[Malamute Lover]

In short, do you really know what you are talking about as opposed to just throwing fancy words around?

I know what I'm talking about: I got it in pictures at an attachment with the post: "Quantum Gravity Follows?" in this forum. You sound like you would be interested in it.

The article cited makes no sense. It is a collection of buzzwords from the field of Physics that is full of problems. The journal in which it was published was originally mostly dedicated to condensed matter physics where it was fairly well respected. But it was too limited a field with too much competition and to stay above water they gradually started publishing anything and everything that even looked like physics and/or mathematics regardless of actual value. In the same issue that the article you linked appears, there was another article that claimed that prime numbers were predictable and one of the sources it cited claimed that Pi was really a rational number.

If you think your work is so groundbreaking why not send it to Physical Review Letters? They publish oddball speculative things. Of course the first thing they do is check your academic background in the field of physics. :

You have yet to demonstrate that you know what you are talking about.

It is improper terminology to say ‘a set with additional structures’

So it's a set endowed with features. The statement is standard in Mathematics.

You did not know that until I told you. Yet it is an important part of your argument.

A circle drawn on representation of a Riemann sphere will generally not look like a circle but will be ‘squashed’ in the direction of the ∞ pole.

No. A circle through the point at infinity will look like a circle, it will just be denser closer to infinity but a denser line is still a line. This is in the standard representation of a sphere in a coordinate system with the z-axis pointing 45 degrees downwards. Maybe in some other projection, it will appear squashed.

A circle through the ∞ pole is not possible because a center point from which it is equidistant cannot be defined.

A geometric line has zero thickness. It cannot be thicker or thinner.

The Riemann sphere is a mapping of the extended complex plane onto a sphere. It consists entirely of surface. There is no z-axis.

A coordinate system based on axes is orthogonal by definition. There is no such thing as an axis ‘pointing 45 degrees downwards’.

In the Riemann sphere, the Origin (where all values are zero) is located at the ‘south pole’. All complex values contained on the complex plane on or within a circle of radius 1 lie on or below the ‘equator’ of the Riemann sphere. All other values are in the ‘northern hemisphere’ with ∞ at the ‘north pole’. There are no finite complex values on this pole and in fact the distinction between real and imaginary vanishes, since the infinitely distant edges of the complex plane are rolled up into a point.

Since all complex numbers with x and/or y values greater than 1 and i respectively are above the ‘equator’ and the ‘north pole’ is ∞, the geometric distance between successive values of a given positive increment becomes smaller and smaller as these values approach the pole. Using the definition of a circle as the set of points equidistant from a given point, a circle drawn on the surface of a Riemann sphere appears more and more squashed as the center point moves ‘northward’ because the geometric distance between value differences on the surface of the sphere becomes smaller and smaller.

Please explain what you intended to be understood by ‘a circle in the Riemann Sphere and how it fits into your explanation. Presumably, you meant ‘on’ and not ‘in’. There is no ‘in’ relative to a Riemann Sphere. It is all surface.

I can talk about a line "in" a plane if the plane is infinitesimally thick (the thickness is non-zero).

The Riemann sphere is a projection of a plane onto a sphere. A plane has zero thickness and so does the surface of a sphere. An infinitesimally small but non-zero thickness makes no sense unless you are trying to introduce non-standard analysis. Which I do understand, so if you are going to import more buzzwords, I am ready for you. Hint: A Large Cardinal is not a Big Bird.
.

By describing a ‘left out spacetime node’ as a hole in the Riemann Sphere. It would appear that you mean some region of the Riemann Sphere cannot be the result of any calculation, that the coordinates interior to this region are forbidden in some fashion.  Can you shed any light on what these coordinates might be and how you came to determine them? And how does the ‘circle’ come into play?

The circle is like a line of longitude. The holes are holes on a line through the center of the sphere and perpendicular to the momentum vector of the particle, where the holes are located where this line intersects the sphere. I determined them by reasoning about the production of an accelerating force on a photon (protophoton: a photon not yet traveling at the speed of light).

