Rotational symmetry of the Tesseract?
George Hammond
The Tesseract is particularly well known as the unit cube
in the 4 dimensional spacetime of Relativity.
While it is well known that an ordinary 3D cube has
13 (rotational) symmetry axes, namely-
The 3 normals of the cube.
The 4 body diagonals of the cube.
The 6 face diagonals of the cube.
Whence, 3+4+6=13 rotational symmetry
axes of the 3D cube.
The QUESTION at hand is: How many rotational
symmetry axes does a 4D cube have?
If anyone knows the answer to this question, or how to
compute the answer, or where to find the answer
I would very much appreciate the help.
PS: Interestingly, there is a simple formula for the total
number of rotations of an n-dimensional cube, namely
(n!)(2)^(n-1) .... see:
http://gregegan.customer.netspace.net.au/APPLETS/29/HypercubeNotes.html
This formula yields 24 rotations for a 3D cube
(which we can manually confirm to be correct
for the above 13 axes....and it gives 192 rotations
for the 4D cube....but it doesn't tell us the number
of individual rotational AXES in the 4D cube.... which
is what I'm trying to find. Any suggestions?
[Hammond]
hagman
First you need to specify what you call rotational axis.
If you want only orientation preserving mappings (i.e. no
reflections),
then you also have to accept rotations around plains (e.g. keeping the
forth dimension fix and rotating the 3D projection in a standard
manner).
Greg Egan
> The QUESTION at hand is: How many rotational
> symmetry axes does a 4D cube have?
Rotations in 4 dimensions don't have axes (i.e. 1-dimensional
invariant subspaces), they have invariant subspaces with either 2
dimensions (i.e. planes) or 0 dimensions (i.e. only the 0 vector is
unchanged by the rotation).
For example, a 4-d rotation might rotate vectors in the x-y plane by
some angle theta, and leave the z and w components of vectors
completely untouched. Then the whole z-w plane is an invariant
subspace of that rotation (just as the z-axis alone would be the
invariant subspace in the 3-d case).
But you can also have a 4-d rotation that rotates in *both* the x-y
plane and the z-w plane, and has *no* (non-zero) invariant vectors!
OK, so what about the rotational symmetries of the 4-d hypercube?
Here's a partial answer. Suppose you pick one of the 4 dimensions as
the direction from the origin to the centre of a cubic hyperface of
your tesseract, and call that vector e_i. Then there will be 13 axes
of rotation from all the 3d rotations of a cube that we can map into
the remaining three dimensions (those that are not e_i), and the 13
distinct planes we get that are spanned by those 13 directions along
with e_i will be preserved by those rotations that preserve our chosen
cubic hyperface. But we can choose e_i four different ways, so that
gives us 52 distinct planes that are preserved by various rotations.
That's not the full answer, though, because we've only accounted for 4
x 23 = 92 of the 191 non-trivial rotations (i.e. the 23 non-trivial
rotations of the cube, for each choice of e_i). What about the other
99 rotations? Do they all have invariant planes? Are some of these
planes ones that we've already counted? I don't know the answers to
these questions off the top of my head ...
schoenf...@gmail.com
> "Tesseract" is the name given to a 4D cube (a hypercube).
> The Tesseract is particularly well known as the unit cube
> in the 4 dimensional spacetime of Relativity.
>
> While it is well known that an ordinary 3D cube has
> 13 (rotational) symmetry axes, namely-
>
> The 3 normals of the cube.
> The 4 body diagonals of the cube.
> The 6 face diagonals of the cube.
>
> Whence, 3+4+6=13 rotational symmetry
> axes of the 3D cube.
>
> The QUESTION at hand is: How many rotational
> symmetry axes does a 4D cube have?
Every N-dimensional rotation can be described by a minimum of N(N-1)/2
parameters.
Proof:
Let R denote the matrix-representation of the arbitrary rotation. R
is an NxN square matrix. It is known that R can always be given by the
exponentiation of an NxN skew-symmetric matrix S. That is, R = exp(S).
