Chapter 5: Integration

### Vectors and Differential Forms

We have repeatedly made use of the obvious correspondence between vector fields and 1-forms, under which $\xhat$ becomes $dx$, etc. It is time to make this correspondence more formal. Throughout the following discussion, it is useful to imagine that we are working in ordinary three-dimensional Euclidean space, with the usual dot product. However, the argument is the same in any dimension and signature.

Suppose $\FF$ is a vector field. Then the corresponding 1-form is given by $$F = \FF\cdot d\rr \label{fdef}$$ This map is clearly an isomorphism of vector spaces (really modules), and therefore has an inverse, which we write in the form $$g(F, d\rr) = \FF \label{gdef}$$ and which can be taken as the definition of $g$. Assuming linearity, and picking out the components of the vectors on both sides of the equality, ($\ref{gdef}$) does indeed reduce to the inner product $g$ on 1-forms that we have been using all along. Furthermore, the construction ($\ref{gdef}$) clearly preserves the inner product, that is $$g(\FF\cdot d\rr,\GG\cdot d\rr) = \FF\cdot\GG \label{gdot}$$ which could also have been taken as the definition of $g$ acting on 1-forms.

All three of these equations, ($\ref{fdef}$), ($\ref{gdef}$), and ($\ref{gdot}$), make clear the fundamental role played by $d\rr$. Among other things, if $d\rr$ is expressed in terms of an orthonormal basis of vectors, $\{\ee_i\}$, that is, if $$d\rr = \sigma^i \,\ee_i$$ for some 1-forms $\sigma^i$, then from ($\ref{gdot}$) we have $$g(\sigma^i,\sigma^j) = g(\ee_i\cdot d\rr,\ee_j\cdot d\rr,) = \ee_i \cdot \ee_j = \pm \delta_{ij}$$ so that $\{\sigma^i\}$ is an orthonormal basis of 1-forms, a fact we have already been using.