On a Riemann sphere, a line of longitude intersects both 0 and ∞. A circle is the set of points equidistant from a given point. On an ordinary sphere, a line of longitude is a pole intersecting Great Circle. There are two possible center points, both on the equator lying on another pole intersecting Great Circle 90° around from the given longitude. However, on a Riemann sphere the length from pole to pole is infinite, from zero to ∞.  There is no unique point or pair of points that can be defined as the center point of the circle. Any point you choose has a varying distance to the different parts of the circle. The equidistant requirement cannot be satisfied. It is not possible to construct a pole intersecting circle on a Riemann sphere. It may look like a circle on a globe but it is not one on a Riemann sphere.

Also, a hole though a Riemann sphere (which does not have an inside anyway) is not a valid concept since the surface does not have a linear metric. A straight line through a Riemann sphere would only be definable at the equator. Otherwise there are no straight lines.

Also why are you representing spacetime as a Riemann Sphere?  The time dimension requires the use of imaginary numbers. But you need three real dimensions to represent spatial coordinates. How do you map these three dimensions onto a Riemann Sphere?

I propose copying one dimension of space and one of time into a Riemann Sphere, resulting in a particle when endowed with charges.

A particle has three translational degrees of freedom in space. You cannot represent the state of a particle in one dimension. Good luck on defining momentum vector. And how do ‘charges’ (in your expanded definition) fit into the picture? Yes, I read your paper. No, it does not answer the question.

And what features are you endowing to the complex (involving imaginary numbers) structure (set with features) of the Riemann Sphere that results in a ‘hole’?

Points of space on the circle versus left out points (infinitesimal breaks in the circle).

We have already seen that a line of longitude on a Riemann sphere is not a circle. So either change your terminology or change your model.

In addition, infinitesimally small holes do not work well with quantum theory.

Also, you need to talk more about what structure(s) you want to endow to the set of points on the surface of the Riemann sphere to build a coherent picture. In particular in what fashion do the required features get endowed?

[Kryptid]

You would just be able to reproduce the proof if you are telepathic, knows Earth's voice, and if your mind knows how to copy spacetime into a Riemann Sphere.

Can you provide any actual evidence for those claims?

[talanum1]

Can you provide any actual evidence for those claims?

Just my experience, so no objective proof. You may ask a Psychic person.

A circle through the ∞ pole is not possible because a center point from which it is equidistant cannot be defined.

The projection allows a circle through the pole: the point at ∞ is projected at a unit distance from the center of the sphere. I've seen a picture of a Riemann sphere. The distance along a longitude line from the equator to the north pole is finite in the projection (not to scale).

There is no such thing as an axis ‘pointing 45 degrees downwards’

The projection of the z-axis onto 2 dimensions is such.

Also, a hole though a Riemann sphere (which does not have an inside anyway) is not a valid concept since the surface does not have a linear metric. A straight line through a Riemann sphere would only be definable at the equator. Otherwise there are no straight lines.

By the projection, one could imagine the sphere to be an ordinary sphere with a linear metric by reassigning distances. There is a one-to-one mapping between a Riemann sphere and an ordinary sphere. The 3-dimensional space it is in allows for straight lines as chords intersecting the Riemann sphere.

A particle has three translational degrees of freedom in space. You cannot represent the state of a particle in one dimension. Good luck on defining momentum vector. And how do ‘charges’ (in your expanded definition) fit into the picture? Yes, I read your paper. No, it does not answer the question.

You have it wrong: my model defines a particle as equivalent to a Riemann sphere with holes. Actually more than one superimposed Riemann sphere for particles other than the photon. I just need 3 dimensions to put the particle in. I simply encode the charges as added or left out points. That way spacetime does not need a table of particle names and properties at each event of spacetime. If you have ever considered what it would take to implement calculated laws of physics this would seem logically expedient.

Also, you need to talk more about what structure(s) you want to endow to the set of points on the surface of the Riemann sphere to build a coherent picture. In particular in what fashion do the required features get endowed?

