# Critique on tensor notation

I am studying tensor algebra for an introductory course on General Relativity and I have stumbled upon an ambiguity in tensor notation that I truly dislike. But I am not sure if I am understanding the situation properly. Suppose I find myself facing the following tensor: $$g_{ij}$$ This is surely a rank 2 tensor; usually, we see a tensor of this kind in expressions like this:
$$X^iY^jg_{ij}$$ In this instance it acts as an application $$A$$ of the following kind: $$A:V \times V \to \mathbb{K}$$ where $$V$$ is a vector space and $$\mathbb{K}$$ is a scalar space. Ok, no problems with that, but suppose I find exactly the same tensor (or at least exactly the same symbol) in this expression: $$X^ig_{ij}$$ According to the rules of tensor algebra, this is equal to a dual vector: $$X^ig_{ij}=Z_j$$ But this implies that exactly the same notation $$g_{ij}$$ also represents an application $$B$$ like the following: $$B:V \to V^*,$$ where, of course, $$V^*$$ is the dual space of $$V$$. If my reasoning is correct this means that the symbol $$g_{ij}$$, and more in general any symbol that we can use to indicate a tensor, says nothing regarding the nature of the tensor itself except for how it varies under a coordinate transformation. Any real information on the nature of the tensor can only be derived from the context. Is this the case or am I missing something?

One last thing: I know that usually a tensor is defined as an application that has a scalar space as its output rather than a vector space, but of course nothing keeps us to use the tensor on a list of inputs to obtain a list of outputs and so we can use tensors to represent application between vector spaces, etc. This use of tensors is very common, so maybe this specification is useless.

Clarification: If I understand tensor notation correctly the symbol $$g_{ij}$$ can represent multiple kinds of applications, depending on the context in which the symbol is spotted, such as shown above. Is this true? If this is really true seems to me that this notation can lead to some confusion for the beginners.

• I've moved some comments which were borderline between answering the question and clarifying it to a chat room. – David Z Jun 20 at 5:25

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.

• Note that this phenomenon is not unique to tensors, but it is a much general concept: whenever you have a function $f: X \times Y \to Z$, you can identify it with a map from $X$ to the space of functions $Y \to Z$ (currying). – Federico Poloni Jun 20 at 11:55
• Wait, so how is $T: {V^*}^m \times V^n \to {V^*}^r \times V^s$, where $m + r = p$ and $n + s = q$? What does that look like in index notation? – Mateen Ulhaq Jun 21 at 4:38
• @MateenUlhaq I made a mistake - it should be $n+r = q$ and $m+s=p$. – Javier Jun 21 at 13:40
• Shouldn't you state that $T: {V^*}^p \times V^q \to \mathbb{R}$ is equivalent to $T: {V^*}^p \times V^q \to {V^*}^r \times V^s$, after which $T: {V^*}^r \times V^s \to \mathbb{R}$? – descheleschilder Jun 21 at 23:06
• @Javier this is completely independent of index notation, though of course indices make it obvious How do indices make it obvious? Can you write it out? :) – descheleschilder Jun 23 at 13:07

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.

• I'd say that tensor is more than a careful arrangement numbers. Think a vector in 2D-space. Any ordered pair of reals defines a vector. Is vector just a pair of reals? Arguably no, as you can get rid of the background coordinate system, and the vector will still be a geometric object. I think this is what the essence of the Q is. The matrix in your example is a thing that acts on a vector to produce another vector. I dunno, but for me that was the only way to intuit GR; matrices are only a calculation props. Not saying your thinking is wrong, just highlighting it's not the only possible. – kkm Jun 20 at 0:18
• @kkm yes thanks of course I agree all that. The matrix example serves merely to illustrate that it is ordinary (and logical) mathematical behaviour for a given object to have different outcomes when you serve up two vectors to it than when you serve up just one. – Andrew Steane Jun 20 at 9:36

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.