# Is there a standard for the definition of the wedge product?

1. Jan 3, 2009

### pellman

Are the factorials a part of the standard definition of the wedge product? Is there an overwhelming majority or does the definition tend to vary from author to author? (As many other things do.) What I am referring to is: given two one-forms $$\sigma$$ and $$\omega$$ does everyone pretty much use

$$\sigma\wedge\omega\equiv\frac{1}{2!}(\sigma\otimes\omega-\omega\otimes\sigma)$$

or do some leave off the 1/2!? I'm reading a book in which it appears so far that the author is using simply $$\sigma\wedge\omega\equiv\sigma\otimes\omega-\omega\otimes\sigma$$ , but I'm not sure if he meant to or just made a mistake.

2. Jan 3, 2009

### Hurkyl

Staff Emeritus
I'm pretty sure I've seen both conventions used for embedding the space of wedge products into the space of tensor products. Incidentally, there's always the option of not using such an embedding at all, and instead working directly with the space of wedge products.

3. Jan 3, 2009

### shoehorn

Well, the definition of the wedge product is very often given explicitly in terms of a sum over the elements of a permutation group of a given order. As a result, it makes sense to include an appropriate factorial term as a denominator. My experience is that the majority of authors, particularly in texts that are oriented towards physics, include the factorial term explicitly. You will, however, encounter many books that take the opposite approach.

Think about it this way: because you're summing over permutations, you're guaranteed to have factorials appear whenever you work with wedge products, particularly if you're fond of writing p-forms with indices as opposed to a more coordinate-free notation. The only difference between the two definitions is whether your factorials appear above or below the line!

4. Jan 3, 2009

### mathwonk

it is not entirely standard. but it depends on what you regard as the proper home for the wedge product.

the more intuitive objects to perform wedge product on are functions. i.e. functions of several variables are sometimes multilinear, and those are further sometimes also alternating.

so in that context the idea is to multiply two linear functions together and get a bilinear alternating function. now another condition you can ask for is "normalized".

i.e. you can ask for your function f1^f2 to have value 1 on the standard object e1^e2.

but then you need the 1/2 in there.

on the other hand, if you just think of wedge products in the abstract as mkaing sense for any modules at all, then instead of the alternating gadgets being a SUBSET of the tensor doo hickeys, they are a QUOTIENT of them.

In that case S^T is merely the same object as StensorT, only modulo a bigger submodule. then you naturally get that S^T "=" StensorT.

i.e. S^T is the residue of StensorT in the module M^N which is a quotient of the module MtensorN.

In the abstract setting it is standard and natural not to have a 1/2 in there.

so to me the difference is whether you are regarding the wedge product of two modules as a subobject, or a quotient object, of the tensor product.

I guess all I'm saying is there is no completely standard rule when M^N is a subobject of MtensorN, but when M^N is a quotient object of MtensorN, then there is a standard rule.

so look closely at the definitions.

mike spivak in his calculus no manifolds makes it a subobject and does introduce the 1/2, to get normalization, as he clearly points out.

i see now that hurkyl ALREADY SAID THIS MORE SUCCINCTLY.

5. Jan 4, 2009

### pellman

Thanks! Much appreciated.

Maybe someone could take a look at this then and see where I'm going wrong.

In the symplectic approach to (non-relativistic) mechanics the two form $$d\theta=dp_i\wedge dq^i$$ plays an important role. Let X be the vector field $$v^i(q,p)\frac{\partial}{\partial q^i}+f_j(q,p)\frac{\partial}{\partial p_j}$$. The space we are working in is the phase space R^2N where the first N coordinates are $$q^i$$ and the second N coordinates are $$p_i$$

So then what is $$d\theta[X]$$?

I assume that

$$dp_i\wedge dq^i=\frac{1}{2}(dp_i\otimes dq^i - dq^i \otimes dp_i)$$

and assume that

$$<dp_i\otimes dq^i, \frac{\partial}{\partial q^j}>=\delta^i_j dp_i$$
$$<dp_i\otimes dq^i, \frac{\partial}{\partial p_j}>=0$$
$$<dq^i\otimes dp_i, \frac{\partial}{\partial q^j}>=0$$
$$<dq^i\otimes dp_i, \frac{\partial}{\partial p_j}>=\delta^j_i dq^i$$

That is, I assume that only the second factor in the tensor product of covectors acts on the vector. Very likely this is where I am going wrong.

So I get $$d\theta[X]=\frac{1}{2}(v^i dp_i - f_i dq^i)$$. The author of the book I am reading gets instead $$d\theta[X]=-(v^i dp_i - f_i dq^i)$$

So what it boils down to is that I don't understand the explicit form of a two-form acting on a vector.

Last edited: Jan 4, 2009
6. Jan 5, 2009

### Doodle Bob

There are two things you need to be careful about when plugging a vector into a 2-form.

The first is what you were initially concerned with: whether the author uses the factorial convention or not. In this case, it looks likes the author doesn't.

The second, though, is something you've elided over: which slot in the 2-form do you put the vector? There is no one way for a 2-form to act on vectors: there are two ways, which have the opposite signs. A 2-form is after all a special 2-tensor: it takes two vectors *in a specific order* and gives back a scalar. So, you need to specify whether X goes into the first or second slots of dtheta. You are putting it into the second slot, so the author must be putting it into the second one.

7. Jan 5, 2009

### pellman

Thanks a heap. Makes perfect sense.

When I get stuck on something like this, I can't tell if it is because there is something I don't understand, or if there is simply nothing to understand--that is, an arbitrary choice by the author.

What a great resource this forum is to an amateur hobbyist like me! Used to be these little mysteries would just have to go unresolved. Each one doesn't mean much but after a while they add up.

8. Jan 5, 2009

### Doodle Bob

I'm glad to help but you really should seek out a graduate school somewhere to study this kind of stuff. Human discourse is indispensable in understanding this level of material. In fact, These kind of mysteries are usually resolved after 5 minutes of conversation.

9. Jan 5, 2009

### gts87

In Bachman's book (A Geometric Approach to Differential Forms) he describes the wedge product as a determinant. First he defines $$\omega\wedge\nu(V_{1},\;V_{2}) = det[\omega(V_{1})\;\omega(V_{2})\;\;,\;\;\nu(V_{1})\;\nu(V_{2})]$$ (sorry about the formatting there -- the first and second rows of the matrix are separated by the comma.) Anyway, that looks to be the same as $$\omega\otimes\nu - \nu\otimes\omega$$. Am I right?

He later defines $$\omega_{1}\wedge\omega_{2}\wedge...\wedge\omega_{n}(V_{1},\;V_{2},\;...\;,V_{n}) = det[\omega_{1}(V_{1}) \;\;... \;\; \omega_{1}(V_{n}) \;,\; ... \;, \;\;\omega_{n}(V_{1}) \;\;... \;\; \omega_{n}(V_{n})]$$ Is this essentially correct? The formal definition of a wedge product looks a lot like the Leibniz formula for the determinant. If the wedge product is essentially a determinant, why not look at it that way?

10. Jan 7, 2009

### pellman

Very interesting, gts87. Thanks.

Don't I know it!