I Definition of Clifford Algebras

  • I
  • Thread starter Thread starter jv07cs
  • Start date Start date
  • Tags Tags
    Clifford algebra
jv07cs
Messages
44
Reaction score
2
I was reading D.J.H. Garling's book "Clifford Algebra: An Introduction" and it defines clifford algebras as follows:

Screenshot from 2024-04-21 10-31-03.png
Screenshot from 2024-04-21 10-31-13.png


But if ##1 \notin j(E)##, how come ##j(E)## generate ##A## since it doesn't generate its identity element?
 
Physics news on Phys.org
You could have elements 1+x and -x in the range of j, and then 1 is contained in the algebra generated by the range
 
Office_Shredder said:
You could have elements 1+x and -x in the range of j, and then 1 is contained in the algebra generated by the range
But if, let's say, ##j(v) = 1 + x## and ##j(w) = -x##, for ##v,w \in E##. The element ##u = v + w## is also in ##E## and since ##j## is linear, we would have ##j(u) = j(v+w) = j(v) + j(w) = 1##, which would imply ##1 \in j(E)##, right?
 
jv07cs said:
But if, let's say, ##j(v) = 1 + x## and ##j(w) = -x##, for ##v,w \in E##. The element ##u = v + w## is also in ##E## and since ##j## is linear, we would have ##j(u) = j(v+w) = j(v) + j(w) = 1##, which would imply ##1 \in j(E)##, right?
Sorry you're right you need multiplication to make things interesting. E.g. you could have some ##x,y## in the range of ##j## such that ##xy=1## (or maybe you need to layer some addition on top of multiplication, I haven't thought carefully about it obviously)
 
Office_Shredder said:
Sorry you're right you need multiplication to make things interesting. E.g. you could have some ##x,y## in the range of ##j## such that ##xy=1## (or maybe you need to layer some addition on top of multiplication, I haven't thought carefully about it obviously)
Oh okay. So by "##j## generates ##A##" we don't limit ourselves to just linear combinations, we should also consider the algebra multiplication right?
 
Generates as an algebra over the given field means the smalest algebra that contains the gwnerating set. It will also contain the field. The same way that the set of one element ##x## generates the algebra of polynomials in ##x##.
 
jv07cs said:
Oh okay. So by "##j## generates ##A##" we don't limit ourselves to just linear combinations, we should also consider the algebra multiplication right?
This is correct
 
When you say ##j(E)## generates ##A##, which is an algebra, then it's already implicit that one considers the smallest subalgebra that contains ##j(E)##. IOW, it's the set of all linear combinations of finite products of the generating set - polynomials, if you will.
 
Back
Top