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Does Clifford algebra solve both QFT and GR 
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#1
Feb1312, 11:27 AM

P: 978

I noticed a few sources that seem to indicate that Clifford algebra may be used in both QFT and GR. I've seen where the Clifford algebra is a type of associative algebra that generalizes the real numbers, complex numbers, quaternions, and octonions, see Wikipedia on Clifford Algebra. And I've seen where the complex numbers, quaternions, and octernions can be used in the description for the U(1), SU(2), SU(3) symmetries of the Standard Model, see this article, and this book. However, I've also seen where the Clifford algebra can be used in an alternative description of differential geometry used in the formulation of GR, see this book for example.
So my question is does this common algebra allow us to derive GR in terms of QFT or visa versa? Or if we could justify the use of the complex numbers, quaternions, octonions, and the Clifford algebra by some other means, could we derive both QFT and GR from that common justification of the algebra? What more information or constraints would be needed to do so? 


#2
Feb1412, 03:54 PM

P: 978

I'm trying to decide whether I should research the Clifford algebra and how it applies to physics. I'm looking to get a sense of how valid is the application of Clifford algebra in physics. Is there a consenses that Clifford algebra, quaternions, and octonions are indeed fundamental or at least have a valuable use?
If the use of these algebras is valid in physics, then I have to wonder... Since the Clifford algebra is used in differential geometry, and quaterions and octonions are used in the SM, then can the quaternions and octonions be used to contrain the Clifford algebra in such a way in differential geometry as to product GR if the SM is assumed? Or is it more the case that the Clifford algebra is used on the metric whereas the quaternion and octonions are used on wavefunctions, which is a totally different animal? Any insight would be appreciated. Thanks. 


#3
Feb2212, 12:46 AM

P: 450

[QUOTE=friend;3764308]I'm trying to decide whether I should research the Clifford algebra and how it applies to physics...QUOTE]
What a strange question! Of course Clifford (geometric) algebra does apply to physics! Just think about spinors, Dirac equation... 


#4
Mar112, 12:51 PM

P: 978

Does Clifford algebra solve both QFT and GR
So when they use the quaternions and octonions in the SM, do they use the algebra to establish commutation relation between fields? How is the Clifford algebra used in GR? Is it used on connection fields perhaps? Does it establish commutation relations, or perhaps the Bianchi identity, what? Thanks. 


#5
Mar112, 01:33 PM

Sci Advisor
P: 1,682

Clifford algebras are particularly necessary for understanding spinor representations and spin groups. Depending on what you're interested in studying, they could be important (a good knowledge of the mathematical underpinnings of SM and QFT will require knowledge of Clifford algebras). I'd say that quaternions (and octonions) are less important: quaternions offer an alternative to the tensor formulation of GR, and are not something I'd concern myself with on an initial read through.
I would certainly get up to speed on your linear algebra and group theory (in particular Lie groups and Lie algebras which are widespread in particle physics) before I'd worry about Clifford algebras and quaternions. 


#6
Mar112, 03:05 PM

P: 978




#7
Mar212, 12:22 AM

P: 450




#8
Mar212, 04:44 PM

P: 450

You get everything you need to beginn with and if you are patient, at the very end of the book, a part of the answer to your initial question. 


#9
Mar312, 11:12 AM

P: 978

But from what I can tell from what I've seen on the Net, clifford algebra is used either for geometry or for the SM in the form of quaternions and octonions. What I don't see is anyone connecting the two. Is this because the algebra operates on two different objects that can't be connected in some way? 


#10
Mar312, 11:25 AM

P: 978

For example, is there a connection between spin foam networks of LQG and the spinors derived using clifford algebra?



#11
Mar312, 11:59 AM

P: 450

But I just guess that the answer is: some where yes. Good luck http://arxiv.org/pdf/1201.2120v1.pdf 


#12
Mar2814, 06:58 AM

P: 10

For Clifford/Geometric Algebra the most relevant reference is 
"Geometric Algebra for Physicsts" by Doran and Lasenby http://www.amazon.com/GeometricAlge.../dp/0521715954 Which covers (among other things) both the Dirac equation and the Gauge Theory of Gravity. I have a set of notes based on Doran and Lasenby and symbolic software in python at https://github.com/brombo/GA 


#13
Mar2914, 01:22 PM

P: 978

What I'm reading so far is that the S0(1,3) symmetry of Lorentz invariance is a "double cover" for the SU(2) symmetry of the Standard Model. And I'm read other places that the spin of a particle is deeply connected to Lorentz invariance. Is this deep connection the double cover?
What does double cover mean anyway? Is this covered in the book you mention? Thanks. 


#14
Mar2914, 03:05 PM

P: 10

It is mentioned in D&L. In GA rotations (in any dimension) are preformed with rotors which are scalars plus bivectors, R, with the condition that R*rev(R) = 1 (rev(R) is the reverse of R and * the geometric product). Then to rotate a vector, x, in the plane that R defines you evaluate
R*x*rev(R) Since rev(R) = rev(R) if you replace R with R you get the same result. This is the double cover. Note that R = exp((theta/2)*B) where B is a normalized bivector defining the plane of rotation. For the Lorentz transformation in Minkowski space if B is a spacelike plane if B**2 = 1 and R = cos(theta/2)+sin(theta/2)*B If the plane is timelike B**2 = 1 and R = cosh(theta/2)+sinh(theta/2)*B also note that rev(R) = exp((theta/2)*B). Multiple Lorentz transformation R1, R2 are evaluated with R2*R1*x*rev(R1)*rev(R2) and the geometric product * is the group operation. 


#15
Mar2914, 03:36 PM

P: 360

Hello friend, there's been lots of good advice here so far. While playing with Clifford algebra is great fun, there's a risk of both overestimating and underestimating its importance in theoretical physics, and it's good to understand its context. Clifford algebra is useful in physics because the antisymmetric Clifford product between Clifford bivector elements is isomorphic to the Lie bracket between corresponding elements of the spin Lie algebra. This means Clifford algebra is hugely important in fundamental physics, but not a replacement for all of differential geometry. One still needs to understand Lie groups and differential forms, and then have even more fun with Clifford valued differential forms, specifically for describing connections (gauge fields). Even more specifically, gravity is well described using a spin connection, which is a spin(1,3)valued 1form, which can also be written as a Cl^{2}(1,3)valued 1form, which usually acts on a Cl^{1}(1,3)valued gravitational frame as well as on spinors, which are defined using matrix representations of the spin group which correspond to matrix representations of Clifford algebras. Clifford algebra is also useful in the Standard Model because u(1)xsu(2)xsu(3) embeds in spin(10) (the Lie algebra of the Spin(10) GUT) or equivalently Cl^{2}(10), with the SM gauge fields all parts of a spin(10)valued 1form, and the SM fermion multiplets living in 16 dimensional spinor representation spaces. There are, of course, other things one can do with Clifford algebras, including how this structure unifies in spin(11,3) and E8, but that's my opinion on their importance in fundamental physics. Hope you find that helpful.



#16
Mar2914, 08:29 PM

P: 978

Remind me again, U(1) is the symmetry of the Electromagnetic force and is represented by the dirac matrix? And SU(2) is the symmetry of the Weak force and is represented by the Pauli matix, and SU(3) is the symmetry of the Strong force and is represented by the what matix? Am I close?



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