Wedge Product

An antisymmetric operation on Differential Forms (also called the Exterior Derivative)

$$dx_i\wedge dx_j = -dx_j\wedge dx_i,$$ (1)

which Implies

$$dx_i\wedge dx_i = 0$$ (2)

$$b_i\wedge dx_j = dx_j\wedge b_i = b_i\,dx_j$$ (3)

$$dx_i\wedge(b_i \,dx_j) = b_i \,dx_i\wedge dx_j$$ (4)

$$\theta_1\wedge\theta_2 = (b_1\,dx_1+b_2\,dx_2)\wedge(c_1\,dx_1+c_2\,dx_2)$$

$$= (b_1c_2-b_2c_1)\,dx_1\wedge dx_2$$

$$= -\theta_2\wedge\theta_1.$$ (5)

The wedge product is Associative

$$(s\wedge t)\wedge u=s\wedge(t\wedge u),$$ (6)

and Bilinear

$$(\alpha _1s_1+\alpha _2s_2)\wedge t=\alpha _1(s_1\wedge t)+\alpha _2(s_2\wedge t)$$ (7)

$$s\wedge(\alpha _1t_1+\alpha _2t_2)=\alpha _1(s\wedge t_1)+\alpha _2(s\wedge t_2),$$ (8)

but not (in general) Commutative

$$s\wedge t=(-1)^{pq} (t\wedge s),$$ (9)

where $s$ is a $p$-form and $t$ is a $q$-form. For a 0-form $s$ and 1-form $t$,

$$(s\wedge t)_\mu = st_\mu.$$ (10)

For a 1-form $s$ and 1-form $t$,

$$(s\wedge t)_{\mu\nu} = {\textstyle{1\over 2}}(s_\mu t_\nu-s_\nu t_\mu).$$ (11)

The wedge product is the ``correct'' type of product to use in computing a Volume Element

$$dV=dx_1\wedge \ldots\wedge dx_n.$$ (12)

See also Differential Form, Exterior Derivative, Inner Product, Volume Element