- Symmetric Spaces
- Volumes of Spheres
- Packings of Spheres
- Poincare Dodecahedral Space
- Exotic Spheres
- Poincare Conjecture
- Differential Structures on Spheres
- Homotopy
- Klein Bottle
- Periodicity
- Spheres, Octonions, and Reflexivity
- Sri Yantra

Such erroneous terminology still exists on my website, with respect to, for example, Spin(7) / G2.

The book Einstein Manifolds by Besse (Springer1987) lists (on page 203) SO(7) / G2 as a compact non-symmetricstrongly isotropy irreducible space, and on pages 179-180 it statesthat Spin(7) / G2 and S7 are related by Spin(7) acting transitivelyon S7 with isotropy subgroup K and a 0-dimensional "space ofG-invariant Riemannian metrics up to homotheties (i.e., isometriesand multiplication by a positive constant)."and that "... G2[subgroup of] SO(7) ... G2 ...[is]... NOT the fullgroup of isometries of any Riemannian metric on the correspondingsphere [S7]. We recall that any SO(n)-invariant metric onS(n-1) is (proportional to) the canonical one, so the full group ofisometries is O(n). Thus even SO(n) is not the full group ofisometries of any Riemannian metric on S(n-1). ...".

Joseph A. Wolf, in his book Spaces of ConstantCurvature (5th ed) (Publish or Perish 1984), says (at page 302): "...Added in proof. A completes structure theory and classification[

] has now been worked out for the isotropyirreducible riemannian manifolds. Unfortunately it is too long andtechnical to summarize here. ...".

The book Clifford Algebras and the ClassicalGroups by Ian Porteous (Cambridge 1995) describes the structure G2-> Spin(7) -> S7 in terms of exact sequences and coset spaces.

Examples of Symmetric Spaces of rank 1 are

- Spin(n+1) / Spin(n) = SO(n+1) / SO(n) = O(n+1) / O(n) - n-spheres Sn
- SU(n+1) / SU(n) = U(n+1) / U(n) - (2n+1)-spheres S(2n+1)
- Sp(n+1) / Sp(n) - (4n+3)-spheres S(4n+3)
- SO(n+1) / O(n) - Real Projective Spaces RPn
- SU(n+1) / S(U(n)xU(1)) - Complex Projective Spaces CPn
- Sp(n+1) / Sp(n)xSp(1) - Quaternionic Projective Spaces QPn
- F4 / Spin(9) - Octonionic Projective Plane OP2
- Spin(4) / U(2)
- Spin(8) / Spin(7) - 7-sphere (round: no torsion, twisting, or squashing)
- Spin(7) / G2 - 7-sphere with torsion
- Spin(6) / SU(3) = SU(4) / SU(3) - 7-sphere twisted
- Spin(5) / Spin(3) = Sp(2) / SU(2) = Sp(2) / Sp(1) - squashed 7-sphere (this leads to an H-space, the Hilton-Roitberg Criminal)

Some examples of Symmetric Spaces of rank 2 are

- Spin(8) / U(4) - compare dimensional reduction of Spin(8) gauge group
- Spin(n+2) / Spin(n)xU(1) - Lie Spheres
- E6 / Spin(10)xU(1) - Rosenfeld's (CxO)P2
- E6 / F4 - set of OP2 in (CxO)P2
- G2 / Spin(4) = G2 / Spin(3)xSpin(3) = G2 / SU(2)xSU(2) = G2 / Sp(1)xSp(1) - Quaternionic subalgebras of Octonions

Some examples of Symmetric Spaces of rank 3 are

- Spin(12) / U(6)
- Spin(n+3) / Spin(n)xSpin(3) - for n at least 6
- E7 / E6xU(1) - set of (CxO)P2 in (QxO)P2

Some examples of Symmetric Spaces of rank 4 are

- Spin(16) / U(8)
- Spin(n+4) / Spin(n)xSpin(4) - for n at least 8
- F4 / Sp(3)xSp(1) - set of QP2 in OP2
- E6 / SU(6)xSp(1) - set of (CxQ)P2 in (CxO)P2
- E7 / Spin(12)xSp(1) - Rosenfeld's (QxO)P2
- E8 / E7xSp(1) - set of (QxO)P2 in (OxO)P2

An example of a Symmetric Space of rank 6 is

- E6 / Sp(4)

An example of a Symmetric Space of rank 7 is

- E7 / SU(8)

Some examples of Symmetric Spaces of rank 8 are

- Spin(n+8) / Spin(n)xSpin(8) - for n at least 16 - compare Cl(n+8) = Cl(n)xCl(8)
- E8 / Spin(16) - Rosenfeld's (OxO)P2

Spheres are related to Lie groups, particularly to the B and D series of Lie groups Spin(n). Here is another list of symmetric spaces.

Conway and Sloane, in Sphere Packings, Lattices, and Groups (3rd ed Springer 1999), say, at pages 9-10: "... the volume of an n-dimensional sphere [that is, a sphere in n-dimensional Euclidean space] of radius r ... is Vn r^n where Vn, the volume of a sphere or radius 1, is given by Vn = pi^(n/2) / (n/2)! [for even n] = 2^n pi^((n-1)/2) ((n-1)/2)! / n! [for odd n] ... The surface area of a sphere of radius r is n Vn r^(n-1) ...". For a unit sphere S(2n+1) in (2n+2)-dimensional space, Conway and Sloane's formula would be (2n+2) pi^(n+1) / (n+1)! For a unit sphere S(2n) in (2n+1)-dimensional space, Conway and Sloane's formula would be (2n+1) 2^(2n+1) pi^n n! / (2n+1)!Clifford Pickover, in his book Surfing Through Hyperspace (Oxford 1999, notes that the dimension in which a sphere of radius r=1 encloses maximum volume is 5, and the maximal volume dimension increases with radius:

For r=1.1 it is 7 and for r=1.2 it is 8. The dimension in which a sphere of radius r=2 encloses maximum volume is 24:

According to aMathWorld web page:

"... the hyper-Surface Area and Content reach Maxima and then decrease towards 0 as [sphere dimension] n increases. ...... The point of Maximal hyper-Surface Area ...[and]... The point of Maximal Content ...[cannot]... be solved analytically for n, but the numerical solutions are n = 7.256295... for hyper-Surface Area and n = 5.25695... for Content. As a result, the 7-D [ S6 in R7 ] and 5-D [ S4 in R5 ] hyperspheres have Maximal hyper-Surface Area and Content, respectively. ..".

