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This is so true!

For machine-learning and data science, tensors are largely what is called n-dimensional array. Numpy is a popular python library built basically around this data type.

Mathematical tensors in a particular base (sorry for the sloppy language, theoretical physicist here) can be displayed with such a data type, depsite the n-dimensional array data type (as available in numpy) lacks the (co)vector properties.

Funny enough, in theoretical physics there are two groups of people: The one prefering to write out complicated objects (i.e. tensors) with indices and the ones who do not (typically called abstract tensor notation in general relativity). The latter are much more close to mathematical physics and prefer to understand tensors as mappings from (co)vector spaces to scalars, similar to as you already wrote.

Most general relativity computer codes are written by the index people, not by the mathematical physicists ;-)

I think that tensors have a much more broad meaning to mathematicians. At least in pure mathematics and algebra, there is much more of a focus on the tensor product operation, a way of taking two (or more) vector spaces and producing a new vector space: their tensor product. Tensors are elements of this new space. In particular they don’t need to be multilinear functions on a product of vector spaces (for finite dimensional spaces there is not much difference, but for infinite-dimensional spaces there is a real difference between these two things).

Tensor products in this generality don’t need to have an (n, m) rank in the sense you’re describing, since they might be put together from completely different spaces. For example it’s perfectly fine to form the tensor product of a 2-dimensional space with a 5-dimensional space, yielding a 10-dimensional space.

That's a fair point about the tensor product being more central in algebra. But I think the 'friendliest' introduction to tensors is the multilinear function viewpoint. IMO it motivates the tensor product.

I'm not familiar with the infinite-dimensional case. Do you have an example of a tensor that isn't a multilinear function on a product of vector spaces? I'd be interested to refine my understanding here.

The easiest way to see it is cardinality: the space of linear functions on a countably-infinite dimensional vector space is uncountably-infinite dimensional. This is the reason why you can’t necessarily swap out vectors with linear functions on an infinite-dimensional vector space (if you restrict to functions with finite support or something, it’s ok).

A familiar example of a tensor product of countably infinite dimensional vector spaces would be polynomials in multiple variables. Say F[x] is the vector space of polynomials in x with coefficients in the field F, and F[x,y] is the space of polynomials in the variables x, y. For example x^2 - x is an element of F[x], and yx^2 - y^2 is an element of F[x,y]. Then it’s not hard to see that as vector spaces, F[x,y] is isomorphic to (F[x] tensor F[y]). A tensor in this new space is precisely a polynomial in two variables.

In the above it’s important to note that a polynomial has finitely many terms, so a power series like 1 + x + x^2 + … is not a polynomial. The space of power series is isomorphic to the space of linear functions on F[x], and is uncountably-infinite dimensional.

> Say F[x] is the vector space of polynomials in x with coefficients in the field F, and F[x,y] is the space of polynomials in the variables x, y. For example x^2 - x is an element of F[x], and yx^2 - y^2 is an element of F[x,y]. Then it’s not hard to see that as vector spaces, F[x,y] is isomorphic to (F[x] tensor F[y]). A tensor in this new space is precisely a polynomial in two variables.

I think I'm missing part of the argument. So we have a polynomial p in F[x,y]. The claim in my previous post basically says that there's always a way to associate p with a linear map that takes two elements of the dual space of F[x] and produces a real number. I don't see why that's impossible here.

That’s definitely possible (the function should take two elements in F[x], not in its dual space), the problem is that there are strictly more linear functions on the set of polynomials than there are polynomials. For example you will have trouble finding a polynomial representing the “evaluate at x=1” linear function F[x] -> F, since such a polynomial would have to have infinitely many terms.

So every polynomial could be represented as a linear function on polynomials, but not every linear function on polynomials is itself a polynomial.

Not trying to be stubborn here - I just don't understand.

So we're talking about the case when tensors don’t need to be multilinear functions on a product of vector spaces.

Each element of V* is (trivially) a tensor and a linear function on V. Each element of V is (trivially) a tensor and a linear function on V*. However, not all linear functions on V* are in V. So V** is bigger than V. No problem so far.

But all elements of V** are functions on (trivial) products of vector spaces, and by definition all functions in V** are linear. So how have we misunderstood each other here?

I guess the problem is that saying "tensor products are spaces of multilinear functions on vector spaces" is tantamount to saying "vector spaces are spaces of multilinear functions on vector spaces", which is simply not true: the second set is strictly smaller than the first. For example, there is no space of linear functions on a vector space which is countably-infinite dimensional: they are all either finite-dimensional or countably infinite dimensional. Said another way, if we're talking about 1-fold tensor products, it is not right to say "a 1-fold tensor product of V is V*", since V itself is a perfectly fine 1-fold tensor product.

