8.1 r-Vectors and r-forms

中文

A tensor of type $(r,0)$ is said to be antisymmetric $\mathrm{if},$ as a multilinear function, it changes sign whenever any pair of its arguments are interchanged,

$$A({\alpha }^1,\dots ,{\alpha }^i,\dots ,{\alpha }^j,\dots ,{\alpha }^r)=-A({\alpha }^1,\dots ,{\alpha }^j,\dots ,{\alpha }^i,\dots ,{\alpha }^r).\text{\hspace{1em}(8.1)}$$

Equivalently, if $\pi$ is any permutation of $1,\dots ,r$ then

$$A({\alpha }^{\pi (1)},{\alpha }^{\pi (2)},\dots ,{\alpha }^{\pi (r)})=(-1{)}^{\pi }A({\alpha }^1,{\alpha }^2,\dots ,{\alpha }^r).$$

To express these conditions in components, let $\{e_i\}$ be any basis of $V$ and $\{{\epsilon }^j\}$ its dual basis. Setting ${\alpha }^1={\epsilon }^{i_1},{\alpha }^2={\epsilon }^{i_2},..$ . in Eq. (8.1), a tensor $A$ is antisymmetric if it changes sign whenever any pair of component indices is interchanged,

$$A^{i_1\dots j\dots k\dots i_r}=-A^{i_1\dots k\dots j\dots i_r}.$$

For any permutation $\pi$ of $1,\dots ,r$ we have

$$A^{i_{\pi (1)}\dots i_{\pi (r)}}=(-1{)}^{\pi }{A}^{i_1\dots i_r}.$$

Antisymmetric tensors of type $(r,0)$ are also called $r$-vectors, forming a vector space denoted ${\Lambda }^r(V)$ . Ordinary vectors of $V$ are 1-vectors and scalars will be called 0-vectors.

A similar treatment applies to antisymmetric tensors of type $(0,r)$ , called $r$-forms. These are usually denoted by Greek letters $\alpha ,\beta$ , etc., and satisfy

$$\alpha (v_{\pi (1)},\dots ,v_{\pi (r)})=(-1{)}^{\pi }\alpha (v_1,\dots ,v_r),$$

or in terms of components

$${\alpha }_{i_{\pi (1)}\dots i_{\pi (r)}}=(-1{)}^{\pi }{\alpha }_{i_1\dots i_r}.$$

Linear functionals, or covectors, are called 1-forms and scalars are 0-forms. The vector space of $r$-forms is denoted ${\Lambda }^{\ast r}(V)\equiv {\Lambda }^r(V^{\ast })$ .

As shown in Eq. (7.33), the total contraction of a 2-form $\alpha$ and a symmetric tensor $S$ of type (2, 0) vanishes,

$${\alpha }_{ij}{S}^{ij}=0.$$

The same holds true of more general contractions such as that between an $r$-form $\alpha$ and a tensor $S$ of type ($s$, 0) that is symmetric in any pair of indices; for example, if $S^{ikl}=S^{lki}$ then

$${\alpha }_{ijkl}{S}^{ikl}=0.$$

The antisymmetrization operator

Let $T$ be any totally contravariant tensor of degree $r$; that is, of type ($r$, 0). Its antisymmetric part is defined to be the tensor $\mathbf{AT}$ given by

$$\mathcal{A}T({\omega }^1,{\omega }^2,\dots ,{\omega }^r)=\frac{1}{r!}\sum _{\sigma }(-1{)}^{\sigma }T({\omega }^{\sigma (1)},{\omega }^{\sigma (2)},\dots ,{\omega }^{\sigma (r)}),\text{\hspace{1em}(8.2)}$$

where the summation on the right-hand side runs through all permutations $\sigma$ of 1, 2, … , $r$. If $\pi$ is any permutation of 1, 2, … , $r$ then

$$\begin{array}{l}\mathcal{A}T({\alpha }^{\pi (1)},{\alpha }^{\pi (2)},\dots ,{\alpha }^{\pi (r)})=\frac{1}{r!}{\sum }_{\sigma }(-1{)}^{\sigma }T({\alpha }^{\pi \sigma (1)},{\alpha }^{\pi \sigma (2)},\dots ,{\alpha }^{\pi \sigma (r)})\\ =\frac{1}{r!}{\sum }_{{\sigma }^{\prime }}(-1{)}^{\pi }(-1{)}^{{\sigma }^{\prime }}T({\alpha }^{{\sigma }^{\prime }(1)},{\alpha }^{{\sigma }^{\prime }(2)},\dots ,{\alpha }^{{\sigma }^{\prime }(r)})\\ =(-1{)}^{\pi }\mathcal{A}T({\alpha }^1,{\alpha }^2,\dots ,{\alpha }^r),\end{array}$$

since ${\sigma }^{\prime }=\pi \sigma$ runs through all permutations of 1, 2, … , $r$ and $(-1{)}^{{\sigma }^{\prime }}=(-1{)}^{\pi }(-1{)}^{\sigma }$ Hence $\mathbf{A}\mathbf{T}$ is an antisymmetric tensor.