Conway and Sloane, in Sphere Packings, Lattices, and Groups(3rd ed Springer 1999), say, at pages 14-16, 128:

"... We see from Fig. 1.5 ....... that the laminated latttices /\n are the densest packings known in dimensions

<29, except for dimensions 10-13. In dimension 12 ... the Coxeter-Todd lattice K12 ... the real form of ..[a]... 6-dimensional complex lattice over the Eisenstein integers ... is the densest known, and in dimensions 10, 11, and 13 there are non-lattice packings ... that are denser than any known lattices. ... Minkowski gave anonconstructiveproof in 1905 that there exist lattices with density ...>zeta(n) / 2^(n-1) where zeta(n) ... is the Riemann zeta-function. ... Many generalizations and extensions ... have been found, although no essential improvement is known for large n. In its general form this bound is known as the Minkowski-Hlawka theorem. We still do not know how to construct packings that are as good as ...[ the bound of the Minkowski-Hlawka theorem]... ".

According to an article by George Szpiro in Nature424 (03 July 2003) 12-13:

"... Just under five years ago, Thomas Hales ... declared that he had used a series of computers to prove an idea that has evaded certain confirmation for 400 years. The subject of his message was Kepler's conjecture, proposed by the German astronomer Johannes Kepler, which states that the densest arrangement of spheres is one in which they are stacked in a pyramid - much the same way as grocers arrange oranges. ... But today, Hales's proof remains in limbo. It has been submitted to the prestigious Annals of Mathematics, but is yet to appear in print. Those charged with checking it say that they believe the proof is correct, but are so exhausted with the verification process that they cannot definitively rule out any errors. So when Hales's manuscript finally does appear in the Annals, probably during the next year, it will carry an unusual editorial note - a statement that parts of the paper have proved impossible to check. At the heart of this bizarre tale is the use of computers in mathematics ... In 1977, for example, a computer-aided proof was published for the four-colour theorem, which states that no more than four colours are needed to fill in a map so that any two adjacent regions have different colours1, 2. No errors have been found in the proof, but some mathematicians continue to seek a solution using conventional methods. ...... Hales, who started his proof at the University of Michigan in Ann Arbor before moving to the University of Pittsburgh, Pennsylvania, began by reducing the infinite number of possible stacking arrangements to 5,000 contenders. He then used computers to calculate the density of each arrangement. Doing so was more difficult than it sounds. The proof involved checking a series of mathematical inequalities using specially written computer code. In all, more than 100,000 inequalities were verified over a ten-year period. ... It was not enough for the referees to rerun Hales's code - they had to check whether the programs did the job that they were supposed to do. Inspecting all of the code and its inputs and outputs, which together take up three gigabytes of memory space, would have been impossible. So the referees limited themselves to consistency checks, a reconstruction of the thought processes behind each step of the proof, and then a study of all of the assumptions and logic used to design the code. A series of seminars, which ran for full academic years, was organized to aid the effort. But success remained elusive. Last July [2002], Fejes Tóth reported that he and the other referees were 99% certain that the proof is sound. They found no errors or omissions, but felt that without checking every line of the code, they could not be absolutely certain that the proof is correct. For a mathematical proof, this was not enough. After all, most mathematicians believe in the conjecture already - the proof is supposed to turn that belief into certainty. The history of Kepler's conjecture also gives reason for caution. In 1993, Wu-Yi Hsiang, then at the University of California, Berkeley, published a 100-page proof of the conjecture in the International Journal of Mathematics ... But shortly after publication, errors were found in parts of the proof. Although Hsiang stands by his paper, most mathematicians do not believe it is valid. ...

...[In January 2003, Hales]... launched the Flyspeck project, also known as the Formal Proof of Kepler. Rather than rely on human referees, Hales intends to use computers to verify every step of his proof. The effort will require the collaboration of a core group of about ten volunteers, who will need to be qualified mathematicians and willing to donate the computer time on their machines. The team will write programs to deconstruct each step of the proof, line by line, into a set of axioms that are known to be correct. If every part of the code can be broken down into these axioms, the proof will finally be verified. ... Pierre Deligne, an algebraic geometer at the Institute for Advanced Study, is one of the many mathematicians who do not approve of computer-aided proofs. "I believe in a proof if I understand it," he says. For those who side with Deligne, using computers to remove human reviewers from the refereeing process is another step in the wrong direction. ...

... Whether or not computer-checking takes off, it is likely to be several years before Flyspeck produces a result. ... Hales estimates that the whole process, from crafting the code to running it, is likely to take 20 person-years of work. Only then will Kepler's conjecture become Kepler's theorem, and we will know for sure whether we have been stacking oranges correctly all these years. ...".

The alternative real division algebras are R, C, Q, and O, the real and complex numbers, quaternions, and octonions, of dimensions 1 = 2^0, 2 = 2^1, 4 = 2^2, and 8 = 2^3 They correspond to the Hopf fibrations of spheres: RP1 -> S1 -> S0 (S1/Z2 can be either RP1 or an orbifold) S1 -> S3 -> S2S3 -> S7 -> S4S7 -> S15 -> S8 Their dimensions are the 0, 1, 2, and 3 powers of 2.

Z3 = {1,2,3=0} is the only field such that all of its elements are real prime integers. All of its subgroups Z0, Z1, and Z2 are prime fields. The sedenions of dimension 16 = 2^4 are not a division algebra, and Z4 decomposes into Z2 x Z2.

