0

In the chapter 7 problems of Hall's Lie Groups, Lie Algerbas and Representations we have the following one:

  1. Show that the real Lie algebra $\mathfrak{s}\mathfrak{o}(3,1)$ is isomorphic to $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})_{\mathbb{R}}$ (If $\mathfrak{g}$ is a complex simple Lie algebra let $\mathfrak{g}_\mathbb{R}$ denote the Lie algebra $\mathfrak{g}$ viewed as a real Lie algebra of twice the dimension). Conclude that $\mathfrak{s}\mathfrak{o}(3,1)$ is simple as a real Lie algebra, but that $\mathfrak{s}\mathfrak{o}(3,1)_\mathbb{C}$ is not simple and is isomorphic to $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})\oplus\mathfrak{s}\mathfrak{l}(2,\mathbb{C})$.

Also says: Hint: First show that $\mathfrak{s}\mathfrak{o}(3,1)_{\mathbb{C}}\cong\mathfrak{s}\mathfrak{l}(2,\mathbb{C})\oplus\mathfrak{s}\mathfrak{l}(2,\mathbb{C})$ and then show that the two copies of $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})$ are conjugates of each other.

So, I'm a physics student and I know how to do this problem using the conmutation relations of $\mathfrak{s}\mathfrak{o}(3,1)$ but now that I'm trying to understand Lie theory formally I dont even know where to start this problem. The hint even confuses me in that I dont know how to show they are conjugates of each other.

Also, in physics we use $\mathfrak{s}{u}(2)$ not $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})$ I know that $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})\cong \mathfrak{s}{u}(2)_{\mathbb{C}}$ but still confuses me. In general I'm very confunsed about all of this.

Also, I want to ask what's the relation of all of this with the Dirac Clifford algebra doesn't have to do with the problem but I know there is, obviously because we use it, but I don't see it at this formal level.

Any light will be welcomed!

olsrcra
  • 33
  • 2
    "I know that $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})_{\mathbb{C}}\cong \mathfrak{s}{u}(2)$"- is plain wrong. – Moishe Kohan Oct 26 '23 at 01:20
  • @MoisheKohan omg sorry I write it down in reverse. I'm editing it sorry and thanks! – olsrcra Oct 26 '23 at 02:07
  • 1
    Cf. https://math.stackexchange.com/q/639749/96384, https://math.stackexchange.com/a/3258221/96384 (especially further links in "Related Questions") , https://math.stackexchange.com/q/1109369/96384, and recently https://math.stackexchange.com/q/4776715/96384 for that "complexification creates two copies" business. – Torsten Schoeneberg Oct 26 '23 at 03:54
  • @TorstenSchoeneberg I have read some of that posts but now I really think i'm even more confused. Are the following isomorphisms true? $(\mathfrak{s}\mathfrak{o}(3,1){\mathbb{C}}){\mathbb{R}} \cong (\mathfrak{s}\mathfrak{l}(2,\mathbb{C}){\mathbb{C}}){\mathbb{R}} \cong \mathfrak{s}\mathfrak{l}(2,\mathbb{C}){\mathbb{R}}\oplus \mathfrak{s}\mathfrak{l}(2,\mathbb{C}){\mathbb{R}}$ and if so how do i get from there the result $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})_{\mathbb{R}} \cong \mathfrak{s}\mathfrak{o}(3,1)$ ? – olsrcra Oct 27 '23 at 01:41
  • @TorstenSchoeneberg I thought of $(\mathfrak{s}\mathfrak{o}(3,1) \oplus \mathfrak{s}\mathfrak{o}(3,1) ){\mathbb{R}} \cong \mathfrak{s}\mathfrak{o}(3,1){\mathbb{R}} \oplus \mathfrak{s}\mathfrak{o}(3,1){\mathbb{R}} \cong \mathfrak{s}\mathfrak{o}(3,1) \oplus \mathfrak{s}\mathfrak{o}(3,1)\cong \mathfrak{s}\mathfrak{l}(2,\mathbb{C}){\mathbb{R}}\oplus \mathfrak{s}\mathfrak{l}(2,\mathbb{C}){\mathbb{R}}$ and then $\mathfrak{s}\mathfrak{o}(3,1) \cong \mathfrak{s}\mathfrak{l}(2,\mathbb{C}){\mathbb{R}}$ but although I get the result I don't think is done properly. – olsrcra Oct 27 '23 at 01:56
  • ah and of course im my first comentary of these 3 im saying that $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})_{\mathbb{C}}=\mathfrak{s}\mathfrak{l}(2,\mathbb{C}) \oplus \mathfrak{s}\mathfrak{l}(2,\mathbb{C})$ assuming that the two $\mathfrak{s}\mathfrak{l}(2,\mathbb{C})$ copies are conjugates of each other but, although I know how this is done using conmutation, realations, I dont how a more abstract, If that's the point, way of doing so. – olsrcra Oct 27 '23 at 02:14
  • Well in your first comment, you just use what you want to prove, for the last isomorphism. -- I do not understand what you say or ask in your latest comment. – Torsten Schoeneberg Oct 29 '23 at 21:31
  • @TorstenSchoeneberg what isomorphism? how can I do it then? – olsrcra Oct 30 '23 at 01:20
  • I mean, it seems you want to conclude from $\mathfrak{s}\mathfrak{o}(3,1)\oplus\mathfrak{s}\mathfrak{o}(3,1)\cong\mathfrak{s}\mathfrak{l}(2,\mathbb{C}){\mathbb{R}}\oplus\mathfrak{s}\mathfrak{l}(2,\mathbb{C}){\mathbb{R}}$ that $\mathfrak{s}\mathfrak{o}(3,1)\cong\mathfrak{s}\mathfrak{l}(2,\mathbb{C})_{\mathbb{R}}$, but how do you show the first isomorphism without already using the second? – Torsten Schoeneberg Oct 30 '23 at 16:00
  • As for "how to prove it then". Well either you follow the methods in https://math.stackexchange.com/q/4759891/96384, or you follow the hint there. For that, the first thing to do is show the complexification of $\mathfrak{so}(3,1)$ is a direct sum of two copies of $\mathfrak{sl}_2(\mathbb C)$. For that, you might want to look at explicit bases and relations, e.g. the $K_j$ and $J_j$ in https://math.stackexchange.com/a/3258221/96384. – Torsten Schoeneberg Oct 30 '23 at 16:01
  • Then, you are supposed to show that complex conjugation flips those two copies. Again, this is probably best seen by continuing using certain basis elements. --- Once one has that, now it is time for a hammer from a big theory. When you are given a complex Lie algebra which you know is given as complexification $\mathbb C \otimes \mathfrak g$ of a real form $\mathfrak g$, and you have how complex conjugation operates on the complex version, then you get the real form by looking at the fixed space of conjugation. Now if conjugation flips those two copies, the fixed space can be identified ... – Torsten Schoeneberg Oct 30 '23 at 16:05

0 Answers0