The structure of a 2-sphere.

In addition, infinitesimally small holes do not work well with quantum theory.

Call them little Planck-length openings then.

[Kryptid]

Just my experience, so no objective proof. You may ask a Psychic person.

No evidence = no science.

[Malamute Lover]

A circle through the ∞ pole is not possible because a center point from which it is equidistant cannot be defined.

The projection allows a circle through the pole: the point at ∞ is projected at a unit distance from the center of the sphere. I've seen a picture of a Riemann sphere. The distance along a longitude line from the equator to the north pole is finite in the projection (not to scale).

There is no center of a Riemann sphere. It consists only of the surface. It is a projection of a plane and such has only 2 dimensions, just like the plane. The length of a line from the 0 point to the ∞ point is infinite. If you think it is finite, then there is some maximum value for x and y, which would mean that the entire complex plane is not projected onto the sphere. The whole point of the Riemann sphere is that ∞ is defined and a bunch of otherwise nasty functions become well behaved without needing to put arbitrary constraints on them.

You do not understand what a Riemann sphere is. Trying to make it into an ordinary sphere fails big time. You are just using impressive sounding words in order to sound impressive.

There is no such thing as an axis ‘pointing 45 degrees downwards’

The projection of the z-axis onto 2 dimensions is such.

You are talking about a picture. Just like with the Riemann sphere, you are looking at a picture and thinking it somehow is the real thing. The placement of the z-axis on the xy plane in such pictures is entirely arbitrary. It is simply a way of showing three dimensions on a two-dimensional page. In fact, there is no such thing as mathematical projecting a third orthogonal axis onto a plane.

Projection of points having non-zero z coordinate values onto the xy plane is a simple matrix operation. For vector (x,y,z) the projection vector is (x,y,0).

Also, a hole through a Riemann sphere (which does not have an inside anyway) is not a valid concept since the surface does not have a linear metric. A straight line through a Riemann sphere would only be definable at the equator. Otherwise there are no straight lines.

By the projection, one could imagine the sphere to be an ordinary sphere with a linear metric by reassigning distances. There is a one-to-one mapping between a Riemann sphere and an ordinary sphere. The 3-dimensional space it is in allows for straight lines as chords intersecting the Riemann sphere.

The picture is simply for visualization purposes. Imagining a Riemann sphere to be an ordinary sphere removes all its unique properties. A Riemann sphere is surface only and is not embedded in three dimensions. Even an ordinary sphere which is a ‘rolled up’ projection of a finite section of a plane still only has a surface and is not embedded in three dimensions.

Stop looking at pictures and learn some math.

A particle has three translational degrees of freedom in space. You cannot represent the state of a particle in one dimension. Good luck on defining momentum vector. And how do ‘charges’ (in your expanded definition) fit into the picture? Yes, I read your paper. No, it does not answer the question.

You have it wrong: my model defines a particle as equivalent to a Riemann sphere with holes. Actually more than one superimposed Riemann sphere for particles other than the photon. I just need 3 dimensions to put the particle in. I simply encode the charges as added or left out points. That way spacetime does not need a table of particle names and properties at each event of spacetime. If you have ever considered what it would take to implement calculated laws of physics this would seem logically expedient.

If you need three dimensions, you cannot use a Riemann sphere. It only has two dimensions just like the plane it maps to.  It consists only of the surface.

Spacetime does not contains names of particles at each point, only the properties as they interact with whatever fields may apply. A field theory predicts what forces will be in effect at each point. It is the field that controls what happens at the points. No need for each point to figure out what should happen. And properties fall nicely in the symmetries of the Standard Model. These properties are in effect regardless of where the particle is. The interaction of these properties with the field(s) determines what happens, including whatever transformations may take place. Spacetime does not need a table of particle names and properties for each point.  The imagined interior has no coordinates assigned to it.

