Re: question from newbie to Clifford algebras
- From: victor_meldrew_666@xxxxxxxxxxx
- Date: Wed, 25 Jul 2007 01:51:49 -0700
On 24 Jul, 19:42, hanrahan...@xxxxxxxxxxx wrote:
Hi, I am a newbie to Clifford algebras and having read Tony Smith's
page at
http://www.valdostamuseum.org/hamsmith/clfpq.html
Smith's page is (deliberately?) rather confusing. A more
lucid account is on Wikipedia:
http://en.wikipedia.org/wiki/Clifford_algebra ,
http://en.wikipedia.org/wiki/Classification_of_Clifford_algebras .
I am even more interested but still confused as to where to start.
Cl(0) looks like the reals.
Cl(1) looks like the complex numbers
Cl(2) looks like the quaternions
In Smith's notation Cl(n) is the Clifford algebra over R
of the quadratic form -x_1^2 - x_2^2 - ... - x_n^2.
These three examples do not just "look like" R, C and H;
they are isomorphic to them.
So numbers in Cl(0) have matrix form (a),
numbers iin Cl(1) have matrix form ((a,-b),(b,a))
and numbers in Cl(2) have matrix form ((a,-b,-c,-d),(b,a,-d,c),(c,d,a,-
b),(d,-c,b,a))
Could someone give me some examples of numbers in Cl(3), Cl(4) and
indeed Cl(n),
and of how to multiply them?
Well, Cl(3) is isomorphic to H x H, the algebra of ordered
pairs of quaternions, while Cl(4) is isomorphic to M_2(H),
the algebra of 2x2 matrices of quaternions.
Can they be expressed in matrix form too? Or would one need to
generalise matrices into something else?
Yes. Indeed as Cl(n) is an algebra of dimension 2^n over R
it has a repesentation (the regular representation)
as an algebra of size 2^n matrices over R. One can produce
this by a recursive process but one that is slightly too fiddly
for me to produce right now :-(
Victor Meldrew
"I don't believe it!"
.
- References:
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- From: hanrahan398
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