Lobachevskii Journal of Mathematics

http://ljm.ksu.ru

Vol. 17, 2005, 229 – 256 \begin{matrix}c \\ ◯ \end{matrix} Vadim V. Shurygin, junior

Vadim V. Shurygin, junior

POISSON STRUCTURES ON WEIL BUNDLES

(submitted by B. Shapukov)

Abstract. In the present paper, we construct complete lifts of covariant and contravariant tensor fields from the smooth manifold M to its Weil bundle T^{A}M for the case of a Frobenius Weil algebra A. For a Poisson manifold (M, w) we show that the complete lift w^{C} of a Poisson tensor w is again a Poisson tensor on T^{A}M and that w^{C} is a linear combination of some ”basic” Poisson structures on T^{A}M induced by w. Finally, we introduce the notion of a weakly symmetric Frobenius Weil algebra A and we compute the modular class of (T^{A}M, w^{C}) for such algebras.

1. Preliminaries

A Weil algebra [5, 16] is an associative commutative algebra A with

unit over the field R of real numbers, which is of the form A = R ⊕ A^{∘},

where A^{∘} is a finite-dimensional maximal ideal, consisting of nilpotent

elements. In what follows we denote n = dim_{R} A^{∘}.

By A^{∘}^{r} we denote the rth power of A^{∘}. Let d_{k}(A) = dim_{R}A^{∘}^{k}/A^{∘}^{k+1}. The number d_{1}(A) is usually called the width of A. The positive integer

q defined by A^{∘}^{q} ≠ 0, A^{∘}^{q+1} = 0 is called the height of A.

The chain of embedded ideals

A ⊃ A^{∘} ⊃ A^{∘}^{2} ⊃ ⋯ ⊃ A^{∘}^{q} ⊃ 0

Weil functor.

Key words and phrases. Poisson structure, modular class, Weil algebra,

can be extended to the chain of ideals called the Jordan-Hölder composition series [16]

A ⊃ A^{∘} = I_{1} ⊃ I_{2} ⊃ … ⊃ I_{n} ⊃ 0,

where I_{a}/I_{a+1} is a 1-dimensional algebra with the zero multiplication. Here

A^{∘}^{k} = I_{1+d_{1}(A)+···+d_{k−1}(A)} for 2 ≤ k ≤ q.

This is the particular case of the general ring construction, see [12]. Using the Jordan-Hölder composition series one can choose the Jordan-Hölder basis

{e_{a}} = {e_{0}, eâ}, a = 0, 1, …, n = dim A^{∘}, â = 1, …, n,  (1.1)

in A such that e_{0} = 1 ∈ R, eâ ∈ Iâ, eâ ∉ I_{â+1}. In general, this basis is not unique. For X = x^{a}e_{a} = x^{0} + xâeâ ∈ A we set X^{∘} = xâeâ, then X = x^{0} + X^{∘}. Let δ^{a} be the coordinates of unit of A, i.e., 1 = δ^{a}e_{a}.

We denote by (γ_{ab}^{c}) the structural tensor of A with respect to the basis (1.1), i.e., e_{a}e_{b} = γ_{ab}^{c}e_{c}. It satisfies γ_{0a}^{b} = δ_{a}^{b} (the Kronecker’s delta) and γ_{ab̂}^{c} = 0 for a ≥ c. Since A is commutative and associative, it also satisfies the conditions γ_{ab}^{c} = γ_{ba}^{c} and

γ_{ab}^{c}γ_{ef}^{b} = γ_{ae}^{b}γ_{bf}^{c}.  (1.2)

The conditions of differentiability of a function f : U ⊂ A → A on a commutative associative algebra A (or, briefly, A-differentiability), usually called Scheffers’ equations, are (see [19]):

∂_{c}f^{b}γ_{ag}^{c} = γ_{ac}^{b}∂_{g}f^{c},  (1.3)

where ∂_{a}f^{b} = ∂f^{b}/∂x^{a}. Scheffers’ equations are equivalent to

∂_{a}f^{b} = γ_{ac}^{b}δ^{g}∂_{g}f^{c}.  (1.4)

For f : U ⊂ A^{m} → A, f : {X^{i} = x^{ia}e_{a}} ↦ f(X^{i}) = f^{b}(x^{ia})e_{b}, where A^{m} = A×⋯×A is the A-module of m-tuples of elements of A, Scheffers’ conditions of A-differentiability are of the form [19]:

∂_{ia}f^{b} = γ_{ac}^{b}δ^{g}∂_{ig}f^{c}.  (1.5)

If a function f satisfies (1.5), its differential can be represented in the form df = f_{i}dX^{i}, where f_{i} = δ^{a}∂_{ia}f is the partial derivative with respect to X^{i} ∈ A. Thus,

f_{i} = \frac{∂f}{∂X^{i}} = δ^{a} \frac{∂f}{∂x^{ia}} .  (1.6)

The functions f_{i}(X^{j}), i = 1, …, m, are also A-differentiable.

The following theorem (see [16]) describes the local structure of an A-differentiable map of the form F : U ⊂ A^{m} → A^{k} for a Weil algebra A. The natural epimorphism π_{0}^{q} : A^{m} → R^{m} determines the canonical A^{∘}^{m}-foliation on A^{m}. Recall that a smooth map f : M → N of a foliated manifold (M, F) is called projectable or basic if f is constant along the leaves of F.

Theorem 1.1 ([16]). 1) Let U ⊂ A^{m} be an open set. Then any A-smooth map Φ : U → A^{k} is of the form

X^{i′} = φ^{i′} + ∑_{|p|=1}^{q} \frac{1}{p!} \frac{D^{p}φ^{i′}}{Dx^{p}} X^{∘}^{p},  (1.7)

(where i = 1, …, m, i′ = 1, …, k, p = (p_{1}, …, p_{m}) is a multiindex of length m and X^{∘}^{p} = (X^{∘}^{1})^{p_{1}} … (X^{∘}^{m})^{p_{m}}) for some basic smooth map φ^{i′} : U → A which is projectable with respect to the canonical A^{∘}^{m}- foliation.

Definition. The map Φ : U → A^{k} given by the formulas (1.7) is called the analytic prolongation of the projectable map φ : U → A^{k}.

The analytic prolongation of a map φ will be denoted by φ^{A}.

Proposition 1.1 ([16]). The analytic prolongation has the following properties:

1^{∘}. (φ + ψ)^{A} = φ^{A} + ψ^{A}.

2^{∘}. (φ · ψ)^{A} = φ^{A} · ψ^{A}.

3^{∘}. (φ^{A} ∘ ψ)^{A} = φ^{A} ∘ ψ^{A}.

4^{∘}. (D^{p}φ/Dx^{p})^{A} = D^{p}φ^{A}/DX^{p} for φ : U ⊂ A^{n} → A.

We denote by Mf the category of smooth manifolds and by FM that of fibered manifolds. To each Weil algebra A there corresponds a functor T^{A} : Mf → FM called the Weil functor which maps a smooth manifold M to the fibered manifold π_{A} : T^{A}M → M called the Weil bundle (see [5, 16, 20]). A.P. Shirokov proved [15] that T^{A}M carries the structure of a smooth manifold over A. Weil functors preserve products, i.e., T^{A}(M × N) ≅ T^{A}M×T^{A}N. Moreover, under some additional conditions (locality and regularity) each product preserving bundle functor F : Mf → FM is equivalent to a Weil functor T^{A} for a Weil algebra A [5].

A Weil algebra A is said to be Frobenius (cf. [19, 2]) if there exists a nondegenerate bilinear form q : A × A → R, satisfying the following condition of associativity:

q(XY, Z) = q(X, Y Z) for any X, Y, Z ∈ A.  (1.8)

Frobenius algebras play an important role in the theory of smooth manifolds over algebras in constructing realizations of tensor operations [8].

With respect to the basis (1.1) the condition (1.8) is written as

q_{bc}γ_{ef}^{c} = γ_{be}^{c}q_{cf}.  (1.9)

We will call q a Frobenius form. A Frobenius form is not unique (if exists). For a Frobenius algebra A we define the Frobenius covector p : A → R by p(X) := q(X, 1). Its coordinates with respect to the basis (1.1) satisfy

p_{a}γ_{bc}^{a} = q_{bc}.  (1.10)

Contracting (1.10) with δ^{c} and taking into consideration that δ^{c} = δ_{0}^{c} (the Kronecker’s delta) with respect to the basis (1.1), we obtain

p_{b} = q_{bc}δ^{c}.  (1.11)

From (1.8) and (1.10) it easily follows that

q(X, Y) = p(XY) for each X, Y ∈ A.  (1.12)

We denote the set of all Frobenius covectors on A by A_{F r}^{∗}.

Example 1.1. The important example of a Frobenius Weil algebra is the algebra of dual numbers D = R(ε) = {x_{0} + x_{1}ε | x_{0}, x_{1} ∈ R, ε^{2} = 0}. To this algebra there corresponds the tangent bundle functor: T^{R(ε)}M = T M.

Example 1.2. Another example is the algebra D^{n} = R(ε^{n}) = {x_{0} + x_{1}ε + ⋯ + x_{n−1}ε^{n−1} | x_{i} ∈ R, ε^{n} = 0}. of plural numbers which is a generalization of the previous one. To this algebra there corresponds the functor of jet bundle of higher order.

If A and B are Frobenius algebras, then A ⊗ B is also a Frobenius algebra (see, e.g. [19]).

In what follows we assume all Weil algebras under consideration to be Frobenius algebras.

2. The structure of a Frobenius Weil algebra

Let A be a Frobenius Weil algebra of height q and let n = dim A^{∘}. Let us choose a Jordan-Hölder basis (1.1) in A.

Lemma 2.1. dim A^{∘}^{q} = 1, i.e., A^{∘}^{q} = I_{n}.

Proof. On the contrary, suppose that dim_{R}A^{∘}^{q} ≥ 2. Then at least two last elements e_{n−1} and e_{n} of a basis (1.1) belong to A^{∘}^{q} and, consequently, for any â = 1, …, n there hold eâe_{n−1} = eâe_{n} = 0. Hence for any c = 0, …, n the matrix ∥γ_{ab}^{c}∥ contains zeros everywhere in two last columns (with the numbers n − 1 and n) except for the first row. Then for each covector (p_{c}) the matrix ∥q_{ab}∥ = ∥p_{c}γ_{ab}^{c}∥ also contains zeros everywhere in two last columns except the first row, and thus is degenerate. Contradiction. By this reason, dim_{R}A^{∘}^{q} = 1, which implies A^{∘}^{q} = I_{n} or, equivalently, A^{∘}^{q} = R · e_{n}. □

Lemma 2.2. For each Frobenius covector p on A, its last component p_{n} is not zero.

