0
$\begingroup$

Let $V=C^n$. How can I describe all decomposable tensors in $\Lambda^2 {V}$ for $n=3$ and $n \ge 4$?

$\endgroup$
  • $\begingroup$ I don't understand what kind of answer you're looking for. I would say that "you can describe all decomposable tensors in $\wedge^2 V$ as $v \wedge w$ for $v,w \in \Bbb C^n$". However, that seems like a restatement of the defintion and therefore not completely useful. $\endgroup$ – Omnomnomnom Apr 16 '17 at 15:53
  • $\begingroup$ It seems for $n=3$ all tensors are decomposable and for $n \ge 4$ tensor $v$ is decomposable iff $v \wedge v=0$. Is it? $\endgroup$ – iou Apr 16 '17 at 15:57
  • $\begingroup$ That's the kind of thing that should be in your original question. In any case: you're right about $n \leq 3$. I'm not sure about your statement for $n \geq 4$. It's clear that the "only if" holds, but I'm not sure about the "if". $\endgroup$ – Omnomnomnom Apr 16 '17 at 16:02
  • $\begingroup$ I have a feeling that your test happens to work for $n = 4,5$ because $(\wedge^2 \Bbb R^n) \wedge (\wedge^2 \Bbb R^n) \cong \wedge^4 \Bbb R^n$ consist only of decomposable tensors. However, this is no longer the case with $n = 6$, and I suspect that at this point your argument will fall apart. $\endgroup$ – Omnomnomnom Apr 16 '17 at 16:10
1
$\begingroup$

Via the Plücker relations. I'll let $e_1,\ldots,e_n$ be the standard basis of $\mathbb{R}^n$ and let $\omega=\sum_{i,j}a_{i,j}e_i\wedge e_j$ be a typical element of $\bigwedge^2\mathbb{R}^n$ where the sum is only over the $i$ and $j$ with $i<j$. Then $\omega$ is decomposable iff $$a_{i,j}a_{k,l}-a_{i,k}a_{j,l}+a_{i,l}a_{j,k}=0$$ for all $i$, $j$, $k$, $l$ with $i<j<k<l$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.