• Non ci sono risultati.

All linear natural operators transforming affinors to tensor fields of type (0, p) on Weil bundles are classified

N/A
N/A
Protected

Academic year: 2021

Condividi "All linear natural operators transforming affinors to tensor fields of type (0, p) on Weil bundles are classified"

Copied!
5
0
0

Testo completo

(1)

The linear natural operators transforming affinors to tensor fields of type (0, p) on Weil bundles

Wlodzimierz M. Mikulski

Institute of Mathematics, Jagiellonian University, Krak´ow, Reymonta 4, Poland, wlodzimierz.mikulski@im.uj.edu.pl

Received: 28 April 1999; accepted: 3 April 2000.

Abstract. All linear natural operators transforming affinors to tensor fields of type (0, p) on Weil bundles are classified.

Keywords:Weil bundles, linear natural operators

MSC 2000 classification:primary 58A20, secondary 53A55

Introduction

Let F :M → FM be a product preserving bundle functor and let A = F (R) be its Weil algebra, [3].

If λ : A→ R is a linear map and Φ is an affinor on an n-manifold M, then we have (trΦ)(λ): F (M )→ R, where trΦ : M → R is the trace of Φ and ( )(λ) is the (λ)-lift of functions to F in the sense of [2].

Clearly, for a given linear map λ : A→ R the correspondence Φ → (trΦ)(λ)is a linear natural operator T|M(1,1)

n  T(0,0)F transforming affinors into functions on F in the sense of [3]. Similarly, for a given linear map λ : A → R the correspondence Φ → d(trΦ)(λ) is a linear natural operator T|M(1,1)

n  T(0,1)F transforming affinors into 1-forms on F .

In this short note we prove

Theorem 1. Let F and A be as above.

1. Every linear natural operator T|M(1,1)

n  T(0,0)F is of the form Φ→ (trΦ)(λ) for a linear map λ : A→ R.

2. Every linear natural operator T|M(1,1)

n  T(0,1)F is of the form Φ→ d(trΦ)(λ) for a linear map λ : A→ R.

3. For p≥ 2 every linear natural operator T|M(1,1)n  T(0,p)F is 0.

(2)

Problem of finding all natural operators of some type on affinors is very difficult. Classifications of base extending natural operators on affinors are un- known. The author knows only the paper of Debecki, cf. [1], where the natural operators TM(1,1)

n  T(p,q) for p = q = 0, 1, 2 and (p, q) = (0, 1) are classified.

Recently Debecki obtained a classification for p = q = 3. It seems that classi- fications of natural operators on affinors would be very useful because affinors play important role in differential geometry.

Throughout this note the usual coordinates on Rn are denoted by x1, . . . , xn and ∂i= ∂xi, i = 1, . . . , n.

All manifolds and maps are assumed to be of class C.

1. A reducibility lemma

The crucial point in our consideration is the following general lemma.

Lemma 1. Let L : T(1,1)|Mn  HF be a linear natural operator, where F : Mn → FM is a natural bundle and H : Mdim(F (Rn)) → VB ⊂ FM is a natural vector bundle. If L(x11⊗ dx1) = 0, thenL = 0.

Proof. At first we prove that

L((x1)p1⊗ dx1) = 0 over 0∈ Rn (1) for p = 0, 1, 2, . . ..

We consider three cases:

1. p = 0. Applying the invariance of L with respect to the translation (x1− 1, x2, . . . , xn) from the assumptionL(x11⊗ dx1) = 0 it follows that L((x1− 1)∂1⊗ dx1) = 0. ThenL(∂1⊗ dx1) = 0 because of the linearity of L.

2. p = 1. The equality (1) for p = 1 is the assumption.

3. p ≥ 2. Applying the invariance of L with respect to the local diffeo- morphism (x1+ (x1)p, x2, . . . , xn)−1 from the assumption it follows that L((x1+ (x1)p)∂1⊗ dx1) = 0 over 0∈ Rn. Then we have (1) because of the same reasons as in case 1.

Next we prove that if n≥ 2, then

L((x1)px21⊗ dx1) = 0 over 0∈ Rn (2) for p = 0, 1, 2, . . ..

Let p∈ {0, 1, 2, . . .}.

