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In this post the cotangent space $T_p^* M$ was defined as the quotient $C^\infty(M)/W_p(M)$ where $W_p(M) \subset C^\infty(M)$ is the subspace of smooth functions with vanishing 1st derivative (i.e. all partial derivatives) at $p$ in some (and so in any smoothly compatible) coordinates. There I argued that $\dim (T_p^* M) = \dim \mathbb R^{*n}$. An element of this space will be denoted $[f]_p$; the class of functions that differ from $f$ by a function whose 1stPDs vanish at $p$.

The trsnsformation law for the cobasis when one changes coordinates as $\psi\circ\phi^{-1}:\phi(U) \to U \subset M$ such that $x \mapsto y(x)$ at $p$ is by $$ [y^i]_p = [y(p) + \frac{\partial y^i}{\partial x^j}(p) x^j + \dots]_p = \frac{\partial y^i}{\partial x^j}(p) [x^j]_p $$ where I used the equivalence relation to get rid of the all terms which have vanishing 1stPD at $p$ and the fact that this quotient map is a linear map. This is exaclty the transformation $dy^i = \frac{\partial y^i}{\partial x^j}dx^j$.

The idea is: Since a tangent vector at a point is a classe of curves through that point, now (in a similar manner) a cotangent vector (or a dual) at a point is class of scalar functions in a neighborhood of that point. The pairing of a curve and scalar function at a point (if differentiable) gives the directional derivative, without any reference to any coordinates. A 1-form is a (smooth) assignment of smooth scalar fields on a neighborhood of each point on the manifold.

In prticular, one can visualise the $dx^i \in T_p^*M$ on the manifold (at $p$) as a scalar field $x^i$ or as any field in its class $[x^i]_p$ in a neighborhood of $p$.

I just wanted to know whether all of that makes sense

Note: The first exterior derivative of $f \in C^\infty(M)$ is now the map $$ p \mapsto [f]_p = \frac{\partial f}{\partial x^j}(p) [x^j]_p, $$ its value at $p$ is a class of functions, not a scalar value as is the case of $f$.

Physor
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    It's not clear that there's a well-defined question here, but everything looks quite standard (although you probably want to quotient not $C^\infty(M)$ but rather the subspace of functions that vanish at $p$). This construction is often called the Zariski cotangent space, and is probably more commonly seen in algebraic geometry than in differential geometry. – EuYu Nov 07 '21 at 08:15
  • I'm aware of the definition you mentioned, but the one mentioned in the post is another one that is linked above and turned out to have been introduced by N. Hitchin https://math.stackexchange.com/questions/4230215/hitchins-definition-of-tangent-space-and-tangent-vectors – Physor Nov 07 '21 at 08:51
  • I don't really see how the two definitions are different. Taylor's theorem implies that a function has first (and zeroth) derivatives vanishing at $p$ if and only if it is a linear combination of products of functions that vanish at $p$. So Hitchin's definition (up to the subtlety that he does not require his functions to vanish at $p$, which should be an error) is completely equivalent to the standard Zariski construction. Just the Zariski definition is more coordinate invariant. – EuYu Nov 07 '21 at 09:07
  • If you think that the definition of Hitchin contains an error then please right an answer here https://math.stackexchange.com/questions/4254933/possible-definition-of-the-cotangent-space and show that my argument in the link is flawed. And I don't know what you mean with equivalent, at the end we all get some $n$-dimensional space and who cares about the vectors. Well the definition is stating explicitly what the vectors are, and different definitions have different concrete vectors, different objects I mean. – Physor Nov 07 '21 at 09:13
  • I wanted the concrete objects on the manifolds, that work as cotangent vectors, Zariski's construction is one answer, but that of Hitchin is another – Physor Nov 07 '21 at 09:17
  • No, Hitchin's construction is fine. I was being silly. But I still don't see any functional difference between it and the usual Zariski definition besides the (in my view) superficial difference of phrasing the quotient in terms of derivatives rather than products. And I meant equivalent in the sense that there is a rather canonical bijection between the two definitions. If $[f]_a$ is your equivalence class and $\langle f \rangle_a$ is the usual Zariski equivalence class, then $[f]_a \leftrightarrow \langle f - f(a)\rangle_a$ is an isomorphism of vector spaces. – EuYu Nov 07 '21 at 09:24
  • I understand that there is no functional difference but the definition of N. Hitchin allows more freedom in considering any smooth function in a neighborhood of $p \in M$ as covector, not only those which vanish at $p$, and it is shorter in that sense than the other definition – Physor Nov 07 '21 at 09:30
  • At this point, it seems like you agree the two definitions are equivalent up to minor difference. After that, it seems like primarily opinion which definition you prefer. I certainly sympathize with the fact that the derivative based definition is much more intuitive, but it is also less canonical and less amenable to generalization. In any case, I'm still not sure what your actual question is. – EuYu Nov 07 '21 at 09:48
  • If you would like to close it as an opinion-based post, you can do that. Essentially it was to check whether the equation above makes sense, based on the definition of equivalence of the 1st derivative and whether it is correctly equivalent to $dy^i = \frac{\partial y^i}{\partial x^j}dx^j$ – Physor Nov 07 '21 at 09:52

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