Kervaire and Milnor (Ann. Math. 77 (1963) 505-537) have calculated the numbers of different (including exotic) differentiable structures for some spheres:

According to Freedman and Quinn (Topology of 4-Manifolds,Princeton University Press, 1990, pages 3-4 and 118-119): "...Manifolds of dimension 2 ... have been largely understood for half acentury. ... in dimension 3 ... Moise in 1952 reduced topological[TOP] theory to the piecewise linear [PL] (orequivalently smooth [DIFF]) category ... higher dimensions (__>__ 5 ) were developed next. The two key events were thedevelopment of the methods of smooth [DIFF] and PL handlebodytheory by Smale in the late 1950s, and the extension to topological[TOP] manifolds by Kirby and Siebenmann in 1969. Dimension 4was last, with the corresponding results published by the authors in1982. ... it is the behavior of 2-dimensional disks which separatesthe dimensions in this way: the generic map of a 2-disk in a3-manifold has 1-dimensional self-intersections; in a 4-manifold theintersections are isolated points; and in dimensions 5 and above2-disks are generically embedded. ... Papakyriakopoulos' breakthroughin dimension 3 was a proof of the Dehn lemma for embedding 2-disks in3-manifolds. In dimension 5 and higher 2-disks are used in theWhitney trick ... New phenomena are ... encountered in dimension 4... In other dimensions the principal methods (handlebody theory orembeddedsurface theory) proceed uniformly in the smooth[DIFF], PL, or topological [TOP] categories ...Differences between the categories can for the most part beunderstood in terms of the associated bundle theories. In dimension 4qualitatively different ... phenomena occur in the smooth[DIFF] category ... the topological [TOP] theory ...results parallel those of higher dimensions. ... the smooth[DIFF] analogs are known to be false by Donaldson's work ...In dimensions less than 4 the categories DIFF, PL, TOP areequivalent. ... [PL and DIFF] are the same up to isotopy, formanifolds of dimension less than 6. ... ".

John Baezsays: "... there are various distinct questions floating around,including:

- A) how many topological manifolds are homotopy-equivalent to the sphere?
- B) how many PL ( = piecewise-linear = combinatorial) manifolds are homeomorphic to the sphere?
- C) how many smooth manifolds are PL equivalent to the sphere?

... In the case of dimension 3, **question A is open**[ However, as of April 2003, after this waswritten by John Baez, it may have been provenby Grisha (Grigori) Perelman. ]**(the Poincare conjecture)**, while questions B and C are solvedand the answer is 1 ... In the case of dimension 4, question A issolved (in the 1980s, by Freedman) and the answer is 1; question C issolved and the answer is 1, but question B is open (the smoothPoincare conjecture in dimension 4) ...".

Why is question B hard to answer in dimension 4 ?As John Baez says: "... in 4 dimensions we have PL = DIFF, so question B) is equivalent to: how many smooth manifolds are homeomorphic to the sphere ...", and question C) is equivalent to how many smooth manifolds are diffeomorphic to the sphere. As Donaldson and Kronheimer say in their book The Geometry of Four-Manifolds (Oxford 1990): "... Smale's h-cobordism theorem: if X and Y are h-cobordant then they are diffeomorphic ... breaks down in four dimensions ... ".

Why can question A be answered in dimension 4 ?Ian Stewart (Nature 320 (20 March 1986) said: "... Steven Smale developed a method of breaking manifolds into nice pieces, called handles ... [and] ... moving handles around ... Another important tool ... was .. the technique of surgery. This is in essence a systematic study of the effect of removing cerain nice pieces of a manifold and putting them back with a specific twist. ... [In] the three-dimensional case [Poincare's original conjecture] and the four-dimensional case ... both handle theory and surgery fail to work. The number of dimensions is high enough to allow complicated behaviour, but too small to leave adequate room for manoeuvre in eliminating those complications. ... But ... Andrew Casson found a way to make handle theory work in four dimensions. In 1982 Michael Freedman used Casson-handles to prove the four dimensional case of the Poincare conjecture (but not the more specialized differentiable case, which remains open). ...".

Why isquestion Ahard to answer in dimension 3 ?Ian Stewart (Nature 325 (12 February 1987) 579-580) said: "... If evey loop in a given [simply-connected, unlike PDS3] three-dimensional manifold can be shrunk to a point, must that manifold be topologically a hypersphere? ... start with a description of a three-dimensional manifold in terms of a system of links. ... assume that every loop in this manifold can be shrunk, and as what constraint this requirement imposes on the system of links. Then ... find ways to modify the links, using those constraints, until at the end of the process ... [there is] ... a system of links corresponding to the hypersphere. Many complicated moves are needed to effect the simplification. ... The sources of technical difficulty are twofold. The first is the combinatorial complexity of the moves, and the rules stating which one should be applied in circumstances to achieve which tiny step in the simplification. The second is that, in order for the proof to proceed rigorously, a certain amount of technical baggage must be carried along during each move. At the end of hte proof this additional superstructure can be discarded, but until the end is reached it is essential. More than one topologist has attempted to prove the Poincare conjecture by this route, but nobody ... [has] ... succeeded fitting all the pieces together and making them all work at once. ...". One way to describe a 3-sphere in terms of loops is the Hopf Fibration.According to a 15 April article by Sara Robinson in The New York Times:

"...Grigori Perelmanof the Steklov Institute of Mathematics of the Russian Academy of Sciences in St. Petersburg ...is reporting that he has proved the Poincare Conjecture... Perelman's work ... relies on ideas pioneered by another mathematician, Richard Hamilton. ... Perelman's personal story has parallels to that of ... Andrew Wiles ... who, without confiding in his colleagues, worked alone in his attic on Fermat's Last Theorem. Though his early work has earned him a reputation as a brilliant mathematician, ... Perelman spent the last eight years sequestered in Russia, not publishing. ... Perelman's ... papers say that he has proved what is known as the Geometrization Conjecture, a complete characterization of the geometry of three-dimensional spaces.Since the 19th century, mathematicians have known that a type of two-dimensional space called a manifold can be given a rigid geometric structure that looks the same everywhere. Mathematicians could list all the possible shapes for two-dimensional manifolds and explain how a creature living on the surface of one can tell what kind of space he is on.

