(This section can be skipped on first reading.)

Tensors are multilinear maps on vectors. Differential forms are a type of tensor.

Consider the 1-form $df$ for some function $f$. How does it act on a vector $\vv$? That's easy: by giving the directional derivative, namely \begin{equation} df(\vv) = \grad f\cdot\vv \end{equation} More generally, if $F=\FF\cdot d\rr$ is a 1-form, then we define the action of $F$ on vectors via \begin{equation} F(\vv) = \FF\cdot\vv \end{equation}

There is a natural product on multilinear maps. If $\alpha$ and $\beta$ are 1-forms, then we can construct the map \begin{equation} (\vv,\ww) \longmapsto \alpha(\vv)\,\beta(\ww) \end{equation} This operation defines a new product on 1-forms, called the tensor product, written as 1) \begin{equation} (\alpha\otimes\beta) (\vv,\ww) = \alpha(\vv)\,\beta(\ww) \end{equation} We say that $\alpha\otimes\beta$ is a rank 2 covariant tensor, because it is a multilinear map taking two vectors to a scalar.

Now consider the 2-form $\alpha\wedge\beta$. Since \begin{equation} g(\alpha,F) = \alpha(\FF) \end{equation} and \begin{equation} g(\alpha\wedge\beta,\gamma\wedge\delta) = g(\alpha,\gamma)g(\beta,\delta) - g(\alpha,\delta)g(\beta,\gamma) \end{equation} it is natural to define \begin{equation} (\alpha\wedge\beta) (\vv,\ww) = \alpha(\vv)\beta(\ww) - \alpha(\ww)\beta(\vv) \end{equation} With this definition, 2-forms are a special case of rank 2 covariant tensors, and in fact 2) \begin{equation} \alpha\wedge\beta = \alpha\otimes\beta - \beta\otimes\alpha \end{equation} With these conventions, $(dx\wedge dy)(\xhat,\yhat)=1$; more generally, $(dx\wedge dy)(\vv,\ww)$ is the oriented area spanned by $\vv$ and $\ww$.

So differential forms are antisymmetric tensor products of 1-forms. Another important special case is that of symmetric tensor products, often written as \begin{equation} \alpha\otimes_S\beta = \frac12 (\alpha\otimes\beta + \beta\otimes\alpha) \end{equation} where the factor of $\frac12$ ensures that \begin{equation} \alpha\otimes_S\alpha = \alpha\otimes\alpha \end{equation} The line element is an example of a symmetric rank 2 tensor, and is often called the metric tensor. Terms such as $dx^2$ in such expressions should really be interpreted as $dx\otimes dx$.

Another example is Killing's equation, involving the dot product of vector-valued 1-forms. Such products involve a dot product on the vector-valued coefficients, but should also be interpreted as implying a symmetrized tensor product of 1-forms. That is, we define \begin{equation} (\alpha\,\uu)\cdot(\beta\,\vv) = (\uu\cdot\vv)\,(\alpha\otimes_S\beta) \end{equation} resulting in a symmetric rank 2 tensor.

We will occasionally need to work with symmetric rank 2 tensors such as the metric tensor and Killing's equation. However, we only rarely need to think of such objects explicitly as multilinear maps, and will therefore omit the symbol $\otimes$. Thus, we simply write “$dx^2$” or “$dx\,dy$” — which is the same as “$dy\,dx$”, since the order doesn't matter due to the symmetrization. It is important to remember that such products are not wedge products, but can instead be manipulated using ordinary algebra.