What's a 0-form and what's not?

1. Aug 12, 2007

quasar987

Is a function from R^n to R^m for aritrary m a considered a 0-form on R^n, or does 0-form refers only to functions from R^n to R ?

2. Aug 12, 2007

dextercioby

What's the definition of a p-form on R^n ?

3. Aug 12, 2007

Hurkyl

Staff Emeritus
If you have a finite dimensional vector space V with scalar field k, then the space of n-forms is isomorphic to the space of alternating multilinear maps Vn --> k.

In particular, a 0-form is an element of k.

In the typical setting of differential geometry, when analyzing a single point, your scalar field is R and your vector space is the tangent space, so a 0-form would simply be a real number. But more exotic things are possible, and sometimes even fruitful.

4. Aug 12, 2007

quasar987

In 'Calculus on manifolds', Spivak defines a k-form on R^n as a function w sending a point p of R^n to an alternating multilinear maps (R^n)^k-->R.

This makes sense only for k>0, so he treats the case k=0 separately by saying that by a 0-form we mean a function f.

I was 90% sure he meant a function f:R^n-->R but wanted to make sure.

5. Aug 12, 2007

Hurkyl

Staff Emeritus
Well, it really does make sense for k=0: an A-valued function of 0 variables is the same thing as an element of A, and it's vacuously true that such a thing is alternating and 0-linear.

So a 0-form on the tangent bundle to R^n is, indeed, a map R^n --> R. This agrees with what I said pointwise -- if f is such a thing, then f(P) is a 0-form on the tangent space at P, which is the same thing as an element of R.

6. Aug 17, 2007

quasar987

Why are forms defined specifically as sending points to alternating tensors? What's wrong with good old arbitrary tensors? Or equivalently, what's so special about alternating ones?

7. Aug 18, 2007

Hurkyl

Staff Emeritus
Integrating along an opposite orientation should give you the opposite answer -- thus the sign change.

From an algebraic perspective, they are trying to capture first-order differential information -- thus you want dx dx = 0. An immediate consequence of this identity is that differentials must be alternating.

8. Aug 18, 2007

quasar987

9. Aug 18, 2007

Hurkyl

Staff Emeritus
Oh, and there's a geometric picture too -- given two vectors, you want to combine them to form a bivector that represents the area swept out by your vectors. So this product too should satisfy v v = 0. And since 1-forms are dual to tangent vectors...

10. Aug 18, 2007

mathwonk

license to steal: salary for answering the same question infinitely many times.