Marseille workshop on loops and spin foams

[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     1                                      1    1     
                 0     1
               2     2     2                              1      2       1
               0     1     2

           3     3     3     3                     1        3        3        1
           0     1     2     3  
        4     4     4     4     4            1         4        6         4       1
        0     1     2     3     4      
                 n pick k                                pascal

                                                 


                    1      1     1     1
                    1      2     3     4
 
                2     2      2     2                   Riemann tensor
                1     2      3     4  

           3     3     3      3     
           1     2     3      4 
 
       4     4     4     4
       1     2     3     4

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