Since S is an NxN matrix, S has N^2 elements. Since S is skew-
symmetric, the N diagonal elements of S must all be 0 leaving us with
N^2-N unique elements.
But the skew-symmetry of S also implies that upper-triangular portion
is defined by the lower-triangular portion leaving us with only (N^2-
N)/2=N(N-1)/2 unique elements in S. It follows trivially that only
N(N-1)/2 elements are necessary to describe an arbitrary N-rotation.
QED
Now, if we wish to also generalize Euler angles into N-space, we can
choose to correlate each of the parameters to rotation angles on the
orthogonal basis planes. That is,
(1) R(x_1, x_2, ..., x_n) = MUL(k=1..n) R_k(x_k)
Where
R_n(x_n) denotes the matrix for rotating on the n'th
orthogonal basis plane.
x_n is the angle to rotate on the n'th orthogonal basis plane
There are many possible sets of orthogonal basis planes to choose
from. In 3-space we could choose from {XY, YZ, ZX} or {XY, XZ, YZ} or
{XY,YZ, XZ}, etc. In N-space the distinction does not really matter
much. The set of planes and the order in which they are applied are
thus left to convention.
For simplicity, we can choose a regime where we permute the basis
vectors sequentially to form our set of basis planes. For example, in
a 4-space with basis (X,Y,Z,W) we construct the set of basis planes as
{XY, XZ, XW, YZ, YW, ZW}.
The canonical rotation matrix R_n(x_n) is constructed by
exponentiation of a skew-symmetric matrix S (as we used in the proof).
For simplicity, we will rename the index 'n' into the set of basis
planes to 'ij' where 'i' denotes some basis vector and 'j' some other
basis vector such that when combined into 'ij' ij'th basis plane is
represented (i.e. the n'th plane in the set of basis planes).
The canonical rotation matrix R_n(x_n) thus becomes R_ij(x_ij) defined
(2) R_ij(x_ij) = exp( x_ij S_ij)
where
(S_ij)[i][j] = 1 and (S_ij)[j][i] = -1 and (S_ij)[_][_] = 0
for all other elements
This convention allows us various possible definitions of the skew-
symmetric matrix S. We have chosen the definition such that it is
consistent with sequential permutation of the basis vectors (as
discussed previously).
Hope that helps.
Greg Egan
> On Mar 26, 3:45 pm, "George Hammond" <Nowhe...@notspam.org> wrote:
>
> > The QUESTION at hand is: How many rotational
> > symmetry axes does a 4D cube have?
> [W]e can choose e_i four different ways, so that
> gives us 52 distinct planes that are preserved by various rotations.
Sorry, there aren't 52 *distinct* planes here, because there is some
duplication. Of the 13 axes of rotation, 3 will be basis vectors
(assuming our cubes and hypercubes are aligned in such a way that each
face is displaced from the centre along a basis direction). So there
are really only 4 x 10 = 40 planes from the other 10 axes, and then
all 6 choices of pairs of basis vectors, giving a total of 46 distinct
planes.
> That's not the full answer, though, because we've only accounted for 4
> x 23 = 92 of the 191 non-trivial rotations (i.e. the 23 non-trivial
> rotations of the cube, for each choice of e_i).
And I double-counted to get those "92" rotations, also; there are
actually just 74 distinct rotations that are produced this way.
I can give you some figures from a brute-force Mathematica
calculation:
(a) there are 6 invariant planes that contain two basis vector, and
which are related to 18 rotations;
(b) there are 40 invariant planes that contain one basis vector, and
which are related to 56 rotations;
[Cases (a) and (b) give the 46 planes and 74 rotations described
previously]
(c) there are 12 invariant planes that contain no basis vector, and
are related to 12 rotations;
(d) the remaining 105 rotations have *no* invariant vectors.
It would be nice to think of an elegant combinatorial or group-
theoretic way of classifying all these cases.
George Hammond
<goo...@von-eitzen.de> wrote:
[Hammond]
Yes, only "orientation preserving mappings".
No "reflections".
>then you also have to accept rotations around plains (e.g. keeping the
>forth dimension fix and rotating the 3D projection in a standard
>manner).
