# Decomposable tensors

Problem:
V a vector space with $dimV \le 3$, then every homogeneous element in $\Lambda(V)$ 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

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fzero
Science Advisor
Homework Helper
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A homogeneous element of $$\Lambda(V)$$ is a form of definite degree, $$p$$. A decomposable form is one that can be written as a single exterior product of two forms of lower degree $$\omega = \mu \wedge \nu$$.

Hello,

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

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

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

Thanks

fzero
Science Advisor
Homework Helper
Gold Member
Hello,

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

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

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