Marseille workshop on loops and spin foams

[nightcleaner]

Ok, I think I see the pattern now.

A zero dimensional simplex is a vertex (or point).
A one dimensional simplex is an edge (or line), composed of two vertices (zero dimensional simplices).
A two dimensional simplex is a triangle (or plane), composed of three edges (one dimensional simplices).
A three dimensional simplex is a tetrahedron, composed of four triangles (two dimensional simplices)
A four dimensional simplex is a hyper-tetrahedron, and should be composed of five tetrahedral ( three dimensional simplices).

The table then would look like this:


N s0 s1 s2 s3 s4
0 , 1 , 0 , 0 , 0 , 0
1 , 2 , 1 , 0 , 0 , 0
2 , 3 , 3 , 1 , 0 , 0
3 , 4 , 6 , 4 , 1 , 0
4 , 5 , ? , ? , 5 , 1

Where N is dimension, s0 is number of vertices, s1 is number of edges, s2 is number of triangles, s3 is number of tetrahedrons, and s4 is number of hyper-tetrahedrons. So, to get the right s2 and s1 for n=4, we have to adjoin five 3-simplices, that is five tetrahedrons. We can do this simply by surrounding one tetrahedron with four more, each adjoined to one of the original tetrahedron’s faces. Then we have a three dimensional model of a four dimensional structure, consisting of five co-joined tetrahedrons.

All four of the original tetrahedron’s faces are now interior to the new structure, and are co-joined to one face each of the outer four tetrahedrons. Since five tetrahedrons would have twenty faces, but eight are now co-joined on the interior of the new structure, we are left with twelve exterior s2 triangular faces.

Five tetrahedrons would have thirty edges, but the lines of the central tetrahedron are all co-joined with two other exterior tetrahedrons. We can neglect these interior lines in our count. Then four exterior tetrahedrons would have twenty-four edges, but twelve of these edges are co-joined with one other exterior tetrahedron each, leaving a count of twenty-four minus six, or eighteen exterior edges. Twelve of these edges are acute, and the other six, the co-joined ones, are oblique.

So the completed table would look like this:


N s0 s1 s2 s3 s4
0 , 1 , 0 , 0 , 0 , 0
1 , 2 , 1 , 0 , 0 , 0
2 , 3 , 3 , 1 , 0 , 0
3 , 4 , 6 , 4 , 1 , 0
4 , 5 , 18 , 12 , 5 , 1


However I now notice that there are stated to be five s0 vertices in the four dimensional simplex. This comes from the rule that the number of vertices is n+1. The described joining of four tetrahedrons to one tetrahedron in the center does not result in a three dimensional structure with five points. Instead, the structure has four external vertices and four internal vertices. Do we throw out the structure or modify the rule?

Is there any way to co-join four tetrahedrons to end up with five points? I haven’t thought of any.

Is there any reason to change the rule? Where did the rule come from? Let’s look at the rule.

We began with a single vertex in otherwise empty space, and noted that it had zero dimension. To obtain one dimension, we had to add another vertex, so that space was no longer empty. This new vertex had to be constrained to be exterior to the original vertex, so that a line was formed.

Then, when we added a third vertex, we had to constrain the placement again, so that the new vertex was placed in the plane, and not on the edge previously constructed. Our new vertex was non-co-linear with the two that formed the definition of the edge.

Then, when we added the fourth vertex, to make a three dimensional simplex, we had to specify that the new vertex was non-co-planar with the existing three. So what is the rule for the fifth vertex? It cannot occupy the same three dimensional space as the existing three vertices. How might this be done?

We might place the new vertex as an offset in time. In other words, we might take our three dimensional simplex, a tetrahedron, and map it onto a tetrahedron extended one instant in time. Each vertex of our t=0 tetrahedron corresponds with a vertex on our t=1 tetrahedron. This would then be eight vertices, four in one instant and four in the offset instant. But we only want five vertices.

Five vertices can be achieved by setting the time interval to infinity. Doing this reduces one of the tetrahedrons to a point. Then, we see, we have five points, four in one instant mapped onto a fifth which, being in another instant, is not in the same space, and so obeys our constraint.

The result is a space-time structure with five vertices. It has four space-like vertices and six space-like edges, and it has four time-like edges which map from the four space-like vertices onto the one time-like vertex. So we see that there are now, as originally predicted, four dimensions, five vertices, and ten edges. But what happened to the five predicted s3 tetrahedral simplices? Are we justified in saying that they are somehow interspersed along the time line, so that we really have a set of five, one in 3space, one at infinite time, and then one each at the half and quarter marks? What is half or a quarter of infinity?

I have, as usual, an alternative proposition, which I think is more elegant. It is this. Any vertex in 4 space has at least four positions in any 3space. It exists at the origin. It exists at infinity. And it exists at least at two points somewhere in between. Those would be the spaces which contain the original tetrahedron we built in 3space at t=0, and the offset tetrahedron we built in another 3space at t=1.

