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Old 2007-11-17, 02:26   #1
ShiningArcanine
 
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Default An Exceptionally Simple Theory of Everything

A physicist in Nevada created a theory that explains everything:

http://science.slashdot.org/article..../11/15/2322225
http://uk.youtube.com/watch?v=-xHw9zcCvRQ
http://en.wikipedia.org/wiki/An_Exce..._of_Everything
http://arxiv.org/abs/0711.0770

Now, would someone explain Group theory, Symmetry groups, Differentiable manifolds and Lie groups to me in that order, so I can understand E8 and hopefully have some chance of understanding "An Exceptionally Simple Theory of Everything?" Someone on slashdot said that people need to learn them in that order to understand it, although I think Differentiable manifolds do not need to be learned as the third item and could be learned as the first or second instead, but I do not know anything about differentiable manifolds, so I am probably wrong.

Last fiddled with by ShiningArcanine on 2007-11-17 at 02:28
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Old 2007-11-17, 15:34   #2
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Quote:
Originally Posted by ShiningArcanine View Post
Now, would someone explain Group theory, Symmetry groups, Differentiable manifolds and Lie groups to me in that order ...
You don't ask for much, do you?

I don't know enough to begin to address that request, but I do know that those are pretty deep topics that would require quite a bit of study to properly understand. Start in the usual places (wikipedia, mathworld, etc), if you don't have access to the right books. Then get the right books.

Last fiddled with by bsquared on 2007-11-17 at 15:51
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Old 2007-11-17, 15:57   #3
Orgasmic Troll
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Quote:
Originally Posted by ShiningArcanine View Post
A physicist in Nevada created a theory that explains everything:

http://science.slashdot.org/article..../11/15/2322225
http://uk.youtube.com/watch?v=-xHw9zcCvRQ
http://en.wikipedia.org/wiki/An_Exce..._of_Everything
http://arxiv.org/abs/0711.0770

Now, would someone explain Group theory, Symmetry groups, Differentiable manifolds and Lie groups to me in that order, so I can understand E8 and hopefully have some chance of understanding "An Exceptionally Simple Theory of Everything?" Someone on slashdot said that people need to learn them in that order to understand it, although I think Differentiable manifolds do not need to be learned as the third item and could be learned as the first or second instead, but I do not know anything about differentiable manifolds, so I am probably wrong.
Groups are mathematical structures that can be used to describe all types of symmetry. Defined axiomatically they are:

A set G with an associative operation * such that:

i) There exists an identity element in G, i.e. x * e = e * x = x
ii) If x is an element of G, then there exists a y such that x * y = y * x = e (i.e. the inverse of x)
iii) If x, y are in G, then x * y is in G (closure)

Examples of groups are the integers with addition, real numbers with addition, the symmetries of a polygon (for example, given a square with vertices labeled, there are 8 ways to pick up the square, rotate it, flip it, whatever, and put it back down in the same place it was)

A manifold is a space that is locally euclidean at each point. For example, surfaces are 2-dimensional manifolds. The surface of a sphere is not euclidean (the fifth postulate is not satisfied) but since we are so small compared to the earth, we consider our neighborhood to be on a flat, 2d surface. So a sphere is a manifold. With a bit more of a headache, you might be able to see that all invertible 2x2 matrices with real entries are a 4-dimensional manifold

A differentiable manifold is a space that is not only locally euclidean, but "smooth" enough to do calculus on.

A Lie group is a differentiable manifold that is also a group. The example of all invertible 2x2 matrices with real entries (It's called GL2(R) ) can be considered a group under matrix multiplication. Therefore, GL2(R) is a Lie group.
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Old 2007-11-18, 11:52   #4
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Is it a coincidence that the properties you listed for groups correspond to the reflexive, symmetric and transitive closures of relations on sets (i.e. equivalence relations)?

If all invertible 2x2 matrices with real entries are 4-dimensional manifolds, why do they claim that the circle be represented by a 2x2 matrice on the following website:

http://aimath.org/E8/representation.html

The only way that I can see that the circle can be represented by that would be to take its determinant, which gives sin^2(x) + cos^2(x), which resembles the equation sin^2(x) + cos^2(x) = 1, which I believe is the sum of the absolute values of its x and y coordinates of a circle with a radius of 1 as a function of its angle measure, although the specialness of that does not seem to really jump out at me and it does not seem to be four dimensional. Do you mean that each entry in the matrix represents a different coordinate in a four dimensional space according to some kind of coordinate system?

Are differentiable manifolds related to Gauss's theorem?

