$su(1,1) \cong su(2)$? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-06-18T13:36:18Z https://physics.stackexchange.com/feeds/question/183414 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/183414 10 $su(1,1) \cong su(2)$? kaiser https://physics.stackexchange.com/users/71894 2015-05-11T15:05:30Z 2018-06-28T07:53:32Z <p>The three generators of $su(2)$ satisfy the commutation relations</p> <p>$$[J_0 , J_\pm] = J_\pm , \quad [J_+, J_- ] = +2J_0 .$$ </p> <p>The three generators of $su(1,1)$ satisfy the commutation relations</p> <p>$$[K_0 , K_\pm] = K_\pm , \quad [K_+, K_- ] = -2K_0 .$$</p> <p>Now, let us define </p> <p>$$K_0 = J_0, \; K_+ = J_+,\; K_- = - J_-.$$</p> <p>It is apparent that so defined $K$'s satisfy the $su(1,1)$ algebra! Does this mean that $su(1,1)$ is actually equivalent to $su(2)$? </p> <p>Where is the argument wrong? </p> https://physics.stackexchange.com/questions/183414/su1-1-cong-su2/183417#183417 13 Answer by Qmechanic for $su(1,1) \cong su(2)$? Qmechanic https://physics.stackexchange.com/users/2451 2015-05-11T15:26:07Z 2018-06-28T07:53:32Z <p>The <a href="http://en.wikipedia.org/wiki/Ladder_operator" rel="nofollow noreferrer">ladder operators</a> do belong to the real <a href="http://en.wikipedia.org/wiki/Lie_algebra" rel="nofollow noreferrer">Lie algebras</a>$^1$ \begin{align} su(1,1)&amp;:=~\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid m^{\dagger}\sigma_3=-\sigma_3m,~ {\rm tr}(m)=0\} ~=~{\rm span}_{\mathbb{R}}\{ \sigma_1, \sigma_2, i\sigma_3 \}\cr ~\cong~sl(2,\mathbb{R}) &amp;:=~\{m\in {\rm Mat}_{2\times 2}(\mathbb{R}) \mid {\rm tr}(m)=0\} ~=~{\rm span}_{\mathbb{R}}\{ \sigma_1, i\sigma_2, \sigma_3 \} ~=~{\rm span}_{\mathbb{R}}\{ \sigma_+, \sigma_-, \sigma_3 \}\cr ~\cong~ so(2,1)&amp;:=~\{m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^{t}\eta =- \eta m\}, \end{align} $$\sigma_{\pm}~:=~\frac{\sigma_1\pm i \sigma_2}{2}, \qquad \eta~=~{\rm diag}(1,1,-1),$$ but they do <em>not</em> belong to the real Lie algebras \begin{align} su(2)&amp;:=~\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid m^{\dagger}=-m,~ {\rm tr}(m)=0\} ~=~{\rm span}_{\mathbb{R}}\{ i\sigma_1, i\sigma_2, i\sigma_3 \} \cr ~\cong~ so(3)&amp;:=~\{m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^{t}=-m\}. \end{align} All the above 5 real Lie algebras have <a href="http://en.wikipedia.org/wiki/Complexification" rel="nofollow noreferrer">complexifications</a> isomorphic to $$sl(2,\mathbb{C})~:=~\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid {\rm tr}(m)=0\} ~=~{\rm span}_{\mathbb{C}}\{ \sigma_1, \sigma_2, \sigma_3 \}.$$</p> <p>--</p> <p>$^1$ Here we follow the mathematical definition of a real Lie algebra. Be aware that in much of the physics literature, the definition of a real Lie algebra is multiplied with a conventional extra factor of the imaginary unit $i$, cf. footnote 1 in my Phys.SE answer <a href="https://physics.stackexchange.com/a/365549/2451">here</a>.</p> https://physics.stackexchange.com/questions/183414/su1-1-cong-su2/183426#183426 14 Answer by Luboš Motl for $su(1,1) \cong su(2)$? Luboš Motl https://physics.stackexchange.com/users/1236 2015-05-11T16:10:33Z 2015-05-12T03:54:39Z <p>You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for the coefficients.</p> <p>A general matrix in the $SU(2)$ group is written as $$M = \exp[ i( \alpha J_+ + \bar\alpha J_- + \gamma J_0 )]$$ where $\alpha\in {\mathbb C}$ and $\gamma\in {\mathbb R}$ while the general matrix in $SU(1,1)$ is given by $$M' = \exp [ i( \alpha_+ J_+ + \alpha_- J_- + \beta J_0 ]$$ where $\alpha_+,\alpha_-,\beta\in {\mathbb R}$ are three different real numbers.</p> <p>To summarize, for $SU(2)$, the coefficients in front of $J_\pm$ are complex numbers conjugate to each other, while for $SU(1,1)$, they are two independent real numbers. (And I apologize that I am not sure whether the $i$ should be omitted in the exponent of $SU(1,1)$ only according to your convention. Probably.)</p> <p>If you allow all three coefficients in front of $J_\pm,J_0$ to be three independent complex numbers, you will obtain the complexification of the group. And as Qmechanic also wrote, the complexification of both $SU(2)$ and $SU(1,1)$ is indeed the same, namely $SL(2,{\mathbb C})$.</p>