Of course t=1 is not t=infinity, but again, what is half of infinity? From 3space, when we try observe the 5-vertice structure which exists in 4space, all we can see at one glance is ten edges between five points. That is the whole structure, as far as our familiar 3space geometry will allow. But we must conclude that it is not the entire 4space simplex. Parts of it are hidden from our 3space view.


Now, (groan) to return to the isomatrix model. There is one sphere in the center, representing any universe you choose. There are twelve spheres around it, representing the fundamental unit of spacetime in multiple dimensions. This structure is extended to infinity, in, out, and in every conceivable direction, both in time and in space. When an observer notes that there has been a change in the universe, it is not because the spacetime structure of the multiverse, that is the frozen river of 4d, has changed, but it is because the observer has moved through the unmoving spacetime structure. When the observer moves, it is in a direction, just one direction, not every direction possible. This movement results in a loss of information from the direction opposite to the motion. Information from the direction opposite to motion in the multiverse cannot catch up to the observer. Instead, the observer can obtain information only about the universe current to the observers instantaneous position (one sphere, the center sphere of the observer) and the three spheres which are annexed to that central sphere and still in the line of motion of the observer, and then one more bit, from the sphere that is just beyond the three spheres that form the possible next instant. The universe as you know it in this instant, the three universes that are possible in the next instant, and the one universe that has to be beyond those three. That’s five, the very same five vertices, or origins, that make up the structure we view as fourth dimensional from our familiar three dimensional universe.

Be well. Comments appreciated. And, yes, Love always,


Richard

3850

 

[nightcleaner]

1998   3
1999   3
2000   5 
2001   4
2002   6
2003   4
2004   4

these are not perfect, they get some they shouldn't and probably miss some, but I've found the links are a good way to check for existence of papers I didnt previously know about in this area. As you can see there is little or no growth as yet. Will be interesting to run the same keyword search in 2005 and see if there's any change

Marcus! This is really great work! If I didn't have an eyestrain, I'd stay up all night reading. Oh well, tomorrow is tomorrow. Thank you very much for these searches.

Richard

 

[marcus]

Richard, do you know about the "N-choose-k" number?

it is relevant here.

it is often written
$$\left(\begin{array}{cc}N\\k\end{array}\right)$$

and sometimes called the "combinations" of size k taken from a set of size N, the language is awkward but the idea is very simple


if you have a set of N things then how many subsets of size k are there?
N-choose-k

If you have a set of 3 things (N=3) then how many subsets of size 2 (k=2) are there?

3-choose-2 is equal to 3


the N-choose-k numbers are those appearing in the "Pascal triangle"

$$\left(\begin{array}{cc}4\\0\end{array}\right) = 1$$

$$\left(\begin{array}{cc}4\\1\end{array}\right) = 4$$

$$\left(\begin{array}{cc}4\\2\end{array}\right) = 6$$

$$\left(\begin{array}{cc}4\\3\end{array}\right) = 4$$

$$\left(\begin{array}{cc}4\\4\end{array}\right) = 1$$

 

[nightcleaner]

No, Marcus, my math is pretty limited. I've had some college calc, but didn't do well in it, and that's been long ago. I was pretty good at physics, in the Life Sciences version, which was light on calclulus. I thought yesterday's problem was interesting, and just tried to think my way through it.

I have considered returning to school to improve my maths. University of Minnesota Duluth is closest, but does not offer much of a curriculum.

Meanwhile I have been reviewing, using a GRE text, and trying to get what I can from internet. Any suggestions would be welcome.

Thanks,

Richard

 

[marcus]

$$\left(\begin{array}{cc}N\\k\end{array}\right) = \frac{N!}{k!(N-k)!}$$

...

for this, you have to know what N! the factorial of N is,
and you should know the convention that 0!, the zero factorial, equals one.

...

(but these, after all, are not terribly hard facts to learn
or, more precisely, to accept)

the N-choose-k numbers are those appearing in the "Pascal triangle"

$$\left(\begin{array}{cc}3\\0\end{array}\right) = 1$$

$$\left(\begin{array}{cc}3\\1\end{array}\right) = 3$$

$$\left(\begin{array}{cc}3\\2\end{array}\right) = 3$$

$$\left(\begin{array}{cc}3\\3\end{array}\right) = 1$$



$$\left(\begin{array}{cc}4\\0\end{array}\right) = 1$$

$$\left(\begin{array}{cc}4\\1\end{array}\right) = 4$$

$$\left(\begin{array}{cc}4\\2\end{array}\right) = 6$$

$$\left(\begin{array}{cc}4\\3\end{array}\right) = 4$$

$$\left(\begin{array}{cc}4\\4\end{array}\right) = 1$$



$$\left(\begin{array}{cc}5\\0\end{array}\right) = 1$$

$$\left(\begin{array}{cc}5\\1\end{array}\right) = 5$$

$$\left(\begin{array}{cc}5\\2\end{array}\right) = 10$$

$$\left(\begin{array}{cc}5\\3\end{array}\right) = 10$$

...


