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Let $L$ be a real Lie algebra. We can consider $L_{\mathbb{C}}:=L \otimes_{\mathbb{R}} \mathbb{C}$ the complexification of $L$. So on $L$ we can define the bracket operation: $[x \otimes z, y \otimes w] := [x,y] \otimes zw$ for all $x,y \in L$ and $z, w \in \mathbb{C}$. I can prove that $L_{\mathbb{C}}$ is a Lie algebra verifying the bilinearity, alternating and Jacobi identities. Is there a better way, which is maybe more elegant?
How can I look at $L$ as complex vector space?

rschwieb
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ArthurStuart
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  • What do you mean by «how can I look at $L$ as complex vector space?» (Presumably, your first sentence should have been «Let $L$ be a real Lie algebra», btw) – Mariano Suárez-Álvarez Feb 17 '13 at 09:16
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    I doubt there is anything more elegant. Just show it works be forget about it: the quest for elegance is pretty silly at times... – Mariano Suárez-Álvarez Feb 17 '13 at 09:16
  • @MarianoSuárez-Alvarez 1) Yes, $L$ is a real Lie algebra. 2) For more elegant I mean a method without a lot of calculations. Infact $$ [[x\otimes z_1, y\otimes w],v\otimes w_1] + [y\otimes w,[x\otimes z_1, v\otimes w_1]]=$$ $$= [[x,y]\otimes z_1w,v\otimes w_1] + [y\otimes w,[x,v]\otimes z_1w_1] = [[x,y],v]\otimes z_1ww_1 + [y,[x,v]]\otimes z_1ww_1 =$$

    $$= [x,[y,v]]\otimes z_1ww_1 = [x\otimes z_1,[y,v]\otimes ww_1] =[x\otimes z_1,[y\otimes w,v\otimes w_1]] $$... is quite long.

     – ArthurStuart Feb 17 '13 at 09:29
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    That is not a lot of calculation. A lot of calculation is 50 pages. – Mariano Suárez-Álvarez Feb 17 '13 at 09:30
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    If the algebra is a real Lie algebra please edit the question so that it says so. Always write questions so that they contain all the relevant information. – Mariano Suárez-Álvarez Feb 17 '13 at 09:31

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I find it easiest to view the complexified Lie algebra $L_{\mathbb{C}}$ as the set of all formal elements $$A+iB$$ such that $A$ and $B$ lie in $L$. This is clearly a real vector space, and become a complex vector space if we define $$i(A+iB)=-B+iA$$

It's then not hard to show that the Lie bracket on $L$ uniquely extends to a Lie bracket on $L_{\mathbb{C}}$, using the $\mathbb{C}$-bilinearity. The calculations are very slightly messy, but quite intuitive.

I agree that the tensor product viewpoint above is quite inelegant. It looks to me like notational overkill for a simple concept!

Edward Hughes
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    I think this "method of complexification" is specially better when you try to understand real forms of Lie algebras (http://en.wikipedia.org/wiki/Real_form_%28Lie_theory%29) – Júlio César Feb 17 '13 at 12:51
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    This is certainly a very good way to get your head around the complexification. The tensor viewpoint above has an advantage, though, of helping you understand the functoral properties of the complexification functor. – rschwieb Feb 17 '13 at 13:11