⊕ I = B′ ⊕ I′, Then every isomorphism of associative algebras τ : B → B′ satisfying τ(f^{[p]}) ≡ τ(f)^{[p]} (mod k) for all f ∈ B extends uniquely to an automorphism θ ∈ G such that θ(I) = I′. If, moreover, τ(f^{[p]}) = τ(f)^{[p]} for all f ∈ B and both I and I′ are closed under the [p]-map then θ ∈ G_{[p]}.

Proof. By proposition 4.2 we have a commutative diagram

B_{2n} ⟶^{σ} F (B) ⟶^{τ_{∗}} F (B′) ⟵^{σ′} B_{2n}
↑↓ ↑↓ ↑↓ ↑↓
B ⟶^{id} B ⟶^{τ} B′ ⟵^{id} B′

where the upward arrows are the inclusion maps, the downward arrows on the sides of the diagram are retractions with kernel I and I′, respectively, the downward arrows in the middle are canonical retractions, and the arrows in the top row are the homomorphisms of Poisson algebras induced by the respective algebra homomorphisms in the bottom row. Both σ and σ′ are isomorphisms by proposition 4.2. The isomorphism τ^{−1} induces τ_{∗}^{−1} : F (B′) → F (B), and it is clear that τ_{∗}^{−1} ∘ τ_{∗} and τ_{∗} ∘ τ_{∗}^{−1} are both identity maps by the uniqueness property. Hence τ_{∗} is also an isomorphism. Now θ = σ^{′−1} ∘τ_{∗} ∘σ is the required element of G, and the uniqueness is also clear.

Suppose that τ commutes with the [p]-map and both I and I′ are closed under the [p]-map. Given g ∈ I, consider the element h = (τ_{∗} ∘ σ)(g^{[p]}) − σ′((θg)^{[p]}) ∈ F (B′). We have π(h) = 0 and

\begin{aligned}[τ(f), h] & = (τ_{∗} ∘ σ)([f, g^{[p]}]) − σ′([θ(f), (θg)^{[p]}]) \\ & = −(τ_{∗} ∘ σ)((D_{g}^{p}(f)) + (σ′ ∘ θ)((D_{g}^{p}(f)) = 0 \end{aligned}

for all f ∈ B. It follows that [B′, h] = 0, whence h ∈ B′ ∩ ker π = 0. We deduce that θ(g^{[p]}) = θ(g)^{[p]} for all g ∈ I. We have θ(f^{[p]}) = θ(f)^{[p]} for all f ∈ B since θ|_{B} = τ. By the Jacobson p-th power formula the equality θ(f^{[p]}) = θ(f)^{[p]} holds then for all f ∈ B_{2n}. Thus θ ∈ G_{[p]}. □

Remark. In this section we never used an explicit expression for the Poisson bracket on B_{2n}. The only properties actually needed are:

(a) there exists a [p]-map making B_{2n} into a p-Lie algebra,

(b) B_{2n} contains no nonzero proper associative ideals which are stable under all D_{f}. One can strengthen theorem 4.3 as follows. Given two Poisson brackets satisfying

(a) and (b), a Lagrangian pair B, I with respect to the first bracket and a Lagrangian pair B′, I′ with respect to the second bracket, every [p]-compatible isomorphism τ : B → B′ extends to an automorphism θ ∈ Aut B_{2n} which transforms one Poisson bracket to another. Given a Poisson bracket satisfying (a) and (b), there exist two n-dimensional tori T_{1}, T_{2} ⊂ B_{2n} such that T_{1}, T_{2} ⊂ 𝔪 and B_{2n} = k ⊕ T_{1} ⊕ T_{2} ⊕ 𝔪^{2}. In fact, it is easy to ascertain the existence of an n-dimensional torus T_{0} ∈ 𝔪^{2} as 𝔪^{2}/𝔪^{3} is a p-Lie algebra operating faithfully on V = 𝔪/𝔪^{2} as sp(V) when p > 2 and as the commutant of sp(V) when p = 2. To obtain T_{1} and T_{2} one has to apply Winter’s switchings of tori (see [24]) taking a certain amount of care. Now T_{1} generates a Lagrangian subalgebra B and T_{2} a Lagrangian ideal I such that B_{2n} = B ⊕ I. If T′_{1}, T′_{2} and B′, I′ are constructed similarly with respect to a second Poisson bracket satisfying (a) and (b) then an isomorphism of tori T_{1} ≅ T′_{1} extends to a [p]-compatible isomorphism τ : B → B′. It follows that any two Poisson brackets satisfying (a) and (b) are conjugate with respect to Aut B_{2n}. This can be also derived from the known results. A Poisson bracket satisfying (a) and (b) corresponds to a Hamiltonian form on W_{2n} which has to be exact by [17], Th. 2.2. According to the classification of Hamiltonian forms on W_{2n} [12], the exact Hamiltonian forms belong to a single orbit with respect to Aut B_{2n}.

5. Cross section of regular classes

In the construction of cross sections in theorem 5.2 we assume that x_{i}^{[p]} = 0 for all i = 1, …, 2n (see lemma 1.1).

Lemma 5.1. Suppose f, h ∈ B_{2n} and g ∈ 𝔪 satisfy [f, g] = 1 and [f, h] ∈ g^{p−1}B_{2n}. If p = 2 assume also that g^{[2]} = 0. Then

(f + g^{p−1}h)^{[p]} = f^{[p]} − h + g^{p−1}h[g, h].

Proof. We will need a precise computation of terms occurring in the formula ([p]_{3}) from section 1. Put y = g^{p−1}h. As g^{p} = 0, we have D_{f}^{i}(y) = (−1)^{i}i!g^{p−1−i}h for i < p. Now s_{1}(f, y) = D_{f}^{p−1}(y) = −h. Note that [y, D_{f}^{i}(y] ∈ [g^{p−1}B_{2n}, g^{2}B_{2n}] = 0 for i < p − 2. In particular, s_{l}(f, y) = 0 for 2 < l ≤ p − 1. If p > 2 then

2s_{2}(f, y) = [y, D_{f}^{p−2}(y)] = [g^{p−1}h, −gh] = (2 − p)g^{p−1}h[g, h].