Last fiddled with by ShiningArcanine on 2007-11-18 at 11:54
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Old 2007-11-18, 20:54   #5
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Quote:
Originally Posted by ShiningArcanine View Post
Is it a coincidence that the properties you listed for groups correspond to the reflexive, symmetric and transitive closures of relations on sets (i.e. equivalence relations)?
I don't think they correspond. So yes.

Quote:
If all invertible 2x2 matrices with real entries are 4-dimensional manifolds, why do they claim that the circle be represented by a 2x2 matrice on the following website:

http://aimath.org/E8/representation.html
Because they aren't using ALL the 2x2 matrices. You can still put a circle into 4 dimensional space.

Quote:
The only way that I can see that the circle can be represented by that would be to take its determinant, which gives sin^2(x) + cos^2(x), which resembles the equation sin^2(x) + cos^2(x) = 1, which I believe is the sum of the absolute values of its x and y coordinates of a circle with a radius of 1 as a function of its angle measure, although the specialness of that does not seem to really jump out at me and it does not seem to be four dimensional. Do you mean that each entry in the matrix represents a different coordinate in a four dimensional space according to some kind of coordinate system?

Are differentiable manifolds related to Gauss's theorem?
That's like asking if curves are related to the fundamental theorem of calculus.

No more questions. You don't seem to be thinking for yourself anymore, and I'm too tired to do it for both of us.
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Old 2007-11-19, 05:40   #6
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I don't like their definition of a representation.

A representation is just a map from a group into the space of nxn matrices so that composition of group elements corresponds to multiplication of their respective matrix representations.

The circle is a group under addition of angles (where everything is working modulo 2pi). You map an angle x to the matrix you have above. Then if you take the matrix associated to an angle x, and multiply it by the matrix associated to an angle y, the result is that same as the matrix you'd get by putting in x+y.
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Old 2007-11-19, 21:20   #7
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just 3 comments :

1/ obviously (?!), the words "exceptional" and "simple" are a "clin d'oeil" to the precise meaning of these terms in group theory, rather than referring to an extremly simple theory.

2/ astonishing that the author does not cite any published paper of himself in the references

but at least, without attempting any other judgement whatsoever, I think it is a nice pedagogical review of the quite well known (group) structures of the usual GUT's, of course assuming the reader has some background on the subject.
- I don't know another as complete and as nicely & colorful illustrated compilation of all this information, which although certainly otherwise available, is so only in very widely scattered sources.
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Old 2007-11-20, 13:37   #8
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Quote:
Originally Posted by m_f_h View Post
1/ obviously (?!), the words "exceptional" and "simple" are a "clin d'oeil" to the precise meaning of these terms in group theory, rather than referring to an extremly simple theory.
THX for this caveat.
"The truth is rarely pure and never simple" or words to that effect.
(Oscar Wilde).
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Old 2007-11-26, 05:17   #9
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At least that's better then "The truth is purely simple, and rarely pure." as the more cynical think.
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Old 2007-11-29, 16:40   #10
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Quote:
Originally Posted by nibble4bits View Post
At least that's better then "The truth is purely simple, and rarely pure." as the more cynical think.
What a profound statement. I would even dare to replace "cynical" by "realistic" (at least! if not : experienced, wise,... or something alike).

PS: didn't you rather mean "... never pure" ?!

Last fiddled with by m_f_h on 2007-11-29 at 16:41
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Old 2010-04-02, 17:40   #11
mdettweiler
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Originally Posted by seena View Post
i like this page and post of you. 'an other project searching for primes for these 3 k's, the rest of the 9 k's are reserved by us. In order to prove the primality of a number we need to perform a primality test called PRP. This test takes very long; hence to reduce the time of the project we look for numbers with small factors and remove these numbers from our primality-testing list. This process is called sieving. The PRP testing method is not 100 % efficient, that is, it can make errors. On the other hand once a factor is found for a number, we can be 100 % sure that, that number is not prime. All numbers that are checked by PRP need to be double-checked. Since the probability of finding a prime is higher than having made an error and missing a prime, we are not currently pursuing the double check of numbers. There is another method that allows us to find factors for numbers called P-1 ("P minus one"). The running time for this method is similar to PRP. Currently we are not pursuing this method either, since it is more efficient to use sieving to find these factors. We do plan to use these methods a bit later in the project.'
Did you post this in the wrong forum by chance? It sounds like it belongs somewhere in the Prime Search Projects forum group, possibly Riesel Prime Search (or maybe one of the now-defunct projects archived at the bottom of the forum since it's referring to PRP tests).
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