# Point symmetry generators

1. Mar 2, 2008

### alikvot

I am taking a first course in Lie algebras and currently working with this problem (see attached file). I understand that the product of the two operators should be regarded as composition. How to explain the final expression?
Regards
Staffan

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2. Mar 2, 2008

### HallsofIvy

Haven't you tried just doing the computation? That's how you learn mathematics- not just by looking at formula and expecting to understand them, but by actually doing the calculations!

You are given that
$$Z_j= \xi_j(x,u)\frac{\partial}{\partial x}+ \chi_j(x,u)\frac{\partial}{\partial u}$$
for j= 1 and 2. In other words, the difference between Z1 and Z2 is the functions multiplying the derivatives.

If f is any function of x and u (any reason for using x and u instead of x and y?) then
$$Z_1 Z_2(f)= \xi_1(x,u)\frac{\partial}{\partial x}+ \chi_1(x,u)\frac{\partial}{\partial u}[ \xi_2(x,u)\frac{\partial f}{\partial x}+ \chi_2(x,u)\frac{\partial f}{\partial u}]$$
$$= \xi_1(x,u)\frac{\partial}{\partial x}[\xi_2(x,u)\frac{\partial f}{\partial x}+ \chi_2(x,u)\frac{\partial f}{\partial u}]+ \chi_j(x,u)\frac{\partial}{\partial u}[\xi_2(x,u)\frac{\partial f}{\partial x}+ \chi_2(x,u)\frac{\partial f}{\partial u}]$$
$$= \xi_1\xi_2\frac{\partial^2 f}+ \xi_1 \frac{\partial \xi_2}{\partial x}\frac{\partial f}{\partial x}+ \cdot\cdot\cdot$$

Finish that, then do the same for $Z_2Z_1$ and subtract. All the second derivative terms (those not involving derivatives of $\xi_1$, $\xi_2$, $\chi_1$, or $\chi_2$) will cancel leaving only first derivative terms.

3. Mar 2, 2008

### alikvot

Yes, I *have* calculated but I didn't manage to get the 1st derivatives right. Thanks a lot!

4. Mar 2, 2008

### HallsofIvy

Yeah, I hate tedious calculations like that!