## Can anyone explain what conjugation is?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

### Can anyone explain what conjugation is?

I've seen quite a lot of people mention "conjugation" on this forum but I never understood what it meant? An image would help as well as a simple explanation.

ubersketch
Trionian

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### Re: Can anyone explain what conjugation is?

I suppose that your "conjugation" stems out of the context of complex numbers.
There you can represent the real 2D space as being identified with the complex 1D space,
called the "Argand plane", by means of (a,b) = a*ex+b*ey --> a*1+b*i = a+ib.

Then you have the complex conjugation, which mainly is an algebraic operation,
which provides you with
Code: Select all
`conj(a+ib) = a-ib`
that is, spoken within the Argand plane, conjugation is nothing but the reflection at the x-axis,
i.e. the reflection on the embedding of the real line.

If you further have some arbitrary complex number z, say z=a+ib, then you can consider |z|.
It represents the vector length, i.e. the distance from that point within the Argand plane to the origin.
Therefore, by means of Pythagoras, that one would be the square root of a2+b2.
But OTOH you have z*conj(z) = (a+ib)*(a-ib) = a2+iab-iab-i2b2 = a2-(-1)*b2,
therefore z*conj(z) = a2+b2 = |z|2, or taken the other way round, conj(z) = |z|2/z.

But, there is also a different connotation to "conjugation" as well.
Again it is meant to be the algebraical conjugate.
That is, the other root of some (usually quadratic) equation.

E.g. you might have some equation like x2-x-1=0. Then you have 2 solutions.
One being x1 = (1+sqrt(5))/2 = 1.618034, the golden mean (sometimes represented by phi, somtimes by tau),
the other one is x2 = -(sqrt(5)-1)/2 = (1-sqrt(5))/2 = -0.618034.

Again you can consider the module of these numbers, that is
ZZ[tau] = { a+b*tau : a,b in ZZ } = { (c+d*sqrt(5))/2 : c,d in ZZ } = (1/2)*ZZ[sqrt(5)],
as a similar 2D representation. In fact you would have z = a+b*tau = ((2a+b)+b*sqrt(5))/2 = (c+d*sqrt(5))/2.

Thus again you can define a mapping on this module, which then extrapolates the
mentioned relationship of those 2 solutions according to
Code: Select all
`conj( (a+b*sqrt(5))/2 ) = (a-b*sqrt(5))/2`
thus you'll then would get: this algebraic conjugation is nothing but the reflection wrt. the x-axis again!

--- rk
Klitzing
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### Re: Can anyone explain what conjugation is?

Seems interesting. What I'm getting is that conjugating a+bi results in a-bi but how does this apply in geometry? For example, the conjugate of the icosahedron is the great icosahedron and the conjugate of the dodecahedron is the great stellated dodecahedron. I'm assuming this has to do with conjugation in terms of complex numbers.

ubersketch
Trionian

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### Re: Can anyone explain what conjugation is?

No, that one is the algebraic conjugation within the modul QQ[sqrt(5)], i.e. the numbers q+p*sqrt(5), p and q in QQ.
(Sorry, don't have a better typographical symbol than QQ for the rationals at hand.)
There you'd have:
Code: Select all
`conj( q+p*sqrt(5) ) = q-p*sqrt(5).`

Then f=|(1+sqrt(5))/2| becomes v=|(1-sqrt(5))/2|.
And the obtuse triangle with sides x (|x|=1), x, and f has a tip angle of 108 degrees,
i.e. the dihedral angle of a corner of a pentagon.
OTOH an acute triangle with sides x, x, and v has a tip angle of 36 degrees,
i.e. the dihedral angle of a corner of a pentagram.

Thus the pentagram happens to be the algebraic conjugate of the pentagon (wrt. the above operation on that module).
Same holds higher dimensional figures.
E.g. conj( {5,5/2} ) = {5/2,5}, conj( {3,5} ) = {3,5/2}, conj( {3,3,5} ) = {3,3,5/2}, etc.

To be more precise, you'd have to take account of the absolute value included above.
That is, you rather have the more precise conjugacy between 5/1 (prograde pentagon) and 5/3 (retrograde pentagram),
Then you will get correctly too x5x (as Dynkin symbol, where x marks the ringed nodes),
which represents the regular decagon, happening to be conjugate to x5/3x, the regular decagram.

BTW there is a similar conjugacy between 4 (prograde square) and 4/3 (retrograde square).
This will take effect when going likewise to its Stott expansion:
x4x represents the regular octagon, while x4/3x represents the regular octagram.

--- rk
Klitzing
Pentonian

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### Re: Can anyone explain what conjugation is?

The process of 'isomorphism' is where one cycles through the roots of various equations. For the class-2 systems, this usually means replacing sqrt(5) as +2.236067 or -2.236067. A similar scheme exists in all other polygon systems, where one transforms various equation-solutions.

If one takes a graph of rays, say the ray graph of eight radiating lines, the steps in order (0,1,2,3,4,5,6,7), give a regular octagon, while (0,5,2,7,4,1,6,3) gives an octagram. If one supposes that 0 points to 1,0 and 2 to 0,1 (in clock-face style), then 1 is at ½r2, ½ r2 (where r2 is sqrt(2) ). But changing sqrt(2), replaces r2 by -r2, and this swaps 1 and 5, and 3 and 7, while leaving 0,2,4,6 alone. This is equivalent to multiplying the point number by 5, modulo 8. The order (0,1,2,3,4,5,6,7) now gives the octagram, while 0,5,2,7,4,1,6,3 gives an octagon.

A clock-face with the numbers running from 0 to 9, can be used to draw a pentagon (0,2,4,6,8), a pentagram (0,4,8,2,6), a decagon (0,1,2,3,4,5,6,7,8,9) and a decagram (0,3,6,9,2,5,8,1,4,7). If you multiply the coordinates by 3 or 7, pentagons and pentagrams swap, and decagons and decagrams swap. It's a bit more complex than just +sqrt(5) <> -sqrt(5), but this is indeed the heart of the matter.

The thing can then be represented as a kind of dual-projection of something in 2n space, where in one case, an octagon appears, and in the other an octagram appears.

The relative importance of all of this, is that if a figure is finite, it must have finite conjugates. This means that while we could put three decagrams around a corner (as x10/3o3o) the conjugate must also be of the same nature (ie x10o3o as three decagons, does not exist), and so this can not be finite. The first is an infinitely wound polyhedron, and the second is a hyperbolic tiling. Information about both tells us a lot more than about either alone. We can see that x10o3o can not contain x10o6o, because the conjucate of the first is a spherical polyhedron, while the second is a dense hyperbolic tiling. Since the conjugates can not contain each other, nor can the original set.

One should also note that every tiling that solves a finite polynomial, with a unit first term, has solutions that are "algebraic integers", every algebraic integer necessarily divides a real integer. Such is usually the unit term on the equation, or the determinate of the defining matrix.
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wendy
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### Re: Can anyone explain what conjugation is?

Is there any sort of image describing how conjugation works because I learn from images much easier than words.

ubersketch
Trionian

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### Re: Can anyone explain what conjugation is?

An quite easy pair of conjugate polyhedra (aka isomorph ones as abstract polytopes) are the couple
--- rk
Klitzing
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