# Unifying Gravity and EM

by sweetser
Tags: gravity, unifying
 P: 361 Hello: A basic property of electromagnetism is that like charges repel and different charges attract. That must be a property of any proposal related to EM. The Reissner-Nordstrom can only accept like charges that repel, since it uses Q2, which always has the same sign (and opposite the M/R term). That looks like a deadly flaw to me, one that prevents me for doing anything further with the metric. In the GEM proposal, the sign of the relative electric charge (ie same sign for source and probe means positive) and the mass charge are different. That way like electric charges can repel, and like mass charges can attract. If the relative electric charge is negative, as happens with a positive and negative charge, then the electric charge contribution would be attractive in exactly the same way as gravitational charge. The two add up. There is no negative mass charge in the GEM proposal because the mass field is about symmetric fields. There are two charges for EM because of the antisymmetric fields. Doug
 P: 190 Obviously with the 421 kg mass in a box or a unit charge it will be easy to tell the difference. The mass will attract all masses equally, while the charge will attract or repel the same whether that has the same or opposite charge. The proton will have negligable attraction to an uncharged mass. The NS metric is for a spacetime determined by a sperically symmetric mass with a charge. One could well enough compute the orbit of a charged particle in this spacetime. Yet what I compare is the fall of a neutral mass around a charged black hole and a charge falling into a neutral black hole. What I find is that the "braking" of a charged particle fall by the emission of radiation by purely classical means is equivalent to the geodesic rate of fall for a neutral particle entering a black hole with the equivalent charge. This result illustrates I must be at least on the right track. It is in line with the "no hair" theorem for black holes is that the details of how a black hole reaches a final state by the acquisition of charge and mass by the infall of matter and fields is independent of the details by which that occurred. A black hole will end up with the same final mass $M~=~M_0~-~(q/r)^2$ indepdendent of which masses brought in what charge. Lawrence B. Crowell
 P: 361 Hello Lawrence: The thought experiment did not concern itself with the ability to tell the difference between the behavior of electric charge or mass charge. Rather, it was an effort to find a condition where the behavior was indistinguishable. > One could well enough compute the orbit of a charged particle in this spacetime. If and only if the charged mass had the same sign as the charge on the test mass would the NS metric work. If th test had the opposite sign, it would fail because the metric must demonstrate attraction. I appreciate that you looked into the consequences of the NS metric, and that other skilled people have worked with it. But it looks unacceptable to me because it cannot model electrical attraction. Doug
P: 190
 Quote by sweetser Hello: A basic property of electromagnetism is that like charges repel and different charges attract. That must be a property of any proposal related to EM. The Reissner-Nordstrom can only accept like charges that repel, since it uses Q2, which always has the same sign (and opposite the M/R term). That looks like a deadly flaw to me, one that prevents me for doing anything further with the metric. Doug
The RN metric uses $(Q/r)^2$ as the self-energy of a charged mass. This can be derived by using the electromagnetic field tensor as a source of the gravitational field. From there the RN metric is a solution to the Einstein field equation. In more advanced settings solutions of this type are BPS black holes, where gauge "charges" are the source of the black hole metric. The charge here is the charge of the black hole. One might then consider the motion of a charged mass in this spacetime, where that charge can have either sign.

This is fairly standard stuff.

Lawrence B. Crowell
P: 190
 Quote by sweetser Hello Lawrence: > One could well enough compute the orbit of a charged particle in this spacetime. If and only if the charged mass had the same sign as the charge on the test mass would the NS metric work. If th test had the opposite sign, it would fail because the metric must demonstrate attraction. Doug
I am not sure where you got this idea. You can well enough compute the orbit of a charge with any value or sign in an RN metric.

