I am currently taking a module on Differential Geometry, and as such I have been mostly looking at tensors as objects following the usual transformation rule. However I would like to understand how that is equivalent to viewing tensors as multilinear maps from direct products of Vector and their Dual spaces to the corresponding field of scalars.
Here is my understanding of the distinction between a covector and a vector(as multilinear maps):
Let $V$ be a vector space over $\mathbb{R}$. The dual space of $V$, denoted $V^*$, is defined as $V^* =\{ϕ:V→\mathbb{R}∣ϕ$ is linear$\}$. $V^*$ is an n-dimensional vector space over the same field. Thus, $V$ is isomorphic to $V^*$. Thus they are, as vector spaces, identical. But we consider them to be different by considering elements of $V^*$(covectors) to act on vectors as linear maps.
My confusion:
I have seen it said that you can consider vectors to also be linear a maps of covectors. Doesn't this contradict the distinction of $V$ and $V^*$?
My attempted explanation for this:
Either $V$ and $V^*$ can be considered the dual space of the other, the choice is arbitrary
It seems to me that when viewing vectors and covectors as objects obeying their respective transformation rule, this confusion is not an issue, since we simply define the operation of contraction to combine a vector and a covector into a scalar.
I really can't make intuitive sense of this, so any insights would be appreciated.
This mainly is a question on linear algebra, and it touches on a subtle point. As you say, two (finite dimensional) vector spaces of the same dimension are isomorphic. But this does not mean that they are (or can be considered as) identical. Indeed, there is a big freedom of choice for the isomorphism between two spaces of the same dimension. In particular, if you look at an abstract vector space $V$, you know that $V^*$ has the same dimension, but there is no "natural" way to associate to $v\in V$ a linear functional in $V^*$. To get such an association you need to make additional choices, e.g. a basis for $V$ or an inner product on $V$. This is related to the situation you meet in differential geometry, where you know that all tangent spaces have the same dimension, but there is no natural way to identify different tangent spaces without making additional choices (say a chart). So you cannot simply consider them as "identical" although they are isomorphic. In the same spirit, you have to distinguish between a tangent space and its dual space.
Still, there is a certain symmetry in the situation between $V$ and $V^*$. This comes from the fact that you can also consider the dual $(V^*)^*$ of $V^*$. Of course, this has again the same dimension as $V^*$ and hence as $V$, but here much more is true: There is a natural way to identify the space $(V^*)^*$ with $V$. (But there is no natural way to identify $(V^*)^*$ with $V^*$.) The natural identification sends a vector $v\in V$ to the linear map $V^*\to\mathbb R$ given by $\phi\mapsto\phi(v)$. This means that in a way you can swap the roles of $V$ and $V^*$ (and this is used frequently when dealing with tensor fields in differential geometry), but this does not mean that you have a way to identify $V$ with $V^*$.
This does not mean that there is no difference between vectors and covectors. From a differential geometry point of view, this is because smooth maps (and diffeomorphisms) act on them differently. Consequently, there is a diffeomorphism invariant linear operation on covectors (the exterior derivative) and no such operation exists on vectors. But you can also interpret vectors in terms of linear functionals.