about GRE review, sounds smart, but I can't advise
Duluth likewise.
maybe selfAdjoint, who also lives in midwest, can give wise and kind counsel
I really cant. all what you say sounds sensible and intelligent
(but in our tangled web of hardship and difficulty how can anyone give advice or encouragement to anyone else besides to say take care)

However whatever you do or do not do in your life, you should understand Pascal triangle and N-choose-k. I seriously insist on this.

the number of triangle faces of a tetrahedron is the number
of THREEPOINTED simplices belonging to a FOURPOINTED
four choose three

[edited to moderate language]

 

[nightcleaner]

Hi Marcus.

Thanks for trying to cheer me up. The world has been, uh, dissappointing. I think it was Dr. Seuss whose last words were "we could have done much better."

I have been sleeping and doing chores all day, and havn't gotten back to this except for a few minutes and then I was interupted. Tonight I have to work. But I will work on the n-choose-k thing.

I am enjoying this conversation.

Pascal triangle, I have read of this somewhere. This board's format makes it hard to show here, but you line the numbers up in centered rows, as I recall. The numbers in each row are the sum of the numbers immediately above them, I see, so (1,1,)(1,2,1)(1,3,3,1)(1,4,6,4,1)(1,5,10,10,5,1) (1,6,15,20,15,6,1) and so on. I don't remember what the connection was but I do know this is familiar.

Four choose three? is four? Four ways to choose three things from a set of four things. I am still looking at the triangle, but my first thought is that if I choose three dimensions to be spatial, then there is one left, which we use for time. So, blessed be, there should be four ways to do this, so four different choices of time line in each instant. Of course, we are coming from one line, so we really have a choice of three time lines at any instant, since going back to the previous node is the same as not going anywhere at all. So the three we can choose from become our three spatial dimensions. The critical point here seems to me to be that the multiverse extends beyond what we have to choose from.

The sceptic view of this idea is that if we cannot choose, cannot even know anything about it, why should we believe it exists? Actually you can get along just fine without all those unwieldy extra universes cluttering up the scenery. Does the Donald know or care about the multiverse?

Well, here is the thing. In the isomatrix model, any instant is surrounded by and in contact with twelve other instants. Three of them are in the immediate future, three in the immediate past, and six co-inhabit the present instant. The expansion pushes out in all twelve directions. We only measure the past-future push. That leaves ten other directions to account for the extra dimensions of string theory. That means that when looking at infinities, like the total expansion of the universe, we find a factor of ten masses unaccounted for. Huh. The missing matter is co-instantaneous but found along other time lines, ones that lead, instantaneously, to other time lines, other dimensions, other universes in the multiverse.

I have to tell you that I have found Rees et al and their knob twiddling to be rather trivial. Please don't be mad. There may be other universes where such things as the electron-muon distance are different from ours, but none of those universes is anywhere close to our region of the multiverse. Those choices were made so long ago that we do not have to make them ever again. All the universes that branch out from here, and all the others that have branched out for several billion years now, are nearly identical to our own in choice of these fundamental numbers.

Of course, one may learn a great deal about how the chosen numbers affect our universe, but the anthropological principle is correct, in so far as we only have to be concerned about the universal constants we find ourselves inhabiting.

Anyway, the reason I have been insisting that you and the others look into the isomatrix is because not all the directions are through the triangles. There are also directions that move in a space-like fashion. The isomatrix has as its fundamental simplex the triangle, but then there are also sets of rectangular planes. In fact, there are four sets of triangular planes, but three sets of rectangular planes. The rectangular planes look just like our usual interpretation of 3space, the Cartesian coordinate system.

By the way, I agree that life is torment, or at least, pain. The Budhists point out that the source of pain is desire. Eliminate desire, and the pain is, well, not gone, but not so important. I would not be here today typing at this screen if I had not fallen off a tall roof and shattered my right femur. That was a painful experience, but it gave me pause for reflection. I decided, while laying in a frozen swamp at the bottom of a hill below the building which was so kind as to try to kill me, that I should not keep all these thoughts about the multiverse to myself any more. After all, the Dalai Lamma, who may well be the only real spiritual leader in the world today, has called for all the libraries to be opened. He has played the drum, he has given breath to the flute, and we in the public have heard these things. It is time to open the libraries, which have been kept so close to preserve them, but now must be flung to the winds if they are to be saved.

In our hundred years of solitude, we must rely, at last, on the wings of butterflies to come and carry us away. Gaia is waiting for us.