Lawrence B. Crowell
 P: 361 Hello Lawrence: I agree, you have represented standard analysis, which is darn good most of the time. If I see a killer flaw, I will point it out. Even better is when I realize I was wrong about the killer flaw. I "get" the standard way, and can pinpoint where I drive a different direction. That is what happened with regards to the Reissner-Nordstrom metric. In general relativity, gravity binds to the stress energy tensor. How much energy will the electric field contribute to the metric? $G Q^2/c^4 R^2[/tex]. If the charge is positive, or the charge is negative, it does not matter, the same amount goes in. For the GEM proposal, the charge coupling term is [itex]-\frac{1}{4 c}(J A + (J A)^* - J^{*1} A^{*2 *3} - (J^{*1} A^{*2 *3})^*)$ With GEM, the effort is to do both gravity and EM. All these conjugates are required to get the phase to have both spin 1 and spin 2 symmetry. The coupling of gravity is not to energy, but instead to the 4-current density. The energy of the electric field makes zero contribution to gravitational effects. What we get in return for giving up the energy connection is a new equivalence principle for particles that attract each other. The attraction might be caused by electric charge or by a mass charge. Attraction looks the same either way. Sure, you could do other tests to tell which one was at work, but one could have videos of a particle being attracted by gravity or by EM that are absolutely identical. That is the bridge between gravity and EM. People who work on EM almost never deal with the tools of differential geometry. Calculate the divergence of the Christoffel symbol of the second kind for this metric: $$d \tau^2 = exp(2(\sqrt{G} Q - GM)/c^2 R) dt^2 - exp (-2(\sqrt{G} Q - GM)/c^2 R) dR^2/c^2 - R^2d\Omega^2/c^2$$ Since this is for a static, spherically symmetric metric, to be logically consistent with EM, the answer had better be both Gauss' and Newton's potential theory for EM and gravity,$\nabla^2 \frac{\sqrt{G} Q - GM)}{R}$, which it is. Lucky? I don't think so. Doug
 P: 361 Hello: In this post, I will apply similar, but not quite identical, approaches to generating the field equations for the fields E, B, e, and b. If one hopes to model particles that travel at the speed of light, that requires that the field theory be invariant under a gauge transformation. This is one of those constraints on finding a solution I have heard, I have accepted, and I don't understand as well I should. In deriving the Maxwell field equations using quaternion operators, the gauge invariance was achieved by noting all the derivatives that make up a gauge are in the first term of $-A \nabla$. The first term was subtracted away using the time honored quaternion trick subtracting the conjugate, q - q* = vector(q). The identical method was applied to the symmetric fields e and b. If were were to just add these two results together, there would be no link between the two sets of equations. This time we will not use the trick of subtracting the conjugate, yet no terms with g appear in the final field equations because the g2 terms in one cancel the g2 terms of the other in the scalar. There is a g in the 3-vector, but that is not used for generating the field equations. Start by taking the derivatives of 4-potentials in two ways. For quaternions written in the Hamilton basis, change the order of the differential operator with the potential, which flips the sign of B. For quaternions written in the Even basis, change which term gets conjugated. Calculate: $$\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))$$ $$=((-\frac{\partial \phi}{\partial t} ~+~ c \nabla . A, -\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~+~ \nabla X A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A , \frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~+~ \nabla X A)$$ $$~+~ (\frac{\partial \phi}{\partial t} - c \nabla . A,\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~-~ \nabla .X2. A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A,-\frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~-~ \nabla .X2. A))$$ $$= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))$$ $$(-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E X B ~-~ e .X2. e ~+~ b .X2. b ~+~ 2 gE ~+~ 2 gb) \quad eq 1$$ The g field is not in the scalar due to a cancelation, but is in the 3-vector. The field equations are generated from the scalar, not the 3-vector, so any choice for the gauge g will not effect the field equations I am about to derive. The current coupling term is complicated by the need to have spin 1 and spin 2 symmetry in the phase. This was worked out earlier in the thread, and here is the solution: $$-\frac{1}{4}(J A ~+~ (J A)^* ~-~ J^{*1} A^{*2 *3} ~-~ (J^{*1} A^{*2 *3})^*) = ~-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 2$$ Write out the Lagrangian by its components, including the current coupling terms: $$\mathcal{L}_{BEbe} ~=~ -c^2 \frac{\partial \Ay}{\partial z} \frac{\partial Az}{\partial y} ~-~ c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x} ~-~ c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}$$ $$-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t} ~-~ c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t} ~-~ c \frac{\partial \phi }{\partial z} \frac{\partial Az}{\partial t}$$ $$-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 3$$ There are fewer terms than in either the Maxwell Lagrangian or the symmetric field Lagrangian because terms between the two cancel. The fields in the field equations will need to do the same. Calculate the first field equation by taking the derivative of $\mathcal{L}_{BEbe}$ with respect to the 4 derivatives of phi. $$\frac{\partial}{\partial x^{\mu}}\left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial \phi}{\partial x^{\mu} \right)}} \right) ~=~ -c \frac{\partial ^2 Az}{\partial t\partial z} ~-~ c\frac{\partial ^2 Ay}{\partial t\partial y} ~-~ c\frac{\partial ^2 Ax}{\partial t\partial x}\right) ~-~ \rho$$ $$=~ \frac{1}{2}\nabla . (E ~-~ e) ~-~ \rho ~=~ 0 \quad eq 4$$ Nice, the E and e terms work together to isolate the A derivatives. And yet, you can spot Gauss' law for EM where like charges repel. There is a Gauss-like law for like charges that attract. Repeat for the derivative with respect to Ax: $$\frac{\partial}{\partial x^{\mu}} \left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial Ax}{\partial x^{\mu} \right)}} \right) ~=~ -c^2 \frac{\partial ^2 Az}{\partial x\partial z} ~-~ c^2 \frac{\partial ^2 Ay}{\partial x\partial y} ~-~ c\frac{\partial ^2\phi }{\partial t\partial x}\right) ~+~ Jx ~=~ \frac{1}{2}(-\nabla X B ~-~ \nabla .X2. b ~+~ \frac{\partial(Ex ~+~ ex)}{\partial t}) ~+~ Jx ~=~ 0 \quad eq 5$$ You should be able to spot Ampere's law. What has been done There is now a formulation of the GEM proposal that uses quaternions exclusively. The standard quaternion algebra inherited from the nineteenth century needed to be extended in two ways. First the idea of a conjugate (or anti-involutive automorphism in fancy jargon, or thingie that flips the sign of all but one part in simple words) had to be expanded to *1, *2, and *3 to get the phase symmetry right for the current coupling term. The second advance in quaternion algebra needed is the Even representation of quaternion multiplication. Here the eigenvectors of the representation must be excluded to make the representation an algebraic field. I have consistently said I am more comfortable when I found out someone else has done this before. The multiplication table is known as the Klein four-group. I will have to see if others have noticed what happens when the eigenvectors are excluded. With these two innovations, the field equations for E, B, e and b have been generated. These field equations are gauge invariant because g may be whatever one chooses since it is canceled out in the process. This is vital since the graviton and proton both travel at the speed of light. What needs to be done I need to develop a non-gauge invariant set of field equations for massive particles, where the gauge symmetry is broken by the mass charge. I have something technical to keep me off the streets. Doug
P: 361
Hello:

