Contra- vs. Co-variant Definition of Tensors?

[Ron Riegert]

(This is a repost of my so far unanswered question of July 1.)
In xAct, after defining an antisymmetric rank-2 tensor F, I do the following: square F, contract over the 4 free indices to form a scalar and multiply by the square root of the metric determinant to form a scalar Lagrangian density. I then use VarD to find the variational derivative with respect to the metric, which is defined as the stress-energy-momentum tensor-density T of the field F. But I obtain two different results depending on whether I initially define F as a covariant versus a contravariant tensor. Only the covariant definition gives the known valid form of T, whereas the contravariant form has a sign error between two terms and fails to be traceless. My questions are:
1) In defining F, why does the position of the indices in DefTensor matter at all in performing this calculation?
2) Since it does seem to matter, for more general tensor fields and Lagrangians, what is the proper way to use DefTensor to guarantee that VarD gives the valid T?
A notebook exhibiting the issue is attached:

[Leo Stein]

Hi Ron,

Suppose you fix what you consider the fundamental field variables in your theory, for example the metric and your 2-form L_{ab} . Moving indices means that there is implicit dependence of the metric or inverse metric, in this case

  L^{ab} = invg^{ac} invg^{bd} L_{cd} .

When we vary the action, we need to vary it completely with respect to all the fundamental fields. So, while the variation of the covariant L_{ab} is just \delta L_{ab}, the variation of the contravariant L^{ab} now also includes dependence on the variation of the inverse metric,

  \delta(L^{ab}) = ... = \delta(invg^{ac}) invg^{bd} L_{cd} + invg^{ac} \delta(invg^{bd}) L_{cd} + invg^{ac} invg^{bd} \delta L_{cd} .

As usual we can express the variation of the inverse metric in terms of the variation of the metric by hitting the identity invg^{ab} g_{bc} = \delta^a_c with a variation [e.g. Carroll's Eq. (4.56)].

The point is that if your fundamental variable is the covariant L_{ab}, then the term L_{ab}L^{ab} actually has dependence on two factors of the inverse metric. Meanwhile if your fundamental variable is the contravariant U^{ab}, then the term U_{ab}U^{ab} has dependendence on two factors of the metric, not the inverse metric. And the variation of the metric and its inverse are different.

As far as what is the "correct" choice to make when defining your theory... I think the only answer is it's up to you (I'm interested to hear anybody else's thoughts!). For your example, the vanishing of the trace of the stress tensor when using L_{ab} reflects the conformal invariance of that theory in d=4, whereas the theory defined in terms of U^{ab} is not conformally invariant in d=4. You can check with xTensor that the EOMs for L_{ab} have a homogeneous conformal transformation if L has conformal weight 0, whereas the EOMs for U^{ab} have a homogeneous conformal transformation when U has conformal weight -4.

Best

Leo

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[Ron Riegert]

Hi Leo--

Thanks for your rapid reply!

I understand your explanation that the differing factors of g_{ab} versus g^{ab} in the two Lagrangian-densities are responsible for the different stress-tensors T, one traceless and one not. 

But that leaves me with the following puzzle: as you correctly observe, in 4D the Lagrangian-density L(U) for U^{ab} can be made conformally-invariant by assigning U^{ab} the conformal-weight -4. In that case, L(U) becomes independent of the metric conformal-factor and so, by my understanding, it should give rise to a traceless T.  Yet, when explicitly calculated it does not. Why is that?

Ron