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I am currently learning about manifolds from Hobson's General Relativity: An Introduction for Physicists.

In Chapter 3 on 'Vector Calculus on Manifolds', pg. 57, the author defined coordinate basis vectors for a point $P$ on a $N$-dimensional manifold with coordinate system $x^a$ as $$\vec{e_a}=\lim_{\delta x^a \rightarrow 0}\frac{\delta \vec{s}}{\delta x^a}$$ where $\delta \vec{s}$ is the infinitesimal vector displacement between point $P$ and a nearby point $Q$ whose coordinate separation from $P$ is $\delta x^a$ along the $x^a$ coordinate curve.

Say if $P$ and $Q$ are points on the $x^1$ coordinate cuve, the coordinate separation $\delta x^1$ between them will be $0$ since on the coordinate curve the $x^1$ coordinate stays constant. If this is correct, the definition for $\vec{e}_a$ will not make sense since it is dividing by $0$. What am I getting wrong?

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  • $\begingroup$ Isn't your question answered by: (1) equation 3.4 that follows on page 57 of Hobson text; (2) basic understanding of limits and differentials from Calculus? $\endgroup$
    – K7PEH
    Commented Aug 27, 2020 at 15:43
  • $\begingroup$ here you are wrong -> "...on the $x^1$ coordinate curve the $x^1$ coordinate stays constant". $\endgroup$
    – Umaxo
    Commented Aug 27, 2020 at 15:53
  • $\begingroup$ @Umaxo Wikipedia says "In two dimensions, if one of the coordinates in a point coordinate system is held constant and the other coordinate is allowed to vary, then the resulting curve is called a coordinate curve." Does that not mean the $x^1$ coordinate curve has the $x^1$ coordinate held constant? $\endgroup$
    – TaeNyFan
    Commented Aug 27, 2020 at 15:56
  • $\begingroup$ @K7PEH I can understand equation 3.4 but not this. $\endgroup$
    – TaeNyFan
    Commented Aug 27, 2020 at 15:57
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    $\begingroup$ @TaeNyFan This works only for two dimensions. For example fixing one coordinate in 3 dimensions would produce surface and not curve. You need to fix all coordinates but the one which has the same name. $\endgroup$
    – Umaxo
    Commented Aug 27, 2020 at 15:58

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Maybe it becomes clear for the example of polar coordinates in the plane:

$$\mathbf e_r = \left(\frac{\partial x}{\partial r} , \frac{\partial y}{\partial r}\right) = (\cos(\theta) , \sin(\theta))$$

$$\mathbf e_{\theta} = \left(\frac{\partial x}{\partial \theta} , \frac{\partial y}{\partial \theta}\right) = (-r\sin(\theta) , r\cos(\theta))$$

The derivatives are the limits mentioned.

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