A gauge-free unified field equation has been constructed using quaternion operators (previous post), a very good thing because the particles that mediate gravity and EM travel at the speed of light. The particles that interacted with these mediators of force - massive and possibly electrically charged particles - do not travel at the speed of light, so the gauge symmetry must be broken.

In a standard approach to EM, gauge symmetry is broken with the Higgs mechanism. This is done by postulating that there is a Higgs field everywhere any particle ever goes. The vacuum state is a false vacuum, actually higher than a nearby state that will give particles like quarks mass without breaking the symmetry needed by EM.

False vacuums remind me of false gods, something akin to the aether that had to be everywhere, no place in the Universe could be without. The Universe is a clumpy place, and it would be a magical coincidence if the Higgs field got to every place it needed to in the right density in order to give every particle the same symmetry breaking experience needed so all protons, neutrons, and electrons have the same mass. These are reasonable skeptical objections, but they get no air time today for a good reason: there is no alternative. One choice makes things simple.

A thought experiment may show that mass charge does break electric charge symmetry. You have a pair of electrons sitting 1 cm apart, and you measure how fast they accelerate away from each other, $a_e = F/m_e[/tex]. Repeat the experiment, but for a pair of protons, [itex]a_p = F/m_p[/tex]. Let's say your experimental system is so good you are able to measure the two accelerations to ten significant digits. The value of the two accelerations is identical. This equivalence indicates that electric charge is universal. Repeat the experiment without changing anything concerning the setup. There are two electrons over here rushing away from each other, two protons doing the same thing. The one difference is the acceleration is now measured to twenty significant digits. Now the two accelerations will not be the same. The reason is the gravitational mass has a trivial effect that will keep the heavier protons from accelerating as fast as the pair of electrons because the protons attract each other gravitationally more than the electrons. Unlike the standard model + Higgs which characterize inertial mass and ignores gravity, the GEM proposal is about gravity and EM working together ever so lightly. We saw that in post 457, eq 1, where the g2 term contributed from EM canceled with the g2 tossed in by gravity. This time, I will combine the method used to generate the Maxwell equations (post 438) with its Even quaternion representation clone (post 442). If you look carefully at the two Lagrangians, you would notice all the cross terms are the same, and they all have a minus sign (but don't look too close or you will notice I got the sign of one term wrong). This means if we subtract one from the other, all the mixed terms drop, leaving only 12 squared terms. When I was using tensors, I found the contraction of the two asymmetric rank 2 tensors, [itex]\nabla_{\mu} A_{\nu} \nabla^{\mu} A^{\nu}$, had these same twelve terms, along with 4 others. The field equations that come out of the asymmetric field strength tensor contraction are drop dead gorgeous (from page 1 of this long thread):

$$J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}$$

 Quote by Feynman lectures, book II, 18-11 What a beautiful set of equations! They are beautiful, first, because they are nicely separated - with the charge density, goes $\phi$; with the current, goes A. Furthermore, although the [right] side looks a little funny - a Laplacian together with a $(\partial/\partial t)^2$ - when we unfold it we see...it has a nice symmetry in the x, y, z, t - the [c's are] necessary because, of course, time and space are different; they have different units.
To get the context correct, Feynman was extolling the virtues of writing the Maxwell equations in the Lorenz gauge. Earlier in this thread we went over how Feynman showed the current coupling term $J^{\mu} A_{\mu}$ has spin 1 symmetry, necessary for EM where like charges attract. I know Feynman did not analyze the spin of $(i J^{\mu} i)^* (k(j A_{\mu} j)^* k)^*$. That has spin 2 symmetry, necessary for gravity where like mass charges attract. That's the strength of this proposal, it keeps getting more subtle.

I think I will stop here tonight and let the goal sink in...

Doug
P: 190
 Quote by sweetser Hello: The field equations that come out of the asymmetric field strength tensor contraction are drop dead gorgeous (from page 1 of this long thread): $$J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}$$ To get the context correct, Feynman was extolling the virtues of writing the Maxwell equations in the Lorenz gauge. Doug
The problem with this wave equation which gives the currents is that it assigns a mass current to an electromagnetic potential. I doubt many are going to cotten on to this idea.

