# I Wedge product of basis vectors

1. Nov 30, 2016

### Kevin McHugh

Is there a set of relationships for the wedge product of basis vectors as there are for the dot product and the cross product?

i.e. e1*e1 = 1
e1*e2 = 0

e1 x e2 = e3

2. Nov 30, 2016

### Staff: Mentor

It's bilinear, associative and $v \wedge v=0$ (if char $\mathbb{F} \neq 2$). In general it's $a \wedge b = (-1)^{nm} b \wedge a$ for $a \in \Lambda^n(V)= \underbrace{V \wedge V \wedge \ldots \wedge V}_{n\ times}$ and $b \in \Lambda^m(V)= \underbrace{V \wedge V \wedge \ldots \wedge V}_{m\ times}$

3. Nov 30, 2016

### Kevin McHugh

Are those components or basis vectors? Can you explain that in English?

4. Nov 30, 2016

### Staff: Mentor

It is meant as for arbitrary vectors. So especially for basis vectors, too, if you like. In coordinates, it would be some equivalence classes of tensor products, but I don't know how to press this equivalence relation into coordinates. It means basically that the wedge-product or better exterior product is a tensor product, but some tensors are considered to be equal, because the relations below have to be met.

If we have a vector space $V$ over a field $\mathbb{F}$ in which $1+1 \neq 0$ holds, then for $a,b,c \in V$ and $\lambda \in \mathbb{F}$ the following is true:
1. $(a+b) \wedge c = a\wedge c + b \wedge c$
2. $\lambda (a \wedge b) = (\lambda a) \wedge b = a \wedge (\lambda b)$
3. $a \wedge (b \wedge c) = (a \wedge b) \wedge c$
4. $a \wedge a = 0$
5. $a_1 \wedge \ldots \wedge a_n \wedge b_1 \wedge \ldots \wedge b_m = (-1)^{nm} b_1 \wedge \ldots \wedge b_m \wedge a_1 \wedge \ldots \wedge a_n$
The first two are called linearity, which together with the fifth becomes multi-linearity (linear in all "factors"), the third one is associativity, the fourth is a special case of the fifth together with the fact, that $1+1 \neq 0$, which is said as the characteristic of $\mathbb{F}$ is not two, and the fifth alone can be called graduated commutativity, i.e. it determines what happens, if we change the order of "factors". I don't know how to put the formulas in other English words as their names are.

Perhaps you want to read the Wikipedia entry on it: https://en.wikipedia.org/wiki/Exterior_algebra

5. Dec 3, 2016

### Stephen Tashi

The first question is whether $a \wedge b$ is something (e.g. a vector or a scalar) in the same vector space that contains $a$ and $b$.

I'd say no. For $a$ and $b$ in a a vector space $V$, in order to define $a \wedge b$, you must define a different mathematical structure than $V$ itself. ( As an analogy, we can use two real numbers x1, x2 to define an interval [x1,x2], but "an interval" is a different thing than a single real number. )

By contrast, the cross product operation (in 3 dimensions) $a \times b$ does produce a result that is also a element of the the same vector space $V$ that contains $a$ and $b$.