(3)

We shall use (1). We have L(∂1 ⊗ dx1) = 0 over 0 ∈ Rn. Then by the invariance of L with respect to the diffeomorphism (x1 − x2, x2, . . . , xn) we derive thatL(∂1⊗ (dx1+ dx2)) = 0 over 0∈ Rn. Then

L(∂1⊗ dx2) = 0 over 0∈ Rn. (3) There is a diffeomorphism ϕ : R→ R such that ϕ × idRn−1 sends the germ of ∂1 at 0 into the germ of ∂1+ (x1)p1 at 0. Then using the invariance of L with respect to ϕ× idRn−1 from (3) we obtain thatL((∂1+ (x1)p1)⊗ dx2) = 0 over 0∈ Rn. Then

L((x1)p1⊗ dx2) = 0 over 0∈ Rn. (4) On the other hand by the invariance ofL with respect to the diffeomorphisms (x1 − τx2, x2, . . . , xn), τ = 0 from (1) for p + 1 instead of p we have L((x1 + τ x2)p+11⊗ (dx1+ τ dx2)) = 0 over 0∈ Rn. The left hand side of this equality is a polynomial in τ . Considering the coefficients at τ1 of this polynomial we get

(p + 1)L((x1)px21⊗ dx1) +L((x1)p+11⊗ dx2) = 0 over 0∈ Rn. Then we have (2) because of (4) for p + 1 instead of p.

We continue the proof of the lemma. By the linearity of L and the base- extending version of Peetre theorem (see Th. 19.9 in [3]) it is sufficient to verify that

L(xαi⊗ dxj) = 0 over 0∈ Rn (5) for any α = (α1, . . . , αn)∈ (N ∪ {0})n and i, j = 1, . . . , n.

Because of (1) we can assume that n ≥ 2. Using the invariance of L with respect to the diffeomorphisms permuting the coordinates we can assume that either i = j = 1 or i = 1 and j = 2.

Consider two cases:

1. i = j = 1. If α2 =· · · = αn= 0, then by (1) for p = α1 we get L(xα1 dx1) = 0 over 0 ∈ Rn. So, we can assume that (α2, . . . , αn) = 0. Then by the invariance of L with respect to the local diffeomorphisms (x1, x2+ (x2)α2· · · (xn)αn, x3, . . . , xn)−1 from (2) for p = α1 we derive that

L((x1)α1(x2+ (x2)α2· · · (xn)αn)∂1⊗ dx1) = 0 over 0∈ Rn. Then L(xα1⊗ dx1) = 0 over 0∈ Rn.

2. i = 1, j = 2. We consider two subcases:

(4)

a. Assume n≥ 3 and (α3, . . . , αn)= 0. Then from the case 1 we have (in particular) that L(x31⊗ dx1) = 0 over 0∈ Rn. Then using the in- variance ofL with respect to the diffeomorphisms (x1−x2, x2, . . . , xn) we obtain L(x31 ⊗ (dx1 + dx2)) = 0 over 0 ∈ Rn. Consequently L(x31⊗ dx2) = 0 over 0∈ Rn.

There is a diffeomorphism ϕ : R → R such that ϕ × idRn−1 sends the germ of ∂1 at 0 into the germ of ∂1 + (x1)α11 at 0. Using the invariance of L with respect to ϕ × idRn−1 from L(x31⊗ dx2) = 0 over 0 ∈ Rn we derive that L(x3(∂1 + (x1)α11)⊗ dx2) = 0 over 0∈ Rn. Then

L((x1)α1x31⊗ dx2) = 0 over 0∈ Rn. (6) There is a diffeomorphism ψ : R → R such that idR× ψ × idRn−2 sends the germ of dx2 at 0 into the germ of dx2 + (x2)α2dx2 at 0.

Using the invariance ofL with respect to idR× ψ × idRn−2 from (6) we deduce thatL((x1)α1x31⊗ (dx2+ (x2)α2dx2)) = 0 over 0∈ Rn. Then

L((x1)α1(x2)α2x31⊗ dx2) = 0 over 0∈ Rn. (7) Then using the invariance of L with respect to the local diffeomor- phism (x1, x2, x3+(x3)α3· · · (xn)αn, x4, . . . , xn)−1from (7) we deduce thatL((x1)α1(x2)α2(x3+(x3)α3· · · (xn)αn)∂1⊗dx2) = 0 over 0∈ Rn. Hence L(xα1⊗ dx2) = 0 over 0∈ Rn.

b. n = 2 or α3 =· · · = αn= 0. By (4), L((x1)α11⊗ dx2) = 0 over 0 Rn. Now, using the invariance ofL with respect to idR× ψ × idRn−2 (as above) we deduce thatL((x1)α11⊗ (dx2+ (x2)α2dx2)) = 0 over 0∈ Rn. ThenL((x1)α1(x2)α21⊗ dx2) = 0 over 0∈ Rn.