The electromagnetic and gravity fields have their sources because of their group structure. Electromagnetism is an abelian gauge field $A\wedge A~=~0[/tex] which has the unitary group U(1). The two roots of this group are the real parts of the circle on the Argand plane [itex]C$ of compex numbers. Those are of course the numbers +1 and -1. Gravitation is the Lorentz group which is the hyperbolic group $SO(3,1)$. This group is similar to $SO(4)~=~SO(2)\times SO(2)$ though the hyperbolic group is noncompact. $SO(3,1)~=~SL(2,C)\times Z_2$ and the special linear group is $SU(2)\times SU(1,1)$. These two parts define the set of three rotations and three boosts which give the six generators of the Lorentz group.

Compact groups have nice properties that a set of transformations of the group generator will converge in a Cauchy sequence. Hyperbolic groups are non-compact and so a sequence is likely to go off to asymptopia and not converge.

If we think of gravity as a gauge-like theory with $F~=~dA~+~A\wedge A$ for nonabelian guage fields the DE's for these on the classical level are nonlinear. Yet we can quantize these, but renormalization is a bit complicated. We can well enough quantize a SO(4) theory obtained in euclideanization. But gravity is a strangely different. Why? The gauge group SU(1,1) is hyperbolic. In the Pauli matrix representation we have that $\tau_z~=~i\sigma_z$. So we form a gauge connection

$$A~=~A^{\pm}\sigma_{\pm}~+~iA^3 \sigma_3$$

and for the group element $g~=~exp(ix\tau_3)$ $$=~exp(-x\sigma_3)[/itex] the connection term transforms as [tex] A'~=~g^{-1}Ag~+~g^{-1}dg~=~e^{-2x}A^{\pm}\sigma_{\pm}~+~iA^3\sigma_3$$

and for $x~\rightarrow~\infty$ this gives $A~\rightarrow~iA^3\sigma_3$. Now $A^{\pm}\sigma_{\pm}$ and $A^3\sigma_3$ have distinct holonomy groups and are thus distinct points (moduli) in the moduli space. But this limit has a curious implication that the field $F~=~dA~+~A\wedge A$ for these two are the same and the moduli are not separable. In other words the moduli space for gravity is not Hausdorff. This is the most serious problem for quantum gravity.

Because of this it is not possible to construct the connection coefficient for gravitation from an internal gauge connection that is the generator of a compact group. The basic point set topologies for the two are different. The point set topology for gravitation is the Zariski topology, and defines certain algebraic varieties and a sheaf bundle system that is fundamentally different from the more trivial principal bundle system for the unitary U(1) group of electromagnetism. Penrose's twistor theory exploits this property and in some ways is a fairly honest approach to quantum gravity, even if it has not proven to be terribly workable in more recent time.

Lawrence B. Crowell
P: 361
Hello Lawrence:

 Quote by Lawrence B. Cronwell The problem with this wave equation which gives the currents is that it assigns a mass current to an electromagnetic potential. I doubt many are going to cotten on to this idea.
Yeah, people will probably quickly jump to the wrong conclusion. I cannot stop that, nor do I care about such a null reaction.

If you want to look at the group theory behind the GEM proposal, you must look at the Lagrange density, not the field equations. The field equations don't say a word about the spin symmetries, or the symmetries of the field strength tensor, for Maxwell or GEM.

You fortunately are smarter than that, and discuse F. That is in the GEM proposal as $- A \nabla + (A \nabla)^* = (0, E + B)=F$. The way we write these - you preferring differential geometry, I choosing to work with quaternions - does not have an impact on group theory: it is the unitary group U(1) as it must be to characterize electromagnetism.

The problem as you note is gravity. We agree on that. You then discuss what the group for gravity is, which is kind of surprizing since we don't have a quantum gravity theory yet. How we know what the group theory for the theory we don't have is beyond me. You do know the limitations of current efforts to apply group theory to the problem of gravity. That was a fun challenge to read.

It would appear like you missed a comment made in post #457 about group theory. There I noted that I am using Klein four group, $Z_2~X~Z_2$, also known as the dihedral group, $Dih_2$. Since the group is finite, it is compact, a good thing as you have pointed out. This avoids the problem you cited about hyperbolic groups. My proposal is a linear theory for gravity because the spin-coupling is between 2 4-vectors, just like in EM. It is far better to quantize a linear theory than a nonlinear one!

If the technically skilled take a brief glance at my proposal and assume I only use U(1), I can wait for the more reflective to think about the advantages and challenges of using $Z_2~X~Z_2$ for gravity and U(1) for EM.

Doug
P: 190
 Quote by sweetser Hello Lawrence: It would appear like you missed a comment made in post #457 about group theory. There I noted that I am using Klein four group, $Z_2~X~Z_2$, also known as the dihedral group, $Dih_2$. Since the group is finite, it is compact, a good thing as you have pointed out. This avoids the problem you cited about hyperbolic groups. My proposal is a linear theory for gravity because the spin-coupling is between 2 4-vectors, just like in EM. It is far better to quantize a linear theory than a nonlinear one! If the technically skilled take a brief glance at my proposal and assume I only use U(1), I can wait for the more reflective to think about the advantages and challenges of using $Z_2~X~Z_2$ for gravity and U(1) for EM. Doug
A dihedral group, or any polytope is a Coxeter-Weyl A-D-E system define a system of roots. These are the discrete symmetries for the bundle framing of a system. The continuous symmetry are the group generators.

