8.4 Interior product

中文

Let $u$ be a vector in $V$ and $\alpha$ an $r$-form. We define the interior product ${\mathrm{i}}_u\alpha$ to be an ($r$ 1)-form defined by

$$({\mathrm{i}}_u\alpha )(u_2,\dots ,u_r)=r\alpha (u,u_2,\dots ,u_r).\text{\hspace{1em}(8.18)}$$

The interior product of a vector with a scalar is assumed to vanish, ${\mathrm{i}}_{u}a=0$ for all $a\in {\Lambda }^{\ast 0}(V)\equiv \mathbb{R}$ . The component expression with respect to any basis $\left\{{\mathbf{e}}_i\right\}$ ofthe interior product of a vector with an $r$-form is given by

$$\begin{array}{rl}& ({\mathrm{i}}_u\alpha {)}_{i_2\dots i_r}=({\mathrm{i}}_u\alpha )(e_{i_2},\dots ,e_{i_r})\\ & =r\alpha (u^i{e}_i,e_{i_2},\dots ,e_{i_r})\\ & =r{u}^i{\alpha }_{i{i}_2\dots i_r}.\end{array}$$

Hence

$${\mathrm{i}}_u\alpha =r{C}_1^1(u\otimes \alpha ),$$

where $C_1^1$ is the (1, 1) contraction operator.

Performing the interior product with two vectors in succession on any $r$-form $\alpha$ has the property

$${\mathrm{i}}_u({\mathrm{i}}_v\alpha )=-{\mathrm{i}}_v({\mathrm{i}}_u\alpha ),\text{\hspace{1em}(8.19)}$$

for

$$\begin{array}{rl}({\mathrm{i}}_u({\mathrm{i}}_v\alpha ))(u_3,\dots ,u_r)& =(r-1)({\mathrm{i}}_v\alpha )(u,u_3,\dots ,u_r)\\ & =(r-1)r\alpha (v,u,u_3,\dots ,u_r)\\ & =-(r-1)r\alpha (u,v,u_3,\dots ,u_r)\\ & =-({\mathrm{i}}_v({\mathrm{i}}_u\alpha ))(u_3,\dots ,u_r).\end{array}$$

It follows immediately that $({\mathrm{i}}_u{)}^2\equiv {\mathrm{i}}_u{\mathrm{i}}_u=0$

Another important identity, for an arbitrary $r$-form $\alpha$ and $s$-form ${\beta }_{;}$ , is the antiderivation law

$${\mathrm{i}}_u(\alpha \wedge \beta )=({\mathrm{i}}_u\alpha )\wedge \beta +(-1{)}^r\alpha \wedge ({\mathrm{i}}_u\beta ).\text{\hspace{1em}(8.20)}$$

Proof

Let $u_1,u_2,...,u_{r+s}$ be arbitrary vectors. By Eqs. Eq. (8.18) and Eq. (8.17)

$$\begin{array}{l}({\mathrm{i}}_{u_1}(\alpha \wedge \beta ))(u_2,\dots ,u_{r+s})=(r+s)\mathcal{A}(\alpha \otimes \beta )(u_1,u_2,\dots ,u_{r+s})\\ =\frac{1}{(r+s-1)!}{\sum }_{\sigma }(-1{)}^{\sigma }\alpha (u_{\sigma (1)},\dots ,u_{\sigma (r)})\beta (u_{\sigma (r+1)},\dots ,u_{\sigma (r+s)}).\end{array}$$

For each $1\le a\le r+s$ let ${\gamma }_a$ be the cyclic permutation (1 2 … $a$). If $\sigma$ is any permutation such that $\sigma (a)=1$ then $\sigma ={\sigma }^{\prime }{\gamma }_a$ where ${\sigma }^{\prime }(1)=\sigma (a)=1$ . The signs of the permutations $\sigma$ and ${\sigma }^{\prime }$ are related by $(-1{)}^{\sigma }=(-1{)}^{{\sigma }^{\prime }}(-1{)}^{a+1}$ , and the sum of permutations in the above equation may be written as

$$\begin{array}{l}\frac{1}{(r+s-1)!}({\sum }_{a=1}^r{\sum }_{{\sigma }^{\prime }}(-1{)}^{{\sigma }^{\prime }}(-1{)}^{a+1}\alpha \left(u_{{\sigma }^{\prime }(2)},\dots ,u_{{\sigma }^{\prime }(a)},u_1\dots ,u_{{\sigma }^{\prime }(r)}\right)\\ \times \beta \left(u_{{\sigma }^{\prime }(r+1)},\dots ,u_{{\sigma }^{\prime }(r+s)}\right)+{\sum }_{b=1}^s{\sum }_{{\sigma }^{\prime }}(-1{)}^{{\sigma }^{\prime }}(-1{)}^{r+b+1}\alpha \left(u_{{\sigma }^{\prime }(2)},\dots ,u_{{\sigma }^{\prime }(r+1)}\right)\\ \times \beta \left(u_{{\sigma }^{\prime }(r+2)},\dots ,u_{{\sigma }^{\prime }(r+b)},u_1,\dots ,u_{{\sigma }^{\prime }(r+s)}\right)).\end{array}$$

By cyclic permutations $u_1$ can be brought to the first argument of $\alpha$ and $\beta$ respectively, introducing factor $(-1{)}^{a+1}$ and $(-1{)}^{b+1}$ in the two sums, to give

$$\begin{array}{l}\frac{1}{(r+s-1)!}(r{\sum }_{{\sigma }^{\prime }}(-1{)}^{{\sigma }^{\prime }}\alpha (u_1,u_{{\sigma }^{\prime }(2)},\dots ,u_{{\sigma }^{\prime }(a-1)},u_{{\sigma }^{\prime }(a+1)},\dots ,u_{{\sigma }^{\prime }(r)})\\ \times \beta (u_{{\sigma }^{\prime }(r+1)},\dots ,u_{{\sigma }^{\prime }(r+s)})+s(-1{)}^r{\sum }_{{\sigma }^{\prime }}(-1{)}^{{\sigma }^{\prime }}\alpha (u_{{\sigma }^{\prime }(1)},\dots ,u_{{\sigma }^{\prime }(r)})\\ \times \beta (u_1,u_{{\sigma }^{\prime }(r+1)},\dots ,u_{{\sigma }^{\prime }(r+b-1)},u_{{\sigma }^{\prime }(r+b+1)},\dots ,u_{{\sigma }^{\prime }(r+s)})),\end{array}$$

where ${\sigma }^{\prime }$ ranges over all permutations of $(2,3,\dots ,r+s)$ . Thus

$$\begin{array}{c}{\mathrm{i}}_{u_1}(\alpha \wedge \beta )(u_2,\dots ,u_{r+s})=\mathcal{A}(({\mathrm{i}}_{u_1}\alpha )\otimes \beta +(-1{)}^r\alpha \otimes ({\mathrm{i}}_{u_1}\beta )\left)\right(u_2,\dots ,u_{r+s})\\ =\left({\mathrm{i}}_{u_1}\alpha \right)\wedge \beta +(-1{)}^r\alpha \wedge ({\mathrm{i}}_{u_1}\beta )(u_2,\dots ,u_{r+s}).\end{array}$$

Equation (8.20) follows on setting $u=u_1$