Also, you need to talk more about what structure(s) you want to endow to the set of points on the surface of the Riemann sphere to build a coherent picture. In particular in what fashion do the required features get endowed?

The structure of a 2-sphere.

 A 2-sphere is a circle. As we have seen, your longitudinal circles do not work on a Riemann sphere. There is no definable center point. A conventional sphere can have longitudinal circles. 

In addition, infinitesimally small holes do not work well with quantum theory.

Call them little Planck-length openings then.

I would make them larger than that. Space at the Planck length is unstable because its Heisenberg Uncertainty self-energy is enough to make a black hole.

[talanum1]

The distance from zero to ∞ of the sphere as you measure it on the projection with a yardstick that doesn't get denser is finite. It is just conceptually equal to ∞. You are caught up in the conceptual. One can make a finite model of the Riemann Sphere (RS).

One could conceptually say that one can line up the RS with events of spacetime such that there is a plane that cuts the RS at a circle, on which (plane) one can draw a line. Spacetime just needs to be locally flat.

The holes must be smaller than a photon. The size of a quantum of spacetime. Yes, a photon has a size in my model.

It seems to me that somehow it is not satisfying to say a particle is a vibration in a field.

I thought about implementing the laws of physics and came to the conclusion that particle properties must be communicated between fields. Just think about an electron scattering off another electron.

[Bored chemist]

Just my experience, so no objective proof.

There aren't any.

Did you somehow think that was the way adults discuss things?

[Kryptid]

What physical experiment could be performed to demonstrate that particles are Riemann spheres? Keep in mind that if it can't be experimentally tested, then it isn't a scientific hypothesis.

[Malamute Lover]

The distance from zero to ∞ of the sphere as you measure it on the projection with a yardstick that doesn't get denser is finite. It is just conceptually equal to ∞. You are caught up in the conceptual. One can make a finite model of the Riemann Sphere (RS

One could conceptually say that one can line up the RS with events of spacetime such that there is a plane that cuts the RS at a circle, on which (plane) one can draw a line. Spacetime just needs to be locally flat. ).

The Riemann sphere IS conceptual. One cannot make a finite model of it. You think that because it is pictured as a globe that it is a globe. The surface is a mapping of the complex plane. There is no inside and the surface is infinite. The image of a globe appears to be the sum total of your knowledge of the Riemann sphere. You are just playing games with impressive sounding words to try to make yourself sound impressive. It is not working.

You have no real knowledge of serious mathematics. Give it up.

The holes must be smaller than a photon. The size of a quantum of spacetime. Yes, a photon has a size in my model.

A quantum of spacetime is not a meaningful concept, at least not the way you want to use it. The concept of a quantum generally involves a fundamental unit, for example, the electric charge of an electron. All electric charges are multiples of this fixed quantum.

At the scale of a Planck volume, the self-energy due to the Uncertainty Principle combined with the extremely small dimensions results in a black hole. One might want to speak of the size of the black hole in Euclidean terms, e.g., the diameter. However, a black hole does not have a diameter. It is all surface. There is no mapping of that region of space into a Euclidean space because, as with macro black holes, division by zero happens. Moreover, near that scale spacetime fluctuations, also due to the Uncertainty Principle, do not allow a fixed length to be defined.

‘The size of a quantum of spacetime’ is not a meaningful term.

You have no real knowledge of serious quantum physics. Give it up.

It seems to me that somehow it is not satisfying to say a particle is a vibration in a field.

Quantum field theorists speak of particles as disturbances or excitations in a field. The term ‘vibration’, although often used in popularized science explanations, can be misleading since it could imply that a particle can be described simply by its wavelength.  All particles with mass have a Compton wavelength but they also have other characteristics.

I thought about implementing the laws of physics and came to the conclusion that particle properties must be communicated between fields. Just think about an electron scattering off another electron.

The idea of interactions of excitations in a field is the basis for field type physical theories, the first serious example of which was Maxwell’s equations formulated over a century and a half ago. Quantum Field Theory (QFT) is a newer and much more complicated example.