I have not looked at your reference, but the relationship between cyclic groups such as Z_n are

$$D_n~=~\{(x_1,~\dots,~x_n)~\in~Z_n: \sum_{i=1}^n x_i = 2Z (even)\}$$

which puts constraints on the $D_n$ group that can be used for Voronoi cell "glue vectors." But at any rate the simple cyclic group $Z_2$ corresponds to the $}D_2~\simeq~SO(4)$ or the Lorentzian version $SO(3,1)$. Each of the $x_i$ corresponds to the rotations in an SU(2). The product of these would correspond to a tetrad where one group is for local transformations of affine connections on the base manifold and the other would correspond to an internal bundle.

I could go on a whole lot more about this and how it leads to $D_4~=~SO(8)$ and how SO(7,1) and how the $spin(6)~\rightarrow~spin(4,2)$ embedded in $SO(7,1)$ is the group for conformal gravity. This group has the cyclotomic ring defined by the Galois field $GF(4)$. The Galois field is $GF(4)~=~(0,~1,~z,~z^2)$ with $z~=~{1\over 2}(i\sqrt{3}~-~1)$ with $z^2~=~z^*$. $GF(4)$ is the Dynkin diagram for the Lie Algebra $D_4~=~spin(8)$. The properties of the basis elements that produce a commutator are

$$z^2~=~z~+~I,~z^3~=~I,~{\bar z}~=~z^2,$$

and defines the hexcode system $C_6$

None of this gets away from the problem that the holonomy groups for the Euclideanized versions of groups do not contain those in the Lorentzian or hyperbolic case. The Berger classification of groups according to their holonomy groups for loop variables of affine variables does not cover groups of the sort $SO(n)\times SL(2,C)$. This is in spite of the fact that $SL(2,C)~=~SU(2)\times SU(1,1)$ and both SU(2) and S(1,1) have the same Dynkin root system.

Lawrence B. Crowell
P: 361
Hello Lawrence:

I am trying to decipher what is good or bad about the specific proposal in this thread. A gentle reader would need a good graduate level class on group theory to appreciate all the relationships you have pointed out. For those readers of this thread further behind than myself, I will point out a few things I have learned so far.

The Klein four-group is abelian, so two elements in the group commute. That is consistent with my definition of the Even representation of quaternion multiplication.

The Klein four-group can be viewed as a subgroup of $A_4$. It does not have a simple graph. Instead the graph has 4 vertices, where only two of them are connected:

.-.
. .

where the dots are the vertices, the dash a connecting edge.

Since I came to this group from a quaternion, it was interesting to read this on wikipedia:

 Quote by http://en.wikipedia.org/wiki/Klein_four-group The Klein four-group is the group of components of the group of units of the topological ring of split-complex numbers.
Quaternions in the usual Hamilton representation are 3 complex numbers that share the same real. In the Even representation, the complex numbers are split. Nice.

A cyclic graph is created by taking powers of the group to the n. The graph for the Dih2 has the identity element in the middle, connected to the three other elements of the group. For the chemists in the audience, it looks like an amino group.

Another element of this puzzle are the finite fields or Galois fields. A way to represent the Klein group is as the set of four elements, {1, 3, 5, 7} modulo 8. So 1*3=3, 3*7 Mod 8=5, 5*7 Mod 8 = 3. The representation is Abelian, since 3*5 = 5*3. The Klein group is isomorphic to GF(4) because there are 4 prime numbers in the representation shown.

So far so good.

One way in which the Even representation of quaternions is different from the Klein group is that the group is modulo the eigen vectors of the 4x4 real matrix representation. This is necessary so that the group is a division algebra, a subject I did not come across in my readings on the topic. The moding out of the eigen vectors might have big implications, because then from one element of the group, there would necessarily be a way to get to any other element. This implies that the graph for Dih2 Mod (eigen vectors) is simply connected.

Now we come to the objection:

 Quote by Lawrence B. Crowell None of this gets away from the problem that the holonomy groups for the Euclideanized versions of groups do not contain those in the Lorentzian or hyperbolic case.
The gentle reader might wonder what a holonomy group is, having never read about it in the funny pages. This is a topic that is central to differential geometry. It is about the relationship between the connection and the manifold. If one moves around the manifold, and does not quite get back to where one started, that is the subject of this topic.

I can now appreciate the problem at hand. If the graph for a group is not simply connected, then one cannot represent the Lorentz group - a picture of flat spacetime - or the small deviations from the Lorentz group needed for gravity to be a dynamic metric theory. If the GEM proposal used the group Dih2 for gravity, it is reasonable to say it could not represent gravity as a smooth metric theory. Although not sure how to write this, I am using a different group, Dih2 Mod (eigen vectors). One can travel anywhere on the manifold with the connection, a necessary thing. At this time, I don't know the impact on parallel transport.

