# What is the difference between $\ker( L \bigwedge L \overset{[-,-]}{\rightarrow} L )$ and $\langle a \wedge b \big| [a,b]=0\rangle$?

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