homework and exercises - Half Witt algebra

I have the following Lie algebra which is generated by $\{L_n|n\geq 0\}.$ It satisfies the following commutation rule $$ \Big[ L_i ,L_j \Big]=\frac18 \frac{(2i+2j-1)(2j-2i)}{(2j+1)(2i+1)}L_{i+j-1}-\frac12 \frac{(2i+2j+1)(2j-2i)}{(2j+1)(2i+1)}L_{i+j}.$$ My question is that if the algebra algebra defined above is a half witt algebra?

I thought we can conclude by seeing the commutator relation of $L_n$ but it can happen there exists a change of basis say $\{V_n| n\geq 0\}$ such that where $$V_n=\sum_{\text{finite sum over some indedx}}L_k $$ such that $$\Big[ V_i ,V_j \Big]=[i-j]V_{i+j}~?$$
How can I conclude it is not a witt algebra etc?

Answer

We can start with the obvious rescaling $$ \tilde{L}_i = (2i + 1)L_i\,, $$ which turns the algebra to $$ [\tilde{L}_i,\tilde{L}_j] = (i-j)\left(\tilde{L}_{i+j}-\frac{1}{4}\tilde{L}_{i+j-1}\right)\,. $$ Then one can tentatively postulate a change of basis of the form $$ V_i = \sum_{m=0}^i a_m\,\tilde{L}_m\,. $$ Let's work out the algebra, let's just say without loss of generality that $i>j$. $$ [V_i,V_j] = \sum_{m=0}^i\sum_{n=0}^j a_ma_n\,(m-n)\left(\tilde{L}_{m+n}-\frac{1}{4}\tilde{L}_{m+n-1}\right)\,, $$ change summation variable to $(m,n) \to (M = m+n, N = m-n)$ $$ \begin{aligned} \,[V_i,V_j] &= \sum_{M=0}^{i+j}\sum_{\substack{N=\max(-M,M-2j)\\N\equiv M \mod 2}}^{\min(M,2i-M)}a_{\frac{M+N}{2}}a_{\frac{M-N}{2}} N\left(\tilde{L}_{M}-\frac{1}{4}\tilde{L}_{M-1}\right)=\\&= \sum_{M=1}^{i+j}\sum_{\substack{N=\max(-M,M-2j)\\N\equiv M \mod 2}}^{\min(M,2i-M)}a_{\frac{M+N}{2}}a_{\frac{M-N}{2}} N\,\tilde{L}_{M}-\frac{1}{4}\sum_{\tilde{M}=0}^{i+j-1}\sum_{\substack{N=\max(-\tilde{M}-1,\tilde{M}+1-2j)\\N\equiv \tilde{M}+1 \mod 2}}^{\min(\tilde{M}+1,2i-\tilde{M}-1)}a_{\frac{\tilde{M}+N+1}{2}}a_{\frac{\tilde{M}-N+1}{2}} N\,\tilde{L}_{\tilde{M}} \end{aligned} $$ Where I replaced $\tilde{M} +1 = M$ in the second term. Now we can group everything by simply dropping the $\sim$ and obtain an equation of the form $$ [V_i,V_j] = \sum_{M=0}^{i+j} b_M^{(i,j)}\,\tilde{L}_M\,, $$ for some coefficients $b_M^{(i,j)}$ that are defined as the whole mess above. Finally the equation seeked it $$ b_M^{(i,j)} = (i-j)\, a_M\,. $$ I have no idea if this equation has solutions or not, but it's more concrete so perhaps one can try and see if there are solutions for the first two or three values of $i,j$. If you find it, there might be a way to prove that it exists always. If you find an obstruction it might be that the change of basis I chose was not general enough or that there are no solutions indeed. The inexistence of solutions implies that the algebras are not isomorphic.

This is not by any means a complete answer, but hopefully it goes in the right direction.