We see also by induction on i that D_{y}^{i}(B_{2n}) ⊂ g^{i(p−2)}B_{2n}. If p > 2 then y^{[p]} = 0 as y ∈ 𝔪^{2} and D_{y}^{3} = 0. Suppose p = 2. We have

D_{gh}^{2} = (gD_{h} + hD_{g})^{2} = (gD_{h})^{2} + [gD_{h}, hD_{g}] + (hD_{g})^{2}
= g^{2}D_{h}^{2} + g[h, g]D_{h} + ghD_{[h,g]} + h^{2}D_{g}^{2} + h[g, h]D_{g}
= D_{gh[g,h]}

since g^{2} = 0 and D_{g}^{2} = 0 by the assumptions on g. If h ∈ 𝔪 then (gh)^{[2]} ∈ 𝔪^{2}, whence (gh)^{[2]} = gh[g, h]. In general, let h = λ + h′ where λ ∈ k and h′ ∈ 𝔪. Then

(gh)^{[2]} = (λg + gh′)^{[2]} = λ^{2}g^{[2]} + λg[g, h′] + gh′[g, h′] = gh[g, h]. □

We adopt the convention that the product over an empty set is equal to 1. For λ = (λ_{1}, …, λ_{n}) ∈ 𝔸^{n} and h ∈ k[x_{1}, …, x_{n}] ⊂ B_{2n} define an element

u_{λ,h} = ∑_{i=1}^{n} ((−1)^{i−1}(λ_{i} + x_{i}) ∏_{j=1}^{i−1} x_{n+j}^{p−1}) + (−1)^{n−1}h ∏_{j=1}^{n} x_{n+j}^{p−1} ∈ B_{2n}

and consider the following affine subspaces of dimension p^{n} + n and p^{n} in B_{2n}:

\begin{aligned}S & = \{ u_{λ,h} | λ ∈ 𝔸^{n} and h ∈ k[x_{1}, …, x_{n}] \}, \\ S_{0} & = \{ u_{λ,h} | λ_{2} = … = λ_{n} = 0 and h ∈ 𝔪 ∩ k[x_{1}, …, x_{n}] \}. \end{aligned}

Theorem 5.2. (1) The invariant functions φ_{1}, …, φ_{p^{n}}, κ_{0}, …, κ_{n−1} (respectively φ̃_{1}, …, φ_{p^{n}−1}, κ) separate the G_{[p]}-orbits (respectively G-orbits) on U.

(2) The morphisms of algebraic varieties φ : S → 𝔸^{p^{n}+n} and φ̃ : S_{0} → 𝔸^{p^{n}} given by functions φ_{1}, …, φ_{p^{n}}, κ_{0}, …, κ_{n−1} and φ̃_{1}, …, φ_{p^{n}−1}, κ, respecctively, are isomorphisms.

(3) One has S ⊂ U, and each G_{[p]}-orbit (respectively G-orbit) on U meets S (respectively S_{0}) at a single point.

(4) The algebra of invariants k[B_{2n}]^{G_{[p]}} (respectively k[B_{2n}]^{G}) is generated by functions φ_{1}, …, φ_{p^{n}}, κ_{0}, …, κ_{n−1} (respectively φ̃_{1}, …, φ_{p^{n}−1}, κ) which are algebraically independent.

Proof. Let f, g ∈ U. Suppose κ(f) = κ(g) and φ̃_{a}(f) = φ̃_{a}(g) for all 1 ≤ a < p^{n}. The elements f^{⟨p^{i}⟩} (respectively g^{⟨p^{i}⟩}) with 0 ≤ i < n lie in 𝔪 and generate the Lagrangian subalgebra 𝔷(f) (respectively 𝔷(g)). It follows that there exists an isomorphism of associative algebras τ : 𝔷(f) → 𝔷(g) such that f^{⟨p^{i}⟩} ↦ g^{⟨p^{i}⟩} for each 0 ≤ i < n. We have τ(f^{⟨a⟩}) = g^{⟨a⟩} for all a < p^{n}. Put

V = {h ∈ 𝔷(f) | τ(h^{⟨p⟩}) = (τh)^{⟨p⟩}}.

The assignment h ↦ h^{[p]} is a p-semilinear map on every Lagrangian subalgebra. The same is true then for the map h ↦ h^{⟨p⟩}. Hence V is a subspace of 𝔷(f). If h = f^{⟨p^{i}⟩} where i < n then h^{⟨p⟩} = f^{⟨p^{i+1}⟩} and τ(h)^{⟨p⟩} = g^{⟨p^{i+1}⟩}. Hence f^{⟨p^{i}⟩} ∈ V for i < n − 1 by the construction of τ. Next,

τ(f^{⟨p^{n}⟩}) = τ(− ∑_{a=1}^{p^{n}−1} φ̃_{a}(f)f^{⟨p^{n}−a⟩}) = − ∑_{a=1}^{p^{n}−1} φ̃_{a}(g)g^{⟨p^{n}−a⟩} = g^{⟨p^{n}⟩},

which shows that f^{[p^{n−1}]} ∈ V too. If h ∈ (𝔪 ∩ 𝔷(f))^{2} then h^{⟨p⟩} = 0 and τ(h)^{⟨p⟩} = 0 by lemma 1.2. Finally, 1 ∈ V since 1^{⟨p⟩} = 0. Thus V = 𝔷(f). As h^{[p]} = h^{⟨p⟩} +κ_{1}(h), we get τ(h^{[p]}) ≡ (τh)^{[p]} (mod k) for all h ∈ 𝔷(f). In view of lemma 1.2 there exist Lagrangian ideals I, I′ ⊂ B_{2n} such that for B = 𝔷(f) and B′ = 𝔷(g) we meet the hypotheses of theorem 4.3. The extension θ ∈ G of τ satisfies θ(f) = θ(f^{⟨1⟩} + κ(f)) = g^{⟨1⟩} + κ(g) = g. Thus f and g are G-conjugate.

