How to understand the conductivity tensor using the more modern language of multi-linear mappings I'm trying to understand tensor analysis from a more modern point of view using Spivak's Calculus on Manifolds. I find myself getting very confused when I try to relate the modern viewpoint to my previous exposure to old-fashioned expositions of Cartesian tensors. For example, the conductivity tensor relates the current vector field to the electric vector field in an anisotropic medium:
$$\bf{J}=\sigma \bf{E}$$
From my previous exposure I'd call $\sigma$ a rank 2 tensor that maps the electric field at a point $p$ to the current field at the same point, i.e.:
$$\sigma: \bf{E} (p) \rightarrow \bf{J} (p)$$ Which I guess is really just
$$\sigma: \mathbb{R}^3\rightarrow\mathbb{R}^3$$
But in Spivak, a tensor $T$ of rank 2 is a multi-linear mapping from two copies of some vector space $V$ to the reals:
$$T: V \times V \rightarrow \mathbb{R}$$
Are these equivalent descriptions of the same mathematical object? If so, can you help me see the equivalency?
 A: There is a difference between a matrix that defines a linear map
$$
M:V\to V, \quad {\bf y}={\bf M}{\bf x}, \quad y^i={M^i}_j x^j
$$
and one that defines a quadratic form
$$
{\bf x}^T{\bf M} {\bf x}= x^i M_{ij}x^j
$$
I suggest you look at section 10.2.2 (p 393) of our book for more details.
A: The key thing to keep in mind is that vectors themselves are (linear) maps: $$v: \mathbb{R}^n \to \mathbb{R} \,.$$
If you plug a single vector $v \in \mathbb{R}^n$ into a tensor $T: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$, you are left with a map $$Tv: \mathbb{R}^n \to \mathbb{R} \,,$$
which is by definition also a vector.
In that way, you can interpret a rank 2 tensor as either a map from two vectors to $\mathbb{R}$ or as a map from one vector to another. Remember that the space of linear maps on $\mathbb{R}^n$ is again $\mathbb{R}^n$, which is why the vector $v$ can be though of as an element of either.
Note that I cheated a bit and identified all vector spaces with $\mathbb{R}^n$. This is only possible once you pick bases in your spaces. In reality you also need differentiate between a vector space and its dual space. But as long as you work in a metric space, this distinction is not relevant, you can always use the metric tensor to move between the spaces.
A: There are three equivalebt notions of tensors.

*

*Transformational - this is usually discussed in textbooks dealing with general relativity. They are written with imdices such as $T_{ijk}$


*Algebraic - this is usually dealt with in mathematics books in abstract algebra. They are written $u \otimes v \otimes w$


*Multilinear - these are usually discussed in more modern books dealing with general relativity.
Unfortunately, there are few books that discuss all three in a coherent and pedagogic fashion.
Your example of $T: V \times V \rightarrow \mathbb{R}$ can be transformed into $T: V \rightarrow V^*$. The rule is you can transfer a vector space factor from one side to the other by decorating with a dual. If there are no factors left, then you leave the ground field, which in this case is $\mathbb{R}$.
Thus your $\sigma:\mathbb{R}^3 \rightarrow  \mathbb{R}^3$ is equivalently, $\sigma:\mathbb{R}^3 \times (\mathbb{R}^3)^* \rightarrow \mathbb{R}$. This shows that it is a mixed tensor of type (1,1).
