Hyperdimensional "E8" structure solved?

[d.m.falk]

Culled from Yahoo! News- Apropos here as this involves 248 dimensions... (No, not a typo- 248!)

http://news.yahoo.com/s/afp/20070319/ts_alt_afp/ussciencemathematicsfrancegermany_070319121747

d.m.f.

[wendy]

In lie groups, each node of the E8 diagram gives a different dimension, between 8 and 248. It produces some rather interesting symmetries in each of the dimensions it appears in.

Since the first instance of E8 is 8 dimensions, it is this that the 'this is not a typo' refers to.

E8 corresponds in simplest terms to the 8d lattice that gosset described, the final cumulation of the 3d triangular prism.

I am still wondering if there is a quasi-tiling of extraordinary density in 124 dimensions, related to [3,3,5]. Still.

Wendy

[bo198214]

what do you mean by extraordinary density?

[wendy]

At that range, about 3 s-units. Enough to push up the minimum solid angle of a simplex to at least 3 tegmic radians.

In eight dimensions, it is possible to tile space, such that a sphere of diam sqrt(2), fits into a unit volume. This is E8. The spheres occupy 1/4 of space.

In 24 dimensions, it is possible to tile space such that a sphere of diameter 2 fits into a unit volume. This is the Leech lattice.

In 124 dimensions, it ought be possible to put a sphere of diameter 4 sqrt(2) into a unit volume. This is a new threshold, and such a tiling will push the simplex vertex to 3 tegmic radians, or 3/(119!) solid radians. Although the space occupied is slight (ie 363/2^120), it is still the maximum packing of spheres into solid space.

W

[bo198214]

and what is an s-unit? and what is a tegmic (which is not mentioned in wikipedia) radian?
And why are 3 of these a threshold?

[wendy]

Just as a square, cubic radian corresponds to the surface of the spherem glome, measured in square, cubic measure of length one radian, the tegmic radian correspinds to the measure of the surface of an n-sphere in terms of a tegum (cross-polytope) of unit diagonal.

It is relatively easy to show that the solid angle of a simplex in n dimensions, lies between 1 and sqrt(n/e) such tegmic radians, since we can contain a volume of 1 solid tegum inside, and sqrt(n/e) outside.

It is also easy to show, that in terms of this unit, the solid angle of a simplex in n dimensions is greater than m dimensions, as n > m.

One method for evaluating this solid angle is to use sphere packings. One supposes that a sphere-packing of simplex-holes is more dense than any other tiling. The densist packing can be expressed in terms of the simplex unit, which is the implied angle if the tiling were made of simplex holes.

Most tilings give a value of s (the implied simplex angle), less than 1. In a few dimensions, like 7, 8, 24, 25, s is greater than 1, though just over in many cases.

In 8 and 24 dimensions, the value is pushed up appreciably, relative to the previous best known limit. For 8d, the value is very close to 4/3 = 1.3333, while 24 dimensions gives a minimum of 1.6

In 120 dimensions, we then see that the effective range of s is then somewhere between 1.6 and 6.5. This means that we can calculate how many simplexes go around a vertex from these numbers.

If the super dense tiling in 124 dimensions exists, then the lower limit is pushed to 3.

W

[bo198214]

Boy, Wendy, it really took me 5 min to decipher your first sentence (I wonder whether you *just* dont care or whether you *especially* dont care whether someone understands you)

Just as a square, cubic radian corresponds to the surface of the spherem glome, measured in square, cubic measure of length one radian, the tegmic radian correspinds to the measure of the surface of an n-sphere in terms of a tegum (cross-polytope) of unit diagonal.

So my deciphered sentence looks then like:

"Just as a square/cubic radian corresponds to the surface of the sphere/glome, measured in square/cubic measure of length, the tegmic radian corresponds to the measure of the surface of an n-sphere in units of the surface of an n-dimensional tegum with diagonal 1 (which is described by the coordinates |x₁|+...+|x_n|<=1)."

Is this right?
So for example if we talk about 2 dimensions we have the normal radians which measures an angle as the length of the corresponding arc with radius 1, which gives 2pi for the circumference of a circle. The tegum with radius one however has a side length of sqrt(2), and hence a circumference of 4sqrt(2). So I would guess that we can convert radians to tegum radians by multiplying with 4sqrt(2)/(2pi). Right or not?

It is relatively easy to show that the solid angle of a simplex in n dimensions, lies between 1 and sqrt(n/e) such tegmic radians, since we can contain a volume of 1 solid tegum inside, and sqrt(n/e) outside.

But then I dont understand this. Take again the 2 dim case. The simplex is the equisided triangle with angle 60 deg which is pi/3 rad. By the above calculation this would correspond to (2/3)sqrt(2) tegum radians. But (2/3)sqrt(2) is smaller than 1. And I anyway dont understand what placing a tegum inside a simplex (what imho is already impossible) has to do with measuring the angle.

It seems we have first to clarify the basics before we can become concerned with your further statements.

[wendy]

I don't know: i have been doing this for thirty years without using any words or symbols.

Tegum-measure is a kind of measure, where the unit of measure is taken to be the cross-polytope of unit diameter. This makes, for example, the volume of a pyramid as the product of the bases and the height.

A notional unit tegum has a diameter of one, not a radius. That is, its surface might be represented by sum(abs(x)) = 0.5

The volume of a unit tegum, then is the same as a pyramid of unit height in each axis, ie 1 * 1/2 * 1/3 * 1/4 .... = 1/n! You are using the edge, but this becomes 1/sqrt(2), not 1.

