Etapas de desarrollo del parque, desde la Plaza de la Concordia al Parque el Lago

In the previous section we defined a scalar field by its transformation law (5.2.12). In this section we generalise this transformation law to vector and tensor fields. As an example of a tensor of the second rank we discuss the properties of the metric tensor. Once again we mention that all calculations are done up to first order inθ, but some of them can be written to all orders [36].

5.3.1

Fields

In analogy with (5.2.12), the transformation law of a covariant vector field is given by δξVµ = ₋ξλ(∂λVµ)₋(∂µξλ)Vλ =₋Xξ?. Vµ−X(?∂µξλ). Vλ (5.3.1) = ₋ξλ?(∂λVµ) + i 2θ ρσ_(∂ρξλ_{)?(∂σ∂λVµ)} −(∂µξλ)? Vλ+ i 2θ ρσ_{(∂ρ∂µξ}λ_{)?(∂σVλ) +. . . ,} where in the last two linesX?

ξ and X(?∂µξλ) are expanded. This we generalise to the transfor-

mation law of an arbitrary covariant tensor δξTµ1...µr = −ξ λ_{(∂λTµ1...µ} r)−(∂µ1ξ λ_)Tλµ2...µ r − · · · −(∂µrξ λ_)Tµ1...µ r−1λ = ₋X_ξ?. Tµ1...µr −X ? (∂µ₁ξλ). Tλµ2...µr − · · · −X ? (∂µrξλ). Tµ1...µr−1λ. (5.3.2)

It is not difficult to check that this transformation close in the algebra (5.2.13).

By demanding that the ?-product of two scalar fields should be a scalar field again we derived the Leibniz rule for the operator δξ (5.2.18) and then abstracted the coproduct (5.2.19). Now we check if this was the correct thing to do. Namely, the coproduct is a representation-independent concept so our result (5.2.19) should also apply to vector and tensor fields. For example, the ?-product of two covariant vector fields should transform as a second rank tensor if we use (5.2.19)

δξ(Vµ? Vν) = (δξVµ)? Vν +Vµ?(δξVν)₋ i 2θ ρσ_(δ (∂ρξ)Vµ)(∂σVν) + (∂ρVµ)(δ(∂σξ)Vν) . Expanding the ?-product in first two terms and cancelling some of the terms coming from that expansion with the nontrivial terms in the coproduct leads to

δξ(Vµ? Vν) = ₋ξλ∂λ(VµVν)₋(∂µξλ)VλVν ₋(∂νξλ)VµVλ −i 2θ ρσ_ξλ_{∂λ (∂ρVµ)(∂σVν)} + (∂µξλ)(∂ρVλ)(∂σVν) + (∂νξλ)(∂ρVµ)(∂σVλ) = ₋ξλ∂λ(Vµ? Vν)₋(∂µξλ)(Vλ? Vν)₋(∂νξλ)(Vµ? Vλ) = ₋X_ξ?.(Vµ? Vν)₋X₍?_∂µξλ₎.(Vλ ? Vν)−X₍?_∂ νξλ).(Vµ? Vλ), (5.3.3)

which we wanted to prove. This means that (5.2.19) is the correct coproduct.

All that has been done for the covariant vectors (tensors) can also be done for the contravariant ones. We just summarise the results

δξVµ = ₋X_ξ?. Vµ+X₍?_∂λξµ₎. Vλ, (5.3.4)

δξTµ1...µr = −Xξ?. Tµ1...µr +X(?∂λξµ1). T

λµ2...µr ₊

5.3 Tensor calculus 77

Again, Vµ _{? V}ν _{transforms like a second rank tensor due to the coproduct (5.2.19). Also,} having covariant and contravariant vectors and tensors one can construct invariants. For example, δξ(Vµ? Vµ) = (δξVµ)? Vµ+Vµ?(δξVµ)₋ i 2θ ρσ_(δ (∂ρξ)Vµ)(∂σV µ_{) + (∂ρVµ)(δ} (∂σξ)V µ₎ = . . . = ₋ξλ∂λ(Vµ? Vµ) =₋X_ξ?.(Vµ? Vµ). (5.3.6) To summarise, we know now how to multiply vector and tensor fields and how to construct invariants under the transformations (5.2.26).