Proof. From the equalities 1 · e_{n} = e_{0}e_{n} = e_{n} and eâe_{n} = 0 for each â = 1, …, n it follows, that for each c = 0, …, n − 1 and for each b there holds

γ_{bn}^{c} = 0,  γ_{0n}^{n} = 1,  γ_{ân}^{n} = 0.

Hence, for each c = 0, …, n − 1 the last column of the matrix ∥γ_{ab}^{c}∥ contains only zeros and the last column of the matrix ∥γ_{ab}^{n}∥ is

T (1, 0, …, 0).

Therefore the last column of ∥q_{ab}∥ = ∥p_{c}γ_{ab}^{c}∥ is

T (p_{n}, 0, …, 0),  (2.1)

hence p_{n} ≠ 0. □

Remark 2.1. One can prove both lemmas without using coordinates. Indeed, denote

Ann A^{∘} := {X ∈ A^{∘} | X · A^{∘} = 0}.

Let 0 ≠ X ∈ Ann A^{∘}, then for each Y = y^{0} + Y^{∘} we have XY = Xy^{0}, whence q(X, Y) = p(XY) = y^{0}p(X). From the degeneracy of q it follows that p(X) ≠ 0. Therefore Ann A^{∘} ∩ ker p = 0 which implies that dim Ann A^{∘} ≤ 1. But, clearly, 0 ≠ A^{∘}^{q} ⊂ Ann A^{∘}, hence dim A^{∘}^{q} =

dim Ann A^{∘} = 1.

Let us denote ∥h_{ab}∥ := ∥γ_{ab}^{n}∥.

Lemma 2.3. The Jordan-Hölder basis (1.1) can be chosen in such a way that the matrix ∥h_{ab}∥ is nondegenerate.

Proof. Let us choose any Jordan-Hölder basis (1.1). If p_{(0)} = (0, …, 0, 1) is a Frobenius covector, then the matrix ∥h_{ab}∥ is nondegenerate. Assume the contrary and consider any p ∈ A_{F r}^{∗}. Without loss of generality we may assume that p_{n} = 1 (otherwise consider \frac{1}{p_{n}} p).

1) We prove that the first component p_{0} of p may be taken to be zero. Indeed, in the matrix ∥q_{ab}∥ only the element q_{00} depends on p_{0}:

∥q_{ab}∥ = ( \begin{matrix}p_{0} & ∗ & … & ∗ & 1 \\ ∗ & ∗ & … & ∗ & 0 \\ ⋮ & ⋮ & ... & ⋮ & ⋮ \\ ∗ & ∗ & … & ∗ & 0 \\ 1 & 0 & … & 0 & 0 \end{matrix} )

(∗ denotes the elements which do not depend on p_{0}.) The cofactor of q_{00} = p_{0} contains only zeros in the last column, hence it is zero itself. Thus, the determinant det ∥q_{ab}∥ does not depend on p_{0} and we may assume that p_{0} = 0.

2) By the assumption, p_{(0)} = (0, …, 0, 1) is not a Frobenius covector, therefore there exists at least one c, 1 ≤ c ≤ n − 1, such that p_{c} ≠ 0. Consider another basis {e′_{a}} in A:

e′_{0} = e_{0} = e_{0}−p_{0}e_{n},  e′_{a} = e_{a}−p_{a}e_{n}, a = 1, 2, …, n−1,  e′_{n} = e_{n}.

(2.2)

One can easily see that {e′_{a}} is also a Jordan-Hölder basis. Since e_{a}e_{n} = 0 for each a ≥ 1, the structural constants γ′_{ab}^{c} will have the following form with respect to this basis: γ′_{ab}^{c} = γ_{ab}^{c} for c = 1, 2, …, n − 1 and γ′_{ab}^{n} = γ_{ab}^{n} + γ_{ab}^{d}p_{d}, where the summation over d is taken from 1 to n − 1. Thus, ∥γ′_{ab}^{n}∥ equals to ∥q_{ab}∥ and therefore is nondegenerate. □

In what follows we will suppose the Jordan-Hölder basis to be chosen in such a way that ∥h_{ab}∥ is nondegenerate and we will call p_{(0)} = (0, …, 0, 1) the standard Frobenius covector.

Remark 2.2. One can also give a noncoordinate proof of Lemma 2.3. Let p^{∘} ∈ A^{∗} be defined by p^{∘}(X) := x^{0}.

1) We show that if p ∈ A_{F r}^{∗} then p̃ := p − p(1)p^{∘} also is a Frobenius covector. Suppose the contrary. Then there exists X ∈ A such that p̃(XY) = 0 for any Y ∈ A. This means that x^{0}p(Y^{∘})+y^{0}p(X^{∘})+p(X^{∘}Y^{∘}) = 0. Let Z ∈ A^{∘}^{q} be an element such that p(Z) = 1 (in terms of the Jordan-Hölder basis, Z = e_{n}) and let X̃ = X −x^{0}Z. Then for any Y ∈ A one has

X̃Y = XY − x^{0}y^{0}Z = x^{0}y^{0} + x^{0}Y^{∘} + y^{0}X^{∘} + X^{∘}Y^{∘} − x^{0}y^{0}Z. One can easily

see that p(X̃Y) = 0, which contradicts to the fact that p is a Frobenius covector.

This means that we can deform any p ∈ A_{F r}^{∗} in such a way that R ⊂ ker p.

2) Now, by Lemma 2.2 or Remark 2.1, one has A = ker p ⊕ A^{∘}^{q}. We define a bilinear form h(X, Y) on A to be the projection of XY onto A^{∘}^{q} along ker p. This form is nondegenerate. Indeed, let X ∈ A be such that h(X, Y) = 0 for any Y ∈ A. We write XY = U + Z, where U ∈ ker p, Z ∈ A^{∘}^{q}. Then 0 = h(X, Y) = Z, hence p(XY) = p(Z) = 0 which contradicts to Lemma 2.2.

Thus, without loss of generality, we may assume the matrix ∥h_{ab}∥ = ∥γ_{ab}^{n}∥ to be nondegenerate. This matrix has the following form:

∥h_{ab}∥ = ( \begin{matrix}0 & 0 & ⋯ & 0 & 1 \\ 0 & & & & 0 \\ ⋮ & & B & & ⋮ \\ 0 & & & & 0 \\ 1 & 0 & ⋯ & 0 & 0 \end{matrix} ) ,  (2.3)

where B denotes the nonsingular square block. The inverse matrix is of the same form:

∥h^{ab}∥ = ( \begin{matrix}0 & 0 & ⋯ & 0 & 1 \\ 0 & & & & 0 \\ ⋮ & & B^{−1} & & ⋮ \\ 0 & & & & 0 \\ 1 & 0 & ⋯ & 0 & 0 \end{matrix} )  (2.4)

This allows us to introduce another basis in A: we put \overline{e}^{a} = h^{ab}e_{b}, then e_{a} = h_{ab}\overline{e}^{b}. Denote by \overline{γ}_{c}^{ab} the structural constants of A with respect to the basis \{\overline{e}^{a}\}, i.e. \overline{e}^{a}\overline{e}^{b} = \overline{γ}_{c}^{ab}\overline{e}^{c}. Clearly,

\overline{γ}_{s}^{ab} = h^{ak}h^{bℓ}h_{cs}γ_{kℓ}^{c}  (2.5)

and

γ_{kℓ}^{c} = h_{ak}h_{bℓ}h^{cs}\overline{γ}_{s}^{ab}.  (2.6)

Since δ^{a} = δ_{0}^{a} with respect to the basis (1.1), we have h_{cs}δ^{s} = γ_{cs}^{n}δ^{s} = γ_{c0}^{n}, which is not zero only for c = n and γ_{n0}^{n} = 1. Therefore \overline{γ}_{0}^{ab} = \overline{γ}_{s}^{ab}δ^{s} = h^{ak}h^{bℓ}γ_{kℓ}^{c}h_{cs}δ^{s} = h^{ak}h^{bℓ}γ_{kℓ}^{n} = h^{ak}h^{bℓ}h_{kℓ} = h^{ak}δ_{k}^{b} = h^{ab}. Thus,

h^{ab} = \overline{γ}_{0}^{ab}.

Moreover, it is clear from (2.4) that \overline{e}^{n} = e_{0}, hence \overline{e}^{a}\overline{e}^{n} = \overline{e}^{a}, which implies that

\overline{γ}_{c}^{an} = δ_{c}^{a}.  (2.7)

From the formula (1.2) it follows that γ_{sr}^{c}h_{cd} = γ_{sr}^{c}γ_{cd}^{n} = γ_{ds}^{c}γ_{cr}^{n} = γ_{ds}^{c}h_{cr} = γ_{dr}^{c}γ_{cs}^{n} = γ_{dr}^{c}h_{cs}. Thus,

γ_{sr}^{c}h_{cd} = γ_{ds}^{c}h_{cr} = γ_{dr}^{c}h_{cs}.  (2.8)

The tensors γ_{ab}^{c} and \overline{γ}_{c}^{ab} are also related with the following formulas. We have γ_{ka}^{c}h^{kℓ} = h^{sc}h_{ba}h_{dk}h^{kℓ}\overline{γ}_{s}^{db} (by (2.6)) = h^{sc}h_{ba}\overline{γ}_{s}^{ℓb}, hence h_{ba}\overline{γ}_{s}^{ℓb} = h_{sc}γ_{ka}^{c}h^{kℓ} = γ_{sa}^{c}h_{ck}h^{kℓ} (by (2.8)) = γ_{sa}^{ℓ}. Thus,

γ_{sa}^{ℓ} = h_{ba}\overline{γ}_{s}^{ℓb},  (2.9)

whence

\overline{γ}_{s}^{bc} = h^{ab}γ_{sa}^{c}.  (2.10)

Let p (p_{n} ≠ 0) be an arbitrary Frobenius covector on A and ∥q_{ab}∥ = ∥γ_{ab}^{c}p_{c}∥. Let us find the explicit form of the inverse matrix ∥q^{ab}∥. Denote ∥\overline{q}^{ab}∥ = ∥\overline{γ}_{c}^{ab}t^{c}∥, where (t^{c}) are to be defined later.

