Double Dual of Vector Space: Is V** Always Same as V?

[Phinrich]

TL;DR Summary

Mathematically, for finite dimensional vector spaces, there is a natural isomorphism between a vector space V and its Double Dual V**. Does this men that V** is always the original V ? Or simply that V** is a vector space with the same properties or structure as V ?

Hi

I believe I understand the concept of a vector space V and its dual V*. I also understand that for V finite dimensional, there is a natural isomorphism between V and V**.

What I am struggling to understand is - Does this natural isomorphism mean that V** is always IDENTICAL to V (identical meaning it IS V) ? In other words have you come right back to the original vector space V ?

In the book General Relativity (by Robert M Wald) he writes the following (page 20) - "Thus taking the double dual gives nothing new. We can naturally identify V** with the original vector space V. This identification will be assumed in the discussion below".

Is this always the case or is he just assuming this is the case in his particular context ?

Thank you !

 

[PeterDonis]

Does this natural isomorphism mean that V** is always IDENTICAL to V (identical meaning it IS V) ?

I'm not sure this question is answerable, because I'm not sure there is any meaningful difference between the statements "the double dual IS V" and "the double dual is isomorphic to V" in the general case.

 

[Phinrich]

Hi

Thanks for your response. My problem is the same as yours. As I understand the mathematical term is "isomorphic" but is this the sae as the english phrase "identical to or the same as ?". I know the english phrase is imprecise while mathematics is very precise.

 

[George Jones]

Some isomorphisms between vector spaces depend on a choice of basis, e.g., between a finite-dimensional space (with no other structure) and its dual. There is a natural, i.e., basis-independent, isomorphism between a finite-dimensional vector space and its double dual.

 

[Phinrich]

Thank you

Yes I understand the relation between V and V* being dependent upon the Basis. If I have a finite dimensional vector space V and its Double Dual V** is it always the case that these two vector spaces are one and the same OR is it possible that they have the same structure (whatever that may mean mathematically) but could be two vector spaces V and (for example) a vector space Y ?

 

[PeterDonis]

is it possible that they have the same structure (whatever that may mean mathematically) but could be two vector spaces V and (for example) a vector space Y ?

I don't think the distinction you are making here is meaningful. The structure of a vector space is what defines it, so I don't think the notion of "two different vector spaces with exactly the same structure" makes sense.

 

[Phinrich]

I think I should relook at my understanding of what a vector space is ?

 

[PeterDonis]

I think I should relook at my understanding of what a vector space is ?

Wald gives a definition in his book, doesn't he?

 

[robphy]

Possibly useful:
https://en.wikipedia.org/wiki/Dual_space#Injection_into_the_double-dual
[supports the comment by @George Jones ]

 

[Phinrich]

I was thinking to read his definition. I only just received a copy of Wald and have not yet read through it other than finding and reading the following

 

[vanhees71]

It's the old question whether you consider realizations of isomorphic mathematical structures as the same or not. E.g. you can realize a 2D real vector space as arrows on an affine plane, ordered pairs of real numbers, the solutions of a 2nd order ODE, the set of real Fibonacci sequences and for sure many more "things". Up to isomorphisms they all form 2D real vector spaces, and in fact you only need to understand the one abstract 2D vector space without any such representation. It's often simpler to first analyze the properties of the abstract mathematical entity and then make use of the theorems to the properties of the very different representations of this entity. In this sense there's only one 2D real vector space with a plethora of quite different realizations.

 

[PeterDonis]

It's the old question whether you consider realizations of isomorphic mathematical structures as the same or not.

That question is actually more general than the question of how to view the isomorphism between a vector space and its double dual.

The dual $V^*$ of a vector space $V$ is the space of bounded linear maps from $V$ into $\mathbb{R}$. (Vector spaces can actually be over any field, not just $\mathbb{R}$, but we'll restrict to vector spaces over $\mathbb{R}$ here since those are the ones relevant to GR and hence to Wald's book.) So the double dual $V^{**}$ is the dual of $V^*$, i.e., the space of bounded linear maps from $V^*$ into $\mathbb{R}$, which is the space of bounded linear maps from (bounded linear maps from $V$ into $\mathbb{R}$) into $\mathbb{R}$. The existence and structure of the double dual, and its isomorphism to $V$, is therefore independent of any specific realization of $V$.

 

[Phinrich]

Hi

Thanks for your I believe that you are right. In trying to understand this I watched a few videos on Youtube and in one the Lecturer was trying to explain what "isomorphism" was and he said something like what you have said. I don't remember all the finer mathematical points but he showed that the vector space of 2x2 matrices was isomorphic to the space of polynomials having 4 coefficients (or something like that) . Please don't quote me on this example as I am sure to have the details incomplete.

 

[vanhees71]

That question is actually more general than the question of how to view the isomorphism between a vector space and its double dual.

The dual $V^*$ of a vector space $V$ is the space of bounded linear maps from $V$ into $\mathbb{R}$. (Vector spaces can actually be over any field, not just $\mathbb{R}$, but we'll restrict to vector spaces over $\mathbb{R}$ here since those are the ones relevant to GR and hence to Wald's book.) So the double dual $V^{**}$ is the dual of $V^*$, i.e., the space of bounded linear maps from $V^*$ into $\mathbb{R}$, which is the space of bounded linear maps from (bounded linear maps from $V$ into $\mathbb{R}$) into $\mathbb{R}$. The existence and structure of the double dual, and its isomorphism to $V$, is therefore independent of any specific realization of $V$.

Yes, and it's even "canonical", i.e., you can establish the isomorphism without reference to the choice of bases of $V$ and $V^{**}$. This is not the case for the isomorphism between $V$ and $V^*$ for a general vector space. There you need a basis and its dual basis to construct an isomorphism. That changes if you have a vector space with a fundamental form (or even a scalar product). For these a canonical isomorphism between $V$ and $V^*$ exists.

 

[Infrared]

I think the study of bundles illustrates, at least intuitively, why the naturality of a (family of) isomorphism(s) might be important.

For example, think about the tangent bundle to a sphere, $TS^2.$ Each of the tangent spaces is abstractly isomorphic to $\mathbb{R}^2$, but there isn't a natural way of choosing an isomorphism $T_xS^2\cong \mathbb{R}^2$ as you vary $x$. If there were, then you could choose the tangent vector in each tangent space corresponding to, say, the vector $\begin{pmatrix}1\\0\end{pmatrix}\in\mathbb{R}^2$. This would give you an everywhere-nonzero vector field on $S^2$, which is impossible. Nontrivial bundles rely on having isomorphisms which cannot be natural.

This is a little bit hand-wavy, because it is technically possible for the natural family of isomorphisms $T_xS^2\cong\mathbb{R}^2$ to be discontinuous in $x$, but in practice, this doesn't really happen for naturally-defined constructions. It's the basis-dependent definitions that you have to watch out for...