5.3.2

Metric tensor

An important example of a tensor is the metric tensor. Classically, it is a symmetric tensor of rank two

δcl_ξgµν =₋ξρ(∂ρgµν)₋(∂µξρ)gρν₋(∂νξρ)gµρ. (5.3.7) Its inverse gµν _{is defined by}

gµνgνρ =δ_ρµ. (5.3.8)

In analogy to the classical case, we define the noncommutative metric tensor Gµν as a symmetric tensor of rank two

ˆ

δξGµν =₋X_ξ?. Gµν ₋X₍?_∂µξρ₎. Gρν−X₍?_∂

νξρ). Gµρ, (5.3.9)

with the condition that it reduces to the classical metric tensor in the θ _→0 limit, Gµν

θ=0 =gµν. (5.3.10)

However, these conditions do not determine Gµν uniquely and in the following we present a few different solutions.

Looking at the transformation law ofGµν we see that the choice Gµν =gµν, that is the noncommutative metric equals the classical metric, is consistent with (5.3.9). The condition (5.3.10) is automatically fulfilled and we obtain the θ-independent metric tensor. Our final aim is the construction of the deformed Einstein-Hilbert action. Varying this action with respect to the metric one should obtain deformed equations of motion. By solving these equations we should obtain the noncommutative metric in terms of the classical one and the θ-dependent corrections. Therefore, starting with the commutative metric and saying later that it becomesθ-dependent might look a little odd8_{. Instead, one can choose from the}

beginning a θ-dependent metric tensor. Then one expands it in orders of the deformation parameter θ

Gµν =gµν +G1_µν+. . . , (5.3.11) where G1

µν is the first order correction which one calculates by solving the equations of motion.

8

However, this is just the problem of interpretation. Starting with the classical fields and finding out later that they have to haveθ-dependent corrections is normally done in the framework of the Seiberg-Witten map, see Chapter 3 and Chapter 4.

On the other hand, we remember that the classical metric tensor can be expressed in terms of the vierbein e a

µ

gµν =ηabe_µae_νb, (5.3.12)

where ηab is the flat Minkowski metric and a and b are local Lorentz indices. This we generalise to the noncommutative metric tensor

Gµν = 1 2

E_µa? E_νb+E_νa? E_µbηab, (5.3.13) where E a

µ is the noncommutative vierbein. In order to fulfil (5.3.9), Eµa has to transform as a vector field (5.3.1) and the coproduct (5.2.19) has to be used. Because of (5.3.10) in the limit θ _→0 it has to reduce to the classical vierbein

E_µa=e_µa+E_µa1+. . . . (5.3.14) Note that one can also start with the classical vierbein (it is consistent with both (5.3.9) and (5.3.10)) and after solving the equations of motion obtain that it becomesθ-dependent. The arguments pro and contra are the same as for choosing gµν as the noncommutative metric and we do not repeat them.

For the moment we do not specify the metric tensor. Instead, we look at the inverse metric. Starting with the noncommutative metric tensorGµν, one can introduce two inverses. The inverse with respect to the pointwise multiplication (classical inverse) we denote byGµν

Gµν _·Gνρ =δ_µρ, (5.3.15)

and the inverse with respect to the ?-multiplication with Gµν?

Gµν? Gνρ? =δ_µρ. (5.3.16)

ExpandingGνρ? _{in the deformation parameterθand inserting the expansion in (5.3.16) gives} the ?-inverse in terms of the classical inverse

Gµν? _{= G}µν ₊ i 2θ

ρσ_(∂

ρGµα)(∂σGαβ)Gβν

= 2Gµν ₋Gµα? Gαβ? Gβν. (5.3.17) This result is valid up to first order inθ. The exact result will of course depend on the choice of Gµν. From (5.3.16), using the comultiplication (5.2.19), it follows that Gµν? _transforms like a tensor of rank two

δξGµν? =−Xξ?. Gµν?+X(?∂ρξµ). G

ρν?_+X?

(∂ρξν). G

µρ?_{. (5.3.18)} Although Gµν is a symmetric tensor, its?-inverse is not symmetric

Gµν?₆=Gνµ?. (5.3.19)