Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 17, 2005, 231 – 258

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.

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

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 $A = R \oplus \overset{\circ}{A}$ , where $\overset{\circ}{A}$ is a ﬁnite-dimensional maximal ideal, consisting of nilpotent elements. In what follows we denote $n = {dim}_{R} \overset{\circ}{A}$ .

By ${\overset{\circ}{A}}^{r}$ we denote the $r$ th power of $\overset{\circ}{A}$ . Let $d_{k} (A) = {d i m}_{R} {\overset{\circ}{A}}^{k} ∕ {\overset{\circ}{A}}^{k + 1}$ . The number $d_{1} (A)$ is usually called the width of $A$ . The positive integer $q$ deﬁned by ${\overset{\circ}{A}}^{q} \neq 0$ , ${\overset{\circ}{A}}^{q + 1} = 0$ is called the height of $A$ .

The chain of embedded ideals

A \supset \overset{\circ}{A} \supset {\overset{\circ}{A}}^{2} \supset \dots \supset {\overset{\circ}{A}}^{q} \supset 0

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

A \supset \overset{\circ}{A} = I_{1} \supset I_{2} \supset \dots \supset I_{n} \supset 0,

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

{\overset{\circ}{A}}^{k} = I_{1 + d_{1} (A) + \dots + d_{k - 1} (A)} f o r 2 \leq k \leq 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}, e_{\hat{a}}}, a = 0, 1, \dots, n = dim \overset{\circ}{A}, \hat{a} = 1, \dots, n,

(1.1)

in $A$ such that $e_{0} = 1 \in R$ , $e_{\hat{a}} \in I_{\hat{a}}$ , $e_{\hat{a}} \notin I_{\hat{a} + 1}$ . In general, this basis is not unique. For $X = x^{a} e_{a} = x^{0} + x^{\hat{a}} e_{\hat{a}} \in A$ we set $\overset{\circ}{X} = x^{\hat{a}} e_{\hat{a}}$ , then $X = x^{0} + \overset{\circ}{X}$ . Let $δ^{a}$ be the coordinates of unit of $A$ , i.e., $1 = δ^{a} e_{a}$ .

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

γ_{a b}^{c} γ_{e f}^{b} = γ_{a e}^{b} γ_{b f}^{c} .

(1.2)

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

\partial_{c} f^{b} γ_{a g}^{c} = γ_{a c}^{b} \partial_{g} f^{c},

(1.3)

where $\partial_{a} f^{b} = \partial f^{b} ∕ \partial x^{a}$ . Scheffers’ equations are equivalent to

\partial_{a} f^{b} = γ_{a c}^{b} δ^{g} \partial_{g} f^{c} .

(1.4)

For $f : U \subset A^{m} \to A$ , $f : {X^{i} = x^{i a} e_{a}} \mapsto f (X^{i}) = f^{b} (x^{i a}) e_{b}$ , where $A^{m} = A \times \dots \times A$ is the $A$ -module of $m$ -tuples of elements of $A$ , Scheffers’ conditions of $A$ -differentiability are of the form [19]:

\partial_{i a} f^{b} = γ_{a c}^{b} δ^{g} \partial_{i g} f^{c} .

(1.5)

If a function $f$ satisﬁes (1.5), its differential can be represented in the form $d f = f_{i} d X^{i}$ , where $f_{i} = δ^{a} \partial_{i a} f$ is the partial derivative with respect to $X^{i} \in A$ . Thus,

f_{i} = \frac{\partial f}{\partial X^{i}} = δ^{a} \frac{\partial f}{\partial x^{i a}} .

(1.6)

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

The following theorem (see [16]) describes the local structure of an $A$ -differentiable map of the form $F : U \subset A^{m} \to A^{k}$ for a Weil algebra $A$ . The natural epimorphism $π_{0}^{q} : A^{m} \to R^{m}$ determines the canonical ${\overset{\circ}{A}}^{m}$ -foliation on $A^{m}$ . Recall that a smooth map $f : M \to N$ 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 $U \subset A^{m}$ be an open set. Then any $A$ -smooth map $Φ : U \to A^{k}$ is of the form

X^{i^{'}} = ϕ^{i^{'}} + \sum_{∣ p ∣ = 1}^{q} \frac{1}{p!} \frac{D^{p} ϕ^{i^{'}}}{D x^{p}} {\overset{\circ}{X}}^{p},

(1.7)

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

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

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

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

$1^{\circ}$ . ${(ϕ + ψ)}^{A} = ϕ^{A} + ψ^{A}$ .

$2^{\circ}$ . ${(ϕ \cdot ψ)}^{A} = ϕ^{A} \cdot ψ^{A}$ .

$3^{\circ}$ . ${(ϕ^{A} \circ ψ)}^{A} = ϕ^{A} \circ ψ^{A}$ .

$4^{\circ}$ . ${(D^{p} ϕ ∕ D x^{p})}^{A} = D^{p} ϕ^{A} ∕ D X^{p}$ for $ϕ : U \subset A^{n} \to A$ .

We denote by $ℳ f$ the category of smooth manifolds and by $ℱ M$ that of ﬁbered manifolds. To each Weil algebra $A$ there corresponds a functor $T^{A} : ℳ f \to ℱ M$ called the Weil functor which maps a smooth manifold $M$ to the ﬁbered manifold $π_{A} : T^{A} M \to M$ 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., $T^{A} (M \times N) ≅ T^{A} M \times T^{A} N$ . Moreover, under some additional conditions (locality and regularity) each product preserving bundle functor $F : ℳ f \to ℱ M$ 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 $q : A \times A \to R$ , satisfying the following condition of associativity:

q (X Y, Z) = q (X, Y Z) f o r a n y X, Y, Z \in 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_{b c} γ_{e f}^{c} = γ_{b e}^{c} q_{c f} .

(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 $p : A \to R$ by $p (X) : = q (X, 1)$ . Its coordinates with respect to the basis (1.1) satisfy

p_{a} γ_{b c}^{a} = q_{b c} .

(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_{b c} δ^{c} .

(1.11)

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

q (X, Y) = p (X Y) f o r e a c h X, Y \in 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 $D = R (ɛ) = {x_{0} + x_{1} ɛ ∣ x_{0}, x_{1} \in R, ɛ^{2} = 0}$ . To this algebra there corresponds the tangent bundle functor: $T^{R (ɛ)} M = T M$ .

Example 1.2. Another example is the algebra $D^{n} = R (ɛ^{n}) = {x_{0} + x_{1} ɛ + \dots + x_{n - 1} ɛ^{n - 1} ∣ x_{i} \in R, ɛ^{n} = 0}$ . 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 $A \otimes B$ 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 $n = dim \overset{\circ}{A}$ . Let us choose a Jordan-Hölder basis (1.1) in $A$ .

Lemma 2.1. $dim {\overset{\circ}{A}}^{q} = 1$ , i.e., ${\overset{\circ}{A}}^{q} = I_{n}$ .

Proof. On the contrary, suppose that ${d i m}_{R} {\overset{\circ}{A}}^{q} \geq 2$ . Then at least two last elements $e_{n - 1}$ and $e_{n}$ of a basis (1.1) belong to ${\overset{\circ}{A}}^{q}$ and, consequently, for any $\hat{a} = 1, \dots, n$ there hold $e_{\hat{a}} e_{n - 1} = e_{\hat{a}} e_{n} = 0$ . Hence for any $c = 0, \dots, n$ the matrix $∥ γ_{a b}^{c} ∥$ 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 $∥ q_{a b} ∥ = ∥ p_{c} γ_{a b}^{c} ∥$ also contains zeros everywhere in two last columns except the ﬁrst row, and thus is degenerate. Contradiction. By this reason, ${d i m}_{R} {\overset{\circ}{A}}^{q} = 1$ , which implies ${\overset{\circ}{A}}^{q} = I_{n}$ or, equivalently, ${\overset{\circ}{A}}^{q} = R \cdot e_{n}$ . $□$

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

Proof. From the equalities $1 \cdot e_{n} = e_{0} e_{n} = e_{n}$ and $e_{\hat{a}} e_{n} = 0$ for each $\hat{a} = 1, \dots, n$ it follows, that for each $c = 0, \dots, n - 1$ and for each $b$ there holds

γ_{b n}^{c} = 0, γ_{0 n}^{n} = 1, γ_{\hat{a} n}^{n} = 0 .

Hence, for each $c = 0, \dots, n - 1$ the last column of the matrix $∥ γ_{a b}^{c} ∥$ contains only zeros and the last column of the matrix $∥ γ_{a b}^{n} ∥$ is

^{T} (1, 0, \dots, 0) .

