MHB Graded Algebra: Understanding Color Dirac Spinors in Space-Time

topsquark
Science Advisor
Homework Helper
Insights Author
MHB
Messages
2,020
Reaction score
843
I just read through a paper on a [math]\mathbb{Z} _ 3[/math] graded Algebra. In this instance we are talking about color Dirac spinors in space-time. It looks like the author is talking about [math]\left ( SU(3) \otimes L^4 \otimes \mathbb{Z}_2 \otimes \mathbb{Z} _2 \right ) \otimes \mathbb{Z} _3[/math]. ( SU(3) is the color group, [math]L_4 [/math] is the Lorentz group, [math]\mathbb{Z} _2 \otimes \mathbb{Z} _2[/math] is the Dirac 4-spinor group, and [math]\mathbb{Z} _3[/math] is the usual group on 3 elements.

I can (mostly) follow the paper assuming the tensor products, but what do they mean by the word "graded?"

Thanks!

-Dan
 
Physics news on Phys.org
In mathematics, an "algebra" is a vector space (so we can add vectors and multiply vectors by scalars) in which we can also multiply two vectors, the result being a vector. An algebra is said to be "graded" if we can divide the vectors into "grades" that are "closed" under addition and scalar multiplication (so the sum of two vectors in one "grade" are also in that "grade" and the product of a scalar and a vector in a given "grade" is in that same "grade") but the product of two vectors in the same "grade" is not necessarily in that "grade".

The simplest example of a "graded algebra" is the algebra of all polynomials. Each "grade" is the vector space of polynomials of degree "n" or less for some positive integer "n". Adding two polynomials of degree n or less gives a polynomial of degree n or less and multiplying a polynomial of degree n or less times a scalar (real or complex number) is a polynomial of degree n or less. But while the product of two polynomials of degree n or less is a polynomial it is not necessarily of degree n or less.
 
Thank you. That actually answers another question I had about the paper as well. I think I've got the idea now.

Thanks again!

-Dan
 
Thread 'Derivation of equations of stress tensor transformation'
Hello ! I derived equations of stress tensor 2D transformation. Some details: I have plane ABCD in two cases (see top on the pic) and I know tensor components for case 1 only. Only plane ABCD rotate in two cases (top of the picture) but not coordinate system. Coordinate system rotates only on the bottom of picture. I want to obtain expression that connects tensor for case 1 and tensor for case 2. My attempt: Are these equations correct? Is there more easier expression for stress tensor...

Similar threads

Replies
3
Views
2K
Replies
0
Views
492
2
Replies
69
Views
8K
Replies
42
Views
10K
2
Replies
93
Views
15K
Replies
12
Views
3K
Back
Top