Note: This question seems like it might have been asked before, but the poster deleted it and I don't have enough reputation to see it.

Let $V$ be a finite-dimensional real vector space. Such an object is, in a canonical way, a smooth manifold. Thus it makes sense to speak of smooth charts $(U,\varphi)$ for such a space (where $U \subseteq V$ is open of course). Let us restrict attention to those charts for which $U=V$.

Then if the chart function $\varphi: V \to \mathbb{R}^n$ is linear, each of its component functions (which I call coordinate functions, since each specifies a different coordinate of the chart) is a linear function $V \to \mathbb{R}$, thus an element of $V^*$.

(Obviously none of this works over arbitrary fields or for infinite-dimensional vector spaces, hence the restrictions I assume/gave above.)

Question: Thus, does it make sense to interpret or even define dual bases to be linear charts from $V$ to $\mathbb{R}^n$, and $V^*$ to be the space of all linear coordinate functions?

In particular, this would provide a ready explanation for why the components of a vector are written with the opposite index placement when using the Einstein summation convention. If $E = x^i E_i$ is a vector in $V$, then the components $x^i$ could be interpreted as short-hand for the evaluation of the linear coordinate functions (i.e. elements of $V^*$) evaluated at $E$ rather than just scalars, i.e. $$E = x^i(E) E_i \,.$$ Since the $x^i$ are elements of $V^*$, rather than scalars in $\mathbb{R}$, it becomes quite obvious why they should have the index placement they do.

Also components/linear coordinates transform the same way that dual vectors do under changes of basis (i.e. they are covariant), so thinking about vectors/dual spaces/tensors the way physicists do, interpreting components/linear coordinate functions as being the same thing as elements of $V^*$ seems to make a certain amount of sense. It might also make dual vector spaces easier to understand for those who tend to think physically or geometrically (e.g.).

However, I have not seen this practice mentioned in any texts on differential geometry or linear algebra which I have ever read, so I imagine that there are probably problems with it which I am not noticing or understanding yet.

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    $\begingroup$ It's not the same question. Somewhat related, but different. But I'm curious, how did you know about the older question and get the link to it? $\endgroup$ – Daniel Fischer Jan 11 '17 at 12:57
  • $\begingroup$ @DanielFischer Haha, I didn't know it at all, it came up on a google search. $\endgroup$ – Chill2Macht Jan 11 '17 at 13:32
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    $\begingroup$ @William So Google has got more reputation than you do, what a shame! $\endgroup$ – Marc van Leeuwen Jan 11 '17 at 13:39

Answering your second question, this notion is quite useful in coordinate-free algebraic geometry.

Given a finite-dimensional vector space $V$ over any field $k$, one may consider the dual space $V^*$ as the space of linear polynomial functions $V \to k$. Concretely, given a basis $e_1,\ldots,e_n$ for $V$, the dual elements $e_i^*$ corresponds to the coordinate functions $x_i : V \to k$. More generally however, the symmetric powers $Sym^d(V^*)$ corresponds to homogeneous polynomial functions of degree $d$. The symmetric algebra $Sym(V^*)$ is the ring of polynomial functions. Given a basis, one may identify it with $k[x_1,\ldots,x_n]$.

  • $\begingroup$ I hadn't considered that this point of view might be useful in algebraic geometry, but now that you explain why it makes a lot of sense. I had heard of the symmetric algebra being identifiable with polynomials before, but this explanation really makes it clear why -- I appreciate you taking the time to explain it to me! $\endgroup$ – Chill2Macht Jan 11 '17 at 13:34

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