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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 pro

by joppy 5y ago

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.

soVeryTired 5y ago
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.
- joppy 5y ago
  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.
  - soVeryTired 5y ago
    > 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.
    - joppy 5y ago
      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.
      - soVeryTired 5y ago
        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?
      - joppy 5y ago
        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.

by joppy 5y ago

soVeryTired 5y ago

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.

joppy 5y ago
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.
- soVeryTired 5y ago
  > 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.
  - joppy 5y ago
    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.
    - soVeryTired 5y ago
      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?
    - joppy 5y ago
      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.