In the 1950's, however, a Russian mathematician proved that the problem was impossible to resolve in four dimensions and that even for three dimensions, the question looked hopelessly complex.

In the early 1970's, ... William P. Thurston, a professor at the University of California at Davis, conjectured that three-dimensional manifolds are composed of many homogeneous pieces that can be put together only in prescribed ways and proved that in many cases his conjecture was correct. ... Thurston won a Fields Medal, the highest honor in mathematics, for his work.

... Perelman's work, if correct, would provide the final piece of a complete description of the structure of three-dimensional manifolds and, almost as an afterthought, would resolve Poincare's famous question.

... Perelman's approach uses a technique known as the Ricci flow, devised by ... Hamilton, who is now at Columbia University. The Ricci flow is an averaging process used to smooth out the bumps of a manifold and make it look more uniform. ... Hamilton uses the Ricci flow to prove the Geometrization Conjecture in some cases and outlined a general program of how it could be used to prove the Geometrization Conjecture in all cases. He ran into problems, however, coping with certain types of large lumps that tended to grow uncontrollably under the averaging process.

... Perelman has ... figure[d] out some new and interesting ways to tame these singularities ...".

As of 15 April 2003, Grisha Perelman has posted two papers: math.DG/0211159 and math.DG/0303109. In the latter paper, he says:

"... This is a technical paper, which is a continuation of [math.DG/0211159]. Here we verify most of the assertions, made in [math.DG/0211159, §13]; the exceptions are (1) the statement that a 3-manifold which collapses with local lower bound for sectional curvature is a graph manifold - this is deferred to a separate paper, as the proof has nothing to do with the Ricci flow, and (2) the claim about the lower bound for the volumes of the maximal horns and the smoothness of the solution from some time on, which turned out to be unjustified, and, on the other hand, irrelevant for the other conclusions.The Ricci flow with surgery was considered by Hamilton ... unfortunately, his argument, as written, contains an unjustified statement (R_MAX = GAMMA, on page 62, lines 7-10 from the bottom), which I was unable to fix. Our approach is somewhat different, and is aimed at eventually constructing a canonical Ricci flow, defined on a largest possible subset of space-time, - a goal, that has not been achieved yet in the present work. For this reason, we consider two scale bounds: the cutoff radius h, which is the radius of the necks, where the surgeries are performed, and the much larger radius r, such that the solution on the scales less than r has standard geometry. The point is to make h arbitrarily small while keeping r bounded away from zero. ...".

**The following table is of Smooth=Differentiable manifolds thatare Combinatorial=Piecewise Linear equivalent to the n-sphere Sn**.For Topological manifolds that are homotopy-equivalent to then-sphere, the answer is unknown for dimension n = 3. ForCombinatiorial=Piecewise-Linear manifolds that are homeomorphic tothe n-sphere, the answer is unknown for dimension n = 4 .

S1 - 1 S2 - 1 S3 - 1 = 2x1 / 2 (but S3 is a subset of any exotic R4#) S4 - 1 S5 - 1 S6 - 1 S7 - 28 = 8x7 / 2 = 2^3 x (2^3 - 1) / 2 S8 - 2 S9 - 8 S10 - 6 S11 - 992 = 32x31 = 2^5 x (2^5 - 1) S12 - 1 S13 - 3 S14 - 2 S15 - 16,256 = 128x127 = 2^7 x (2^7 - 1) S16 - 2 S17 - 16 S18 - 16 The number of differentiable structures for S7 is 28 = 2^2 x (2^3 - 1) for S11 is 992 = 2 x 496 = 2 x 2^4 x (2^5 - 1) for S15 is 16,256 = 2 x 8,128 = 2 x 2^6 x (2^7 - 1) It is interesting that 28, 496, and 8,128 are even perfect numbers of the form 2^(n-1) x (2^n - 1) and are the sums of consecutive odd cubes 28 = 1^3 + 3^3, 496 = 1^3 + 3^3 + 5^3 + 7^3, and 8,128 = 1^3 + 3^3 + 5^3 + 7^3 + 9^3 + 11^3 + 13^3 + 15^3and that they are related to three Mersenne Primes: 2^3 - 1 = 7,2^5 - 1 = 31, and 2^7 - 1 = 127.(see The Loom of God, by Clifford Pickover (Plenum 1997)) These interesting facts are not coincidental. Kervaire and Milnor give the formula DIFF(4n-1) = 2^(2n-4) (2^(2n-1) - 1) P(4n-1) Bn an / n where, for k unequal to 3 or 4, DIFF(k) is the number of diffeomorphism classes of differentiable structures on the topological k-sphere, which, by Smale, Ann. Math. 74 (1961) 391-406, is the number of h-cobordism classes of homotopy k-spheres P(k) is the order of the stable homotopy group of the k-stem of p-spheres, for p greater than k+1, Bk is the kth Bernoulli number, in the sequence 1/6, 1/30, 1/42, 1/30, 5/66, 691/2730, 7/6, ... Bernoulli numbers come from Bernoulli polynomials. (Bernoulli numbers, as well as surreal numbers, may be useful in describing the allspaces of Many-Worlds Quantum Theory.) ak is 1 or 2 according to whether k is even or odd.

Another page deals with the Milnor spheres S(4k-1)#, of dimension 4k-1 for k=2 or greater, which are homeomorphic to normal spheres S(4k-1) but not diffeomorphic to them.