Therefore the last column of $∥ q_{a b} ∥ = ∥ p_{c} γ_{a b}^{c} ∥$ is

^{T} (p_{n}, 0, \dots, 0),

(2.1)

hence $p_{n} \neq 0$ . $□$

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

Ann \overset{\circ}{A} : = {X \in \overset{\circ}{A} ∣ X \cdot \overset{\circ}{A} = 0} .

Let $0 \neq X \in Ann \overset{\circ}{A}$ , then for each $Y = y^{0} + \overset{\circ}{Y}$ we have $X Y = X y^{0}$ , whence $q (X, Y) = p (X Y) = y^{0} p (X)$ . From the degeneracy of $q$ it follows that $p (X) \neq 0$ . Therefore $Ann \overset{\circ}{A} \cap ker p = 0$ which implies that $dim Ann \overset{\circ}{A} \leq 1$ . But, clearly, $0 \neq {\overset{\circ}{A}}^{q} \subset Ann \overset{\circ}{A}$ , hence $dim {\overset{\circ}{A}}^{q} = dim Ann \overset{\circ}{A} = 1$ .

Let us denote $∥ h_{a b} ∥ : = ∥ γ_{a b}^{n} ∥$ .

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

Proof. Let us choose any Jordan-Hölder basis (1.1). If $p_{(0)} = (0, \dots, 0, 1)$ is a Frobenius covector, then the matrix $∥ h_{a b} ∥$ is nondegenerate. Assume the contrary and consider any $p \in A_{F r}^{*}$ . 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 $∥ q_{a b} ∥$ only the element $q_{00}$ depends on $p_{0}$ :

∥ q_{a b} ∥ = (\begin{matrix} p_{0} & * & \dots & * & 1 \\ * & * & \dots & * & 0 \\ ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ * & * & \dots & * & 0 \\ 1 & 0 & \dots & 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 $det ∥ q_{a b} ∥$ does not depend on $p_{0}$ and we may assume that $p_{0} = 0$ .

2) By the assumption, $p_{(0)} = (0, \dots, 0, 1)$ is not a Frobenius covector, therefore there exists at least one $c$ , $1 \leq c \leq n - 1$ , such that $p_{c} \neq 0$ . 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, \dots, 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 $a \geq 1$ , the structural constants ${γ^{'}}_{a b}^{c}$ will have the following form with respect to this basis: ${γ^{'}}_{a b}^{c} = γ_{a b}^{c}$ for $c = 1, 2, \dots, n - 1$ and ${γ^{'}}_{a b}^{n} = γ_{a b}^{n} + γ_{a b}^{d} p_{d}$ , where the summation over $d$ is taken from $1$ to $n - 1$ . Thus, $∥ {γ^{'}}_{a b}^{n} ∥$ equals to $∥ q_{a b} ∥$ and therefore is nondegenerate. $□$

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

Remark 2.2. One can also give a noncoordinate proof of Lemma 2.3. Let $\overset{\circ}{p} \in A^{*}$ be deﬁned by $\overset{\circ}{p} (X) : = x^{0}$ .

1) We show that if $p \in A_{F r}^{*}$ then $\tilde{p} : = p - p (1) \overset{\circ}{p}$ also is a Frobenius covector. Suppose the contrary. Then there exists $X \in A$ such that $\tilde{p} (X Y) = 0$ for any $Y \in A$ . This means that $x^{0} p (\overset{\circ}{Y}) + y^{0} p (\overset{\circ}{X}) + p (\overset{\circ}{X} \overset{\circ}{Y}) = 0$ . Let $Z \in {\overset{\circ}{A}}^{q}$ be an element such that $p (Z) = 1$ (in terms of the Jordan-Hölder basis, $Z = e_{n}$ ) and let $\tilde{X} = X - x^{0} Z$ . Then for any $Y \in A$ one has $\tilde{X} Y = X Y - x^{0} y^{0} Z = x^{0} y^{0} + x^{0} \overset{\circ}{Y} + y^{0} \overset{\circ}{X} + \overset{\circ}{X} \overset{\circ}{Y} - x^{0} y^{0} Z$ . One can easily see that $p (\tilde{X} Y) = 0$ , which contradicts to the fact that $p$ is a Frobenius covector.

This means that we can deform any $p \in A_{F r}^{*}$ in such a way that $R \subset ker p$ .

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

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

∥ h_{a b} ∥ = (\begin{matrix} 0 & \dots & 1 \\ ⋮ & B & ⋮ \\ 1 & \dots & 0 \end{matrix}),

(2.3)

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

∥ h^{a b} ∥ = (\begin{matrix} 0 & \dots & 1 \\ ⋮ & B^{- 1} & ⋮ \\ 1 & \dots & 0 \end{matrix})

(2.4)

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

{\bar{γ}}_{s}^{a b} = h^{a k} h^{b ℓ} h_{c s} γ_{k ℓ}^{c}

(2.5)

and

γ_{k ℓ}^{c} = h_{a k} h_{b ℓ} h^{c s} {\bar{γ}}_{s}^{a b} .

(2.6)

Since $δ^{a} = δ_{0}^{a}$ with respect to the basis (1.1), we have $h_{c s} δ^{s} = γ_{c s}^{n} δ^{s} = γ_{c 0}^{n}$ , which is not zero only for $c = n$ and $γ_{n 0}^{n} = 1$ . Therefore ${\bar{γ}}_{0}^{a b} = {\bar{γ}}_{s}^{a b} δ^{s} = h^{a k} h^{b ℓ} γ_{k ℓ}^{c} h_{c s} δ^{s} = h^{a k} h^{b ℓ} γ_{k ℓ}^{n} = h^{a k} h^{b ℓ} h_{k ℓ} = h^{a k} δ_{k}^{b} = h^{a b}$ . Thus,

h^{a b} = {\bar{γ}}_{0}^{a b} .

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

{\bar{γ}}_{c}^{a n} = δ_{c}^{a} .

(2.7)

From the formula (1.2) it follows that $γ_{s r}^{c} h_{c d} = γ_{s r}^{c} γ_{c d}^{n} = γ_{d s}^{c} γ_{c r}^{n} = γ_{d s}^{c} h_{c r} = γ_{d r}^{c} γ_{c s}^{n} = γ_{d r}^{c} h_{c s}$ . Thus,

γ_{s r}^{c} h_{c d} = γ_{d s}^{c} h_{c r} = γ_{d r}^{c} h_{c s} .

(2.8)

The tensors $γ_{a b}^{c}$ and ${\bar{γ}}_{c}^{a b}$ are also related with the following formulas. We have $γ_{k a}^{c} h^{k ℓ} = h^{s c} h_{b a} h_{d k} h^{k ℓ} {\bar{γ}}_{s}^{d b} (by (2.6)) = h^{s c} h_{b a} {\bar{γ}}_{s}^{ℓ b}$ , hence $h_{b a} {\bar{γ}}_{s}^{ℓ b} = h_{s c} γ_{k a}^{c} h^{k ℓ} = γ_{s a}^{c} h_{c k} h^{k ℓ} (by (2.8)) = γ_{s a}^{ℓ}$ . Thus,

γ_{s a}^{ℓ} = h_{b a} {\bar{γ}}_{s}^{ℓ b},

(2.9)

whence

{\bar{γ}}_{s}^{b c} = h^{a b} γ_{s a}^{c} .

(2.10)

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

From (2.7) it follows that the last column of $∥ {\bar{q}}^{a b} ∥$ is

^{T} (t^{0}, t^{1}, \dots, t^{n}) .

(2.11)

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

q_{a b} t^{b} = γ_{a b}^{k} p_{k} t^{b} = δ_{a}^{n} .

(2.12)

Equivalently, $t = (t^{b})$ is deﬁned by $q (t, \cdot) = p_{(0)} (\cdot)$ . Therefore, by (2.11), the last column of the matrix $∥ q_{a b} {\bar{q}}^{b c} ∥$ 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 $c = 0, 1, \dots, n - 1$ ,

q_{a b} {\bar{q}}^{b c} = δ_{a}^{c} .

