Proving Non-Decomposability of Wedge Product in Higher Dimensional Vector Spaces

[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

 

[tiny-tim]

hi facenian!

(use "\wedge" in latex )

I have this problem(from Tensor Analysis on Manyfolds by Bishop and Goldberg): prove that
$e_1\wedge e_2 + e_3\wedge e_4$ 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

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 ​

 

[facenian]

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 ​

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

 

[tiny-tim]

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

 

[facenian]

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

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

 

[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 x_ie_i ​

 

[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

 

[tiny-tim]

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 $\sum x_i e_i\wedge e_j\wedge e_k=0$ …

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

yes ​

 

[facenian]

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