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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?

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  • $\begingroup$ 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
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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.

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