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I already know how to do the complexification of a real Lie algebra $\mathfrak{g}$ by the usual process of taking $\mathfrak{g}_\Bbb{C}$ to be $\mathfrak{g} \oplus i\mathfrak{g}$. Now suppose I take the approach of trying to complexify things using tensor products. I look at $\mathfrak{g} \otimes_\Bbb{R} \Bbb{C}$ with the $\Bbb{R}$ - linear map

$$\begin{eqnarray*} f : &\mathfrak{g}& \longrightarrow \mathfrak{g} \otimes_\Bbb{R} \Bbb{C} \\ &v&\longmapsto v \otimes 1. \end{eqnarray*}$$

Now suppose I have an $\Bbb{R}$ - linear map map $h : \mathfrak{g} \to \mathfrak{h}$ where $\mathfrak{h}$ is any other complex Lie algebra. Then I can define a $\Bbb{C}$ - linear map $g$ from the complexification $\mathfrak{g} \otimes_\Bbb{R} \Bbb{C}$ to $\mathfrak{h}$ simply by defining the action on elementary tensors as

$$g(v \otimes i) = ih(v).$$

I have checked that $g$ is a $\Bbb{C}$ - linear map. Now my problem comes now in that my $f,g,h$ have to somehow be compatible with the bracket on $[\cdot,\cdot]_\mathfrak{g}$ of $\mathfrak{g}$ and $[\cdot,\cdot]_\mathfrak{h}$ of $\mathfrak{h}$. This is because I don't want them to just be linear maps but also Real/Complex Lie algebra homomorphisms.

My question is: How do we define the bracket on the complexification? A reasonable guess would be $[v \otimes i,w \otimes i] = \left([v,w] \otimes [i,i]\right)$ but this is zero.

Edit: Perhaps I should add, in the usual way of defining the complexification, the bracket on $\mathfrak{g}$ extends uniquely to one on the complexification $\mathfrak{g} \oplus i\mathfrak{g}$. Should it not be the case now that my bracket on $\mathfrak{g}$ extends uniquely to one on the tensor product then?

Edit: How do we know that the Lie Bracket defined by MTurgeon is well-defined? Does it follow from the fact that we are tensoring vector spaces, and so there is one and only one way to represent a vector in here?

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2  
Have you tried $[v\otimes \lambda,w\otimes \mu]=[v,w]\otimes \lambda\mu$? –  M Turgeon Aug 30 '12 at 12:35
    
@MTurgeon I think that should work. Guess I should not have asked this question so quickly. –  user38268 Aug 30 '12 at 12:38
    
@BenjaLim When we say "uniquely" we usually mean "uniquely satisfying some conditions". Do you have such conditions in mind? Extending implies that it should "restrict to the old bracket on $\mathfrak{g}$, but shouldn't there be a restriction about how it behaves on the $\mathbb{C}$-half of the tensor? My feeling could be completely wrong... –  rschwieb Aug 30 '12 at 13:56

3 Answers 3

up vote 4 down vote accepted

First of all, it seems the right extension is the following: $$[v\otimes\lambda,w\otimes\mu]:=[v,w]\otimes\lambda\mu.$$

This satisfies bilinearity, and Jacobi's identity. However, how can we show that this is the unique extension of the Lie bracket? We have the following result (taken, for example, from Bump's Lie groups):

Proposition: If $V$ and $U$ are real vector spaces, any $\mathbb R$-bilinear map $V\times V\to U$ extends uniquely to a $\mathbb C$-bilinear map $V_{\mathbb C}\times V_{\mathbb C}\to U_{\mathbb C}$.

