Wedge product question

1. Nov 30, 2008

Diffy

1. The problem statement, all variables and given/known data
Please, I just am trying to understand the question. I wish to prove it on my own, but the way the question is phrased makes no sense.

So here it is:

Let us define the linear map
$$\phi : V^{*} \otimes \bigwedge^{i} V \rightarrow \bigwedge^{i-1} V$$

by the formula

$$\ell \otimes v_1 \wedge ... \wedge v_i \mapsto \sum_{s=1}^{i} (-1)^{s-1} \ell (v_s) v_1 \wedge ... \wedge \hat{v_s} \wedge ... \wedge v_s$$

Prove that the map $$\phi$$ is well defined and does not depend on the choice of basis.

2. Relevant equations
Well all the usual definition of exterior algebras, and tensor products are needed.

3. The attempt at a solution

As I stated, I haven't started solving yet, I am simply trying to understand the question. I don't see how it goes to wedge i-1. What exactly is v hat sub s? Does that make i wedges?
I don't think this formula is going to i-1 wedges. Please help me to understand what is going on here.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 30, 2008

Dick

What they are trying to notate by v hat sub s, is that the vector v_s is NOT included in the wedge product. All the other v_i's are. That's why it's an i-1 form.

3. Nov 30, 2008

Diffy

Now it makes sense!

Now I just have to solve the problem. Any hints are appreciated.

4. Nov 30, 2008

lurflurf

Well there are two main ways to prove something is invariant
1 show it does not change with form
ie
let x and x' be different forms of x
if
f(x)=f(x')
f is invariant

2 provide an invariant definition
ie
define
f(x)
so that the form of x is not a factor

Since you have been provided a noninvariant definition
let Av be a change of basis
you want to show
l(v1)^v2^...^vn=l(Av1)^Av2^...^Avn=det(a)*l(v1)^v2^...^vn
with det(A)=1