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As six generators of the real Lie algebra $\mathfrak sl(2,\Bbb C)_\Bbb R$ I can use the Pauli matrixes as follow:

$X_1=\frac{1}{2} \sigma_1, X_2=\frac{1}{2} \sigma_2, X_3=\frac{1}{2} \sigma_3$

$Y_1=\frac{1}{2}i \sigma_1, Y_2=\frac{1}{2}i \sigma_2, Y_3=\frac{1}{2}i \sigma_3$

cause it's easy to see that they have null traces and they are linearly independent on $\Bbb R$ span.

Based on Hall "Lie Groups, Lie Algebras, and Representations" page 66, a real Lie algebra $\mathfrak g $ of complex $n \times n $ matrix can be complexified only if $iX \notin \mathfrak g $ for every $X \in \mathfrak g$.

The generators above obviously doesn't satisfy that conditions, cause $iX_i = Y_i \in \mathfrak sl(2,\Bbb C)_\Bbb R$.

So it seems I can't complexify $\mathfrak sl(2,\Bbb C)_\Bbb R$ using the generators above. However, on some books of QFT the complexification of $\mathfrak sl(2,\Bbb C)_\Bbb R$ is mentioned using as generators the standard generators of the rotations and boosts, and it's mentioned also in the following Wiki:

https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group

So my questions are:

  1. Does the complexification exist or doesn't exist depending on the generators used?
  2. Is $\mathfrak sl(2,\Bbb C)_\Bbb C \cong sl(2,\Bbb C)_\Bbb R \oplus sl(2,\Bbb C)_\Bbb R$ ?
Andrea
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  • Please quote exactly what Hall writes there. If by complexification of a real Lie algebra $L$ we mean $\mathbb C \otimes_{\mathbb R} L$ (and I don't think I have ever seen it mean anything else), then of course one can complexify every real Lie algebra. – Torsten Schoeneberg Oct 08 '21 at 15:39

1 Answers1

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As regards question 2, you seem to be using subscripts for two different things. If subscript $\mathbb C$ stands for complexification but subscript $\mathbb R$ stands for scalar restriction to $\mathbb R$, then a) please use a different notation and b) it is still not true. Cf. answers to Are Lie algebra complexifications $\mathfrak g_{\mathbb C}$ equivalent to Lie algebra structures on $\mathfrak g\oplus \mathfrak g$? and Is the complexification of $\mathfrak{sl}(n, \mathbb{C})$ itself?.

What is true is

$$sl(2, \mathbb C) \otimes_{\mathbb R} \mathbb C\simeq sl(2, \mathbb C) \oplus sl(2, \mathbb C)$$

where both sides are complex Lie algebras and the isomorphism is one of complex LAs. If you view both sides as real Lie algebras via scalar restriction (and we denote that by the subscript $\mathbb R$), then this implies, of course, that there is an isomorphism of real Lie algebras

$$(sl(2, \mathbb C) \otimes_{\mathbb R} \mathbb C)_\mathbb R \simeq sl(2, \mathbb C)_{\mathbb R} \oplus sl(2, \mathbb C)_{\mathbb R}.$$

[So the left hand side is the complexification but viewed as a real algebra again, i.e. complexification followed by scalar restriction; if you insist on sticking to your confusing notation -- which you shouldn't -- you should write the left hand side as $(sl(2, \mathbb C)_{\mathbb C})_{\mathbb R}$.]

Torsten Schoeneberg
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  • I am using the same $_\Bbb C$ notation as the wiki page https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group, in "The Lie Algebra" section, my understanding is that it means complexification, but I may be wrong and it means the $\Bbb C$ scalar, not the complexification – Andrea Oct 08 '21 at 16:21
  • However, Hall book seems to use $\Bbb C$ to mean complexification, cause it says as an example " The complex Lie algebra $ \mathfrak g\Bbb C$ is called the complexification of the real Lie algebra $ \mathfrak g$." (pag 65), now I'm really confused – Andrea Oct 08 '21 at 16:32
  • Yes, both sources use the subscript for complexification (and that is quite standard, and so do I myself in the answers to the linked questions). But in your question you use the subscript $\mathbb R$ for scalar restriction. I am sure one can find some other source that also does it. But is it not clear that it's a bad idea to use it for both? – Torsten Schoeneberg Oct 09 '21 at 03:48
  • Yes, you are tight, I was confused cause I thought complexification of real LA was equivalent to the complex LA (that Is the LA on $\Bbb C$ field instead of $\Bbb R$), but after reading your answer I get I was wrong – Andrea Oct 09 '21 at 07:03
  • Actually I'am also confused by the use of $\oplus$ symbol, as far as I know direct sum of vector space is defined when $V_1 \cap V_2 = \emptyset$, since LAs are vector spaces I don't understand the meaning of $sl(2,\Bbb C) \oplus sl(2,\Bbb C)$. What am I missing or thinking wrong? – Andrea Oct 09 '21 at 07:22
  • There is an abstract notion of direct sum of vector spaces. You can write $\mathbb R^2 \simeq \mathbb R \oplus \mathbb R$, can you not? To be super precise, one can write $\simeq \mathbb R\pmatrix{1\0} \oplus \mathbb R \pmatrix{0\1}$ or something. Those are "two distinct copies" of $\mathbb R$, and in your situation it are two distinct copies of $sl(2, \mathbb C)$. They are structurally identical (like clones), but not the same, they indeed would have empty intersection. – Torsten Schoeneberg Oct 09 '21 at 15:54
  • Actually by reading the following wiki https://en.wikipedia.org/wiki/Complexification it seems that it can been written more precise as $sl(2,\Bbb C) \oplus isl(2,\Bbb C)$ – Andrea Oct 09 '21 at 19:26
  • Yes and no. One has an iso $L_{\mathbb C} \simeq L \oplus iL$ of real vector spaces for any real Lie algebra $L$. However, notice that the second summand is not closed under the Lie bracket, and neither is a complex vector space. What is remarkable in the case $L = sl(2, \mathbb C)$ viewed as a real Lie algebra is that there is also an isomorphism of complex Lie algebras $sl(2, \mathbb C) \simeq sl(2, \mathbb C) \oplus sl(2, \mathbb C)$ i.e. there exist ... – Torsten Schoeneberg Oct 10 '21 at 01:49
  • ... there exist two distinct complex subalgebras, with intersection $0$, each of them $\simeq sl(2, \mathbb C)$, in the LHS. Those are not the real subspaces $1 \otimes sl(2, \mathbb C)$ and $i \otimes sl(2, \mathbb C)$ in the other decomposition. Rather, those complex $sl(2, \mathbb C)$-factors lie rather "skew" inside the tensor product. It's agood exercise to find them explicitly. --- Note that the comments and my answer to the first linked post discuss very similar issues. – Torsten Schoeneberg Oct 10 '21 at 01:51