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I am trying to understand the Wikipedia article about covariance and contravariance of vectors. It says the following:

A contravariant vector has components that "transform as the coordinates do" under changes of coordinates (and so inversely to the transformation of the reference axes), including rotation and dilation.

That seems intuitive enough. For example, if I have a set of cartesian axes in the plane and obtain a new set by rotating them an angle $\theta$ about the origin, then, in the new coordinates, it would seem as though all points in the plane have rotated the opposite way.

By contrast, a covariant vector has components that change oppositely to the coordinates or, equivalently, transform like the reference axes. For instance, the components of the gradient vector of a function $$\nabla f = \frac{\partial f}{\partial x^1}\hat x^1 + \frac{\partial f}{\partial x^2}\hat x^2 + \frac{\partial f}{\partial x^3}\hat x^3$$ transform like the reference axes themselves.

This part I cannot quite visualize. Let's take the simple example of a topographic map, with $x$, $y$ describing eastness and northness, and $f(x,y)$ describing elevation. We are standing at the origin, on the slope of a hill that rises to our north, so, say $\nabla f = (0,1)$ here. Now we define new (primed) coordinate axes rotated by $\theta = \pi/2$, so that $x'$, $y'$ denote northness and westness, respectively. Obviously we still have the hill to our north, so $(\nabla f)' = (1,0)$, where the prime denotes representation in the primed coordinate basis. Hence it seems like $\nabla f$ "transforms as the coordinates do", contradicting the quoted statement from Wikipedia.

Is the statement from Wikipedia misleading, or am I making a mistake?

ummg
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  • Thank you! Unfortunately I have a bit of trouble following that answer because I am not familiar with the framework of differential forms. Also, it does not directly address the apparent contradiction in my example. Still, I think I might have picked up a few of the points made. Maybe you could clarify? I think the issue in my particular case might be that $\nabla f$ is a vector field, and therefore transforms contravariantly, and that when one says that the gradient is covariant one actually means something else. Maybe that the gradient operator transforms covariantly? – ummg Apr 01 '21 at 22:08
  • The trouble is that under some transformations, namely rotations, covariant and contravariant components will transform in the same way. Try the same simple example, but with axes which are related by a scaling or shearing transformation instead, and things might be more clear. – Kajelad Apr 11 '21 at 02:16

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