I am confused by your phrase “communicated between fields”. The field is the underlying background in which the excitations (particles) interact.  In QFT the field incorporates classical mechanics and electromagnetism with quantum mechanics (including QED and QCD) and Special Relativity. No one has yet figured out how to get General Relativity to play nice with the other children.

So what does “communicated between fields (plural)” mean?

Or is it that you have no real knowledge of what a field theory is, either?

[talanum1]

What physical experiment could be performed to demonstrate that particles are Riemann spheres?

Orientate an Electron such that it's axis (pointing from the point 0 to the point ∞) points in the "up" direction. Then it will not emit a photon in the "down" direction.

Also: particles seen from their northern hemispheres will look different as seen from their southern hemispheres.

So what does “communicated between fields (plural)” mean?

Think of the Feynman diagram of an electron scattering off an electron. What happens is: one electron must tell the electromagnetic field that it wants to scatter of another electron following a path λ, and it must specify its momentum. This must be read off the electron field interacting with spacetime. Then the electron must specify it's expected momentum change. Then the electromagnetic field must compute the virtual photon direction and wavelength and the positions for invoking the creation and annihilation operators, then the virtual photon must communicate the change of direction of momentum to the other electron. Only then can the virtual photon be emitted.

This is the story that must be considered when trying to simulate the scattering on a computer.

The Riemann sphere IS conceptual.

Wrap your conceptual head around this: the RS is conceptually a distortion of the Complex plane such that the point at ∞ sits at a sphere's north pole. As such it sits at a finite distance as measured with an undistorted measuring rod.

[Malamute Lover]

What physical experiment could be performed to demonstrate that particles are Riemann spheres?

Orientate an Electron such that it's axis (pointing from the point 0 to the point ∞) points in the "up" direction. Then it will not emit a photon in the "down" direction.

Also: particles seen from their northern hemispheres will look different as seen from their southern hemispheres.

Wrong. The direction of photon emission when an electron drops from excited state to ground state is random with respect to the electron's spin state, which is the only way of talking about an electron’s axis.  The direction of photon emission is determined in a probabilistic manner by multiple factors including the prevailing magnetic field, of which the nearby nucleus is a major component, and preserving energy momentum conservation over event accumulation.

In addition, electron spin is a matter of angular momentum, in effect spinning clockwise or counterclockwise. It is not a matter of which way a pole is pointing. Electrons with opposing spin values are not ‘upside down’ with respect to each other. They are spinning in opposite directions. To drive that home, the only formulation for spin states that matches measurements is the formalism of angular momentum.  For that to be the case, the clockwise or counterclockwise spin must be with respect to a fixed axis. That is, your hypothetical 0 and ∞ axis must always point in the same direction, negating your argument.

Now explain what differences there may be between the two hemispheres of an electron. Oceans, continents, ice caps? Your insistence that the metrics of a Riemann sphere are those of an ordinary sphere and not the varying metrics of a complex plane mapping argues against any difference.

Also, since the question Kryptid asked is about a physical experiment, how would you determine which photon came from which electron and how to determine the axis orientation of the individual electrons?

So what does “communicated between fields (plural)” mean?

Think of the Feynman diagram of an electron scattering off an electron. What happens is: one electron must tell the electromagnetic field that it wants to scatter of another electron following a path λ, and it must specify its momentum. This must be read off the electron field interacting with spacetime. Then the electron must specify it's expected momentum change. Then the electromagnetic field must compute the virtual photon direction and wavelength and the positions for invoking the creation and annihilation operators, then the virtual photon must communicate the change of direction of momentum to the other electron. Only then can the virtual photon be emitted.

Wow, where do I start?

To begin with, you are using two different meanings of the term ‘field’. Originally you used the term with respect to a particle being a ‘vibration’ in a field, which would imply field theory, a field being a mathematical apparatus for determining the properties at any given point. Now you are using the term to refer to the electromagnetic fields of electrons. Same word, different meanings.

