I was still not really sure what is meant by the statement that partials form a coordinate basis. Then if they do, is their character (null, timelike or spacelike) related to the metric? At last I asked on Physics forums and PeterDonis had the final word. So now I do know what is meant by partials forming a coordinate basis and there is some relationship to the metric.

Carroll writes$$

\frac{d}{d\lambda}=\frac{dx^\mu}{d\lambda}\partial_\mu

$$"Thus the partials ##\left\{\partial_\mu\right\}## do indeed represent a good basis for the vector space of the directional derivatives, which we can therefore safely identify with the tangent space."

\frac{d}{d\lambda}=\frac{dx^\mu}{d\lambda}\partial_\mu

$$"Thus the partials ##\left\{\partial_\mu\right\}## do indeed represent a good basis for the vector space of the directional derivatives, which we can therefore safely identify with the tangent space."

We have a parabola parameterized by ##\lambda## given by$$t=\frac{\lambda^2}{10}\ ,\ \ x=\lambda$$so$$\frac{dt}{d\lambda}=\frac{\lambda}{5}\ \ ,\ \ \ \frac{dx}{d\lambda}=1$$and the tangent vector ##V=d/d\lambda## has components ##\left(dt / d\lambda , dx/ d \lambda\right)## at ##\left(t,x\right)##.

In plane polar coordinates ##\left(r,\theta\right)## the ##\partial_\theta## basis vector is along lines with parameters ##\left(k_r,\lambda\right)##. So the ##\theta## basis vectors lie on concentric circles around the origin. Similarly ##r## basis vectors are radial.

All you need for plane polar coordinates is a line segment to measure ##\theta## from and one end to serve as the origin. The line is normally drawn horizontally. That is not essential.

See how it all hangs together: Commentary 2.3 Coordinates and basis vectors.pdf

And the thread on physics forums here.

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