# Critique on tensor notation

You are correct: a tensor can be viewed as a linear function in many different ways. We define a $(p,q)$ tensor $T$ as a function that takes $p$ covectors and $q$ vectors and returns a number:

$$T: {V^*}^p \times V^q \to \mathbb{R},$$

where the exponent indicates the number of Cartesian products. But this is equivalent to a function that takes $m$ covectors and $n$ vectors and returns a $(r,s)$ tensor, with $m+s = p$ and $n+r = q$, because you just leave some slots open.

You'll notice that this is completely independent of index notation, though of course indices make it obvious. Whether this is an advantage or a disadvantage is subjective. Like many other notations, it's at the same time confusing for beginners and versatile for experts. In most cases, it's better to not have to worry about what kind of function you're dealing with; the indices make equivalent things equal.

Forget tensors for a moment and just think about our old friend the matrix. Take a square matrix for example. You can multiply it onto a column vector and get back a column vector. Or you can put a row vector to the left and a column vector to the right, multiply them all together, and the outcome is a scalar. This is not normally felt to be a problem. It is just what happens.

Tensors are very similar, as you will have immediately seen no doubt.

In fact the tensor result is rather elegant. The index notation takes care of all the details and everything is consistent and logical.

This is why mathematicians have a tendency to define tensors using essentially a category-theoretical way, if $V,W$ are finite dimensional vector spaces over the same field, the tensor product is a pair $(V\otimes W,p)$ where $V\otimes W$ is a vector space and $p:V\times W\rightarrow V\otimes W$ is a bilinear map such that for any $A:V\times W\rightarrow X$ bilinear map ($X$ is some auxiliary vector space), there is a *unique* linear map $A^\otimes:V\otimes W\rightarrow X$ such that $$ A=A^\otimes\circ p. $$

One can then prove that this pair $(V\otimes W,p)$ has the *universal factorization property*, namely that if there is any other pair $(V\otimes^\prime W,p^\prime)$ satisfying this then there is a *natural* isomorphism $\imath:V\otimes W\rightarrow V\otimes^\prime W$ such that $p^\prime=\imath\circ p$, so the tensor product is unique up to natural isomorphism and afterwards one can prove existence by constructing an explicit representation.

This definition is nice because it shows that while the notion of a tensor can denote multiple different kind of maps and objects, they are all essentially equivalent.

On the other hand, I see the index notation being good because this property is actually manifest there. In the index notation, we don't really care what kind of map a tensor realizes, unlike the usual parenthetical/map notation where this is given from the start.

To give an explicit example, the curvature tensor of a linear connection $\nabla$ is usually defined using the "map" approach to be a bilinear map $$ R_x:T_xM\times T_xM \rightarrow \mathrm{Hom}(T_xM), (u,v)\mapsto R_x(u,v)=\nabla_U\nabla_V-\nabla_V\nabla_U-\nabla_{[U,V]}, $$ where $U,V$ are smooth extensions of $u,v\in T_xM$ into locally defined vector fields, and the covariant derivatives are evaluated at $x$.

However the curvature tensor is also a trilinear map $T_xM\times T_xM\times T_xM\rightarrow T_xM, (u,v,w)\mapsto R_x(u,v)w$ and can also be taken to be a quadrilinear map $$ T^\ast_xM\times T_xM\times T_xM\times T_xM\rightarrow M,\ (\omega,u,v,w)\mapsto \omega(R_x(u,v)w), $$ and the list doesn't end here.

But in the index notation, we simply write $R^\kappa_{\ \lambda\mu\nu}$, and from this notation it is clear what kind of maps do $R$ represent: Any map that can be given by any possible contraction of its indices with any other tensor field.