Suppose now that φ_{a}(f) = φ_{a}(g) for all 1 ≤ a ≤ p^{n} and κ_{i}(f) = κ_{i}(g) for all 0 ≤ i < n. According to proposition 3.5 φ̃_{a} belongs to the subalgebra generated by φ_{1}, …, φ_{p^{n}}, κ_{0}, …, κ_{n−1}. Hence φ̃_{a}(f) = φ̃_{a}(g) for all 1 ≤ a < p^{n}, and we can apply our previous construction of θ. Note that κ_{n}(f) = κ_{n}(g) by (5) of proposition 3.5. Hence τ(f^{[p^{i}]}) = τ(f^{⟨p^{i}⟩} + κ_{i}(f)) = g^{⟨p^{i}⟩} + κ_{i}(g) = g^{[p^{i}]} for 0 ≤ i ≤ n. This means that the equality τ(h^{[p]}) = (τh)^{[p]} holds for h = f^{[p^{i}]} with 0 ≤ i < n. As it holds also for h ∈ (𝔪 ∩ 𝔷(f))^{2} by lemma 1.2 and for h = 1 since 1^{[p]} = 0, it is fulfilled on the whole 𝔷(f). Since I and I′ can be chosen closed under the [p]-map, we get θ ∈ G_{[p]} by theorem 4.3. Thus f and g are G_{[p]}-conjugate, which completes the proof of (1).

Consider the element u = u_{λ,0} ∈ S. We prove by induction on r that

u^{[p^{r}]} = ∑_{i=r+1}^{n} ((−1)^{i−1−r}(λ_{i} + x_{i}) ∏_{j=r+1}^{i−1} x_{n+j}^{p−1}) for all r = 0, …, n

(note that for r = n the formula should be understood as u^{[p^{n}]} = 0). Denote by w_{r} the right hand side of the formula. If r > 0 then w_{r−1} = f − g^{p−1}w_{r} where f = λ_{r} + x_{r} and g = x_{n+r}. Note that f, g together with h = −w_{r} satisfy the hypotheses of lemma 5.1, and furthermore [f, h] = [g, h] = 0 in this case. Hence w_{r−1}^{[p]} = w_{r}, as required.

An arbitrary element v = u_{λ,h} ∈ S can be written as

v = u + (−1)^{n−1}h ∏_{j=1}^{n} x_{n+j}^{p−1} where h ∈ k[x_{1}, …, x_{n}].

Put I = x_{n+1}B_{2n} + … + x_{2n}B_{2n}. From the computations above we see at once that u^{[p^{r−1}]} ≡ λ_{r} + x_{r} (mod I) for each r = 1, …, n. In particular, u^{[p^{r−1}]} − λ_{r} ∈ 𝔪. Denote by

h^{u} = h(u − λ_{1}, u^{[p]} − λ_{2}, …, u^{[p^{n−1}]} − λ_{n}) ∈ k[u, u^{[p]}, …, u^{[p^{n−1}]}]

the image of h under the algebra homomorphism k[x_{1}, …, x_{n}] → B_{2n} such that x_{i} ↦ u^{[p^{i−1}]} − λ_{i} for all i = 1, …, n. We claim that

v^{[p^{r}]} = u^{[p^{r}]} + (−1)^{n−1−r}h^{u} ∏_{j=r+1}^{n} x_{n+j}^{p−1} for all r = 0, …, n − 1.

When r = 0, the formula follows from the fact that h^{u} ≡ h (mod I). Suppose that the formula is valid for r = s − 1 where 0 < s < n. Thus

v^{[p^{s−1}]} = u^{[p^{s−1}]} + x_{n+s}^{p−1}h′ where h′ = (−1)^{n−s}h^{u} ∏_{j=s+1}^{n} x_{n+j}^{p−1}.

Apply lemma 5.1 with f = u^{[p^{s−1}]}, g = x_{n+s} and h′ in place of h. By our previous calculations f = λ_{s} + x_{s} + g^{p−1}w where w lies in the associative subalgebra of B_{2n} generated by the elements x_{i}, x_{n+i} with i > s. It follows [f, g] = [x_{s}, x_{n+s}] = 1 and [f, x_{n+j}] = g^{p−1}[w, x_{n+j}] for j > s. As [f, h^{u}] = 0, we get [f, h′] ∈ g^{p−1}B_{2n}. Thus the hypotheses of the lemma are fulfilled, and so

v^{[p^{s}]} = u^{[p^{s}]} − h′ + h′x_{n+s}^{p−1}[x_{n+s}, h′].  (∗)

Observe that h′ ∈ x_{2n}^{p−1}B_{2n}, whence [x_{n+s}, h′] ∈ x_{2n}^{p−1}B_{2n}, and h′[x_{n+s}, h′] = 0. It follows v^{[p^{s}]} = u^{[p^{s}]} − h′, which is the required formula for v^{[p^{r}]} with r = s.

For s = n we can repeat the arguments above until the formula (∗) is reached. In this case h′ = h^{u}. Since u^{[p^{n}]} = 0, we deduce v^{[p^{n}]} ≡ −h^{u} ≡ −h (mod I).

