# Wedge product property

1. Nov 4, 2006

### ak416

This has to do with the wedge product ^ on alternating tensors. I cant seem to prove w ^ n = (-1)^kl * n ^ w. where w is a k alternating tensor and n is an l alternating tensor.

I know w ^ n = (k+l)!/(k! l!) * Alt(w x n) where x is the tensor product.
Now, Alt(wxn) (v_1,...,v_k+l) = 1/(k+l)! * (sum over all s in Sk) ( sgn(s) * (w x n) (v_s(1),...,v_s(k+l) )
= 1/(k+l)! * (sum over all s in Sk) ( sgn(s) * (n x w) (v_s(k+1),...,v_s(k+l),v_s(1),...,v_s(k)) )

Note Sk is the set of all permutations of the numbers 1,...,k.

Now if I let s' be the permutation such that
Alt(w x n) = 1/(k+l)! * (sum over all s in Sk) ( sgn(s) * (n x w) (v_s'(1),...,v_s'(k+l)) )
then shouldnt Alt(w x n) = 1/(k+l)! * (sum over all s' in Sk) ( (-1)^m * sgn(s') * (n x w) (v_s'(1),...,v_s'(k+l)) )

where m is the number of transpositions needed to transform (k+1,...,k+l,1,...,k) to (1,....,k+l). Because if this is true then doesnt the property imply that m = l*k. And this isnt true. So what am i doing wrong here?

2. Nov 5, 2006

### mathwonk

isn that more or less the definition of alternating?

3. Nov 5, 2006

### ak416

definition of an alternating tensor is that when you swap any of its input vectors, you get the same result but times negative one. (like the determinant).

4. Nov 6, 2006

### gvk

It's not right to use w ^ n = (k+l)!/(k! l!) * Alt(w x n), because w and n are already forms, or alternative tensors (not the tensor products!).
It is easy to prove your formula using just the bases of rank k and l and the simple relation dx^i ^ dx^j=-dx^j ^ dx^i.

Last edited: Nov 6, 2006
5. Nov 6, 2006

### mathwonk

think about it, if you have a pair of vectors of form (a1,...,ak,.b1,...,bl) and you rearrange it to be of form (b1,...bl,a1,...ak) how many interchanges jhave you made? if the aNSWER IS KL, THEN YOU ARE DONE.

6. Nov 9, 2006

### ak416

ya i tried that. But take for example k = 3, l = 7.
so you have (1,2,3,4,5,6,7,8,9,10,11) you want (5,6,7,9,9,10,11,1,2,3,4)
step 1: transform this to (8,9,10,11,5,6,7,1,2,3,4) (4 interchanges)
step 2: transform to (5,6,7,11,8,9,10,1,2,3,4) (3 interchanges)
step 3: transform to (5,6,7,8,11,9,10,1,2,3,4) (1 interchange)
step 3: transform to (5,6,7,8,9,11,10,1,2,3,4) (1 interchange)
step 3: transform to (5,6,7,8,9,10,11,1,2,3,4) (1 interchange)

total: 10 interchanges
k*l = 21

what is wrong with this counterexample?

7. Nov 10, 2006

### coalquay404

Suppose that you have indices $(a_1,\ldots,a_k,b_1,\ldots,b_l)$ and that we want to swap around the indices $b_1$ and $a_k$. Because of the antisymmetry we have

$$(a_1,\ldots,a_k,b_1,\ldots,b_l) = (-1)(a_1,\ldots,b_1,a_k,\ldots,b_l)$$

Agreed? Now suppose that we want to go one further and move $b_1$ into the position of $a_{k-1}$:

$$(a_1,\ldots,a_k,b_1,\ldots,b_l) = (-1)(a_1,\ldots,a_{k-1},b_1,a_k,\ldots,b_l) = (-1)^2(a_1,\ldots,b_1,a_{k-1},a_k,\ldots,b_l)$$

