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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).2012-04-08
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    @QiaochuYuan Thanks a lot, that's it. I got confused about that basis.2012-04-08
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    @QiaochuYuan Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-18

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