Showing that the gradient of a scalar field is a covariant vector

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SUMMARY

The discussion centers on proving that the covariant gradient of a scalar field, represented as ##\nabla f=\frac{\partial f}{\partial x^{i}} g^{i j} \mathbf{e}_{j}##, is a covariant vector. Participants emphasize the importance of the transition matrix ##C## and its role in transforming basis vectors. The conversation highlights the need to correctly apply the transformation rules for tensors, particularly the relationship between covariant and contravariant components. The conclusion suggests that a clearer understanding of the Jacobian and its application in the transformation of derivatives is essential for resolving the proof.

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AndersF
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Homework Statement
Prove that the covariant gradient of a scalar field is a covariant vector
Relevant Equations
##\nabla f=\frac{\partial f}{\partial x^{i}} g^{i j} \mathbf{e}_{j}##
In a general coordinate system ##\{x^1,..., x^n\}##, the Covariant Gradient of a scalar field ##f:\mathbb{R}^n \rightarrow \mathbb{R}## is given by (using Einstein's notation)

##
\nabla f=\frac{\partial f}{\partial x^{i}} g^{i j} \mathbf{e}_{j}
##

I'm trying to prove that this covariant gradient ##\nabla f## is indeed a covariant vector. To do so, I'm trying to show that it transforms as a 1-covariant tensor under a change of basis.

Let ##C## be the transition matrix from a basis ##\{\mathbf e_i\}## to a basis ##\{\tilde {\mathbf e}_i\}##, that is, ##\tilde {\mathbf e}_i= \mathbf e_iC^i_j##.

The covariant derivative increases the contravariant tensor order of the tensor by one unit. Since the partial derivative of a scalar field is indeed a covariant derivative, the object ##\frac{\partial f}{\partial x^{i}}## will therefore be a 1-covariant tensor which I will call ##F_i##.

On the other hand, the contraction between the dual metric tensor ##g^{ij}## and ##F_i## will raise the subscript ##i## of ##F_i##, and the resulting object will be a 1-contravariant tensor: ##g^{ij}F_i\equiv H^j##.

But then, ##\frac{\partial f}{\partial x^{i}} g^{i j}=H^j## will transform as the contravariant components of a contravariant vector ##\mathbf{v}=H^j\mathbf{e}_{j}##: ##\tilde H^j=(C^{-1})^j_kH^k##, which is just the opposite of what I have to prove...

Where is my mistake? How could this be proved?
 
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I don't recognise your notation. I'm more familiar with things like:
$$g^{ij} = g^{k'l'}\frac{\partial x^i}{\partial x^{k'}}\frac{\partial x^j}{\partial x^{l'}}$$If you do the same for the basis vector and partial derivatives, you can express the components of ##\nabla' f## in terms of the components of ##\nabla f## and derive/confirm the relevant transformation rule.
 
PS the result should drop out in a couple of lines.
 
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PeroK said:
PS the result should drop out in a couple of lines.
I too think that the demonstration should be shorter than what I have tried, but I don't find the problem...

By the way, the components of the transition matrix ##(C^i_j)## of my notation are the terms ## \frac{\partial x^i}{\partial x'^j}## of your notation, aren't they?
 
AndersF said:
I too think that the demonstration should be shorter than what I have tried, but I don't find the problem...
If I understood what you were doing, I'd try to help!
 
AndersF said:
Homework Statement:: Prove that the covariant gradient of a scalar field is a covariant vector
Examine the chain rule: ##\partial / \partial x^j = (\partial x'^i/\partial x^j) \partial / \partial x'^i##. The Jacobian ##\partial x'^i/\partial x^j## is your transition matrix.
 

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