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Previously I didn't understand why $\frac{\partial }{\partial x^i}$ can be a basis for a vector field $X$. For example, in integration of curve length, if we have a curve $\in \mathbb{R}^2$,$\lambda=(x^1(t), x^2(t))$ or $(t,y(t))$, then $\frac{dx^1(t)}{dt}, \frac{dx^2(t)}{dt}$ or $(1,y'(t))$ is vector along the curve and so (part of) the vector field. It seems the basis should instead be something like $\frac{dx^i}{dt}$ or just $dx^i$ (since the denominator is a real number and don't change direction of the basis vector).

Then I find an explanation for $\frac{\partial }{\partial x^I}$: In differential equation, taking a case of $\mathbb{R}^2$ as an example, we have vector field of an integral curve, so we may say vector field $X$ is a function of an integral curve $\lambda$, $X(\lambda)$. And it equals differential of the curve, say $\frac{d \lambda}{dt}$. So we have $X=\frac{d }{dt}$. It seems therefore the basis for $X$ should be $\frac{\partial }{\partial x^i}$.

The two kinds of basis both seem to make sense, but obviously they are completely different.

Though I know that $dx^i$ is the basis for dual/covariant vector, and I may well accept that conclusion, I still can't see what actually goes wrong in my thinking. So what happens?

(Edited to add:)

Here is a note for myself: I see this question seems to be more complicated than it appears and is extensively discussed in the chapter about tangent bundle in Spivak's book. It's about equivalence between different tangent bundles (candidates) and there is, furthermore, a theorem to prove such equivalence. I see relations between these candidates are the thing I'm currently considering, and relation between tangent bundle to cotangent bundle is another issue (and should not be mixed up with this question).

Charlie Chang
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    What do you mean by "the denominator is a real number"? The notation $\frac{dx^i}{dt}$ does not stand for a fraction.

    It may be better to first understand first the bases of the tangent space at a point. By definition, the elements of the tangent space at a point are differential operators, so a basis for that space necessarily consists of certain differential operators. I suggest you try to fully understand the tangent spaces first and once that is done you can start with the cotangent spaces.

     – Jackozee Hakkiuz Jul 28 '20 at 18:21
  • well, I mean when we try to differentiate a variable in a space $S$ against a real number variable, the result is a variable belong to the same space $S$, I here use 'denominator' in a quite loose way, I guess 'the denominator' (as I call it) $dt$ here is to better to be regarded as the independent variable of derivative ($df$) of the function (say $f$) that's to be differentiated; if so we see $\lim_{dt\rightarrow 0} \frac{df}{dt}$ is roughly in the same space where $df$ is, and (not always) in the same space where $f$ is in (And I suspect what I said may go wrong in this case). – Charlie Chang Jul 28 '20 at 18:33
  • I mention cotangent space for comparison, and my question is mainly about tangent space actually. I will reread the part of tangent space. – Charlie Chang Jul 28 '20 at 18:35
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    Ok. Then what "two kinds of bases" are you referring to? – Jackozee Hakkiuz Jul 28 '20 at 18:54
  • I meant the two basis {$A= \frac{dx^i}{dt}$}, (or perhaps more clearly, ${{\frac{dx^i}{dt}}_p, p\in$ a manifold, $x^i$ as coordinates of any curve passing through $p$ on the manifold$}$), $B={\frac{\partial }{\partial x^i}}$ (or more clearly, {$\frac{\partial }{\partial x^i}_p, p\in$ a manifold $}$), as described in the post, both seems to be basis of tangent space. – Charlie Chang Jul 28 '20 at 20:28
  • Now w new notations I see there are some problems here. For basis A we don’t know/specify what curve to differentiate, and so it can be many (> n) or tangent to each other at a point, in which cases won’t give the basis for $M$. Or we can say it will only give basis for a curve itself as 1-manifold(if it is), and so actually it becomes the 1-manifold special case, and its basis can be written as $\frac{d}{dt}$, which is consistent with B definition,... – Charlie Chang Jul 28 '20 at 20:29
  • ... but there are two points here: 1. we need to understand the basis as a set of operators acting on the manifold, otherwise it’s just operator not usual vectors (because of that it’s still unnatural for me to call them basis, particularly for it doesnt specify what it’s to act on; though if we regard the usual basis of vector space as a transposed vector and therefore also an operator, things seem to be the same, and the basis will look like a covariant vector)... – Charlie Chang Jul 28 '20 at 20:29
  • ... 2. The acted-on manifold $M$ is of the same dim as that of basis, but $M$ can be denoted in another coordinate system $c$ (different from the chart on the neighborhood) and can look like having more dim than that of the basis—perhaps often so when $M$ is immersed/embedded in higher dim. ... – Charlie Chang Jul 28 '20 at 20:30
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    $\left(\frac{dx^i}{dt}\right)_p$ is not a vector, it's just a real number. – Jackozee Hakkiuz Jul 28 '20 at 20:30
  • Let us continue this discussion in chat. – Jackozee Hakkiuz Jul 28 '20 at 20:30
  • ...Another example for that is a surface $N$, say {$ (x^1, x^2, f(x^1, x^2))$}, then if we choose the chart to be (in a very loose way) cood sys (x,y), applying the corresponding basis $\frac{\partial }{\partial x^i}$ to $N$, (notice we apply to a point, however many components it has in $c$, a basis operator a time) we get $(1,0, f_{x^1})(0,1, f_{x^2})$ which are a basis for $M$, this example also illustrates the 1st point. – Charlie Chang Jul 28 '20 at 20:30
  • Well it’s because my reply is a bit too long so I sent it separately. I see there are some points to be reconsidered. – Charlie Chang Jul 28 '20 at 20:36

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