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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 $$ \Big[V_i,V_j\Big]=(i-j)V_{i+j}$$

Some progress we made about this question appears here

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.

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  • $\begingroup$ You might condense your algebra by reducing to the minimal Presentation picture, or the smoother nonminimal ones. $\endgroup$ – Cosmas Zachos Feb 24 at 14:42
  • $\begingroup$ Thanks, I am reading the paper you suggested. My indices are all greater than zeros. $\endgroup$ – GGT Feb 24 at 22:59
  • $\begingroup$ 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? $\endgroup$ – GGT Mar 12 at 1:49

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