A way to look at spheres is by their Homotopy Groups PI(k)(Sn), which is roughly the number of ways you can wrap a k-sphere around an n-sphere. For example, PI(n)(Sn) is the infinite cyclic group Z, and each element of Z corresponds to a winding number of a wrapping of Sn around Sn. As noted by John Baez, Homotopy Groups can be related to the 4-dimensional 24-cell, as well as the 24-dimensional Leech lattice, since the (n+3) Homotopy Group of the n-sphere Sn is PI(n+3)(Sn) = Z24 for n greater than 4. They are also related to the 8-dimensional Witting polytope, since PI(n+7)(Sn) = Z240 for n greater than 8. Homotopy Groups have Periodicity relations related to Clifford Periodicity. Some other interesting Homotopy Group relations are: PI(5)(S2) = Z2 PI(6)(S3) = Z12 PI(7)(S4) = Z + Z12 PI(15)(S8) = Z + Z120PI(n)(S1) is trivial for n greater than 1PI(k)(Sn) is trivial if k is less than nPI(3)(S2) is infinite cyclic Z PI(n+1)(Sn) = Z2 if n is greater than 2PI(3)(S2) = Z2 PI(n+2)(Sn) = Z2 if n is greater than or equal to 2PI(n+k)(Sn) is independent of n for sufficiently large nPI(n+11)(Sn) = Z504 for n greater than 12. PI(n+15)(Sn) = Z2 + Z480 for n greater than 16. PI(n+19)(Sn) = Z2 + Z264 for n greater than 20. PI(n+20)(Sn) = Z24 for n greater than 21.

John Baez, in his week 104 and week 105, looks at octonions, PI(k)(O(infinity)), and Bott periodicity PI(n+8)(O(infinity)) = PI(n)(O(infinity))where O(infinity) is the orthogonal group for infinite-dimensional real space which contains as subgroups all orthogonal groups O(n) for all finite n. The Clifford algebra reflection group Pin(n) double covers O(n). The spin group Spin(n) double covers SO(n). For k = 0, PI(0)(O(infinity)) = Z/2, but PI(0)(SO(infinity)) = 0. For k = 1, PI(1)(O(infinity)) = PI(1)(SO(infinity)) = Z/2, but PI(1)(Spin(infinity)) = 0. For k at least 2, PI(k)(O(infinity)) = PI(k)(SO(infinity)) = PI(k)(Spin(infinity)). The following table summarizes the Bott Periodicity Theorem results: R PI(0)(O(infinity)) = Z/2 Pin(n)/Spin(n) connected components O(n)/SO(n) and reflections C PI(1)(O(infinity)) = Z/2 Pin(n)/O(n) spinor spin Spin(n)/SO(n) PI(2)(O(infinity)) = 0 Q PI(3)(O(infinity)) = Z S3s in SO(2n) Lie group rank S3=Spin(3)=SU(2)=Sp(1) PI(4)(O(infinity)) = 0 PI(5)(O(infinity)) = 0 PI(6)(O(infinity)) = 0 O PI(7)(O(infinity)) = Z S7s in Cl(8n) dimension of S7 products give Spin(8) vector space Spin(8) is in Cl(8) Cl(8n) = Cl(8) x...x Cl(8) (n-fold tensor product) For all n, PI(n+8)(O(infinity)) = PI(n)(O(infinity))John Baez notes that the fact that PI(7)(O(infinity)) = Z corresponds to dimension can also be seen by: forming the classifying space BO(infinity) of O(infinity); noting that PI(n)(G) = PI(n+1)(BG) for any Lie group G so that PI(7)(O(infinity)) = PI(-1)(O(infinity)) = PI(0)(BO(infinity)); and noting that PI(0)(BO(infinity)) = KS0 where KS0 is the real K-theory of S0 which counts the difference in dimension between the two fibres of a vector bundle over the 0-sphere S0. As John Baez notes, PI(0), PI(1), PI(3), and PI(7) correspond to the alternative real division algebras: real R, complex C, quaternion Q, and octonion O, as their imaginaries are the spheres S0, S1, S3, and S7.

PI(0)(O(infinity)) = Z/2 Pin(n)/Spin(n) connected components O(n)/SO(n) and reflections

As described in The Topology of Fibre Bundles, by Norman Steenrod (Princeton 1951), the facts that O(k+1) is the group of a k-sphere bundle whose fibre is Sk, that sphere bundles over Sn correspond 1-1 with elements of PI(n-1)(O(k+1)) up to equivalence under the operations of PI(0)(O(k+1)), and that PI(0)(O(k+1)) has 2 elements,imply that there are two Sk bundles over S1, the untwisted S1 x Sk and the twisted generalized Klein Bottle. The twisted Klein(1,k) Bottle is constructed by forming the product of Sk with an interval and matching the ends with an orientation reversing transformation, so that the Klein(1,k) bottle is not orientable. Since RP1 is homeomorphic to S1, the untwisted S1 x Sk corresponds to the Shilov boundary manifolds RP1 x S3 and RP1 x S7 used in the D4-D5-E6-E7-E8 physics model. The corresponding twisted manifolds are Klein(1,3) Bottle and Klein(1,7) Bottle WHAT IF PHYSICAL SPACETIME WERE TWISTED KLEIN(1,3) BOTTLE RATHER THAN UNTWISTED RP1 x S3 ? Igor Kulikov has shown that: for untwisted fields in S1 x R3, decreasing the topological parameter, equivalent to increasing finite temperature, causes the coupling constant to decrease; while for twisted fields in S1 x R3, decreasing the topological parameter, equivalent to increasing finite temperature, causes the coupling constant to increase; and massless untwisted fields get dynamical mass with decreasing topological parameter, equivalent to increasing finite temperature; while effective mass for twisted fields is eliminated at a critical topological parameter, equivalent to a critical finite temperature.

Eightfold periodicity is related to the 8 trigrams of the I Ching.

