Pages 36-37 of Loring Tu's Introduction to Manifolds says:

The wedge product of a $k$-form $\omega$ and an $l$-form $\tau$ on an open set $U$ is defined pointwise: $$(\omega\wedge \tau)_p=\omega_p\wedge \tau_p, \hspace{.75cm} p\in U. $$ In terms of coordinates, if $\omega=\sum_I a_Idx^I$ and $\tau=\sum_J b_J dx^J$, then $$\omega \wedge \tau=\sum_{I,J}(a_Ib_J)dx^I\wedge dx^J\tag{$\ast$}.$$ In this sum, if $I$ and $J$ are not disjoint on the right hand side, then $dx^I\wedge dx^J=0.$ Hence, the sum is actually over disjoint multi-indices: $$ \omega\wedge\tau=\sum_{I,J\text{ disjoint}}(a_Ib_J)dx^I\wedge dx^J, $$ which shows that the wedge product of two $C^\infty$ forms is $C^\infty$.

Here, he uses the capitol letters $I$ and $J$ to denote strictly increasing sets of indices $I=(i_1<\dots<i_k)$ and $J=(j_1<\dots<j_l)$ of lengths $k$ and $l$, respectively, from the set of indices $\{1,\dots,n\}$. So, for example $dx^I=dx^{i_1}\wedge\dots\wedge dx^{i_k}$ where $i_1,\dots,i_k\in \{1,\dots,n\}$.

My question is on the equation in $(\ast)$. Essentially, why does it make sense that the pointwise wedge product of $\omega$ and $\tau$ yields the equation in $(\ast)$? I understand the last half of the text above, and I am even able to use the definition of $\omega\wedge \tau$ given in $(\ast)$ to compute a specific example given in the exercises. But how does one arrive at this equation by defining the wedge product of $\omega$ and $\tau$ pointwise?

  • $\begingroup$ Presumably he's told you earlier how to define the wedge product of $\sum a_I dx^I$ and $\sum b_J dx^J$ when $a_I$ and $b_J$ are (fixed) scalars? $\endgroup$ – Ted Shifrin Jan 9 '17 at 22:29
  • $\begingroup$ Thats what I've been searching for. Thus far, I haven't found anything, and this is given in the first chapter, so it's not like I have a lot to look through. He gives the definition of the wedge product on page 26 as: $f\wedge g(v_1,\dots,v_n) =1/(k!\ell!) \sum_{\sigma\in S_{k+\ell}}(\text{sgn}\sigma)f(v_{\sigma(1)},\dots,v_{\sigma(k)})\cdot g(v_{\sigma(k+1)},\dots,v_{\sigma_(k+\ell)})$. $\endgroup$ – Nicholas Camacho Jan 9 '17 at 22:34
  • $\begingroup$ I presume you have a list of algebraic properties once you have the definition, and then what he says above will follow immediately. $\endgroup$ – Ted Shifrin Jan 9 '17 at 22:38
  • $\begingroup$ You got me on the right track @TedShifrin! Thanks. I was able to answer my own question. See the answers section. $\endgroup$ – Nicholas Camacho Jan 10 '17 at 0:04
  • $\begingroup$ That's always my goal ... It's best if you figure things out for yourself!! :) Keep me posted as you progress! $\endgroup$ – Ted Shifrin Jan 10 '17 at 0:16

I suppose I will answer my own question since I think I've figured it out. It comes down to the fact that the functions $a_I$ and $b_J$ are technically 0-forms. Then we use algebraic properties of the wedge product and look at what those algebraic properties tell us when a 0-form is involved.

First, by the anticommutativity of $\wedge$, which says that for a $k$-form $f$ and an $\ell$-form $g$ $$f\wedge g=(-1)^{k\ell}g\wedge f.$$ So, for a 0-form $f$ and an $\ell$-form $g$, we get $f\wedge g=(-1)^{0\cdot \ell} g\wedge f=g\wedge f$.

Second, Tu remarks on page 37 that the wedge product of a 0-form $f$ and an $\ell$-form $\omega$ is actually regular multiplication; that is $f\wedge \omega=f\omega$. So, at a point $p$ we have $(f\wedge\omega)_p=f(p)\omega_p$.

Therefore, if we assume for a moment that $\omega=a_Idx^I$ and $\tau=b_J dx^J$, then we can view this as $\omega=a_I\wedge dx^I$ and $\tau=b_J\wedge dx^J$. Then $$ \omega\wedge \tau=a_I\wedge dx^I\wedge b_J\wedge dx^J=a_I\wedge b_J\wedge dx^I\wedge dx^J=(a_Ib_J)\wedge dx^I\wedge dx^J=(a_Ib_J) dx^I\wedge dx^J. $$ Here's an example: On Problem 4.3 at the end of the section, we are asked to compute $dx\wedge dy$ where $x=r\cos\theta$ and $y=r\sin\theta$. We get: $$ dx=\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta \\ dy=\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta, $$ and so $$ dx=\cos\theta dr-r\sin\theta d\theta, \\ dy=sin\theta dr+r\cos\theta d\theta. $$ Remember that there is a $\wedge$ between a 0-form and a 1-form; i.e., $\frac{\partial x}{\partial r}dr=\frac{\partial x}{\partial r}\wedge dr$. Now, \begin{align*} dx\wedge dy&= \left(\frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta\right)\wedge\left(\frac{\partial y}{\partial r}dr+\frac{\partial y}{\partial \theta}d\theta\right)\\ &= \left(\frac{\partial x}{\partial r}dr\wedge \frac{\partial y}{\partial r}dr\right) +\left(\frac{\partial x}{\partial r}dr\wedge \frac{\partial y}{\partial\theta}d\theta\right) +\left(\frac{\partial x}{\partial\theta}d\theta\wedge\frac{\partial y}{\partial r}dr\right) +\left(\frac{\partial x}{\partial \theta}d\theta\wedge \frac{\partial y}{\partial\theta}d\theta\right)\\ &= \left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial r}dr\wedge dr\right) +\left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial\theta}dr\wedge d\theta\right) +\left(\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}d\theta\wedge dr\right) +\left(\frac{\partial x}{\partial \theta} \frac{\partial y}{\partial\theta}d\theta\wedge d\theta\right)\\ &=0+\left(\frac{\partial x}{\partial r} \frac{\partial y}{\partial\theta}dr\wedge d\theta\right) +\left(\frac{\partial x}{\partial\theta}\frac{\partial y}{\partial r}d\theta\wedge dr\right)+0\\ &=r dr\wedge d\theta. \end{align*} The first equal sign is by definition; the second is because $\wedge$ is distributive over addition; the third is by the anticommutativity of $\wedge$, where, for example we have $dr\wedge \frac{\partial y}{\partial r}=\frac{\partial y}{\partial r}\wedge dr=\frac{\partial y}{\partial r}dr$ in the first parenthesis; the fourth equal sign is by the fact that $dr\wedge dr=0=d\theta\wedge d\theta$; the fifth equal sign follows after plugging in the partials and simplifying.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.