From (2.7) it follows that the last column of ∥\overline{q}^{ab}∥ is

T (t^{0}, t^{1}, …, t^{n}).  (2.11)

The system of linear equations q_{ab}x^{b} = δ_{a}^{n} on (x^{b}) has the unique solution. We define (t^{b}) to be this solution:

q_{ab}t^{b} = γ_{ab}^{k}p_{k}t^{b} = δ_{a}^{n}.  (2.12)

Equivalently, t = (t^{b}) is defined by q(t, ·) = p_{(0)}( ·). Therefore, by (2.11), the last column of the matrix ∥q_{ab}\overline{q}^{bc}∥ coincides with the last column of the unit matrix.

Let us show that this is true for any other column, i.e., that for each c = 0, 1, …, n − 1,

q_{ab}\overline{q}^{bc} = δ_{a}^{c}.  (2.13)

We need to check that p_{k}γ_{ab}^{k}t^{s}\overline{γ}_{s}^{bc} = p_{k}γ_{ab}^{k}γ_{sr}^{c}h^{br}t^{s} = δ_{a}^{c}. Contracting the left-hand side with h_{cd} and using (2.8) yields p_{k}γ_{ab}^{k}γ_{sr}^{c}h^{br}t^{s}h_{cd} = p_{k}t^{s}h^{br}γ_{ab}^{k}γ_{ds}^{c}h_{cr} = p_{k}t^{s}γ_{ab}^{k}γ_{ds}^{b} = p_{k}t^{s}γ_{bs}^{k}γ_{ad}^{b} = δ_{b}^{n}γ_{ad}^{b} = γ_{ad}^{n} = h_{ad}. Since the contraction of δ_{a}^{c} with h_{cd} also gives h_{ad} and h_{ad} is nondegenerate, the relation (2.13) holds true. Thus, the inverse matrix ∥q^{ab}∥ has the form ∥\overline{γ}_{c}^{ab}t^{c}∥, where t = (t^{c}) is defined by (2.12).

We also show that (p_{a}) is defined uniquely by (t^{c}). It is sufficient to prove that det ∥γ_{ab}^{c}t^{b}∥ ≠ 0. Contracting with h^{ad} gives t^{b}γ_{ab}^{c}h^{ad} = t^{b}\overline{γ}_{b}^{cd} = \overline{q}^{cd}. The last matrix is nondegenerate, hence the result follows. It follows also that if det ∥γ_{ab}^{c}t^{b}∥ ≠ 0 for some t = (t^{b}) then the corresponding covector p ∈ A_{F r}^{∗}. Indeed, in this case ∥\overline{q}^{cd}∥ = ∥q^{cd}∥ is nondegenerate, hence ∥q_{cd}∥ is also nondegenerate.

The last row of ∥q_{ab}∥ has the form (p_{n}, 0, …, 0). Its product with the last column of ∥q^{bc}∥ equals 1 by the definition of the inverse matrix and also equals p_{n}t^{0} by (2.1). Therefore, p_{n}t^{0} = 1, in particular, t^{0} ≠ 0.

Thus, we proved the following

Proposition 2.1. For any Frobenius covector p (p_{n} ≠ 0) on A the matrix ∥q^{ab}∥ is of the form q^{ab} = \overline{γ}_{c}^{ab}t^{c}, where the vector t = (t^{c}) and covector p uniquely define each other by (2.12), moreover, t^{0}p_{n} = 1.

In particular, if p = p_{(0)} is the standard Frobenius covector, i.e., p_{a} = δ_{a}^{n}, then t^{c} = δ_{0}^{c}. Indeed, in this case γ_{ab}^{c}p_{c} = γ_{ab}^{n} = h_{ab}, hence, h_{ab}t^{b} = δ_{n}^{a} by (2.12). Denote by t′ the column ^{T} (t^{1}, …, t^{n−1}). From (2.3) we obtain t^{n} = 0, Bt′ = 0, t^{0} = 1. Since det B ≠ 0, each of t^{1}, …, t^{n−1} is equal to zero.

One can represent the ”multiplication table” of A with respect to the basis (1.1) as follows. By A^{∘}^{s}, s = 1, 2, …, q − 1, in the first column and the first row we denote the d_{s}(A) elements of (1.1) which lie in A^{∘}^{s} \ A^{∘}^{s+1} (or, equivalently, project to the basis of A^{∘}^{s}/A^{∘}^{s+1} under the natural epimorphism A^{∘}^{s} → A^{∘}^{s+1}). The product of two such elements of (1.1), one of them lying in the A^{∘}^{k}-column, another in the A^{∘}^{ℓ}-row, belongs to A^{∘}^{k+ℓ}, thus we write A^{∘}^{k+ℓ} in the intersection of A^{∘}^{k}-column and A^{∘}^{ℓ}-row. The whole table now has the following block structure:

1  A^{∘}  A^{∘}^{2}  … A^{∘}^{k} … A^{∘}^{q−2} A^{∘}^{q−1} e_{n}
1  1  A^{∘}  A^{∘}^{2}  … A^{∘}^{k} … A^{∘}^{q−2}  A^{∘}^{q−2} A^{∘}^{q}
A^{∘}  A^{∘}  A^{∘}^{2}  A^{∘}^{3}  … A^{∘}^{k+1} … A^{∘}^{q−1}  A^{∘}^{q}  0
A^{∘}^{2}  A^{∘}^{2}  A^{∘}^{3}  A^{∘}^{4}  … A^{∘}^{k+2} … A^{∘}^{q}  0  0
⋮  …  …  …  . . . . . . ...  ...  ...  ⋮
A^{∘}^{q−k} A^{∘}^{q−k} A^{∘}^{q−k+1} A^{∘}^{q−k+2} … A^{∘}^{q}  ...  ...  …  ⋮
⋮  …  …  …  ... ... ... . . .  …  ⋮
A^{∘}^{q−2} A^{∘}^{q−2}  A^{∘}^{q−1}  A^{∘}^{q}  0  ... . . .  0  0  0
A^{∘}^{q−1} A^{∘}^{q−1}  A^{∘}^{q}  0  ... . . . . . .  0  0  0
e_{n}  A^{∘}^{q}  0  0  . . . . . . . . .  0  0  0

The secondary diagonal of this table contains blocks A^{∘}^{q} consisting of real multiples of e_{n}. Hence all the matrices ∥γ_{ab}^{c}∥, c = 0, 1, …, n − 1, contain zeros in these blocks and the matrix ∥γ_{ab}^{n}∥ = ∥h_{ab}∥ has the following block structure:

1  A^{∘}  A^{∘}^{2}  … A^{∘}^{q−2}  A^{∘}^{q−1}  e_{n}
1  0  0  0  …  0  0  1
A^{∘}  0  *  *  …  *  B_{1,q−1}  0
A^{∘}^{2}  0  *  *  ... B_{2,q−2}  0  0
⋮  ⋮  …  ...  ...  ...  ⋮  ⋮
A^{∘}^{q−2}  0  *  B_{q−2,2}  ...  0  0  0
A^{∘}^{q−1}  0  B_{q−1,1}  0  …  0  0  0
e_{n}  1  0  0  …  0  0  0

Definition. We will say that the Frobenius Weil algebra A of height q is weakly symmetric, if d_{k}(A) = d_{q−k}(A) for each k = 1, 2, …, q − 1.

For a weakly symmetric algebra all the blocks B_{k,q−k}, k = 1, 2, …, q−1, are squares. One can easily see that in this case

det ∥q_{ab}∥ = det ∥p_{n} · γ_{ab}^{n}∥ = p_{n}^{n+1} det B_{1,q−1} · · · · · det B_{q−1,1}, (2.14)

therefore, all the blocks B_{k,q−k}, k = 1, 2, …, q − 1, are nondegenerate. From (2.14) it follows that for this algebra each p ∈ A^{∗} such that p_{n} ≠ 0, is a Frobenius covector.

Example 2.1. The simplest example of weakly symmetric Weil algebra is the algebra of plural numbers R(ε^{n}). For this algebra q = n and the Jordan-Hölder basis is e_{a} = ε^{a}, a = 0, 1, …, n. Therefore d_{s}(R(ε^{n})) = 1 for each s = 1, …, n and all the blocks B_{k,n−k} consist of one element each. Clearly,

∥γ_{ab}^{n}∥ = ( \begin{matrix}0 & 0 & … & 0 & 1 \\ 0 & 0 & … & 1 & 0 \\ ⋮ & ... & ... & ... & ⋮ \\ 0 & 1 & 0 & … & 0 \\ 1 & 0 & … & … & 0 \end{matrix} ) .

Example 2.2. It is clear that every Frobenius Weil algebra A of height q = 2 is weakly symmetric. For the elements {e_{a}}, a = 0, …, n, of Jordan-Hölder basis we have e_{1} …, e_{n−1} ∈ A^{∘}, e_{n} ∈ A^{∘}^{2}. Therefore, e_{a}e_{b} = λ_{ab}e_{n}, a, b = 1, …, n − 1, and e_{a}e_{n} = 0, a = 1, …, n. Hence, for any Frobenius covector p = (p_{a}) (p_{n} ≠ 0)

∥q_{ab}∥ = ( \begin{matrix}p_{0} & p_{1} & p_{2} & … & p_{n−1} & p_{n} \\ p_{1} & λ_{11} p_{n} & λ_{12} p_{n} & … & λ_{1,n−1} p_{n} & 0 \\ p_{2} & λ_{21} p_{n} & λ_{22} p_{n} & … & λ_{2,n−1} p_{n} & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ p_{n−1} & λ_{n−1,1} p_{n} & λ_{n−1,2} p_{n} & … & λ_{n−1,n−1} p_{n} & 0 \\ p_{n} & 0 & … & … & 0 & 0 \end{matrix} )

and ∥q_{ab}∥ is nondegenerate if and only if Λ = ∥λ_{ab}∥ is nondegenerate. We will denote this Weil algebra by A(Λ, n).

Proposition 2.2. Let A and B be two weakly symmetric Frobenius Weil algebras. Then A ⊗ B is also weakly symmetric.

Proof. If {e_{a}} and {f_{α}} are Jordan-Hölder bases of A and B respectively, then {e_{a} ⊗f_{α}} is the Jordan-Hölder basis of A⊗B. It follows that the height of A ⊗ B equals q_{1} + q_{2}, where q_{1} and q_{2} are the heights of A and B, respectively. It can be easily seen that d_{k}(A ⊗ B) = d_{k}(A) + d_{k−1}(A)d_{1}(B)+⋯+d_{1}(A)d_{k−1}(B)+d_{k}(B). Moreover, d_{q_{1}+q_{2}−k}(A⊗B) = d_{q_{1}−k}(A) + d_{q_{1}−k+1}(A)d_{q_{2}−1}(B) + ⋯ + d_{q_{1}−1}(A)d_{q_{2}−k+1}(B) + d_{q_{2}−k}(B), which coincides with d_{k}(A ⊗ B). □

3. Complete lifts of tensor fields

Let A be a Weil algebra of height q and let n = dim A^{∘}. In what follows we will assume that A is a Frobenius algebra. As before, we denote the Frobenius covector by p = (p_{c}) and the Frobenius form by q = (q_{ab}).

Let M be an m-dimensional smooth manifold. Then T^{A}M is an mdimensional A-smooth manifold and for each x ∈ T^{A}M the tangent space T_{x}T^{A}M is an m-dimensional A-module. Thus, we can consider A-tensors at any point x ∈ T^{A}M and A-smooth tensor fields on T^{A}M (see [19]). In what follows, we assume all the manifolds and the maps between manifolds to be of class C^{∞}.

We denote the algebra of smooth functions on M by C^{∞}(M), the space of covariant tensors on M by T^{∗}(M) and the space of skew-symmetric contravariant tensors (multivector fields) on M by V^{∗}(M). By | · | we denote the degree of a tensor field, i.e., |ξ| = s if ξ ∈ T^{s}(M) and |u| = s if u ∈ V^{s}(M).

In this part of the paper we construct complete lifts of covariant and contravariant tensor fields from M to the Weil bundle T^{A}M.

Let (x^{i}) = (x^{1}, …, x^{m}) be local coordinates on M. We will enumerate the corresponding local coordinates on T^{A}M by the double index ia: (x^{ia}), i = 1, …, m, a = 0, …, n, where we identify x^{i0} = x^{i}.

Let ξ ∈ T^{k}(M) be a tensor field of type (k, 0) on M. In local coordinates

ξ = ξ_{i_{1}...i_{k}}dx^{i_{1}} ⊗ ⋯ ⊗ dx^{i_{k}}.

Let Ξ_{i_{1}...i_{k}} = (ξ_{i_{1}...i_{k}})^{A} be analytic prolongations of the functions ξ_{i_{1}...i_{k}}. We multiply these A-valued functions by e_{a_{1}} … e_{a_{k}}, where {e_{a}} is a Jordan-Hölder basis (1.1). Let Ξ_{i_{1}...i_{k}}e_{a_{1}} … e_{a_{k}} = Ξ_{i_{1}a_{1}...i_{k}a_{k}}^{b}e_{b}, where Ξ_{i_{1}a_{1}...i_{k}a_{k}}^{b} ∈ R. Denote

ξ_{i_{1}a_{1}...i_{k}a_{k}} := Ξ_{i_{1}a_{1}...i_{k}a_{k}}^{b}p_{b}.  (3.1)

We define the complete lift ξ^{C} of ξ by

ξ^{C} = ξ_{A}^{C} := ξ_{i_{1}a_{1}...i_{k}a_{k}}dx^{i_{1}a_{1}} ⊗ ⋯ ⊗ dx^{i_{k}a_{k}}.

If {e_{a′}} is another basis in A and e_{a′} = τ_{a′}^{a}e_{a}, then, obviously, dx^{ia′} = τ_{a}^{a′}dx^{ia}. Thus, our definition does not depend on a choice of a basis in A.

Proposition 3.1. ξ^{C} is a well-defined tensor field of type (k, 0) on T^{A}M. If φ : N → M is a smooth map, then

(T^{A}φ)^{∗}(ξ^{C}) = (φ^{∗}ξ)^{C}.

Proof. Let N be another smooth manifold with local coordinates (y^{α}) and φ : N → M be a smooth map which has the form x^{i} = φ^{i}(y^{α}) with respect to these coordinates. Denote θ = φ^{∗}ξ, then

θ_{α_{1}...α_{k}} = \frac{∂φ^{i_{1}}}{∂y^{α_{1}}} … \frac{∂φ^{i_{k}}}{∂y^{α_{k}}} ξ_{i_{1}...i_{k}}.

Let Θ_{α_{1}...α_{k}} be analytic prolongations of the components θ_{α_{1}...α_{k}} and let Φ^{i} be the analytic prolongations of the maps φ^{i}. Note that Φ^{i}(Y^{α}) = φ^{ia}(Y^{α})e_{a} = φ^{ia}(y^{αb})e_{a} is the local representation of the map T^{A}φ : T^{A}N → T^{A}M, considered as an A-smooth map and the functions φ^{ia}(y^{αb}) are the local representations of T^{A}φ, considered as a map between real smooth manifolds. From Proposition 1.1 it follows that

Θ_{α_{1}...α_{k}} = \frac{∂Φ^{i_{1}}}{∂Y^{α_{1}}} … \frac{∂Φ^{i_{k}}}{∂Y^{α_{k}}} Ξ_{i_{1}...i_{k}}.

Since δ^{a} = δ_{0}^{a} with respect to the Jordan-Hölder basis, (1.6) implies that

\frac{∂Φ^{i_{ℓ}}}{∂Y^{α_{j}}} = \frac{∂Φ^{i_{ℓ}}}{∂y^{α_{j}0}} = \frac{∂Φ^{i_{ℓ}}}{∂y^{α_{j}}} .

Thus

Θ_{α_{1}...α_{k}}e_{b_{1}} … e_{b_{k}} = \frac{∂φ^{i_{1}c_{1}}}{∂y^{α_{1}}} e_{c_{1}} … \frac{∂φ^{i_{k}c_{k}}}{∂y^{α_{k}}} e_{c_{k}}U_{i_{1}...i_{k}}e_{b_{1}} … e_{b_{k}}.  (3.2)

But e_{b_{1}}e_{c_{1}} = γ_{b_{1}c_{1}}^{a_{1}}e_{a_{1}}, …, e_{b_{k}}e_{c_{k}} = γ_{b_{k}c_{k}}^{a_{k}}e_{a_{k}}, therefore the right-hand side of (3.2) takes the form

\frac{∂φ^{i_{1}c_{1}}}{∂y^{α_{1}}} γ_{b_{1}c_{1}}^{a_{1}} … \frac{∂φ^{i_{k}c_{k}}}{∂y^{α_{k}}} γ_{b_{k}c_{k}}^{a_{k}}Ξ_{i_{1}...i_{k}}e_{a_{1}} … e_{a_{k}}.

From Scheffers’ equations (1.5) it follows that

\frac{∂φ^{i_{1}c_{1}}}{∂y^{α_{1}}} γ_{b_{1}c_{1}}^{a_{1}} = \frac{∂φ^{i_{1}a_{1}}}{∂y^{α_{1}b_{1}}} ,  …,  \frac{∂φ^{i_{k}c_{k}}}{∂y^{α_{k}}} γ_{b_{k}c_{k}}^{a_{k}} = \frac{∂φ^{i_{k}a_{k}}}{∂y^{α_{k}b_{k}}} ,

hence,

Θ_{α_{1}...α_{k}}e_{b_{1}} … e_{b_{k}} = \frac{∂φ^{i_{1}a_{1}}}{∂y^{α_{1}b_{1}}} … \frac{∂φ^{i_{k}a_{k}}}{∂y^{α_{k}b_{k}}} Ξ_{i_{1}...i_{k}}e_{a_{1}} … e_{a_{k}}.

Then, by (3.1), we have

θ_{α_{1}b_{1}...α_{k}b_{k}} = \frac{∂φ^{i_{1}a_{1}}}{∂y^{α_{1}b_{1}}} … \frac{∂φ^{i_{k}a_{k}}}{∂y^{α_{k}b_{k}}} ξ_{i_{1}a_{1}...i_{k}a_{k}},  (3.3)

i.e. (T^{A}φ)^{∗}(ξ^{C}) = θ^{C} = (φ^{∗}ξ)^{C}.

In particular, if φ is a local diffeomorphism (coordinate transformation) of M, i.e., x^{j′} = φ^{j′}(x^{i}), then

ξ_{i_{1}a_{1}...i_{k}a_{k}} = \frac{∂φ^{j′_{1}b_{1}}}{∂x^{i_{1}a_{1}}} … \frac{∂φ^{j′_{k}b_{k}}}{∂x^{i_{k}a_{k}}} ξ_{j′_{1}b_{1}...j′_{k}b_{k}}.

This implies that ξ_{i_{1}a_{1}...i_{k}a_{k}} is a well-defined tensor field of type (k, 0) on T^{A}M. □

In order to construct the complete lifts of contravariant tensor fields we need to take another basis in A: e^{a} = q^{ab}e_{b}, then e_{a} = q_{ab}e^{b}.

Let now u ∈ V^{k}(M) be a skew-symmetric contravariant tensor field on M; with respect to the local coordinates,

u = u^{i_{1}...i_{k}} \frac{∂}{∂x^{i_{1}}} ∧ ⋯ ∧ \frac{∂}{∂x^{i_{k}}} .

Let U^{i_{1}...i_{k}} = (u^{i_{1}...i_{k}})^{A} be the analytic prolongations of u^{i_{1}...i_{k}} and let U^{i_{1}...i_{k}}e^{a_{1}} … e^{a_{k}} = U_{b}^{i_{1}a_{1}...i_{k}a_{k}}e^{b}, where U_{b}^{i_{1}a_{1}...i_{k}a_{k}} ∈ R. Denote

u^{i_{1}a_{1}...i_{k}a_{k}} := U_{b}^{i_{1}a_{1}...i_{k}a_{k}}δ^{b}.  (3.4)

We define the complete lift u^{C} of u by

u^{C} = u_{A}^{C} := u^{i_{1}a_{1}...i_{k}a_{k}} \frac{∂}{∂x^{i_{1}a_{1}}} ∧ ⋯ ∧ \frac{∂}{∂x^{i_{k}a_{k}}} .

One can easily check that this definition also does not depend on the choice of a basis in A.

Proposition 3.2. u^{C} is a well-defined skew-symmetric contravariant tensor field of degree k on T^{A}M. If φ : M → N is a smooth map and u is φ-related to a tensor filed v on N, then u^{C} is (T^{A}φ)-related with v^{C}.

