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In the book An introduction to manifolds by Tu, Loring W, covector is defined as follows:

If $V$ and $W$ are real vector spaces, we denote by $\operatorname{Hom}(V, W)$ the vector space of all linear maps $f: V \rightarrow W$. Define the dual space $V^{\vee}$ of $V$ to be the vector space of all real-valued linear functions on $V$ : $$ V^{\vee}=\operatorname{Hom}(V, \mathbb{R}) $$ The elements of $V^{\vee}$ are called covectors or $1$-covectors on $V$.

However, in the book Quick Introduction to Tensor Analysis by R. A. Sharipov, the covector is defined as follows:

Let's denote our hypothetical object by $\mathbf{a},$ and denote by $a_{1}, a_{2}, a_{3}$ that three numbers which represent this object in the basis $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$. By analogy with vectors we shall call them coordinates. But in contrast to vectors, we intentionally used lower indices when denoting them by $a_{1}, a_{2}, a_{3}$. Let's prescribe the following transformation rules to $a_{1}, a_{2}, a_{3}$ when we change $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$ to $\tilde{\mathbf{e}}_{1}, \tilde{\mathbf{e}}_{2}, \tilde{\mathbf{e}}_{3}$: $$ \tilde{a}_{j} =\sum_{i=1}^{3} S_{j}^{i} a_{i} \tag{8.1} $$ $$ a_{j} =\sum_{i=1}^{3} T_{j}^{i} \tilde{a}_{i}\tag{8.2} $$ Here $S$ and $T$ are the same transition matrices as in case of the vectors in $(6.2)$ and $(6.5)$. Note that $(8.1)$ is sufficient, formula $(8.2)$ is derived from $(8.1)$.

DEFINITION 8.1. A geometric object a in each basis represented by a triple of coordinates $a_{1}, a_{2}, a_{3}$ and such that its coordinates obey transformation rules $(8.1)$ and $(8.2)$ under a change of basis is called a covector.

My question is, are the two definitions equivalent? What's the relation between them? I cannot figure out the relation between "obeying transformation rules" and linear maps $f: V \rightarrow W$.

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    $\begingroup$ the definition of Loring Tu is the general one, the other could be a particular case of covectors in $\Bbb R^3$ $\endgroup$
    – Masacroso
    Jan 16, 2021 at 9:52
  • $\begingroup$ Could you expain concretely about how the other one be a particular case? $\endgroup$
    – FFjet
    Jan 16, 2021 at 10:19
  • $\begingroup$ Presumably those transformation rules give rise to the identity $a\cdot v=\tilde{a}\cdot\tilde{v}$ where $v$ and $\tilde{v}$ both refer to the same vector expressed in the different bases. Then the operation (dot product with $a$ or $\tilde{a}$) is independent of basis and can be regarded as a linear functional $V\to\mathbb{R}$ à la Tu. $\endgroup$
    – Ali
    Jan 16, 2021 at 11:20
  • $\begingroup$ Ok I've deleted my addition notes. $\endgroup$
    – FFjet
    Jan 16, 2021 at 15:17
  • $\begingroup$ @FFjet maybe the notes refer to this. Then it seems that the covectors are a case of covariant vectors. $\endgroup$
    – Masacroso
    Jan 16, 2021 at 15:19

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The idea is that $V^\vee:=\operatorname{Hom}(V,\mathbf{R})$ is a vector space, and thus vectors in $V^\vee$ can have components with respect to a basis. If we have two different bases for $V^\vee$, then we will get two different sets of components for the same vector, and the way these components are related to each other is what we call a transformation law, or a change of basis, or something along those lines.

Let's look at the vector space $\mathbf{R}^3$. If we have two bases $E=\{e_1,e_2,e_3\}$ and $F=\{f_1,f_2,f_3\}$, we may express some vector $v\in\mathbf{R}^3$ as $v=a^1e_1+a^2e_2+a^3e_3$ or $b^1f_1+b^2f_2+b^3f_3$. How are the $a^i$'s and $b^i$'s related? Well, it's enough to know how $E$ and $F$ are related; so let $A$ be the matrix sending $e_i$ to $f_i$. That is, $f_i=Ae_i=A^1_ie_1+A^2_ie_2+A^3_ie_3$. Let's write $\epsilon^j$ for the map that takes a vector to its $j$-th component in the basis $E$, so $v=\epsilon^1(v)e_1+\epsilon^2(v)e_2+\epsilon^3(v)e_3$. Notice that $\epsilon^i$ is linear, and $\epsilon^j(Ae_i)=A^j_i$. We then compute \begin{align*} a^j=\epsilon^j(b^1f_1+b^2f_2+b^3f_3) &=b^1\epsilon^j(f_1)+b^2\epsilon^j(f_2)+b^3\epsilon^j(f_3)\\ &=b^1\epsilon^j(Ae_1)+b^2\epsilon^j(Ae_2)+b^3\epsilon^j(Ae_3)\\ &=b^1A^j_1+b^2A^j_2+b^3A^j_3. \end{align*}

So $a^j=\sum_ib^iA^j_i$. Roughly speaking, if $F$ is the old basis system and $E$ is the new one, this formula tells us that we must take the matrix $A$ sending new basis vectors $e_i$ to old ones $f_i$ and multiply it with the old components $b^i$ to get the new components $a^i$. As such, we say that the components of a vector transform contravariantly. (Later you'll find that the components of a covector transform covariantly. And a vector transforms covariantly as well, except people sometimes will say vector when they mean the components of the vector. And so they'll say a vector transforms contravariantly.)

So we can think of a vector as an abstract element of the vector space $\mathbf{R}^n$ subject to the vector space axioms, or as an equivalence class of $n$-tuples as subject to the constraints of transformation laws. Both ways end up imposing the same structure. Similarly for your question on covectors being linear maps from $V$ to $\mathbf{R}$. You should be working on a higher level of abstraction and thinking of covectors as elements of vector spaces, and thus admitting transformation laws with respect to different bases.

Later on you will see vector spaces, vectors and covectors generalized to tensors, and you will notice that mathematicians and physicists tend to treat them rather differently. To understand the interplay between the mathematician's abstract, coordinate-free, element of tensor space perspective and the coordinates, transformation laws perspective can be rather tricky. For that I recommend Michael Spivak's A Comprehensive Introduction to Differential Geometry, as well as section 7 of Keith Conrad's Tensor Products.

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