Geometry of matrix Dirac algebra

[bayakiv]

TL;DR Summary

Questions about the embedding of the Minkowski space in the Dirac algebra and the construction of the corresponding geometric algebra do not interest me. I'm wondering why an 8-space with a neutral metric generates a Dirac algebra.

Indeed, if we take a vector field which dual to the covector field formed by the gradient from a quadratic interval of an 8-dimensional space with a Euclidean metric, then the Lie algebra of linear vector fields orthogonal (in neutral metric) to this vector field is isomorphic to the $sl(4,\mathbb{C})$. Why is this so, is it not connected with the fact that this covector field is the field of accelerations of moving matter (in other words, ether) on the seven-dimensional sphere $S^7$?

 

[jambaugh]

If you take a n-state fermion algebra, generated by its n creation operators and n annihilation operators, these 2n operators generate a Clifford algebra with neutral metric. Note that once you complexify, the signature becomes moot.

$$\sqrt{2}\gamma_k = a_k + a^\dagger_k, \quad \sqrt{2}\gamma_{k+n} = a_k - a^\dagger_k$$
$$\{a_j \pm a^\dagger_j ,a_k \pm' a^\dagger_k\} = \{a_j,a_k\} \pm' \{a_j,a^\dagger_k\}\pm\{a^\dagger_j,a_k\} \pm\pm' \{a^\dagger_j,a^\dagger_k\}=...$$
$$= \pm' \{a_j,a^\dagger_k\} \pm \{a^\dagger_j,a_k\} =(\pm' 1\pm 1)\delta_{jk}$$
You get 2 if + +' and -2 if - -' for j=k else 0. Hence:
$$\{\gamma_j,\gamma_k\} = \delta_{jk}, \{\gamma_{j+n},\gamma_{k+n}\}=-\delta_{jk}$$
for j,k = 1,2,3...n.

This reflects the fact that the Lie algebra $gl(n,\mathbb{R})$ is a subalgebra of $so(n,n)$.
I leave it to you to see this in general but for $SO(1,1) = GL(1,\mathbb{R})$ we see this in the fact that:
$\cosh(\beta)+ \sinh(\beta)=e^\beta, \cosh(\beta)-\sinh(\beta) = e^{-\beta}$

$$\left(\begin{array}{cc} \cosh(\beta) & \sinh(\beta)\\ \sinh(\beta) & \cosh(\beta)\end{array}\right) = e^\beta\left(\begin{array}{cc} +1/2&+1/2\\ +1/2 & +1/2\end{array}\right) + e^{-\beta}\left(\begin{array}{cc} +1/2&-1/2\\-1/2 & +1/2\end{array}\right)$$
Note the two component matrices are commuting idempotent matrices.

To generalize you need only consider the representation of $GL(n,\mathbb{R})$ by noting that the matrix:
$$e^M\otimes\left(\begin{array}{cc} +1/2&+1/2\\ +1/2 & +1/2\end{array}\right) + e^{-M^T}\otimes\left(\begin{array}{cc} +1/2&-1/2\\-1/2 & +1/2\end{array}\right)$$
is orthogonal in the neutral metric.

 

[bayakiv]

Thanks for the interesting comment. I will only add that the neutral metric in the case of the Dirac algebra works on a par with the Euclidean metric, but in general the role of the neutral metric is great, with its help one can even form the exceptional Lie algebra e(8) (link in personal message).

 

[bayakiv]

Summary:: Questions about the embedding of the Minkowski space in the Dirac algebra and the construction of the corresponding geometric algebra do not interest me. I'm wondering why an 8-space with a neutral metric generates a Dirac algebra.

Why is this so, is it not connected with the fact that this covector field is the field of accelerations of moving matter (in other words, ether) on the seven-dimensional sphere?

