Some says covectors are linear map that maps $ V \mapsto R $ (which means it's just a row vector considering vectors are $ n $ x $ 1 $ matrix and mapping is matrix multiplication), while some say it's a certain value whose components follow $ v'_i = \frac{\partial x^j}{\partial x'^i} v_j $ under coordinate transformation from $ x $ coordinate to $ x' $ coordinate.
Let's see gradient for example.
$ \nabla $ itself is not a map $ \nabla : V \mapsto R $, but divergence, $ \nabla \cdot $ is. So $ \nabla $ is not a covector of first definition.
However, considering $ \nabla $'s components are $ \frac{\partial}{\partial x^i} $, they actually follow the transformation rule by $ \frac{\partial x^i}{\partial x'^j} \frac{\partial}{\partial x^i} = \frac{\partial}{\partial x'^j} $. So it's a covector of the second example.
Which one's right? Are they actually the same definition and I'm misunderstanding something or are they just different things I'm talking about?
Best Answer
REPLY TO COMMENT:
A vector field with respect to coordinates $(x^1, \dots, x^n)$ is of the form $$ V = v^i\frac{\partial}{\partial x^i}. $$ Therefore, if we do a change of coordinates, where we view the old coordinates $(x^1, \dots, x^n)$ as functions of the new coordinates $(\bar{x}^1, \dots, \bar{x}^n)$, then coordinate vector fields are related by $$ \frac{\partial}{\partial \bar{x}^j} = \frac{\partial x^i}{\partial\bar{x}^j}\frac{\partial}{\partial x^i}.$$ Therefore, $$ V = v^i\frac{\partial}{\partial x^i} = \bar{v}^j\frac{\partial}{\partial\bar{x}^j}, $$ where $$ v^i = \bar{v}^j\frac{\partial x^i}{\partial\bar{x}^j}. $$ On the other hand, with respect to the coordinates $(x^1, \dots, x^n)$, the gradient of $f$ is the vector field $$\nabla f = \frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^i} $$ and the gradient of $f$ with respect to $(\bar{x}^1,\dots,\bar{x}^n)$ is $$\nabla f = \frac{\partial f}{\partial\bar{x}^j}\frac{\partial}{\partial \bar{x}^j} = \frac{\partial f}{\partial \bar x^i}\frac{\partial x^i}{\partial\bar{x}^j}\frac{\partial}{\partial x^j}.$$ Observe that in the general, the two formulas for $\nabla f$ conflict. They do not give the same vector field. On the other hand, if you compute $df$ with respect to each set of coordinates, you find that the two formulas do match.
ORIGINAL ANSWER:
The point of distinguishing a covector from a vector is to avoid any dependence on coordinates. Why do this? Because the geometric and physical properties of space should not depend on coordinates (e.g., whether you use inches or meters to determine distance or position).
One way to do this is to write everything initially in terms of coordinates and then show that the formulas change in the right way under a change of coordinates.
An alternative way is to not use coordinates at all. We also initially try to avoid using the dot product (this is a special case of what it known as a Riemannian metric). The differential versus gradient of a function is a good example.
You should think of the differential of a function as a way to represent the directional derivatives of a function. To simplify things but emphasize the coordinate independence, let's view $\mathbb{R}^n$ as an abstract vector space.
The directional derivative of a function $f: V \rightarrow \mathbb{R}$ at $x \in V$ in a direction $v$ is defined to be the following: Given any curve $c: (-\delta,\delta) \rightarrow V$ such that $c(0) = x$ and $c'(0) = v$, let $$ D_vf(x) = \left.\frac{d}{dt}\right|_{t=0}f(c(t)). $$ There are the following facts: This definition does not depend on the curve $c$ used, and, if $f$ and $x$ are held fixed, the function $$ v \mapsto D_vf(x) $$ is a linear function from $V$ to $\mathbb{R}$. It therefore is a covector, which is called the differential of $f$ at $x$ and denoted $$df(x): V \rightarrow \mathbb{R}.$$ In particular, $df(x)$ is an element in the dual space $V^*$. The map $x \mapsto df(x)$ is an example of what is known as either a covector field or $1$-form.
Note that everything above used no coordinates or basis on $V$.
Note that there is no analogous way to define the gradient $\nabla f$ in a similar coordinate-independent way. You have to either choose a basis of $V$ or an inner product on $V$. Either way, you get a linear isomorphism $V^* \rightarrow V$. This allows you to map $df(x)$ to an element in $V$, which is known as the gradient of $f$ at $x$. The gradient defines a map $x \mapsto \nabla f(x)$, which is an example of a vector field.
You don't always want to do this, so that's why modern abstract differential geometry carefully distinguishes vectors and vector fields from covectors and $1$-forms. Later, when Riemannian metrics are introduced, then the gradient of a function reappears.