Thinking of basis elements of the dual space as coordinate functions

50 Views Asked by At

I'm a student of Physics self learning representation theory from Peter Woit's book and I am having a bit of trouble understanding a few concepts related to the dual space $V^{*}$ of a vector space $V$. I have seen dual spaces before but I am a bit rusty it seems.

I understand that it is the space of all linear maps defined on the vector space $V$ to some field (let's say $\mathbb{R}$) given some basis $\{e_i\}$ spanning $V$ one can define a dual basis $\{e^{*}_j\}$ by $e^{*}_i(e_j) = \delta_{ij}$, the Kronecker delta. In section 14.3 (page 168) titled Symplectic Geometry, author says that the basis elements of the vector space $V$ can be identified with the partial derivatives \begin{equation} e_i \longleftrightarrow \frac{\partial}{\partial v_i} \end{equation} However, this would mean that the 'coordinate functions' $v_i$ are elements of the dual space $V^{*}$ and I am not sure how that is consistent with the fact that \begin{equation} e^{*}_i(\vec{v}) = v_i, \end{equation} i.e. the one-form bases 'extract' the $i$th coordinate with the action of $e^{*}_i$ on any vector $\vec{v}$. Thanks for your help.

1

There are 1 best solutions below

0
On BEST ANSWER

Let $v = v_1 e_1 + \cdots + v_n e_n$ and $f \in V^*$. This means that $f$ is linear so there exists $a_1,\ldots,a_n \in \mathbb R$ such that

$$ f(v) = a_1 v_1 + \cdots + a_n v_n.$$

Note in particular that $\frac{\partial f}{\partial v_i} = a_i$. That is, for all $v \in V$

$$ f(v) = \sum_i a_i v_i = \sum_i \frac{\partial f}{\partial v_i} \cdot v_i.$$

That is,

$$ f = \sum_{i = 1}^n \frac{\partial}{\partial v_i}(f) \cdot e_i^*.$$

Note that this means that $\frac{\partial}{\partial v_i}(f)$ is the coordinate of $f$ in the basis $e_1^*,\ldots,e_n^*$ which means that the elements $$ \frac{\partial}{\partial v_1}, \ldots, \frac{\partial}{\partial v_n}$$ are precisely the dual basis of the coordinate functions $e_1^*,\ldots, e_n^*$.

Using the identification $(V^*)^* \cong V$ we can identify $e_j$ with $\frac{\partial }{\partial v_j}$ which the book writes as

$$ e_j = \frac{\partial}{\partial x_j}.$$