Definition of Tensor and.... Cotensor?

[kiuhnm]

Why are there (at least) two definitions of a tensor? For some people a tensor is a product of vectors and covectors, but for others it's a functional. While it's true that the two points of view are equivalent (there's an isomorphism) I find having to switch between them confusing, as a beginner.
In a sense, the two points of view seem one the dual of the other because $V$ becomes $V^*$ and $V^*$ becomes $V^{**}\sim V$, i.e. a tensor in $V\otimes \cdots \otimes V \otimes V^*\otimes \cdots \otimes V^*$ becomes a tensor in $V^*\otimes \cdots \otimes V^* \otimes V\otimes \cdots \otimes V$ seen as a functional which, basically, takes tensors of the first definition: $$
T(v_1, \ldots, v_p, w^1, \ldots, w^q) \sim
T(v_1\otimes\cdots\otimes v_p\otimes w^1\otimes\cdots\otimes w^q)
$$ Moreover, $T(e_I\otimes (e^*)^J) = T_I^J$ i.e. the functional gives the components for the tensor in the first definition.
If this is correct, shouldn't we really talk about cotensors?

 

[jambaugh]

The best (imnsho) overarching definition of tensors is as elements of a representation of the group of linear transformations GL(V) on the base vector space V. The subdivision into irreducible representations defines the rank structure. I find this best as a topmost definition because it does not depend on construction methods. But for practical application we must get into those details. What I like about the master definition is that it helps me see how to link the multiple constructive definitions.

• I know how scalars transform under linear transformations (i.e. trivially).
  
• I know how vectors transform under linear transformations (this IS the base definition of linear transformations).
• I thus know how covectors transform since they contract with vectors to yield scalars.
• I know how tensor products of any pair of representations transform in terms of how each of those representations transform.
• The duality mappings give me an inter-consistency between types of tensors. I can contract a rank (n,m) tensor with a rank (m,n) tensor to form a scalar. So each must transform dually to the other and that's all that the rank is keeping track of. So I can also say that any mathematical object that linearly maps a tensor to a scalar is a tensor of dual rank. Cotensors are also (isomorphic to and thus can be identified with) tensors.

I don't know if I've addressed your specific issue but this is how I resolved that problem for myself.

 

[fresh_42]

Why are there (at least) two definitions of a tensor? For some people a tensor is a product of vectors and covectors, but for others it's a functional. While it's true that the two points of view are equivalent (there's an isomorphism) I find having to switch between them confusing, as a beginner.
In a sense, the two points of view seem one the dual of the other because $V$ becomes $V^*$ and $V^*$ becomes $V^{**}\sim V$, i.e. a tensor in $V\otimes \cdots \otimes V \otimes V^*\otimes \cdots \otimes V^*$ becomes a tensor in $V^*\otimes \cdots \otimes V^* \otimes V\otimes \cdots \otimes V$ seen as a functional which, basically, takes tensors of the first definition: $$
T(v_1, \ldots, v_p, w^1, \ldots, w^q) \sim
T(v_1\otimes\cdots\otimes v_p\otimes w^1\otimes\cdots\otimes w^q)
$$ Moreover, $T(e_I\otimes (e^*)^J) = T_I^J$ i.e. the functional gives the components for the tensor in the first definition.
If this is correct, shouldn't we really talk about cotensors?

This question could be discussed on so many levels, e.g. "covectors are vectors, too", "what is a cotensor?", "confusing usage of co- and contravariant in mathematics and physics", "mathematical and physical point of view in general", "duality", "ranks", "area of application", ect., that it is per se difficult.

Let me start with a formal answer.

The tensor product is a solution to a couniversal mapping problem, and as such a couniversal object. This implies that the term cotensor is nonsense. What you probably have meant are the co- and contravariant parts of a tensor product: vectors and covectors. The necessity to distinguish them is driven by its application, resp. its purpose. As I mentioned above, $V^*$ is a vector space, too, so if we want to deal with it as such or as a space of homomorphisms depends on what we want to do with it. The tensor functor and the homomorphism functor are what has to be distinguished. And we have the same problem with both of them: There are also homomorphisms $V^* \longrightarrow W^*$ and the same question occurs.

A mathematical answer I tried to give is here: https://www.physicsforums.com/insights/what-is-a-tensor/
And to see the multiple ways a certain construction can be viewed as, you should have a look at https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/ where I listed (more or less for fun) the many different ways a differentiation can be viewed as.

However, it is generally important to know, whether we speak of a vector such as an electric field or a linear transformation such as a rotation of frames. Simply because we want to consider how an electric field behaves when coordinates are rotated, so better not to mix them up.

 

[kiuhnm]

Thank you both!