Anyway, I am still working on the meaning of the factorial formula. I have to go get ready for the menial labor which keeps me paying my bills, now. But I will copy the factorial formula onto paper and carry it with me to think about while scrubbing. I believe in the virtue of physical work. Especially physical work in service at the root of things. That is where the real differences are made. But I am getting tired. I don't know how much longer I can keep this up. For a while, at least, I guess. Only I feel sorry for my hands, which are gradually becoming lumps covered with a net of scars. They were beautiful, once, I guess.

Be well, Marcus, whoever you are. Why don't you email me and tell me about yourself? I am mostly harmless.

Richard.

 

[Lacy33]

Hi Marcus.

After all, the Dalai Lamma, who may well be the only real spiritual leader in the world today, has called for all the libraries to be opened. It is time to open the libraries, which have been kept so close to preserve them, but now must be flung to the winds if they are to be saved.
Richard.

3899

Richard I hope that one of the books you intend to "fling into the wind" , I'm guessing is teach to the world is not the Sefer Yetzirah which you took from this library.
Anyone suffering from mental illness such as the "deep, serious, permanent depression" you have now shared with the world is a most dangerous combination to the Kabbalah and any works and attempts to comprehend these concepts as I see much of your offerings are directly connected can cause irreversible damage.
Before you go any further with your calculations in the areas you are exploring now and have been. For fear of danger, do what ever you can to make yourself permanently positive and reasonably happy. You are exploring things of a potentially explosive nature.
It is true, "they are out to get you" ...But I am not one of them.
I believe in your work.
Your work however far from being organized is something that may one day make a substantial breakthough in physical science simply because it IS beginning to reveal the mysteries and this I state as your arguments are revealing the secrets of the ancient books of Kabbalah thousands of years old, that I know you had not seen previously.
Suzanne

 

[marcus]

...
... mostly harmless.

a pleasure to hear Douglas Adams quoted. the HH's Guide verdict on our planet

...
I have to tell you that I have found Rees et al and their knob twiddling to be rather trivial. Please don't be mad.

certainly not angry
it is a bit trivial from my perspective too, but I would like some better
articles found or written about everyday life interpretation of basic constants-----havent seen Rees: he may be watered down which would make it blah

...
This board's format makes it hard to show here, but you line the numbers up in centered rows, as I recall. The numbers in each row are the sum of the numbers immediately above them, I see, so (1,1,)(1,2,1)(1,3,3,1)(1,4,6,4,1)(1,5,10,10,5,1) (1,6,15,20,15,6,1) and so on.
...

You should learn the "code" format for writing TABLES and MATRICES here at PF. it is elementary and easy
just look at this post, where I will write a table, and press "quote" and it will show you how it is typed

in essence you just say [kode] at the beginning and [/kode] at the end, but spell code right, with a c.

1     1
1     2     1
1     3     3    1

all that happens is that it is forced to take it seriously when you type spaces between things. It hears the spacebar.

you can use "code" format to make pascal triangle, for example

             1
         1     1
     1      2     1
1       3      3      1

 

[marcus]

from the Babylonian Talmud:

"It is never advisable for anyone to speculate on these four questions:

What is above?
What is below?
What was there before the world?
What will there be after it?

It would have been better for him had he never been born."


this is from the part of the Babylonian Talmud called
Hagigah 2:1, I have just seen this as the reference. I am not an expert about this and have not read it.

this is a bitter wisdom

if everybody would pay attention to this there would be no cosmologists because cosmologists are always asking these unwise questions

One of the young Quantum Gravity researchers has chosen this quote from the babylonian talmud to begin his thesis.
He is French and he has used the French version of this Hagigah 2:1 passage.
His name is Etera Livine and his thesis is here:
http://arxiv.org/gr-qc/0309028

the french version is this:
Quiconque s’est jamais avisé de spéculer sur ces 4 questions :
– Qu’y a-t-il au-dessus ?
– Qu’y a-t-il en-dessous ?
– Qu’y avait-il avant le monde ?
– Qu’y aura-t-il après ?
Il aurait mieux valu pour lui qu’il ne fut jamais né.

Talmud de Babylone

the title of Livine's thesis, in case anyone is curious about that, is
"Boucles et Mousses de Spin en Gravité Quantique"

 

[nightcleaner]

Hi Marcus. Hi S.

Well, I tried to work out the factorial formula, but muffed it. I'll have a look at it again in the daylight. Thanks, Marcus, for the code. That will no doubt turn out to be useful. And thanks, Shoshanna, for the book. That is still protected, unopened, under covers three. It is a dirty world, after all. Lots of work for a night cleaner.

So I should be afraid that asking these questions will make me more depressed? Not likely. Entertainments don't lift my spirit, and focused thought does not drag it down. It is what it is. I am not trying to change it, but to accept it as a gift. And it is not a matter of being negative or even of being unhappy. I am happy enough. And I always try to take what I find ugly and make it better somehow. Not to hide it. Just to make it better.

