Dual Vector Spaces

Let ${\mathbb V}$ be a linear vector space(LVS) over a field ${\mathbb F}$ — in physical applications typically we work with ${\mathbb F}={\mathbb R}$ or ${\mathbb C}$. The dual vector space, usually denoted by ${\mathbb V}^*$, defined to be the LVS of linear maps from ${\mathbb V}$ to ${\mathbb F}$ . The dimension of the dual vector space is the same as that of the vector space, ${\mathbb V}$ — this is easily established using the definitions as shown in the lectures.

The key point is that there exists no natural map between ${\mathbb V}$ and ${\mathbb V}^*$ — this is the mathematical statement that encodes the fact that any map that can be constructed will be basis-dependent. Rather than dwell abstractly on this, we will consider an example to illustrate the issue. Choose ${\mathbb V}={\mathbb R}^2$. Consider two different bases:

Basis 1 — $(e_1,e_2)$ and Basis 2 — $(\tilde{e}_1,\tilde{e}_2)=(e_1, \sqrt3\ e_1 + e_2)$,

where we have indicated the linear transformation that relates the two bases. Recall that there is no notion of length of a vector (as yet) in the definition of a LVS. So do not picture the basis as some orthonormal basis. The basis, $(f_1,f_2)$, dual to $(e_1,e_2)$ is given by the linear maps from ${\mathbb R}^2 \rightarrow {\mathbb R}$

(1)
\begin{align} f_i(e_j)=\delta_{ij}\ . \end{align}

Similarly, the dual basis, $(\tilde{f}_1,\tilde{f}_2)$, to $(\tilde{e}_1,\tilde{e}_2)$ is given by the linear maps from ${\mathbb R}^2 \rightarrow {\mathbb R}$

(2)
\begin{align} \tilde{f}_i(\tilde{e}_j)=\delta_{ij}\ . \end{align}

Exercise: Since we know how the two basis for ${\mathbb V}$ are related, we can work out how the two dual bases are related. Show that

(3)

Let us consider a vector $v=e_1\in {\mathbb V}$. In basis 1, it is natural to identify it with the linear map $f_1\in {\mathbb V}^*$ while in basis 2 it is natural to identify it with $\tilde{f}_1$. However, we have just seen that $\tilde{f}_1=f_1 - \sqrt3\ f_2\neq f_1$. Thus, a change in basis maps a given vector in ${\mathbb V}$ to a different vector in ${\mathbb V}^*$. Thus the answer is basis-dependent and thus is not natural (in maths terminology). We leave it as an exercise to see that this is true for an arbitrary vector $v=v^1\ e_1 + v^2\ e_2$.

Exercise: However, show that $(\mathbb{V}^*)^*=\mathbb{V}$.

However, if the LVS is endowed with an inner product, one can construct linear maps without choosing a basis. Let us denote the inner product between two vectors $v$ and $w$ by $\langle v, w\rangle$. Then, given a vector $v$, one can construct the linear map which we denote by $\langle v, \phantom{w}\rangle$ (or the bra vector $\langle v|$ in Dirac notation)

(4)
\begin{eqnarray} \langle v, \phantom{w}\rangle:\quad {\mathbb V} &\rightarrow& {\mathbb F} \\ w &\mapsto & \langle v,w\rangle \textrm{ or } \langle v|w\rangle \textrm{ in Dirac notation} \end{eqnarray}

It is obvious that no basis was specified in constructing the linear map.
Exercise: Verify that one does indeed get a linear map.
For ${\mathbb V}={\mathbb R}^n$ with the standard inner product, elements of ${\mathbb V}$ are called contravariant vectors and elements of the dual are called covariant vectors.

To be added Einstein summation convention, raising and lowering of indices.