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**wik_chick88** consider the polynomial algebra $\displaystyle C[x_1, x_2, x_3]$ with $\displaystyle x_1, x_2, x_3$ 3 independent real variables

show that the differential operators

$\displaystyle a_{ij} = x_{i}\frac{\partial}{\partial x_{j}}$

satisfy the gl(3) commutation relations

does this mean i have to show that $\displaystyle [a_{ij}, a_{kl}] = \delta_{kj}a_{il} - \delta_{il}a_{kj}$?

this is my working (assuming thats what i have to do):

$\displaystyle [a_{ij}, a_{kl}]$

$\displaystyle = [x_i\frac{\partial}{\partial x_{j}}, x_k\frac{\partial}{\partial x_{l}}]$

$\displaystyle = x_i\frac{\partial}{\partial x_{j}} \times x_k\frac{\partial}{\partial x_{l}} - x_k\frac{\partial}{\partial x_{l}} \times x_i\frac{\partial}{\partial x_{j}}$

$\displaystyle = \frac{\partial x_k}{\partial x_j} \times x_i\frac{\partial}{\partial x_{l}} - \frac{\partial x_i}{\partial x_l} \times x_k\frac{\partial}{\partial x_{j}}$

but for this to equal $\displaystyle \delta_{kj}a_{il} - \delta_{il}a_{kj}$, that means that $\displaystyle \delta_{kj}$ has to equal $\displaystyle \frac{\partial x_k}{\partial x_j}$ and $\displaystyle \delta_{il}$ has to equal $\displaystyle \frac{\partial x_i}{\partial x_l}$...is this correct?