# Lie Bracket

1. Mar 8, 2006

### Oxymoron

Could someone check if I have done this right.

$$R_1 = x^2\partial_3 - x^3\partial_2$$
$$R_2 = x^3\partial_1 - x^1\partial_3$$
$$R_3 = x^1\partial_2 - x^2\partial_1$$

Where $x^i$ are coordinates.

I need to calculate the commutator $[R_1,R_2]$.

$$[R_1,R_2] = x^2\partial_3x^3\partial_1 - x^3\partial_1x^2\partial_3 + x^3\partial_2x^1\partial_3 - x^1\partial_3x^3\partial_2 - (x^3\partial_2x^3\partial_1-x^3\partial_1x^3\partial_2) - (x^2\partial_3x^1\partial_3-x^1\partial_3x^2\partial_3)$$
$$.\quad\quad\quad = x^2x^3\partial_3\partial_1 - x^2x^3\partial_1\partial_3 + x^1x^3\partial_2\partial_3-x^1x^3\partial_2\partial_3 - x^3x^3\partial_2\partial_1 + x^3x^3\partial_1\partial_2-x^1x^2\partial_3\partial_3+x^1x^2\partial_3\partial_3$$
$$.\quad\quad\quad = x^2\partial_1-x^2\partial_1+x^1\partial_2-x^1\partial_2 -x^3x^3\partial_2\partial_1 + x^3x^3\partial_1\partial_2 - x^1x^2\partial_3\partial_3 + x^1x^2\partial_3\partial_3$$
$$.\quad\quad\quad = 0$$

And as a result,

$$[R_1,R_2] = [R_2,R_3] = [R_3,R_1] = 0$$

by cyclically permuting the indices.

Last edited: Mar 8, 2006
2. Mar 9, 2006

### dextercioby

I think you're wrong...Check the angular momentum so(3) algebra...

Daniel.

3. Mar 9, 2006

### Oxymoron

You're right. But I dont know where I went wrong. I would have expected

$$[R_1,R_2] = R_3[/itex] Last edited: Mar 9, 2006 4. Mar 9, 2006 ### Oxymoron I think my problem is evaluating the differentials. Here is an example of what Im doing in a simpler problem. let $A=\partial_x$, $B=\partial_y$, and $C=x\partial_y-y\partial_x$. Im going to compute $[A,C]$ and let me know if Ive done something wrong... [tex][A,C] = [\partial_x,x\partial_y-y\partial_x]$$
$$= \partial_x(x\partial_y - y\partial_x)-(x\partial_y-y\partial_x)\partial_x$$
$$= \partial_xx\partial_y-\partial_xy\partial_x - x\partial_y\partial_x+y\partial_x\partial_x$$

Now Im pretty sure $\partial_xx = 1$ so the first term is $\partial_y$.

The second term $\partial_xy\partial_x = \partial_x\partial_xy$ since the partial derivatives commute. The second term equals the fourth term so they cancel.

This leaves the third term. $x\partial_y\partial_x=\partial_y(\partial_xx) = \partial_y(1) = 0$

So I would say the answer is

$$[A,C] = \partial_y = B[/itex] how does this look? Last edited: Mar 9, 2006 5. Mar 9, 2006 ### Oxymoron To be honest I dont think Ive done this right at all. My basic question is what does [tex]\partial_yy\partial_x$$

equal? And is it the same thing as

$$y\partial_x\partial_y$$

EDIT: Could

$$\partial_yy\partial_x = \partial_y(y)\partial_x + y\partial_x\partial_y$$?

If this calculation is right could someone tell me how.

Last edited: Mar 9, 2006
6. Mar 9, 2006

### topsquark

What $$\partial_yy\partial_x$$ means depends on how you are applying the $$\partial_y$$. From a commutator, for example:
$$[\partial_y, y\partial_x] = \partial_y(y\partial_x)-y\partial_x(\partial_y)$$
In this case the y operator acts on the whole expression so you use the product rule.

-Dan

7. Mar 14, 2006

### Oxymoron

Ah, of course! So in fact

$$[R_1,R_2] = -R_3$$

and...

$$[R_2,R_3] = -R_1$$
$$[R_3,R_1] = -R_2$$

Last edited: Mar 14, 2006