[Hammond]
No.... no "planar rotations", only simple "axial
rotations"... i.e. rotation of the entire 4D structure
around a simple axis-line in 4D space.... entirely analogous
to the rotational axes described above for a 3D cube.
Sylvain Croussette
But that's not possible in 4D. It's been already explained in a
previous answer, the only possible invariants in 4D are 0D (a point),
2D (a plane), or 4D (the whole space). You can't have a line (1D)
axis in 4D. It's only in odd-dimension spaces that you can have a
line axis, and even then it's not always the case, as it is in 3D.
All this follows from the properties of rotations and their
eigenvalues/eigenvectors.
JEMebius
> "Tesseract" is the name given to a 4D cube (a hypercube).
> The Tesseract is particularly well known as the unit cube
> in the 4 dimensional spacetime of Relativity.
(*)
>
> While it is well known that an ordinary 3D cube has
> 13 (rotational) symmetry axes, namely-
>
> The 3 normals of the cube.
> The 4 body diagonals of the cube.
> The 6 face diagonals of the cube.
>
> Whence, 3+4+6=13 rotational symmetry
> axes of the 3D cube.
>
> The QUESTION at hand is: How many rotational
> symmetry axes does a 4D cube have?
>
> If anyone knows the answer to this question, or how to
> compute the answer, or where to find the answer
> I would very much appreciate the help.
>
> PS: Interestingly, there is a simple formula for the total
> number of rotations of an n-dimensional cube, namely
> (n!)(2)^(n-1) .... see:
> http://gregegan.customer.netspace.net.au/APPLETS/29/HypercubeNotes.html
> This formula yields 24 rotations for a 3D cube
> (which we can manually confirm to be correct
> for the above 13 axes....and it gives 192 rotations
> for the 4D cube....but it doesn't tell us the number
> of individual rotational AXES in the 4D cube.... which
> is what I'm trying to find. Any suggestions?
> [Hammond]
>
>
Consult the Ph.D. thesis of Salomon Levi van Oss:
Record from online catalog of Delft University of Technology Library
http://library.tudelft.nl/ws/index.htm >Catalogue >Access to the catalogue
Title
Die Bewegungsgruppen der regelmässigen Gebilde von vier Dimensionen
Author Oss, Salomon L. van
Publisher Utrecht : z. uitg., 1894
Descr. 38 S. M. Taf., 8 o
Gen. note Giessen, Dr.-Ing.-Diss. 1894
Classific. BTA / Meetkunde (algemeen) / Geometry (general)
Shelf number CB 06164502
I made an excerpt, which I do not have at hand at this moment. Will post
it to the newsgroup when I will have found it among my paperwork.
An approach different from Van Oss's uses the several different
homomorphisms and isomorphisms of the 4D rotation group SO(4), the unit
quaternion group S^3 and its direct product with itself, and the 3D
rotation group SO(3) and its direct product with itself.
I did not do this exercise. Presumably it will amount to squaring the
order (24) of the rotational congruence group of the 3D cube, and
dividing this number (576) by two each time a 2:1 group homomorphism
turns up. Or something similar to this...
(*) BTW, the tesseract does not play any role in the geometry of Special
Relativity. It is an object from Euclidean 4D space with signature ++++,
whereas the signature of Einstein-Minkowski-Lorentz space is -+++ (or
+---, depending on the convention on signs).
Cheers: Johan E. Mebius
mensa...@aol.com
Gee, is this going to make Hammond's Structural Model of Personality
and its mapping to the Sephirot fall apart? What a shame.
>
> Cheers: Johan E. Mebius
George Hammond
<greg...@netspace.net.au> wrote:
>On Mar 26, 3:45 pm, "George Hammond" <Nowhe...@notspam.org> wrote:
>
>> The QUESTION at hand is: How many rotational
>> symmetry axes does a 4D cube have?
>
>Rotations in 4 dimensions don't have axes
[Hammond]
Thanks Greg.... that statement says it all... I won't forget
where I first heard that either!
