Is Gravity Complex?

[Dubbelosix]

 

I've had quite a lot of reading to do, basically getting primers on geometric algebra. I have read what may have appeared to be conflicting evidence that it is a general rule that you cannot add a scalar with a vector, and essentially that is what this does:

 

[math]uv = u \cdot v + u \wedge v[/math]

 

How do you determine that this is a product of a symmetric and skew symmetric matrix? A general thing to remember is that skew symmetries are expressed using cross products, It is also of interest to define the dyad:

 

''The dot product takes in two vectors and returns a scalar, while the cross product returns a pseudovector. Both of these have various significant geometric interpretations and are widely used in mathematics, physics, and engineering. The dyadic product takes in two vectors and returns a second order tensor called a dyadic in this context. A dyadic can be used to contain physical or geometric information, although in general there is no direct way of geometrically interpreting it. The dyadic product is distributive over vector addition, and associative with scalar multiplication. Therefore, the dyadic product is linear in both of its operands. In general, two dyadics can be added to get another dyadic, and multiplied by numbers to scale the dyadic. However, the product is not commutative; changing the order of the vectors results in a different dyadic.''

 

from wiki. For two vectors [math](u.v)[/math] the geometric produict is the sum of a symmetric product and antisymmetric product;

 

[math]ub = \frac{1}{2}(uv + vu) + \frac{1}{2}(uv - vu)[/math]

 

The inner product of the vectors is

 

[math]u \cdot v := g(u,v)[/math]

 

So that a symmetric product is expressed as

 

[math]\frac{1}{2}(uv - vu) = \frac{1}{2}((u + v) - u - v)^2 = u \cdot v[/math]

 

It is stated in what I have read, that conversely [math]g[/math] is completely determined by the algebra and the antisymmetric part of the exterior product of the two vectors is

 

[math]u \wedge v := \frac{1}{2}(uv - vu) = -(v \wedge u)[/math]

 

then by addition we obtain

 

[math]uv = u \cdot v + u \wedge v[/math]

 

which is the vector form of the geometric product. Taking the dot product of the Pauli vector (which is a dyad) is in fact the same thing

 

[math](\sigma \cdot u)(\sigma \cdot v) = u \cdot v + u \wedge v[/math]

 

An equivalent form of this, of course involves the cross product, which as explained, is given by its antisymmetric parts:

 

[math]u \cdot v + i \sigma \cdot (u \times v)[/math]

 

 

And so using the information I have gathered, it is clear the Einstein equation can also be written as:

 

 

[math]\mathbf{G}_{\mu \nu} = \mathbf{T}(\gamma_\mu \gamma_{\nu}) = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\cdot \gamma_{\nu} + \gamma_{\mu} \wedge \gamma_{\nu}) = \frac{1}{2}\mathbf{T}[\gamma_{\mu} \cdot \gamma_{\nu} + i \sigma \cdot (\gamma_{\mu} \times \gamma_{\nu})][/math]

Edited December 17, 2018 by Dubbelosix

[Dubbelosix]

It's difficult to interpret the cross product

 

[math]\gamma_{\mu} \times \gamma_{\nu}[/math]

 

because it implies a cross product in four dimensional space. Once again, the conflicting material on the internet is causing me a headache. It seems, there is literature stating you cannot have a cross product in anything other than 3 or 7 dimensional space, that means there is no cross product in four dimensions.

 

https://en.wikipedia.org/wiki/Seven-dimensional_cross_product


https://www.jstor.org/stable/2323537?seq=1#page_scan_tab_contents
 

 

But then I come across other material which is very on topic with the algebra I am using stating it is possible to express a cross product in four dimensions

 

https://www.sciencedirect.com/science/article/pii/B9780080507552500282

 

I haven't been able to access this paper though through sci-hub.

:

[Dubbelosix]

At least it should be possible in principle? Perhaps I just don't understand the arguments enough. In four dimensions the equivalent would be to take three vectors and return a fourth that is perpendicular to the remaining three.

 

 

[Dubbelosix]

Seems I found the paper eventually

 

http://sci-hub.tw/http://www.sciencedirect.com/science/article/pii/B9780080507552500282

[Dubbelosix]

Ok so I have read the paper, it doesn't display a four dimensional case for cross products but does explain why four dimensional cases are no good. I suppose if you were willing to give up on the three dimensionality of [math]\gamma_{\mu}[/math] and treat it as only one vector, then sure, [math]\gamma_{\mu} \times \gamma_{\nu}[/math] is possible, but right now I am unsure how to proceed.

[Dubbelosix]

I think I can answer this now.

In the 3D space [math]R^3[/math] the outer product of any four vectors [math](a,b,c,d)[/math] is zero, an automatic consequence of the outer product properties.

In three dimensions only three vectors can be independent and therefore a fourth must be expressible as a weighted sum of the other three:

[math]d = \alpha a + \beta b + \gamma c[/math]

Associativity, distributivity and antisymmetry make the outer product of the four vectors zero:

[math] a \wedge b \wedge c \wedge d = a \wedge b \wedge c \wedge (\alpha a + \beta b + \gamma c) = 0[/math]

Where [math]\gamma[/math] in this case is not a gamma matrix and [math]a \wedge b \wedge c \wedge d[/math] is known as a quadvector.