The relationship between homotopy periodicity and Clifford algebra periodicity is shown by this table, in which O represents O(infinity) real Lie group rotations, Sp represents Sp(infinity) quaternionic Lie group rotations, U represents U(infinity) complex Lie group rotations, Cl(n) is the real Clifford algebra Cl(0,n), and Clc(n) is the complex Clifford algebra over complex n-space: n Cl(n) PI(n)(O) PI(n)(Sp) Clc(n) PI(n)(U) 7=-1 R+R Z Z C+C Z 0 R Z/2 0 C 0 1 C Z/2 0 C+C Z 2 Q 0 0 C 0 3 Q+Q Z Z C+C Z 4 Q 0 Z/2 C 0 5 C 0 Z/2 C+C Z 6 R 0 0 C 0 What patterns emerge from this table? 1 - For each field K, if the Clifford algebra n is of the form K+K, the origin of the first K of the K+K can be shifted relative to the origin of the second K of the K+K by any integral distance, so that the homotopy PI(n) is Z. 2 - For each field K, if the Clifford algebra n is of the form K and preceding Clifford algebra n-1 is of the form K+K, then the homotopy PI(n) has the same order as the number of different multiplicative products of K, because there is a K+K from n-1 inside the K of the n, and the product of the first K of the K+K can be reconciled with the product of the second K of the K+K in that many different ways. For K = C, the complex numbers, there is only one product because i fixes +1 and -1, and PI(n) = 0. For K = R, the real numbers, and for K = Q, the quaternions, there are two different multiplicative products because for R the unit can be +1 or -1 and for Q the product ij can be +k or -k so that the two products are mirror images of each other, and PI(n) = Z/2. Such a "first Z/2" represents reflections. 3 - For each field K, if the Clifford algebra n is of the form K and preceding Clifford algebra n-1 is of the form K+K, then the homotopy PI(n+1) has the order of spin structures on manifolds over the field K. For K = C, these are Z, rather than Z/2, because for n+1 = 1 and 5 these are complexifications of real Clifford algebras of type C and, since CxC = C+C, are complex Clifford algebras of type C+C as in pattern number 1 above. For K= R and K = Q, these are Z/2. Such a "second Z/2" for K = R and K = Q represents spinors.

WHAT ABOUT the 3-sphere S3, of dimension 2^k + 1 for k=1? It is accurate to say that there is no exotic S3, in the sense that anything homeomorphic to normal S3 is also diffeomorphic to normal S3. HOWEVER, the 3-dimensional analogue S3# of the exotic Milnor sphere DOES exist, but it is not only not diffeomorphic to normal S3, it is NOT even homeomorphic to normal S3. That is because S3# is NOT SIMPLY CONNECTED. WHAT IS S3# ?

Poincare, because it would be a counterexample to the Poincare conjecture in 3 dimensions if it were simply connected. Dodecahedral, because it has dodecahedral/icosahedral symmetry of the 120-element binary icosahedral group, which double covers the simple 60-element icosahedral group. A reference is Topology and Geometry, by Glen Bredon, Springer (1993). WHAT DOES S3# LOOK LIKE? Here are some images from the WWW pages of Richard Hawkins, who calls S3# the Mayan Time Star. His pages contain many more images and movies that help you understand how S3# looks, and also how a lot of other things look.

How did Richard Hawkins find out about the Time Star?Krsanna Duransays: "... I wrote an article about what the Sirians told me about five interpenetrated tetrahedra embodying and unifying all prime geometries which was published in January, 1995. Richard Hawkins read the article and and sent an email to Gerald de Jong about it. Gerald de Jong constructed a computer model of the five interpenetrated tetrahedra to discover that it did all the things I said it did with extraordinary elegance. ...".

The Time Star is one of my favorite Archetypes.

Start with a dodedecahedron. Five tetrahedra fit inside the dodecahedron:

The alternating permutation group of the 5 tetrahedra is the 60-element icosahedral group. Now, to see things clearly, look at just one tetrahedron.You can symmetries more clearly when you put an octhedron inside the tetrahedron and a cuboctahedron inside the octahedron:

Take the one tetrahedron and put it inside a cube, with one edge of the tetrahedron in each face of the cube. Now rotate the cube around inside the dodecahedron, while you also rotate each of the 6 edges of the tetrahedron each of the 6 faces of the cube. Start like this:

then, 36 degrees later, it looks like this:

The tetrahedra edges now are parallel to the cube edges. 36 more degrees, after 72 degrees total rotation, the edges will have again formed a tetrahedron. Keep rotating. After 360 degrees, you have made 5 tetrahedra (one each 72 degrees), and this is what you have:

The cube is back like it was, BUT THE TETRAHEDRON IS ORIENTED OPPOSITELY with respect to the cube from its original position. YOU HAVE TO ROTATE 720 degrees TO GET BACK LIKE YOU STARTED. That means that, to make S3#, instead taking the quotient of SO(3) by the 60-element icosahedral group, you should take the quotient of S3 = Spin(3) = SU(2), the double cover of SO(3), by the 120-element binary icosahedral group. Therefore, S3# is a natural spinor space, and 5-fold Golden Ratio Icosahedral Symmetryis a manifestation in 3 and 4 dimensions of the Milnor sphere structure of 7 and 8 dimensions.

