TITLE: What is a covector and what is it used for?
QUESTION [74 upvotes]: From what I understand, a covector is an object that takes a vector and returns a number. So given a vector $v \in V$ and a covector $\phi \in V^*$, you can act on $v$ with $\phi$ to get a real number $\phi(v)$. Is that "it" or is there more to it? 
I find this simple realization hard to reconcile with an example given in my textbook: "Geometric Measure Theory" by Frank Morgan, where he explains that given $\mathbb{R}^n$, the dual space $\mathbb{R}^{n*}$ has basis $dx_1, dx_2 \dots$. So lets say we are in $\mathbb{R}^2$, so $dx$ is a function on any vector, that will return a scalar? I can't imagine what $dx([1,2])$ is. Can someone explain to me this concept of covectors reconciled with the infinitesimal case?
Edit: It has something to do with biorthogonality? I am really stumped on these concepts, like I'm ramming my head against a wall.

REPLY [16 votes]: The answers here are all excellent and in depth, but I might also admit, a little abstract and intimidating. I'd like to present a concrete reason for covectors to exist: to make the concept of dot product invariant to coordinate transformations.
Let's say we have some point $p$ on a manifold $M$ with a vector $V$ coming off of $p$. This vector has a length, that is independent of what local coordinate system you represent it with around $p$. We might be doing computations on that length as we "transport" the vector along points on the manifold, therefore we want to base our computational machinery in operations that ignore what the local coordinates are doing and only focus on absolute quantities. To normally compute the length of a vector, you might use a dot product with itself and take the square root, but I'll show here that the dot product is vulnerable to changes in coordinate systems. That will be why I bring in covectors to solve this problem - though the difference is very subtle.
Definition of vector
Let $\mathbf p$ be the position of a point $p$. We can set up a local coordinate basis by taking the vectors that result from varying $\mathbf p$ by varying each coordinate $x_i$ while holding the others constant, i.e. taking the partial derivative $\frac{\partial \mathbf p}{\partial x_i}$. A vector at $p$ can therefore be written in terms of that basis
$$ V = \sum_i v_i \frac{\partial \mathbf p}{\partial x_i} .$$
If we switch local coordinate systems to $x_i'$ then via the chain rule we get the relationship:
$$ V = \sum_i v_i \frac{d \mathbf p}{d x_i} = \sum_i v_i \sum_j \frac{dx_j'}{dx_i} \frac{d \mathbf p}{d x_j'} = \sum_i \left ( \sum_j v_j\frac{dx_i'} {dx_j}\right) \frac{d \mathbf p}{d x_i'} =\sum_i v_i' \frac{d \mathbf p}{d x_i'} $$
which means that $$v_i' = \sum_j \frac{dx_i'}{dx_j} v_j ,$$ so we've essentially multiplied by the Jacobian matrix $J$ of the transformation. Writing $v$ and $v'$ as column vectors we get the relation
$$ v' = Jv.$$
Problem with length of vector as we change coordinate systems
Let's say we have a vector $V = 1 \hat x + 1 \hat y$. The length of this vector is $\sqrt{V \cdot V} = \sqrt{1+1} = \sqrt{2}$. Now lets scale the local coordinate system down by 2 so that the vector becomes $V' = 2 \hat x' + 2 \hat y'$. The new length is $\sqrt{V' \cdot V'} = \sqrt{4+4} = 2\sqrt{2}$. The length has changed, because we are measuring our vector in terms of local coordinates, but we do not want that. We want a way of computing the length of the vector that is invariant to re-specifications of the local coordinate system. 
Introduction to covector
A covector is simply a linear function from vectors to real numbers, $\alpha : V \to \mathbb R$. For an example of a covector, we have these functions $dx_i$. As has been said in the previous answers, $dx_i$ is not a length but a function that takes vectors and picks out the $i$th coordinate component, for example in $ \mathbb R^3 $:
$$ dx_1( A\hat e_1 + B\hat e_2 + C \hat e_3) = A.$$
$$ dx_2( A\hat e_1 + B\hat e_2 + C \hat e_3) = B.$$
$$ dx_3( A\hat e_1 + B\hat e_2 + C \hat e_3) = C.$$
These functions form a basis in the space of covectors. Every linear function from vectors to the real numbers can be written as a linear combination of these functions:
$$ \alpha(v) = \sum_i a_i dx_i(v).$$
Now if we do a coordinate change to $x_i'$ we have to take contributions from all the different 
$$ \alpha(v) = \sum_i a_i dx_i(v) = \sum_i \sum_j a_i \frac{dx_i}{dx_j'} dx_j'(v) = \sum_i \left ( \sum_j a_j \frac{dx_j}{dx_i'} \right ) dx_i'(v) = \sum_i a_i' dx_i'(v) .$$
We see now that 
$$ a_i' = \sum_j a_j \left ( \frac{dx_j}{dx_i'} \right )$$
Now comes the important part. If we take a covector and apply it to a vector, we basically get a dot-product:
$$ \alpha(v) = \sum_i a_i v^i = av .$$
However, when we make a coordinate transform to $x_i'$ then:
$$ \alpha(v)' = a'v' = a \frac{dx}{dx'} \frac{dx'}{dx} v = av.$$
The answer stays the same. Thus, we've found a way to compute a "dot product" that is invariant to local coordinate transformations. 
Revisiting problem
When we scale the coordinate system for $V$ as in the previous problem, now we treat this as a covector being applied to a vector. Therefore upon the transform, the covector components actually half, so that the answer stays the same:
$$ |V| = \sqrt{\alpha(V)} = \sqrt{1/2*2 + 1/2 *2} = \sqrt{2} $$.
tldr covectors implement a "fix" to the dot product in differential geometry so that the result is invariant to changes of the local coordinate system.