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Decomposable tensors

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  • #1
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Problem:
V a vector space with [itex]dimV \le 3[/itex], then every homogeneous element in [itex]\Lambda(V)[/itex] is decomposable.


So, this exercise doesn't sound very difficult. My problem is, that i don't know the definition of homogeneous and decomposable. Can you please help me?
Thank you
 

Answers and Replies

  • #2
fzero
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A homogeneous element of [tex]\Lambda(V)[/tex] is a form of definite degree, [tex]p[/tex]. A decomposable form is one that can be written as a single exterior product of two forms of lower degree [tex]\omega = \mu \wedge \nu[/tex].
 
  • #3
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Hello,

Thank you for your help!
are these examples correct?:
For dimV=2 We have [itex]\Lambda(V) =[/itex] IK [itex]\times T^1(V) \times T^2(V)[/itex].

-Then a homogeneous elm. would be all elm. in [itex]v\in \IK[/itex] or [itex]v\in T^1(V)[/itex] or [itex]v\in T^2(V)[/itex].
Where T is the Alternator.

-Is it allowed to use the single exterior product of two form of equal degree, i.e. [itex]v=\mü \wedge \tau[/itex] with [itex]v,\mü,\tau \in T^1(V)[/itex] for instance?


Thanks
 
  • #4
fzero
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Hello,

Thank you for your help!
are these examples correct?:
For dimV=2 We have [itex]\Lambda(V) =[/itex] IK [itex]\times T^1(V) \times T^2(V)[/itex].

-Then a homogeneous elm. would be all elm. in [itex]v\in \IK[/itex] or [itex]v\in T^1(V)[/itex] or [itex]v\in T^2(V)[/itex].
Where T is the Alternator.
Yes a homogeneous element would be in [tex]T^p(V)[/tex] for some [tex]p[/tex].

-Is it allowed to use the single exterior product of two form of equal degree, i.e. [itex]v=\mü \wedge \tau[/itex] with [itex]v,\mü,\tau \in T^1(V)[/itex] for instance?
Yes it is possible that the decomposition involves forms of equal degree. However, in your example, if [tex]\mu, \tau \in T^1(V)[/tex], then [tex] v= \mu \wedge\tau\in T^2(V)[/tex].
 

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