24, the Monster, and quantum gravity

https://plus.google.com/101584889282878921052/posts/9sKMLRJYjna

by msvan • 12 years ago

151 52 12 years ago

plus.google.com

sekasi• 12 years ago

I always wonder why I find things I don't understand so fascinating. It's the same reason I read in depth articles on cryptography and play the 'wikipedia rabbit hole' game.

Because it describes most primes, the ignorant part of me can't help but wonder if it does have anything to do with the magic that is crypto.. but I digress. Wikipedia might tell me more, brb losing eight hours!

Thanks for the share.

coder23• 12 years ago

Is there any importance of the number 24.

Surely there are infinite such cases for square primes.

prime^2 = ( n * m ) + c

Where n and c are constants.

madaxe_again• 12 years ago

All sorts - read the cited wikipedia articles for starters - the most important being that it's a small finite number that grants exceptional packing efficiency in the Leech lattice, which is then the basis for the CFT tie-in.

Apart from anything else, this sort of thing is just beautiful mathematically.

logicallee2• 12 years ago

for the general case, the article states there are not infinite such cases:

"A mathematician might state this property of the number 24 as follows:

If m is relatively prime to 24, then m^2 is congruent to 1 modulo 24.

One might ask if any numbers other than 24 have this property. The answer is “yes”, but the only other numbers that exhibit this property are 12, 8, 6, 4, 3, 2 and 1; in other words, the factors of 24."

mmaldacker• 12 years ago

http://en.wikipedia.org/wiki/Interesting_number_paradox

gajomi2• 12 years ago

So there are lots of facts here. And the facts are connected together at various points. And I like to hear about interesting connections. But it seems to me that unless you have looked at these things in depth (and I have not for the most part) that you would have only a vague idea of what is being talked about here.

But as I said, since I like connections, I am interested in moving beyond vagaries. In particular I am wondering about this connection to quantum gravity, and I have a few questions to this effect.

If this is about symmetries in a field theory then what is the field in this case? If I see a representation of a permutation group or a special orthogonal group factoring out of operations in a field theory I have some intuition about what this is. So what about this Monster group and what, if anything does this have to do with quantum gravity? Is it a gravity thing? Is it a quantum thing. Both?

noobermin• 12 years ago

Usually, requiring that the state vector in a quantum theory be invariant under a unitary transformation require other fields (what we call gauge fields) be added that end up being representing physical interactions.

For example, consider the state vector |y>. The norm <y|y> is invariant under local U(1) transformations

|y> -> e^(-I* T(x,t))|y>

For

<y|y> -> <y|e^(I* T(x,t))e^(-I* T(x,t))|y> = <y|1|y> = <y|y>.

(local, as in a function of space-time, is key). Write the state as |y>=v|0> where v is some operator, then saying <y|y> is invariant is like saying v* v is invariant when v -> v e^(-I* T(x,t)).

In physics, a term ~v* v would represent the potential energy of a spring. To represent the kinetic energy of a spring, we need a term like (d^{\mu}v* )d_{\mu}v where the d_{\mu} represent derivatives w.r.t. space and time (\mu is an index which runs from 0 to 3, 1,2,3 are x,y,z, 0 is time). Here, the U(1) transformation does not leave this term valid, because the derivative acts on the e^(-I T(x,t)) too...

However, if one were to make the replacement

d^{\mu} -> d^{\mu} - I e A^{\mu}(x,t)

where when

v -> v e^(-I* T(x,t))

we have that

A^{\mu} -> A^{\mu} - d^{\mu} T(x,t)

one can show (this doesn't have to be obvious!) that then the Lagrangian

~(D^{\mu}v)* D_{\mu}v - v* v

(where D is the new "covariant derivative" we described above) is now invariant under local U(1) transformations. It turns out that this Lagrangian, completely written out, looks a lot like classical electromagnetism, and it is, and A^{\mu} is the "4-vector potential". In fact, A^0 represents the good old electric potential V that I'm sure you nerds are quite familiar with. I'm not sure how enlightening that was, but at least you've seen the start of QED :)

Now, for this "monster group" ... from here [0] it seems that apparently, gravity in 2+1 dimensions seems dual to this particular group, that is, take the quantum state with that operator |y>=v|0> (v is what we call a creation operator in QFT, it "creates" particles mathematically; in string theory, it creates "vibrations on the string") that under monster group transforms, apparently whatever Lagrangian Witten made out of the v's in conformal field theory (something I should note is not my field, so I may be off here) the gauge group that might be needed would represent gravity.

