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1-forms

Name: Anonymous 2008-04-02 22:38

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.

Name: Anonymous 2008-04-19 9:26

>>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?

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