We have v^{[p^{r−1}]} ≡ u^{[p^{r−1}]} ≡ λ_{r} + x_{r} (mod I) for each r = 1, …, n. In particular, v, v^{[p]}, …, v^{[p^{n−1}]} are linearly independent modulo k + 𝔪^{2}, and so v ∈ U. By proposition 2.2 𝔷(v) coincides with the associative subalgebra of B_{2n} generated by v, v^{[p]}, …, v^{[p^{n−1}]}. Let π : B_{2n} → k[x_{1}, …, x_{n}] be the retraction with kernel I. It is clear that π maps 𝔷(v) isomorphically onto k[x_{1}, …, x_{n}]. Denote by

h^{v} = h(v − λ_{1}, v^{[p]} − λ_{2}, …, v^{[p^{n−1}]} − λ_{n}) ∈ 𝔷(v)

the image of h under the algebra homomorphism k[x_{1}, …, x_{n}] → B_{2n} such that x_{i} ↦ v^{[p^{i−1}]} − λ_{i} for all i = 1, …, n. Then π(h^{v}) = h. As v^{[p^{n}]} ∈ 𝔷(v) and π(v^{[p^{n}]}) = −h, we conclude that v^{[p^{n}]} + h^{v} = 0.

By the above calculations κ_{i−1}(v) = λ_{i} for i = 1, …, n and φ_{a}(v) is equal to the coefficient of v^{[p^{n}−a]} in the expansion of h^{v} for a = 1, …, p^{n}. Given (μ_{1}, …, μ_{n}) ∈ 𝔸^{n} and (ν_{1}, …, ν_{p^{n}}) ∈ 𝔸^{p^{n}}, there exists a unique v ∈ S such that κ_{i−1}(v) = μ_{i} for all i = 1, …, n and φ_{a}(v) = ν_{a} for all a = 1, …, p^{n}. In fact we must take μ_{i} = λ_{i} and h = ∑_{a+b=p^{n},a>0} ν_{a} ∏_{i=1}^{n}(x_{i} + λ_{i})^{b_{i}} (recall that we denote by b_{i} the p-adic coefficients of b). Thus φ is bijective, and the inverse map 𝔸^{p^{n}+n} → S is a morphism of algebraic varieties.

We have v^{⟨p^{i−1}⟩} = v^{[p^{i−1}]} − λ_{i} for i = 1, …, n. If h ∈ 𝔪, then h^{v} ∈ 𝔪, and it follows that v^{⟨p^{n}⟩} = v^{[p^{n}]}, so that v^{⟨p^{n}⟩} + h^{v} = 0. Hence φ̃_{a}(v) is the coefficient of v^{⟨p^{n}−a⟩} in the expansion of h^{v} as a linear combination of v^{⟨b⟩} with 0 < b < p^{n}. We conclude as before that φ̃ is an isomorphism. The proof of (2) is complete.

We have checked already that S ⊂ U. The remainder of (3) is immediate from (1) and (2). Let k[U] be the algebra of regular functions on U. The restriction homomorphisms k[U]^{G} → k[S_{0}] and k[U]^{G_{[p]}} → k[S] are injective by (3) and surjective by (2). Since B_{2n} ∖ U has codimension at least 2 in B_{2n}, there is an equality k[B_{2n}] = k[U]. That gives (4). □

Remark. If N denotes the null fiber of the morphism B_{2n} → 𝔸^{p^{n}+n} (respectively B_{2n} → 𝔸^{p^{n}}) given by invariant functions of theorem 5.2, then it can be shown that the codimension of N ∩ 𝔪^{3} in B_{2n} is less than p^{n} + n (respectively p^{n}) provided that either p or n is big enough. This means that k[B_{2n}] is a free module neither over k[B_{2n}]^{G_{[p]}} nor over k[B_{2n}]^{G}.

Corollary 5.3. If y = x_{1}^{p−1} ⋯ x_{2n}^{p−1} then φ_{1}(y) = (−1)^{n−1} and φ_{a}(y) = 0 for a > 1.

Proof. Let v = u_{0, th} ∈ S where t ∈ k and h = x_{1}^{p−1} ⋯ x_{n}^{p−1}. Then v = u+(−1)^{n−1}ty where u = u_{0,0}. By calculations in the proof of the theorem φ_{a}(v) is equal to the coefficient of v^{[p^{n}−a]} in the expansion of th^{v} = t ∏_{i=0}^{n−1}(v^{[p^{i}]})^{p−1} = tv^{[p^{n}−1]}. Hence φ_{1}(v) = t and φ_{a}(v) = 0 for a > 1. As φ_{a} is homogeneous of degree a, we have φ_{a}(v) = (−1)^{(n−1)a}t^{a}φ_{a}(y) modulo a polynomial in t of degree < a. Comparing the coefficients of t^{a}, we find φ_{a}(y). □

We conclude this section with a result which has a bearing on the representation theory of the Poisson algebra in view of [16], Th. 5.4. I will investigate the representations more thoroughly in a forthcoming paper.

Corollary 5.4. Suppose that f ∈ U, and let K denote the asymptotic cone of the orbit G_{[p]}f. Then K ∩ U is nonempty.

Proof. We may assume that f = u_{λ,h} ∈ S. For 0 ≠ t ∈ k define θ_{t} ∈ Aut B_{2n} setting θ_{t}(x_{i}) = t^{−p^{i−1}}x_{i} and θ_{t}(x_{n+i}) = t^{p^{i−1}}x_{n+i} for i = 1, …, n. One checks readily that θ_{t} ∈ G_{[p]}. We have

θ_{t}(f) = t^{−1}u + ∑_{i=1}^{n} ((−1)^{i−1}λ_{i}t^{p^{i−1}−1} ∏_{j=1}^{i−1} x_{n+j}^{p−1}) + (−1)^{n−1}t^{p^{n}−1}θ_{t}(h) ∏_{j=1}^{n} x_{n+j}^{p−1}

where u = u_{0,0} = ∑_{i=1}^{n} (−1)^{i−1}x_{i} ∏_{j=1}^{i−1} x_{n+j}^{p−1}. Since h ∈ k[x_{1}, …, x_{n}], the map t ↦ θ_{t}(h) is polynomial in t^{−1} of degree < p^{n}. Hence θ_{t}(f) = t^{−1}u + ψ(t) where ψ(t) is a polynomial map in t.