Proof. Let N be another smooth manifold with local coordinates (y^{α}) and φ : M → N be a smooth map which is given by y^{α} = φ^{α}(x^{i}) with respect to the local coordinates. Let v ∈ V^{k}(N) be another tensor field and let u and v be φ-related ([13, 14]). Then, in local coordinates we have

v^{α_{1}...α_{k}} = \frac{∂φ^{α_{1}}}{∂x^{i_{1}}} … \frac{∂φ^{α_{k}}}{∂x^{i_{k}}} u^{i_{1}...i_{k}}

Denote the analytic prolongations of v_{α_{1}...α_{k}} by V_{α_{1}...α_{k}} and the analytic prolongations of the maps φ^{i} by Φ^{i}. Then

V^{α_{1}...α_{k}} = \frac{∂Φ^{α_{1}}}{∂X^{i_{1}}} … \frac{∂Φ^{α_{k}}}{∂X^{i_{k}}} U^{i_{1}...i_{k}},

and

V^{α_{1}...α_{k}}e^{b_{1}} … e^{b_{k}} = \frac{∂φ^{α_{1}c_{1}}}{∂x^{i_{1}}} e_{c_{1}} … \frac{∂φ^{α_{k}c_{k}}}{∂x^{i_{k}}} e_{c_{k}}U^{i_{1}...i_{k}}e^{b_{1}} … e^{b_{k}}.

We have e^{b_{r}}e_{c_{r}} = q^{b_{r}s_{r}}e_{s_{r}}e_{c_{r}} = q^{b_{r}s_{r}}γ_{s_{r}c_{r}}^{d_{r}}e_{d_{r}} = q^{b_{r}s_{r}}γ_{s_{r}c_{r}}^{d_{r}}q_{a_{r}d_{r}}e^{a_{r}} for each r = 1, …, k. Let us show that q^{bs}γ_{sc}^{d}q_{ad} = γ_{ac}^{b}. Indeed, the contraction of the left-hand side with q_{bf} gives q^{bs}γ_{sc}^{d}q_{ad}q_{bf} = γ_{fc}^{d}q_{ad}, which coincides with γ_{ac}^{b}q_{bf} by (1.9). Thus,

V^{α_{1}...α_{k}}e^{b_{1}} … e^{b_{k}} = \frac{∂φ^{α_{1}c_{1}}}{∂x^{i_{1}}} e_{c_{1}} … \frac{∂φ^{α_{k}c_{k}}}{∂x^{i_{k}}} e_{c_{k}}U^{i_{1}...i_{k}}e^{b_{1}} … e^{b_{k}} =

= γ_{a_{1}c_{1}}^{b_{1}} \frac{∂φ^{α_{1}c_{1}}}{∂x^{i_{1}}} … γ_{a_{k}c_{k}}^{b_{k}} \frac{∂φ^{α_{k}c_{k}}}{∂x^{i_{k}}} U^{i_{1}...i_{k}}e^{a_{1}} … e^{a_{k}} =

= \frac{∂φ^{α_{1}b_{1}}}{∂x^{i_{1}a_{1}}} … \frac{∂φ^{α_{k}b_{k}}}{∂x^{i_{k}a_{k}}} U^{i_{1}...i_{k}}e^{a_{1}} … e^{a_{k}},

hence, by (3.4),

v^{α_{1}b_{1}...α_{k}b_{k}} = \frac{∂φ^{α_{1}b_{1}}}{∂x^{i_{1}a_{1}}} … \frac{∂φ^{α_{k}b_{k}}}{∂x^{i_{k}a_{k}}} u^{i_{1}a_{1}...i_{k}a_{k}}.

Therefore, u^{C} is (T^{A}φ)-related to v^{C}.

In particular, if φ is a local diffeomorphism of M, i.e. x^{j′} = φ^{j′}(x^{i}), then

u^{j′_{1}b_{1}...j′_{k}b_{k}} = \frac{∂φ^{j′_{1}b_{1}}}{∂x^{i_{1}a_{1}}} … \frac{∂φ^{j′_{k}b_{k}}}{∂x^{i_{k}a_{k}}} u^{i_{1}a_{1}...i_{k}a_{k}}.

This means that u_{i_{1}a_{1}...i_{k}a_{k}} is a well-defined tensor field of type (0, k) on T^{A}M. Since u is skew-symmetric and the multiplication in A is commutative, u^{C} is also skew-symmetric. □

Let

[·, ·] : V^{k}(M) × V^{ℓ}(M) → V^{k+ℓ−1}(M),

be the Schouten-Nijenhuis bracket on the multivector fields (a generalization of Lie bracket of vector fields [9, 10, 14, 18]).

Proposition 3.3. The complete lift is compatible with the Schouten-Nijenhuis bracket, i.e.,

[u, v]^{C} = [u^{C}, v^{C}].  (3.5)

Proof. First let us derive an auxiliary formula. Let f be a real-valued function on M and F = F^{a}e_{a} = F_{b}e^{b} be its analytic prolongation. Then F_{b} = q_{ab}F^{a}. From the Scheffers’ conditions (1.4) it follows that ∂_{c}F_{b} = γ_{cd}^{a}q_{ab}δ^{g}∂_{g}F^{d} = γ_{cd}^{a}q_{ab}δ^{g}q^{dr}∂_{g}F_{r}. Contracting with δ^{b}, by (1.11), we obtain ∂_{c}(δ^{b}F_{b}) = δ^{b}q_{ab}δ^{g}γ_{cd}^{a}q^{dr}∂_{g}F_{r} = p_{a}γ_{cd}^{a}q^{dr}δ^{g}∂_{g}F_{r} = q_{cd}q^{dr}δ^{g}∂_{g}F_{r} = δ^{g}∂_{g}F_{c}. Thus,

∂_{c}(δ^{b}F_{b}) = δ^{g}∂_{g}F_{c}.  (3.6)

Let u ∈ V^{g}(M), v ∈ V^{h}(M) be multivector fields. With respect to the local coordinates,

\begin{aligned}[u, v]^{k_{2}...k_{g+h}} = & \frac{1}{(g − 1)!h!}ε_{i_{2}...i_{g}j_{1}...j_{h}}^{k_{2}...k_{g+h}}u^{ri_{2}...i_{g}} \frac{∂}{∂x^{r}} v^{j_{1}...j_{h}}+ \\ & +\frac{(−1)^{g}}{g!(h − 1)!}ε_{i_{1}...i_{g}j_{2}...j_{h}}^{k_{2}...k_{g+h}}v^{rj_{2}...j_{h}} \frac{∂}{∂x^{r}} u^{i_{1}...i_{g}}, \end{aligned}  (3.7)

where ε_{i_{1}...i_{s}}^{j_{1}...j_{s}} = δ_{i_{1}}^{j_{1}} … δ_{i_{s}}^{j_{s}} (see, e.g., [9]). By Theorem 1.1, the same formula holds for the analytic prolongations:

\begin{aligned}[U, V]^{k_{2}...k_{g+h}} = & \frac{1}{(g − 1)!h!}ε_{i_{2}...i_{g}i_{g+1}...i_{g+h}}^{k_{2}...k_{g+h}}U^{ri_{2}...i_{g}} \frac{∂}{∂X^{r}} V^{i_{g+1}...i_{g+h}}+ \\ & +\frac{(−1)^{g}}{g!(h − 1)!}ε_{i_{1}...i_{g}i_{g+2}...i_{g+h}}^{k_{2}...k_{g+h}}V^{ri_{g+2}...i_{g+h}} \frac{∂}{∂X^{r}} U^{i_{1}...i_{g}}, \end{aligned}

(3.8)

Let us multiply both sides of (3.8) by e^{a_{2}} … e^{a_{g+h}} and then contract with δ^{m}. In the left-hand side we get ([u, v]^{C})^{k_{2}a_{2}...k_{g+h}a_{g+h}}. Consider the first summand in the right-hand side (without ε):

\frac{1}{(g − 1)!h!}U^{ri_{2}...i_{g}}e^{a_{2}} … e^{a_{g}} \frac{∂}{∂X^{r}} V^{i_{g+1}...i_{g+h}}e^{a_{g+1}} … e^{a_{g+h}} =

\frac{1}{(g − 1)!h!}U^{ri_{2}...i_{g}}e^{a_{2}} … e^{a_{g}}δ^{b} \frac{∂}{∂x^{rb}} V^{a_{g+1}i_{g+1}...a_{g+h}i_{g+h}} by (1.6) =

\frac{1}{(g − 1)!h!}U^{ri_{2}...i_{g}}e^{a}e^{a_{2}} … e^{a_{g}}δ^{b} \frac{∂}{∂x^{rb}} V_{a}^{a_{g+1}i_{g+1}...a_{g+h}i_{g+h}} =

\frac{1}{(g − 1)!h!}U^{ri_{2}...i_{g}}e^{a}e^{a_{2}} … e^{a_{g}} \frac{∂}{∂x^{ra}} (δ^{b}V_{b}^{a_{g+1}i_{g+1}...a_{g+h}i_{g+h}}) by (3.6) =

\frac{1}{(g − 1)!h!}U_{m}^{ra i_{2}a_{2}...i_{g}a_{g}}e^{m} \frac{∂}{∂x^{ra}} ((v^{C})^{a_{g+1}i_{g+1}...a_{g+h}i_{g+h}}).

Contracting with δ^{m} we get

\frac{1}{(g − 1)!h!}(u^{C})^{ra i_{2}a_{2}...i_{g}a_{g}} \frac{∂}{∂x^{ra}} ((v^{C})^{a_{g+1}i_{g+1}...a_{g+h}i_{g+h}}).

The commutativity of multiplication in A yields that

ε_{i_{2}...i_{g+h}}^{k_{2}...k_{g+h}} = ε_{i_{2}a_{2}...i_{g+h}a_{g+h}}^{k_{2}a_{2}...k_{g+h}a_{g+h}}.

