3 ms·

> 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 o

> 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?

JadeNB 5y ago
> 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.)

JadeNB 5y ago

> 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.)