4 ms·

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

by 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.
- soVeryTired 5y ago
  Got you. The countability argument is interesting. Thanks for the discussion, I feel like I learned something!

by joppy 5y ago

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.

soVeryTired 5y ago
Got you. The countability argument is interesting. Thanks for the discussion, I feel like I learned something!