gl(3) commutation relations question

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?

am i just completely on the wrong path or am i headed in the right direction?? please someone help!!!

Re: gl(3) commutation relations question

Quote:

Originally Posted by

**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?

That is correct, assuming your variables are independent.

Quote:

am i just completely on the wrong path or am i headed in the right direction?? please someone help!!!

Re: gl(3) commutation relations question

Quote:

Originally Posted by

**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?

am i just completely on the wrong path or am i headed in the right direction?? please someone help!!!

the product rule for differentiation gives us: $\displaystyle a_{ij}a_{k \ell}=x_i \frac{\partial}{\partial x_j}x_k \frac{\partial}{\partial x_{\ell}} =x_i \left( \delta_{jk} \frac{\partial}{\partial x_{\ell}} + x_k \frac{\partial^2}{\partial x_j \partial x_{\ell}} \right).$ thus

$\displaystyle [a_{ij},a_{k \ell}]=a_{ij}a_{k \ell} - a_{k \ell}a_{ij}=\delta_{jk}x_i \frac{\partial}{\partial x_{\ell}} - \delta_{j \ell}x_k \frac{\partial}{\partial x_j}= \delta_{jk}a_{i \ell} - \delta_{i \ell}a_{kj}. $

Re: gl(3) commutation relations question

thankyou. they are specified as being independent so now i can see how that can work.

now i have to use ado's theorem to find expressions for the o(3) generators $\displaystyle L_1, L_2, L_3$ in terms of (*) and show that $\displaystyle L^2 = x^2\nabla^2 - \vec x.(\vec x.\nabla + 2)\nabla$

any idea how im supposed to start this?

i know that ado's theorem states that every finite dimensional Lie Algebra over a fiel of characteristic 0 has a faithful finite dimensional representation, which means it can be viewed as a Lie Algebra of square matrices under the commutator bracket.

im not exactly sure what the o(3) generators are, only that they have the basis $\displaystyle \{{\alpha^i}_j = e_{ij} - e_{ji} : i, j = 1, 2, 3\}$