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Let $L$ be a finite dimensional Lie algebra. We view the Lie bracket as a linear map on the exterior square: $$\pi:L \bigwedge L \rightarrow L$$

Define $$\bigwedge L := \langle a \wedge b \big| [a,b]=0\rangle$$

Why is in general $\bigwedge L \neq \ker(\pi)$ ?

If $(x_i)$ is a basis of $L$ then $L \bigwedge L$ has a basis $x_i \wedge x_j$ where $i \neq j$, so can't we just write $$a \wedge b = \sum_{i \neq j} \lambda_{ij} (x_i \wedge x_j)$$ and $$[a,b] = \sum_{i \neq j} \lambda_{ij}[x_i,x_j] = \pi(a \wedge b)$$ thus it would follow that $$\langle a \wedge b \big| [a,b]=0\rangle = \ker \pi$$

What am I missing?

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You're missing the fact that the exterior square doesn't consist entirely of pure tensors (rather it is spanned by pure tensors). –  Qiaochu Yuan Apr 8 '12 at 17:03
    
@QiaochuYuan Thanks a lot, that's it. I got confused about that basis. –  crt Apr 8 '12 at 17:56
    
@QiaochuYuan Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. –  Julian Kuelshammer Jun 18 '13 at 17:26
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You're missing the fact that the exterior square doesn't consist entirely of pure tensors (rather it is spanned by pure tensors).

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