Understanding Composition of Point Symmetry Generators in Lie Algebras

[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

 

[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.

 

[alikvot]

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

 

[HallsofIvy]

Yeah, I hate tedious calculations like that!