Let Z ⊂ B_{2n} ⊕ k be the closure of the set of elements (tg, t) with t ∈ k ∖ {0} and g ∈ G_{[p]}f. Denote by η : Z → k the restriction of the projection B_{2n} ⊕ k → k onto the second summand. Then K = η^{−1}(0) by [6], Ch. II, 4.2. We have (u+tψ(t), t) = (tθ_{t}(f), t) ∈ Z for t ≠ 0, whence (u, 0) ∈ Z as well. It follows u ∈ K, and it remains to notice that u ∈ S ⊂ U. □

6. Semisimple and nilpotent elements

Denote by Z_{s} the closure of the set of [p]-semisimple elements and by N the closed set of [p]-nilpotent elements in B_{2n}.

Lemma 6.1. (1) φ_{a}(f^{[p]}) = φ̃_{a}(f^{[p]}) = φ̃_{a}(f)^{p} if a = p^{n} − p^{i}, i = 0, …, n, and φ_{a}(f^{[p]}) = φ̃_{a}(f^{[p]}) = 0 otherwise.

(2) f^{[p^{n+1}]} is [p]-semisimple for any f ∈ B_{2n}.

Proof. If c ≠ 0 and c ≠ p^{i} for i = 0, …, n then f^{⟨c⟩} ∈ 𝔫^{2}, where 𝔫 is the maximal ideal of the associative subalgebra k[f, f^{[p]}, …, f^{[p^{n−1}]}] ⊂ B_{2n}. We have (f^{⟨c⟩})^{[p]} = 0 for such integers c (see lemma 1.2), and it follows

∑_{i=0}^{n} φ_{p^{n}−p^{i}}(f)^{p}f^{[p^{i+1}]} = ( ∑_{b+c=p^{n}} φ̃_{b}(f)f^{⟨c⟩})^{[p]} = 0.

Taking the projection onto 𝔪 with respect to the decomposition B_{2n} = k⊕𝔪, we get a zero linear combination of f^{⟨p⟩}, …, f^{⟨p^{n+1}⟩} with the same coefficients. If f ∈ U_{s} then f^{[p]} ∈ U_{s} by proposition 2.2. In this case φ_{a}(f^{[p]}) (respectively φ̃_{a}(f^{[p]})) is the coefficient of f^{[b]} (respectively f^{⟨b⟩}) where b = p^{n+1} − pa in the above linear combinations, which yields (1). Since U_{s} is open in B_{2n}, (1) holds for all f.

Let i be the smallest integer among 0, …, n such that φ_{p^{n}−p^{i}}(f) ≠ 0. Then f^{[p^{i+1}]} is a linear combination of elements f^{[p^{j+1}]} with i < j ≤ n, and so f^{[p^{i+1}]} is [p]-semisimple. This proves (2). □

Remark. The proof of the lemma also describes the minimal p-polynomial for the Poisson algebra, as defined in [14].

Proposition 6.2. (1) Z_{s} is an irreducible subvariety of codimension p^{n} −n in B_{2n}. (2) Z_{s} ∩ U = {f ∈ U | φ_{a}(f) = 0 for all 1 ≤ a ≤ p^{n}, a ≠ p^{n} − p^{i}}.

(3) Z_{s} ∩ U contains only nonsingular points of Z_{s}.

(4) f^{[p]} ∈ Z_{s} for all f ∈ B_{2n}.

Proof. Consider the morphism q : B_{2n} → B_{2n}, f ↦ f^{[p^{r}]}, where r > n. Its image coincides with the set of [p]-semisimple elements in B_{2n}. As q(B_{2n}) is irreducible, so is its closure Z_{s} too. Let f ∈ U_{s} and g ∈ q^{−1}(q(f)). Then g^{[p^{r}]} = f^{[p^{r}]} = f_{s}^{[p^{r}]}, and so g ∈ 𝔷(f_{s}^{[p^{r}]}) = 𝔷(f_{s}). By proposition 2.2 𝔷(f_{s}) is a Lagrangian subalgebra of B_{2n}. If T ⊂ B_{2n} denotes the torus generated by the semisimple element f_{s}, then dim T = n and 𝔷(f_{s}) = k + T + 𝔫^{2} where 𝔫 = 𝔪 ∩ 𝔷(f_{s}). By lemma 1.2 the [p]-map on k + 𝔫^{2} is zero. Now g_{s} ∈ T. As g_{s}^{[p^{r}]} = g^{[p^{r}]} = f_{s}^{[p^{r}]}, we get g_{s} = f_{s}. It follows q^{−1}(q(f)) = f_{s} +k +𝔫^{2}. In particular, dim q^{−1}(q(f)) = p^{n} −n. Since U_{s} is an open subset of B_{2n}, we get (1). Furthermore, (4) holds for f ∈ U_{s} since f^{[p]} is semisimple in this case. Then (4) is fulfilled everywhere.