Doug
 P: 190 Doug, For some reason it took a long time to get on here, so I don't have much time. I have a Schubertiade to atttend to. The Wiki-p page indicates what I said. This is involved with D_2, or its roots. These structures are useful in deriving group structures, and I am actively involved with work along these lines, though more up the ladded to Leecha lattices, Conway groups and up to the Fischer-Griess "monster." Yet at the end this does not directly address the issue of non-compact group structure and holonomies. There is in my thinking a system of projective varieties over these in the form of quantum codes, which give light cone structure. Penrose's twistors are related to this type of structure. Anyway, I will try to address more directly what it is that you say above later this weekend. Lawrence B. Crowell
 P: 361 Hello: In this post I will propose what graphs describe the Even and Hamilton representations of quaternions. A graph has a number of vertices and then edges which are made of pairs of vertices. The Hamilton and Even representations both have the same 4 vertices: (e, i, j, k). To keep this possibly related to physics, I like to think that in flat spacetime, the absolute value of each of these is equal to 1, but in the curved spacetime of GEM theory, the absolute value of e is the inverse of the absolute value of i, j, and k. For the Even representation of quaternions, all the vertices are connected to each other. The graph looks like a box with an X: .-. |X| ._. vertices: (e, i, j, k) edges: ((e, i), (e, j), (e, k), (i, j), (i, k), (j, k) name: K4:6 This is known as a complete graph because every pair of distinct vertices is connected to an edge. Here was a bit of fun I had mixing my physics in with pure math. I was thinking about the labels for the edges. For the edge (e, i), I thought I would just use i, similarly for j and k. For the edges connect two 3-vectors, it would be the third 3-vector. The problem is that no labels use e. That didn't sound good to me, ignoring e. What if I labeled each edge as i/e, j/e, or k/e? The physics of GEM suggest the absolute value of this will be one in curved or flat spacetime, so the edge is invariant. Nice. If one is at a vertex and wants to get to another vertex, read the label, and form the product, such as e (j/e) = j. What about the Hamilton representation of quaternions? We still have four vertices. The part of the graph that connects with e is exactly the same. What changes are the edges that connect the 3-vectors. Because the 3-vectors do not commute, the edges must be directional. The edge that connects i to j has a label of -k/e because -i k/e = j. The directional edge from j to i has the label k/e. Here is the graph for the Hamilton representation of quaternions: .=. |X() ._. vertices: (e, i, j, k) edges: ((e, i), (e, j), (e, k)) directed edges: ((i, j), (j, i), (i, k), (k, i), (j, k), (k, j)) name: K4:9 ? Note: the \ of the X should be two directed lines, the limitations of ASCII graphs. I am not certain if the K classification allows for including directed edges. It is significant that the graphs for the Hamilton and Even representations of quaternions are not the same. Gravity is not the same as EM, in fact, gravity is a little bit simpler (one charge, only attracts), and it is gratifying that the graph for the Even representation is a little simpler than that for the Hamilton representation. This was a lot of fun, I hope you enjoyed. I have never applied graph theory to anything before in my life. Doug
 P: 190 This is in part a test. I am having trouble getting a post sent here.
 P: 190 What about the Hamilton representation of quaternions? We still have four vertices. The part of the graph that connects with e is exactly the same. What changes are the edges that connect the 3-vectors. Because the 3-vectors do not commute, the edges must be directional. The edge that connects i to j has a label of -k/e because -i k/e = j. The directional edge from j to i has the label k/e. Here is the graph for the Hamilton representation of quaternions: .=. |X() ._. vertices: (e, i, j, k) edges: ((e, i), (e, j), (e, k)) directed edges: ((i, j), (j, i), (i, k), (k, i), (j, k), (k, j)) name: K4:9 ? Note: the \ of the X should be two directed lines, the limitations of ASCII graphs. Doug[/QUOTE] It is best to go all the way and consider the 120 icosian of quaterions, and its extension to the 240 cell with the 128 elements which give the 240 roots plus the 8 Cartan center elements. The basic group system for gauge theory is the heterotic group $E_8$, which is one of the Heterotic groups. There has been a lot of activity with this, and the Vogan DeCloux group found its system of root representations and possible impacts on elementary particles and gravitation. There is also a business called supersymmetry, which has 2^{N} elements in its representation. For something called N = 8 this has 256 elements. There is then the Clifford algebra, noted as $CL(16)~=~CL(8)\times CL(8)$ which gives a relationship between these 256 elements and the 240 roots of $E_8$ and its 8 weights (weights are the Cartan centers or eigen-matrices of the roots). The roots are what define the physical states. Now "half" of this CL(8) is defined by $$CL(8)~=~1~+~8~+~28~+~56~+~70~+~56~+~28~+~8~+~1$$ for spins 2, 3/2, 1, 1/2, 0, -1/2, -1, -3/2, -2. The 248-dim Lie algebra $E_8$ = 120-dim adjoint $Spin(16)$ + 128-dim half-spinor $spin(16)$ is rank 8, and has 240 root vectors that form the vertices of an 8-dim polytope called the Grosset polytope. Now the Clifford basis decomposition of the 256 there are the following elements $$CL(8)~=~1~+~8~+~(24+*4*)~+~(24+4+28)~+~(32+3+3+32)~+~(28+4+24)~+~ (24+*4*)~+~8~+~1,$$ The * * eclosed give the 4+4 = 8 that corresponds to the 8 $E_8$ Cartan subalgebra elements that are not represented by root vectors (they are the eigen-matrices), and the black non-underlined 1+3+3+1 = 8 correspond to the 8 elements of 256-dim Cl(8) that do not directly correspond elements of 248-dim $E8$. Now for the spins $\pm 1/2$ there is a 24 = 8 + 8 + 8, which are the three generations of fermions. These are the long roots associated with the 24, which corresponds to the elements of gauge fields which are the SU(3)xSU(2)xU(1) of the standard model. Now the 8 + 8 + 8 are the short roots which complement the roots of the SU(3), and define a triality condition on the fermionic sector. This triality condition appears to be a very central aspect of group theory and irreducible representations (ir-rep). I have been doing some calculations on a three fold structure with mutually unbiased bases on Jordan algebras (an ir-rep of E_8 elements with octonionic $OP^2$ structure), so that the heterotic E_8 is embeded in a three fold structure within the Leech lattice and a general system of modular forms. But to bring things to a more common level, the $E_8$ is a rich structure on symmetries with the numbers 2, 3, 5, 8. The two is on the helicity match up above, the 3 is on the triality condition of subgroup