5 Matching Annotations
1. Jun 2021
2. math.stackexchange.com math.stackexchange.com
1. If you have a vector space, any vector space, you can define linear functions on that space. The set of all those functions is the dual space of the vector space. The important point here is that it doesn't matter what this original vector space is. You have a vector space V

One of the better "simple" discussions of dual spaces I've seen:

If you have a vector space, any vector space, you can define linear functions on that space. The set of all those functions is the dual space of the vector space. The important point here is that it doesn't matter what this original vector space is. You have a vector space V, you have a corresponding dual V∗.

OK, now you have linear functions. Now if you add two linear functions, you get again a linear function. Also if you multiply a linear function with a factor, you get again a linear function. Indeed, you can check that linear functions fulfill all the vector space axioms this way. Or in short, the dual space is a vector space in its own right.

But if V∗ is a vector space, then it comes with everything a vector space comes with. But as we have seen in the beginning, one thing every vector space comes with is a dual space, the space of all linear functions on it. Therefore also the dual space V∗ has a corresponding dual space, V∗∗, which is called double dual space (because "dual space of the dual space" is a bit long).

So we have the dual space, but we also want to know what sort of functions are in that double dual space. Well, such a function takes a vector from V∗, that is, a linear function on V, and maps that to a scalar (that is, to a member of the field the vector space is based on). Now, if you have a linear function on V, you already know a way to get a scalar from that: Just apply it to a vector from V. Indeed, it is not hard to show that if you just choose an arbitrary fixed element v∈V, then the function Fv:ϕ↦ϕ(v) indeed is a linear function on V∗, and thus a member of the double dual V∗∗. That way we have not only identified certain members of V∗∗ but in addition a natural mapping from V to V∗∗, namely F:v↦Fv. It is not hard to prove that this mapping is linear and injective, so that the functions in V∗∗ corresponding to vectors in V form a subspace of V∗∗. Indeed, if V is finite dimensional, it's even all of V∗∗. That's easy to see if you know that dim(V∗)=dimV and therefore dim(V∗∗)=dimV∗=dimV. On the other hand, since F is injective, dim(F(V))=dim(V). However for finite dimensional vector spaces, the only subspace of the same dimension as the full space is the full space itself. However if V is infinite dimensional, V∗∗ is larger than V. In other words, there are functions in V∗∗ which are not of the form Fv with v∈V.

Note that since V∗∗again is a vector space, it also has a dual space, which again has a dual space, and so on. So in principle you have an infinite series of duals (although only for infinite vector spaces they are all different).

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3. math.stackexchange.com math.stackexchange.com
1. There are some very beautiful and easily accessible applications of duality, adjointness, etc. in Rota's modern reformulation of the Umbral Calculus. You'll quickly gain an appreciation for the power of such duality once you see how easily this approach unifies hundreds of diverse special-function identities, and makes their derivation essentially trivial. For a nice introduction see Steven Roman's book "The Umbral Calculus".

Note to self: Look at [[Steven Roman]]'s book [[The Umbral Calculus]] to follow up on having a more intuitive idea of what a dual space is and how it's useful

2. Dual spaces also appear in geometry as the natural setting for certain objects. For example, a differentiable function f:M→R

Dual spaces also appear in geometry as the natural setting for certain objects. For example, a differentiable function f:M→R where M is a smooth manifold is an object that produces, for any point p∈M and tangent vector v∈TpM, a number, the directional derivative, in a linear way. In other words, ==a differentiable function defines an element of the dual to the tangent space (the cotangent space) at each point of the manifold.==

3. This also happens to explain intuitively some facts. For instance, the fact that there is no canonical isomorphism between a vector space and its dual can then be seen as a consequence of the fact that rulers need scaling, and there is no canonical way to provide one scaling for space. However, if we were to measure the measure-instruments, how could we proceed? Is there a canonical way to do so? Well, if we want to measure our measures, why not measure them by how they act on what they are supposed to measure? We need no bases for that. This justifies intuitively why there is a natural embedding of the space on its bidual.
4. The dual is intuitively the space of "rulers" (or measurement-instruments) of our vector space. Its elements measure vectors. This is what makes the dual space and its relatives so important in Differential Geometry, for instance.

A more intuitive description of why dual spaces are useful or interesting.