Denote by Z the zero set of p^{n} − n functions φ_{a} with 1 ≤ a ≤ p^{n}, a ≠ p^{n} − p^{i}. All semisimple elements are contained in Z by lemma 6.1. Hence Z_{s} ⊂ Z. All irreducible components of Z have codimension ≤ p^{n} − n in B_{2n}. If f ∈ U ∩ Z then f is a nonsingular point of Z and the irreducible component of Z containing f has codimension p^{n} − n by lemma 3.3. Put Z′ = {f ∈ Z | φ_{p^{n}−1}(f) = 0}. All irreducible components of Z′ ∩ U have, similarly, codimension p^{n} − n + 1. Hence every irreducible component of Z ∩ U contains a point outside of Z′. If f ∈ Z then φ_{p^{n}−1}(f) = φ_{p^{n}−1}(f) by (3) of proposition 3.5 since \binom{p^{n}−2}{p^{n}−p^{i}−1} ≡ 0 (mod p) for 0 < i < n. If f ∈ Z ∖ Z′ then φ_{p^{n}−1}(f) ≠ 0, and (1) of proposition 3.5 shows that f is a linear combination of elements f^{[p^{i}]} with 1 ≤ i ≤ n, so that f is semisimple. Thus Z ∖ Z′ ⊂ Z_{s}, and we conclude Z ∩ U = Z_{s} ∩ U. □

Denote by χ_{D}(t) the characteristic polynomial of D ∈ W_{2n} as a linear transformation of B_{2n}. As was proved by Premet [15],

χ_{D}(t) = t^{p^{2n}} + ∑_{i=0}^{2n−1} ψ_{i}(D)t^{p^{i}}

where ψ_{i} is a polynomial function of degree p^{2n} − p^{i} on W_{2n}.

Lemma 6.3. If f ∈ B_{2n} then ψ_{i}(D_{f}) = 0 and ψ_{n+i}(D_{f}) = φ_{p^{n}−p^{i}}(f)^{p^{n}} for all 0 ≤ i < n.

Proof. It suffices to check the equalities for f ∈ U. Under this assumption 𝔷(f) ≅ B_{n} as associative algebras. Let 𝔷(f) = J_{0} ⊃ J_{1} ⊃ … ⊃ J_{p^{n}} = 0 be a chain of associative ideals such that dim J_{a−1}/J_{a} = 1 for all a = 1, …, p^{n}. Then J_{a}B_{2n} is stable under D_{f} for each a. Since J_{1}J_{a−1} ⊂ J_{a} and B_{2n} is a free 𝔷(f)-module,

J_{a−1}B_{2n}/J_{a}B_{2n} ≅ J_{a−1}/J_{a} ⊗_{𝔷(f)} B_{2n} ≅ J_{a−1}/J_{a} ⊗ \overline{B} ≅ \overline{B}

where \overline{B} = B_{2n}/J_{1}B_{2n} ≅ B_{n}. Denote by D ∈ Der \overline{B} ≅ W_{n} the derivation induced by D_{f}. The linear transformation induced by D_{f} in J_{a−1}B_{2n}/J_{a}B_{2n} is equivalent to D for each a. It follows χ_{D_{f}} (t) = χ_{D}(t)^{p^{n}} where χ_{D} is the characteristic polynomial of D. By Premet’s result χ_{D}(t) = t^{p^{n}} + ∑_{i=0}^{n−1} λ_{i}t^{p^{i}} for some λ_{0}, …, λ_{n−1} ∈ k. We have D^{p^{n}} + ∑_{i=0}^{n−1} λ_{i}D^{p^{i}} = 0 by Cayley-Hamilton theorem. If 0 ≤ c ≤ p^{n} and c ≠ p^{i} for i = 0, …, n then f^{⟨c⟩} ∈ J_{1}^{2}, whence [f^{⟨c⟩}, B_{2n}] ⊂ J_{1}B, so that D_{f^{⟨c⟩}} induces a zero derivation of \overline{B}. Clearly D_{f^{⟨p^{i}⟩}} = D_{f}^{p^{i}} induces D^{p^{i}} on \overline{B}. It

follows now from proposition 3.5(1) that D^{p^{n}} + ∑_{i=0}^{n−1} φ_{p^{n}−p^{i}}(f)D^{p^{i}} = 0. Note that J_{1}B is a Lagrangian ideal of B_{2n} generated by f^{⟨1⟩}, f^{⟨p⟩}, …, f^{⟨p^{n−1}⟩}. Hence D, D^{p}, …, D^{p^{n−1}} are linearly independent by lemma 1.3. We can conclude that λ_{i} = φ_{p^{n}−p^{i}}(f) for each i, and the required formulas are immediate. □

Theorem 6.4. (1) N is an irreducible normal complete intersection of codimension n in B_{2n}.

(2) The ideal I = {φ ∈ k[B_{2n}] | φ(N) = 0} is generated by n functions φ_{p^{n}−p^{i}} with i = 0, …, n − 1.

(3) N ∩ U contains only nonsingular points of N and codim_{N} N ∖ U ≥ 2.

(4) N = {f ∈ B_{2n} | f^{[p^{n+1}]} = 0}.

Proof. An element f ∈ B_{2n} is [p]-nilpotent if and only if D_{f} is nilpotent, if and only if χ_{D_{f}} (t) = t^{p^{2n}}, if and only if ψ_{i}(D_{f}) = 0 for all i = 0, …, 2n−1. In view of lemma 6.3 we get

N = {f ∈ B_{2n} | φ_{p^{n}−p^{i}}(f) = 0 for all i = 0, …, n − 1}.

