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On B.C.Hall book of Lie Algebra, it's clearly stated that $su(2)$ is not isomorphic to $sl(2, \Bbb R)$, the proof is left as an exercise.

My proposed proof use the fact that Hermitian matrix can be diagonalized through unitary matrix to show that there's no isomorphism that preserve the Lie bracket:

$X,Y,Z \in sl(2,\Bbb R)$

where the usual brackets are:

$[X,Y] = Z\,,\,[X,Z] = -2X\,,\,[Y,Z] = 2Y$

The isomorphism $\phi:sl(2,\Bbb R) \to su(2)$ must preserve the brackets, let's use the last one

$\phi(2Y) = \phi([Y,Z]) = [\phi(Y),\phi(Z)]$

$A\equiv \phi(Y)\in su(2)\,,\,B\equiv\phi(Z)\in su(2)$

$[A,B] = 2A$

Since $A$ is Hermitian we have

$A = W^*\Lambda W\,,\,W^*W=I$, where $\Lambda$ is a diagonal matrix

Substituting in $[A,B] = 2A$ and easy calculations show that

$[\Lambda,H] = 2\Lambda$

where $H\equiv WBW^* \in su(2)\,,\,\Lambda \in su(2)$

So we have that:

$\Lambda = \left[ \begin{array}{cc} s&0\\ 0&-s \end{array} \right]\,,\,H = \left[ \begin{array}{cc} r&\gamma^*\\ \gamma&-r \end{array} \right]\,,\,r,s\in\Bbb R$ and $\gamma \in \Bbb C $

and sobstituing in $[\Lambda,H] = 2\Lambda$ it's easy to see that matrix equation is satisfied only if $s = 0$

which would imply $\phi(Y)=A=\Lambda = 0$ and then $Y=0$ (cause $\phi$ is an isomorfism), which is absurd, so the brackets can't be preserved.

Is that proof correct?

I am a bit confused cause in the book M.D.Schwartz "Quantum field theory" on pag.162 he states $sl(2,\Bbb R) = su(2)$ , which seems incorrect to me, cause that equation would have sense only if there is a Lie algebra isomorphism between $sl(2,\Bbb R)$ and $su(2)$.

Andrea
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2 Answers2

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  1. First of all, when considering real Lie algebras it is cleanest to use the mathematical convention, i.e. $su(2)$ consists of traceless anti-hermitian $2\times 2$ matrices.

  2. OP has already shown that $sl(2,\mathbb{R})$ (which is the set of traceless real $2\times 2$ matrices) contains 2 ladder operators $\sigma_{\pm}$, such that $[\sigma_3,\sigma_{\pm}]=\pm 2\sigma_{\pm}.$

  3. One way to show that the real Lie algebras $su(2)$ and $sl(2,\mathbb{R})$ are not isomorphic, is to show that $su(2)$ cannot contain a ladder operator. This is not hard.

  4. See also this related Phys.SE post.

Qmechanic
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No, they are not isomorphic, you can prove it by showing there is a 2-dimensional subgroup in SL(2,$\mathbb{R}$) but not in SU(2).

The 2-dimensional subgroup in SL(2,$\mathbb{R}$):
\begin{bmatrix} m & n \\ 0 & 1/m \end{bmatrix} you can prove that SU(2) cannot have a 2-dimensional subgroup by proving SO(3) cannot have a 2-dimensional subgroup, it is not hard to prove it by using the commutation relations in SO(3): $$ [\vec{a}\cdot\vec{J},\vec{b}\cdot\vec{J}] = (\vec{a}\times\vec{b})\cdot\vec{J} $$ thus two infinitesimal rotation will create a rotation belonging to another dimension:
$$ \mathrm{R}(\hat{a},\theta_1)\mathrm{R}(\hat{b},\theta_2)\mathrm{R}(\hat{a},\theta_1)^{-1}\mathrm{R}(\hat{b},\theta_2)^{-1}=\mathrm{R}(\hat{a}\times\hat{b},\theta_1\theta_2) $$

Diego Zhao
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