The link states so the highest order element that can exist in the subspace algebra of [math]R^3[/math] is a trivector.

Apparently this has nothing to do with a limitation of the outer product but rather it satisfies a geometric uselessness of the construction of elements of higher dimension than [math]R^3[/math] - it can however be interpreted that the element [math]0[/math] as the empty subspace of any dimensionality so this one element $0$ is the zero scalar, the zero bivector and so on. There is no algebraic or geometric reason to distinguish between those, for the empty subspace has no orientation or weight.


https://books.google.co.uk/books?id=-1-zRTeCXwgC&pg=PA32&redir_esc=y#v=onepage&q&f=false

Edited December 17, 2018 by Dubbelosix

[Dubbelosix]

Polymath, I cannot speak one to one right now, but this geometric approach, if you like could be seen as evidence of an approach to a seven dimensional spacetime outside of string theory. I wouldn't pounce on it just yet though. It could be that for gravity the second term simply vanishes.

[Dubbelosix]

more precisely, M-theory or hyperspace theory is the theory of seven dimensional space. String theory embraced it and developed into 11 dimensions.

[Dubbelosix]

I'm starting to ask new questions about the mysterious cross product term arising in what should be a four dimensional theory of gravity. The cross product presented problems because it would seem to suggest it prefers a seven dimensional case: I also suggested it may also be true that the terms simply vanish in a true theory of gravity - in which case, the cross product term appears very similar to Bianchi identities? Let's put that one to the side, we know already the formalism in the second term has [something] to do with spin. Unlike gravity, I tend to associate spin may more fundamentally related to magnetism, such as it is, magnetism affects the spin of a quantum system. ''Can we describe a spin in four dimensions(?)'' became my next question...

 

A four dimensional spin in a Euclidean space and this presents an interesting case for a six dimensional structure - spin requires extra degrees of freedom so physicist tend to build these theories on higher dimensional spacetime. Mathematically, it is related to the system being non-commutative. We already know this could be possible [since] it involves the cross product and it is non-commutative from principle.

 

https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space

 

If gravity was being described six dimensionally, we still cannot make sense however of the cross product in dimensions other than [math]3[/math] or [math]7[/math], both are at odds with a commonsense view... but quantum mechanics doesn't necessarily have to have a commonsense view from historical reference.  In the latter case, we would have to invoke that it belongs in the odd-dimensional rotation group and so would not have a central inversion and itself, part of the simple group.

 

https://en.wikipedia.org/wiki/Simple_group

 

There does exist in literature however, a notion of spin-7 manifolds on an even worse idea, an eight dimensional spacetime:

 

http://paperity.org/p/52474693/spin-7-manifolds-in-compactifications-to-four-dimensions

 

There exists more literature though on higher spin gauge theory in four dimensions:

 

https://arxiv.org/pdf/1102.3297.pdf

 

And maybe the presence of the spin on that term may require such a correction? The approach has diffeomorphisms will reduce instead to gauge transformations, which it states has been vital to the construction of topological field theories and suggested important for a gravitational interpretation to boot. The interesting fact for me is that we loose the definition of the diffomorphism for gauge transformations and I wonder if this difference allows us to find a description of time, since differmorphism invariance is not something that will lead to time because they do not generate real time evolutions within Einstein's theory.

 

I have no answers yet on how to proceed with the strange result but hopefully this gives you an idea on the massive reading required to do to maybe find possible solutions.

[Dubbelosix]

It seems there is a special case from gauge theory in which gravity is totally described by (2 + 1) dimensions.

 

https://www.researchgate.net/publication/222440566_Chern-Simons_gravity_from_3_1-dimensional_gravity

 

https://www.sciencedirect.com/science/article/abs/pii/S0370269302018233

 

https://en.wikipedia.org/wiki/Chern%E2%80%93Simons_form

 

https://en.wikipedia.org/wiki/Chern%E2%80%93Simons_theory

 

So perhaps a three dimensional theory of gravity is still on the table at least in theory.

[Dubbelosix]

[math]\mathbf{G}_{\mu \nu} = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\cdot \gamma_{\nu} + \gamma_{\mu} \wedge \gamma_{\nu} - i \sigma \cdot (\gamma_{\mu} \times \gamma_{\nu}))  = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma_{\mu}) = \frac{1}{2}\mathbf{T}[\gamma_{\mu} \cdot \gamma_{\nu}][/math]

 

So in a true four dimensional vacuum, all terms in:

 

[math]\gamma_{\mu} \wedge \gamma_{\nu} = i \sigma \cdot (\gamma_{\mu} \times \gamma_{\nu}) = 0[/math]

 

But the term should be non-vanishing for D=7. If we had a simple theory where [math]\gamma_{\mu}[/math] relates to only one spatial dimension, then perhaps we can consider this a case only for dimensions of 2; as would be the case in the holographic principle. Not only this, but as from literature I studied before, you can even consider a spacetime uncertainty principle as predicted from scattering experiments. For non-vanishing terms, it is fair to not only call it a geometric approach, but also as a bivector theory of gravity - in fact, it is possible to retrieve a form for the four dimensional case in which we take the product of two bivectors!