> (i.e. 1-dimensional
>invariant subspaces), they have invariant subspaces with either 2
>dimensions (i.e. planes) or 0 dimensions (i.e. only the 0 vector is
>unchanged by the rotation).
>
>For example, a 4-d rotation might rotate vectors in the x-y plane by
>some angle theta, and leave the z and w components of vectors
>completely untouched. Then the whole z-w plane is an invariant
>subspace of that rotation (just as the z-axis alone would be the
>invariant subspace in the 3-d case).
>
>But you can also have a 4-d rotation that rotates in *both* the x-y
>plane and the z-w plane, and has *no* (non-zero) invariant vectors!
>
[Hammond]
OK... I'm get the picture.... in 4D space there are no
"single rotational symmetry axes" but only "planes full of
rotational symmetry axes" (i.e. every line in said plane is
a rotational symmetry axis).... hmmmm
>OK, so what about the rotational symmetries of the 4-d hypercube?
>
>Here's a partial answer. Suppose you pick one of the 4 dimensions as
>the direction from the origin to the centre of a cubic hyperface of
>your tesseract, and call that vector e_i. Then there will be 13 axes
>of rotation from all the 3d rotations of a cube that we can map into
>the remaining three dimensions (those that are not e_i), and the 13
>distinct planes we get that are spanned by those 13 directions along
>with e_i will be preserved by those rotations that preserve our chosen
>cubic hyperface. But we can choose e_i four different ways, so that
>gives us 52 distinct planes that are preserved by various rotations.
>
>That's not the full answer, though, because we've only accounted for 4
>x 23 = 92 of the 191 non-trivial rotations (i.e. the 23 non-trivial
>rotations of the cube, for each choice of e_i). What about the other
>99 rotations? Do they all have invariant planes? Are some of these
>planes ones that we've already counted? I don't know the answers to
>these questions off the top of my head ...
ok... so far so good... I'm going to read your other post
now.
George Hammond
<greg...@netspace.net.au> wrote:
>On Mar 26, 8:39 pm, "Greg Egan" <grege...@netspace.net.au> wrote:
>> On Mar 26, 3:45 pm, "George Hammond" <Nowhe...@notspam.org> wrote:
>>
>> > The QUESTION at hand is: How many rotational
>> > symmetry axes does a 4D cube have?
>[snip]
>
>> [W]e can choose e_i four different ways, so that
>> gives us 52 distinct planes that are preserved by various rotations.
>
>Sorry, there aren't 52 *distinct* planes here, because there is some
>duplication. Of the 13 axes of rotation, 3 will be basis vectors
>(assuming our cubes and hypercubes are aligned in such a way that each
>face is displaced from the centre along a basis direction).
[Hammond]
Roger that....
> So there
>are really only 4 x 10 = 40 planes from the other 10 axes, and then
>all 6 choices of pairs of basis vectors, giving a total of 46 distinct
>planes.
>
ok...
>> That's not the full answer, though, because we've only accounted for 4
>> x 23 = 92 of the 191 non-trivial rotations (i.e. the 23 non-trivial
>> rotations of the cube, for each choice of e_i).
>
>And I double-counted to get those "92" rotations, also; there are
>actually just 74 distinct rotations that are produced this way.
>
>I can give you some figures from a brute-force Mathematica
>calculation:
>
whew....
>(a) there are 6 invariant planes that contain two basis vector, and
>which are related to 18 rotations;
>
that's interesting.... and highly suspicious (e.g.
significant) for my particular application...
>(b) there are 40 invariant planes that contain one basis vector, and
>which are related to 56 rotations;
>
>[Cases (a) and (b) give the 46 planes and 74 rotations described
>previously]
>
>(c) there are 12 invariant planes that contain no basis vector, and
>are related to 12 rotations;
>
>(d) the remaining 105 rotations have *no* invariant vectors.
>
that's highly significant in this application...
>It would be nice to think of an elegant combinatorial or group-
>theoretic way of classifying all these cases.
Well... you're not going to believe this but I'll tell
you anyway. Computerized Factor Analysis of psychology
(personality structure) has discovered that the underying
model is a 3D CUBE which is traced back to the fact that the
cleavage geometry of the brain is actually Cartesian (aka
"cubic").
