# A new classification proposed for Crystal Classes

Last Updated: 26th Feb 2014## By Giuliano Bettini

**A new classification proposed for Crystal Classes**

I decided that the best place to post this article was this "temple" of mineralogy (Mindat), where it is possible

that someone has some idea.

(Further details may be found in my paper http://vixra.org/abs/1101.0052).

The idea here is the classification of the 32 crystal classes with 5 bits, bits the meaning of which

I have not yet fully understood.

Here it’s argued that bits should be identified with five basic unknown symmetries or properties,

generating these 32 groups. Probably it is not merely a coincidence that 32 means 5 bits.

And probably is it not merely a coincidence that each complete subset of bits (properties) means

the holohedry of a crystal system; and each new bit means a new crystal system; neither is a coincidence

the fact that the proposed classification matches ALMOST perfectly the physical properties of the

various crystals.

The purpose of this article is of course not to draw a conclusive theory, but to suggest ideas that,

we hope, will be useful for researchers in mathematics, group theory and crystallography.

**Premise**

As is well known to any mineralogist or mathematician there are only 32 crystal classes.

These are mathematically 32 point groups, generated by symmetry operations: reflection, rotation,

inversion, and the combined operation rotoinversion.

Usually an axis of rotational symmetry is represented by the capital letter A: A2, A3, A4, A6.

The number of such axes present is indicated by a numeral preceding the capital A. For example,

1A2, 2A3, and 3A4 represent one 2-fold axis of rotation, two 3-fold axes, and three 4-fold axes

respectively.

A center of inversion is noted by the letter “c” or “i”.

A mirror plane is denoted by “m”. A numeral preceding the m indicates how many mirror planes

are present.

An axis of rotary inversion (a compound symmetry operation which is produced by performing a

rotation followed by an inversion) is represented by the capital letter R.

A rotoinversion axis is sometime replaced by the equivalent rotations and reflections. For example,

a 2-fold rotoinversion axis is equivalent to reflection through a mirror plane perpendicular to the

rotation axis.

The reflection, rotation, inversion, and rotoinversion symmetry operations may be combined in a

variety of different ways. There are 32 different possible combinations of these symmetry

operations, or 32 point groups.

The different symmetry operations are a lot.

For example, some authors use these EIGHT symmetry operations:

i, A2, m, R4, A4, A3, R6, A6.

But....here the idea.

**The idea**

Question: why so much symmetry operations and / or combinations of symmetry operations?

Or: what is the minimum number of symmetry operators required to create all possible combinations

(i.e. the 32 point groups)?

To me the answer is: five

As a matter of fact, it is inevitable to associate “32” to “5 bits”.

This means making the assumption that there are only five basic properties or only five basic

symmetries generating these 32 groups.

Their presence or absence will automatically determine which class or point group of origin.

I think it’s useful to note that 32 classes can be classified with 5 bit.

This is not a conjecture: this is A FACT.

At the same time, it’s possible give a name, a label, to each bit.

This is also is a fact.

The problem is: has each bit a physical meaning?

Does it correspond to some “property”?

Some “quality”?

Let’s assume the following.

Bit c, less significant bit

0000c in position 00001

The bit c is in some way related to the property “center”.

Bit m

000m0 in position 00010

The bit m is related to the property “plane”.

Bit 2

00200 in position 00100

The bit 2 is related to the property “axis 2”.

Bit 4

04000 in position 01000

The presence of bit 4 corresponds to the presence of a property that provisionally identify in a more

subtle way: a symmetry that corresponds to half of a possible and / or existing symmetry.

This bit then appears to identify axis 4, 90° as half of 180°, or even an axis 6, 60° as half of 120°.

Bit 3, must significant bit

30000 in position 10000

The bit 3, the most significant bit, is related to the property “axis 3”.

**A property of the bit counting**

If you count 32 wit 5 bits (ie fro 0 to 31 or from 1 to 32) each bit appears 16 times.

Each pair of bits appears 8 times etc.

This is well known, and in any case one can make a direct verification.

So 32 is naturally divided into groups of 16, groups of 8, 4, groups of 12 = 8 + 4 and so on.

In fact, let’s count from 1 to 32 (or 0 to 31).

00000

**00001**

00010

**00011**

00100

00101

00110

**00111**

01000

01001

01010

01011

01100

01101

01110

**01111**

10000

10001

10010

10011

10100

10101

10110

**10111**

11000

11001

11010

**11011**

11100

11101

11110

**11111**

But also the 32 Crystal Classes are naturally divided into groups of 16, groups of 8, groups of 12

and so on (or.........“quasi-naturally” divided......).

Let’s see how.

The number of Classes in each System is

Triclinic 2

Monoclinic 3

Orthorombic 3

Tetragonal 7

Hexagonal 12

Isometric 5

Sometime the Hexagonal System is divided into the hexagonal and rhombohedral or trigonal

divisions.

Anyway:

- the whole Hexagonal System has exactly 12 Classes;

- there are exactly 8 Classes from Triclinic to Orthorombic;

- there are 7+5=12 Classes in the Tetragonal plus Isometric System.

