# Wedge product

by facenian
Tags: wedge
 P: 208 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
HW Helper
Thanks
P: 26,148
hi facenian!

(use "\wedge" in latex )
 Quote by facenian 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
P: 208
 Quote by 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
helo tiny-tim, thanks for your prompt response and yes I did what you suggested and it led me to what I explained

 Sci Advisor HW Helper Thanks P: 26,148 Wedge product how about $a\wedge (e_1\wedge e_2 + e_3\wedge e_4)$ ?
P: 208
 Quote by tiny-tim 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
 Sci Advisor HW Helper Thanks P: 26,148 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
 P: 208 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
 Quote by facenian 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 $\sum x_i e_i\wedge e_j\wedge e_k=0$ …