Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

(Note: I'm using the word "natural" to mean "without the need to choose a basis." I'm aware that there is a precise category-theoretic meaning of this word, but I don't have great intuition for it yet and am hoping, perhaps naively, it's not necessary to understand the following.)

There exists a natural injection $V\rightarrow (V^{*})^{*}$ defined by sending $v\in V$ to a functional $\mu_{v}$ on $V^{*}$ such that $\mu_{v}(f)=f(v)$ for all $f\in V^{*}$. When $V$ is finite dimensional, this map is an isomorphism by comparing dimensions, so there is also an injection $(V^{*})^{*}\rightarrow V$. Is there a "natural" (again, in this context I understand this to mean basis-free) way to write down this injection, other than as simply the reverse of the first one?

share|cite|improve this question
A natural map that way would work also for infinite dimensional spaces, and that cannot be. (I am using the word natural in the nontechnical sense, so this statement is not provable... but if we turn to the standard technical meaning, I think one can prove it) – Mariano Suárez-Alvarez Jan 26 '12 at 7:27
You can think of the difficulty this way. What you want to achieve does not exist if the space is not finite dimensional. So any construction you give must be based on the property of being finite dimensional, which is defined by the existence of a finite basis (or at least generating set). But you want to give a proof without using a basis. This cannot work, unless you cheat by using a property specific for finite dimensional vector spaces, which hides the use of a basis in its proof; using that the first injection is bijective is an example of such a property. – Marc van Leeuwen Jan 26 '12 at 7:56
there is, however, a natural map from the triple dual down to the first dual, which is an isomorphism for finite-dimensional spaces. So if you are willing to use the fact that every finite-dimensional space is the dual of some other space, then you do at least get a natural map, but it won't be an injection unless one is in the finite-dimensional setting – user16299 Jan 30 '12 at 8:57
up vote 1 down vote accepted

For the sake of having an answer: no. Any good definition of "natural" would imply that this map also existed for infinite-dimensional vector spaces, which it doesn't. You shouldn't be able to do any better than "the inverse, when it exists, of the natural map $V \to (V^{\ast})^{\ast}$."

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.