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I'm writing some code that does simulations of quantum field theory, and I came to the sudden realization that I hadn't yet written the part for the SU(n) Lie group. I know this group is a subgroup of U(n) and GL(n), but GL is already infinite and U is also infinite as far as I know. Computers don't deal well with infinite sets, so I was wondering if there was a way I could compute the generators of the n-th SU group without having to examine all of GL

Cheers!

richyfeynman
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  • I don't know if this will help you. Dr. Todor Milev has worked on Lie algebras and wrote computer programs to study them. Search for him on Google. However, I can't promise whether he can help or he will respond. – Batominovski May 23 '17 at 19:51
  • Rarely seen "nightmares" in the title of a question, but why not ? – Jean Marie May 23 '17 at 19:52
  • I don't understand the question. Anyway $SU(n)$ is generated by transvections, which are relatively simple. – user420261 May 23 '17 at 20:01

1 Answers1

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Well, after much research, I've figured it out. I have written a program that finds the roots of the special unitary group's Cartan matrix. I can use this information to find the generator matrices.

Special thanks to Batominovski, I got talking with Dr. Milev and we were able to solve the problem.

Cheers!

richyfeynman
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