What is SU*(N)? Definition and Explanation

William Nelso
Messages
20
Reaction score
1
I've run across a Lie group notation that I am unfamiliar with and having trouble googling (since google won't seem to search on * characters literally).

Does anyone know the definition of the "star groups" notated e.g. SU*(N), SO*(N) ??
The paper I am reading states for example that SO(5,1) is isomorphic to SU*(4).
(Ref Kugo+Townsend, Nuc. Phys. B221, p. 357, "Supersymmetry and the division algebras")

In fact it has a "definition" of these groups, however I am not able to understand it. It appears to say that
SU*(N) consists of elements X of SL(N,C) such that
X = B-1X* B
where B is the spinor conjugation operator (and N is the dimension of a spinor rep of the SO(D,1) that the paper is talking about)
 
Physics news on Phys.org
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC). Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
Back
Top