I am using these supergravity lecture notes by Gary W. Gibbons. On page 18, the author claims that geodesics and autoparallels coincide for a theory with totally antisymmetric torsion, and proves it by using the following identity (which he states without proof):
Γρμν={ρμν}+Kρμν,
where
Kμρν=−12(Tμρν+Tρμν+Tρνμ).
Here, Γρμν are the coefficients of an arbitrary connection with totally antisymmetric torsion and {ρμν} are the coefficients for the Levi-Civita connection, while Tρνμ is the torsion. Is there a proof of this identity somewhere?
Answer
OP's question is probably spurred by the fact that Ref. 1 forgets to mention that:
The other connection ∇:Γ(TM)×Γ(TM)→Γ(TM) with lower Christoffel symbols Γij,k := Γℓij gℓk
is still assumed to be compatible with the metric ∇g=0⇔∂kgij=Γki,j+Γkj,i.The contorsion tensor K ∈ Γ(T∗M⊗⋀2(T∗M)) is defined1 as the difference between the metric-compatible ∇ and the Levi-Civita connection ∇LC, i.e. Ki,jk := Γij,k−ΓLCij,k (2)= −Ki,kj.
The torsion tensor T ∈ Γ(⋀2(T∗M)⊗T∗M) is defined1 as T(X,Y) := ∇XY−∇YX−[X,Y]
⇔Tij,k := Γij,k−Γji,k (3)= Ki,jk−Kj,ik.One may show that the inverse relation to (4) is Ki,jk (4)= 12(Tij,k−Tjk,i+Tki,j),
and vice-versa.Let us decompose the contorsion tensor Ki,jk = 12(K+i,jk+K−i,jk),
in components K±i,jk := Ki,jk±Kj,ik,K+i,jk = Tki,j+Tkj,i,K−i,jk = Tij,k,that are symmetric/antisymmetric wrt. the first two indices i↔j.The geodesic equation reads 0 = ∇LC˙γ˙γ⇔¨γℓ = −ΓLC,ℓij˙γi˙γj
⇔−gkℓ¨γℓ = ΓLCij,k˙γi˙γj.In contrast, the auto-parallel equation 0 = ∇˙γ˙γ⇔¨γℓ = −Γℓij˙γi˙γj
⇔−gkℓ¨γℓ = Γij,k˙γi˙γj = (ΓLCij,k+12K+i,jk)˙γi˙γjcan detect the symmetric part K+i,jk of the contorsion.Observation: Tij,k is totally antisymmetric⇔Ki,jk is totally antisymmetric
⇔K+i,jk = 0.In the affirmative case, we have Tij,k=2Ki,jk, and the geodesic & auto-parallel equations (9) & (10) are the same.
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1 Pertinent applications of the musical isomorphism are implicitly implied from now on.
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