Equivalence of two algebras

Question

I am a theoretical physicist and in the context of my research I find myself with two algebras which should both correspond to the $\mathfrak{sl}(2,\mathbb{R})$ Lie algebra, but I can't seem to find the isomorphism between the two. I use the following form of the algebra: \begin{equation} [L_-,L_+] = 2L_0, \quad [L_0,L_+] = L_+, \quad [L_0,L_-] = -L_-. \end{equation} The "other" algebra I have has the following form \begin{equation} [K_-,K_+] = 4K_0, \quad [K_0,K_+] = 2 K_+, \quad [K_0,K_-]= -2K_-. \end{equation} The statement is that the second algebra is also the special linear one. However scaling the $K_0$ generator to obtain the correct first commutation relation scales the other ones in the opposite sense $(K_0 \rightarrow \frac12\tilde{K}_0)$ \begin{equation} [K_-,K_+] = 2\tilde{K}_0, \quad [\tilde{K}_0,K_+] = 4 K_+, \quad [\tilde{K}_0,K_-]= -4K_-. \end{equation} Either my brain is failing me and there is a simple isomorphism or the two algebras are surprisingly not equivalent. Any help would be greatly appreciated!

Answer 1

Make $\tilde K_- = \frac12 K_-$, $\tilde K_+ = \frac12 K_+$, and $\tilde K_0 = \frac12 K_0$, and the "tilde $K$'s" satisfy the exact same relations as the corresponding $L$'s.