Wedge product of 0-form with 1-form

Question

What is the wedge product $\wedge$ of a $0$-form $f(x_1,...,x_n)$ with a $1$-form $\displaystyle\sum_{i=1}^{n} a_i dx_i$? According to the theory, it should be a (0+1=1)-form.

Answer 1

A $0$-form is nothing but a function. The exterior product is in this case the usual multiplication, i.e. the product is

$$f(x)\wedge\sum_{i=1}^n a_i(x) dx_i=\sum_{i=1}^n a_i(x)f(x) dx_i$$

As you can see, this really is a $1$-form.

Answer 2

For any $0$-form such as $f(x_1, x_2, . . . , x_n)$ and any $k$-form $\beta$, the operation $f \wedge \beta$ of wedge multiplication of $\beta$ by $f$ is simply ordinary multiplication by $f$: $f \wedge \beta = f\beta$, a $k$-form. Thus

$f \wedge \sum_1^n a_i dx_i = \sum_1 ^n fa_i dx_i, \tag{1}$

clearly another $1$-form. Details may be found at http://www.sjsu.edu/faculty/watkins/difforms0.htm.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!