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I stumbled across exercises asking to prove the following isomorphisms:

  1. $\mathfrak{sl}_2(\mathbb{R}) \cong \mathfrak{so}_{2,1}(\mathbb{R})$
  2. $\mathfrak{sl}_2(\mathbb{C}) \cong \mathfrak{so}_{3,1}(\mathbb{R})$
  3. $\mathfrak{so}_{2,2}(\mathbb{R}) \cong \mathfrak{sl}_2(\mathbb{R}) \oplus \mathfrak{sl}_2(\mathbb{R})$

I actually tried finding the isomorphisms. For the first one, I chose the basis $$i = \begin{pmatrix} 0&1&0\\-1&0&0\\0&0&0 \end{pmatrix}, \, j = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 1 & 0 & 0 \end{pmatrix}, \, k=\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}$$ and computed $[i,j] = -k$, $[i,k] = j$ and $[j,k]=i$. Of course the aim is to relate this to the commutators of the usual $\mathfrak{sl}_2$-triple. By adding $i$ and $j$, I get $2[i+j,k] = i+j$ which already looks promising, but I struggle to go on.

I don't really see the point in doing these calculations and I don't think spending hours trying to figure out some nice way to add the basis elements such that all works out is a very valuable exercise.

Could anyone give the isomorphisms explicitly or at least know a reference of it?

user302234
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  • There are many ways to prove it, explicitly or by clever arguments, but I want to stress that these isomorphisms are absolutely fundamental, especially in physics. Among other consequences, they're the reason why spin-1/2 particles such as the electron (with the spin up or down) may exist in the 3D space. – Luboš Motl Jan 19 '16 at 13:24
  • @LubošMotl: I don't know much about physics and this is my first introduction to Lie groups/Lie algebras. I'm very interested in reading about or finding those clever arguments. Could you maybe give a reference or at least a hint? – user302234 Jan 19 '16 at 13:29
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    Note that the Killing form of $\mathfrak{sl}_2(\Bbb R)$ is proportional to the determinant $$\det : \pmatrix{a & b\c & -a} \mapsto -a^2 - bc .$$ It is nondegenerate (this is also a consequence of the semisimplicity of $\mathfrak{sl}_2(\Bbb R)$), and its negative, $-\det$, has signature $(2, 1)$. – Travis Willse Jan 19 '16 at 13:34
  • By the way, can you choose your own explicit representation for $\mathfrak{so}_{2, 1]}(\Bbb R)$, or are you stuck with $\langle i, j, k\rangle$? Some are a little easier to use for this purpose than others. – Travis Willse Jan 19 '16 at 13:37
  • @Travis: This is not homework or to be handed in, so any argument that is within my scope is fine for me. – user302234 Jan 19 '16 at 13:38
  • In that case, might I suggest using the representation of $O(2, 1)$ corresponding to the antidiagonal bilinear form $$\pmatrix{0&0&1\0&1&0\1&0&0} ?$$ – Travis Willse Jan 19 '16 at 13:49
  • See also this MSE question. – Dietrich Burde Jan 19 '16 at 16:28

1 Answers1

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Let's start with the first example and use your $i,j,k$ for the $SO(2,1,R)$ matrices. The analogous $SL(2,R)$ matrices with the same commutators are multiples of the Pauli matrices $$ j = \frac 12 \pmatrix{ 0&1\\1&0 },\,\, i = \frac 12 \pmatrix{ 0&-1\\1&0 },\,\, k = \frac 12 \pmatrix{ 1&0\\0&-1 }, $$ You may easily verify that these three matrices anticommute with each other and the same commutators hold $$ [i,j]=-k,\,\,[i,k]=j,\,\,[j,k]=i $$ which is the only nontrivial part of the isomorphism.

The second isomorphism may be obtained simply by realizing that $SL(2,C)$ is the complexification of $SL(2,R)$ – we allow the coefficients to be complex, not real. And similarly $SO(3,1)$ is isomorphic to the complexification of $SO(3,R)$ or $SO(2,1,R)$, too. The latter statement is easily seen if we write the generators of $SO(3,1,R)$ as $J_{ij}$ which are $ij$-antisymmetric and write $$ K^\pm_3 =\frac 12 ( J_{12} \pm J_{34}) $$ and the 123-cyclic permutations of this expression. With this definition, $K^\pm_{1,2,3}$ form two separate algebras that are isomorphic to $SO(3)$ or $SO(2,1)$, one only has to distinguish the signs and reality conditions.

For $SO(3,1)$, the generators $K^+_i$ are the Hermitian conjugates to $K^-_i$, so the coefficients have to be complex conjugate to each other as well, and we get one complexified $SL(2,C)$ algebra. For $SO(4,R)$ or $SO(2,2,R)$, which is your last cases, one finds out that $K^+$ and $K^-$ are independent of each other, even through Hermitian conjugation, and finds out that these two algebras are isomorphic to $SU(2) \times SU(2)$ or $SO(2,1,R)\times SO(2,1,R)$, respectively (these two cases differ by some signs already apparent in the 3-dimensional algebras).

It's useful to go through the calculation using a particular form of the matrices at least once. But the final result may be expressed in many ways and is independent of the choice of the bases etc. And it may be seen to hold in many ways. There are really not too many 3- or 6-dimensional Lie algebras.

Those results are basics of Lie group/algebra theory and their importance especially in physics cannot be overstated.

Luboš Motl
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  • Thanks for your detailed answer. I don't quite understand what you mean by $J_{ij}$ for the second isomorphism. If I haven't made a mistake, the "obvious" basis for $\mathfrak{so}{3,1}(\mathbb{R})$ consists of three antisymmetric and 3 symmetric matrices. $J{12}, J_{13}, J_{23}$ would be the antisymmetric ones (with entry $1$ at $ij$ and $-1$ at $ji$). Did I misunderstand the meaning of $\mathfrak{so}{3,1}(\mathbb{R})$ or do you mean something else by $J{ij}$? – user302234 Jan 20 '16 at 13:15
  • Hi @user302234 - yes, I meant exactly that matrix, $1$ at $ij$ and $-1$ at $ji$, otherwise zeroes. The funny thing is that these matrices have the commutators $[J_{ij},J_{kl}] = J_{il}\delta_{jk}-...-...+...$, four similar terms in total, and one may find completely different matrices of different size that have exactly the same commutators. For example, one may find them in terms of the "Dirac gamma matrices" and their products. The Dirac matrices are $2^N\times 2^N$ matrices of various sorts, tensor products of many copies of Pauli matrices and the identity. They anticommute with each other. – Luboš Motl Jan 21 '16 at 08:17
  • This successful localization of "very different matrices" with the same commutator algebras is equivalent to finding "very different representations" of the Lie algebra (and therefore the Lie group, too). The Dirac matrices are matrices expressing the generators of the Lie algebra with respect to "spinor representations". The dimension of the spinor representation is the power of two. Instead of just two nonzero entries, $J_{ij}$ in the spinor representation have one nonzero entry in each row (and in each column). – Luboš Motl Jan 21 '16 at 08:19