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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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  • Possible duplicate: $su(1,1) \cong su(2)$? – Qmechanic Sep 27 '21 at 19:23
  • It doesn't seem the same question, my post Is about su(2) and sl(2, R), not about their complexification. Moreover I have referenced two books that appears to be contractidory on the same subject – Andrea Sep 27 '21 at 19:39
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    Well, Schwartz is just wrong at this point/implicitly talking about the complexification. – Qmechanic Sep 27 '21 at 20:04
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    I agree, the statement in Schwartz is probably a typo or he meant implicitly the complexifications (which in physics text are often not written explicitly). $\mathfrak{sl}(2,\mathbb{R})$ is isomorphic to $\mathfrak{so}(1,2)$, the Lie algebra of the 3d Lorentz group, while $\mathfrak{su}(2)$ is isomorphic to the Lie algebra of the "Euclidean version" of the 3d Lorentz group, i.e. $\mathfrak{so}(3)$. – G. Blaickner Sep 27 '21 at 20:41
  • He is talking about the complexified algebras. – Prahar Sep 28 '21 at 09:10
  • The complexified $sl(2;\Bbb R)$ is isomorphic to $sl(2; \Bbb C)$, so I'm wondering why it's not merely written as $sl(2; \Bbb C)$ if he's talking about the complexified algebra – Andrea Sep 28 '21 at 14:23

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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  • I believe that actually my proof shows that su(2) doesn't have ladder operators, cause It shows that equality [A,B]=2A can't be true. Do you know amy different way to prove the same? – Andrea Sep 28 '21 at 03:50
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there is a 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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