Sphere angle is always surface, not volume, so we have 4pi steradians, not (4pi/3). It is also measured as a euclidean surface, so the squares are taken as the limit as side -> 0, and not spherical quadralaterals.

The relation between tegum-measure and normal (prism) measure is that prism^n = n! * tegum ^ n. This is because the two use different geometric realisations of number multiplication.

The measure of angle in radians is always taken to be the surface, so for a circle, we have 2pi P1-radians or 2pi T1-radians, because the surface is 1d. So:

circle = 1d 2pi P1 radians = 2pi T1 radians
sphere = 2d 4pi P2 radians = 8pi T2 radians
glome = 3d 2pi^2 P3 radians = 12pi T3 radians.

An alternate measure is to take all-space ( C ) as 1. One might write, eg C/20 or 0.05 C.

The solid angle for the first three simplexes (2-4d) is then

2d c/6 c(2) / 6 = 1.046...
3d c/22.794 = c(3) * (1.5 acos[1/3] - 1/4) = 1.10257119
4d c/102.2 = c(4) * (acos[1/4] - 1/5) = 1.1588520414

The acos function is to be reduced to full circles (eg divide by 2pi or 360)

c(2), c(3), c(4) is the surface of the sphere in 2, 3, 4 dimensions in tegum units. For the more usual prism units, one has 2pi, 4pi, 2pi^2. To convert to tegums, one multiplies by 1, 2, 6 respectively.

The value shown in the table above corresponds to the vertex angle, measured in terms of tegum-radians of surface (or volume).

Tegum measure is a measure, not a shape. Just because a shape is of volume 1, it does not mean you can stick a unit cube in it. Anyway, although we can't stick a cross-polytope of unit volume a 60-degree sector (such as a simplex vertex), we can get a volume equal to a unit-diagonal cross-polytope in it, in every dimension.

Anyway, the next step can be illistrated in eight dimensions.

The density of E8 is the same as packing a sphere of diameter sqrt(2) into a unit cube. In eight dimensions, the volume of a sphere is pi^4 r^8 / 24 or pi ^4 d^8 / 24 / 256. Since d^8 = 16, we have

pi^4 / 24 / 16 = pi^4 / 384. Putting pi^4 as 97.409090, we get a density equal to 0.2536695 of all space.

We now take a simplex of edge 2. The volume of this is e^8 sqrt(9) 2^(n/2)n!, gives a volume of 3*256/16/3 = 48 n! prism8 units, and since one prism8 unit = n! tegum8 units, 48 tegum8 units.

A simplex of edge 2 contains 9 simplex angles, and some space, so for each simplex-angle, we have 48/9 tegum-units, or 5.33333 t8 units.

You then see that the greatest density of a tiling, assuming every hole is a simplex, is then 3s/16, where s is unknown. But we have already a tiling where the density is 0.2536659, or 4.05871, so s( 8 ) > 1.3529.

Using much subtler arguements, we have 1.41421 > s( 8 ) > 1.400000.

[papernuke]

What is the E8 structure? what does it mean?

[houserichichi]

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You don't have permission to access /forum/posting.php on this server.

Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request.

wtf??

Just try this link
http://golem.ph.utexas.edu/category/2007/03/news_about_e8.html

[bo198214]

404
wtf??

Have a look at this thread.

You simply have to search your post for the string "rm" followed by a space. For example as in the word "form" or the word "harm" followed by a space. And replace them by suitable substitues.

Btw, your link is as always an excellent source of information. Strange type this John Baez. Someone whos profession seems to be to make comments (even based on understanding;)

[bo198214]

I don't know: i have been doing this for thirty years without using any words or symbols.

I have doing this for some days and to understand I need clear words and symbols.
And if you want to communicate those ideas (as it indeed looks for me) with others (for example me) you *must* use (write and read) words and symbols, since most people including me are not that telepathicly skilled.

The volume of a unit tegum, then is the same as a pyramid of unit height in each axis, ie 1 * 1/2 * 1/3 * 1/4 .... = 1/n!

Ok, this I can follow now: The n-dim pyramid has the volume V=A*h/n, where A is the (n-1) volume of the base and h is its hight. Now the n-dim tegum are two pyramids with the (n-1) dim tegum as base and height 0.5. So the Volume of the n-dim tegum T_n is
T₁ = 1
T_n = 2*T_n-1*0.5/n = T_n-1/n = 1/n!

Sphere angle is always surface, not volume, so we have 4pi steradians. It is also measured as a euclidean surface, so the squares are taken as the limit as side -> 0,

I thought it was just the cut out area of the surface of a sphere with radius 1? No limits needed?

The relation between tegum-measure and normal (prism) measure is that prism^n = n! * tegum ^ n.

Thats something clear, I can relate to. And quite sure not deducable from your original "definition". Though it looks for me that the ^n is wrong, if you mean the power to n instead of the n-dim prism/tegum measure.

An alternate measure is to take all-space ( C ) as 1. One might write, eg C/20 or 0.05 C.

With all-space you mean the surface of the full sphere?

So can you now (that I know what you mean by tegum-radians) again explain your:

It is relatively easy to show that the solid angle of a simplex in n dimensions, lies between 1 and sqrt(n/e) such tegmic radians, since we can contain a volume of 1 solid tegum inside, and sqrt(n/e) outside.

The density of E8 is the same as packing a sphere of diameter sqrt(2) into a unit cube.

??? How is the density of En declared?