(2.13)

We need to check that $p_{k} γ_{a b}^{k} t^{s} {\bar{γ}}_{s}^{b c} = p_{k} γ_{a b}^{k} γ_{s r}^{c} h^{b r} t^{s} = δ_{a}^{c}$ . Contracting the left-hand side with $h_{c d}$ and using (2.8) yields $p_{k} γ_{a b}^{k} γ_{s r}^{c} h^{b r} t^{s} h_{c d} = p_{k} t^{s} h^{b r} γ_{a b}^{k} γ_{d s}^{c} h_{c r} = p_{k} t^{s} γ_{a b}^{k} γ_{d s}^{b} = p_{k} t^{s} γ_{b s}^{k} γ_{a d}^{b} = δ_{b}^{n} γ_{a d}^{b} = γ_{a d}^{n} = h_{a d}$ . Since the contraction of $δ_{a}^{c}$ with $h_{c d}$ also gives $h_{a d}$ and $h_{a d}$ is nondegenerate, the relation (2.13) holds true. Thus, the inverse matrix $∥ q^{a b} ∥$ has the form $∥ {\bar{γ}}_{c}^{a b} t^{c} ∥$ , 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 $det ∥ γ_{a b}^{c} t^{b} ∥ \neq 0$ . Contracting with $h^{a d}$ gives $t^{b} γ_{a b}^{c} h^{a d} = t^{b} {\bar{γ}}_{b}^{c d} = {\bar{q}}^{c d}$ . The last matrix is nondegenerate, hence the result follows. It follows also that if $det ∥ γ_{a b}^{c} t^{b} ∥ \neq 0$ for some $t = (t^{b})$ then the corresponding covector $p \in A_{F r}^{*}$ . Indeed, in this case $∥ {\bar{q}}^{c d} ∥ = ∥ q^{c d} ∥$ is nondegenerate, hence $∥ q_{c d} ∥$ is also nondegenerate.

The last row of $∥ q_{a b} ∥$ has the form $(p_{n}, 0, \dots, 0)$ . Its product with the last column of $∥ q^{b c} ∥$ 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, $t^{0} \neq 0$ .

Thus, we proved the following

Proposition 2.1. For any Frobenius covector $p$ ( $p_{n} \neq 0$ ) on $A$ the matrix $∥ q^{a b} ∥$ is of the form $q^{a b} = {\bar{γ}}_{c}^{a b} t^{c}$ , 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 $γ_{a b}^{c} p_{c} = γ_{a b}^{n} = h_{a b}$ , hence, $h_{a b} t^{b} = δ_{n}^{a}$ by (2.12). Denote by $t^{'}$ the column $^{T} (t^{1}, \dots, t^{n - 1})$ . From (2.3) we obtain $t^{n} = 0$ , $B t^{'} = 0$ , $t^{0} = 1$ . Since $det B \neq 0$ , 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 ${\overset{\circ}{A}}^{s}$ , $s = 1, 2, \dots, q - 1$ , in the ﬁrst column and the ﬁrst row we denote the $d_{s} (A)$ elements of (1.1) which lie in ${\overset{\circ}{A}}^{s} ∖ {\overset{\circ}{A}}^{s + 1}$ (or, equivalently, project to the basis of ${\overset{\circ}{A}}^{s} ∕ {\overset{\circ}{A}}^{s + 1}$ under the natural epimorphism ${\overset{\circ}{A}}^{s} \to {\overset{\circ}{A}}^{s + 1}$ ). The product of two such elements of (1.1), one of them lying in the ${\overset{\circ}{A}}^{k}$ -column, another in the ${\overset{\circ}{A}}^{ℓ}$ -row, belongs to ${\overset{\circ}{A}}^{k + ℓ}$ , thus we write ${\overset{\circ}{A}}^{k + ℓ}$ in the intersection of ${\overset{\circ}{A}}^{k}$ -column and ${\overset{\circ}{A}}^{ℓ}$ -row. The whole table now has the following block structure:


	1	$\overset{\circ}{A}$	${\overset{\circ}{A}}^{2}$	…	${\overset{\circ}{A}}^{k}$	…	${\overset{\circ}{A}}^{q - 2}$	${\overset{\circ}{A}}^{q - 1}$	$e_{n}$

1	1	$\overset{\circ}{A}$	${\overset{\circ}{A}}^{2}$	…	${\overset{\circ}{A}}^{k}$	…	${\overset{\circ}{A}}^{q - 2}$	${\overset{\circ}{A}}^{q - 2}$	${\overset{\circ}{A}}^{q}$

$\overset{\circ}{A}$	$\overset{\circ}{A}$	${\overset{\circ}{A}}^{2}$	${\overset{\circ}{A}}^{3}$	…	${\overset{\circ}{A}}^{k + 1}$	…	${\overset{\circ}{A}}^{q - 1}$	${\overset{\circ}{A}}^{q}$	0

${\overset{\circ}{A}}^{2}$	${\overset{\circ}{A}}^{2}$	${\overset{\circ}{A}}^{3}$	${\overset{\circ}{A}}^{4}$	…	${\overset{\circ}{A}}^{k + 2}$	…	${\overset{\circ}{A}}^{q}$	0	0

$⋮$	…	…	…	…	…	$. . .$	$. . .$	$. . .$	⋮

${\overset{\circ}{A}}^{q - k}$	${\overset{\circ}{A}}^{q - k}$	${\overset{\circ}{A}}^{q - k + 1}$	${\overset{\circ}{A}}^{q - k + 2}$	…	${\overset{\circ}{A}}^{q}$	$. . .$	$. . .$	…	⋮

$⋮$	…	…	…	$. . .$	$. . .$	$. . .$	…	…	⋮

${\overset{\circ}{A}}^{q - 2}$	${\overset{\circ}{A}}^{q - 2}$	${\overset{\circ}{A}}^{q - 1}$	${\overset{\circ}{A}}^{q}$	0	$. . .$	…	0	0	0

${\overset{\circ}{A}}^{q - 1}$	${\overset{\circ}{A}}^{q - 1}$	${\overset{\circ}{A}}^{q}$	0	$. . .$	…	…	0	0	0

$e_{n}$	${\overset{\circ}{A}}^{q}$	0	0	…	…	…	0	0	0

The secondary diagonal of this table contains blocks ${\overset{\circ}{A}}^{q}$ consisting of real multiples of $e_{n}$ . Hence all the matrices $∥ γ_{a b}^{c} ∥$ , $c = 0, 1, \dots, n - 1$ , contain zeros in these blocks and the matrix $∥ γ_{a b}^{n} ∥ = ∥ h_{a b} ∥$ has the following block structure:


	1	$\overset{\circ}{A}$	${\overset{\circ}{A}}^{2}$	…	${\overset{\circ}{A}}^{q - 2}$	${\overset{\circ}{A}}^{q - 1}$	$e_{n}$

1	0	0	0	…	0	0	1

$\overset{\circ}{A}$	0	*	*	…	*	$B_{1, q - 1}$	0

${\overset{\circ}{A}}^{2}$	0	*	*	$. . .$	$B_{2, q - 2}$	0	0

$⋮$	$⋮$	…	$. . .$	$. . .$	$. . .$	$⋮$	$⋮$

${\overset{\circ}{A}}^{q - 2}$	0	*	$B_{q - 2, 2}$	$. . .$	0	0	0

${\overset{\circ}{A}}^{q - 1}$	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 $k = 1, 2, \dots, q - 1$ .

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

det ∥ q_{a b} ∥ = det ∥ p_{n} \cdot γ_{a b}^{n} ∥ = p_{n}^{n + 1} det B_{1, q - 1} \cdot \dots \cdot det B_{q - 1, 1},

(2.14)

therefore, all the blocks $B_{k, q - k}$ , $k = 1, 2, \dots, q - 1$ , are nondegenerate. From (2.14) it follows that for this algebra each $p \in A^{*}$ such that $p_{n} \neq 0$ , is a Frobenius covector.

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

∥ γ_{a b}^{n} ∥ = (\begin{matrix} 0 & 0 & \dots & 0 & 1 \\ 0 & 0 & \dots & 1 & 0 \\ ⋮ & . . . & . . . & . . . & ⋮ \\ 0 & 1 & 0 & \dots & 0 \\ 1 & 0 & \dots & \dots & 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 ${e_{a}}$ , $a = 0, \dots, n$ , of Jordan-Hölder basis we have $e_{1} \dots, e_{n - 1} \in \overset{\circ}{A}$ , $e_{n} \in {\overset{\circ}{A}}^{2}$ . Therefore, $e_{a} e_{b} = λ_{a b} e_{n}$ , $a, b = 1, \dots, n - 1$ , and $e_{a} e_{n} = 0$ , $a = 1, \dots, n$ . Hence, for any Frobenius covector $p = (p_{a})$ ( $p_{n} \neq 0$ )

∥ q_{a b} ∥ = (\begin{matrix} p_{0} & p_{1} & p_{2} & \dots & p_{n - 1} & p_{n} \\ p_{1} & λ_{11} p_{n} & λ_{12} p_{n} & \dots & λ_{1, n - 1} p_{n} & 0 \\ p_{2} & λ_{21} p_{n} & λ_{22} p_{n} & \dots & λ_{2, n - 1} p_{n} & 0 \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ p_{n - 1} & λ_{n - 1, 1} p_{n} & λ_{n - 1, 2} p_{n} & \dots & λ_{n - 1, n - 1} p_{n} & 0 \\ p_{n} & 0 & \dots & \dots & 0 & 0 \end{matrix})

and $∥ q_{a b} ∥$ is nondegenerate if and only if $Λ = ∥ λ_{a b} ∥$ 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 $A \otimes B$ is also weakly symmetric.

