1-forms

Can anyone give me a somewhat simplistic explanation of 1-forms?  For some reason I'm really stuck on this, I understand that they're a linear transformation mapping point and a vector to R.  (I'm guessing they actually map to all of C but for sake of learning I'd appreciate it simplified as my understanding of complex analysis is virtually nonexistant at this point in time.)  I don't even understand their purpose, except that a differential is a type of 1-form.  The wikipedia article was a bit over my head & the wolfram article wasn't much help either.

I'm going to bump this question once before I let it die because I would really like to know.  Any insight at all would be appreciated.

looks like 1-form is just a name for a linear function that sends vectors to the field of scalars for the space.

like, if you had vectors of real numbers, and your scalars are real numbers, f would be a 1-form if it was a function from the set of vectors to the real numbers which preserved addition and scalar multiplication.

e.g. you'd be looking at things like this in R^3:
f(<a,b,c>) = a+b+c

It's linear because:
f(<a,b,c> + <d,e,f>) = f(<a+d,b+e,c+f>) = a+d+b+e+c+f = a+b+c+d+e+f = f(<a,b,c>) + f(<d,e,f>)
&
f(s<a,b,c>) = f(<sa,sb,sc>) = sa+sb+sc = s(a+b+c) = s*f(<a,b,c>)

and it's a function that sends a vector to a real number.

a 2 form would be a function sending pairs of vectors to a scalar, 3 form would be a function sending triplets of vectors to a scalar, etc.

Thanks for responding.  Could you give me an explanation of what a differential is, without recursing? (ie, I've seen a lot of people shrug it of with df = dx + dy + dz or something similar & that explains nothing for me)

>>4
Long explanation: Take differential geometry
Short explanation: A differential 0-form is simply a function of n variables. A differential 1-form is f1*dx1 + f2*dx2 + ... + fn*dxn, where f1, f2, ... fn are functions of n variables. A 2-form has dxi^dxj (1 <= i < j <= n) as the basis and, once again, functions of n variables as the coefficients. And so on.

>>5
note dx1 = df1/dx1 + df2/dx1 + ... + dfn/dx
where the d's on the left hand side should be partial derivatives not normal derivatives

the 1 form defined as such takes a vector function (at a point) to the reals. if you don't have a point, it just gives a real valued function. it is linear because differentiation is linear.

you really should take geometry or advanced calc to understand this properly

>>5
>>6
I'm currently taking elementary differential geometry.  I've been able to understand most of what's going on but I've been really unsatisfied with the explanations I've been given on this particular subject.  I really like to understand the math I'm learning, & not just know how to apply it.  Given the responses here it sounds like I'm going to need a lot more background for this (or maybe I'm just stupid?), but thanks for responding.

>>7
How's your linear algebra background?

I notice with some math classes, there's a lag in the 'understanding' of a few months, for me at least. Like, I'll cover something, and be able to use it, but not really feel that I understand it until I've digested it for a while. Maybe that's just me though.

>>8
Exactly.  Often I feel like I'm taking classes so I can really grasp the prerequisites.  I have a decent understanding of elementary linear algebra, but it's not really second nature to me yet so sometimes I have to spend a little time thinking for eigenvalue problems & some other stuff that I really should have down by now.  The major exception is determinants, I've pretty much completely given up on understanding those, I know a couple of the ways to find them but honestly I have no idea what they are.  I've been told volume of a parallelepiped (sp?), but I don't understand how that comes about given the formulae for deriving determinants, & I rarely ever understand the relation when determinants are invoked.  Also I have no concept of the significance the sign/orientation.  I think textbooks really ought to focus on determinant-free proofs when possible, the other students I talk to don't seem to know either, it just seems to be accepted as proof by intimidation.  Whoops, off-topic...

>>9
Linear algebra by Kenneth Hoffman and Ray Kunze explains where determinants come from and why they behave like they do.

The same book says:
"A multilinear function on V^r [V a vector space] will also be called an r-linear form on V or a multilinear form of degree r on V. Such functions are sometimes called r-tensors on V."

>>10
Thanks, I'm thinking of picking up an intermediate linear text for the summer as I won't be taking classes at the time.  I'm leaning towards Sheldon Axler's "..Done Right" text, is Hoffman/Kunze a better option (for linear in general, not determinants exclusively)?  As for the forms, thanks, but that still doesn't clear up much for me.  Unless there's a significant difference between linear & multilinear, I haven't heard the word multilinear used often but my assumption has always been that the difference is negligable.

>>11
I hear kunze is a more rigorous, 'mathematical' (ie. for math majors) text than axler.

Speak of the devil, I just received Axler's book as a gift.  Hopefully I'll have more time to read it soon.

>>11
I haven't read Axler's but I've heard it's pretty good. At my uni they used Lay, but luckily I had Kunze lying around at home. Kunze is pretty rigourous, the first page has the axioms for a field. I wish all my books were like that.

To answer your question:
A linear function is 1-linear, a multilinear function is linear in all its arguments. If a,b are scalars and u,v,w,x are vectors in some vector space V over the field F. f a bilinear form on V (f: V^2 -> F) means:

f(au + v, bw + x) = f(au, bw + x) + f(v, bw + x) = f(au,bw) + f(au,x) + f(v,bw) + f(v,x) = abf(u,w) + af(u,x) + bf(v,w) + f(v,x)

(strictly speaking we only need a commutative ring and a module over it, but I'm sure if you know what that means you'll be able to generalize what I said)

The scalar product, or any inner product for that matter, is bilinear. Wouldn't you agree that there's a significant difference between this kind of linearity and the f(ax + y) = af(x) + f(y) type?

One thing I didn't quite catch in the definition of a form is whether it has to be into the field (or ring) that the vectors are over, or whether it can be into any set.

The vector product on R^3 is also bilinear, but its codomain is R^3 and not R, so is it a bilinear form on R^3?

>>14
I'm pretty sure forms take to the ring. At least that's what I remember seeing in most of the definitions.