wedge product


by facenian
Tags: wedge
facenian
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#1
Jun24-11, 08:18 AM
P: 205
I have this problem(from Tensor Analysis on Manyfolds by Bishop and Goldberg): prove that
[itex]e_1^ e_2 + e_3^e_4[/itex] 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
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tiny-tim
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#2
Jun24-11, 08:56 AM
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hi facenian!

(use "\wedge" in latex )
Quote Quote by facenian View Post
I have this problem(from Tensor Analysis on Manyfolds by Bishop and Goldberg): prove that
[itex]e_1\wedge e_2 + e_3\wedge e_4[/itex] 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 [itex]a\wedge b[/itex] where a and b are 1-forms
so express a and b in terms of the basis
facenian
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#3
Jun24-11, 09:02 AM
P: 205
Quote Quote by tiny-tim View Post
hi facenian!

(use "\wedge" in latex )


you need to prove that it cannot equal [itex]a\wedge b[/itex] 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

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#4
Jun24-11, 09:50 AM
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wedge product


how about [itex]a\wedge (e_1\wedge e_2 + e_3\wedge e_4)[/itex] ?
facenian
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#5
Jun28-11, 10:34 AM
P: 205
Quote Quote by tiny-tim View Post
how about [itex]a\wedge (e_1\wedge e_2 + e_3\wedge e_4)[/itex] ?
you mean, let [itex]a=\sum_{i<j} x_{ij} e_i\wedge e_j[/itex] and then conclude tha [itex]a[/itex] must be null? Please let me know if that's what you meant and/or if I'm correct
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#6
Jun28-11, 10:43 AM
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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
facenian
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#7
Jun28-11, 10:56 AM
P: 205
I'm sorry I did not explained it correctly I should have said:

you mean, let [itex]a=\sum_i x_{i} e_i[/itex] and then conclude tha [itex]a[/itex] must be null because we are left with a linear conbination of basic vectors of the form [itex] \sum x_i e_i\wedge e_j\wedge e_k=0[/itex] .Please let me know if that's what you meant and/or if I'm correct
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Jun28-11, 11:00 AM
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Quote Quote by facenian View Post
you mean, let [itex]a=\sum_i x_{i} e_i[/itex] and then conclude tha [itex]a[/itex] must be null because we are left with a linear conbination of basic vectors of the form [itex] \sum x_i e_i\wedge e_j\wedge e_k=0[/itex]
which has to be 0, because a ∧ (a ∧ b) = 0
yes
facenian
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#9
Jun28-11, 11:03 AM
P: 205
thank you very much tiny-tim your method is much better than mine!


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