More: in order to have much more symmetry, let's imagine that, in fact, a Class has to be

moved from the Isometric to the Tetragonal System.

(Why? For which reason? Which Class? Don't know.....let's imagine).

The lowest symmetry class in the Isometric System is the Tetartoidal Class. Imagine to move

that Class from the Isometric to the Tetragonal System in order to have:

-exactly 4 Isometric Classes;

-exactly 8 Tetragonal Classes.

This way the situation is much more suited to a 5 bit classification.

**The proposed classification**

Now, just as an example, let’s examine the first 8 Classes, which of course may be classified

with 3 bit, the less significant bits.

They are:

Bit c, “center”.

0000c in position 00001

Bit m, “plane”.

000m0 in position 00010

Bit 2, “axis 2”.

00200 in position 00100

Taking into account the properties center, plane, axis 2 exhibited by the 3 Classes, we must “match”

the physical properties of the crystals with the bits. The best classification I’ve found is:

1 00000 0 triclinic

**1 0000c 1 triclinic**

m 000m0 2 monoclinic

**2/m 000mc 3 monoclinic**

2 00200 4 monoclinic

222 0020c 5 orthorhombic

mm 002m0 6 orthorhombic

**mmm 002mc 7 orthorhombic**

Readers who have experience of the crystal Classes will easily recognize that the classification

with the 3 bits c, m, 2, matches ALMOST perfectly the physical properties of the various Classes

(“center”, “plane”, “axis 2”).

(Note: the first column is the Hermann Mauguin notation.

The class which possesses the highest possible symmetry within each crystal system is termed the

holomorphic class of that system or the holohedry of the system. The holomorphic class of each crystal

system is indicated here in by bold type. For example, the mmm class is the holomorphic class of the

orthorombic system. The 2/m class is the holomorphic class of the monoclinic system.

Each complete subset of bits (ie 00111 or 00011) means the holohedry of a System.

Note also that the classification with the 3 bits c, m, 2, works ALMOST perfectly. But note also

that I do not know the exact meaning of these bits c, m, 2. I said they are somewhat RELATED to

the physical properties “center”, “plane”, “axis 2”).

Well, without boring the reader, I jump immediately to illustrate the overall classification, the

best classification that I was able to find.

It is not entirely satisfactory, because there are still some ambiguities that I have not solved.

The classification is as follows

**5 bit Tentative Classification**

1 00000 0 triclinic

**1 0000c 1 triclinic, holohedry triclinic**

m 000m0 2 monoclinic

**2/m 000mc 3 monoclinic, holohedry monoclinic**

2 00200 4 monoclinic

222 0020c 5 orthorhombic

mm 002m0 6 orthorhombic

**mmm 002mc 7 orthorhombic, holohedry orthorombic**

4 04000 8 tetragonal

4 0400c 9 tetragonal

23 040m0 10 isometric

4/m 040mc 11 tetragonal

422 04200 12 tetragonal

42m 0420c 13 tetragonal

4mm 042m0 14 tetragonal

**4/mmm 042mc 15 tetragonal, holohedry tetragonal**

3 30000 16 exagonal

3 3000c 17 exagonal

6 300m0 18 exagonal

3m 300mc 19 exagonal

6 30200 20 exagonal

32 3020c 21 exagonal

62m 302m0 22 exagonal

3m 302mc 23 exagonal

622 34000 24 exagonal

6/m 3400c 25 exagonal

6mm 340m0 26 exagonal

**6/mmm 340mc 27 exagonal, holohedry exagonal**

432 34200 28 isometric

m3 3420c 29 isometric

43m 342m0 30 isometric

**m3m 342mc 31 isometric, holohedry isometric**

Again, readers who have experience of the crystal Classes will easily recognize that the classification

with the 5 bits c, m, 2, 4, 3 matches ALMOST perfectly the physical properties of the various Classes

(“center”, “plane”, “axis 2”, “axis 4”, “axis 3”).

Note that this classification is still, most likely, WRONG, but it is the best classification for now I was

able to find. Ambiguities and inconsistencies (example: the strange position of the Tetartoidal Class 23)

may only be resolved by further studies.

**Conclusion**

Summing up, the 5 bit classification works ALMOST perfectly.

How may I summarize?

I have submitted a conjecture, which may be called 32 point groups of three dimensional crystal cells

described by 5 bits. To my knowledge this conjecture has not been discussed elsewhere, and

therefore may be useful for further research, possibly in the area of mathematics, group theory and

crystallography.

I would like to point out:

it’s possible to identify 32 crystal classes with 5 bit. This is not a conjecture: this is A FACT.

But: is the correspondence a one-to-one correspondence?

From a mathematical point of view: is there a

**bijection**between the 32 classes and 5 bits?

(In mathematics, a bijection, or bijective function, is a function between the elements

of two sets. Every element of one set is paired with exactly one element of the

other set, and every element of the other set is paired with exactly one element of the first set).

At the same time, it’s possible give a name, a label, to each bit.

This also is A FACT.

But: has each bit a physical meaning?

