Assuming a vector space V and it's basis set $\{\vec{e}_\nu\}$. A vector $\vec{v}$ can be written as: $\vec{v}=x^\nu\vec{e}_\nu$ where $x^\nu$ is the corresponding contravariant coordinate. We can choose another basis set $\{\vec{e'}_\mu\}$ and perform a transformation in order to find the coordinates of the vector $\vec{v}$ in the new basis. The transformation is:
$$x'^\mu=\frac{\partial x'^\mu}{\partial x^\nu} x^\nu \tag I $$
$\frac{\partial x'^\mu}{\partial x^\nu}$ is a (1,1) tensor living in the tensor product space spanned by :$\ e_{\mu}\otimes e^{\nu}$. Let's call the tensor $T$. The components of $T$ can be written as: ${T^{\mu}}_{\nu}$. I would like to know what happens formally when we "apply" the tensor on the vector in order to perform the transformation. Is the word "apply" even correct?
Physicists only worry about the components. Contraction is simply removing indices which are appear both as lower and upper. I want to know what happens to the basis. The contraction results in: $x'^\mu={T^{\mu}}_{\nu} \ x^{\nu}$. What I don't understand is what is happening to the basis. Are we formally forming a bigger tensor product space by multiplying $\ e_{\mu}\otimes e^{\nu}$ by $e_{\nu}$ i.e. $\ e_{\mu}\otimes e^{\nu}\otimes e_{\nu}$? Or is $\ e_{\mu}\otimes e^{\nu}$ acting as an operator on $e_{\nu}$, so we have: $\ e_{\mu}\otimes e^{\nu}(e_{\nu})$? and then using the formula $e^{\mu}(e_{\nu})={\delta^{\mu}}_{\nu}$ to finally have $e_{\mu}$ left only? If so, isn't delta itself a (1,1) tensor? So how is the dimension of the tensor product being reduced to 1?
Part of the confusion arises from the Kronecker delta itself. In Wikipedia (http://en.wikipedia.org/wiki/Kronecker_delta) it is said to be a (1,1) tensor which can act as one of the three cases: 1) Identity mapping $V \rightarrow V$ or $V^{*} \rightarrow V^{*}$ 2) Trace or contraction $V \otimes V^{*} \rightarrow K$ 3) Scalar multiplication $K \rightarrow V \otimes V^{*}$ (please see Generalization of Kronecker delta in the wiki link provided). Can someone clarify how can a single object be considered as various mappings?
Best Answer
I'm no physicist so I can't speak exactly to what they do or how they think of things, but here is some related math. As you suggested, $(1,1)$ tensor is an element of $V \otimes V^*$ for some finite dimensional vector space $V$. So let's talk about this space $V \otimes V^*$ and some of the structure it has, I will address the different actions you mentioned in a somewhat different order.
-First and foremost, the definition of a dual vector space means that we have a bilinear map from $V \times V^*$ to the ground field $K$. If $\vec{e}$ is a vector in $V$ and $f \in V^*$ is a linear map from $V$ to $K$ this bilinear map just sends $(\vec{e},f)$ to $f(\vec{e})$. Bilinear maps from a pair of vector spaces are the same thing as linear maps from the tensor product, this is how tensor products are (usually) defined by mathematicians. Hence we can view this pairing as a linear map $V \otimes V^* \rightarrow K$, this is the trace or contraction map, it is also often called the evaluation map by mathematicians.
-Next, there is a natural map from $V \otimes V^*$ into $End(V)$ the space of linear maps from $V$ to itself. In order to define this map it will be enough to say how a $(1,1)$ tensor acts on a vector $\vec{v}$ and extend linearly. This is just given by $(\vec{e}\otimes f)(\vec{v}) = f(\vec{v})\vec{e}$. More coordinate freely this map is given by the evaluation map from above applied to the $V^*$ component along with the copy of $V$ corresponding to the input vector $\vec{v}$, followed by the scalar multiplication map from $V \otimes K$ to $V$. This map is defined and injective for any vector space $V$, but for infinite dimensional vector spaces it is not surjective, its image is the space of finite rank linear maps from $V$ to $V$. So this is how we get a linear map from $V$ to $V$, and since $V$ and $V^*$ play essentially the same role (for finite dimensional vector spaces) we can also get a linear map from $V^*$ to itself essentially the same way. Also note that if we choose a basis $\vec{e_1},...,\vec{e_n}$ for $V$ then $End(V)$ is can be identified with $n \times n$ matrices, and the evaluation map from above is just taking trace.
-Finally, we get to the scalar multiplication map which is often called coevaluation by mathematicians. From the previous part, if $V$ is finite dimensional we have an isomorphism between $V\otimes V^*$ and $End(V)$ the space of linear maps from $V$ to itself. Inside here there is a distinguished linear map given by the identity map $\vec{v} \rightarrow \vec{v}$. Coevaluation is just the map $K \rightarrow V \otimes V^*$ that sends $1$ to the identity map under the identification of $V \otimes V^*$ with $End(V)$. Again this part relies heavily on $V$ being finite dimensional as the identity map only has finite rank (and is hence in the image of $V \otimes V^*$) in this case.
Note that all of these maps are canonically defined (i.e. independent of the basis you choose), which means that if you do any calculation only involving these maps the answer you get won't depend on how you choose coordinates.