# Simultaneous shifted diagonalization of bunch of operators

I have the following Lie algebra which is generated by $$\{L_n|n\geq 0\}.$$ It satisfies the following commutation rule

$$\Big[ L_i ,L_j \Big]= (i-j)(L_{i+j}-\frac14 L_{i+j-1}).....*$$ My question is the above algebra is still Half witt. That is if there exist a basis $$V_n$$ which are finite linear combination of $$L_n$$ such that $$\Big[V_i,V_j\Big]=(i-j)V_{i+j}$$

Half Witt algebra

We made the problem more concrete. Our question reduces to the existence of a solution of infinitely many quadratic equations. Though I still could not show the existence or disprove it. I have the following idea which I want to make some progress. If someone can verify help it will be great.

Lets $$V$$ be the vector space with the ordered basis $$\Big\langle v_i \Big\rangle_{i=1}^{\infty}$$having a commutation relation given by $$*$$ imply there exists infinitely many operators $$T_{i}:V\rightarrow V \hspace{1cm} \forall i\in {1,\ldots,\infty}$$ such that $$T_i(v_j)=(i-j)(v_{i+j}+\lambda v_{i+j-1})$$ where $$\lambda$$ is a constant.

Notice $$V$$ is infinite dimensional. Let write the matrix of $$T_i$$ wrt to the basis $$v_i$$. The $$i-th$$ column of the matrix is consist of all zeroes and the $$j-th$$ column will have $$(i-j)\lambda$$ in the $$(i+j-1,j)$$ position and $$(i-j)$$ in the $$(i+j,j)$$. These matrix look like matrix in Jordon canonical form.

Let define our idealistic operator as $$T_{i}^{*}:=V\rightarrow V$$ such that $$T_{i}^{*}(v_j)=(i-j)v_{i+j}$$. The matrix of this operator look like a shifted diagonal matrix.

If there exists a change of basis such that we can transform $$T_i$$ to $$T_i^{*}$$ simultaneously.

Each $$T_i$$ is diagonalisable.(I might be wrong.) as the Jordan canonical form have different eigenvalues. Also, this matrix is not finite not sure how much I use the theorem of the finite dimensional matrix.

If some progress can be made through this I would be very curious to know.

• You might condense your algebra by reducing to the minimal Presentation picture, or the smoother nonminimal ones. – Cosmas Zachos Feb 24 at 14:42
• Thanks, I am reading the paper you suggested. My indices are all greater than zeros. – GGT Feb 24 at 22:59
• My indices are greater than 0 so I started with the generating set $L_0, L_1 , L_2$ and there is finitely many conditions that can generate the algebra I am looking for now is there any way I can show it's isomorphic with respect to a change of basis? – GGT Mar 12 at 1:49