The angular structure of the 13 rotational symmetry axes
turn out to be the angular arrangement of the (13) Factors
(eigenvectors) of the personality correlation matrix!
However, while the Structural Model of Personality is a 3D
cube, when you add Intelligence (IQ) as a 4th dimension, you
get a 4D space and the model becomes a 4D CUBE.... the
Tesseract.
Again, one would suppose that the data should reflect the
rotational symmetry axes.... or in this case the "rotational
symmetry planes" of the hypercube.
Interestingly, while the 3D cubic structure of
Personality data is easily empirically confirmed (Hammond
1994), no one has ever bothered to Factor the 4D correlation
data since, as in Relativity, a "3+1" analysis is far
simpler than a 4D analysis.
However.... it is now suspected that the hypercube and
its symmetries HAVE in fact surfaced in something known as
the (800 year old) "Sepheriot" diagram which Hammond has now
suggested may be a "graph" of the Tesseract (e.g. a 2D map),
OR that in fact the Sephirot may be a 2D graph (plan) of the
"symmetry axes" (symmetry planes) of the Tesserat.
Will keep you posted if I stumble on anything looking
like a historic breakthrough.
In the meantime, thanks for teaching me the basics of
higher dimensional rotation!
[Hammond]
George Hammond
<sylvaincr...@yahoo.ca> wrote:
[Hammond]
Roger that.... Greg Egan straightened me out on that
point... see my replies to his posts below.
[Hammond]
Uncle Al
[snip crap]
> [Hammond]
If you know you are an idiot, and we know you are an idiot, and each
knows the other knows you are an idiot, why do you further pursue the
point?
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
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> >George Hammond wrote:
> >[snip crap]
> >> [Hammond]
>
> >If you know you are an idiot, and we know you are an idiot, and each
> >knows the other knows you are an idiot, why do you further pursue the
> >point?
>
> >--
> >Uncle Al
> >http://www.mazepath.com/uncleal/
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schoenf...@gmail.com
Since when is a "Minkowski space" an "Einstein-Minkowski-Lorentz
space"? Einstein did not discover the 4D spacetime, he was initially
baffled by it. Lorentz was not (directly) involved in its discovery
either, and was baffled by other concepts already established by
Einstein.
Uncle Al
[snip Hammond crap]
Hey Georgie Hammond, Spam King of Kings of the halfling faggot
god-on-a-stick: Uncle Al had a nice conversation with Yahweh. HE is
pulling all your accounts - verson, hotmail, google... - under all
your pseudonyms. All of them. You are to be permanently disconnected
across all of the Web and Usenet for chronic abuse.
Have a nice Rapture.
Eric Gisse
> George Hammond wrote:
>
> [snip crap]
>
> > [Hammond]
>
> If you know you are an idiot, and we know you are an idiot, and each
> knows the other knows you are an idiot, why do you further pursue the
> point?
One can never be too careful.
>
> --
> Uncle Alhttp://www.mazepath.com/uncleal/
JanPB
> ==================
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>
> "JanPB" <film...@gmail.com>
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> wrote:
>
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> >From: "JanPB" <film...@gmail.com>
> >Newsgroups: sci.math,sci.physics,sci.physics.relativity
> >Subject: Re: Rotational symmetry of the Tesseract?
> >Date: 26 Mar 2007 19:34:27 -0700
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Actually, you have zero chance.
--
Jan Bielawski
victor_me...@yahoo.co.uk
> The QUESTION at hand is: How many rotational
> symmetry axes does a 4D cube have?
"rotational symmetry axes"? That is SUCH a three-dimensional
mode of thought :-(
Victor Meldrew
schoenf...@gmail.com
> Since when is a "Minkowski space" an "Einstein-Minkowski-Lorentz
> space"?
Never. Just as I thought.
hagman
> On 26 Mar 2007 05:39:08 -0700, "Greg Egan"
>
> >On Mar 26, 3:45 pm, "George Hammond" <Nowhe...@notspam.org> wrote:
>
> >> The QUESTION at hand is: How many rotational
> >> symmetry axes does a 4D cube have?