QED

2. The proof of Theorem 1

We are now in position to prove the theorem. Let F and A be as in the Introduction. Let a1, . . . , ak∈ A be a basis of A, and let a1, . . . , ak be the dual basis.

1. Consider a linear natural operatorL : T|M(1,1)n  T(0,0)F . Since the (xi)(aν) for i = 1, . . . , n and ν = 1, . . . , k form a coordinate system on F (Rn) (see [2]), we can write L(x11⊗ dx1) = f ((xi)(aν)) for some f : RN → R, with N ={1, . . . , n} × {1, . . . , k}. By the invariance of L with respect to

(5)

the diffeomorphisms (x1, tx2, . . . , txn), t = 0, we deduce that L(x11 dx1) = f ((x1)(aν)) for some f : R{1,...,k}→ R. Now, by the linearity and the invariance of L with respect to the diffeomorphisms (tx1, x2, . . . , xn), t= 0, f is homogeneous of weight 1. Then by the homogeneous function theorem, cf. [3], f is linear. Hence L(x11⊗ dx1) = (x1)(λ) = (tr(x11 dx1))(λ) for some linear λ : A→ R. Applying Lemma we end the proof of part 1.

2. Consider a linear natural operator L : T|M(1,1)n  T(0,1)F . We can write L(x11⊗dx1) =n

j=1

k

µ=1f((xi)(aν))d(xj)(aµ)for some functions f : R{1,...,n}×{1,...,k} → R. By the linearity and the invariance of L with re- spect to the homotheties (tx1, tx2, . . . , txn), t = 0, we deduce that the functions f are constants. Now, by the invariance of L with respect to the diffeomorphisms (x1, tx2, . . . , txn), t= 0, we deduce that f = 0 for j = 2, . . . , n. Hence L(x11⊗ dx1) = d(x1)(λ) = d(tr(x11⊗ dx1))(λ) for some linear λ : A→ R. Applying Lemma we end the proof of part 2.

3. Consider a linear natural operator L : T|M(1,1)n  T(0,p)F , where p ≥ 2.

Similarly as above, from the linearity and the invariance of L with respect to the homotheties it follows that L(x11⊗ dx1) = 0. Applying Lemma we finish the proof.

References

[1] J. Debecki: Natural transformations of affinors into linear forms, Suppl. Rend. Circolo Mat. Palermo32 (II) 1993, 49–59.

[2] J. Gancarzewicz, W. Mikulski, Z. Pogoda: Lifts of some tensor fields and linear connections to product preserving functors, Nagoya Math. J.,135 (1994), 1–41.

[3] I. Kol´r, P. M. Michor, J. Slov´ak: Natural operations in differential geometry, Springer-Verlag, Berlin 1993.

Riferimenti

Documenti correlati

Keywords: delay differential equations, semigroup of solution operators, pseudospectral methods, spectral methods.. All the arguments developed in the sequel apply as well to

Ciascuna delle due onde interagisce con il bersaglio situato in campo lontano, e scattera un'onda il cui stato di polarizzazione può essere espresso

stained with conventional histopathology (hematoxylin-eosin) and immuno-hystochemistry (S-100 and HMB-45). The pattern of SLN involvement was defined as macro- or microscopic on

In particular, the following contents are presented in the next sections: the structural and dimensional characteris- tics of the building; the mechanical properties of the

While in the first testing session the users were asked to move the virtual object on the display one time for each of the three target points, in the second testing session the

As a contribution to the World Environment Day 2017, the Accademia Nazionale dei Lincei promoted the meeting “Changes and Crises in the Mediterranean Sea” devoted to the effects

La discontinuità del lavoro e il difficile accesso all’acquisto o a tradizio- nali forme di affitto della casa - dovuto in alcuni paesi ai costi proibitivi - sono fattori che