I am not ashamed of my depression. It is pain. Should I be ashamed of my broken leg? Thanks to medicine and surgery, I can walk, and still want to. Without help, I would not be here. Should I be ashamed of that? Then who would be here if not for help? Only a blind fool thinks he or she stands alone in this world. None of us got here on our own. I am neither proud nor ashamed of myself. On the whole, however, I am ashamed of the Human species. We could have done much better. For human individuals, I feel a great love, and a great pride, but it is not for their humanity, it for their transcendence. We find ourselves in the muck. We try to rise. It is no sin to fail to rise, but it is terrible not to try.

So, onward and upward. To paraphrase Oscar Wilde, 'I may be lying in the gutter, madam, but while you are looking down on me, I am looking upward, at the stars.' And of course, the next day, he was sober, although I am sure he wanted to get drunk again, while she, never having been drunk, was still stupid and ugly.

Thank you for the warnings, but I think I know the danger of star gazing. I already think it would have been better, for me at least, not to have been born, so what have I to lose? I am here. I will do the best I can with what I have been given. And among the things given to me was a curiosity about what is above the above, and what is beneath that which is below. That curse, if it is one, has already been cast.

As for the explosions, there was a time when I thought I should keep what I know to myself, for fear of the big W. Not Dubya, no, but Weaponization. I thought it would be terrible if someone found out how to use my ideas to make a more terrible bomb with which to destroy the world. But what use was that thought to me? And, what use was it to the world? I am a lowly servant of idiots, so it has done me no good to hide. And as for the world, there is already enough explosive power to destroy us all. If we are going to blow ourselves up, then it will happen without my little bit of wisdom. What we need now is not to hide the light under a basket, but to bring it out and use it to make tools for our common survival.

My mind is still open, so is my heart. I will not let fear shut either down. If you want me to shut up, give me good reason, and put away your bug-a-bears, I am not frightened. Time will shut me up soon enough. While we are here, let's converse a while. Let's learn what we can from each other. Tell me if I am wrong, I will not be angry.

Let us be well,

in the name of Love,

Richard

 

[Lacy33]

Hi Marcus. Hi S.
My mind is still open, so is my heart. I will not let fear shut either down. If you want me to shut up, give me good reason, and put away your bug-a-bears, I am not frightened. Time will shut me up soon enough. While we are here, let's converse a while. Let's learn what we can from each other. Tell me if I am wrong, I will not be angry.

Let us be well,

in the name of Love,

Richard

Good day Marcus and Richard,
Marcus, The Talmud says the things you say but it is also written that if a person seeks out the mysteries and can not leave it alone than "They belong to him". Seeking the mysteries are discouraged to test a persons strength and need to engage. It is not the Rabbis, who wrote the Talmud, who discourage the seeking of the mysteries, but the nature of the mysteries themselves. Certianaly we could fill pages of Talmud here arguments for both.
The Mystic will find their way through while the Scholars of the Talmud go back and forth and the Scholars of Blessed Memory know this!
Richard's work, by his own intuitition has led him to the same findings of the Mystics of long ago, (in that it hints at what is written in the Kabbalah). Again it is not complete in the way it resembles the Kabbalah nor do I think it is a complete work in physics. (That part I can not comment on as I am not a physicsist). BUT it does have Merit.

Richard, we note the way you choose to represent youself personally on this open physics forum.
I would not add one way or the other if personal truth and intellectual honesty are requirements here.
Again I want to say that I support your efforts to communicate your vision. This seems to be a safe place to do that as you do not claim any professionalism and you have only to gain from the educated and tolerant people willing to help. As we all know, finding assistance for creative thinkers is at best maddening.
I will conclude with more Talmud as you seem to be communicating with someone who has studied such and is willing to post it on a physics forum... "Three went into the Garden only one came out unharmed". This means to say that three were smart enough to get into the garden or receive information, but only one was fortified enough to endure the light.
Suzanne

 

[marcus]

$$\left(\begin{array}{cc}N\\k\end{array}\right) = \frac{N!}{k!(N-k)!}$$

for this, you have to know what N! the factorial of N is,
and you should know the convention that 0!, the zero factorial, equals one.


the N-choose-k numbers are those appearing in the "Pascal triangle"

$$\left(\begin{array}{cc}3\\0\end{array}\right) = 1$$

$$\left(\begin{array}{cc}3\\1\end{array}\right) = 3$$

$$\left(\begin{array}{cc}3\\2\end{array}\right) = 3$$

$$\left(\begin{array}{cc}3\\3\end{array}\right) = 1$$



$$\left(\begin{array}{cc}4\\0\end{array}\right) = 1$$

$$\left(\begin{array}{cc}4\\1\end{array}\right) = 4$$

$$\left(\begin{array}{cc}4\\2\end{array}\right) = 6$$

$$\left(\begin{array}{cc}4\\3\end{array}\right) = 4$$

$$\left(\begin{array}{cc}4\\4\end{array}\right) = 1$$



$$\left(\begin{array}{cc}5\\0\end{array}\right) = 1$$

$$\left(\begin{array}{cc}5\\1\end{array}\right) = 5$$

$$\left(\begin{array}{cc}5\\2\end{array}\right) = 10$$

$$\left(\begin{array}{cc}5\\3\end{array}\right) = 10$$

$$\left(\begin{array}{cc}5\\4\end{array}\right) = 5$$

$$\left(\begin{array}{cc}5\\5\end{array}\right) = 1$$



the number of triangle faces of a tetrahedron is the number
of THREEPOINTED simplices belonging to a FOURPOINTED
is four choose three, namely 4.

