In this section we will assume \(V\) is a real \(n\)-dimensional vector space with a symmetric non-degenerate inner product (metric), \(g(\cdot,\cdot):V\times V\mapto\RR\). In such a vector space we can always choose an orthonormal basis, \(\{e_i\}\), and know from the classification result, Theorem~\ref{in prod class}, that such spaces are characterised up to isometry by a pair of integers, \((n,s)\), where \(s\) is the number of \(e_i\) such that \(g(e_i,e_i)=-1\).
We have seen that the dimensions of the spaces \(\Lambda^r(V)\) are given by the binomial coefficients, \({n \choose r}\). In particular, simply by virtue of having the same dimension, this means that the spaces \(\Lambda^r(V)\) and \(\Lambda^{n-r}(V)\) are isomorphic. In fact, as we shall see, the metric allows us to establish an essentially natural isomorphism between these spaces called Hodge duality.
Take any pair of pure \(r\)-vectors in \(\Lambda^r(V)\), \(\alpha=v_1\wedge\dots\wedge v_r\) and \(\beta=w_1\wedge\dots\wedge w_r\), with \(v_i,w_i\in V\). Then we can define an inner product on \(\Lambda^r(V)\) as
\begin{equation}
(\alpha,\beta)=\det(g(v_i,w_j)),
\end{equation}
where \(g(v_i,w_j)\) is regarded as the \(ij\)th entry of an \(r\times r\) matrix, and extended bilinearly to the whole of \(\Lambda^r(V)\). Since the determinant of a matrix and its transpose are identical, the inner product is symmetric. Given our orthonormal basis, \(\{e_i\}\), of \(V\), consider the inner product of the corresponding basis elements, \(e_{i_1}\wedge\dots\wedge e_{i_r}\), where \(1\leq i_1
Now whenever we have a symmetric non-degenerate inner product on some space \(U\), there is a natural isomorphism, \(U\cong U^*\), which associates to every linear functional, \(f\), on \(U\) a unique vector, \(v_f\in U\), such that \(f(u)=(v_f,u)\) for all \(u\in U\). Choose a normalised basis vector, \(\sigma\), for \(\Lambda^n(V)\) and notice that to any \(\lambda\in\Lambda^r(V)\) is associated a linear functional on \(\Lambda^{n-r}(V)\), \(f_\lambda\), according to \(\lambda\wedge\mu=f_\lambda(\mu)\sigma\). But to \(f_\lambda\) we can uniquely associate an element of \(\Lambda^{n-r}(V)\), call it \(\star\lambda\), according to \(f_\lambda(\mu)=(\star\lambda,\mu)\). \(\star\lambda\) is called the Hodge dual of \(\lambda\) and we may write,
\begin{equation}
\lambda\wedge\mu=(\star\lambda,\mu)\sigma.
\end{equation}
As a map, \(\star:\Lambda^r(V)\mapto\Lambda^{n-r}(V)\) is clearly linear.
Example Consider the 2-dimensional vector space \(\RR^2\) with the usual inner (scalar) product which we’ll here denote \(g(\cdot,\cdot)\). Denoting it’s standard basis vectors by \(\mathbf{e}_1\) and \(\mathbf{e}_2\), we have \(g(\mathbf{e}_i,\mathbf{e}_j)=\delta_{ij}\) and a basis for \(\Lambda^2(\RR^2)\) is \(\mathbf{e}_1\wedge\mathbf{e}_2\) with \((\mathbf{e}_1\wedge\mathbf{e}_2,\mathbf{e}_1\wedge\mathbf{e}_2)=1\). Clearly, we must then have
\begin{equation}
\star1=\mathbf{e}_1\wedge\mathbf{e}_2,
\end{equation}
and
\begin{equation}
\star(\mathbf{e}_1\wedge\mathbf{e}_2)=1.
\end{equation}
\(\star\mathbf{e}_1\) must be such that \((\star\mathbf{e}_1,\mathbf{e}_1)=0\) and \((\star\mathbf{e}_1,\mathbf{e}_2)=1\), that is,
\begin{equation}
\star\mathbf{e}_1=\mathbf{e}_2,
\end{equation}
and \(\star\mathbf{e}_2\) must be such that \((\star\mathbf{e}_2,\mathbf{e}_1)=-1\) and \((\star\mathbf{e}_2,\mathbf{e}_2)=0\), so
\begin{equation}
\star\mathbf{e}_2=-\mathbf{e}_1.
\end{equation}
Notice that if we had chosen \(\mathbf{e}_2\wedge\mathbf{e}_1=-\mathbf{e}_1\wedge\mathbf{e}_2\) as the basis for \(\Lambda^2(\RR^2)\) then \(\star1=-\mathbf{e}_1\wedge\mathbf{e}_2\), \(\star(-\mathbf{e}_1\wedge\mathbf{e}_2)=1\), \(\star\mathbf{e}_1=-\mathbf{e}_2\) and \(\star\mathbf{e}_2=\mathbf{e}_1\).
