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Gravity and Strings

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12 Differential geometry<br />

a metric-compatible connection of a Riemann–Cartan spacetime always has the form<br />

Ɣµν ρ =<br />

ρ<br />

µν<br />

<br />

+ Kµν ρ . (1.50)<br />

Observe that the symmetric part of the contorsion tension does not vanish, but<br />

<br />

Tµρν + Tνρµ = 0. (1.51)<br />

K(µν) ρ = 1<br />

2<br />

This means that the presence of torsion implies not only that the connection has a nonvanishing<br />

antisymmetric part, but also that the symmetric part is not fully determined by<br />

the metric but<br />

Ɣ(µν) ρ <br />

ρ<br />

= + K(µν)<br />

µν<br />

ρ <br />

ρ<br />

= . (1.52)<br />

µν<br />

The curvature, Ricci, <strong>and</strong> Einstein tensors of a metric-compatible connection satisfy further<br />

identities. On contracting the γ <strong>and</strong> σ indices in the third Bianchi identity Eqs. (1.30)<br />

<strong>and</strong> using the metric postulate, we find the so-called contracted Bianchi identity<br />

∇αGµ α + 2Tµαβ R βα − Tαβγ Rµ γαβ = 0. (1.53)<br />

Furthermore, by applying the Ricci identity to the metric <strong>and</strong> using the metric postulate,<br />

one can prove a fourth Bianchi identity:<br />

Rαβ(γ δ) = 0. (1.54)<br />

If we modify the connection according to Eq. (1.31) <strong>and</strong> Ɣ is metric-compatible, the<br />

Ricci scalar is<br />

R( ˜Ɣ) = R(Ɣ) − Tµν ρ τρ µν + 2∇µτν µν + τµ µλ τν ν λ + τν µρ τµρ ν . (1.55)<br />

If ˜Ɣ is not a metric-compatible connection, then τ contains all the contributions of<br />

the non-metricity tensor <strong>and</strong> the above formula allows us to work in the framework of a<br />

Riemann–Cartan spacetime with non-metric-compatible connections.<br />

If both ˜Ɣ <strong>and</strong> Ɣ are metric-compatible connections <strong>and</strong> ˜Ɣ has torsion but Ɣ = Ɣ(g), then<br />

τ = ˜K , the contorsion tensor of ˜Ɣ, <strong>and</strong> the above formula takes a simpler form:<br />

R( ˜Ɣ) = R[Ɣ(g)] + 2∇µ ˜Kν µν + ( ˜Kµ µλ ) 2 + ˜Kν µρ ˜Kµρ ν . (1.56)<br />

Now, this formula allows us to work with torsion in a Riemann spacetime. Particularly<br />

interesting is the case in which the contorsion ˜Kµνρ is a completely antisymmetric tensor<br />

(proportional to the Kalb–Ramond field strength Hµνρ, for instance). Then we have, if<br />

<br />

d d x <br />

|g| R( ˜Ɣ) =<br />

˜Kµνρ = 1<br />

√ 12 Hµνρ, (1.57)<br />

d d x <br />

|g| R[Ɣ(g)] + 1<br />

2 · 3! Hµνρ H µνρ<br />

<br />

. (1.58)

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