Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For precise definitions of the spaces referenced below, please refer to this question.

For $X \subset \mathbb{R}^n$, I understand that the $n$-dimensional tangent space $T_pX$ has a natural/canonical basis $((e_1)_p, \dots, (e_n)_p)$ where each $(e_i)_p$ is the standard basis vector $e_i$ translated to the point $p$. I also understand from linear algebra that the $n$-dimensional cotangent space $T^*_pX$ has a basis which is dual to the canonical basis given above and that the elements of this basis are typically denoted by $(dx^1)_p, \dots, (dx^n)_p$. In the cotangent space, these basis elements can be interpreted as the differentials of the canonical projection functions. Now, in this context, I am trying to parse the following claim from a text where, for notional ease, the point $p$ is suppressed:


The list $(e_1, \dots, e_n)$ is a module basis for $\mathcal{V}^1(X)$ and the list $(dx^1, \dots dx^n)$ is a module basis for $\Omega^1(X)$


I'm not sure how to make sense of this statement because, by definition, $$ \mathcal{V}^1(X) = C^1(X,\mathbb{R}^n) \;\;\; \text{and} \;\;\; \Omega^1(X)= C^1(X, (\mathbb{R}^n)^{\prime}) $$ where $(\mathbb{R}^n)^{\prime}$ denotes the continuous dual space of $\mathbb{R}^n$. These seem to me (and indeed, later in the text it is demonstrated) that both of these are infinite dimensional vector spaces.

So my question is, Is there a way to interpret the above statement so it makes sense, particularly in light of the fact that these spaces are infinite dimensional (!) ?

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

As $\mathbb{R}$-modules (real vector spaces) both $\mathcal{V}^1(X)$ and $\Omega^1(X)$ are usually infinite dimensional, but as modules over the ring $C^1(X,\mathbb{R})$ they are finitely generated and free (with basis as you've listed).

It is easy to see why: take any $\omega \in \mathcal{V}^1(X)$, and note that there are $\{ \omega_1, \omega_2, \ldots, \omega_n \} \subseteq C^1(X,\mathbb{R})$ such that $\omega(x) = (\omega_1(x), \omega_2(x), \ldots, \omega_n(x))$ for all $x \in X$, i.e. $\omega = \sum_{i =1}^n \omega_i e_i$. The proof for $\Omega^1(X)$ is similar.

share|improve this answer
    
Thanks for the explanation; I was not properly discerning the distinction between these spaces when being considered as $C^1(X, \mathbb{R})$-modules versus $\mathbb{R}$-modules. –  ItsNotObvious Apr 4 '12 at 18:15
add comment

Your Answer

 
discard

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.