Given two bases of a vector space \(V\), \(\{e_i\}\) and \(\{f_i\}\), we say that they share the same orientation if the determinant of the change of basis matrix relating them is positive. Bases of \(V\) thus belong to one of two equivalence classes. From a slightly different perspective, given the bases \(\{e_i\}\) and \(\{f_i\}\) we can form the vectors \(e_1\wedge\dots\wedge e_n\) and \(f_1\wedge\dots\wedge f_n\) both of which belong the the 1-dimensional space \(\Lambda^n(V)\) and so we must have
\begin{equation}
f_1\wedge\dots\wedge f_n=ce_1\wedge\dots\wedge e_n.
\end{equation}
We know that we must be able to express the \(f_i\) in terms of the \(e_i\) as \(f_i=T_i^je_j\) where \(T_i^j\) are the elements of the change of basis linear operator defined by \(Te_i=f_i\). But we know that,
\begin{equation}
f_1\wedge\dots\wedge f_n=T^{\wedge n}(e_1\wedge\dots\wedge e_n)=\det Te_1\wedge\dots\wedge e_n,
\end{equation}
so \(c=\det T\). In other words given a basis \(\{e_i\}\) of \(V\), another basis \(f_i\) shares the same orientation if the corresponding top exterior powers are related by a positive constant. The Hodge dual thus depends on both the metric and the orientation of a given vector space.
Example Consider the 3-dimensional space \(\RR^3\) equipped with the usual inner product, with standard basis vectors \(\mathbf{e}_1\), \(\mathbf{e}_2\) and \(\mathbf{e}_3\) and \(\mathbf{e}_1\wedge\mathbf{e}_2\wedge\mathbf{e}_3\) as our prefered top exterior product. Then,
\begin{align}
\star1&=\mathbf{e}_1\wedge\mathbf{e}_2\wedge\mathbf{e}_3\\
\star\mathbf{e}_1&=\mathbf{e}_2\wedge\mathbf{e}_3\\
\star\mathbf{e}_2&=\mathbf{e}_3\wedge\mathbf{e}_1\\
\star\mathbf{e}_3&=\mathbf{e}_1\wedge\mathbf{e}_2\\
\star(\mathbf{e}_1\wedge\mathbf{e}_2)&=\mathbf{e}_3\\
\star(\mathbf{e}_2\wedge\mathbf{e}_3)&=\mathbf{e}_1\\
\star(\mathbf{e}_3\wedge\mathbf{e}_1)&=\mathbf{e}_2\\
\star(\mathbf{e}_1\wedge\mathbf{e}_2\wedge\mathbf{e}_3)&=1.
\end{align}
Let us now establish some general properties of the Hodge dual. We take an orthonormal basis of the \(n\)-dimensional space \(V\) to be \(\{e_i\}\) with top exterior form \(\sigma=ae_1\wedge\dots\wedge e_n\) with \(a=\pm1\). Then consider the pure \(r\)-vector \(e_I=e_1\wedge\dots\wedge e_r\) (no loss of generality will be incurred choosing \(I=(1,\dots,n)\)), we must then have that
\begin{equation}
\star e_I=ce_{r+1}\wedge\dots\wedge e_n=ce_J,
\end{equation}
for \(c=\pm1\) and \(J=(r+1,\dots,n)\). Of course \(c\) depends on our original choice \(a\) according to,
\begin{equation}
c=a(e_J,e_J).
\end{equation}
Consider now, \(\star e_J\), clearly
\begin{equation}
\star e_J=de_I,
\end{equation}
for some \(d=\pm1\) but since \(e_J\wedge e_I=(-1)^{r(n-r)}e_I\wedge e_J\), we have,
\begin{equation}
d=a(-1)^{r(n-r)}(e_I,e_I).
\end{equation}
We may therefore conclude that,
\begin{equation}
\star\star e_I=(-1)^{r(n-r)}(e_I,e_I)(e_J,e_J)e_I,
\end{equation}
but assuming \((\sigma,\sigma)=(-1)^s\) this is then,
\begin{equation}
\star\star e_I=(-1)^{r(n-r)+s}e_I,
\end{equation}
and by linearity we may conclude that for any \(\lambda\in\Lambda^r(V)\),
\begin{equation}
\star\star\lambda=(-1)^{r(n-r)+s}\lambda.
\end{equation}
Notice that for \(\lambda,\mu\in\Lambda^r(V)\), \(\lambda\wedge\star\mu=(\star\lambda,\star\mu)\sigma=(\star\mu,\star\lambda)\sigma=\mu\wedge\star\lambda\), that is,
\begin{equation}
\lambda\wedge\star\mu=\mu\wedge\star\lambda.
\end{equation}
But \(\mu\wedge\star\lambda=(-1)^r(n-r)\star\lambda\wedge\mu=(-1)^s(\lambda,\mu)\sigma\), that is,
\begin{equation}
\lambda\wedge\star\mu=\mu\wedge\star\lambda=(-1)^s(\lambda,\mu)\sigma.
\end{equation}