So, let the velocity field be equal to
$$\begin{equation}
\textbf{v}(x)=\sum_1^8 v_{i}(x)
\end{equation}$$
and the acceleration field
$$\begin{equation}
\dot{\textbf{v}}(x)=\sum_1^8 \dot{v}_{i}(x)
\end{equation}$$
Then if
$$\begin{equation}
\sum_1^8 \dot{v}_{i}^2(x) = \sum_1^8 x_i^2 = \frac{\mathrm{e}^{\tau} + \, \mathrm{e}^{-\tau}}{2}
\end{equation}$$
and
$$\begin{equation}
\dot{\textbf{v}}^2(x) = \sum_1^4 \dot{v}_{i}^2(x) - \sum_5^8 \dot{v}_{i}^2(x) = \sum_1^4 x_i^2 - \sum_5^8 x_i^2 = 0
\end{equation}$$
then
$$\begin{equation}
\begin{cases}
\sum\limits_1^4 \dot{v}_{i}^2(x) = \sum_1^4 x_i^2 = \frac{\mathrm{e}^{\tau} + \, \mathrm{e}^{-\tau}}{4} \\
\sum\limits_5^8 \dot{v}_{i}^2(x) = \sum_5^8 x_i^2 = \frac{\mathrm{e}^{\tau} + \, \mathrm{e}^{-\tau}}{4}
\end{cases}
\end{equation}$$
Thus, the set of radius vectors of the static zero acceleration field belongs to the Clifford torus $S^3\times S^3$ lying on the seven-dimensional sphere, and therefore the velocity field corresponding to the zero acceleration field will correspond to the Clifford torus $S^3\times S^3$ and the sphere $S^7$. It should be noted that in this case, the Clifford torus is symmetrical at all times of evolution, since the radii of its spheres are always the same. If we turn to a static non-zero acceleration field, then assuming
$$\begin{equation}
\dot{\textbf{v}}^2(x) = \sum_1^4 \dot{v}_{i}^2(x) - \sum_5^8 \dot{v}_{i}^2(x) = \sum_1^4 x_i^2 - \sum_5^8 x_i^2 = \frac{\mathrm{e}^{\tau} - \,\mathrm{e}^{-\tau}}{2}
\end{equation}$$
obtained system
$$\begin{equation}
\begin{cases}
\sum\limits_1^4 \dot{v}_{i}^2(x) = \sum_1^4 x_i^2 = \mathrm{e}^{\tau} \\
\sum\limits_5^8 \dot{v}_{i}^2(x) = \sum_5^8 x_i^2 = \mathrm{e}^{-\tau}
\end{cases}
\end{equation}$$
So, the geometry of the static vacuum vector acceleration field in evolution (i.e., the dynamic vector field) is described by the changing geometry of the Clifford torus $S^3 \times S^3$. And since at the present moment of evolution the radius of one of the spheres of the Clifford torus is negligible, it can be assumed that the vector field of accelerations is orthogonal to the 3-sphere, which sweeping out the circle drifts along the surface of the 7-sphere. Thus, the Minkowski space embedded in the Dirac algebra should be associated with its compact image $S^3\times S^1$.

 

[bayakiv]

If you will, I will show how you can compactify an 8-dimensional space with a neutral metric. So let
$$\begin{equation}
\begin{cases}
\sum\limits_1^4 x_i^2 = R^2 = \mathrm{e}^{\tau_1} \\
\sum\limits_5^8 x_i^2 = r^2 = \mathrm{e}^{\tau_2}
\end{cases}
\end{equation}$$
where $(\tau_1,\tau_2)$ isotropic coordinates of the pseudo-Euclidean plane, that is $\tau'_1\tau'_2=\tau_1\tau_2=\mathrm{const}$
We now compactify the pseudo-Euclidean plane $(\tau_1,\tau_2)$ into a torus $S^1\times S^1$
$$\begin{equation}
\begin{split}
\tau_1 \rightarrow \mathrm{e}^{\mathrm{i}2\pi\tau_1} \\
\tau_2 \rightarrow \mathrm{e}^{\mathrm{i}2\pi\tau_2}
\end{split}
\end{equation}$$
Thus, we have constructed an image of a space with a neutral metric in $S^3\times S^3\times S^1\times S^1$ and hence the doublet of Minkowski spaces now has a compact image. Moreover, the isotropic cone of space with a neutral metric also compactifies into the product $S^3\times S^3\times P\mathbb{R}^1$ whose symmetries can be associated with the group of unitary symmetries.