and to take another example the number of edges of a tetrahedron, the number of
TWOPOINTED simplices belonging to a FOURPOINTED
is four choose two, namely 6.
==============

now we go to 4 dimensions. the basic simplex in 4D, the socalled "4-simplex" is a FIVEPOINTED simplex

how many of ITSELF does the fivepoint simplex have?
five choose five, namely 1.

how many tetrahedrons does it have, as its threedimensional "sides"?
well a tetrahedron is a FOURPOINTED so to specify one you have to choose 4 points from the 5
so it is five choose four, namely 5

and how many triangles?
five choose three, namely 10

and how many edges?
five choose two, namely 10

and how many vertices?
we already said,
five choose one, namely 5

just as a game, to complete the sequence, let us ask
a further question
and how many NOTHINGS does the basic fivepoint simplex have?
five choose zero, namely 1

 

[nightcleaner]

Hi Marcus and Shoshana

Thanks for the wisdom, S., always appreciated, even when not well taken. And Marcus, always appreciated even when not understood. I am still trying. I have to go make coffee and do chores, but will return to Pascal and the N! formulation this afternoon offline.

Last night I tried to fit the N!'s into the formula to come out with the triangle, and it worked some of the time, but not for every trial. Probably I don't remember some of the details of factorials and will have to go to Wickipedia for a refresher. I thought about the zero factorial, and also about how to make a factorial of a negative.

3!=1+2=3
4!=3!+3=6
5!=4!+4=10

so,
2!=1!+1=1
1!=0!+0=0

0!=-1!-1 so -1! should equal 1?

Hmmmm.

No that can't be right. Will have to have coffee.

By the way, just finished reading "The man who loved only numbers" about Paul Erdos, a famous mathematician whom I had never heard of. I guess from the reading that he is only famous among mathematicians. An inspiring story for impoverished seekers of truth. However the title is misleading. The way i read it, Erdoes loved many people, and loved to talk, but numbers were the only thing that made any sense to him. Apparently he often began conversations, even with old friends whom he had not seen for long periods of time, with something like "Hello. Let n be an even integer..." He was a brilliant, kind, and funny man who was a beloved pain in the ass to most everyone he knew, especially to those who understood him.

btw, Marcus i think Rees is the real stuff and the others are the imitators, but I could be wrong. Sir Martin Rees, Lord High Astronomer to the British Crown or some such fooferall. He is at Cambridge, I think, and was one of Brian Greene's professors.

Be well. Coffee calls.

nc

 

[marcus]

0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120

to get the next you always multiply the number you are at.

so, start at 0!= 1 and to get the next, namely 1!, multiply by one
then to get the next, namely 2!, multiply what you have already by two
then to get the next, namely 3!, multiply what you have already by three

Act III of "Lady Windermere's Fan"
We are all in the gutter, but some of us are looking at the stars.
brevity and generality

 

[nightcleaner]

Oh yeah, do I feel dumb. I just went to Wikipedia and still have not gotten coffee. Ugh. So I'll try it again with multiples instead of adds. Cheese. BRB

Ahhh, the bitter and the sweet.

0! = 1    
1! = 1     =0!x1
2! = 2     =1!x2    
3! = 6     =2!x3
4! = 24   =3!x4
5! = 120  =4!x5

So we could imagine 0!= 1 = -1!x0. So -1! is the number which, when multiplied by zero, equals one. Now there is an insistant explosion for you.

5! = 120  =4!x5
4! = 24   =3!x4
3! = 6     =2!x3
2! = 2     =1!x2 
1! = 1     =0!x1
0! = 1     =-1!x0
-1!= n     =-2!x-1

comments?

We could multiply both sides of the last equation by -1, then 1!=-n=2!. Not so good. Could be a problem with multiplication. Maybe -1! x -1 does not equal 1!

-1! = n = -2! x -1
-1 x (-1!) = -n = -1 x (-2! x -1)
-1 x (-1!) = -n = -1 x (-2!) x -1 = -2!

I suppose we could say that n is infinity, so 0 x infinity is unity. Unity is no infinities? Sort of seems to make sense. One is to infinity as zero is to one?

