# Help understanding Clifford/Geometric Algebra on curved spacetimes

Geometric Algebra on curved spacetimes. I seem to come across a lot of mentions of “spacetime algebra” (especially by Hestenes). As I understand it, this is simply the Clifford algebra $$Cl(3,1)$$. I'm more interested in General Relativity where, in general, one needs a more complicated structure.

Say, for example, I'm looking at a spacetime with a $$S^{3}\times S^{1}$$ topology. We still have that the tangent space at each point is described by a $$Cl(3,1)$$ Clifford algebra, but what about the manifold as a whole? The standard way seems to use an embedding, so I might say:

$$S^{3}\times S^{1}\in\mathbb{R}^{4,2}$$

So if I want to write the general Clifford algebra that covers the whole spacetime, I'd say it's $$Cl(4,2)$$ with point-dependent constraints. At some particular point $$P$$ then we would have that the elements of the Clifford algebra $$Cl(3,1)$$ which form the tangent space at P are made from a particular reduction of $$Cl(4,2)$$. Am I thinking about this correctly?

Another way I've thought about looking at it is from the perspective of the oriented orthonormal frame bundle. In this case the oriented orthonormal frame bundle looks something like $$SO(4,2)$$ with constraints, which at each point reduce to $$SO(3,1)$$. However, the Clifford algebra seems more general, since I can form the latter groups from it.

Basically I'm asking if I'm getting this correctly and if not can someone explain what I'm attempting to go over above?

• "In this case the oriented orthonormal frame bundle looks something like SO(4,2) with constraints" Could you explain what you mean by this? There's no need to refer to a flat space in which the manifold is embedded in order to describe the frame bundle. Mar 9, 2021 at 8:06
• @octonion I realize there is no need; however I was looking for an analogy to the Clifford algebra case. For what I meant consider a circle embedded in 2d space. I can write the unit tangent vector on the circle at a given point as a weighted sum of unit the unit vectors in 2d. In this case the weights are our "constraints" which depend upon position, usually chosen to be $cos(\theta)$ and $sin(\theta)$ Mar 9, 2021 at 9:29
• @octonion See Kajelad's comments to my question here for why I went this route: math.stackexchange.com/questions/3835284/… Mar 9, 2021 at 9:33
• I don't know what "tangent space at each point described by a Clifford algebra" is supposed to mean. You can associate to each tangent space $T_x M$ a Clifford algebra $\mathrm{Cl}(T_x M, g_x)$ (for $g_x$ the metric at that point), but how does it "describe" the tangent space? What is "the Clifford algebra that covers the whole spacetime"? A Clifford algebra is just a construction for an inner product space, it is not a notion for differential geometry/manifolds. Why are you doing this? Are you perhaps looking for something like the spinor bundle? Mar 9, 2021 at 16:47
• @ACuriousMind. That is exactly what I'm looking for! Mar 9, 2021 at 18:57

In this case the oriented orthonormal frame bundle looks something like SO(4,2) with constraints, which at each point reduce to SO(3,1).

This sort of idea has been explored by someone call Roger Penrose, who happened to have been awarded a Nobel prize last year. Mathematically, his projective twistor space can be most naturally understood as the space of chiral (Weyl) spinors for the conformal group $$SO(4,2)/\mathbb{Z}_{2}$$ of Minkowski space. He wrote a two-volumn tome about it: Spinors and Space-time.

The connection of twistor space with Clifford/Geometric algebra has been explored here.

If you are only interested in curved spacetimes in terms of $$Cl(3,1)$$, you can take a look at: Gravity, gauge theories and geometric algebra.

Added note: There are some nice introductions to Clifford Algebra in Curved Space-Time. See e.g.:

• Excellent! I have those books! Yes I know the double cover the. Is the same group Penrose and Rindler used. Thank you, I'll read this more after work! Mar 9, 2021 at 20:02
• @R.Rankin, is it your intention for twistor (or $Cl(4,2)$) to cover both electroweak $U(2) \times U(1)$ and gravity? Peter Woit has been toying with this sort of topics, but he seems to be hitting dead ends, see here: math.columbia.edu/~woit/wordpress/?p=11899 Mar 9, 2021 at 20:58
• In a sense yes.. from the topological perspective, all of the constraints will come from cosmological factors, like the scale/hubble parameters, and the whole setup can be considered the real part of the complexification Cl(6), which naturally has U(3) as well. Maybe look at Fureys work, which doesn't consider the physical origin of the structures she's considering Mar 9, 2021 at 21:09
• Yes that would be the dream though! Mar 9, 2021 at 21:09
• Fureys and alike used the complexification version of $\mathbb{Cl}(6)$. Actually the complexification route is not necessary, since either $Cl(6,0)$ or $Cl(0,6)$ can accommodate $SU(4)$ which subsumes $U(3)$. And not quite obvious to you at first sight, $Cl(3,1)\subset Cl(6, 0)$ or $Cl(1,3) \subset Cl(0,6)$. Mar 9, 2021 at 22:18