In order to have V* = V for an infinite-dimensional vector space V, you need to redefine V^* to some kind of restricted dual, rather than defining it as the set of all linear functions. In the polynomial example, if we take the space of all linear maps g: F[x] -> F such that g(x^n) = g(x^(n+1)) = ... = 0 for some n >> 0, then this restricted dual is isomorphic to F[x] again. But the evaluation map g(f) = f(1) is not in this restricted dual.

There are more reasons why confusing a vector space with its dual is a bad idea. For example you cannot cook up a map V -> V* without extra knowledge, for example a choice of basis of V or something. There are many examples in abstract algebra where there is a perfectly good vector space V, and absolutely no good choice of basis for V, so trying to identify elements of V with V* is unnatural. We may still be able to speak perfectly well of vectors in V or V*, but trying to identify V with V* is still unnatural. A good example is V = (functions R -> R). I can speak easily of elements of V (for example, x + sin(x)), and of elements of V* (for example, f -> integral of xf(x)), but trying to figure out which element in the dual either of these corresponds to is hopeless. We're better off just accepting at some point that there is a real difference between a vector space and its dual.

Got you. The countability argument is interesting. Thanks for the discussion, I feel like I learned something!

> For mathematicians, a tensor is a function F. We pass the function an ordered tuple of N vectors and M covectors (or, to keep things simple, N column vectors and M row vectors), and the function returns a scalar. The function F needs to be linear in each of the N vectors and M covectors. In this view, a matrix is a tensor with N=1, M=1. The operations used by machine learning types arise naturally once you crank the mathematical handle a little bit.

As with almost all non-technical statements one can make about math (including my amendment—recursive nerd-sniping away), this is neither quite true nor quite false.

A tensor is an element of a tensor product—that's it. (But what tensor product? There are a lot of notions. I'll use the bare tensor product of non-topologised vector spaces.)

One kind of thing you can tensor is copies of a single vector space, and/or its dual. This is the sort of tensor product you likely have in mind. Tensoring copies of the dual allows you to feed in elements of the original vector space. Tensoring copies of the vector space allows you to feed in elements of the dual vector space (there is mathematically no intrinsic difference between a vector and a co-vector; having a favourite vector space in mind, you can talk about the elements of that vector space or of its dual, and—which is where the vector vs. co-vector terminology in physics comes from—about how coefficient vectors change when you change the basis of V), although, importantly, if your original vector space is infinite dimensional then the natural embedding into its double dual is not an isomorphism.

Which is to say, yes, the objects you describe all arise as mathematical tensors; but mathematical tensors can also describe much more.

> if your original vector space is infinite dimensional then the natural embedding into its double dual is not an isomorphism

Ok, so this feels like the crux of the matter. So how does the fact that V** is not isomorphic to V make the tensor product construction a more general concept than the linear function construction?

> Ok, so this feels like the crux of the matter. So how does the fact that V* is not isomorphic to V make the tensor product construction a more general concept than the linear function construction?

This doesn't really have anything to do with tensor products per se; it can already be seen with tensor products involving only a single factor, which are just vector spaces. Thinking of tensors only as functions means that, for example, one can never think of the original vector space V itself, only of its image in the double dual V^{**} (a fancy way of saying that you can evaluate a vector v \in V on an element v^* of the dual vector space V^* by evaluating v^* at v: in confusing but suggestive notation, v(v^*) = v^*(v)).

It's certainly true that you can do this, and, given the axiom of choice, you don't lose any information; you know everything about a vector v \in V by knowing its value on elements of V^* (which is to say, by knowing the values of elements of V^* on V). However, if V is infinite dimensional, then you are forcing yourself to carry around extra, possibly unwanted information: if you are taking bare algebraic duals, not topological duals, then V^{**} is inconceivably larger than V, which is to say that there are way more linear functionals on V^* than just those coming from evaluation at a fixed element of V.

You can fix some of this inconceivable largeness by knowing a little more structure carried by V, and by forcing your dual to reflect that structure—usually you know the topology, and ask that the dual consist of continuous functionals; and once there's topology, you start asking things of the tensor product, too. (For example, you probably don't want to take the vector-space tensor product of Hilbert spaces, but rather its completion in some suitable sense.) But, even with a more refined notion of duality, it's still only the nice spaces V that are identified with their double duals via the canonical map V \to V^{**}; the terminology is 'reflexive'.

(I think I caught all the asterisks that the Markdown parser ate the first time through.)