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

by quasar987
Tags: 0form
 Sci Advisor HW Helper PF Gold P: 4,768 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 ?
 Sci Advisor HW Helper P: 11,866 What's the definition of a p-form on R^n ?
Emeritus
PF Gold
P: 16,101
 Quote by 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 ?
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.

HW Helper
PF Gold
P: 4,768

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

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.
Emeritus
PF Gold
P: 16,101
 Quote by 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.
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.
 Sci Advisor HW Helper PF Gold P: 4,768 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?
 Emeritus Sci Advisor PF Gold P: 16,101 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.