The antisymmetrization operator $\mathcal{A}:V^{(r,0)}\to {\Lambda }^r(V)\subseteq V^{(r,0)}$ is clearly a linear op erator on $V^{(r,0)}$

$$\mathcal{A}(aT+bS)=a\mathcal{A}(T)+b\mathcal{A}(S),$$

and since the antisymmetric part of an $r$-vector $A$ is always $A$ itself, it is idempotent

$${\mathcal{A}}^2=\mathcal{A}.$$

Thus is aprojection operator (see Problem 3.6). This property generalizes to the following useful theorem:

Theorem 8.1 · Antisymmetrization and tensor products

If $T$ is a tensor of type ($r$, 0) and $S$ a tensor of type ($s$, 0), then

$$\mathcal{A}(\mathcal{A}T\otimes S)=\mathcal{A}(T\otimes S),\mathcal{A}(T\otimes \mathcal{A}S)=\mathcal{A}(T\otimes S).$$

Proof

We will prove the first equation, the second being essentially identical. Let ${\omega }^1,{\omega }^2,\dots ,{\omega }^{r+s}$ be any $r+s$ covectors. Then

$$\mathcal{A}T\otimes S({\omega }^1,{\omega }^2,\dots ,{\omega }^{r+s})=\frac{1}{r!}\sum _{\sigma }(-1{)}^{\sigma }T({\omega }^{\sigma (1)},\dots ,{\omega }^{\sigma (r)}\left)S\right({\omega }^{r+1},\dots ,{\omega }^{r+s}).$$

Treating each permutation $\sigma$ in this sum as a permutation ${\sigma }^{\prime }$ of $1,2,\dots ,r+s$ that leaves the last $s$ numbers unchanged, this equation can be written

$$\mathcal{A}T\otimes S\left({\omega }^1,{\omega }^2,\dots ,{\omega }^{r+s}\right)=\frac{1}{r!}\sum _{{\sigma }^{\prime }}(-1{)}^{{\sigma }^{\prime }}T\left({\omega }^{{\sigma }^{\prime }(1)},\dots ,{\omega }^{{\sigma }^{\prime }(r)}\right)S\left({\omega }^{{\sigma }^{\prime }(r+1)},\dots ,{\omega }^{{\sigma }^{\prime }(r+s)}\right).$$

Now for each permutation ${\sigma }^{\prime }$ , as $\rho$ ranges over all permutations of $1,2,\dots ,r+s$ , the product $\pi =\rho {\sigma }^{\prime }$ also ranges over all such permutations, and $(-1{)}^{\pi }=(-1{)}^{\rho }(-1{)}^{{\sigma }^{\prime }}$ . Hence

$$\begin{array}{l}\mathcal{A}(\mathcal{A}T\otimes S)({\omega }^1,{\omega }^2,\dots ,{\omega }^{r+s})\\ =\frac{1}{(r+s)!}{\sum }_{\rho }(-1{)}^{\rho }\frac{1}{r!}{\sum }_{{\sigma }^{\prime }}(-1{)}^{{\sigma }^{\prime }}T({\omega }^{\rho {\sigma }^{\prime }(1)},\dots ,{\omega }^{\rho {\sigma }^{\prime }(r)})S({\omega }^{\rho {\sigma }^{\prime }(r+1)},\dots ,{\omega }^{\rho {\sigma }^{\prime }(r+s)})\\ =\frac{1}{r!}{\sum }_{{\sigma }^{\prime }}\frac{1}{(r+s)!}{\sum }_{\pi }(-1{)}^{\pi }T({\omega }^{\pi (1)},\dots ,{\omega }^{\pi (r)}\left)S\right({\omega }^{\pi (r+1)},\dots ,{\omega }^{\pi (r+s)}),\end{array}$$

since there are $r$! permutations of type ${\sigma }^{\prime }$ , each making an identical contribution. Hence

$$\begin{array}{l}\mathcal{A}(\mathcal{A}T\otimes S)({\omega }^1,{\omega }^2,\dots ,{\omega }^{r+s})\\ =\frac{1}{(r+s)!}{\sum }_{\pi }(-1{)}^{\pi }T({\omega }^{\pi (1)},\dots ,{\omega }^{\pi (r)}\left)S\right({\omega }^{\pi (r+1)},\dots ,{\omega }^{\pi (r+s)})\\ =\mathcal{A}(T\otimes S)({\omega }^1,{\omega }^2,\dots ,{\omega }^{r+s}),\end{array}$$

as required.

The same symbol $\mathbf{A}$ can also be used to represent the projection operator $\mathcal{A}:V^{(0,r)}{\Lambda }^{\ast r}$ defined by

$$\mathcal{A}T(u_1,u_2,\dots ,u_r)=\frac{1}{r!}\sum _{\sigma }(-1{)}^{\sigma }T(u_{\sigma (1)},u_{\sigma (2)},\dots ,u_{\sigma (r)}).$$

Theorem 8.1 has a natural counterpart for tensors $T$ of type (0, $r$) and $S$ of type (0, $s$)