Proof. If ${e_{a}}$ and ${f_{α}}$ are Jordan-Hölder bases of $A$ and $B$ respectively, then ${e_{a} \otimes f_{α}}$ is the Jordan-Hölder basis of $A \otimes B$ . It follows that the height of $A \otimes B$ 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 $d_{k} (A \otimes B) = d_{k} (A) + d_{k - 1} (A) d_{1} (B) + \dots + d_{1} (A) d_{k - 1} (B) + d_{k} (B)$ . Moreover, $d_{q_{1} + q_{2} - k} (A \otimes B) = d_{q_{1} - k} (A) + d_{q_{1} - k + 1} (A) d_{q_{2} - 1} (B) + \dots + d_{q_{1} - 1} (A) d_{q_{2} - k + 1} (B) + d_{q_{2} - k} (B)$ , which coincides with $d_{k} (A \otimes B)$ . $□$

3. Complete lifts of tensor ﬁelds

Let $A$ be a Weil algebra of height $q$ and let $n = dim \overset{\circ}{A}$ . 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 $q = (q_{a b})$ .

Let $M$ be an $m$ -dimensional smooth manifold. Then $T^{A} M$ is an $m$ -dimensional $A$ -smooth manifold and for each $x \in T^{A} M$ the tangent space $T_{x} T^{A} M$ is an $m$ -dimensional $A$ -module. Thus, we can consider $A$ -tensors at any point $x \in T^{A} M$ 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 $C^{\infty}$ .

We denote the algebra of smooth functions on $M$ by $C^{\infty} (M)$ , 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 $∣ \cdot ∣$ we denote the degree of a tensor ﬁeld, i.e., $∣ ξ ∣ = s$ if $ξ \in T^{s} (M)$ and $∣ u ∣ = s$ if $u \in V^{s} (M)$ .

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

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

ξ = ξ_{i_{1} \dots i_{k}} d x^{i_{1}} \otimes \dots \otimes d x^{i_{k}} .

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

ξ_{i_{1} a_{1} \dots i_{k} a_{k}} : = Ξ_{i_{1} a_{1} \dots 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} \dots i_{k} a_{k}} d x^{i_{1} a_{1}} \otimes \dots \otimes d x^{i_{k} a_{k}} .

If ${e_{a^{'}}}$ is another basis in $A$ and $e_{a^{'}} = τ_{a^{'}}^{a} e_{a}$ , then, obviously, $d x^{i a^{'}} = τ_{a}^{a^{'}} d x^{i a}$ . 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 $ϕ : N \to M$ 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 $ϕ : N \to M$ 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} \dots α_{k}} = \frac{\partial ϕ^{i_{1}}}{\partial y^{α_{1}}} \dots \frac{\partial ϕ^{i_{k}}}{\partial y^{α_{k}}} ξ_{i_{1} \dots i_{k}} .

Let $Θ_{α_{1} \dots α_{k}}$ be analytic prolongations of the components $θ_{α_{1} \dots α_{k}}$ and let $Φ^{i}$ be the analytic prolongations of the maps $ϕ^{i}$ . Note that $Φ^{i} (Y^{α}) = ϕ^{i a} (Y^{α}) e_{a} = ϕ^{i a} (y^{α b}) e_{a}$ is the local representation of the map $T^{A} ϕ : T^{A} N \to T^{A} M$ , considered as an $A$ -smooth map and the functions $ϕ^{i a} (y^{α b})$ 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} \dots α_{k}} = \frac{\partial Φ^{i_{1}}}{\partial Y^{α_{1}}} \dots \frac{\partial Φ^{i_{k}}}{\partial Y^{α_{k}}} Ξ_{i_{1} \dots i_{k}} .

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

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

Thus

Θ_{α_{1} \dots α_{k}} e_{b_{1}} \dots e_{b_{k}} = \frac{\partial ϕ^{i_{1} c_{1}}}{\partial y^{α_{1}}} e_{c_{1}} \dots \frac{\partial ϕ^{i_{k} c_{k}}}{\partial y^{α_{k}}} e_{c_{k}} U_{i_{1} \dots i_{k}} e_{b_{1}} \dots 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{\partial ϕ^{i_{1} c_{1}}}{\partial y^{α_{1}}} γ_{b_{1} c_{1}}^{a_{1}} \dots \frac{\partial ϕ^{i_{k} c_{k}}}{\partial y^{α_{k}}} γ_{b_{k} c_{k}}^{a_{k}} Ξ_{i_{1} \dots i_{k}} e_{a_{1}} \dots e_{a_{k}} .

From Scheffers’ equations (1.5) it follows that

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

hence,

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

Then, by (3.1), we have

θ_{α_{1} b_{1} \dots α_{k} b_{k}} = \frac{\partial ϕ^{i_{1} a_{1}}}{\partial y^{α_{1} b_{1}}} \dots \frac{\partial ϕ^{i_{k} a_{k}}}{\partial y^{α_{k} b_{k}}} ξ_{i_{1} a_{1} \dots 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} \dots i_{k} a_{k}} = \frac{\partial ϕ^{j_{1}^{'} b_{1}}}{\partial x^{i_{1} a_{1}}} \dots \frac{\partial ϕ^{j_{k}^{'} b_{k}}}{\partial x^{i_{k} a_{k}}} ξ_{j_{1}^{'} b_{1} \dots j_{k}^{'} b_{k}} .

This implies that $ξ_{i_{1} a_{1} \dots i_{k} a_{k}}$ 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 $A$ : $e^{a} = q^{a b} e_{b}$ , then $e_{a} = q_{a b} e^{b}$ .

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

u = u^{i_{1} \dots i_{k}} \frac{\partial}{\partial x^{i_{1}}} \land \dots \land \frac{\partial}{\partial x^{i_{k}}} .

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

u^{i_{1} a_{1} \dots i_{k} a_{k}} : = U_{b}^{i_{1} a_{1} \dots 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} \dots i_{k} a_{k}} \frac{\partial}{\partial x^{i_{1} a_{1}}} \land \dots \land \frac{\partial}{\partial 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 $ϕ : M \to N$ 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 $ϕ : M \to N$ be a smooth map which is given by $y^{α} = ϕ^{α} (x^{i})$ with respect to the local coordinates. Let $v \in V^{k} (N)$ be another tensor ﬁeld and let $u$ and $v$ be $ϕ$ -related ([13, 14]). Then, in local coordinates we have

v^{α_{1} \dots α_{k}} = \frac{\partial ϕ^{α_{1}}}{\partial x^{i_{1}}} \dots \frac{\partial ϕ^{α_{k}}}{\partial x^{i_{k}}} u^{i_{1} \dots i_{k}}

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

V^{α_{1} \dots α_{k}} = \frac{\partial Φ^{α_{1}}}{\partial X^{i_{1}}} \dots \frac{\partial Φ^{α_{k}}}{\partial X^{i_{k}}} U^{i_{1} \dots i_{k}},

and

V^{α_{1} \dots α_{k}} e^{b_{1}} \dots e^{b_{k}} = \frac{\partial ϕ^{α_{1} c_{1}}}{\partial x^{i_{1}}} e_{c_{1}} \dots \frac{\partial ϕ^{α_{k} c_{k}}}{\partial x^{i_{k}}} e_{c_{k}} U^{i_{1} \dots i_{k}} e^{b_{1}} \dots 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 $r = 1, \dots, k$ . Let us show that $q^{b s} γ_{s c}^{d} q_{a d} = γ_{a c}^{b}$ . Indeed, the contraction of the left-hand side with $q_{b f}$ gives $q^{b s} γ_{s c}^{d} q_{a d} q_{b f} = γ_{f c}^{d} q_{a d}$ , which coincides with $γ_{a c}^{b} q_{b f}$ by (1.9). Thus,

