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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?

Example--

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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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$ \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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