The structure of the dual space

An element \({\varphi \colon V\to\mathbb{R}}\) of \({V^{*}}\) is called a 1-form. Given a pseudo inner product on \({V}\), we can construct an isomorphism between \({V}\) and \({V^{*}}\) defined by \({v\mapsto\left\langle v,\;\right\rangle }\), i.e. \({v\in V}\) is mapped to the element of \({V^{*}}\) which maps any vector \({w\in V}\) to \({\left\langle v,w\right\rangle}\). This isomorphism then induces a corresponding pseudo inner product on \({V^{*}}\) defined by \({\left\langle \left\langle v,\;\right\rangle ,\left\langle w,\;\right\rangle \right\rangle \equiv\left\langle v,w\right\rangle }\).

An equivalent way to set up this isomorphism is to choose a basis \({e_{\mu}}\) of \({V}\), and then form the dual basis \({\beta^{\lambda}}\) of \({V^{*}}\), defined to satisfy \({\beta^{\lambda}(e_{\mu})=\delta^{\lambda}{}_{\mu}}\). The isomorphism between \({V}\) and \({V^{*}}\) is then defined by the correspondence \({v=v^{\mu}e_{\mu}\mapsto (\eta_{\mu\lambda}v^{\mu})\beta^{\lambda}\equiv v_{\lambda}\beta^{\lambda}}\), corresponding to the isomorphism induced by the pseudo inner product on \({V}\) that makes \({e_{\mu}}\) orthonormal. Note that if \({\left\langle e_{\mu},e_{\mu}\right\rangle =-1}\) then \({e_{\mu}\mapsto-\beta^{\mu}}\). This isomorphism and its inverse (usually in the context of Riemannian manifolds) are called the musical isomorphisms, where if \({v=v^{\mu}e_{\mu}}\) and \({\varphi=\varphi_{\mu}\beta^{\mu}}\) we write

\begin{aligned} v^{\flat} & \equiv\left\langle v,\;\right\rangle \\
& =\left(\eta_{\mu\lambda}v^{\lambda}\right)\beta^{\mu}\\
& \equiv v_{\mu}\beta^{\mu}\\
\varphi^{\sharp} & \equiv\left\langle \varphi,\;\right\rangle \\
& =\left(\eta^{\mu\lambda}\varphi_{\lambda}\right)e_{\mu}\\
& \equiv\varphi^{\mu}e_{\mu}

and call the \({v^{\flat}}\) the flat of the vector \({v}\) and \({\varphi^{\sharp}}\) the sharp of the 1-form \({\varphi}\).

A 1-form acting on a vector can thus be viewed as yielding a projection. Specifically, \({\varphi(v)/\left\Vert \varphi^{\sharp}\right\Vert}\) is the length of the projection of \({v}\) onto the ray defined by \({\varphi^{\sharp}}\).

It is important to note that there is no canonical isomorphism between \({V}\) and \({V^{*}}\), i.e. we cannot uniquely associate a 1-form with a given vector without introducing extra structure, namely an inner product or a preferred basis. Either structure will do: a choice of basis is equivalent to the definition of the unique inner product on \({V}\) that makes this basis orthonormal, which then induces the same isomorphism as that induced by the dual basis.

In contrast, a canonical isomorphism \({V\cong V^{**}}\) can be made via the association \({v\in V\leftrightarrow\xi\in V^{**}}\) with \({\mathbb{\xi}\colon V^{*}\to\mathbb{R}}\) defined by \({\xi\left(\varphi\right)\equiv\varphi\left(v\right)}\). Thus \({V}\) and \({V^{**}}\) can be completely identified (for a finite-dimensional vector space), and we can view \({V}\) as the dual of \({V^{*}}\), with vectors regarded as linear mappings on 1-forms.

Note that since \({\beta^{\lambda}(e_{\mu})=\delta^{\lambda}{}_{\mu}}\) and \({\left\langle e_{\mu},e_{\lambda}\right\rangle =\eta^{\mu\lambda}}\) we have

\begin{aligned}\varphi(v)&=\varphi_{\lambda}\beta^{\lambda}(v^{\mu}e_{\mu})\\&=\varphi_{\mu}v^{\mu}\\&=\eta_{\mu\lambda}\varphi^{\lambda}v^{\mu}\left\langle e_{\mu},e_{\lambda}\right\rangle \\&=\left\langle \varphi^{\sharp},v\right\rangle .\end{aligned}

Vector components are often viewed as a column vector, which means that 1-forms act on vector components as row vectors (which then are acted on by matrices from the right). Under a change of basis we then have the following relationships:

Index notation Matrix notation
Basis \({e_{\mu}^{\prime}=A^{\lambda}{}_{\mu}e_{\lambda}}\) \({e^{\prime}=eA}\)
Dual basis \({\beta^{\prime\mu}=(A^{-1})^{\mu}{}_{\lambda}\beta^{\lambda}}\) \({\beta^{\prime}=A^{-1}\beta}\)
Vector components \({v^{\prime\mu}=(A^{-1})^{\mu}{}_{\lambda}v^{\lambda}}\) \({v^{\prime}=A^{-1}v}\)
1-form components \({\varphi_{\mu}^{\prime}=A^{\lambda}{}_{\mu}\varphi_{\lambda}}\) \({\varphi^{\prime}=\varphi A}\)

Notes: A 1-form will sometimes be viewed as a column vector, i.e. as the transpose of the row vector (which is the sharp of the 1-form under a Riemannian signature). Then we have \({(\varphi^{\prime})^{T}=(\varphi A)^{T}=A^{T}\varphi^{T}}\).

An Illustrated Handbook