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 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
687
Replies
26
Views
5K
Replies
2
Views
1K
  • · Replies 69 ·
3
Replies
69
Views
8K
Replies
15
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
10K
  • · Replies 93 ·
4
Replies
93
Views
15K
  • · Replies 12 ·
Replies
12
Views
3K