Chapter 2: Symmetries

Tensors

(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.

1) The parentheses around $\alpha\otimes\beta$ are usually omitted.
2) Some authors insert factors of $n!$ into the relationship between the tensor and wedge products (with $n=2$ in this case).