# Decomposable tensors in $\Lambda^2 {\mathbb{C^n}}$

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

• 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. – Omnomnomnom Apr 16 '17 at 15:53
• 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? – iou Apr 16 '17 at 15:57
• 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". – Omnomnomnom Apr 16 '17 at 16:02
• 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. – Omnomnomnom Apr 16 '17 at 16:10

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 $i<j<k<l$.