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 Clifford Freaking Algebra 
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Joined: 07 May 2009, 16:45
Posts: 62
Post Clifford Freaking Algebra
My latest confusion comes from Clifford Algebra. Penrose says \gamma\textsuperscript{2 }=-1, if you know what I mean, but tutorials I find on the net (like http://www.mrao.cam.ac.uk/~clifford/introduction/intro/intro.html) say \gamma\textsuperscript{2 }=1. Can you explain?


15 Jul 2009, 18:35

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Clifford Freaking Algebra
I reply to my own post.

Still confused about these \gammas, but I have this to add:

Setting each \gamma\textsuperscript{2 }=1 instead of \gamma\textsuperscript{2 }=-1 does not invalidate the "secondary entities" equivalent to i, j, and k of the quaternions as described on page 209.

In his Introduction to Clifford Algebra (http://www.av8n.com/physics/clifford-intro.htm) John Denker says the appropriate Clifford Algebra basis for general relativity (Minkowski spacetime) is

\gamma\textsubscript{0}\gamma\textsubscript{0}=-1
\gamma\textsubscript{1}\gamma\textsubscript{1}=\gamma\textsubscript{2}\gamma\textsubscript{2}=\gamma\textsubscript{3}\gamma\textsubscript{3}=1

Where those indices on those gammas are supposed to be subscripts, but the LaTex code doesn't seem to work.

The gamma sub 0 represents the time dimension and gamma sub 1 through 3 the space dimensions.

So I am guessing Penrose is defining his Clifford Algebra with each \gamma\textsuperscript{2 }=-1 appropriate for this "spin space" he talks about, maybe. Or maybe he will add a \gamma\textsubscript{0}\gamma\textsubscript{0}=1 for spacetime and have a system kind of opposite what Denker has, but maybe these two basis are not equivalent.

I don't know.


17 Jul 2009, 17:38
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Clifford Freaking Algebra
Use ^ for superscript, it's a lot shorter! (although your text superscript seems to work) and _(underscore) for subscripts.
\gamma_{_0}\gamma_{_0}=-1 which is coded as gamma_{_0}\gamma_{_0}=-1 makes them even smaller and lower.

I have quoted and edited your original text below using a single _ only.

As for your problem with the maths (I have not studied this part of the book yet), I looked in the index and read something on page 619 which seems to address your problem: He talks about positive and negative squares and how he is following the standard physicists convention. Have a look and let me know what you think.
Vasco


DimBulb wrote:
I reply to my own post.

Still confused about these \gammas, but I have this to add:

Setting each \gamma^{2 }=1 instead of \gamma\textsuperscript{2 }=-1 does not invalidate the "secondary entities" equivalent to i, j, and k of the quaternions as described on page 209.

In his Introduction to Clifford Algebra (http://www.av8n.com/physics/clifford-intro.htm) John Denker says the appropriate Clifford Algebra basis for general relativity (Minkowski spacetime) is

\gamma_0\gamma_0=-1
\gamma_1\gamma_1=\gamma_2\gamma_2=\gamma_3\gamma_3=1

The gamma sub 0 represents the time dimension and gamma sub 1 through 3 the space dimensions.

So I am guessing Penrose is defining his Clifford Algebra with each \gamma\textsuperscript{2 }=-1 appropriate for this "spin space" he talks about, maybe. Or maybe he will add a \gamma_0\gamma_0=1 for spacetime and have a system kind of opposite what Denker has, but maybe these two basis are not equivalent.

I don't know.


18 Jul 2009, 07:33

Joined: 07 May 2009, 16:45
Posts: 62
Post Re: Clifford Freaking Algebra
Thanks, great.

Yes, it seems there are two ways about this. As you point out, apparently the basis on page 209 is the standard physics way of presenting it, with \gamma_i^2 = -1, even though that is not Penrose's own preference as he states in note 10 of chapter 24. I guess the two Clifford Algebra Introductions I found on the net (see above) show the mathematical way of presenting the thing with \gamma_i^2 = 1. Perhaps Penrose presents the standard physics way because it flows more directly from the discussion of quaternions, and because two reflections equal a 2\pi rotation, which equates to -1 for a spinor object. But, the mathematical way seems a bit nicer to me. as I explain below, basically repeating something from the first web site in my initial post.

If you have a two dimensional space with two orthonormal basis vectors \gamma_1 and \gamma_2, with \gamma_1^2 = 1 and \gamma_2^2 = 1, two general elements \boldsymbol{A} and \boldsymbol{B} of the Clifford algebra could be represented as follows

\boldsymbol{A}=a_0+a_1\boldsymbol\gamma_1+a_2\boldsymbol\gamma_2+a_3\boldsymbol\gamma_1\boldsymbol\gamma_2

\boldsymbol{B}=b_0+b_1\boldsymbol\gamma_1+b_2\boldsymbol\gamma_2+b_3\boldsymbol\gamma_1\boldsymbol\gamma_2

where a_i, b_i are scalars

The product \boldsymbol{A}\boldsymbol{B} is found by multiplying the components distributively and simplifying based on

\gamma_i^2 = 1
\gamma_1\gamma_2 = -\gamma_2\gamma_1

so, for example

\gamma_1\gamma_2\gamma_2 = \gamma_1
\gamma_1\gamma_2\gamma_1 = -\gamma_1\gamma_1\gamma_2 = -\gamma_2
\gamma_1\gamma_2\gamma_1\gamma_2 = -\gamma_1\gamma_1\gamma_2\gamma_2 = -1

So,

\boldsymbol{A}\boldsymbol{B} = p_0+p_1\boldsymbol\gamma_1+p_2\boldsymbol\gamma_2+p_3\boldsymbol\gamma_1\boldsymbol\gamma_2

where the scalars p_i are

p_0 = a_0b_0+a_1b_1+a_2b_2-a_3b_3
p_1 = a_0b_1+a_1b_0+a_3b_2-a_2b_3
p_2 = a_0b_2+a_2b_0+a_1b_3-a_3b_1
p_3 = a_0b_3+a_3b_0+a_1b_2-a_2b_1

However, if you start out saying \gamma_i^2 = -1, then

p_0 = a_0b_0-a_1b_1-a_2b_2-a_3b_3
p_1 = a_0b_1+a_1b_0+a_2b_3-a_3b_2
p_2 = a_0b_2+a_2b_0+a_3b_1-a_1b_3
p_3 = a_0b_3+a_3b_0+a_1b_2-a_2b_1

Which is not quite as symmetrical, and lead me to think that maybe the two ways were not equivalent


18 Jul 2009, 20:50
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