Defining the derivative of a vector field component

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I'm reading 'Core Principles of Special and General Relativity' by Luscombe, specifically the introductory section on problems with defining usual notion of differentiation for tensor fields. I'll quote the relevant part:
The second way (to see whether the partial derivative of a tensor is a tensor) is to look at the definition of derivative, $$\frac{\partial T^i}{\partial x^j}=\lim_{dx^j\to 0}\frac{T^i(x+dx^j)-T^i(x)}{dx^j}$$ The numerator is not in general a vector! We're comparing (subtracting) vectors from different points, yet the transformation properties of tensors are defined at a point.
Since the equation above is a notational mess, here's my attempt to interpret it:
$$\bigg(\frac{\partial T^i}{\partial x^j}\bigg)_p=\partial_j(T^i\circ x^{-1})(x(p))=\lim_{h\to 0}\frac{(T^i\circ x^{-1})(x(p)+[0,\ldots,h,\ldots,0])-(T^i\circ x^{-1})(x(p))}{h}$$ where ##[0,\ldots,h,\ldots,0]\in\mathbb{R}^n## has ##h## as its ##j##-th coordinate.

Is my above interpretation correct? If so, what's the issue with defining the derivative of a vector field component in this way?
 

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  • #2
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Let's look at an example. Let ##V## be the constant vector field ##\partial _1## (in the usual coordinate system) in ##\mathbb{R}^2.## Then the derivatives of its coordinates are zero.

On the other hand, in polar coordinates ##V=cos(\theta)\partial_r-\frac{1}{r}\sin(\theta)\partial_{\theta}## and then the derivatives of these components in polar coordinates are nonzero.

A tensor can't be zero in one coordinate system and nonzero in another.
 
  • #3
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Let's look at an example. Let ##V## be the constant vector field ##\partial _1## (in the usual coordinate system) in ##\mathbb{R}^2.## Then the derivatives of its coordinates are zero.

On the other hand, in polar coordinates ##V=cos(\theta)\partial_r-\frac{1}{r}\sin(\theta)\partial_{\theta}## and then the derivatives of these components in polar coordinates are nonzero.

A tensor can't be zero in one coordinate system and nonzero in another.
As far as I know, vector field components at some point can very well differ. At some point $p$, it's possible for, e.g., the first component to be zero in one chart and non-zero in the other.

So your point is that the derivative (defined as I did in the OP) of a component of the field is identically zero in one chart and not everywhere zero in the other chart. Am I correct in saying so?
 
  • #4
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If ##T## is a tensor and all components of ##T## vanish at ##p## in some coordinate system, then all components of ##T## must also vanish at ##p## in every coordinate system.

This doesn't happen in my example (where ##T^i_j## is the derivative of the ##i##th component of ##V## in the direction of ##x^j##).
 
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If ##T## is a tensor and all components of ##T## vanish at ##p## in some coordinate system, then all components of ##T## must also vanish at ##p## in every coordinate system.

This doesn't happen in my example (where ##T^i_j## is the derivative of the ##i##th component of ##V## in the direction of ##x^j##).
I'll unpack your example in the notation I'm comfortable with as a beginner , so please bear with me - just summarizing:

You defined the vector field ##V=V^1\partial_x + V^2\partial_y## where ##V^2=0## and ##V^1=1## for all points in the chart ##(U,\phi)## corresponding to the usual coordinate system.

So then ##V^1_1=\partial_1(V^1\circ\phi^{-1})(\phi(p))=\partial_1(1)=0##. Similarly ##V^1_2=0## and ##V^2_1=V^2_2=0## for all points ##p## in the chart.