So, we've moved $b_1$ two places to the left and obtained a factor of $(-1)^2$. In fact, we can move $b_1$ $k$ places to the left to get

$$(a_1,\ldots,a_k,b_1,\ldots,b_l) = (-1)^k(b_1,a_1,\ldots,a_k,b_2,\ldots,b_l)$$

We've now managed to move one of the $b$ indices completely to the left. However, we'd also like to move the rest of them. It doesn't take a great deal of thought to convince yourself that if you also now move $b_2$ to the left by $k$ places you obtain

$$(a_1,\ldots,a_k,b_1,\ldots,b_l) = (-1)^{2k}(b_1,b_2,a_1,\ldots,a_k,b_3,\ldots,b_l)$$

So if you move two of the $b$ indices to the left you get a factor of $(-1)^{2k}$. In fact, if you move all $l$ of the $b$ indices to the left (keeping them in their original order), it should then be straightforward to see that you get

$$(a_1,\ldots,a_k,b_1,\ldots,b_l) = (-1)^{l\cdot k}(b_1,\ldots,b_l,a_1,\ldots,a_k)$$

Does this help you to see what's going on when you swap around the indices?

If you now return to the original question that you asked, suppose that we denote the space of $k$-forms over some space $V$ by $\Lambda^k(V)$. Then, given $\alpha\in\Lambda^k(V)$ and $\beta\in\Lambda^l(V)$, we know from the definition of the wedge product that $\alpha\wedge\beta\in\Lambda^{k+l}(V)$. The important thing about differential forms is that they are simply machines into which you feed vectors. For example, if you have a $k$-form $\alpha$, then you can get a number out of it by letting it act on $k$ vectors $X_{(1)},\ldots,X_{(k)}$:

$$\alpha:(X_{(1)},\ldots,X_{(k)})\mapsto\alpha(X_{(1)},\ldots,X_{(k)}) \in \mathbb{F}$$

where I've denoted a typical field by $\mathbb{F}$. In physics, we're usually interested in the case where $\mathbb{F}=\mathbb{R}$. However, the differential form has the important property that if you swap around the order of any two of the input vectors, then you get a factor of (-1). For example,

$$\alpha(X_{(1)},X_{(2)},\ldots,X_{(k)}) =(-1) \alpha(X_{(2)},X_{(1)},\ldots,X_{(k)})$$

This is exactly analogous to what I said at the start about swapping around indices. Knowing this, it's really easy to see that you have the desired result:

$$\alpha\wedge\beta = (-1)^{k\cdot l}\beta\wedge\alpha$$

Last edited: Nov 10, 2006
8. Nov 10, 2006

### gvk

It's not quite right
you have k = 3, l = 8 or (1,2,3)^(4,5,6,7,8,9,10,11) you want (4,5,6,7,8,9,10,11)^(1,2,3).
Remember (1,2,3) and (4,5,6,7,8,9,10,11) are 3-form and 7-form, or (1,2,3) is (1^2^3) and (4,5,6,7,8,9,10,11) is (4^5^6^7^8^9^10^11).

Step 1 : you move 4 to the left throught the form (1^2^3) and each time you have to change the sign. The sign will be (-1)^3 (here "^" is power)
Step 2 : repeat this procedure for 5,6,....,11. The total sing will be (-1)^(3 *8). In other words (-1)^(k*l).

9. Nov 10, 2006

### gvk

Look,
coalquay404 has discribed it in more general way.

Last edited: Nov 10, 2006
10. Nov 11, 2006

### ak416

whoops did i just say 7+3 = 11. Ok i understand now coolquay. My method isnt really a counterexample because 8*3 = 24 and it is still even like 10. Your way of interchanging only the entries beside eachother is good because it proves that its l*k interchanges. thanks.

11. Nov 12, 2006

### mathwonk

if you have a pair of vectors of form (a1,...,ak,.b1,...,bl) and you rearrange it to be of form (b1,...bl,a1,...ak), then each one of the l b's has moved past the k a's. done.

12. Nov 12, 2006

### coalquay404

What do you mean by a "pair" of vectors? Each of the indices represents one vector input into the $(k+l)$-form so there are $k+l$ vectors involved, not just a pair.