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In an earlier post to math.stackexchange I asked a question beginning with:

Let $\alpha$ be the $1$-form and $\beta$ the $2$-form on $\mathbb{R}^3$ given by

$$\alpha=(x+y)\,dy+(x^2-y^2)\,dz$$

$$\beta=z\,dx\wedge dy+xz\,dx\wedge dz$$

The wedge product I got was $\alpha\wedge\beta(x,y,z)=-yz(x+y)\,dx\wedge dy\wedge dz$

My understanding was that $\alpha$ would be in terms of one of the components in $\mathbb{R}^3$ and $\beta$ would be in terms of two of the components in $\mathbb{R}^3$, but this is not the case at all.

It made sense for $\alpha\wedge\beta$ to be a three form to me since it is in terms of $x,y$ and $z$

But this did not hold for $\alpha$ or $\beta$, so this cannot be the relationship.

One other thing I noticed was that $\alpha$ has just single $dy$ and $dz$ terms, whereas $\beta$ has two differential forms wedged, and $\alpha\wedge\beta$ the three form has three differential forms wedged, is this the relationship?

It makes sense to me for this to be the relationship, but I cannot find a source to verify this for me.

Michael Hardy
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Sam Houston
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    If you are expressing $k$-forms in local coordinates $x_1,\ldots,x_n$, then they are going to be sums with terms of the form $$ f(x_1,\ldots,x_n),x_{i_1}\wedge\cdots\wedge x_{i_k} $$ where $i_1,\ldots,i_k$ are integers lying in ${1,\ldots,n}$ and $f$ is a function. – Ian Coley Jan 01 '15 at 21:03

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In $\Bbb R^3,$ every $1-$form writes as some combination $$\alpha = f \ {\rm d}x + g \ {\rm d}y + h \ {\rm d}z,$$ $2-$forms writes as combinations $$\beta = f \ {\rm d}x \wedge {\rm d}y+ g \ {\rm d}x \wedge {\rm d}z+ h \ {\rm d}y\wedge {\rm d}z,$$ and $3-$forms are $$\omega = f \ {\rm d}x \wedge {\rm d}y\wedge {\rm d}z.$$ You can find proofs about this in Do Carmo's "Differential Forms and Applications", and a little in O'Neill's Elementary Differential Geometry. The wedge product of a $k-$ form with a $s-$form is a $(k+s)-$form. Since the wedge product is anti-commutative, repeats kill. So in your example: $$\alpha \wedge \beta =(x+y)xz \ {\rm dy} \wedge {\rm d}x \wedge {\rm d}z + (x^2-y^2)z \ {\rm d}z \wedge {\rm d}x \wedge {\rm d}y \\ = (x^2z-y^2z-x^2z-xyz) \ {\rm d}x \wedge {\rm d}y \wedge {\rm d}z = -(y^2z+xyz) \ {\rm d}x \wedge {\rm d}y \wedge {\rm d}z, $$ so your result is correct.

Ivo Terek
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  • Perfect answer, just what I was looking for. – Sam Houston Jan 01 '15 at 21:09
  • Glad to help! ${}{}{}$ – Ivo Terek Jan 01 '15 at 21:09
  • Been working on this question, thought you may find it interesting. http://math.stackexchange.com/q/1088375/169389 – Sam Houston Jan 02 '15 at 22:11
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    Yes, interesting. Good that you've already managed to solve it. When I'm dealing with complex numbers and using "i" for something else, I use \rm i for $\rm i$, the imaginary unit, and $i$ for whatever the other object is. I even saw somewhere (that I don't remember now), that in $\LaTeX$, the "right" way to denote the imaginary unit is indeed $\rm i$, but no one does it, 'cause of laziness (even I only do it when there's risk of confunding it with something else) – Ivo Terek Jan 02 '15 at 22:29