\begin{matrix} V^{α_{1} \dots α_{k}} e^{b_{1}} \dots e^{b_{k}} & = \frac{\partial ϕ^{α_{1} c_{1}}}{\partial x^{i_{1}}} e_{c_{1}} \dots \frac{\partial ϕ^{α_{k} c_{k}}}{\partial x^{i_{k}}} e_{c_{k}} U^{i_{1} \dots i_{k}} e^{b_{1}} \dots e^{b_{k}} = \\ = γ_{a_{1} c_{1}}^{b_{1}} \frac{\partial ϕ^{α_{1} c_{1}}}{\partial x^{i_{1}}} \dots γ_{a_{k} c_{k}}^{b_{k}} \frac{\partial ϕ^{α_{k} c_{k}}}{\partial x^{i_{k}}} U^{i_{1} \dots i_{k}} e^{a_{1}} \dots e^{a_{k}} = \\ = \frac{\partial ϕ^{α_{1} b_{1}}}{\partial x^{i_{1} a_{1}}} \dots \frac{\partial ϕ^{α_{k} b_{k}}}{\partial x^{i_{k} a_{k}}} U^{i_{1} \dots i_{k}} e^{a_{1}} \dots e^{a_{k}}, \end{matrix}

hence, by (3.4),

v^{α_{1} b_{1} \dots α_{k} b_{k}} = \frac{\partial ϕ^{α_{1} b_{1}}}{\partial x^{i_{1} a_{1}}} \dots \frac{\partial ϕ^{α_{k} b_{k}}}{\partial x^{i_{k} a_{k}}} u^{i_{1} a_{1} \dots 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} \dots j_{k}^{'} b_{k}} = \frac{\partial ϕ^{j_{1}^{'} b_{1}}}{\partial x^{i_{1} a_{1}}} \dots \frac{\partial ϕ^{j_{k}^{'} b_{k}}}{\partial x^{i_{k} a_{k}}} u^{i_{1} a_{1} \dots i_{k} a_{k}} .

This means that $u_{i_{1} a_{1} \dots i_{k} a_{k}}$ 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

[\cdot, \cdot] : V^{k} (M) \times V^{ℓ} (M) \to 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 $F_{b} = q_{a b} F^{a}$ . From the Scheffers’ conditions (1.4) it follows that $\partial_{c} F_{b} = γ_{c d}^{a} q_{a b} δ^{g} \partial_{g} F^{d} = γ_{c d}^{a} q_{a b} δ^{g} q^{d r} \partial_{g} F_{r}$ . Contracting with $δ^{b}$ , by (1.11), we obtain $\partial_{c} (δ^{b} F_{b}) = δ^{b} q_{a b} δ^{g} γ_{c d}^{a} q^{d r} \partial_{g} F_{r} = p_{a} γ_{c d}^{a} q^{d r} δ^{g} \partial_{g} F_{r} = q_{c d} q^{d r} δ^{g} \partial_{g} F_{r} = δ^{g} \partial_{g} F_{c}$ . Thus,

\partial_{c} (δ^{b} F_{b}) = δ^{g} \partial_{g} F_{c} .

(3.6)

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

\begin{matrix} {[u, v]}^{k_{2} \dots k_{g + h}} = & \frac{1}{(g - 1)! h!} ɛ_{i_{2} \dots i_{g} j_{1} \dots j_{h}}^{k_{2} \dots k_{g + h}} u^{r i_{2} \dots i_{g}} \frac{\partial}{\partial x^{r}} v^{j_{1} \dots j_{h}} + \\ + \frac{{(- 1)}^{g}}{g! (h - 1)!} ɛ_{i_{1} \dots i_{g} j_{2} \dots j_{h}}^{k_{2} \dots k_{g + h}} v^{r j_{2} \dots j_{h}} \frac{\partial}{\partial x^{r}} u^{i_{1} \dots i_{g}}, \end{matrix}

(3.7)

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

\begin{matrix} {[U, V]}^{k_{2} \dots k_{g + h}} = & \frac{1}{(g - 1)! h!} ɛ_{i_{2} \dots i_{g} i_{g + 1} \dots i_{g + h}}^{k_{2} \dots k_{g + h}} U^{r i_{2} \dots i_{g}} \frac{\partial}{\partial X^{r}} V^{i_{g + 1} \dots i_{g + h}} + \\ + \frac{{(- 1)}^{g}}{g! (h - 1)!} ɛ_{i_{1} \dots i_{g} i_{g + 2} \dots i_{g + h}}^{k_{2} \dots k_{g + h}} V^{r i_{g + 2} \dots i_{g + h}} \frac{\partial}{\partial X^{r}} U^{i_{1} \dots i_{g}}, \end{matrix}

(3.8)

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

\begin{matrix} \frac{1}{(g - 1)! h!} U^{r i_{2} \dots i_{g}} e^{a_{2}} \dots e^{a_{g}} \frac{\partial}{\partial X^{r}} V^{i_{g + 1} \dots i_{g + h}} e^{a_{g + 1}} \dots e^{a_{g + h}} = \\ \frac{1}{(g - 1)! h!} U^{r i_{2} \dots i_{g}} e^{a_{2}} \dots e^{a_{g}} δ^{b} \frac{\partial}{\partial x^{r b}} V^{a_{g + 1} i_{g + 1} \dots a_{g + h} i_{g + h}} b y (1.6) = \\ \frac{1}{(g - 1)! h!} U^{r i_{2} \dots i_{g}} e^{a} e^{a_{2}} \dots e^{a_{g}} δ^{b} \frac{\partial}{\partial x^{r b}} V_{a}^{a_{g + 1} i_{g + 1} \dots a_{g + h} i_{g + h}} = \\ \frac{1}{(g - 1)! h!} U^{r i_{2} \dots i_{g}} e^{a} e^{a_{2}} \dots e^{a_{g}} \frac{\partial}{\partial x^{r a}} (δ^{b} V_{b}^{a_{g + 1} i_{g + 1} \dots a_{g + h} i_{g + h}}) b y (3.6) = \\ \frac{1}{(g - 1)! h!} U_{m}^{r a i_{2} a_{2} \dots i_{g} a_{g}} e^{m} \frac{\partial}{\partial x^{r a}} ({(v^{C})}^{a_{g + 1} i_{g + 1} \dots a_{g + h} i_{g + h}}) . \end{matrix}

Contracting with $δ^{m}$ we get

\frac{1}{(g - 1)! h!} {(u^{C})}^{r a i_{2} a_{2} \dots i_{g} a_{g}} \frac{\partial}{\partial x^{r a}} ({(v^{C})}^{a_{g + 1} i_{g + 1} \dots a_{g + h} i_{g + h}}) .

The commutativity of multiplication in $A$ yields that

ɛ_{i_{2} \dots i_{g + h}}^{k_{2} \dots k_{g + h}} = ɛ_{i_{2} a_{2} \dots i_{g + h} a_{g + h}}^{k_{2} a_{2} \dots k_{g + h} a_{g + h}} .

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

\begin{matrix} {({[u, v]}^{C})}^{k_{2} a_{2} \dots k_{g + h} a_{g + h}} = \\ \frac{1}{(g - 1)! h!} ɛ_{i_{2} a_{2} \dots i_{g} a_{g} i_{g + 1} a_{g + 1} \dots i_{g + h} a_{g + h}}^{k_{2} a_{2} \dots k_{g + h} a_{g + h}} {(u^{C})}^{r a i_{2} a_{2} \dots i_{g} a_{g}} \frac{\partial}{\partial x^{r a}} ({(v^{C})}^{a_{g + 1} i_{g + 1} \dots a_{g + h} i_{g + h}}) + \\ + \frac{{(- 1)}^{g}}{g! (h - 1)!} ɛ_{i_{1} a_{1} \dots i_{g} a_{g} i_{g + 2} a_{g + 2} \dots i_{g + h} a_{g + h}}^{k_{2} a_{2} \dots k_{g + h} a_{g + h}} {(v^{C})}^{r a i_{g + 2} a_{g + 2} \dots i_{g + h} a_{g + h}} \frac{\partial}{\partial x^{r a}} ({(u^{C})}^{i_{1} a_{1} \dots i_{g} a_{g}}), \end{matrix}

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 $u = u^{i_{1} \dots i_{k}} \frac{\partial}{\partial x^{i_{1}}} \land \dots \land \frac{\partial}{\partial x^{i_{k}}} \in V^{k} (M)$ . We deﬁne the vertical lift $u^{V} \in V^{k} (T^{A} M)$ of $u$ by

u^{V} = u_{A}^{V} : = u^{i_{1} \dots i_{k}} \frac{\partial}{\partial x^{i_{1} n}} \land \dots \land \frac{\partial}{\partial 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}^{'} \dots i_{k}^{'}} = \frac{\partial x^{i_{1}^{'}}}{\partial x^{i_{1}}} \dots \frac{\partial x^{i_{k}^{'}}}{\partial x^{i_{k}}} u^{i_{1} \dots i_{k}} .

Thus it suffices to prove that

\frac{\partial}{\partial x^{i_{s}^{'} n}} = \frac{\partial x^{i_{s}}}{\partial x^{i_{s}^{'}}} \frac{\partial}{\partial x^{i_{s} n}} .