I am getting cross-eyed. Maybe there just are no factorials of negative numbers, as Wikipedia said. Anyway I promised to work on Pascal and the dimensional stuff for Marcus. I'll go offline and do that now instead of trying to invert infinities.

nc

 

[Lacy33]

Hi Marcus and Shoshana
Will have to have coffee.
c

This is very good Richard. You go have some more coffee and I will take the trash out.

 

[marcus]

$$\left(\begin{array}{cc}N\\k\end{array}\right) = \frac{N!}{k!(N-k)!}$$

suppose we want to use this formula to find out what 4-choose-2 is

(this is the number of edges on a 3-simplex, i.e. a fourpointer, a tetrahedron,
or you can think it's the number of pairs of points you can pick from a batch of four)

so N=4 and k = 2 and we use the formula

$$\left(\begin{array}{cc}4\\2\end{array}\right) = \frac{4!}{2!(4-2)!} = = \frac{4!}{2!2!}$$

$$\left(\begin{array}{cc}4\\2\end{array}\right) = \frac{4!}{2!2!}= \frac{4*3*2*1}{2*1*2*1} = \frac{4*3}{2*1} = \frac{12}{2} = 6$$

so a tetrahedron has 6 edges. but you knew that already. one can visualize. it was just to check to see if the formula works!

but how about a 4-simplex living in 4D space, they have 5 points
lets see how many edges
it would be 5-choose-2

$$\left(\begin{array}{cc}5\\2\end{array}\right) = \frac{5!}{2!3!}= \frac{5*4*3*2*1}{2*1*3*2*1} = \frac{5*4}{2*1} = \frac{20}{2} = 10$$


$$\left(\begin{array}{cc}5\\0\end{array}\right) = 1$$

$$\left(\begin{array}{cc}5\\1\end{array}\right) = 5$$

$$\left(\begin{array}{cc}5\\2\end{array}\right) = 10$$

$$\left(\begin{array}{cc}5\\3\end{array}\right) = 10$$

...

 

[nightcleaner]

                    0                                           1 

                 1    [COLOR=DarkOrange] 1    [/COLOR]                                  1    1     
                 0     [COLOR=DarkOrange]1[/COLOR]
               2     [COLOR=DarkOrange]2     2  [/COLOR]                            1      2       1
               0     [COLOR=DarkOrange]1     2[/COLOR]

           3     [COLOR=DarkOrange]3     3     3  [/COLOR]                   1        3        3        1
           0     [COLOR=DarkOrange]1     2     3[/COLOR]  
        4     [COLOR=DarkOrange]4     4     4     4   [/COLOR]         1         4        6         4       1
        0     [COLOR=DarkOrange]1     2     3     4  [/COLOR]    
                 n pick k                                pascal

                                                 


                [COLOR=DarkOrange]    1 [/COLOR]     1     1     1
                [COLOR=DarkOrange]    1  [/COLOR]    2     3     4
 
                [COLOR=DarkOrange]2     2 [/COLOR]     2     2                   Riemann tensor
              [COLOR=DarkOrange]  1     2 [/COLOR]     3     4  

           [COLOR=DarkOrange]3     3     3  [/COLOR]    3     
           [COLOR=DarkOrange]1     2     3  [/COLOR]    4 
 
       [COLOR=DarkOrange]4     4     4     4
       1     2     3     4[/COLOR]

It now appears to me that the six elements outside the orange triangle could be the redundant elements of the Riemann Tensor?

I got this form of the Riemann tensor from page 41 of Michio Kaku's Hyperspace.

Then I wonder what it would mean, if anything, to extend the Pascal and n-pick-k triangles to cover the other six Riemann terms? Could we talk meaningfully about n-pick-k where k is larger than n? If I have three objects and I pick four of them, do I have one imaginary object? Or do I have one of the three objects twice? Or do I have to go into fractional objects? As if I have three apples to divide among four picknickers.

And what, if anything, lies beyond the edges of the Pascal triangle?

I admit i don't know how to read the math in the Riemann tensor pages of Wikipedia. However a search of Wikipedia for Pascal's triangle turned up the following page, which may be interesting.

http://www.4dsolutions.net/ocn/urner.html

This is from a paper published in 1998 and shows a relationship between the Pascal triangle and the cubeoctahedron or isomatix, a favorite topic for geomancers such as myself.

So if the Riemann tensor is folded over the edge of a 4d Pascal simplex...

 

[nightcleaner]

http://www.4dsolutions.net/ocn/urner.html

I repeat this here. It is a most interesting approach to 4d visualization. I would like to discuss this with anyone interested. I intend to study this and follow the links given, and report what I find here.