ir-reps, the 5 is a symmetry on an icosahedral system "dual" to the triality symmetry, and the 8 the whole system. This naturally embeds in a further three fold system called the Steiner system with the [5, 8, 24] structure for a "code." So the reason there are three families of quarks, three families of leptons (8 + 8 + 8) etc is due to this rich algebraic structure. Why there are four forces is again an aspect of group decomposition. The Cartan center for the $E_8$ is related to the other heterotic groups $G_2,~F_4,~E_6~ and~ E_7$. A basic representation is $F_4~=~C_{E_8}(G_2)$ (the Cartan center weighted on $G_2$), where $G_2~=~SU(3)~+~3~+~{\bar 3}~+~1$, which gives the nuclear force plus a hypercharge "1". The $F_4$ is a $D_4$ group $\sim~SO(7,1)$ which embeds gravity and the weak interactions plus the 8 + 8 + 8. Yet this structure operates equally for Euclidean and Lorentzian signatures. There is a loss of information, which is being glossed over. This does not solve the problem of ambiguities in holonomic structure for non-compact groups. In quantum field theory it is a common practice to Euclideanize time by letting time go to $t~\rightarrow it$. This time turns out to be related to temperature by $$t~=~\frac{\hbar}{kT}$$ This time is not exactly the same as the Lorentzian time, what we measure on a clock (or might we say what a clock “produces”), but is really a measure of the time where quantum fluctuations may be observed. Sometimes the term quantum fluctuations causes trouble, so it really is more the distance in a Euclidean 4-dim space where an instanton (a tunnelling state etc) with a certain magnitude can appear. As this temperature becomes very small the fluctuation time becomes large and the strength of the fluctuation, if we qualitatively invoke the Heisenberg uncertainty principle and consider this instanton time t as this uncertainty in time. As the temperature heats up it also means that the fluctuation is stronger or its coupling is made larger and a phase transition will ensue. This gets into some fascinating stuff! The Lorentzian time implies that the moduli space is not separable. Two moduli, points in the space of “gauge equivalent connections,” are not separable in a Hausdorff point-set topological definition. The topology is Zariski. I can go write more about this if needed, but this gets us into some rather serious stuff. So there is the Lorentzian time and there is this Euclidean time and there is a Wick rotational map between the two. So distinct fluctuation in the Euclidean case which have moduli that are Hausdorff, separable and “nice” correspond to a set of moduli in the Lorentzian case which are not. Physically this means there is some scale invariant physics (again to go into would require a bit of writing work) of phase transitions associated with the correspondence between fluctuations at various pseudo-time scales (or temperatures) and their Lorentzian versions. This correspondence and the phase transition is a quantum critical point, which have been observed with High temp superconductors and in the physics of Landau electron fluids in metallic crystals in the actinide plus transition range. This also shares some aspects of physics with the Hagedorn temperature of strings. Lawrence B. Crowell
 P: 361 Hello Lawrence: Thanks for all the information on the group E8, the largest exceptional Lie group. Garrett Lisi made international news back in November for a unification proposal that used E8 without resorting to strings. Unfortunately, I am not able to implement your suggestion. A brief description of icosian make it clear one is dealing with "a non-commutative algebraic structure". That does work of the Hamilton representation of a quaternion. I am using the word quaternion to mean a 4D division algebra. Many people work productively with the assumption that the only implementation of a 4D division algebra is the one Hamilton developed so many years ago (Gauss got there first by the way). In this thread, I found it necessary to formulate a new representation of a 4D division algebra where the elements commute. To make the Hamilton representation a mathematical field, one only needs to exclude the additive inverse 0 from the quaternions. For the Even representation, there are more quaternions that need to be excluded (0 and the eigen vectors) but it can be done. The Even representation of quaternions is not a Clifford algebra. Clifford algebras have orthogonal basis vectors, $e_i e_j = -e_j e_i$. For the Even representation, $e_i e_j = e_j e_i$. Oops. I am going to go to the 8th International Conference on Clifford Algebras and their Applications to Mathematical Physics at the end of May in Brazil, and claim the generalization of quaternions that Clifford Algebras represent is not enough to unify gravity and light (that will not go over well!). Let's think about the two graphs again, something I have done all weekend with great amusement. My central thesis is that we have yet to do the math of 4D like Nature practices it. In the graph for the Hamilton and Even representations, both have 4 vertices. Sometimes I think of them as events - a t, x, y, and z - but even better might be a difference between two events - a dt, dx/c, dy/c, dz/c. Recall the labels I used, which with this application would be dx/dt c, or $\beta_x$. The Even and Hamilton representations also share three edges, those for (e, i), (e, j), and (e, k). These seven shared elements indicate an overlap between the two representations. So how are they different? When one wants to connect two vertices that are in the 3-vector, there are two choices: use and edge or a directed edge. The Even representation uses an edge, .-., while the Hamilton representation uses two directed edges, .=. (please imagine arrows on both lines in opposing directions). I got so excited by this simple, direct message, that I designed and made a button with the graph for gravity and light last Saturday night, wearing it to a Boston swing dance. Fresh math! If anyone reading these notes would like a similar button, just email me off this thread with your snail mail address, and I will put one in the mail in a few weeks (I should reward those that are doing the work of trying to follow these nerdly riffs). doug sweetser@alum.mit.edu
P: 190
 Quote by sweetser Hello Lawrence: Thanks for all the information on the group E8, the largest exceptional Lie group. Garrett Lisi made international news back in November for a unification proposal that used E8 without resorting to strings.
It is an framing scheme which puts the particles into the $F_4$ and $G_2$ exceptional subgroups of the $E_8$. It is a neat idea in some ways, but there are few ambiguities, in particular how $spin(4,2)$ fits into the scheme. The heterotic string has $E_8\times E_8$ for two chiralities and in this one can supersymmetrize the theory. One does not need to have string theory explicitely in here, but the Jordan subalgebra naturally contains stringy structure, such as the 26-dimensional bosonic string.