John Baez,in a 2003-01-23 post to the sci.physics.research thread "The magic of8", said:

"... "even unimodular lattices" and "invertible symmetric integer matrices with even entries on the diagonal" ...[are]... secretly the same thing as long as your matrix is positive definite ... a "lattice" is a subgroup of R^n that's isomorphic to Z^n. If we give R^n its usual inner product, an "even" lattice is one such that the inner product of any two vectors in the lattice is an even integer. A lattice is "unimodular" if the volume of each cell of the lattice is 1. To get from an even unimodular lattice to a matrix, pick a basis of vectors in the lattice and form the matrix of their inner products. This matrix will then be symmetric, have determinant +-1, and have even entries down the diagonal. ...[a famous 8 x 8 invertible integer matrix with even entries on the diagonal and signature +8 is]...2 -1 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 -1 2 -1 0 0 0 0 0 0 -1 2 -1 0 -1 0 0 0 0 -1 2 -1 0 0 0 0 0 0 -1 2 0 0 0 0 0 -1 0 0 2It's called E8, and it leads to huge wads of amazing mathematics. For example, suppose we take 8 dots and connect the ith and jth dots with an edge if there is a "-1" in the ij entry of the above matrix. We get this:

o----o---o----o----o----o----o | | oNow, make a model with one ring for each dot in the above picture, where the rings link if the corresponding dots have an edge connecting them. We get this:

/\ /\ /\ /\ /\ /\ /\ / \ / \ / \ / \ / \ / \ / \ / \ \ \ \ \ \ \/ / \ / \ / \ / \ / \ / \ \ \ \ / \ / \ / \ / \ / \ / / \ \ \ \ \ /\ \ \ / \ / \ / \ / \ / \ \ \ / \ / \/ \/ \/ \/ / \/ \ \/ \/ / \ \ / \ / \ / \/Next imagine this model as living in the 3-sphere. Hollow out all these rings: actually delete the portion of space that lies inside them! We now have a 3-manifold M whose boundary dM consists of 8 connected components, each a torus. Of course, a solid torus also has a torus as its boundary. So attach solid tori to each of these 8 components of dM, but do it via this attaching map: (x,y) -> (y,-x+2y) where x and y are the obvious coordinates on the torus, numbers between 0 and 2pi, and we do the arithmetic mod 2pi. We now have a new 3-manifold without boundary. This manifold is called

the "Poincare homology sphere". Poincare invented it asa counterexample to his own conjecture that any 3-manifold with the same homology groups as a 3-sphere must *be* the 3-sphere. But he didn't invent it this way. Instead, he got it bytaking a regular dodecahedron and identifying its opposite faces in the simplest possible way, namely by a 1/10th turn. So, we've gone from E8 to the dodecahedron!...

The fundamental group of the Poincare homology sphere has 120 elements. In fact, we can describe it as follows. The rotational symmetry group of the dodecahedron has 60 elements. Take the "double cover" of this 60-element group, namely the 120-element subgroup of SU(2) consisting of elements that map to rotational symmetries of the dodecahedron under the double cover p: SU(2) -> SO(3) This is the fundamental group of the Poincare homology ...[sphere]... Now, this 120-element group has finitely many irreducible representations. One of them just comes from restricting the 2-dimensional representation of SU(2) to this subgroup: call that R. There are 8 others: call them R(i) for i = 1,...,8. Draw a dot for each one, and draw a line from the ith dot to the jth dot if the tensor product of R and R(i) contains R(j) as a subrepresentation. We get this picture:o----o---o----o----o----o----o | | oVoila! Back to E8. ...".

The Unit Spheres of the Division Algebras K are related to the Hopf Fibrations of the form Sn = Sm / KP1. For the Real Numbers, the Complex Numbers, and the Quaternions, their Unit Spheres directly form Lie groups: REAL NUMBERS S0 = S1 / RP1 = Z2 is a discrete group COMPLEX NUMBERS S1 = S3 / CP1 = Spin(2) = U(1) is a Lie group QUATERNIONS S3 = S7 / QP1 = Spin(3) = SU(2) = Sp(1) is a Lie group For the Octonions, the Unit Sphere S7 does NOT form a Lie group. However, the Unit Sphere S7 does EXPAND to form the Lie group Spin(8). If the process of expansion is continued upward from S7 to Sp(2) to SU(4) to Spin(8), and you note that Spin(8) is in the Clifford Algebra Cl(8), then you see that the upward expansion can continue indefinitely to Cl(8N) for arbitrary N. If you go downward from Spin(8), based on Real Numbers, to SU(4), based on Complex Numbers, to Sp(2), based on Quaternions, to S7, based on Octonions, then you see that the downward expansion can continue indefinitely to the ZeroDivisor algebra ZD(2^N) for arbitrary N. THE INTERPENETRATION OF THE UPWARD-BASED WEDGE AND THE DOWNWARD-BASED WEDGE SHOWS OCTONION REFLEXIVITY. OCTONIONS S7 = S15 / OP1 is NOT a Lie group .........Cl(8N) = Cl(8) (x)...(N times)...(x) Cl(8).........Cl(32) = Cl(8)^4 \ / \ / Leech Lattice in Cl(24) \ / \ / Spin(16) and E8 in Cl(16) \ / \ / Spin(8) in Cl(8) \ o / R8 G0 G1 G2 G3 G4 G5 G6 G7 \ / \ /SU(4) \ / \ / C4 G0-iG1 G2-iG3 G4-iG5 G6-iG7 ---------X-----X--------- Sp(2) / \ / \ Q2 G0-iG1-jG2-kG3 G4-iG5-jG6-kG7 / \ / \ S7 / o \ O G0-iG1-jG2-kG3-EG4-IG5-JG6-KG7 / \ / \ ND(16) / \ / \ ND(32) / \ / \ ND(64) ... ... ... OCTONION REFLEXIVITY IS INHERITED BY THE SEDENIONS: SEDENIONS S15 (S7=S15/OP1) is NOT a Lie group .........Cl(8N) = Cl(8) (x)...(N times)...(x) Cl(8).........Cl(40) = Cl(8)^5 \ / \ / Cl(32) = Cl(8)^4 \ / \ / Leech Lattice in Cl(24) \ / \ / Spin(16) and E8 in Cl(16) \ o / R16 \ / \ /SU(8) \ / \ / C8 ---------X-----X--------- Sp(4) / \ / \ Q4 / \ / \ S15 (S7=S15/OP1) / o \ O2 / \ / \ ND(32) / \ / \ ND(64) / \ / \ ND(128) ... ... ... OCTONION REFLEXIVITY IS ALSO INHERITED BY THE LEECH LATTICE, BASED ON A FIBRATION OF S23 BY A MANIFOLD M7 THAT IS NOT A SPHERE: 24-dim LEECH LATTICE S23 (M7=S23/OP2) is NOT a Lie group .........Cl(8N) = Cl(8) (x)...(N times)...(x) Cl(8).........Cl(48) = Cl(8)^6 \ / \ / Cl(40) = Cl(8)^5 \ / \ / Cl(32) = Cl(8)^4 \ / \ / Spin(24) in Cl(24) \ o / R24 \ / \ /SU(12) \ / \ / C12 ---------X-----X--------- Sp(6) / \ / \ Q6 / \ / \ S23 (M7=S23/OP2) / o \ O3 / \ / \ ND(48) / \ / \ ND(96) / \ / \ ND(192) ... ... ... AS OP2 IS THE HIGHEST DIMENSIONAL OCTONION PROJECTIVE SPACE, THERE ARE NO HIGHER DIMENSIONAL STRUCTURES OF THAT TYPE.

