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I want to prove: No Lie algebra over $\Bbb R$ or $\Bbb C$ can have a unit element.

Now I am not sure how to take this in regard to the Lie bracket. I.e. I have now idea where to start. $[x,e]=[e,x]=x$ or something? That doesn't make sense to me, since we are just looking at an abstract bracket it seems.

So many hats
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    The zero Lie algebra has a unit element, namely zero. – ಠ_ಠ Aug 10 '15 at 02:59
  • What is the zero Lie algebra? @ಠ_ಠ – So many hats Aug 10 '15 at 03:03
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    @LieAlgebra This gets into a little bit of a weird territory in linear algebra. You can define the vector space of one element ($0$) and define it so that all operations to it give $0$. It's a zero-dimensional space since it has no basis which is a bit weird. It's not a useful structure to consider and it's probably omitted entirely (subtly). – Cameron Williams Aug 10 '15 at 03:09

3 Answers3

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If $[x,e]=[e,x]$, then what does the antisymmetry of the Lie bracket tell you?

ajd
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It would also contradict Jacobi's identity. If your Lie algebra is commutative, then clearly you can't have that $[x,e]=x$ unless $x=0$ since $[x,e]$ would have to be $0$. If it is non-commutative (meaning that for some $x,y$ in the Lie algebra we have that $[x,y]\neq 0$), then Jacobi's identity would say that

$$0 = [x,[y,e]] + [y,[e,x]] + [e,[x,y]] = [x,y] + [y,x] + [x,y] = [x,y]$$

by anti-symmetry. This holds for all $x,y$ in the Lie algebra which is a contradiction.

Cameron Williams
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  • Hmmm I am confused. Commutativity here means $[x,y]=[y,x]$ and by anti commutativity with commutativity we get $[x,y]=-[y,x]=[y,x]=-[x,y]$ so then $[x,y]=0$ always when the Lie algebra is commutative. Is that right? Is there any point to commutative Lie algebras if so?

    Other than that question above your answer looks awesome

     – So many hats Aug 10 '15 at 03:02
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    There is absolutely no reason to study commutative Lie algebras, honestly. You lose the Lie structure altogether, but I needed to include it to get rid of that case to get a proper contradiction in the second part. Commutative Lie algebras are just vector spaces with an operation that always evaluates to zero.. which is of course useless structure. – Cameron Williams Aug 10 '15 at 03:04
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    $\large\color{green}{\checkmark}$ - Thanks very much! – So many hats Aug 10 '15 at 03:09
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    Neat check mark. I've never seen that done in $\rm\LaTeX$ before. – Cameron Williams Aug 10 '15 at 03:11
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    @CameronWilliams There is a reason to study commutative algebras satisfying the Jacobi identity. They turn out to be Jordan algebras, see here. – Dietrich Burde Aug 10 '15 at 09:07
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    @DietrichBurde that's interesting but it's a little bit different since the Jacobi identity there isn't one coming from the Lie bracket. I'll read some more though. The authors' writing style is very crisp. – Cameron Williams Aug 10 '15 at 12:59
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If $[e,x]=x$ for all $x$, what can you conclude from $0=[e,e]$?

Jyrki Lahtonen
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