[0] http://en.wikipedia.org/wiki/Monstrous_moonshine#Conjectured...

gajomi2• 12 years ago

Thanks for the response. I think this is a good answer to the question of "why would a physicist be working on the Monster group" that addresses the core idea of required invariants, and one along the lines that I have heard before (I suppose at this point I should confess to be an ex-physicist, although I never studied QED at anything but the most elementary level).

But, having heard it, I feel wanting. As you have deftly illustrated, starting with some basic requirements about invariants (from special relativity) you can end up at the Lagrangian associated with QED. And you can then (after the fact!) point out that the field in question is the electromagnetic one. Having studying electromagnetism enough to know what you are talking about, I can hear this kind of argument and it seems clever to me. There is some difference in semantics of the states, of course (classical solution of of the confined electromagnetic field coming in quanta, but perhaps not understood as "quantum observables" proper). But if I had not studied these things I would probably just be confused.

And I think this is how I feel when I hear about this advanced quantum gravity stuff (once again, I don't mean to complain about your excellent response, I think it just addresses a different issue then the one I am talking about).

If you are willing to entertain me, I'll try to ask a clarifying question in terms of analogy. If someone asked me "why does the Lagrangian for QED looks like it does", I might say to them that "the speed of light (light being made of electromagnetic fields) is constant in all inertial reference frames (lets hope they understand Galilean relativity) and that this fact forces QED to look that way". This explanation makes reference to the "stuff" that the field is made out of and the observations about invariants (stated in ordinary language) that it must satisfy.

So, now if I ask "why does the Lagrangian for quantum gravity looks like it does", you would say...

bvv• 12 years ago

It's a little complicated. The idea is that one uses this Leech lattice to define a 24-dimensional space, and we throw in little quantum strings that live on that space. These strings are inspired by (but not exactly the same as) the strings we know from standard 'string theory'.

The quantum field theory is then the one living on the worldsheet of the strings, it is the one that describes their embedding in this peculiar space. The monster group turns out to be a symmetry group of this field theory.

This is all conjecturally related to quantum gravity (according to Witten's paper) through the so-called AdS/CFT correspondence which states that field theories are sometimes equivalent to theories of quantum gravity. It has its problems and it remains to be seen if it is really true. (AdS/CFT is fine in itself, the trouble of the conjecture is whether it holds in this particular example.)

A word of warning based on previous experience: this is highly technical stuff and what is mentioned here are just words. Please don't think you can make contributions to this field without seriously studying the equations.

wolfwyrd• 12 years ago

It's also 42 backwards. Important I feel

rkaplan• 12 years ago

For the curious: the reason that this property with 24 holds is because 24 = 2^3 * 3. For any prime number p:

p^2 - 1 = (p+1)(p-1)

And p+1 and p-1 must both be multiples of 2 because p is odd. Furthermore, one of p+1 or p-1 is also a multiple of 4 (because they are both multiples of 2 and only 2 apart). So, we can see where the 2^3 factor comes from in the magic number 24. The remaining factor, 3, comes from the fact that p is prime and not a multiple of 3, so either p+1 or p-1 must be a multiple of 3 (otherwise p-1, p, and p+1 would be three consecutive numbers, none of which are divisible by 3, which is impossible).

As a result, for any prime p > 3, (p+1)(p-1) is divisible by 24, so p^2 - 1 is also divisible by 24.

logicallee2• 12 years ago

you've just taken a ton of magic out of this article!

It seems downright obvious that p^2 - 1 would have to be divisible by 24 for any prime other than 2 or 3, and in fact completely unremarkable, after you point out the factorization.

After factoring p^2 -1 into (p+1)(p-1) you could have given the rest of the proof as an exercise to the reader.

leorocky• 12 years ago

The wikipedia article on the monster group doesn't even mention the number 24.

shasta• 12 years ago

Factoring p^2-1 into (p+1)(p-1) when you're trying to prove something about the factors of p^2 - 1 isn't exactly magic either. The entire proof is a pretty easy exercise. The only hard step is "stop reading and think for a moment".

acjohnson55• 12 years ago

Haha, I feel so stupid now! Number theory always seems to impart this feeling to me. So many remarkable looking "coincidences" that have perfectly logical reasoning to explain them, if you dig a little.

anigbrowl• 12 years ago

I don't feel it's any the less remarkable for being logical...

