# How does a 2form act?

1. Oct 13, 2006

### SeReNiTy

How does a 2from w, act on vector fields X,Y?

example w(X,Y) =?

2. Oct 14, 2006

### SeReNiTy

Come on, one of the math whizzes here must know!

3. Oct 16, 2006

### ObsessiveMathsFreak

How it acts on vector fields is a tricky question to answer. How it acts on individual vectors is I think the question you mean to ask.

Answering the question succintly isn't really possible, but here's an honest try. Take two vectors, $$\vec{a}$$ and $$\vec{b}$$. You're asking the question, what is the effect of applying the two form $$\acute{\omega}$$ to these two vectors. What is $$\acute{\omega}(\vec{a},\vec{b})$$?

Any two vectors, regardless of dimension, both lie in some plane, and they both define a parallelogram whose edges are the two vectors. This parallelogram has a certain area, call it A. One thing to note here is that this area is "signed", i.e., it can be negative. This isn't really important, it's just something that may come up.

Now, lets look of a two-form in a two dimensional vector space. Lets say that $$\acute{\omega}(\vec{a},\vec{b})= c d\acute{x}\wedge d\acute{y}(\vec{a},\vec{b})$$

This two form computes the area of the parallelogram spanned by $$\vec{a}$$ and $$\vec{b}$$, and then multiplies it by the value c. So the final answer will be cA. That's in two dimensions.

In general, c does not have to be a constant. It can be a function of (x,y), which are the "base points" of the two vectors. So in general you'll have;
$$\acute{\omega}(x,y,\vec{a},\vec{b})= c(x,y) d\acute{x}\wedge d\acute{y}(\vec{a},\vec{b})$$

A lot of authors leave out the x and y in the descrition of $$\omega$$. It's best left in.

That was all in two dimensions. But what happens in three dimensions? Or higher if one were so inclined?

OK, consider two vectors in three dimensions. Again they define a parallelogram with area A. This parallelogram is in general going to lie in three dimensions. But what we can do is project it down onto each of the three flat planes at the axes. When we do, each of the projections will also be a parallelogram again with its own area. Let $$A_{xy} , A_{yz} , A_{zx}$$ be the area of each of the projections of the parallelogram onto respectively the xy plane, the yz plane and the zx plane.

Now consider the two form defined by;
$$\acute{\omega}(\vec{a},\vec{b})= c_{xy} d\acute{x}\wedge d\acute{y}(\vec{a},\vec{b}) + c_{yz} d\acute{y}\wedge d\acute{z}(\vec{a},\vec{b}) + c_{zx} d\acute{z}\wedge d\acute{x}(\vec{a},\vec{b})$$
What this two form does is, compute each of the projected parallelogram areas, $$A_{xy} , A_{yz} , A_{zx}$$, and then multiplies each by one of the respective constants $$c_{xy} , c_{yz} , c_{zx}$$. So the final answer is;
$$\acute{\omega}(\vec{a},\vec{b})= c_{xy} A_{xy} + c_{yz} A_{yz} + c_{zx} A_{zx}$$

Again, ingeneral, each of the constants can instead be a function of the base points of the vectors. The point lies in 3D, so $$\omega$$ is in general defined by;
$$\acute{\omega}(x,y,z,\vec{a},\vec{b})= c_{xy}(x,y,z) d\acute{x}\wedge d\acute{y}(\vec{a},\vec{b}) + c_{yz}(x,y,z) d\acute{y}\wedge d\acute{z}(\vec{a},\vec{b}) + c_{zx}(x,y,z) d\acute{z}\wedge d\acute{x}(\vec{a},\vec{b})$$

You can extend this to higher dimensions by again thinking of the projection of the 2D parallelogram in n-D space being projected onto each of the 2-D planes at the axis. However, for n dimensions there are n(n-1)/2 possible planes to project down onto. Do in 4-D, 3 spaces axes plus time say, you have 6 planes to project onto, meaning sixe constants.

I hope this helps somewhat.

Last edited: Oct 16, 2006
4. Oct 17, 2006

### gvk

It's simple. The 2-form can be regarded as a tensor $$T_{i,j}$$ of type (0, 2). Since if X,Y are two vectors then the scalar
$$T_{i,j} X^i Y^j$$
is the result of your acting.

Last edited: Oct 17, 2006