>
> >Rotations in 4 dimensions don't have axes
>
> [Hammond]
> Thanks Greg.... that statement says it all... I won't forget
> where I first heard that either!
>
> > (i.e. 1-dimensional
> >invariant subspaces), they have invariant subspaces with either 2
> >dimensions (i.e. planes) or 0 dimensions (i.e. only the 0 vector is
> >unchanged by the rotation).
>
> >For example, a 4-d rotation might rotate vectors in the x-y plane by
> >some angle theta, and leave the z and w components of vectors
> >completely untouched. Then the whole z-w plane is an invariant
> >subspace of that rotation (just as the z-axis alone would be the
> >invariant subspace in the 3-d case).
>
> >But you can also have a 4-d rotation that rotates in *both* the x-y
> >plane and the z-w plane, and has *no* (non-zero) invariant vectors!
>
> [Hammond]
> OK... I'm get the picture.... in 4D space there are no
> "single rotational symmetry axes" but only "planes full of
> rotational symmetry axes" (i.e. every line in said plane is
> a rotational symmetry axis).... hmmmm
No. The lines in said plane are not rotational symmetry axes,
it is only the full plane that behaves as a symmetry "axis".
Compare in 3D: If you have a rotation with an axis then you cannot
say that every single point of that line is a centre of symmetry.
Greg Egan
cube (followed by a classification scheme for all the symmetries of an
n-cube of arbitrary dimension) at:
http://www.gregegan.net/APPLETS/29/HypercubeNotes.html#ROT
One interesting feature I hadn't realised before was that some
symmetries involve planes of rotation that don't include the centres
of any lower-dimensional parts of the n-cube. This doesn't happen for
the three-dimensional cube; in that case, all the planes of rotation
contain vertices, edge-centres or face-centres. The *simple*
rotations of an n-cube (those that rotate in only a single plane)
conform to that pattern, and break down into 4 basic kinds, where the
planes of rotation can be defined by the kind of lower-dimensional
centres they contain. However, the rotations in more than one plane
-- which first become possible in 4 dimensions -- include cases like
this:
http://www.gregegan.net/APPLETS/29/NonSimple4E.gif
where the two planes of rotation intersect the 4-cube in two regular
octagons, whose vertices in each case lie off-centre in eight 2-faces
of the 4-cube.
In retrospect, though, it's kind of obvious that something like this
had to happen, because the cosine of the angle between two k-cube-
centres of an n-cube (both with the same k) must be a rational number,
and so they can only be separated by angles of 60, 90, 120 or 180
degrees. But there are obviously going to be some group elements of
the symmetries of a 4-cube with order 8, which implies a rotation by
45 degrees. So the relevant planes of rotation can't contain any k-
cube centres at all.
usti...@gmail.com
could you please describe the symmetry group of the tesseract in terms of Lie algebra, group theory.
Thanks,
Ustin
воскресенье, 22 апреля 2007 г., 4:34:53 UTC+3 пользователь Greg Egan написал:
Robin Chapman
> Greg,
> could you please describe the symmetry group of the tesseract in terms of Lie algebra, group theory.
The symmetry group of the n-hypercube is the "Weyl group of type B_n",
sometimes known as the signed symmetric group or the hyperoctahedral
group: https://en.wikipedia.org/wiki/Hyperoctahedral_group .
In group-theoretic terms it is the wreath product of the two-element
group by the symmetric group S_n.
abu.ku...@gmail.com
of four axes that are ne'er perpendicular.
in the general case, the four-ary axes of the hexahedron is
> The symmetry group of the n-hypercube is the "Weyl group of type B_n",
> sometimes known as the signed symmetric group or the hyperoctahedral
> group: https://en.wikipedia.org/wiki/Hyperoctahedral_group .
> In group-theoretic terms it is the wreath product of the two-element
> group by the symmetric group S_n.
abu.ku...@gmail.com
the pointwise spatial axes are the biperpendiculars
of pairs of opposite edges, which are perpedcicular