JackFr2• 12 years ago

As an exercise to the for the reader show that any pair of twin primes surrounds a multiple of 6.

JoshTriplett• 12 years ago

Not quite true: the first twin primes, 3 and 5, surround 4, not divisible by 6. However, for any other twin primes, both will not have factors of 2 or 3, so the number between them must have a factor of 2 and a factor of 3 (the latter by the same logic as in the parent post: otherwise three numbers in a row would not have a factor of 3).

kremlin3• 12 years ago

Any pair of consecutive odds both of which are not divisible by 3 surround a multiple of 3

woofuls• 12 years ago

I was just curious, why is 24 x 10000+1 = 107 x 2243 not prime? I was under the impression for any n, 24 x n + 1 is prime.

cecilpl• 12 years ago

If that was the case, generating prime numbers would be much easier than it is.

gatehouse• 12 years ago

24x1 + 1 = 25, so we're off to a bad start.

TrainedMonkey2• 12 years ago

For prime generating sequence you would be much better with Mersenne: http://en.wikipedia.org/wiki/Mersenne_prime

AFAIK there is no sequence that lists all primes, because of the property of prime numbers themselves (not divisible by anything except 1 and themselves). This essentially means that you can't use multiplication to generate primes.

Moreover checking if number is prime is hard in itself, because you need to check for divisibility by all previous prime numbers up until square root of prime candidate you are trying to check. Thus you can think of prime numbers as recursive series.

Someone2• 12 years ago

For any prime number p: p^2 - 1 = (p+1)(p-1) That's true for every number, including for example 3.4, PI, and 3+5i

dnautics• 12 years ago

That's why the next sentence begins with the word "and"

TrainedMonkey2• 12 years ago

Which is itself is just special case of difference of squares property:

x^2 - y^2 = (x - y) * (x + y)

In this case y is 1 and y^2 is 1, so this gives us aforementioned:

x^2 - 1 = (x - 1) * (x + 1)

Here is wikipedia page with the proof: http://en.wikipedia.org/wiki/Difference_of_two_squares

kremlin3• 12 years ago

brilliantly clear explanation

mathattack• 12 years ago

Yes - the magic is stripped away by very clear algebra.

capcah• 12 years ago

Just as a remark, proving that given a prime n, n^2 = 1(mod 24)[equivalent to n^2-1 is multiple of 24] is pretty easy: i) prove that n^2 = 1(mod 3). Enumerating, n = {-1,1} (mod 3) (mod 3) [since n is prime, n != 0 (mod 3)]. n^2 = 1 (mod 3) n^2 - 1 = 0 (mod 3). Exists m such that n^2-1 = 3m.

ii) prove that n^2 = 1(mod 8). Enumerating, n = {1,3,-3,-1} (mod 8) [ n != pair (mod 8) since then, it would be divisible by 2]. n^2 = {1,9,9,1} (mod 8) => n^2 = {1,8+1}(mod 8) => n^2 = 1 (mod 8). Exists k such that n^2 - 1 = k8.

iii) There exists 2 integers m,k such that k8 = m3. m must have an 8 factor and k a 3 factor. Then, there exists j such that j = m/8 = k/3 = (n^2-1)/24. qed.

seanalltogether• 12 years ago

Interesting that if we used a base-12 numeral system then this would be immediately obvious. I wonder what other mathematical concepts would be more obvious if we used a different base system.

GotAnyMegadeth3• 12 years ago

Multiply and divide by N, where N is the base

trhway• 12 years ago

using p-adic systems makes a lot of things look different - like smooth things become edgy and edgy (in particular fractals) - smooth, and a lot of issues promise to look simpler :) While they've been known for some time the real power of that tool is only started to be explored.

GotAnyMegadeth3• 12 years ago

Does that not mean it is "trivial" to find all of the prime numbers because you can just go through the 24 times table?

chopin• 12 years ago

I'd say that it makes it at best only 24 times easier. Which cuts the problem down approx. 1.4 order of magnitudes which doesn't matter much when the problem scales exponentially.