(3.10)

Let us determine the corresponding change of coordinates $x^{i^{'} a} = x^{i^{'} a} (x^{i b})$ on $T^{A} M$ . By (1.7),

X^{i^{'}} = x^{i^{'}} + x^{i^{'} \hat{a}} e_{\hat{a}} = x^{i^{'}} + \sum_{∣ p ∣ = 1}^{q} \frac{1}{p!} \frac{D^{p} x^{i^{'}}}{D x^{p}} {\overset{\circ}{X}}^{p} .

(3.11)

Hence, for $T^{A} M$ we have

\begin{matrix} x^{i^{'}} = x^{i^{'}} (x^{i}), \\ x^{i^{'} \hat{a}} = f^{i^{'} \hat{a}} (x^{i b}), \hat{a} = 1, \dots, n . \end{matrix}

Let us show that

\frac{\partial f^{i^{'} \hat{a}}}{\partial x^{i b}} = 0 f o r b > \hat{a} .

Indeed, the coefficients $\frac{1}{p!} \frac{D^{p} x^{i^{'}}}{D x^{p}}$ in (3.11) depend only on $x^{i} = x^{i 0}$ , while the expression ${({\overset{\circ}{X}}^{1})}^{p_{1}} \dots {({\overset{\circ}{X}}^{m})}^{p_{m}}$ contains $x^{i b}$ only as a coefficient at $e_{b}$ in ${\overset{\circ}{X}}^{i}$ . Since $γ_{c \hat{d}}^{s} = 0$ , for $c > s$ the coefficient at $e_{\hat{a}}$ which appears when we expand brackets in ${({\overset{\circ}{X}}^{1})}^{p_{1}} \dots {({\overset{\circ}{X}}^{m})}^{p_{m}}$ does not depend on $x^{i b}$ . Moreover, it can depend on $x^{i \hat{a}}$ only for the summand corresponding to $∣ p ∣ = 1$ (because for $∣ p ∣ \geq 2$ the expression $x^{i \hat{a}} e_{\hat{a}}$ will be multiplied by some element of $\overset{\circ}{A}$ and thus after the expanding brackets the coefficient at $e_{\hat{a}}$ does not depend on $x^{i \hat{a}}$ ). For the case $∣ p ∣ = 1$ we obtain the only summand $\frac{\partial x^{i^{'}}}{\partial x^{i}} x^{i \hat{a}} e_{\hat{a}}$ depending on $x^{i \hat{a}}$ . Therefore,

\frac{\partial x^{i^{'} \hat{a}}}{\partial x^{i \hat{a}}} = \frac{\partial x^{i^{'}}}{\partial x^{i}} f o r a n y \hat{a} = 1, \dots, n .

Hence, the Jacobi matrix of the coordinate transformation $x^{i^{'} a} = x^{i^{'} a} (x^{i b})$ on $T^{A} M$ has the following block structure:


$\frac{\partial x^{i^{'}}}{\partial x^{i}}$	0	0	$\dots$	0	0

$*$	$\frac{\partial x^{i^{'}}}{\partial x^{i}}$	0	$\dots$	0	0

$*$	$*$	$⋱$	$⋱$	$⋮$	$⋮$

$⋮$	$⋱$	$⋱$	$⋱$	$⋱$	$⋮$

$*$	$\dots$	$⋱$	$*$	$\frac{\partial x^{i^{'}}}{\partial x^{i}}$	0

$*$	$*$	$\dots$	$*$	$*$	$\frac{\partial x^{i^{'}}}{\partial x^{i}}$

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

Proposition 3.5. For any $u, v \in V^{*} (M)$ there holds

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

(3.12)

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

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

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

\frac{\partial v^{j_{1} b_{1} \dots j_{ℓ} b_{ℓ}}}{\partial x^{r n}} = 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 ${,} : C^{\infty} (M) \times C^{\infty} (M) \to C^{\infty} (M)$ , satisfying the Leibniz rule

{f, g h} = {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^{i j} \frac{\partial}{\partial x^{i}} \land \frac{\partial}{\partial x^{j}} \in V^{2} (M)

(4.1)

such that

{f, g} = i (w) (d f \land d g)

(4.2)

for any $f, g \in C^{\infty} (M)$ , where $i (w) : Ω^{m} (M) \to Ω^{m - 2} (M)$ is the inner product; in local coordinates

{(i (w) α)}_{i_{1} \dots i_{m - 2}} = w^{j k} α_{j k i_{1} \dots 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 $C^{\infty} (M)$ satisﬁes the Jacobi identity if and only if

[w, w] = 0 .

(4.3)

In local coordinates this is written as

w^{j s} \frac{\partial w^{k ℓ}}{\partial x^{s}} + w^{k s} \frac{\partial w^{ℓ j}}{\partial x^{s}} + w^{ℓ s} \frac{\partial w^{j k}}{\partial x^{s}} = 0 .

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

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

X_{f}^{w} = {f, \cdot}_{w} = \frac{\partial f}{\partial x^{i}} w^{i j} \frac{\partial}{\partial x^{j}} (f \in C^{\infty} (M)) .

(4.4)

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

σ = σ_{w} : V^{k} (M) \to V^{k + 1} (M),

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

{(- 1)}^{∣ u ∣ \cdot ∣ v ∣} [[v, y], u] + {(- 1)}^{∣ v ∣ \cdot ∣ y ∣} [[y, u], v] + {(- 1)}^{∣ y ∣ \cdot ∣ u ∣} [[u, v], y] = 0,

one has $σ \circ σ = 0$ . Thus the cohomology spaces

H_{L P}^{k} (M, w) : = \frac{ker σ : V^{k} (M) \to V^{k + 1} (M)}{im σ : V^{k - 1} (M) \to 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 $f \in C^{\infty} (M)$ , 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] \in H_{L P}^{*} (M, w) \mapsto [u^{C}] \in H_{L P}^{*} (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 $w_{s}^{i j}$ be the components of the analytic prolongation $W^{i j} = {(w^{i j})}^{A}$ , i.e., $W^{i j} = \sum_{s \geq 0} w_{s}^{i j} e_{s}$ . In what follows we will omit the sign of summation over $s$ . Then $W^{i j} e^{a} e^{b} = w_{s}^{i j} e_{s} e^{a} e^{b} = w_{s}^{i j} q^{a c} q^{b k} e_{s} e_{c} e_{d} = w_{s}^{i j} q^{a c} q^{b k} γ_{s c}^{d} γ_{k d}^{ℓ} e_{ℓ} = w_{s}^{i j} q^{a c} q^{b k} γ_{s c}^{d} γ_{k d}^{ℓ} q_{ℓ r} e^{r}$ . Contracting with $δ^{r}$ , we obtain

\begin{matrix} w^{i a j b} = w_{s}^{i j} q^{a c} q^{b k} γ_{s c}^{d} γ_{k d}^{ℓ} q_{ℓ r} δ^{r} = w_{s}^{i j} q^{a c} q^{b k} γ_{s c}^{d} γ_{k d}^{ℓ} p_{ℓ} = \\ w_{s}^{i j} {\bar{γ}}_{r}^{a c} t^{r} {\bar{γ}}_{f}^{b k} t^{f} γ_{s c}^{d} γ_{k d}^{ℓ} p_{ℓ} = w_{s}^{i j} t^{r} t^{f} h^{a g} γ_{g r}^{c} h^{e b} γ_{f e}^{k} γ_{s c}^{d} γ_{k d}^{ℓ} p_{ℓ} (b y (2.5)) = \\ w_{s}^{i j} t^{r} t^{f} h^{a g} h^{e b} γ_{g r}^{c} γ_{d e}^{k} γ_{k f}^{ℓ} γ_{s c}^{d} p_{ℓ} (b y (1.2)) = w_{s}^{i j} t^{r} h^{a g} h^{e b} γ_{g r}^{c} γ_{d e}^{k} δ_{k}^{n} γ_{s c}^{d} = \\ (b y (2.12)) = \\ w_{s}^{i j} t^{r} h^{a g} h^{e b} γ_{g r}^{c} h_{d e} γ_{s c}^{d} = w_{s}^{i j} t^{r} h^{a g} δ_{d}^{b} γ_{g r}^{c} γ_{s c}^{d} = w_{s}^{i j} t^{r} h^{a g} γ_{g r}^{c} γ_{s c}^{b} = \\ w_{s}^{i j} t^{r} {\bar{γ}}_{r}^{a c} γ_{s c}^{b} . \end{matrix}

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} = \sum_{s = 0}^{n} w_{s}^{i j} {\bar{γ}}_{k}^{a c} γ_{c s}^{b} \frac{\partial}{\partial x^{i a}} \land \frac{\partial}{\partial x^{j b}} .

