# Wedge product

1. Jun 24, 2011

### facenian

I have this problem(from Tensor Analysis on Manyfolds by Bishop and Goldberg): prove that
$e_1^ e_2 + e_3^e_4$ is not decomposable when the dimension of the vector space is greater than 3 and e_i are basis vectors.
I solved it by mounting a set of 6 equations with 8 unknows and studying the different posibilities cheking that each one is not solvable.
Is there any nicer way to tackle this problem? if so please let me know

2. Jun 24, 2011

### tiny-tim

hi facenian!

(use "\wedge" in latex )
you need to prove that it cannot equal $a\wedge b$ where a and b are 1-forms …

so express a and b in terms of the basis

3. Jun 24, 2011

### facenian

helo tiny-tim, thanks for your prompt response and yes I did what you suggested and it led me to what I explained

4. Jun 24, 2011

### tiny-tim

how about $a\wedge (e_1\wedge e_2 + e_3\wedge e_4)$ ?

5. Jun 28, 2011

### facenian

you mean, let $a=\sum_{i<j} x_{ij} e_i\wedge e_j$ and then conclude tha $a$ must be null? Please let me know if that's what you meant and/or if I'm correct

6. Jun 28, 2011

### tiny-tim

hi facenian!

no, i'm using the same a as before (in a∧b, which you're trying to prove it isn't)

so let a = ∑i xiei

7. Jun 28, 2011

### facenian

I'm sorry I did not explained it correctly I should have said:

you mean, let $a=\sum_i x_{i} e_i$ and then conclude tha $a$ must be null because we are left with a linear conbination of basic vectors of the form $\sum x_i e_i\wedge e_j\wedge e_k=0$ .Please let me know if that's what you meant and/or if I'm correct

Last edited: Jun 28, 2011
8. Jun 28, 2011

### tiny-tim

… which has to be 0, because a ∧ (a ∧ b) = 0

yes

9. Jun 28, 2011

### facenian

thank you very much tiny-tim your method is much better than mine!