2

Let $\omega$ be a 1-form such that: $\omega = a\,dx + b\,dy$ and $\eta$ is a 0-form.

I've seen a lot of times where $\eta \wedge \omega$ is written as $\eta \omega$... which kind of makes sense as $\eta \wedge \omega = (\eta a)\,dx + (\eta b)\,dy$. But is the first term, $(\eta a)\,dx$, the same thing as $\eta (a\,dx)$?

Also, is $a\,dx = a \wedge dx$? Is $dx$ a 1-form? In general, if $\eta$ is a 0-form, does that mean $\eta \wedge \omega = \eta \omega$?

I'm just really confused about the algebraic structure of differential forms, and this notation doesn't really help. I've been told to treat the wedge product of two forms like multiplying polynomials with the $dx_i$ being "wedged" and the "coefficients" simply being multiplied together, so:

$$(a\,dx+b\,dy) \wedge (e\,dx+f\,dy) = ae \,dx \wedge dx + af\, dx \wedge dy + be\, dy \wedge dx + bf\, dy \wedge dy$$ Can't we derive this somehow by considering $a\,dx$ as $a \wedge dx$ along with anticommutativity and associativity, instead of defining this case separately?

What exactly is the interplay between this multiplication implicit in $a\,dx$ and the wedge product? Is $dx$ itself a 1-form, or do we just treat it as just some "thing"? Does it really make a difference in anything?

Also, if someone could give me a reference to any text regarding this, that would be great.

Michael Hardy
  • 1
Luke
  • 344
  • 2
    I changed $adx$ to $a,dx$ etc. Legibility. ${}\qquad{}$ – Michael Hardy Dec 07 '14 at 00:41
  • 1
    0-forms are smooth functions. The wedge product of a 0-form with any other form does not require the wedge symbol, but sometimes people put it there. Also, since the symbols $x$ and $y$ are actually functions, $dx$ and $dy$ are derivatives 0-forms and therefore are 1-forms. – ndruiven Dec 07 '14 at 00:41
  • Actually, that was another thing that confused me -- how are $x$ and $y$ functions exactly? Can we simply define them that way for the sake of the algebra? And since $a$ in $a, dx$ is a 0-form, is it a function or a function evaluated at a point, like $a(x,y)$ (and therefore a scalar)? I've seen it written both ways... – Luke Dec 07 '14 at 18:56

3 Answers3

3

The algebra of differential forms is an exterior alegbra with product $\wedge$. This means that products involving $\wedge$ are associative, scalar functions can be pulled out to the front from any term of a product, and the most important rule is that for any variable $x_i$, $$dx_i\wedge dx_i=0$$ This means in particular that for any $x,y$ we have $$dx\wedge dy=-dy\wedge dx$$ This is because $$0=(dx+dy)\wedge (dx+dy)=dx\wedge dx+dx\wedge dy+dy\wedge dx+dy\wedge dy=\\dx\wedge dy+dy\wedge dx$$

Matt Samuel
  • 58,164
  • I read somewhere at the rule $df \wedge dg = -dg \wedge df$ could be derived, by "linearity", from just $df \wedge dg = -dg \wedge df$. How is this so? – Luke Dec 07 '14 at 18:50
  • @user144339 $0=(df+dg)\wedge (df+dg)=df\wedge df+df\wedge dg+dg\wedge df+dg\wedge dg=df\wedge dg+dg\wedge df$, thus $df\wedge dg=-dg\wedge df$. – Matt Samuel Dec 07 '14 at 18:51
  • @user144339 I've put this in my answer. – Matt Samuel Dec 07 '14 at 18:55
  • 1
    Is this what is meant in page 7 of http://www.cefns.nau.edu/~schulz/diff.pdf at the part where they say "It is sufficient to postulate this rule for the coordinate differentials only"? – Luke Dec 07 '14 at 22:27
  • 1
    @user144339 What is meant there is that since $df$ and $dg$ are sums of the form $\sum_i{f_idx_i}$ and $\sum_i{g_idx_i}$, we can deduce that $df\wedge dg=-dg\wedge df$ from the fact that $dx_i\wedge dx_j=-dx_j\wedge dx_i$ for all $i,j$ because we can expand the product using linearity of $\wedge$. – Matt Samuel Dec 07 '14 at 22:30
2

You could try the book by Manfredo do Carmo: differential forms and applications!

Bohrer
  • 459
1

The exterior algebra is constructed so that it reduces to ordinary multiplication of functions at the lowest level. The wedge is not necessary in that case as the wedge product reduces to ordinary multiplication. The reason for the wedge product to be explicitly denoted is to keep track of forms of genuinely differing degree. There is no need for such a notation when faced with products of zero forms as $0+0+ \cdots +0=0$, or in the case you are more interested $0+1=1$. As to how derive the rules for the wedge product, there are various approaches, but, one nice place you might read is Curtis' Abstract Linear Algebra. In that text he gives a set of axioms from which the exterior algebra may be formally derived. In a more concrete direction, one might study multilinear mappings and tensor products then pass to the subspace of completely antisymmetric tensors which permits this beautiful $\wedge$.

James S. Cook
  • 16,755
  • 1
    for a more clumsy account (but free) see Chapter 7 (page 155...) of my http://www.supermath.info/AdvancedCalculus2011.pdf – James S. Cook Dec 07 '14 at 02:10