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

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Discussion Overview

The discussion revolves around proving the non-decomposability of the wedge product \( e_1 \wedge e_2 + e_3 \wedge e_4 \) in higher-dimensional vector spaces, specifically when the dimension is greater than 3. Participants explore various approaches to tackle this problem, including algebraic methods and the use of basis vectors.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant describes a method involving a set of 6 equations with 8 unknowns to show that the problem is not solvable.
  • Another participant suggests expressing the forms \( a \) and \( b \) in terms of the basis to prove that \( e_1 \wedge e_2 + e_3 \wedge e_4 \) cannot equal \( a \wedge b \).
  • A different approach is proposed involving the expression \( a \wedge (e_1 \wedge e_2 + e_3 \wedge e_4) \) to explore the implications of the wedge product.
  • Participants discuss the implications of setting \( a \) as a linear combination of basis vectors and conclude that it must be null under certain conditions.
  • There is a recognition that the method suggested by one participant is more efficient than the initial approach described.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a single method to prove the non-decomposability, as multiple approaches are discussed and refined without settling on one definitive solution.

Contextual Notes

Some methods rely on specific assumptions about the forms and their linear combinations, which may not be universally applicable without further clarification.

facenian
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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
 
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hi facenian! :smile:

(use "\wedge" in latex :wink:)
facenian said:
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 :wink:
 
tiny-tim said:
hi facenian! :smile:

(use "\wedge" in latex :wink:)


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 :wink:


helo tiny-tim, thanks for your prompt response and yes I did what you suggested and it led me to what I explained
 
how about a\wedge (e_1\wedge e_2 + e_3\wedge e_4) ? :wink:
 
tiny-tim said:
how about a\wedge (e_1\wedge e_2 + e_3\wedge e_4) ? :wink:

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
 
hi facenian! :smile:

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 :wink:
 
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:
facenian 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 …

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

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

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