2
$\begingroup$

I am wrapping my head around the arithmetic properties of the wedge product. I understand that constants do distribute over the wedge product, i.e. for $c_1,c_2\in\mathbb{R}$, it holds \begin{equation} c_1 \omega_1 ∧ c_2 \omega_2 = (c_1\cdot c_2) \omega_1 ∧ \omega_2 \end{equation} Does this still apply, when the coefficients are functions? Does the following equation hold? \begin{equation} (\cos(x) dx) ∧ (y dz) = (\cos(x)\cdot y) dx∧dz \end{equation} Thank you very much in advance!

$\endgroup$

2 Answers 2

2
$\begingroup$

They do. If you have the wedge product as an assignment of some member of the (nth) exterior product over the cotangent bundle to each point on your space ($\mathbb{R}^n$ or whatever it may be), then something like $xdy\wedge ydz$ is just a representation of this assignment which just so happens can be represented by the exactly same expression everywhere. The form is just a fixed function at any given point in the space, $x,y,z$ etc are not changing and so act exactly the same as constants in the written expression.

$\endgroup$
2
$\begingroup$

It's worth remembering that functions are 0-forms. So when we write $f\omega$, we're really writing $f \wedge \omega$, where $f$ is a 0-form and $\omega$ is a $k$-form, and the product is a (new) $k$-form. Now let's look at $$ \alpha = (f \omega) \wedge (g \eta) $$ where $\eta$ is a $p$-form, and $g$ is again a real-valued function. Writing this out with wedges \begin{align} \alpha &= (f \wedge \omega) \wedge (g \wedge \eta)\\ &= f \wedge \omega \wedge g \wedge \eta & \text {because wedge is associative}\\ &= f \wedge g \wedge \omega \wedge \eta & \text{applying product rule to $\omega \wedge g$}\\ &= (fg) (\omega \wedge \eta) \end{align}

The product rule says that $\phi \wedge \tau = (-1)^s \tau \wedge \phi$, where remembering the rule for $s$ is the tricky bit. It turns out to be the product of the degree of $\tau$ and the degree of $\phi$. In the case where $\phi$ is a $0$-form, one of these degrees is zero, so $s = 0$, and so $0$-forms commute with other forms without any sign-change. Yay!

Of course, one has to prove distributivity and the product rule, but that was in the cards anyhow.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .