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I'm confused about the index representation, If the index up represents a contravariant tensor I can do this with the derivative and get it contravariant derivative?Or when it is said that, it is covariant or contravariant by "essence"(before the contravariant and covariant formulation like the gradient Which is transformed covariant).

I can't get an image because I don't have a score or whatever then I'll try to write,

(Contravariant) $T^{i'}=\dfrac{\partial x^{i'}}{\partial x_j}T^j$

(covariant). $V_{i'}=\dfrac{\partial x^j}{\partial x^{i'}}V_j$

Then $g^{ij}\partial_j=\partial^i$, I can transformed it contravariant, For it has index above, But the derivative is always covariant, im confused, what is $\partial ^j$.

amWhy
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Ian
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    Welcome to MSE. This thread might answer your question: https://math.stackexchange.com/questions/1908008/why-isnt-there-a-contravariant-derivative-or-why-are-all-derivatives-covarian?rq=1 – Adam T Jul 18 '21 at 13:23
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    The answer by SSD in this duplicate sums it up. There are contravariant and covariant tensors (index up index down) but there is only a covariant derivative often denoted by $\nabla,.$ You are rightfully confused now because $\nabla^\mu V^\nu$ is a contravariant rank 2 tensor and $\nabla_\mu V_\nu$ a covariant one. It cannot be helped. The nomenclature is not ideal. It works better to just worry about when an index is up or down. – Kurt G. Jul 18 '21 at 16:07
  • I deleted the one I had sent now, if you could answer it? if The derivative is always covariant which is ∂^μ That transform contravariant , and what the metric tensor does by lowering or lifting the Indices, because from what I've seen index above are transformed contravariant, And it is said that derivatives are transformed covariantly, then I ask again what it is ∂^μ – Ian Jul 19 '21 at 04:23
  • I will write a detailed answer. By 'covariant/contravariant transform' of a derivative do you mean by the Lorentz transformation? It would help if you use more formulas (MathJax is teriffic) and/or cite your references. – Kurt G. Jul 19 '21 at 04:27
  • I still do not know how to mess with mathjax, not necessarily lorentz transformation, a vector with index up is said contravariant, And it also transform contravariant , and I think I can not show image yet, but This appears in several videos of tensors like in the" khan ",That's what it says, but if we use the metric tensor we can lower indexes, and it left me confused, I don't know if it's the notation but I'm confused about, because the transformations are according to the index and follows this structure to the derivative, so if we apply the metric tensor what happened,contravariant? – Ian Jul 19 '21 at 04:44
  • I modified it, now you understand what I didn't understand? – Ian Jul 19 '21 at 05:10

1 Answers1

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There is -as we know- a covariant derivative but not a contravariant one. A vector with upper/lower index is called contravariant/covariant, and -as we know- indices are raised/lowered by the metric: $V^\mu=g^{\mu\nu}V_\nu\,.$ This works for the partial derivative as well: $\partial^\mu=g^{\mu\nu}\partial_\nu\,$ and also for the covariant derivative $\nabla^\mu=g^{\mu\nu}\nabla_\nu\,.$ There is a clash in the nomenclature: $V^\mu$ is a called contravariant vector, but $\partial^\mu,\nabla^\mu$ are not called contravariant derivatives. An acceptable name for $\nabla^\mu$ is covariant derivative with an upper index.

The covariant derivative is related to the partial derivative by $$ \nabla_\mu V^\nu=\partial_\mu V^\nu+\Gamma^\nu_{\mu\lambda}V^\lambda $$ (Carroll (2.19) on p. 45, (3.2) on p. 56) and we have the transformation laws $$ V^{\nu'}=\frac{\partial x^{\nu'}}{\partial x^{\nu}} V^\nu\,,\quad\quad \nabla_{\mu'}=\frac{\partial x^\mu}{\partial x^{\mu'}} \nabla_\mu\,,\quad\quad \nabla_{\mu'}V^{\nu'}=\frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^{\nu'}}{\partial x^{\nu}}\nabla_\mu V^\nu\,. $$ Because the primed $\partial x^{\nu'}$ is in the numerator of the first law this is called a contravariant transformation. Likewise, the second law is a covariant transformation. It is very instructive to work this out explicitly by looking at a Lorentz transformation. The third law is a combination of the first and second law. It is neither covariant nor contravariant (rather mixed) but: because the derivatives operator $\nabla$ can be pulled through the transformation $$ \frac{\partial x^\mu}{\partial x^{\mu'}}\frac{\partial x^{\nu'}}{\partial x^{\nu}} $$ it acts as if that term was a constant. The plain partial derivative can't do that unless the Christoffel symbols $\Gamma$ are all zero (no curvature). This invariance of $\nabla$ is the reason why it is called (somehow unfortunately) covariant derivative. Some people prefer to call it invariant derivative instead. Other people today avoid to call $V_\mu$ covariant, resp. $V^\mu$ contravariant, or are just happy with the understanding that it tells where the index is sitting.

This link, or for example, H. Weyl's book Space Time Matter give some intuition and historical background about these nomenclatures.

Kurt G.
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