In the same manner we deal with the second summand in (3.8) and then obtain. Therefore

([u, v]^{C})^{k_{2}a_{2}...k_{g+h}a_{g+h}} =

\frac{1}{(g−1)!h!}ε_{i_{2}a_{2}...i_{g}a_{g}i_{g+1}a_{g+1}...i_{g+h}a_{g+h}}^{k_{2}a_{2}...k_{g+h}a_{g+h}}(u^{C})^{ra i_{2}a_{2}...i_{g}a_{g}} \frac{∂}{∂x^{ra}} ((v^{C})^{a_{g+1}i_{g+1}...a_{g+h}i_{g+h}})+

+\frac{(−1)^{g}}{g!(h−1)!}ε_{i_{1}a_{1}...i_{g}a_{g}i_{g+2}a_{g+2}...i_{g+h}a_{g+h}}^{k_{2}a_{2}...k_{g+h}a_{g+h}}(v^{C})^{ra i_{g+2}a_{g+2}...i_{g+h}a_{g+h}} \frac{∂}{∂x^{ra}} ((u^{C})^{i_{1}a_{1}...i_{g}a_{g}}),

which coincides with (3.5). □

The construction of the vertical lift of multivector fields to the tangent bundle (see [23]) also may be generalized to Weil bundles in the following way. Let u = u^{i_{1}...i_{k}} \frac{∂}{∂x^{i_{1}}} ∧ ⋯ ∧ \frac{∂}{∂x^{i_{k}}} ∈ V^{k}(M). We define the vertical lift u^{V} ∈ V^{k}(T^{A}M) of u by

u^{V} = u_{A}^{V} := u^{i_{1}...i_{k}} \frac{∂}{∂x^{i_{1}n}} ∧ ⋯ ∧ \frac{∂}{∂x^{i_{k}n}} .  (3.9)

Proposition 3.4. The vertical lift u^{V} is a well-defined multivector field on T^{A}M.

Proof. Let x^{i′} = x^{i′}(x^{i}) be a coordinate change on M, then

u^{i′_{1}...i′_{k}} = \frac{∂x^{i′_{1}}}{∂x^{i_{1}}} … \frac{∂x^{i′_{k}}}{∂x^{i_{k}}} u^{i_{1}...i_{k}}.

Thus it suffices to prove that

\frac{∂}{∂x^{i′_{s}n}} = \frac{∂x^{i_{s}}}{∂x^{i′_{s}}} \frac{∂}{∂x^{i_{s}n}} .  (3.10)

Let us determine the corresponding change of coordinates x^{i′a} = x^{i′a}(x^{ib}) on T^{A}M. By (1.7),

X^{i′} = x^{i′} + x^{i′â}eâ = x^{i′} + ∑_{|p|=1}^{q} \frac{1}{p!} \frac{D^{p}x^{i′}}{Dx^{p}} X^{∘}^{p}.  (3.11)

Hence, for T^{A}M we have

\begin{matrix}x^{i′} = x^{i′}(x^{i}), \\ x^{i′â} = f^{i′â}(x^{ib}),  â = 1, …, n. \end{matrix}

Let us show that

\frac{∂f^{i′â}}{∂x^{ib}} = 0 for b > â.

Indeed, the coefficients \frac{1}{p!} \frac{D^{p}x^{i′}}{Dx^{p}} in (3.11) depend only on x^{i} = x^{i0}, while

the expression (X^{∘}^{1})^{p_{1}} … (X^{∘}^{m})^{p_{m}} contains x^{ib} only as a coefficient at e_{b} in X^{∘}^{i}. Since γ_{cd̂}^{s} = 0, for c > s the coefficient at eâ which appears

when we expand brackets in (X^{∘}^{1})^{p_{1}} … (X^{∘}^{m})^{p_{m}} does not depend on x^{ib}. Moreover, it can depend on x^{iâ} only for the summand corresponding to |p| = 1 (because for |p| ≥ 2 the expression x^{iâ}eâ will be multiplied by some element of A^{∘} and thus after the expanding brackets the coefficient at eâ does not depend on x^{iâ}). For the case |p| = 1 we obtain the only summand \frac{∂x^{i′}}{∂x^{i}} x^{iâ}eâ depending on x^{iâ}. Therefore,

\frac{∂x^{i′â}}{∂x^{iâ}} = \frac{∂x^{i′}}{∂x^{i}} for any â = 1, …, n.

Hence, the Jacobi matrix of the coordinate transformation x^{i′a} = x^{i′a}(x^{ib}) on T^{A}M has the following block structure:

\begin{matrix}\frac{∂x^{i′}}{∂x^{i}} & 0 & 0 & ⋯ & 0 & 0 \\ ∗ & \frac{∂x^{i′}}{∂x^{i}} & 0 & ⋯ & 0 & 0 \\ ∗ & ∗ & ... & ... & ⋮ & ⋮ \\ ⋮ & ... & ... & ... & ... & ⋮ \\ ∗ & ⋯ & ... & ∗ & \frac{∂x^{i′}}{∂x^{i}} & 0 \\ ∗ & ∗ & ⋯ & ∗ & ∗ & \frac{∂x^{i′}}{∂x^{i}} \end{matrix} ,

where ∗ denotes the blocks which are unessential for our consideration. Now (3.10) is obvious. □

Proposition 3.5. For any u, v ∈ V^{∗}(M) there holds

[u^{V} , v^{V}] = 0.  (3.12)

Proof. This follows easily from (3.7). Indeed,

(u^{V})^{i_{1}a_{1}...i_{k}a_{k}} = 0,

if at least one of indices a_{1}, …, a_{k} is not equal to n. But then

\frac{∂v^{j_{1}b_{1}...j_{ℓ}b_{ℓ}}}{∂x^{rn}} = 0.

Thus, all the summands in the right-hand side of (3.7) are zero. □

4. Poisson structures on T^{A}M

Recall that a Poisson bracket on a smooth manifold M is a bilinear skew-symmetric mapping { , } : C^{∞}(M)×C^{∞}(M) → C^{∞}(M), satisfying the Leibniz rule

{f, gh} = {f, g}h + g{f, h}

and the Jacobi identity

{{f, g}, h} + {{g, h}, f} + {{h, f}, g} = 0.

A Poisson manifold is a smooth manifold M endowed with a Poisson bracket. The Poisson bracket on M uniquely defines a bivector field

w = w^{ij} \frac{∂}{∂x^{i}} ∧ \frac{∂}{∂x^{j}} ∈ V^{2}(M)  (4.1)

such that

{f, g} = i(w)(df ∧ dg)  (4.2)

for any f, g ∈ C^{∞}(M), where i(w) : Ω^{m}(M) → Ω^{m−2}(M) is the inner product; in local coordinates

(i(w)α)_{i_{1}...i_{m−2}} = w^{jk}α_{jki_{1}...i_{m−2}}.

This bivector field is usually called the Poisson bivector. It is known (see, e.g., [7, 9, 18]) that the bracket (4.2) on C^{∞}(M) satisfies the Jacobi identity if and only if

[w, w] = 0.  (4.3)

In local coordinates this is written as

w^{js} \frac{∂w^{kℓ}}{∂x^{s}} + w^{ks} \frac{∂w^{ℓj}}{∂x^{s}} + w^{ℓs} \frac{∂w^{jk}}{∂x^{s}} = 0.

In what follows we will denote the Poisson manifold by (M, w).

To each function f ∈ C^{∞}(M) there corresponds the Hamiltonian vector field X_{f} = X_{f}^{w} ∈ V^{1}(M) defined by X_{f}(g) := {f, g}. Locally, Hamiltonian vector fields on (M, w) have the form [14, 11, 18]

X_{f}^{w} = \{f, ·\}_{w} = \frac{∂f}{∂x^{i}} w^{ij} \frac{∂}{∂x^{j}}  (f ∈ C^{∞}(M)).  (4.4)

For a Poisson manifold (M, w), the Lichnerowicz-Poisson coboundary operator

σ = σ_{w} : V^{k}(M) → V^{k+1}(M),

is defined by σu := [w, u]. Because the Schouten-Nijenhuis bracket satisfies the super-Jacobi identity (see [9, 14])

(−1)^{|u|·|v|}[[v, y], u] + (−1)^{|v|·|y|}[[y, u], v] + (−1)^{|y|·|u|}[[u, v], y] = 0,

one has σ ∘ σ = 0. Thus the cohomology spaces

H_{LP}^{k} (M, w) := \frac{ker σ : V^{k}(M) → V^{k+1}(M)}{im σ : V^{k−1}(M) → V^{k}(M)},

called the Lichnerowicz-Poisson cohomology spaces of Poisson manifold (M, w) are defined. The problem of computing this cohomology is very difficult (see, e.g., [3, 17, 22]).

One can easily see that [w, f] = X_{f} for any f ∈ C^{∞}(M), hence Hamiltonian vector fields form the space of 1-coboundaries of σ [14].

Let (M, w) be a Poisson manifold and A be a Frobenius Weil algebra. Consider the complete lift w^{C} of w to T^{A}M. By (3.5) and (4.3), w^{C} is a Poisson bivector on T^{A}M.

Proposition 4.1. The complete lift of multivector fields induces a homomorphism of Lichnerowicz-Poisson cohomology

[u] ∈ H_{LP}^{∗} (M, w) ⟼ [u^{C}] ∈ H_{LP}^{∗} (T^{A}M, w^{C}).  (4.5)

Proof. From (3.5) it follows that

(σ_{w}u)^{C} = σ_{w^{C}} u^{C},

which implies that (4.5) is a homomorphism. □

Let us find how the complete lift w^{C} does depend on the choice of a Frobenius covector p on A. We denote by t the corresponding vector defined by (2.12). Let w_{s}^{ij} be the components of the analytic prolongation W^{ij} = (w^{ij})^{A}, i.e., W^{ij} = ∑_{s≥0} w_{s}^{ij}e_{s}. In what follows we will omit the sign of summation over s. Then W^{ij}e^{a}e^{b} = w_{s}^{ij}e_{s}e^{a}e^{b} = w_{s}^{ij}q^{ac}q^{bk}e_{s}e_{c}e_{d} = w_{s}^{ij}q^{ac}q^{bk}γ_{sc}^{d}γ_{kd}^{ℓ}e_{ℓ} = w_{s}^{ij}q^{ac}q^{bk}γ_{sc}^{d}γ_{kd}^{ℓ}q_{ℓr}e^{r}. Contracting with δ^{r}, we obtain

w^{iajb} = w_{s}^{ij}q^{ac}q^{bk}γ_{sc}^{d}γ_{kd}^{ℓ}q_{ℓr}δ^{r} = w_{s}^{ij}q^{ac}q^{bk}γ_{sc}^{d}γ_{kd}^{ℓ}p_{ℓ} =

w_{s}^{ij}\overline{γ}_{r}^{ac}t^{r}\overline{γ}_{f}^{bk}t^{f}γ_{sc}^{d}γ_{kd}^{ℓ}p_{ℓ} = w_{s}^{ij}t^{r}t^{f}h^{ag}γ_{gr}^{c}h^{eb}γ_{fe}^{k}γ_{sc}^{d}γ_{kd}^{ℓ}p_{ℓ} (by (2.5)) =

w_{s}^{ij}t^{r}t^{f}h^{ag}h^{eb}γ_{gr}^{c}γ_{de}^{k}γ_{kf}^{ℓ}γ_{sc}^{d}p_{ℓ} (by (1.2)) = w_{s}^{ij}t^{r}h^{ag}h^{eb}γ_{gr}^{c}γ_{de}^{k}δ_{k}^{n}γ_{sc}^{d} = (by (2.12)) =

w_{s}^{ij}t^{r}h^{ag}h^{eb}γ_{gr}^{c}h_{de}γ_{sc}^{d} = w_{s}^{ij}t^{r}h^{ag}δ_{d}^{b}γ_{gr}^{c}γ_{sc}^{d} = w_{s}^{ij}t^{r}h^{ag}γ_{gr}^{c}γ_{sc}^{b} =
w_{s}^{ij}t^{r}\overline{γ}_{r}^{ac}γ_{sc}^{b}.