 

[jambaugh]

Have you read Porteous' Clifford Algebras and the Classical Groups?

 

[bayakiv]

Alas, I have not read this apparently wonderful book. In general, I have gaps in my education, and therefore I constantly have to study. As for the further development of the topic touched upon in this thread, it would be interesting to look at the geometry that is generated in the 8-dimensional Euclidean space by a pair of harmonic functions through the intersection of their level surfaces. At the very least, if one of the functions is the square of the metric interval of a Euclidean space, and the other is the square of the metric interval of a space with a neutral metric, then the Clifford torus lies at the intersection. But this is the geometry of the vacuum solution, and I would like to get some other solution.

 

[jambaugh]

You should consider the more categorical approach. That is to look at the symmetries (automorphisms) of the structure you are considering and how that changes as you add to the structure.

An 8-dimensional differentiable manifold has principally only topological structure. Your symmetries are the diffeomorphisms. Surfaces reflect (and are reflected by) the subgroup of diffeomorphisms for which they are sets of fixed points, or for which they are invariant sets (orbits). You impose and Euclidean metric and you get your (inhomogeneous) orthogonal subgroup. You impose a neutral metric and you get a different orthogonal group. Your torus manifests with the intersecting symmetry group of the two, a direct product of orthogonal groups.

When you speak of vacuum and non-vacuum solutions you are then considering representations of these groups. It is the representations which manifest the observables associated with symmetries à la Noether's theorem. Note that while Noether's theorem links conserved quantities to symmetries of the dynamics, it can be viewed, in light of a range of possible dynamics, as linking all observables to potential symmetries since, in order to preserve the observable for the sake of consistent repeatable measurement, we must be able to impose a dynamic for which it is conserved for a period of time.

This is what has made group theory and in particular Lie theory so vital to modern physics. Note that Bell inequality violation specifically demonstrates the futility of attempting to understand quantum systems in term of objective states of being. We therefore transition the language of "what is" to one of "what happens". It is a superior paradigm since you can still utilize it to describe classical systems but it provides a richer language capable of describing non-objective systems such as quanta.

Now, I bring up these generalities as it seems to me you are trying to find an objective visualization of some physical system. I'm still not clear on your aim. You are constructing a compactification. Clearly we can find homeomorphisms (topological isomorphisms) from a non-compact manifold to a compact one. The stereographic projection of the plane onto the Riemann sphere (sans the pole at "infinity") is the classic example. While there is a great deal of mathematical utility in this it is not physically relevant once one properly invokes the relevant relativity principles of the system in question.

Using the above example, a "Flat Earther" can go round and round about how since your spherical globes can be mapped to their flat maps, their position is valid. For their position to be valid they must also construct the dynamics of that "flat Earth" where-in empirically measured distances undergo some strange phenomenon they'd have to explain as a consequence of, say, flight times for airline routes between cities all over the globe. In the end one must realize that "flat vs round" are only mathematical constructs which model the phenomenology of the natural world. It's like a question of choices of coordinates and exactly why, in modern physics, the first thing we look for is the transformation rules when we change such choices.

In another example, Eisntein's strong equivalence principle points out that we cannot locally distinguish a dynamic gravitational force from a non-inertial reference frame absent that force. What this then means globally is that we cannot identify the dividing line between dynamical gravitational forces and space-time geometry. We therefore can eliminate the force by choosing an appropriate geometry but this is another of those choices of convention...
but

. The statement "Gravity is just geometry" is false. The correct statement is rather, "Gravity and geometry are indistinguishable." I think failure to recognize this distinction has been one of the major obstructions to a successful progress in quantizing the gravitational interaction. (Were I younger and smarter I'd dive headlong into the exploration of this thesis.)

So, my point to you is, acknowledging my lack of understanding of your aim here, look closely at the questions you are asking with these thoughts in mind.