GotAnyMegadeth3• 12 years ago

Ahh, I was thinking that p^2 - 1 = 24n held for any n as well as any p, but I can see that's not necessarily the case. Thanks

Aardwolf2• 12 years ago

I think such a test is similar to skipping multiples of low primes like 2,3,5,7,... But who knows, maybe it might be more efficient: only one modulo division needed, rather than many :)

jimwalsh2• 12 years ago

Can someone link to this story or copy/paste it somewhere? Linking to social media sites on Hacker News is bad news for people that work at companies during the day that block all social media but would still like to read the news here.

gregschlom• 12 years ago

Here you go:

24, the Monster, and quantum gravity

By Richard Green

Think of a prime number other than 2 or 3. Multiply the number by itself and then subtract 1. The result is a multiple of 24. This observation might appear to be a curiosity, but it turns out to be the tip of an iceberg, with far-reaching connections to other areas of mathematics and physics.

This result works for more than just prime numbers. It works for any number that is relatively prime to 24. For example, 25 is relatively prime to 24, because the only positive number that is a factor of both of them is 1. (An easy way to check this is to notice that 25 is not a multiple of 2, or 3, or both.) Squaring 25 gives 625, and 624=(24x26)+1.

A mathematician might state this property of the number 24 as follows: If m is relatively prime to 24, then m^2 is congruent to 1 modulo 24. One might ask if any numbers other than 24 have this property. The answer is “yes”, but the only other numbers that exhibit this property are 12, 8, 6, 4, 3, 2 and 1; in other words, the factors of 24.

The mathematicians John H. Conway and Simon P. Norton used this property of 24 in their seminal 1979 paper entitled Monstrous Moonshine. In the paper, they refer to this property as “the defining property of 24”. The word “monstrous” in the title is a reference to the Monster group, which can be thought of as a collection of more than 8x10^53 symmetries; that is, 8 followed by 53 other digits. The word “moonshine” refers to the perceived craziness of the intricate relationship between the Monster group and the theory of modular functions.

The existence of the Monster group, M, was not proved until shortly after Conway and Norton wrote their paper. It turns out that the easiest way to think of M in terms of symmetries of a vector space over the complex numbers is to use a vector space of dimension 196883. This number is close to another number that is related to the Leech lattice. The Leech lattice can be thought of as a stunningly efficient way to pack unit spheres together in 24 dimensional space. In this arrangement, each sphere will touch 196560 others. The closeness of the numbers 196560 and 196883 is not a coincidence and can be explained using the theory of monstrous moonshine.

It is now known that lying behind monstrous moonshine is a certain conformal field theory having the Monster group as symmetries. In 2007, the physicist Edward Witten proposed a connection between monstrous moonshine and quantum gravity. Witten concluded that pure gravity with maximally negative cosmological constant is dual to the Monster conformal field theory. This theory predicts a value for the semiclassical entropy estimate for a given black hole mass, in the large mass limit. Witten's theory estimates the value of this quantity as the natural logarithm of 196883, which works out at about 12.19. As a comparison, the work of Jacob Bekenstein and Stephen Hawking gives an estimate of 4π, which is about 12.57.

Relevant links Wikipedia on the Monster group: http://en.wikipedia.org/wiki/Monster_group Wikipedia on the Leech lattice: http://en.wikipedia.org/wiki/Leech_lattice Wikipedia on Monstrous Moonshine: http://en.wikipedia.org/wiki/Monstrous_moonshine A 2004 survey paper about Monstrous Moonshine by Terry Gannon: http://arxiv.org/abs/math/0402345

jimwalsh2• 12 years ago

Thanks!

adam704• 12 years ago

Here is an accessible explanation of what the monster group is in more detail: http://youtu.be/jsSeoGpiWsw.

Aardwolf2• 12 years ago

Is there any other number than 24 with this property, or is 24 the only one?

Well, 2 also has the property: multiply any prime number other than 2 with itself, subtract one, and it's a multiple of 2. This one is quite obvious, all those prime numbers are odd, so of course if you subtract one of their square (which is also odd), it's even and a multiple of 2.

But is there any other than 24 and 2? Is there one larger than 24?

mmusson• 12 years ago

The other numbers are 3 and 4. This is what yields 24. I think the comment above [0] explains this well.

[0] https://news.ycombinator.com/item?id=7839666

AnimalMuppet• 12 years ago

The article also mentions 6, 8, and 12 (all factors of 24).

24, the Monster, and quantum gravity

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