(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 $k, ℓ = 0, \dots, n$ . 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 $p_{(0)} = (0, \dots, 0, 1)$ is a Frobenius covector. Recall that the corresponding vector $t_{(0)}$ is $(1, 0, \dots, 0)$ . Since the determinant $det ∥ γ_{a b}^{c} t^{b} ∥$ 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 $ɛ > 0$ such that this determinant is nonzero for each of the vectors $t_{(k)} = (1, 0 \dots, 0, ɛ, 0, \dots, 0)$ ( $ɛ$ at the $k$ th place), $t_{(k -)} = (1, 0 \dots, 0, - ɛ, 0, \dots, 0)$ ( $- ɛ$ at the $k$ th place) and

t_{(k, ℓ)} = (1, 0 \dots, 0, ɛ, 0, \dots, 0, ɛ, 0, \dots, 0)

( $ɛ$ at the $k$ th and $ℓ$ th places), $k, ℓ = 1, \dots, n$ . 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 $w_{0}^{C} \in V^{2} (T^{A} M)$ and $[w_{0}^{C}, w_{0}^{C}] = 0$ . Now substitute $t = t_{(k)}$ and $t = t_{(k -)}$ , which gives $w_{0}^{C} + ɛ w_{k}^{C} \in V^{2} (T^{A} M)$ , therefore $w_{k}^{C} \in V^{2} (T^{A} M)$ and

[w_{0}^{C} + ɛ w_{k}^{C}, w_{0}^{C} + ɛ w_{k}^{C}] = [w_{0}^{C} - ɛ w_{k}^{C}, w_{0}^{C} - ɛ w_{k}^{C}] = 0 .

Expanding these equations yields $2 ɛ [w_{k}^{C}, w_{0}^{C}] + ɛ^{2} [w_{k}^{C}, w_{k}^{C}] = - 2 ɛ [w_{k}^{C}, w_{0}^{C}] + ɛ^{2} [w_{k}^{C}, w_{k}^{C}] = 0$ , 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 $k, ℓ = 0, \dots, n$ .

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 ${\bar{e}}^{n} = e_{0}$ and that each of ${\bar{e}}^{a}$ , $a = 0, 1, \dots, n - 1$ , does not contain $e_{0}$ in its decomposition. Hence the only nonzero component of ${\bar{γ}}_{n}^{a c}$ is ${\bar{γ}}_{n}^{n n} = 1$ . Then $γ_{n d}^{b}$ is 1 only for $b = n$ , $d = 0$ and $γ_{n d}^{b} = 0$ for any other values of $b$ and $d$ . Consequently,

w_{n}^{C} = w^{i j} \frac{\partial}{\partial x^{i n}} \land \frac{\partial}{\partial x^{j n}} .

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 $R (ɛ^{n})$ . Choose a Jordan-Hölder basis $e_{0} = 1$ , $e_{a} = ɛ^{a}$ , $a = 1, \dots, n$ in it. The explicit form of the analytic prolongations $W^{i j} = \sum_{s \geq 0} w_{s}^{i j} ɛ^{s}$ can be found by (1.7), for instance,

\begin{matrix} w_{0}^{i j} = w^{i j}, \\ w_{1}^{i j} = x^{k 1} \frac{\partial w^{i j}}{\partial x^{k}}, \\ w_{2}^{i j} = x^{k 1} x^{ℓ 1} \frac{\partial^{2} w^{i j}}{\partial x^{k} \partial x^{ℓ}} + x^{k 2} \frac{\partial w^{i j}}{\partial x^{k}}, e t c . \end{matrix}

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

w_{0}^{C} = (\begin{matrix} 0 & 0 & \dots & 0 & 0 & w^{i j} \\ 0 & 0 & \dots & 0 & w^{i j} & w_{1}^{i j} \\ ⋮ & ⋮ & . . . & . . . & . . . & ⋮ \\ 0 & 0 & . . . & . . . & \dots & \dots \\ 0 & w^{i j} & w_{1}^{i j} & \dots & \dots & w_{n - 1}^{i j} \\ w^{i j} & w_{1}^{i j} & \dots & \dots & w_{n - 1}^{i j} & w_{n}^{i j} \end{matrix}),

w_{k}^{C} = (\begin{matrix} 0 & 0 & \dots & \dots & 0 & 0 & \dots & 0 & 0 \\ 0 & 0 & \dots & \dots & 0 & 0 & \dots & ⋮ & ⋮ \\ ⋮ & ⋮ & . . . & . . . & . . . & . . . & . . . & \dots & 0 \\ 0 & 0 & . . . & . . . & 0 & 0 & \dots & 0 & w^{i j} \\ 0 & 0 & . . . & . . . & 0 & \dots & 0 & w^{i j} & w_{1}^{i j} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ & . . . & . . . & . . . & ⋮ \\ 0 & 0 & \dots & \dots & . . . & . . . & . . . & \dots & ⋮ \\ 0 & 0 & \dots & 0 & w^{i j} & w_{1}^{i j} & \dots & \dots & w_{n - k - 1}^{i j} \\ 0 & \dots & 0 & w^{i j} & w_{1}^{i j} & \dots & \dots & w_{n - k - 1}^{i j} & w_{n - k}^{i j} \end{matrix}), k = 1, \dots, n - 1,

w_{n}^{C} = (\begin{matrix} 0 & 0 & \dots & 0 & 0 \\ 0 & 0 & \dots & 0 & 0 \\ ⋮ & ⋮ & . . . & ⋮ & ⋮ \\ 0 & 0 & \dots & \dots & 0 \\ 0 & 0 & \dots & 0 & w^{i j} \end{matrix}) .

In particular, for the case $A = R (ɛ)$ , 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^{i j} \frac{\partial}{\partial x^{i}} \land \frac{\partial}{\partial y^{j}} + \frac{1}{2} y^{k} \frac{\partial w^{i j}}{\partial x^{k}} \frac{\partial}{\partial y^{i}} \land \frac{\partial}{\partial 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 $λ^{a b}$ .

By (1.7), denote

w_{a}^{i j} = x^{k a} \frac{\partial w^{i j}}{\partial x^{k}}, w_{n}^{i j} = x^{k n} \frac{\partial w^{i j}}{\partial x^{k}} + λ_{a b} x^{k a} x^{ℓ b} \frac{\partial^{2} w^{i j}}{\partial x^{k} \partial x^{ℓ}} .

Then, for this algebra (4.7) implies that

w_{0}^{C} = (\begin{matrix} 0 & 0 & 0 & \dots & 0 & w^{i j} \\ 0 & λ^{11} w^{i j} & λ^{12} w^{i j} & \dots & λ^{1, n - 1} w^{i j} & w_{1}^{i j} \\ 0 & λ^{21} w^{i j} & λ^{22} w^{i j} & \dots & λ^{2, n - 1} w^{i j} & w_{2}^{i j} \\ ⋮ & \dots & \dots & \dots & \dots & ⋮ \\ 0 & λ^{n - 1, 1} w^{i j} & λ^{n - 1, 2} w^{i j} & \dots & λ^{n - 1, n - 1} w^{i j} & w_{n - 1}^{i j} \\ w^{i j} & w_{1}^{i j} & w_{2}^{i j} & \dots & w_{n - 1}^{i j} & w_{n}^{i j} \end{matrix}),

w_{b}^{C} = (\begin{matrix} 0 & 0 & \dots & \dots & 0 & 0 \\ ⋮ & ⋮ & \dots & \dots & \dots & ⋮ \\ ⋮ & ⋮ & \dots & \dots & \dots & w^{i j} \\ ⋮ & ⋮ & \dots & \dots & \dots & ⋮ \\ 0 & \dots & \dots & \dots & 0 & 0 \\ 0 & \dots & w^{i j} & \dots & 0 & \sum_{a = 1}^{n - 1} λ_{a b} w_{a}^{i j} \end{matrix}), b = 1, \dots, n - 1, w^{i j} at the b th place

w_{n}^{C} = (\begin{matrix} 0 & \dots & 0 & 0 \\ ⋮ & \dots & \dots & ⋮ \\ 0 & \dots & \dots & 0 \\ 0 & \dots & 0 & w^{i j} \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 ${d i v}_{μ} X$ of a vector ﬁeld $X$ is deﬁned by

ℒ_{X} μ = ({d i v}_{μ} X) μ

and one has

{d i v}_{μ} (f X) = f {d i v}_{μ} X + X f, f \in C^{\infty} (M) .