 

[bayakiv]

Thanks for not dropping this thread. I will think about your questions, but for now I will only note that the dynamics of this model is based on the generalized principle of least action, where the action is the angle of rotation of the acceleration vector in 8-dimensional Euclidean space. As for the quantum gravity of this model, the metric geometry here is generated by the dynamics of the vector field, and quantum mechanics - by winding the action on a circle.

 

[bayakiv]

Let's match:

absolute time and action
$$
\begin{equation}
\begin{split}
\tau=\frac{1}{2}\ln (x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}+x_{7}^{2}+x_{8}^{2})\\
\varphi=\arctan\sqrt{ \frac{x_{5}^{2}+x_{6}^{2}+x_{7}^{2}+x_{8}^{2}}{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}}
\end{split}
\end{equation}
$$
space-time and dual space-time
$$
\begin{align}
x&=\frac{1}{2}\ln (x_{1}^{2}+x_{5}^{2}), &
\varphi_{x}&=\arctan \frac{x_5}{x_1},\\
y&=\frac{1}{2}\ln (x_{2}^{2}+x_{6}^{2}), &
\varphi_{y}&=\arctan \frac{x_6}{x_2},\\
z&=\frac{1}{2}\ln (x_{3}^{2}+x_{7}^{2}), &
\varphi_{z}&=\arctan \frac{x_7}{x_3},\\
t&=\frac{1}{2}\ln (x_{4}^{2}+x_{8}^{2}), &
\varphi_{t}&=\arctan \frac{x_4}{x_8}
\end{align}
$$
4-momentum
$$
\begin{equation}
\begin{split}
p_{x}=\frac{d\varphi_{x}}{d\tau},\\
p_{y}=\frac{d\varphi_{y}}{d\tau},\\
p_{z}=\frac{d\varphi_{z}}{d\tau},\\
p_{t}=\frac{d\varphi_{t}}{d\tau}
\end{split}
\end{equation}
$$
If you will allow, then next time it will be possible to touch upon the gauge symmetries of this model.

 

[bayakiv]

Preliminary information

Let be $n>1$, $ST(n)=\mathrm{diag}_n(\mathrm{e}^{\mathrm{i}\varphi_{j}})$, where $\sum^{n}\varphi_{j} = 2\pi z$ and $u=abc$, $u\in ABC(n)$, where $b\in SO(n,\mathbb{R})$ and $a,c\in ST(n)$

And since $u\in SU(n)$ and $\mathrm{Dim}SU(n)=\mathrm{Dim}ABC(n)$, then $ABC(n)=SU(n)$

Thus, the mapping
$$S^1\times S^1 \to S^2\setminus (S,N)$$
has the symmetries of the group $SU(2)$ and the mapping
$$S^1\times S^1\times S^1 \to S^3\setminus (S,N)$$
has the symmetries of the group $SU(3)$

On the other hand, if in space $(t,x,y,z,\varphi_{t},\varphi_{x},\varphi_{y},\varphi_{z})$ the square of the metric interval is
$$t\varphi_{t} -x\varphi_{x} - y\varphi_{y} - z\varphi_{z}$$
then the compactified part of the isotropic cone is $S^1\times S^3$ and hence the gauge symmetry group is generated by the compactification of an 8-dimensional space with a neutral metric.

 

[bayakiv]