The differentials of n functions φ_{p^{n}−p^{i}} with i = 0, …, n−1 are linearly independent at all f ∈ U by proposition 3.5, and so the points in N ∩ U are nonsingular. We can find a linear subspace E ⊂ B_{2n} such that dim E = n + 2 and E ∩ N ⊂ {0} ∪ U. Take E to be the linear span of elements u, v, s_{1}, …, s_{n} where

u = ∑_{i=1}^{n} ((−1)^{i−1}x_{i} ∏_{j=1}^{i−1} x_{n+j}^{p−1}), v = ∑_{i=1}^{n} ((−1)^{i−1}x_{n+i} ∏_{j=1}^{i−1} x_{j}^{p−1}),

and s_{i} = x_{i}x_{n+i} for i = 1, …, n. Let T = {θ_{t} | t ∈ k^{∗}} ⊂ G_{[p]} be the onedimensional torus such that θ_{t}(x_{i}) = t^{p^{i−1}}x_{i} and θ_{t}(x_{n+i}) = t^{−p^{i−1}}x_{n+i} for i = 1, …, n. Then B_{2n} decomposes as a direct sum of weight spaces V_{m} = {f ∈ B_{2n} | θ_{t}(f) = t^{m}f}, m ∈ ℤ, with respect to T. We have f^{[p]} ∈ V_{pm} for all f ∈ V_{m}. Note that u ∈ V_{1}, v ∈ V_{−1} and s_{i} ∈ V_{0}. Write f ∈ E as λu + s + μv where λ, μ ∈ k and s is a linear combination of s_{1}, …, s_{n}. Using Jacobson’s formula, we deduce

f^{[p^{r}]} ≡ λ^{p^{r}}u^{[p^{r}]} + μ^{p^{r}}v^{[p^{r}]} (mod ∑_{a=1−p^{r}}^{p^{r}−1} V_{a})

for all r ≥ 0. Note that 𝔪 and 𝔪^{2} are stable under T, and the weights of T on 𝔪/𝔪^{2} are p^{±i} with i = 0, …, n − 1. Hence V_{m} ⊂ k + 𝔪^{2} for all other m. As we have seen in the proof of theorem 5.2, u^{[p^{r}]} ≡ x_{r+1} (mod 𝔪^{2}) for r = 0, …, n − 1. Now v = τ(u) where τ ∈ G_{[p]} sends x_{i} to x_{n+i} and x_{n+i} to −x_{i} for i = 1, …, n. It follows that v^{[p^{r}]} = τ(u^{[p^{r}]}) ≡ x_{n+r+1} (mod 𝔪^{2}), and

f^{[p^{r}]} ≡ λ^{p^{r}}x_{r+1} + μ^{p^{r}}x_{n+r+1} (mod k + 𝔪^{2}).

If either λ ≠ 0 or μ ≠ 0 then f, f^{[p]}, …, f^{[p^{n−1}]} are linearly independent modulo k+𝔪^{2}, i.e., f ∈ U. If λ = μ = 0 then f = s is [p]-semisimple. Thus E∩N ⊂ {0}∪U, as required. Now (1) and (2) follow from lemma 1.5. If f ∈ N then f^{[p^{n+1}]} is [p]- nilpotent. As the latter element is [p]-semisimple by lemma 6.1, it has to be zero. □

7. Central elements in the universal enveloping algebra

Let U(B_{2n}) be the universal enveloping algebra of the Lie algebra structure on B_{2n}. It has a canonical increasing filtration U_{m}(B_{2n}), m ≥ 0, such that gr U(B_{2n}) ≅ S(B_{2n}), the symmetric algebra of B_{2n}. Endow the center Z of U(B_{2n}) with the induced filtration. Then gr Z ⊂ S(B_{2n})^{L}, the subalgebra of L-invariants in S(B_{2n}). The nondegenerate invariant bilinear form on B_{2n} (see corollary 3.6) yields an isomorphism of L- and G-modules B_{2n} ≅ B_{2n}^{∗}. Hence an L- and G-equivariant isomorphism of algebras S(B_{2n}) ≅ S(B_{2n}^{∗}) ≅ k[B_{2n}]. Denote by φ_{a}^{∨}, κ_{i}^{∨} ∈ S(B_{2n}) the images of φ_{a}, κ_{i} ∈ k[B_{2n}] under this isomorphism (see section 3). Thus each φ_{a}^{∨} is L- and G_{[p]}-invariant, and each κ_{i}^{∨} is L_{0}- and G_{[p]}-invariant. We are going to prove that φ_{a}^{∨} ∈ gr Z whenever a ≢ 0 (mod p), 0 < a < p^{n}.

For a ≥ 0 define ξ_{a} : B_{2n} → k[B_{2n}] by the formula ξ_{a}(g)(f) = φ_{1}(f^{[a]}g) for f, g ∈ B_{2n}. Then ξ_{a} is a homomorphism of L-modules and G_{[p]}-modules. In fact

Hom(B_{2n}, k[B_{2n}]) ≅ k[B_{2n}] ⊗ B_{2n}^{∗} ≅ k[B_{2n}] ⊗ B_{2n} ≅ Pol(B_{2n}, B_{2n})

as L-modules and G_{[p]}-modules. Under this isomorphism ξ_{a} corresponds to the Land G_{[p]}-invariant polynomial map B_{2n} → B_{2n} given by f ↦ f^{[a]}.

Denote by ξ_{a}^{∨} : B_{2n} → S(B_{2n}) the composite of ξ_{a} and the invariant isomorphism k[B_{2n}] ≅ S(B_{2n}) considered earlier, and put y = x_{1}^{p−1} ⋯ x_{2n}^{p−1}.

Lemma 7.1. (1) ξ_{a}(y) = (−1)^{n−1} ∏_{i≥0} κ_{i}^{a_{i}} and ξ_{a}^{∨}(y) = (−1)^{n−1} ∏_{i≥0}(κ_{i}^{∨})^{a_{i}}.

(2) ξ_{a}^{∨}(y) ∈ S(𝔪^{(2n−j)(p−1)}) when a < p^{j+1}, 0 ≤ j ≤ 2n.

Proof. (1) Since 𝔪y = 0 and φ_{1}(y) = (−1)^{n−1} by corollary 5.3, we have

φ_{1}(f^{[a]}y) = κ(f^{[a]})φ_{1}(y) = (−1)^{n−1} ∏_{i≥0} κ_{i}(f)^{a_{i}},

which gives the first equality. Take the images in S(B_{2n}) to get the second one.

