Wedge Product of Vectors

1. May 29, 2013

Mandelbroth

Consider $\vec{a}=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$ and $\vec{b}=\begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix}$.

Is any part of the following NOT true?

$$\vec{a}\wedge\vec{b}=\frac{1}{2}(\vec{a}\otimes\vec{b}-\vec{b}\otimes\vec{a}) = \frac{1}{2}\begin{bmatrix} 0 & a_1b_2-a_2b_1 & a_1b_3-a_3b_1 \\ a_2b_1-a_1b_2 & 0 & a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 & a_3b_2-a_2b_3 & 0 \end{bmatrix}$$
(Edited for accuracy to make me feel better :tongue:)

I wasn't sure if that is actually an equality or not. If it is an equality, it makes defining the cross product rather easy. It would just be the Hodge dual of the wedge product of a with b, because the inner product of the cross product of a and b with any basis of the matrix would be 0...

Last edited: May 29, 2013
2. May 29, 2013

WannabeNewton

$v\wedge w = \frac{1}{2}(v\otimes w - w\otimes v)$ by definition i.e. $(v \wedge w)_{ab} = v_{[a}w_{b]} = \frac{1}{2}(v_{a}w_{b} - v_{b}w_{a})$.

3. May 29, 2013

Mandelbroth

So it's one half of what I put for the second and third parts?

4. May 29, 2013

WannabeNewton

If by second and third parts you mean the two expressions following the wedge product (the alternating tensor product and the matrix) then yes.

5. May 29, 2013

Ben Niehoff

WBN has the factor of 2 wrong.

$$v \wedge w \equiv v \otimes w - w \otimes v$$

6. May 29, 2013

WannabeNewton

7. May 29, 2013

WannabeNewton

Ok, based off of one of my texts, $(v\wedge w)_{ab} = 2v_{[a}w_{b]} = (v\otimes w)_{ab} - (v\otimes w)_{ba}$ which makes more sense because then we get that $^{*}(v\wedge w)_{a} = (v\times w)_{a}$ as usual but the notes linked above seem to say otherwise. Where are they getting the half factor from then?

8. May 30, 2013

Mandelbroth

Either way, the dot product of a basis of the matrix with the cross product is 0, so my direction is going to be fairly clear either way.

I think they forgot something. The second and third parts of the final equality (VERY bottom) are equivalent.

Suppose that $\omega: V^k \rightarrow K$ and $\eta:V^m\rightarrow K$ are two anti-symmetric maps where K is the base field. Then, the exterior product is of the form

$$\omega\wedge\eta = \frac{\Gamma(k+m+1)}{\Gamma(k+1)\Gamma(m+1)}\operatorname{Alt}(\omega \otimes \eta)\implies \vec{a}\wedge\vec{b}=\frac{\Gamma(3)}{\Gamma(2)^2}\operatorname{Alt} (\vec{a} \otimes \vec{b})$$

I could be egregiously wrong, but...

Last edited: May 30, 2013
9. May 30, 2013

Ben Niehoff

It's easiest to think of $\omega \wedge \eta$ as an antisymmetric map. Then for any vector $X \in V$, $(\omega \wedge \eta)(X)$ is an antisymmetric map $V^{k+m-1} \to \mathbb{R}$ given by

$$(\omega \wedge \eta)(X) = \omega(X) \wedge \eta + (-1)^k \, \omega \wedge \eta(X)$$
where the notation $\eta(X)$ means that $X$ goes into the first "slot", leaving the remaining $m-1$ slots free.

You'll notice if you follow this idea through, then you don't get the factor of 1/2 that WBN gave.

Last edited: May 30, 2013
10. May 30, 2013

WannabeNewton

Is this a convention issue? In Lee "Smooth Manifolds", for example, the wedge product is a map $\wedge :\Lambda ^{k}(V)\times \Lambda ^{l}(V)\rightarrow \Lambda ^{k+l}(V)$ defined by $\eta\wedge \omega = \frac{(k+l)!}{k!l!}\text{Alt}(\eta\otimes \omega)$. So in this case if $\eta,\omega\in \Lambda ^{1}(V)$ then the $\frac{1}{2}$ factor will not appear i.e. we will just get $\eta\wedge \omega = \eta\otimes \omega-\omega\otimes \eta$. But then why in that pdf, and in this thread: https://www.physicsforums.com/showthread.php?t=150289 do they have factors of $\frac{1}{2}$?

EDIT: Ok micromass just told me that it's a conventional issue in the definition. There is another convention where the $\frac{1}{2}$ does come in. See page 302 of Lee "Smooth Manifolds".

11. May 30, 2013

micromass

His answer isn't wrong. He simply uses a different convention. Many authors use the convention

$$\omega\wedge \eta = Alt(\omega \otimes \eta)$$

This is called the determinant convention, for example the paper wbn linked uses it.

Other authors, like Lee, use the Alt convention:

$$\omega\wedge \eta = \frac{(k+l)!}{k! l!}Alt(\omega \otimes \eta)$$

I guess it doesn't matter as long as you are consistent.

12. May 30, 2013

Ben Niehoff

I think the confusion comes from the distinction between an n-form and the components of an n-form. Generically, we can write the n-form $F$ as

$$F = \frac{1}{n!} F_{a_1 \ldots a_n} \, e^{a_1} \wedge \ldots \wedge e^{a_n}$$
and you'll notice that there is a $1/n!$ here. But this does NOT imply that the components of $F$ are $\frac{1}{n!} F_{a_1 \ldots a_n}$. Recall the definition of what "components" are, namely

$$F_{a_1 \ldots a_n} \equiv F(e_{a_1}, \ldots, e_{a_n})$$
where $e_i$ are the basis of $V$ given by

$$e^i (e_j) = \delta^i_j$$
(remember that $e^i$ are a basis of $V^*$). One could eliminate the $1/n!$ factor above by using the tensor product instead of the wedge product

$$F = F_{a_1 \ldots a_n} \, e^{a_1} \otimes \ldots \otimes e^{a_n}$$
assuming, of course, that $F_{a_1 \ldots a_n}$ is antisymmetric.

13. May 30, 2013

Ben Niehoff

What is the rule for distributing the interior product over wedge products in this convention? I.e., what is

$$i_X (\omega \wedge \eta)$$

14. May 30, 2013

WannabeNewton

Here's the relevant section in Lee, by the way, if you're interested:

15. May 30, 2013

micromass

The definitions on both cases seem to differ. In Lee, we have

$$i_X\omega(X_1,...,X_{n-1}) = \omega(X,X_1,...,X_{n-1})$$

while in Morita, we have

$$i_X\omega(X_1,...,X_{n-1}) = n \omega(X,X_1,...,X_{n-1})$$

I think that this causes that the distribution rule is the same in both cases:

$$i_X(\omega\wedge \eta) = i_X(\omega)\wedge \eta + (-1)^n \omega\wedge i_X(\eta)$$

16. May 30, 2013

Ben Niehoff

How perverse! Mathwonk seems to dislike this $1/n!$ convention as well, he goes on about it for some time in this thread: https://www.physicsforums.com/showthread.php?t=150289