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

Δ_{μ} : f \in C^{\infty} (M) \mapsto {d i v}_{μ} X_{f} \in C^{\infty} (M)

is deﬁned, where $X_{f}$ is a Hamiltonian vector ﬁeld of $f$ . Easy computations show that $Δ_{μ}$ is a derivation on $C^{\infty} (M)$ 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 $σ Δ_{μ} = 0$ [6]. If we replace $μ$ with any other volume form $a μ$ , where $a \in C^{\infty} (M)$ is a positive function, then the modular vector ﬁeld changes to $Δ_{a μ} = Δ_{μ} + H_{- log a}$ [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 $H_{L P}^{1} (M, w)$ . This cohomology class is called the modular class of the Poisson manifold $(M, w)$ .

Let $g = g_{i j} d x^{i} \otimes d x^{j}$ be a Riemannian metric on an $m$ -dimensional oriented manifold $M$ . Then

d V_{g} = \sqrt{det g} d x^{1} \land \dots \land d x^{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 $G_{i j} = g_{i j}^{s} e_{s}$ be the analytic prolongations of $g_{i j}$ . Then $G_{i j} e_{a} e_{b} = G_{i a j b}^{c} e_{c}$ and $g_{i a j b} = G_{i a j b}^{c} p_{c} = G_{i a j b}^{n}$ . Clearly, $g_{i a j b} = g_{j b i a}$ for all $a, b, i, j$ . If $s > 0$ , then $e_{s} \in \overset{\circ}{A}$ , hence the component $g_{i a j b}$ contains $g_{i j}^{s}$ only if $e_{a} e_{b} \notin {\overset{\circ}{A}}^{q}$ . Therefore, if $e_{a} e_{b} \in {\overset{\circ}{A}}^{q}$ , then $g_{i a j b}$ depends only on $g_{i j}^{0} = g_{i j}$ and the matrix $∥ g_{i a j b} ∥$ has the following block structure:


*	*	*	$\dots$	*	*	$∥ g_{i j} ∥$

*	$\dots$	$\dots$	$. . .$	*	${\hat{B}}_{1}$	0

*	$\dots$	$. . .$	$. . .$	${\hat{B}}_{2}$	0	0

$⋮$	$. . .$	$. . .$	$. . .$	$. . .$	$. . .$	$⋮$

*	*	${\hat{B}}_{q - 2}$	$. . .$	$. . .$	$\dots$	0

*	${\hat{B}}_{q - 1}$	0	$. . .$	$\dots$	$\dots$	0

$∥ g_{i j} ∥$	0	0	$\dots$	$\dots$	$\dots$	0

where ${\tilde{B}}_{k} = B_{q - k, k} \otimes ∥ g_{i j} ∥$ and the symbol $\otimes$ denotes the tensor (Kronecker) product of matrices.

The determinant of this matrix is the product of the determinants of diagonal blocks: $det ∥ g_{i a j b} ∥ = det ∥ g_{i j} ∥ \cdot det (B_{1, q - 1} \otimes ∥ g_{i j} ∥) \cdot \dots \cdot det (B_{q - 1, 1} \otimes ∥ g_{i j} ∥) \cdot det ∥ g_{i j} ∥$ . For any two matrices $S$ and $T$ of dimensions $k \times k$ and $ℓ \times ℓ$ respectively, one has

det (S \otimes T) = {(det S)}^{k} {(det T)}^{ℓ} .

We have $d_{1} (A) + \dots + d_{q - 1} (A) = n - 1$ . Hence, $det ∥ g_{i a j b} ∥ = M {(det g)}^{n + 1}$ , where $M = {(det B_{1, q - 1})}^{d_{1} (A)} \cdot \dots \cdot {(det B_{q - 1, 1})}^{d_{q - 1} (A)}$ . $□$

Let

Φ = d V_{g^{C}} = \sqrt{det g^{C}} d x^{1} \land \dots d x^{m} \land \dots \land d x^{1 n} \land \dots \land d x^{m n}

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

Proposition 5.2. Let $A$ $(dim \overset{\circ}{A} = n)$ 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_{d V_{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_{d V_{g}} = \sum_{j = 1}^{m} (\frac{\partial w^{i j}}{\partial x^{j}} + w^{i j} \frac{\partial ln \sqrt{det g}}{\partial x^{j}}) \frac{\partial}{\partial x^{i}} .

Then (5.1) implies that

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

Let $w_{k}^{i a j b}$ denote the components of $w_{k}^{C}$ . At ﬁrst, show that

$\begin{matrix} \frac{\partial w_{0}^{i a j b}}{\partial x^{j b}} = \{\begin{matrix} \frac{\partial w^{i j}}{\partial x^{j}}, & a = n, \\ 0, & a = 0, 1 \dots, n - 1, \end{matrix} \\ \frac{\partial w_{k}^{i a j b}}{\partial x^{j b}} = 0, k = 1, \dots, n . \end{matrix}$	(5.4)

By (4.7), $w_{k}^{i a j b} = \sum_{s = 0}^{n} w_{s}^{i j} {\bar{γ}}_{k}^{a c} γ_{c s}^{b}$ . The arguments similar to the proof of Proposition 3.4 show that $\frac{\partial w_{s}^{i j}}{\partial x^{j b}} = 0$ for $s < b$ and that $\frac{\partial w_{b}^{i j}}{\partial x^{j b}} = \frac{\partial w^{i j}}{\partial x^{j}}$ . Moreover, $γ_{c s}^{b} = 0$ $s > b$ . Hence, the only nonzero summand in $\frac{\partial w_{k}^{i a j b}}{\partial x^{j b}}$ corresponds to $s = b$ . Therefore $c = 0$ , otherwise $γ_{c s}^{b} = 0$ . But ${\bar{γ}}_{k}^{a 0}$ is not zero only for $a = n$ and $k = 0$ by virtue of (2.4) (since ${\bar{e}}^{0} = e_{n}$ , ${\bar{e}}^{n} = 1$ ). Hence $\frac{\partial w_{k}^{i a j b}}{\partial x^{j b}} = 0$ for $k = 1, \dots, n$ and $\frac{\partial w_{0}^{i a j b}}{\partial x^{j b}} = 0$ for $a \neq n$ . As for $a = n$ , we have

\frac{\partial w_{0}^{i n j b}}{\partial x^{j b}} = \frac{\partial w_{b}^{i j}}{\partial x^{j b}} = \frac{\partial w^{i j}}{\partial x^{j}} .

This completes the proof of (5.4).

Now, show that

$\begin{matrix} w_{0}^{i a j 0} = \{\begin{matrix} w^{i j}, & a = n, \\ 0, & a = 0, 1 \dots, n - 1, \end{matrix} \\ w_{k}^{i a j 0} = 0, k = 1, \dots, n . \end{matrix}$	(5.5)

Indeed, $w_{k}^{i a j 0} = \sum_{s = 0}^{n} w_{s}^{i j} {\bar{γ}}_{k}^{a c} γ_{c s}^{0}$ , thus $c = s = 0$ (otherwise $γ_{0}^{c s} = 0$ ) which implies $a = n$ and $k = 0$ as before.

It remains to prove (5.2). We have

Δ_{Φ} = \sum_{j b} (\frac{\partial w^{i a j b}}{\partial x^{j b}} + w^{i a j b} \frac{\partial ln \sqrt{det g^{C}}}{\partial x^{j b}}) \frac{\partial}{\partial x^{i a}} .

From (5.4) it follows that

\sum_{j = 1}^{m} \frac{\partial w^{i a j b}}{\partial x^{j b}} \frac{\partial}{\partial x^{i a}} = (n + 1) \frac{\partial w^{i j}}{\partial x^{j}} \frac{\partial}{\partial x^{i n}},

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

In the summand

\sum_{j b} w^{i a j b} \frac{\partial ln \sqrt{det g^{C}}}{\partial x^{j b}} \frac{\partial}{\partial x^{i a}}

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

\sum_{j b} w^{i a j b} \frac{\partial ln \sqrt{det g^{C}}}{\partial x^{j b}} \frac{\partial}{\partial x^{i a}} = (n + 1) w^{i j} \frac{\partial ln \sqrt{det g}}{\partial x^{j}} \frac{\partial}{\partial x^{i n}} .

$□$

Corollary 5.3. For a weakly symmetric Frobenius Weil algebra $A$ the modular class of the Poisson manifold $(T^{A} M, w^{C} = w_{0}^{C} + \sum_{k = 1}^{n} t^{k} w_{k}^{C})$ 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 $Δ_{d V_{g}}$ . As in [11], we have

{(σ_{w} f)}^{V} = σ_{w^{C}} (f \circ π_{A}), f \in C^{\infty} (M), π_{A} : T^{A} M \to M .

This immediately proves the Corollary. $□$

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Kazan State University

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