Now the same field can be expressed in the polar chart ##(U,\psi)## as ##V=V^1\partial_r + V^2\partial_{\theta}##. Here ##(V^1\circ\psi^{-1})(\psi(p))=\cos(\psi^2(p))## and ##(V^1\circ\psi^{-1})(\psi(p))=-\sin(\psi^2(p))/\psi^1(p)##. And now I'll do a couple of obvious calculations in detail just to highlight consistency with the partial derivative definition in the OP:
$$V^1_1=\partial_1(V^1\circ\psi^{-1})(\psi(p))=\lim_{h\to 0}\frac{(V^1\circ\psi^{-1})(\psi(p)+[h,0])-(V^1\circ\psi^{-1})(\psi(p))}{h}$$ $$=\lim_{h\to 0}\frac{(V^1\circ\psi^{-1})(\psi(p)+[h,0])-(V^1\circ\psi^{-1})(\psi(p))}{h}=\lim_{h\to 0}\frac{\cos(\psi^2(p)+0)-\cos(\psi^2(p))}{h}=0$$
Similarly
$$V^1_2=\partial_2(V^1\circ\psi^{-1})(\psi(p))=\lim_{h\to 0}\frac{(V^1\circ\psi^{-1})(\psi(p)+[0,h])-(V^1\circ\psi^{-1})(\psi(p))}{h}$$ $$=\lim_{h\to 0}\frac{\cos(\psi^2(p)+h)-\cos(\psi^2(p))}{h}=-\sin(\psi^2(p))$$
and so on for ##V^2_1,V^2_2##.

Effectively, given two different charts ##\phi## (usual coordinate system) and ##\psi## (polar) with chart overlap region ##U##, we see that ##V^i_j=0## for all ##i,j## and for all points in ##U## under the ##\phi## chart.

Conversely, ##V^1_1=0##, ##V^1_2\neq 0, V^1_2\neq 0, V^1_2\neq 0## for most points in ##U## under the ##\psi## chart.

If ##V^i_j## had been a ##(1,1)##-tensor, since all its components vanish for all points in ##U## in one coordinate system, they should've vanished under the polar system as well (due to tensor component transformation laws). So it's not a tensor. Hope I got that correct.
 
  • #6
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for all points ##p## in the chart.
This isn't necessary. If the components of a tensor all vanish at a single point ##p## in some coordinate system, then they do in all coordinate system. This does fail at every point ##p## in my example, but a single point would have been enough.

I think you also mis-wrote the coordinate expressions for ##V## in polar coordinates.
 
  • #7
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This isn't necessary. If the components of a tensor all vanish at a single point ##p## in some coordinate system, then they do in all coordinate system. This does fail at every point ##p## in my example, but a single point would have been enough.

I think you also mis-wrote the coordinate expressions for ##V## in polar coordinates.
Understood, thanks! So one thing I'm trying to understand is, why do we want the partial derivative of a tensor field to be a tensor field at all? Probably a silly question but still.

I've looked at a great many sources just for this and there are mainly two motivations for introducing covariant derivative. One is that we can't compare vectors at different points on a curved manifold the same way we can in flat space. Fair enough - and the definition I gave in the OP addresses the comparison issue.

Now based on the book quote and also from what you say, the problem with this definition is that the derivative components defined in that way can't be components of a tensor, i.e. they don't obey the typical tensor transformation laws. Why do we need the derivative to be a tensor? The answer is probably captured in the following Wiki quote:

The primary difference from the usual directional derivative is that ##\nabla_{\mathbf{u}}\mathbf{v}## must, in a certain precise sense, be independent of the manner in which it is expressed in a coordinate system.

The above quote is vague: in what "precise sense" is a tensor independent of the manner of expression in coordinate systems?
 
  • #8
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I think that the idea (as it's in the beginning of the book, although I'm not familiar with this particular book), is to show that ##\partial_{\nu} T^{\mu}## is NOT a tensor, and to motivate the derivation of the covariant derivative.

Why would you want the derivative a position vector X to be a vector?

I'm not an expert on differential geometry, but I think you can imagine a type of maths with different axioms where we didn't require tensors to be independent of the coordinate system. We would lose a lot of really nice properties and identities that have been used and abused for ages. It would be like not using the metric tensor to solve for the covariant derivative. You could have theoretically picked any arbitrary tensor and set the covariant derivative of that tensor to zero and then solved for the covariant derivative, but you would lose geodesics, levi cevita and a lot of other nice things.
 
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