The term λ (lambda) does not appear in Feynman diagrams. That term is used in General Relativity for the Cosmological Constant. What does appear in Feynman diagrams is the term γ (gamma), which is the symbol for the photon. In Feynman diagrams, it is the label applied to the wavy line that indicates the exchange of virtual photons. This diagram show electron-electron scattering.



In Quantum ElectroDynamics (QED) for which Feynman diagrams were first invented, the path is not first chosen. In theory, any path at all might be used, each having its own probability amplitude. This is a wavelike entity. As such, probability amplitudes constructively and destructively interfere with each other. The effective path is the one with the greatest surviving probability amplitude, and usually the only survivor with a resulting probability (square modulus of amplitude) of unity. Paths happen. They are not chosen. There are experiments demonstrating multiple paths being used such as the famous double-slit example.


There are people here who are familiar with deep Physics and deep Mathematics. You cannot make it up as you go along and not get called out.

[Malamute Lover]

The Riemann sphere IS conceptual.

Wrap your conceptual head around this: the RS is conceptually a distortion of the Complex plane such that the point at ∞ sits at a sphere's north pole. As such it sits at a finite distance as measured with an undistorted measuring rod.

Wrap your head around this. The Riemann sphere IS the complex plane with ∞ defined. ∞ does not exist on the complex plane. Having ∞ defined allows the use of functions that might for some values involve division by zero as long as the end result of the evolution of the function does not result in division by zero or otherwise carry forward any infinite term, since this is an unusable result.

The Riemann sphere is NOT a distortion of the complex plane. If it were it would yield different answers than the complex plane and that would be pointless. The distance from (x=1, y=0i) to (x=2, y=0i) is the same as the distance from (x=1000001, y=0i) to (x=1000002, y=0i) regardless of whether we are talking about the Riemann sphere or the complex plane. There is no distortion. Thinking the Riemann sphere acts like an ordinary geometric sphere does not work. If you want to look at the Riemann sphere as an ordinary sphere, then the metric is not constant. If you want to insist there is a constant metric, then you cannot have ∞ at the pole. And if you do not have ∞ at the pole then you do not have a Riemann sphere.

Your problem is that you need to see things as pictures rather than trying to grasp the concepts. This is how you ended up with the idea that the z-axis is at 45° because you looked at a picture and misunderstood it.  You want your particles to be spheres with circles drawn on them and holes in the circles. You saw a picture representing the Riemann sphere and thought that was an impressive sounding thing that you could get away with because nobody would understand it. Surprise!

[talanum1]

The Riemann sphere is NOT a distortion of the complex plane.

I'll ask a Mathematician on the other forum and refer you.

Now explain what differences there may be between the two hemispheres of an electron.

The northern hemisphere would have a higher density of spacetime events than the southern hemisphere.

Also, since the question Kryptid asked is about a physical experiment, how would you determine which photon came from which electron and how to determine the axis orientation of the individual electrons?

Isolate a single electron in a magnetic trap with a known magnetic field imposed. They can do the double-slit experiment with one electron at a time.

Your insistence that the metrics of a Riemann sphere are those of an ordinary sphere and not the varying metrics of a complex plane mapping argues against any difference.

No, I was wrong. It must be the RS metric. The particle lives in locally flat space so there is a line intersecting the RS perpendicular to the momentum, where the holes are located.

[Malamute Lover]

The Riemann sphere is NOT a distortion of the complex plane.

I'll ask a Mathematician on the other forum and refer you.

A Riemann sphere is not a distortion. It is an identity map plus a ∞ point.  f(z) = 1/z. A distortion would imply that it gives different answers than the complex plane and in some irregular manner.

But then you do not understand how a Riemann sphere works. It was just an impressive sounding phrase that you did not think anyone could challenge you on.

I am waiting for that referral.

Now explain what differences there may be between the two hemispheres of an electron.

The northern hemisphere would have a higher density of spacetime events than the southern hemisphere.