 

[math]AB = A \cdot B + A \times B + A \wedge B[/math]

 

which gives rise to the ''grade 4'' exterior product [math]A \wedge B[/math] suitable for four dimensions. This appears to be the way out of this problem because as wiki states in a different way to the way we have worded the problem:

 

''If the bivector is simple the last term is zero and the product is the scalar valued A · A, which can be used as a check for simplicity. In particular the exterior product of bivectors only exists in four or more dimensions, so all bivectors in two and three dimensions are simple.^[1]''

 

https://en.wikipedia.org/wiki/Bivector

[Dubbelosix]

Are you ever going to get to a conclusion, you appear to be rambling mathematically.

 

Does this link amuse you http://web.mit.edu/sahughes/www/8.022/lec24.pdf A very brief introduction to gravity, by Scott Hughes

 

The conclusion was made a while back, which was, it was possible to formulate Einstein's equations under geometric algebra. Everything that has followed are not mathematical ramblings, but we are studying what we have suggested, a grade 4 exterior product for a successful bivector theory of gravity.

[Dubbelosix]

Also I don't know why you would think that paper ''amuses me.'' It's just standard gravity as we know it.

[Dubbelosix]

Call me old fashioned I like to see some sort of direction

 

What you asked for was conclusions, not so much direction. There is no direction when you are literally learning your own theory for the first time.

[Dubbelosix]

[math]\mathbf{G}_{\mu \nu} = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\cdot \gamma_{\nu} + \gamma_{\mu} \wedge \gamma_{\nu} - i \sigma \cdot (\gamma_{\mu} \times \gamma_{\nu}))  = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma_{\mu}) = \frac{1}{2}\mathbf{T}[\gamma_{\mu} \cdot \gamma_{\nu}][/math]

 

So in a true four dimensional vacuum, all terms in:

 

[math]\gamma_{\mu} \wedge \gamma_{\nu} = i \sigma \cdot (\gamma_{\mu} \times \gamma_{\nu}) = 0[/math]

 

But the term should be non-vanishing for D=7. If we had a simple theory where [math]\gamma_{\mu}[/math] relates to only one spatial dimension, then perhaps we can consider this a case only for dimensions of 2; as would be the case in the holographic principle. Not only this, but as from literature I studied before, you can even consider a spacetime uncertainty principle as predicted from scattering experiments. For non-vanishing terms, it is fair to not only call it a geometric approach, but also as a bivector theory of gravity - in fact, it is possible to retrieve a form for the four dimensional case in which we take the product of two bivectors!

 

[math]AB = A \cdot B + A \times B + A \wedge B[/math]

 

which gives rise to the ''grade 4'' exterior product [math]A \wedge B[/math] suitable for four dimensions. This appears to be the way out of this problem because as wiki states in a different way to the way we have worded the problem:

 

''If the bivector is simple the last term is zero and the product is the scalar valued A · A, which can be used as a check for simplicity. In particular the exterior product of bivectors only exists in four or more dimensions, so all bivectors in two and three dimensions are simple.^[1]''

 

https://en.wikipedia.org/wiki/Bivector

 

Let's propose the four dimensional covariant case then explain what it means geometrically:

 

[math]\mathbf{G}_{\mu \nu \rho \sigma} = g_{\rho \eta} \mathbf{T}^{\eta}_{\sigma \mu \nu} = \mathbf{T}(A_{\mu \nu} \cdot B_{\rho \sigma} + A_{\mu \nu} \times B_{\rho \sigma} + A_{\mu \nu} \wedge B_{\rho \sigma})[/math]

Edited December 20, 2018 by Dubbelosix

[Dubbelosix]

The geometric meaning of the product of two bivectors follows from two non-zero bivectors that can be split into symmetric and antisymmetric parts. One formulation suggests there are two such antisymmetric parts

[math]AB = A \cdot B + A \times B + A \wedge B[/math]

As stated before, the quantity [math]A \cdot B[/math] is the scalar valued interior product while [math]A \wedge B[/math] is the grade 4 exterior product that arises in four dimensions and over. Like vectors, the terms have magnitudes:

[math]|A \cdot B| = |A||B|\ \cos \theta[/math]

[math]|A \times B| = |A||B|\ \sin \theta[/math]

Where [math]\theta[/math] is the usual angle between the planes.

 

Edited December 20, 2018 by Dubbelosix

[Dubbelosix]

Well that's where you are wrong then... because we have deviated from standard interpretation of gravity a while ago, maybe you would like to get aquainted with general relativity, specifically Einstein's equations before you run your mouth off. You are critical clearly about something you are not even following properly. A standard theory of gravity (looks nothing like) of the product of a two-vector, which arises as a dual of a two-form product! There is nothing even in literature I can properly track that has approached it like I have done.

 

Secondly, the paper you provided, really was really standard to anyone who studies this thing. I have seen papers like the one you posted many times and so I did not get anything from it.