Understanding Composition of Point Symmetry Generators in Lie Algebras

alikvot
Messages
2
Reaction score
0
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
 

Attachments

Physics news on Phys.org
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.
 
Yes, I *have* calculated but I didn't manage to get the 1st derivatives right. Thanks a lot!
 
Yeah, I hate tedious calculations like that!
 

Similar threads

Back
Top