Thus, for any Frobenius covector p, the complete lift w^{C} is the linear combination

w^{C} = t^{k}w_{k}^{C},  (4.6)

where

w_{k}^{C} = ∑_{s=0}^{n} w_{s}^{ij}\overline{γ}_{k}^{ac}γ_{cs}^{b} \frac{∂}{∂x^{ia}} ∧ \frac{∂}{∂x^{jb}} .  (4.7)

Let us show that each of w_{k}^{C} is a multivector field and that

[w_{k}^{C}, w_{k}^{C}] = [w_{k}^{C}, w_{ℓ}^{C}] = 0  (4.8)

for any k, ℓ = 0, …, n. To this end we will express each of w_{k}^{C} in terms of complete lifts w^{C} corresponding to different Frobenius covectors. We will assume the basis (1.1) to be chosen in such a way that p_{(0)} = (0, …, 0, 1) is a Frobenius covector. Recall that the corresponding vector t_{(0)} is (1, 0, …, 0). Since the determinant det ∥γ_{ab}^{c}t^{b}∥ is a continuous function in t and does not vanish at t = t_{(0)}, this determinant does not vanish in a neighborhood of this point. Therefore one can find ε > 0 such that this determinant is nonzero for each of the vectors t_{(k)} = (1, 0 …, 0, ε, 0, …, 0) (ε at the kth place), t_{(k−)} = (1, 0 …, 0, −ε, 0, …, 0) (−ε at the kth place) and

t_{(k,ℓ)} = (1, 0 …, 0, ε, 0, …, 0, ε, 0, …, 0)

(ε at the kth and ℓth places), k, ℓ = 1, …, n. Hence the corresponding covectors p_{(k)} p_{(k,ℓ)} are Frobenius covectors. For each vector t under consideration the complete lift w^{C} is a Poisson bivector, which implies

[w^{C}, w^{C}] = [t^{k}w_{k}^{C}, t^{ℓ}w_{ℓ}^{C}] = 0.  (4.9)

Substituting t = t_{(0)} into (4.6) and (4.9), we see that w_{0}^{C} ∈ V^{2}(T^{A}M) and [w_{0}^{C}, w_{0}^{C}] = 0. Now substitute t = t_{(k)} and t = t_{(k−)}, which gives w_{0}^{C} + εw_{k}^{C} ∈ V^{2}(T^{A}M), therefore w_{k}^{C} ∈ V^{2}(T^{A}M) and

[w_{0}^{C} + εw_{k}^{C}, w_{0}^{C} + εw_{k}^{C}] = [w_{0}^{C} − εw_{k}^{C}, w_{0}^{C} − εw_{k}^{C}] = 0.

Expanding these equations yields 2ε[w_{k}^{C}, w_{0}^{C}]+ε^{2}[w_{k}^{C}, w_{k}^{C}] = −2ε[w_{k}^{C}, w_{0}^{C}]+ ε^{2}[w_{k}^{C}, w_{k}^{C}] = 0, which implies [w_{0}^{C}, w_{k}^{C}] = [w_{k}^{C}, w_{k}^{C}] = 0. Finally, substituting t = t_{(k,ℓ)} into (4.9) one gets [w_{k}^{C}, w_{ℓ}^{C}] = 0. Thus, the following theorem is valid.

Theorem 4.1. Let (M, w) be a Poisson manifold and T^{A}M be its Weil bundle for a Frobenius Weil algebra A. Then for each Frobenius covector p on A the complete lift w^{C} of Poisson bivector w to T^{A}M is of the form

w^{C} = t^{k}w_{k}^{C},

where t = (t^{k}) and w_{k}^{C} are defined by (2.12) and (4.7) respectively. Moreover,

[w_{k}^{C}, w_{k}^{C}] = [w_{k}^{C}, w_{ℓ}^{C}] = 0

for any k, ℓ = 0, …, n.

Remark 4.1. One can easily see that w_{n}^{C} = w^{V}. Indeed, in this case k = n in (4.7). But (2.4) implies that \overline{e}^{n} = e_{0} and that each of \overline{e}^{a}, a = 0, 1, …, n − 1, does not contain e_{0} in its decomposition. Hence the only nonzero component of \overline{γ}_{n}^{ac} is \overline{γ}_{n}^{nn} = 1. Then γ_{nd}^{b} is 1 only for b = n, d = 0 and γ_{nd}^{b} = 0 for any other values of b and d. Consequently,

w_{n}^{C} = w^{ij} \frac{∂}{∂x^{in}} ∧ \frac{∂}{∂x^{jn}} .

Thus, Theorem 4.1 (as well as Proposition 3.5) implies that w^{V} is also a Poisson bivector on T^{A}M.

Example 4.1. Let A be the algebra of plural numbers R(ε^{n}). Choose a Jordan-Hölder basis e_{0} = 1, e_{a} = ε^{a}, a = 1, …, n in it. The explicit form of the analytic prolongations W^{ij} = ∑_{s≥0} w_{s}^{ij}ε^{s} can be found by (1.7), for instance,

w_{0}^{ij} = w^{ij},

w_{1}^{ij} = x^{k1} \frac{∂w^{ij}}{∂x^{k}} ,

w_{2}^{ij} = x^{k1}x^{ℓ1} \frac{∂^{2}w^{ij}}{∂x^{k}∂x^{ℓ}} + x^{k2} \frac{∂w^{ij}}{∂x^{k}} , etc.

The corresponding bivectors w_{k}^{C} are (here w_{k}^{ij} means the square block ∥w_{k}^{ij}∥)

w_{0}^{C} = ( \begin{matrix}0 & 0 & … & 0 & 0 & w^{ij} \\ 0 & 0 & … & 0 & w^{ij} & w_{1}^{ij} \\ ⋮ & ⋮ & ... & ... & ... & ⋮ \\ 0 & 0 & ... & ... & … & … \\ 0 & w^{ij} & w_{1}^{ij} & … & … & w_{n−1}^{ij} \\ w^{ij} & w_{1}^{ij} & … & … & w_{n−1}^{ij} & w_{n}^{ij} \end{matrix} ) ,

w_{k}^{C} = ( \begin{matrix}0 & 0 & … & … & 0 & 0 & … & 0 & 0 \\ 0 & 0 & … & … & 0 & 0 & … & ⋮ & ⋮ \\ ⋮ & ⋮ & ... & ... & ... & ... & ... & … & 0 \\ 0 & 0 & ... & ... & 0 & 0 & … & 0 & w^{ij} \\ 0 & 0 & ... & ... & 0 & … & 0 & w^{ij} & w_{1}^{ij} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & ... & ... & ... & ⋮ \\ 0 & 0 & … & … & ... & ... & ... & … & ⋮ \\ 0 & 0 & … & 0 & w^{ij} & w_{1}^{ij} & … & … & w_{n−k−1}^{ij} \\ 0 & … & 0 & w^{ij} & w_{1}^{ij} & … & … & w_{n−k−1}^{ij} & w_{n−k}^{ij} \end{matrix} ) , k = 1, …, n−1,

w_{n}^{C} = ( \begin{matrix}0 & 0 & … & 0 & 0 \\ 0 & 0 & … & 0 & 0 \\ ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & … & … & 0 \\ 0 & 0 & … & 0 & w^{ij} \end{matrix} ) .

In particular, for the case A = R(ε), which corresponds to the tangent bundle T M and the standard Frobenius covector p = (0, 1), the complete lift w^{C} is

w^{C} = w^{ij} \frac{∂}{∂x^{i}} ∧ \frac{∂}{∂y^{j}} + \frac{1}{2}y^{k} \frac{∂w^{ij}}{∂x^{k}} \frac{∂}{∂y^{i}} ∧ \frac{∂}{∂y^{j}} .

This Poisson bivector was studied by many authors, see, e.g., [1, 4, 11].

Example 4.2. Consider the algebra A(Λ, n). We will denote the elements of the inverse matrix Λ^{−1} by λ^{ab}.

By (1.7), denote

w_{a}^{ij} = x^{ka} \frac{∂w^{ij}}{∂x^{k}} , w_{n}^{ij} = x^{kn} \frac{∂w^{ij}}{∂x^{k}} + λ_{ab}x^{ka}x^{ℓb} \frac{∂^{2}w^{ij}}{∂x^{k}∂x^{ℓ}} .

Then, for this algebra (4.7) implies that

w_{0}^{C} = ( \begin{matrix}0 & 0 & 0 & … & 0 & w^{ij} \\ 0 & λ^{11} w^{ij} & λ^{12} w^{ij} & … & λ^{1,n−1} w^{ij} & w_{1}^{ij} \\ 0 & λ^{21} w^{ij} & λ^{22} w^{ij} & … & λ^{2,n−1} w^{ij} & w_{2}^{ij} \\ ⋮ & … & … & … & … & ⋮ \\ 0 & λ^{n−1,1} w^{ij} & λ^{n−1,2} w^{ij} & … & λ^{n−1,n−1} w^{ij} & w_{n−1}^{ij} \\ w^{ij} & w_{1}^{ij} & w_{2}^{ij} & … & w_{n−1}^{ij} & w_{n}^{ij} \end{matrix} ) ,

w_{b}^{C} = ( \begin{matrix}0 & 0 & … & … & 0 & 0 \\ ⋮ & ⋮ & … & … & … & ⋮ \\ ⋮ & ⋮ & … & … & … & w^{ij} \\ ⋮ & ⋮ & … & … & … & ⋮ \\ 0 & … & … & … & 0 & 0 \\ 0 & … & w^{ij} & … & 0 & ∑_{a=1}^{n−1} λ_{ab}w_{a}^{ij} \end{matrix} ) , b = 1, …, n−1, w^{ij} at the bth place

w_{n}^{C} = ( \begin{matrix}0 & … & 0 & 0 \\ ⋮ & … & … & ⋮ \\ 0 & … & … & 0 \\ 0 & … & 0 & w^{ij} \end{matrix} ) .

