# Torsion Scalar and Symmetries of Torsion Tensor

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1. Jul 13, 2015

### nanuba

I've started f(T) theory but I have a simple question like something that i couldn't see straightforwardly.
In Teleparallel theories one has the torsion scalar:

And if you take the product you should obtain

But there seems to be the terms like
.
How does this one vanish?

because we know that Torsion tensor is not symmetric in first two indices but antisymmetric in the last two indices.

2. Jul 13, 2015

### bcrowell

Staff Emeritus
Often one is interested in torsion tensors that are totally antisymmetric. Is that the case here?

3. Jul 14, 2015

### eurus

Oscillatory behavior

4. Jul 14, 2015

### nanuba

Nope. These are defined in terms of Wietzenböck konnections:
and are not totally antisymmetric.

5. Jul 14, 2015

### samalkhaiat

Will you buy me strong black coffee if I do it for you? Okay, start by multiplying the two tensor
$$S^{\rho\mu\nu}T_{\rho\mu\nu} = A + B , \ \ \ \ \ (1)$$
where
$$A = \frac{1}{4} T^{\rho\mu\nu}T_{\rho\mu\nu} + \frac{1}{4} T^{\mu\rho\nu}T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu}T_{\rho\mu\nu} , \ \ \ \ (2)$$ and $$B = -\frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} + \frac{1}{2}T_{\alpha}{}^{\nu\alpha} \ T^{\rho}{}_{\rho\nu} ,$$ or
$$B = -\frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} - \frac{1}{2}T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} = - T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\mu\rho} ,$$ which is the same as
$$B = T_{\alpha}{}^{\mu\alpha} \ T^{\rho}{}_{\rho\mu} = T^{\alpha\mu}{}_{\alpha} \ T^{\rho}{}_{\rho\mu} . \ \ \ \ \ (3)$$ That is all for $B$. To work on $A$, we start with the identity
$$T^{\mu\nu}{}_{\rho} \left( T^{\rho}{}_{\nu\mu} + T^{\rho}{}_{\mu\nu} \right) = 0 .$$ In the first term, if we make $\rho \to \nu$, $\mu \to \rho$ and $\nu \to \mu$, we find
$$T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\mu\rho} + T^{\mu\nu}{}_{\rho} \ T^{\rho}{}_{\mu\nu} = 0 .$$ Let us rewrite this as
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} + T_{\mu\nu\rho} \ T^{\rho\mu\nu} = 0 .$$ Using the antisymmetry in the second term, we get
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = T^{\rho\mu\nu} \ T_{\mu\rho\nu} .$$ Now, on the RHS we make $\rho \leftrightarrow \mu$ and get
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = T^{\mu\rho\nu} \ T_{\rho\mu\nu} .$$ Add $T^{\rho\mu\nu}T_{\nu\mu\rho}$ to both sides,
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = \frac{1}{2}\left( T^{\mu\rho\nu} \ T_{\rho\mu\nu} + T^{\rho\mu\nu} \ T_{\nu\mu\rho} \right) .$$ In the second term on the RHS, lower the indices on the first tensor and raise them on the second tensor
$$T^{\rho\mu\nu} \ T_{\nu\mu\rho} = \frac{1}{2}\left( T^{\mu\rho\nu} \ T_{\rho\mu\nu} + T_{\rho\mu\nu} \ T^{\nu\mu\rho} \right) .$$ Using the antisymmetry and dividing by 2, we obtain
$$\frac{1}{4} T^{\mu\rho\nu} T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu} T_{\rho\mu\nu} = \frac{1}{2} T^{\rho\mu\nu}T_{\nu\mu\rho} = - \frac{1}{2} T^{\rho\mu\nu} T_{\nu\rho\mu}.$$ And finally, we rewrite this identity in the form
$$\frac{1}{4} T^{\mu\rho\nu} T_{\rho\mu\nu} - \frac{1}{4} T^{\nu\rho\mu} T_{\rho\mu\nu} = - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} .$$ Now, substituting this identity in (2) leads to $$A = \frac{1}{4} T^{\rho\mu\nu} T_{\rho\mu\nu} - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} . \ \ \ (4)$$ Now, put (3) and (4) in (1), you get
$$S^{\rho\mu\nu}T_{\rho\mu\nu} = \frac{1}{4} T^{\rho\mu\nu} T_{\rho\mu\nu} - \frac{1}{2} T^{\rho\mu}{}_{\nu} \ T^{\nu}{}_{\rho\mu} + T^{\alpha\mu}{}_{\alpha} \ T^{\rho}{}_{\rho\mu} .$$
Okay, don’t forget my coffee.
Sam

6. Jul 28, 2015

### nanuba

I owe you a black coffee :)