(2) For r ≥ 2 we have 𝔪^{r} = {g ∈ 𝔪^{2} | [g, 𝔪] ⊂ 𝔪^{r−1}}. In fact, if g ∉ 𝔪^{r} then [g, x_{i}] ∉ 𝔪^{r−1} for some i = 1, …, 2n. If now g ∈ 𝔪^{r} then g^{[p]} ∈ 𝔪^{2} and [g^{[p]}, 𝔪] ⊂ D_{g}^{p−1}(𝔪^{r−1}) ⊂ 𝔪^{r−1}. Hence 𝔪^{r} is closed under the [p]-map. Using Jacobson’s formula, we find that (f + g)^{[p]} ≡ f^{[p]} (mod 𝔪^{r−p+1}) whenever g ∈ 𝔪^{r} and r ≥ p. Iterating yields (f + g)^{[p^{i}]} ≡ f^{[p^{i}]} (mod 𝔪^{r−i(p−1)}) whenever g ∈ 𝔪^{r} and r > i(p − 1). Thus

κ_{i}(f + g) = κ((f + g)^{[p^{i}]}) = κ(f^{[p^{i}]}) = κ_{i}(f) when g ∈ 𝔪^{i(p−1)+1},

and so we may regard κ_{i} as a polynomial function on B_{2n}/𝔪^{i(p−1)+1}. For each 0 ≤ s ≤ 2n(p − 1) the orthogonal complement of 𝔪^{s} with respect to the invariant bilinear form on B_{2n} coincides with 𝔪^{2n(p−1)−s+1}. This is clear since φ_{1} vanishes on all the monomials in x_{1}, …, x_{2n} of degree less than 2n(p−1). Hence the isomorphism B_{2n} → B_{2n}^{∗} maps 𝔪^{(2n−i)(p−1)} onto the subspace of linear functions vanishing on 𝔪^{i(p−1)+1}. It follows κ_{i}^{∨} ∈ S(𝔪^{(2n−i)(p−1)}), and it remains to apply (1). □

Theorem 7.2. Let Z^{G_{[p]}} ⊂ Z be the subalgebra of G_{[p]}-invariant elements. Then φ_{a}^{∨} ∈ gr Z^{G_{[p]}} whenever a ≢ 0 (mod p), 0 < a < p^{n}.

Proof. Let u(L) and u(L_{0}) be the restricted universal enveloping algebras of L and L_{0}, respectively. Clearly ky = 𝔪^{2n(p−1)} is an L_{0}- and a G-submodule of B_{2n}. Consider the induced L-module Ind ky = u(L) ⊗_{u(L_{0})} ky. As G operates on L by automorphisms and L_{0} is stable under G, there is a canonical action of G on Ind ky. The embedding ky → B_{2n} extends to a homomorphism of L- and G-modules Ind ky → B_{2n}. As L = L_{0} + k∂_{1} + … + k∂_{2n}, the elements ∂_{1}^{r_{1}} ⋯ ∂_{2n}^{r_{2n}} ⊗ y with 0 ≤ r_{i} < p constitute a basis for Ind ky. Furthermore, the elements ∂_{1}^{r_{1}} ⋯ ∂_{2n}^{r_{2n}}(y) give all possible monomials in x_{1}, …, x_{2n}. In other words, Ind ky ≅ B_{2n}.

Put I = 𝔪^{(n+1)(p−1)}. Then [I, I] ⊂ 𝔪^{2(n+1)(p−1)−2} ⊂ 𝔪^{2n(p−1)}. Since [B_{2n}, B_{2n}] has zero intersection with 𝔪^{2n(p−1)}, we see that I is an abelian Lie subalgebra of B_{2n}. Hence there is a canonical isomorphism U(I) ≅ S(I) between the universal enveloping and the symmetric algebras of I. Suppose that a < p^{n}, so that ξ_{a}^{∨}(y) lies in S(I) by lemma 7.1. The embedding ξ_{a}^{∨} : ky → S^{a}(I) ↪ U_{a}(B_{2n}) of L_{0}- and G_{[p]}-modules extends then to a homomorphism B_{2n} ≅ Ind ky → U_{a}(B_{2n}) of Land G_{[p]}-modules. Denote by z_{a} the image of 1 ∈ B_{2n} under this homomorphism. Clearly z_{a} is L- and G_{[p]}-invariant. In particular z_{a} ∈ Z. Now the composite B_{2n} → U_{a}(B_{2n}) → S^{a}(B_{2n}) is a homomorphism of L-modules extending the same embedding ky → S^{a}(B_{2n}). It follows that this composite coincides with ξ_{a}^{∨}, and so the image of z_{a} in S^{a}(B_{2n}) equals ξ_{a}^{∨}(1). If a ≢ 0 (mod p) then ξ_{a}(1) = a^{−1}φ_{a} by proposition 3.2 and the definition of ξ_{a}, whence ξ_{a}^{∨}(1) = a^{−1}φ_{a}^{∨}. □

Remark. We thus have the formula z_{a} = ∂_{1}^{p−1} ⋯ ∂_{2n}^{p−1} · ξ_{a}^{∨}(y), computed in U(B_{2n}), for central elements in U(B_{2n}). If a < p then ξ_{a}^{∨}(y) = (κ_{1}^{∨})^{a} where κ_{1}^{∨} = y as an element of U(B_{2n}). Other elements are much less obvious. Koreshkov [9] constructed central elements up to degree p. He wrote them up explicitly as complicated linear combinations in a PBW basis, and it is difficult to compare his elements with our z_{a}. Using [18], Cor. 3.4, it is possible to prove that Hom_{L}(B_{2n}, S(B_{2n})) is a free S(B_{2n})^{L}-module with basis elements ξ_{a}^{∨} where 0 ≤ a < p^{n}. Since ξ_{a}^{∨}(1) = 0 when p|a, the method we used cannot give the remaining hypothetical central elements in U(B_{2n}).

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