For each Pair of Interpenetrating Triangles, each Single Triangle corresponds to Mt. Meru: o / \ / \ / \ /_______\ 0 Cl(0) 1 1 R a^2 = 1Cl(1) 1 1 2 C a^2 = -1Cl(2) 1 2 1 4 QCl(3) 1 3 3 1 8 OCl(4) 1 4 6 4 1 16 S ab=0Cl(5) 1 5 10 10 5 1 32 SC a0=bCl(6) 1 6 15 20 15 6 1 64 M(8,R)Cl(7) 1 7 21 35 35 21 7 1 128 M(8,R)+M(8,R) a^2=0Cl(8) 1 8 28 56 70 56 28 8 1 256 M(16,R) a^4=0 Each Pair of Interpenetrating Triangles corresponds to a Mogan David:

/\____/__\____\ / \ /\/ \//\ /\/__\____/__\\ /\/

A Mogan David expandsto form a SriYantra:

This Sri Yantra is a more symmetrical modification of Sri Yantrasfrom twodifferentweb pages. It has 4 Sides, corresponding to the 4 dimensions ofphysical spacetime of the D4-D5-E6-E7-E8physics model and to Octonioncoassociative squares. The 2 Border lines of the 4 Sidescorrespond to the 2 Quaternionic 4-dimensional Spaces that form the8-dimensional Octonions, and so to the 8 Directions.

The Sri Yantra has a Center, which, combined with the 4 Sides,corresponds to the Five Elements.

The Sri Yantra has an Outer Lotus of 16 Petals, corresponding t otwo half-spinor representations of the Spin(8) Liealgebra and to the first-generation fermions of the D4-D5-E6-E7-E8physics model. The 3 Border Rings beyond the Outer Lotus Petalscorrespond to the 3 generation of fermions of the D4-D5-E6-E7-E8physics model.

The Sri Yantra has an Inner Lotus of 8 Petals, corresponding tothe vector representation of Spin(8) and to the 8-dimensionalspacetime of the D4-D5-E6-E7-E8 physicsmodel prior to dimensional reduction.

The Sri Yantra has 9 Triangles, with each triangle correspondingto an Octonion associativetriangle.

The first 6 triangles, 3 pairs, correspond to:

the Octonions O (red pair ofreflexive interpenetrating triangles),

the Sedenions OxO (red and green pairsof reflexive interpenetrating triangles), and

the Leech Lattice OxOxO (red,green, and blue pairs of reflexive interpenetrating triangles),

which in turn correspond to the Liealgebras E6, E7, and E8, and to the 3generations of fermions in the D4-D5-E6-E7-E8physics model;

The remaining 3 triangles, one gold pair and one purple triangle,correspond not to pairs of reflexive interpenetrating triangles, butto Octonion associativetriangles. Each triangle therefore represents an entire Octonionalgebra containing that associative triangle, and the 3 trianglestogether represent how the 3 Octonions of OxOxO are related to oneanother.

The gold pair of triangles corresponds to the two mirror imageOctonion half-spinor representations of Spin(8).

The purple triangle corresponds to the Octonion vectorrepresentation of Spin(8).

The Sri Yantra has a Central Vertex.

The 9 Triangles have 27 Vertices, corresponding to the22 Hebrew letters plus 5 Finals.

The 27 Triangle Vertices correspond to the 27-lineConfiguration whose symmetry group is the Weyl Group of the78-dimensional Liealgebra E6.

The 27 Triangle Vertices also correspond to the 27-complex-dimensionalspace E7 / (E6xU(1)).

The 27 Triangle Vertices plus the 1 Central Vertex correspond tothe 28-dimensional adjoint representation of Spin(8), which in turncorresponds to the 12+16 = 28 gauge bosons, Higgs mechanism, andpropagator phase of the D4-D5-E6-E7-E8physics model.

The 28 Vertices also correspond to the 28-quaternionic-dimensionalspace E8 / (E7xSU(2)).

The 28 Vertices plus the 16+8 = 24 Lotus Petals form the52-dimensional Lie algebra F4, to which theLie algebras E6, E7, and E8 are related by the Freudenthal-TitsMagic Square.

The construction of this Sri Yantra is based on the Golden Ratioof the Great Golden Pyramid:

It is so constructed that the distance from the lower corners ofthe 2x1 rectangle to the top and bottom vertices of the Sri Yantra isPHI = (1/2)(diagonal of rectangle + 1) = (1/2)(sqrt(5) + 1).

Since PHI^2 = PHI + 1, the radius of the circle is sqrt(PHI^2 -1^2) = sqrt ((PHI + 1) - 1) = sqrt(PHI).

Let D be the distance between the base lines of the two largesttriangles. Then the highest and lowest base lines are at distance Dabove and below the base lines of the two largest triangles.

Everthing else is constructed symmetrically from those basicelements. The construction method is a more symmetrical modificationof the method shown on a GoldenSri Yantra web page.

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