5. Modular classes of Poisson structures on T^{A}M

In the final part of the present paper we compute the modular classes of Poisson structures w_{k}^{C} for the case of weakly symmetric Frobenius Weil algebras.

Recall that if μ is a volume form on the oriented manifold M then the divergence div_{μ}X of a vector field X is defined by

L_{X}μ = (div_{μ}X)μ

and one has

div_{μ}(fX) = f div_{μ}X + Xf, f ∈ C^{∞}(M).

Therefore for a Poisson manifold (M, w) with the volume form μ the operator

Δ_{μ} : f ∈ C^{∞}(M) ⟼ div_{μ}X_{f} ∈ C^{∞}(M)

is defined, where X_{f} is a Hamiltonian vector field of f. Easy computations show that Δ_{μ} is a derivation on C^{∞}(M) and, hence, a vector field on M [21]. This vector field is called the modular vector field of oriented Poisson manifold (M, w, μ).

The modular vector field satisfies σΔ_{μ} = 0 [6]. If we replace μ with any other volume form aμ, where a ∈ C^{∞}(M) is a positive function, then the modular vector field changes to Δ_{aμ} = Δ_{μ} + H_{− log a} [21]. As far as Hamiltonian vector fields are 1-coboundaries of σ, this implies that the set of modular vector fields for all volume forms on M is an element of H_{LP}^{1} (M, w). This cohomology class is called the modular class of the Poisson manifold (M, w).

Let g = g_{ij}dx^{i} ⊗ dx^{j} be a Riemannian metric on an m-dimensional oriented manifold M. Then

dV_{g} = \sqrt{det g} dx^{1} ∧ ⋯ ∧ dx^{m}

is a volume form on M. Let g^{C} be a complete lift of g.

Proposition 5.1. For a weakly symmetric Frobenius Weil algebra A the complete lift g^{C} is a metric on T^{A}M, moreover,

det g^{C} = M(det g)^{n+1},  (5.1)

where M is some constant (depending only on A).

Proof. We choose the standard Frobenius covector p_{(0)}. Let G_{ij} = g_{ij}^{s}e_{s} be the analytic prolongations of g_{ij}. Then G_{ij}e_{a}e_{b} = G_{iajb}^{c}e_{c} and g_{iajb} = G_{iajb}^{c}p_{c} = G_{iajb}^{n}. Clearly, g_{iajb} = g_{jbia} for all a, b, i, j. If s > 0,

then e_{s} ∈ A^{∘}, hence the component g_{iajb} contains g_{ij}^{s} only if e_{a}e_{b} ∉ A^{∘}^{q}.

Therefore, if e_{a}e_{b} ∈ A^{∘}^{q}, then g_{iajb} depends only on g_{ij}^{0} = g_{ij} and the matrix ∥g_{iajb}∥ has the following block structure:

\begin{matrix}* & * & * & ⋯ & * & * & ∥g_{ij}∥ \\ * & ⋯ & ⋯ & ... & * & B̂_{1} & 0 \\ * & ⋯ & ... & ... & B̂_{2} & 0 & 0 \\ ⋮ & ... & ... & ... & ... & ... & ⋮ \\ * & * & B̂_{q−2} & ... & ... & ⋯ & 0 \\ * & B̂_{q−1} & 0 & ... & ⋯ & ⋯ & 0 \\ ∥g_{ij}∥ & 0 & 0 & ⋯ & ⋯ & ⋯ & 0 \end{matrix}

where B̃_{k} = B_{q−k,k} ⊗ ∥g_{ij}∥ and the symbol ⊗ denotes the tensor (Kronecker) product of matrices.

The determinant of this matrix is the product of the determinants of diagonal blocks: det ∥g_{iajb}∥ = det ∥g_{ij}∥·det(B_{1,q−1}⊗∥g_{ij}∥)·· · ··det(B_{q−1,1}⊗ ∥g_{ij}∥) · det ∥g_{ij}∥. For any two matrices S and T of dimensions k × k and ℓ × ℓ respectively, one has

det(S ⊗ T) = (det S)^{k}(det T)^{ℓ}.

We have d_{1}(A)+⋯+d_{q−1}(A) = n−1. Hence, det ∥g_{iajb}∥ = M(det g)^{n+1}, where M = (det B_{1,q−1})^{d_{1}(A)} · · · · · (det B_{q−1,1})^{d_{q−1}(A)}. □

Let

Φ = dV_{g^{C}} = \sqrt{det g^{C}} dx^{1} ∧ … dx^{m} ∧ ⋯ ∧ dx^{1n} ∧ ⋯ ∧ dx^{mn}

be the corresponding volume form on T^{A}M.

Proposition 5.2. Let A (dim A^{∘} = n) be a weakly symmetric Frobenius Weil algebra, (M, w) a Poisson manifold, and T^{A}M its Weil bundle.

Then for a Poisson structure w_{0}^{C} on T^{A}M its modular vector field is

Δ_{Φ}^{T^{A}M} = (n + 1)(Δ_{dV_{g}}^{M})^{V} ,  (5.2)

where V means the vertical lift. For each of the Poisson structures w_{1}^{C}, …, w_{n}^{C}, the modular vector fields are zero.

Proof. From (4.4) it follows that the modular vector field of (M, w) is [11]

Δ_{dV_{g}} = ∑_{j=1}^{m} (\frac{∂w^{ij}}{∂x^{j}} + w^{ij} \frac{∂ ln \sqrt{det g}}{∂x^{j}} ) \frac{∂}{∂x^{i}} .

Then (5.1) implies that

\frac{∂ ln \sqrt{det g^{C}}}{∂x^{jb}} = \{ \begin{matrix}(n + 1)\frac{∂ ln \sqrt{det g}}{∂x^{j}} , & b = 0, \\ 0, & b = 1, 2 …, n. \end{matrix}  (5.3)

Let w_{k}^{iajb} denote the components of w_{k}^{C}. At first, show that

\begin{matrix}\frac{∂w_{0}^{iajb}}{∂x^{jb}} = \{ \begin{matrix}\frac{∂w^{ij}}{∂x^{j}} , & a = n, \\ 0, & a = 0, 1 …, n − 1, \end{matrix} \\ \frac{∂w_{k}^{iajb}}{∂x^{jb}} = 0,  k = 1, …, n. \end{matrix}  (5.4)

By (4.7), w_{k}^{iajb} = ∑_{s=0}^{n} w_{s}^{ij}\overline{γ}_{k}^{ac}γ_{cs}^{b}. The arguments similar to the proof of Proposition 3.4 show that \frac{∂w_{s}^{ij}}{∂x^{jb}} = 0 for s < b and that \frac{∂w_{b}^{ij}}{∂x^{jb}} = \frac{∂w^{ij}}{∂x^{j}}. Moreover, γ_{cs}^{b} = 0 s > b. Hence, the only nonzero summand in \frac{∂w_{k}^{iajb}}{∂x^{jb}} corresponds to s = b. Therefore c = 0, otherwise γ_{cs}^{b} = 0. But \overline{γ}_{k}^{a0} is not zero only for a = n and k = 0 by virtue of (2.4) (since \overline{e}^{0} = e_{n}, \overline{e}^{n} = 1). Hence \frac{∂w_{k}^{iajb}}{∂x^{jb}} = 0 for k = 1, …, n and \frac{∂w_{0}^{iajb}}{∂x^{jb}} = 0 for a ≠ n. As for a = n, we have

\frac{∂w_{0}^{injb}}{∂x^{jb}} = \frac{∂w_{b}^{ij}}{∂x^{jb}} = \frac{∂w^{ij}}{∂x^{j}} .

This completes the proof of (5.4).

Now, show that

\begin{matrix}w_{0}^{iaj0} = \{ \begin{matrix}w^{ij}, & a = n, \\ 0, & a = 0, 1 …, n − 1, \end{matrix} \\ w_{k}^{iaj0} = 0,  k = 1, …, n. \end{matrix}  (5.5)

Indeed, w_{k}^{iaj0} = ∑_{s=0}^{n} w_{s}^{ij}\overline{γ}_{k}^{ac}γ_{cs}^{0}, thus c = s = 0 (otherwise γ_{0}^{cs} = 0) which implies a = n and k = 0 as before.

It remains to prove (5.2). We have

Δ_{Φ} = ∑_{jb} (\frac{∂w^{iajb}}{∂x^{jb}} + w^{iajb} \frac{∂ ln \sqrt{det g^{C}}}{∂x^{jb}} ) \frac{∂}{∂x^{ia}} .

From (5.4) it follows that

∑_{j=1}^{m} \frac{∂w^{iajb}}{∂x^{jb}} \frac{∂}{∂x^{ia}} = (n + 1)\frac{∂w^{ij}}{∂x^{j}} \frac{∂}{∂x^{in}} ,

since the index b can take n + 1 distinct values from 0 to n.

In the summand

∑_{jb} w^{iajb} \frac{∂ ln \sqrt{det g^{C}}}{∂x^{jb}} \frac{∂}{∂x^{ia}}

the only possibility is b = 0 by virtue of (5.3), whence, by (5.5), we obtain

∑_{jb} w^{iajb} \frac{∂ ln \sqrt{det g^{C}}}{∂x^{jb}} \frac{∂}{∂x^{ia}} = (n + 1)w^{ij} \frac{∂ ln \sqrt{det g}}{∂x^{j}} \frac{∂}{∂x^{in}} .

□

Corollary 5.3. For a weakly symmetric Frobenius Weil algebra A the modular class of the Poisson manifold (T^{A}M, w^{C} = w_{0}^{C} + ∑_{k=1}^{n} t^{k}w_{k}^{C}) is represented by (n + 1)Δ_{μ}^{V}, for every modular vector field Δ_{μ} of the base manifold (M, w).

Proof. By Proposition 5.2 the result is true for the field Δ_{dV_{g}}. As in [11], we have

(σ_{w}f)^{V} = σ_{w^{C}}(f ∘ π_{A}),  f ∈ C^{∞}(M),  π_{A} : T^{A}M → M.

This immediately proves the Corollary. □

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256 VADIM V. SHURYGIN, JUNIOR

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Kazan State University, Kremlevskaya, 18, Kazan, 420008, Russia

E-mail address: 1Vadim.Shurygin@ksu.ru

Received April 8, 2005