Covariant derivative of the metric

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Discussion Overview

The discussion revolves around the properties of the covariant derivative, specifically whether the covariant derivative of the metric tensor yields zero, and how it applies to the exponential of a scalar field. The scope includes theoretical aspects of differential geometry and tensor calculus.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions whether the covariant derivative of the metric tensor gives zero even when one of its indices is raised.
  • Another participant suggests that the covariant derivative of the exponential of a scalar field, φ, is defined by a partial derivative and will only be zero if φ is constant.
  • A participant inquires if applying the covariant derivative to a Kronecker delta results in zero.
  • Another participant asserts that the covariant derivative acting on a (1,1)-type tensor can be explicitly written down and may yield zero, assuming the connection is symmetric.

Areas of Agreement / Disagreement

Participants express differing views on the behavior of the covariant derivative with respect to the metric and the Kronecker delta, indicating that the discussion remains unresolved with competing perspectives.

Contextual Notes

There are assumptions regarding the symmetry of the connection and the constancy of the scalar field φ that are not fully explored or agreed upon.

vitaniarain
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hello!
just a quick question, does the covariant derivative of the metric give zero even when the indices(one of the indices) of the metric are(is) raised?
also another question not entirely related, does the covariant deriv. of exp(2 phi) where phi is the field, also give zero or not necessarily? thanks
 
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Act with the covariant derivative on a kronecker delta. You know that

<br /> g_{ab}g^{bc} = \delta_a^c<br />

The covariant derivative of your exponential of scalar field is given by a partial derivative per definition; this will only be zero for constant phi.
 
if u act on a kronecker delta with the covariant deriv will that give zero? :S
 
Well, you know how a covariant derivative acts on a (1,1)-type tensor, right? Just write it down explicitly and check :) I would say that you'll find that it indeed is zero, provided the connection is symmetric.
 

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