Suppose we have a finite dimensional Lie algebra with basis:
$X_1 = \partial_t,\;X_2 = \beta_2\,\partial_T-\beta_1\,\partial_C,\;X_3 = x_3\,\partial_p+\frac{1}{\beta_2\rho_0 g}\,\partial_C$
$X_4 = 2t\,\partial_t+\sum_i^3 \left(x_i\,\partial_{x_i}-u_i\,\partial_{u_i}\right)-2p\,\partial_p-3T\,\partial_T-3C\,\partial_C$
$X_5 = x_1\,\partial_{x_2}-x_2 \,\partial_{x_1}+u_1\,\partial_{u_2}-u_{2}\,\partial_{u_{1}}, $
and infinite dimensional Lie algebra with basis:
$H_i\left(f_i(t)\right) = f_i(t)\,\partial_{x_i}+f_{i}^{\prime}(t)\,\partial_{u_{i}}-\rho_0\,x_i\,f_{i}^{\prime\prime}(t)\,\partial_p,\quad i= 1, 2, 3,$
$H_{0}(f_{0}(t)) = f_{0}(t)\,\partial_{p}, \quad \text{here} \;.^{\prime} = \frac{d}{dt}$
If we introduced following notations
$f(t) = \left(f_{t}(t), f_{2}(t), f_{3}(t)\right)$ and $H(f(t)) = H_{1}(f_{1}(t))+H_{2}(f_{2}(t))+H_{3}(f_{3}(t))$
Then we can prove following Lie commutations
$[H(f),X_1] = H(-f^{\prime}(t)),\; [H(f),X_4] = H(-2tf^{\prime}(t)-2f_{0}(t))$ etc.
Now my question is, can we assume these commutations as closed with respect to infinite dimensional Lie algebra given above? If yes, then how ?
In my opinion as results of these commutations do not directly belongs to infinite dimensional Lie algebra, so how we can assume these commutations to be closed ?
For reference, the Table $1$ of article "On the normalizers of subalgebras in an infinite Lie algebra" can be seen.