Proof: This basically follows from the properties of tensor products. Any $\mathbb R$-bilinear map $V\times V\to U$ corresponds to a unique $\mathbb R$-linear map $V\otimes_{\mathbb R} V\to U$. But any $\mathbb R$-linear map extends uniquely to a $\mathbb C$-linear map of the complexified vector spaces (this is easy to prove). Hence, we have a $\mathbb C$-linear map $(V\otimes_{\mathbb R} V)_{\mathbb C}\to U_{\mathbb C}$. But we have the following isomorphism: $$(V\otimes_{\mathbb R} V)_{\mathbb C}\cong V_{\mathbb C}\otimes_{\mathbb C} V_{\mathbb C};$$ on the left-hand side, the tensor product is over $\mathbb R$, and on the right-hand side, it is over $\mathbb C$. Finally, our $\mathbb C$-linear map $V_{\mathbb C}\otimes_{\mathbb C} V_{\mathbb C}\to U_{\mathbb C}$ corresponds to a unique $\mathbb C$-bilinear map $V_{\mathbb C}\times V_{\mathbb C}\to U_{\mathbb C}$.

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Thanks. Am I right in saying above you are doing $V \otimes_\Bbb{R} V$? –  user38268 Aug 30 '12 at 14:46
    
@BenjaLim Yes. I will add the subscript so it is no longer ambiguous. –  M Turgeon Aug 30 '12 at 14:47
    
I have edited my answer above concerning the isomorphism that you quoted. –  user38268 Aug 30 '12 at 14:59
    
Actually don't we have to check well-definiteness of the bracket that you defined above? Or is this immediate from the fact that we have bases to work with? –  user38268 Aug 31 '12 at 11:20

Another guess would be this?

$[v \otimes i,w \otimes i] = [v,w] \otimes i$

or $[v \otimes i,w \otimes i] = [v,w] \otimes -1$?

I'm not quick enough to verify. In any case it seems you'd want to confine the defined bracket's working parts to the Lie algebra, and avoid involving the commutative ring.

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I got three votes almost instantly but I have no idea which of my gueses is right :P –  rschwieb Aug 30 '12 at 12:39
    
Please see my edit above. –  user38268 Aug 30 '12 at 12:55
2  
I don't think your first suggestion satisfies bilinearity. As for myself, I voted for your second suggestion :) –  M Turgeon Aug 30 '12 at 13:00
    
@MTurgeon How do we know that the bracket extends uniquely to the tensor product? I guess that should have been my main question. –  user38268 Aug 30 '12 at 13:04
    
@BenjaLim I suggest you have a look at Chapter 11 in Daniel Bump's Lie groups –  M Turgeon Aug 30 '12 at 14:10

Using the suggestion of MTurgeon we check that $g$ is a compatible with the Lie bracket on the complexification:

$$\begin{eqnarray*} g[v\otimes \lambda, w\otimes \mu] &=& g([v,w]_\mathfrak{g}\otimes \lambda\mu)\\ &=&\lambda\mu g([v,w]_\mathfrak{g} \otimes 1)\\ &=&\lambda \mu h([v,w]_\mathfrak{g}) \\ &=&\lambda\mu \big[h(v),h(w)\big]_\mathfrak{h}\\ &=& \big[ \lambda h(v),\mu h(w)\big]_\mathfrak{h}\\ &=&\bigg[g(v \otimes \lambda),g(w \otimes \mu) \bigg]_\mathfrak{h}. \end{eqnarray*}$$

Edit: For those curious about MTurgeon's isomorphism below, recall this is coming from a more general result concerning isomorphisms involving extensions of scalars. The relevant isomorphism I will put here is Theorem 6.15 (2) in here.

Let $M$ be an $R$ - module and $N$ and $S$ - module with $f : R\to S$ a ring homomorphism. Then the $S$ - module $M \otimes_R N$ is isomorphic to $(S \otimes_R M)\otimes_S N$ by sending $1 \otimes m$ to $(1 \otimes m) \otimes n$.

MTurgeon's isomorphism now falls out applying $R= \Bbb{R}$, $S = \Bbb{C}$, $M = V$ and $ N = V \otimes_\Bbb{R} \Bbb{C}$. Recall that $N$ is a $\Bbb{C}$ - module by extension of scalars with complex multiplication defined on elementary tensors as

$$\alpha( v \otimes \beta) = v \otimes (\alpha\beta)$$

for all $\alpha,\beta \in \Bbb{C}$ and $v \in V$.

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