The basic idea is starting with the Pascal Triangle. As the link above says, there is a form of the Pascal triangle called the pegboard. It is used as in the game pachinko, where a triangulated pegboard is used to vector balls which are inserted at the apex. The balls fall through the pegs, bouncing off sucessive layers of pegs in a random walk manner, and end up in collection tubes at the bottom of the board. When balls are inserted at the apex, they fall through the triangulated peg board and end up in one of the tubes at the bottom. The distribution into the tubes follows a Gaussian pattern, the familiar Bell curve. You really have to go look at this link, it says it much better and gives pictures.

Anyway, if we require a three dimensional version of pachinko, we can use the Kepler stack. Insert a very small ball into the interstices between the balls in the Kepler stack, and it will bounce down through the spaces between the balls in the Kepler stack in a three dimensional random walk. We could collect the small balls in tubes at the bottom of the stack and they would show a two dimensional Gaussian distribution.

So, the Kepler stack is shown to act as a three dimensional machine to filter random events into a two dimensional analog of the Bell curve. The link gives a three dimensional Pascal triangle which I will try to transcribe below, since I have not had any luck trying to copy it from the link given above to this forum.

     1          1          1         1             1       
               1  1       2 2       3 3           4 4     
                         1 2 1     3 6 3        6 12 6
                                  1 3 3 1      4 12 12 4
                                             1  4  6  4  1

It was a lot of finnicky word getting the code to come out right, but I think I got it. You see each layer fits on top of the one to the right, making a pyramid. The sides of the pyramid are the Pascal triangles.

I have chores today and work tonight, so will be gone until probably tomorrow. I am not sure how the authors got 12 in the center of the fifth layer, but will study it some more at first opportunity. Comments?

be well,

nc

ps oh yeah, the twelve comes from adding the three numbers from the layer above, so 6+3+3=12. Whats the next layer?

 

[selfAdjoint]

1 5 10 10 5 1. It's called the binomial coefficients because it's the numbers you multiply by the successive terms when you expand $$(a+b)^5 = a^5 + 5a^4b + 10 a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$$. And in general the n-th row of your triangle is the binomial coefficients for $$(a+b)^n$$.

 

[nightcleaner]

1 5 10 10 5 1. It's called the binomial coefficients because it's the numbers you multiply by the successive terms when you expand $$(a+b)^5 = a^5 + 5a^4b + 10 a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$$. And in general the n-th row of your triangle is the binomial coefficients for $$(a+b)^n$$.

Hi selfAdjoint.

1 5 10 10 5 1 is the next edge on the surface triangle, but this is a three dimensional Pascal tetrahedron we are building, so it looks like:

                     1
                   5   5
                 10    10
               10        10
             5             5
           1  5  10  10  5  1

with some numbers in the middle. This is the sixth layer down from the apex. Each number in the interior is the sum of the three numbers above it. That is, above in the sense of in the next triangle up from the one shown here. That would be the triangle with edges 1 4 6 4 1, and the three twelves for interior numbers. I'll work on it a little and bring it back here.

How are you doing in Madison? It is very cold here on the North Shore.

Be well,

Richard

 

[selfAdjoint]

I'd like to see the simplex in 3-d.

BTW I live in the Fox Cities area north of Lake Winnebago. It's around 0^o Fahrenheit here with a wind chill of -10^o. There's a good deal of snow on the ground, from storms Monday and Tuesday.

 

[nightcleaner]

          1
        5   5
      10 20 10
    10 30 30 10
  5  20 30 20  5
1  5  10  10  5  1

I think this is the sixth layer. But I have been multi tasking and am still not convinced. It seems to become rather simple after this, the interior settling into multiples of ten.

Now, what would a Pascal hypertetrahedron look like?

It is bitter cold here in Two Harbors, but I have a warm house to live in for the Winter so I am fine. But tonight I have work to do in Duluth, so will have to drive the twenty five or so miles, and back. I'd rather study math.

We live fairly close together. Maybe we should have a cup of coffee one day. I have always wanted to see if they have tours at Fermilab. And now that I know about physics in Madison and Milwaulkee, that looks like it could be reason to put together a road trip. I could drive down to Madison in less than a day. Do you know anyone at Fermilab?

Of course i am presuming. I certainly have no call to draw on your hospitality. But maybe in the spring or summer i could justify a trip somehow.

I would like to study the math. I suppose at 54 i am too old to think of graduate school. Who wants an old man as a graduate assistant?

Anyway, selfAdjoint, you have been very kind, and good internet company. I wish you the best always.

Be well,

Richard

 

[marcus]

Baez photos of the Marseille conference and scenery

http://math.ucr.edu/home/baez/marseille/

Baez started this thread telling about the May 2004 conference
at marseille where Renate Loll gave that paper on DT
"Emergence of a 4D World from Causal Quantum Gravity"
http://arxiv.org/hep-th/0404156

there is some nice scenery, a bit like Big Sur
some cliffs overlooking the mediterranean
and people from the conference hiking
I already posted some of the photos of people at the conference
and tried to identify some, but Baez includes several I didnt and does a better job of identifying