Since you previously said that the surface of a Riemann sphere is really linear, as you have said earlier and as is clear from your use of the term ‘higher density’, then please tell me at what latitude the size of a spacetime event goes below the Planck area. And how you manage to have a realized infinity at the ∞ pole with an infinite number of spacetime events at one point. If you leave out ∞ then you do not have a Riemann sphere, that being the real power of a Riemann sphere, that 1/0 is defined. BTW that is complex ∞ at the pole. What meaning do you assign to that?

Earlier you referred to an electron as a Riemann sphere. Are there more spacetime events in the northern hemisphere than in the southern hemisphere. What does it mean to have spacetime events on the surface of an electron? And how do you reconcile representing an electron being a Riemann sphere with your definition of a particle as the holes in a Riemann sphere?

Please give a concise explanation of how your model works. Never mind justifications or ramifications. Just describe the model itself clearly and concisely.

Also, since the question Kryptid asked is about a physical experiment, how would you determine which photon came from which electron and how to determine the axis orientation of the individual electrons?

Isolate a single electron in a magnetic trap with a known magnetic field imposed. They can do the double-slit experiment with one electron at a time.

That has nothing to do with the direction a photon would go from an electron in an excited state. The magnetic field would take part in determining the direction, not some hypothetical ∞ pole on the electron. How do you know which way the pole is pointing? If you do not know that, you have no way to verify your claim.

Were you talking about the photon or the electron going through the slits? Presumably the photon, which of course would require that you know which direction the photon is going. However, unless you know which way your hypothetical ∞ pole is pointing – something you have not yet addressed – it is unrelated to your claim. Regardless, as I said earlier, the magnetic field is going to influence the direction.

On top of all this, whether the accumulation of data in the double slit experiment will yield a shotgun (no self-interference) pattern or a concentric circles (self-interference) pattern on the receiver depends on whether or not there are detectors in the slits.

What I see here is that you have grabbed another impressive sounding term to add to your collection, ‘double slit experiment’, with no comprehension of what it means.

Your insistence that the metrics of a Riemann sphere are those of an ordinary sphere and not the varying metrics of a complex plane mapping argues against any difference.

No, I was wrong. It must be the RS metric. The particle lives in locally flat space so there is a line intersecting the RS perpendicular to the momentum, where the holes are located.

When you clearly and concisely explain your model as asked above, be sure to include how the above terms fit in. And once again, explain how with two dimensions you manage to depict three spatial dimensions, with six degrees of freedom, and time. And mass, frequency etc.

And does this mean that things you said earlier in this post and in earlier posts are wrong?

[talanum1]

But then you do not understand how a Riemann sphere works. It was just an impressive sounding phrase that you did not think anyone could challenge you on.

I got it drawn see below.

Since you previously said that the surface of a Riemann sphere is really linear

I was wrong it should have the metric of the Riemann sphere, but live in Euclidean spacetime even if events of spacetime then don't align with the events on the particle.

What does it mean to have spacetime events on the surface of an electron? And how do you reconcile representing an electron being a Riemann sphere with your definition of a particle as the holes in a Riemann sphere?

You got it wrong: the RS is the particle, the holes encode its charges. Actually an electron is two superimposed RSs superimposed with two anti-RS, with another particle (a gravitron) superimposed on that.

The whole model is posted at  "Quantun Gravity Follows?" in this forum (see the newest reply attachment: "Physics from Axioms clean,pdf" and "Physics from Axioms Advanced.pdf").

[talanum1]

You seem to suggest space can compute at the same time as outputting a particle and that the computation takes an infinitesimally long time to execute. I don't think this is the case.

[talanum1]

I have done some experiments with my mind and now it seems an electron is a single RS, not a superposition of RSs.

[Bored chemist]

I have done some experiments with my mind

Did you think that actually meant something?

[talanum1]

Did you think that actually meant something?

I believe so, I can tell if any information is not from my mind.

[Bored chemist]

I can tell if any information is not from my mind.

No, you can not.
That was sorted out a few hundred years ago.