 Quote by sweetser Unfortunately, I am not able to implement your suggestion. A brief description of icosian make it clear one is dealing with "a non-commutative algebraic structure". That does work of the Hamilton representation of a quaternion. I am using the word quaternion to mean a 4D division algebra. Many people work productively with the assumption that the only implementation of a 4D division algebra is the one Hamilton developed so many years ago (Gauss got there first by the way).
The quaternionic division algebra is noncommutative by definition. The wiki-p site just discusses the Hamilton vertex permutation on the icosian. There is a whole lot more structure to this system.

 Quote by sweetser In this thread, I found it necessary to formulate a new representation of a 4D division algebra where the elements commute. To make the Hamilton representation a mathematical field, one only needs to exclude the additive inverse 0 from the quaternions. For the Even representation, there are more quaternions that need to be excluded (0 and the eigen vectors) but it can be done.
This is a contradition in definition. The Cayley numbers 1, 2, 4, 8 lead to the reals, complexes, quaterions and octonins. Quaternions as a division algebra are noncommutative.

 Quote by sweetser The Even representation of quaternions is not a Clifford algebra. Clifford algebras have orthogonal basis vectors, $e_i e_j = -e_j e_i$. For the Even representation, $e_i e_j = e_j e_i$. Oops. I am going to go to the 8th International Conference on Clifford Algebras and their Applications to Mathematical Physics at the end of May in Brazil, and claim the generalization of quaternions that Clifford Algebras represent is not enough to unify gravity and light (that will not go over well!).
This appears related to a graded structure. The even quaterionic structure involves elements $e_i~=~\xi_a e^a_i$, for $\xi^a$ a Grassmannian element. The quaterions then define supermanifold coordinates with

$$y^i~=~x^i~+~{\bar\theta}\sigma^i\theta.$$

You will then have $\{e^a_i,~e^b_j\}~=~\eta^{ab}g_{ij}$, which are fermionic correspondences with quaterionic elements. The quaterionic fields are framed within a Clifford basis, such as a connection form

$${\cal A}~=~A~+~\Gamma\cdot\psi$$

So that in the case of QCD the SU(3) is combined with the $3~+~\bar 3$ in $G_2$. Similarly the fields can be again framed with Grassmannian elements.

This will then extend these elements to a graded algebraic correspondence between bosons and fermions. I have had some communciations with Garrett Lisi and others on just this issue of extending the framing system to a graded Lie algebraic structure.

Lawrence B. Crowell

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