I'm having trouble grasping the use of the wedge product in (what I think is) an exterior derivative. I found this following equation on the wikipedia page (https://en.wikipedia.org/wiki/Exterior_derivative), but am not sure if I was applying it correctly: enter image description here

Is this the proper way to solve an exterior derivatives problem, like the first example below? And how would you solve the second part?


Consider the following differential form fields on $ℝ^3$: $𝛼 = 𝑥^3 𝑑𝑥−4𝑦x^2 𝑑𝑧$ and $𝛽= 𝑧^3 𝑑𝑥∧𝑑𝑦 +sin 𝑧 𝑑𝑥∧𝑑𝑧$

  1. Find the derivative $𝑑𝛽$.

I attempted the following, but it doesn't make much sense to me:

$𝑑𝛽= \frac{∂}{∂x}(z^3) 𝑑𝑥∧𝑑𝑥∧𝑑𝑦 + \frac{∂}{∂y}(z^3) 𝑑y∧𝑑𝑥∧𝑑𝑦 + \frac{∂}{∂x}(sinz) 𝑑𝑥∧𝑑𝑥∧𝑑𝑦 + \frac{∂}{∂z}(sinz) 𝑑z∧𝑑𝑥∧𝑑z$

so $𝑑𝛽= 0$..?

$𝑑𝛽= 0 + 0 + 0+ (cosz)∧𝑑𝑥∧𝑑z$

$𝑑𝛽= (cosz)∧𝑑𝑥∧𝑑z$

Is that correct? If so what does that result mean?

  1. Find $𝛼∧𝛽$.

I'm not sure how to go about solving this part.

  • 1
    $\begingroup$ We can't tell if you apply it correctly, if you don't show your work. $\endgroup$ Jan 19, 2022 at 23:02
  • $\begingroup$ apologies, updated. $\endgroup$
    – figbar
    Jan 19, 2022 at 23:23
  • 1
    $\begingroup$ You're missing a term in $d\beta$. What's more, you are forgetting the $dx_i$ that goes with each $\partial/\partial x_i$ $\endgroup$
    – Astyx
    Jan 19, 2022 at 23:26
  • 1
    $\begingroup$ As for computing $\alpha\wedge\beta$, you can do it term by term, and put the coefficient functions as factors (just like you would do with a tensor product) $\endgroup$
    – Astyx
    Jan 19, 2022 at 23:31
  • 1
    $\begingroup$ Rather than trying to learn from wikipedia, you might take a look at the appropriate lectures in my course posted on YouTube (link in my profile). $\endgroup$ Jan 21, 2022 at 1:07

1 Answer 1


$ \def\red#1{\color{red}{#1}} \def\green#1{\color{limegreen}{#1}} \def\blue#1{\color{blue}{#1}} \def\orange#1{\color{orange}{#1}} $

Let $\alpha = x^3dx − 4x^2ydz$ and $\beta= z^3 dx\wedge dy +\sin (z) dx\wedge dz$ as in your question.

First to compute $d\beta$:

$$ \begin{align} d\beta &= d(z^3dx\wedge dy) + d(\sin(z)dx\wedge dz)\\ &=\frac{\partial z^3}{\partial x}dx\wedge dx\wedge dy + \frac{\partial z^3}{\partial y}dy\wedge dx\wedge dy+\frac{\partial z^3}{\partial z}dz\wedge dx\wedge dy\\ &\hspace{.5cm}+\frac{\partial \sin(z)}{\partial x}dx\wedge dx\wedge dz+\frac{\partial \sin(z)}{\partial y}dy\wedge dx\wedge dz+\frac{\partial \sin(z)}{\partial z}dz\wedge dx\wedge dz\\ &=3z^2dz\wedge dx\wedge dy\,,\\ d\beta&= 3z^2 dx\wedge dy\wedge dz\,. \end{align}$$

Now let's look at $\alpha\wedge \beta$: $$\begin{align} \alpha\wedge\beta&=(\red{x^3dx}-\green{4x^2ydz})\wedge(\blue{z^3dx\wedge dy} + \orange{\sin(z)dx\wedge dz})\\ &=\red{x^3dx}\wedge(\blue{z^3dx\wedge dy}+\orange{\sin(z) dx\wedge dz})\\ &\hspace{.5cm}-\green{4x^2ydz}\wedge(\blue{z^3dx\wedge dy}+\orange{\sin(z) dx\wedge dz})\\ &=\red{x^3}\blue{z^3}\red{dx}\wedge \blue{dx\wedge dy} +\red{x^3}\orange{\sin(z)}\red{dx}\wedge\orange{ dx\wedge dz}\\ &\hspace{.5cm}-\green{4x^2y}\blue{z^3}\green{dz}\wedge \blue{dx\wedge dy}+\green{4x^2y}\orange{\sin(z)}\green{dz}\wedge \orange{dx\wedge dz}\\ &=-\green{4x^2y}\blue{z^3dx\wedge dy}\wedge \green{dz} \end{align}$$


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