It's time to talk about the classical limit of our model and about the gravitational potential.
Let be
$$
\begin{align}
X&=\frac{1}{2}\ln (x_{1}^{2}+x_{2}^{2}), &
\varphi_{X}&=X^*=\arctan \frac{x_2}{x_1}, \\
Y&=\frac{1}{2}\ln (x_{3}^{2}+x_{4}^{2}), &
\varphi_{Y}&=Y^*=\arctan \frac{x_4}{x_3}, \\
Z&=\frac{1}{2}\ln (x_{5}^{2}+x_{6}^{2}), &
\varphi_{Z}&=Z^*=\arctan \frac{x_6}{x_5}, \\
T&=\frac{1}{2}\ln (x_{7}^{2}+x_{8}^{2}), &
\varphi_{T}&=T^*=\arctan \frac{x_8}{x_7}
\end{align}
$$
and quadratic metric form
$$
\begin{equation}
S^{2}= TT^{*} + XX^{*} + YY^{*} + ZZ^{*},
\end{equation}
$$
which by changing variables
$$
\begin{equation}
\begin{split}
T=t+t^*,&\quad T^{*}=t - t^*, \\
X=x^* + x,&\quad X^{*}=x^* - x, \\
Y=y^* + y,&\quad Y^{*}=y^* - y, \\
Z=z^* + z,&\quad Z^{*}=z^* - z,
\end{split}
\end{equation}
$$
reduced to form
$$
\begin{equation}
S^2=t^2-x^2-y^2-z^2-t^{*2}+x^{*2}+y^{*2}+z^{*2}
\end{equation}
$$
The classical limit of our model arises when the angular coordinates are zeroed. Then
$$
\begin{equation}
\begin{split}
T=t+t^*,&\quad T^{*}=t - t^*=0, \\
X=x^* + x,&\quad X^{*}=x^* - x=0, \\
Y=y^* + y,&\quad Y^{*}=y^* - y=0, \\
Z=z^* + z,&\quad Z^{*}=z^* - z=0,
\end{split}
\end{equation}
$$
and therefore $t=t^∗$, $x^∗=x$, $y^∗=y$, $z^∗=z$ and consequently
$$
\begin{equation}
S^2=t^2-x^2-y^2-z^2
\end{equation}
$$
Finally, the gravitational potential of this model is expressed by the formula
$$
\begin{equation}
\varphi= \frac{1}{2}\ln \frac{x_{1}^{2}+x_{3}^{2}+x_{5}^{2}+x_{7}^{2}}{x_{2}^{2}+x_{4}^{2}+x_{6}^{2}+x_{8}^{2}}
\end{equation}
$$

 

[PeterDonis]

@bayakiv I fixed some formatting in your post #12. Please note that you need two backslashes, not one, to separate lines in an align block in LaTeX.

 

[bayakiv]

@PeterDonis, thanks

 

[bayakiv]

A small addition, reminding us that we still have a dynamic model, in which the global geometry is easily transferred to the local level. In particular, the local breaking of the symmetry of the Clifford torus entails a local change in the metric, which is responsible for the gravitational interaction of our model. For example, with a local change in the ratio of the diameters of the spheres of the Clifford torus caused by the hyperbolic angle of deviation of the vector field of accelerations
$$
\begin{equation}
\varphi=\frac{1}{2}\ln\frac{\dot{v}_{1}^2(x)+\dot{v}_{3}^2(x)+\dot{v}_{5}^2(x)+\dot{v}_{7}^2(x)}{\dot{v}_{2}^2(x)+\dot{v}_{4}^2(x)+\dot{v}_{6}^2(x)+\dot{v}_{8}^2(x)}
\end{equation}
$$
it can be related to the gravitational potential.

However, the model is remarkable in many ways. For example, it easily includes quantum mechanics as a Markov process of a random walk of a singularity of a vector acceleration field, where the angular action (angle of rotation on a sphere) of the singularity's free path acts as a random variable.

 

[bayakiv]

The concept of particle mass fits well into the model. Indeed, since the vector field of accelerations of the flow of matter is initially (before the compactification procedure) specified in space with a neutral metric, then the features of the vector field, the streamlines of which are tangential to isotropic lines, can be associated with massless particles, and the singularities with non-isotropic streamlines can be associated with a mass depending on hyperbolic slope of the streamline. In order to formalize this phenomenon of mass gain, one can reduce the model to the original pseudo-Euclidean plane $(T,\varphi_{T})$, and then wind it on a sphere with punctured poles. Here a formula appears that connects the Euclidean angle of inclination of the secant circle of the sphere and the hyperbolic angle of the anisotropic line corresponding to this circle $$\tan\left(\frac{\vartheta}{2} - \frac{\pi}{4}\right) = \mathrm{e}^{-2\psi}$$ If the particles are correlated with torus knots, then, for example, for a trefoil it is possible to calculate a well-defined Euclidean angle of inclination of the secant circle. Does this have anything to do with the Weinberg corner?