Does it correspond to some “property”?

Some “quality”?

For the moment I do not know any theoretical justification for a possible identification with five

basic symmetry properties, if not a guess, and trust that "Essentia non sunt moltiplicanda praeter

necessitatem". However, is it merely a coincidence?

For ALMOST all classes the proposed 5 bit classification is entirely satisfactory but some cases seem

to me still ambiguous. So the purpose of this article was of course not to draw a conclusive theory,

but to suggest further study of this proposed conjecture.

(Further details may be found in my paper http://vixra.org/abs/1101.0052).

ADDENDUM.

My last tentative result (updated).

monohedron no symmetry 1 00000

parallellohedron i 1 0000c

dome 1m m 000m0

prism i, 1A2, 1m 2/m 000mc

sphenoid 1A2 2 00200

rhombic disphenoid 3A2 222 0020c

rhombic pyramid 1A2, 2m mm 002m0

rhombic dipyramid i, 3A2, 3m mmm 002mc

tetragonal pyramid 1A4 4 04000

tetragonal disphenoid 1R4 4 0400c

ditetragonal pyramid 1A4, 4m 4mm 040m0

tetragonal dipyramid i, 1A4, 1m 4/m 040mc

tetragonal trapzohedron 1A4, 4A2 422 04200

gyroid 3A4, 4A3, 6A2 432 0420c

tetragonal scalenohedron 1R4, 2A2, 2m 42m 042m0

ditetragonal dipyramid i, 1A4, 4A2, 5m 4/mmm 042mc

trigonal pyramid 1A3 3 30000

rhombohedron i,1A3 3 3000c

ditrigonal pyramid 1A3,3m 3m 300m0

trigonal trapezohedron 1A3, 3A2 32 300mc

hexagonal pyramid 1A6 6 30200

trigonal dipyramid 1R6 6 3020c

hexagonal trapezohedron 1A6, 6A2 622 302m0

hexagonal scalenohedron i, 1A3, 3A2, 3m 32/m 302mc

dihexagonal pyramid 1A6, 6m 6mm 34000

hexagonal dipyramid i, 1A6, 1m 6/m 3400c

ditrigonal dipyramid 1R6, 3A2, 3m 6m2 340m0

dihexagonal dipyramid i, 1A6, 6A2, 7m 6/mmm 340mc

tetartoid 3A2, 4A3 23 34200

diploid i, 3A2, 4A3, 3m m3 3420c

hextetrahedron 3A2, 4A3, 6m 43m 342m0

hexoctahedron i, 3A4, 4A3, 6A2, 9m m3m 342mc

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# Comments

My last tentative result (updated).

I hope I have not made mistakes in writing.

I may just add: Leroy Jethro Gibbs Rule 39: "There is no such thing as coincidence."

monohedron no symmetry 1 00000

parallellohedron i 1 0000c

dome 1m m 000m0

prism i, 1A2, 1m 2/m 000mc

sphenoid 1A2 2 00200

rhombic disphenoid 3A2 222 0020c

rhombic pyramid 1A2, 2m mm 002m0

rhombic dipyramid i, 3A2, 3m mmm 002mc

tetragonal pyramid 1A4 4 04000

tetragonal disphenoid 1R4 4 0400c

ditetragonal pyramid 1A4, 4m 4mm 040m0

tetragonal dipyramid i, 1A4, 1m 4/m 040mc

tetragonal trapzohedron 1A4, 4A2 422 04200

gyroid 3A4, 4A3, 6A2 432 0420c

tetragonal scalenohedron 1R4, 2A2, 2m 42m 042m0

ditetragonal dipyramid i, 1A4, 4A2, 5m 4/mmm 042mc

trigonal pyramid 1A3 3 30000

rhombohedron i,1A3 3 3000c

ditrigonal pyramid 1A3,3m 3m 300m0

trigonal dipyramid 1R6 6 300mc

hexagonal pyramid 1A6 6 30200

trigonal trapezohedron 1A3, 3A2 32 3020c

ditrigonal dipyramid 1R6, 3A2, 3m 6m2 302m0

hexagonal scalenohedron i, 1A3, 3A2, 3m 32/m 302mc

hexagonal trapezohedron 1A6, 6A2 622 34000

hexagonal dipyramid i, 1A6, 1m 6/m 3400c

dihexagonal pyramid 1A6, 6m 6mm 340m0

dihexagonal dipyramid i, 1A6, 6A2, 7m 6/mmm 340mc

tetartoid 3A2, 4A3 23 34200

diploid i, 3A2, 4A3, 3m m3 3420c

hextetrahedron 3A2, 4A3, 6m 43m 342m0

hexoctahedron i, 3A4, 4A3, 6A2, 9m m3m 342mc

I hope I have not made mistakes in writing.

I may just add: Leroy Jethro Gibbs Rule 39: "There is no such thing as coincidence."

monohedron no symmetry 1 00000

parallellohedron i 1 0000c

dome 1m m 000m0

prism i, 1A2, 1m 2/m 000mc

sphenoid 1A2 2 00200

rhombic disphenoid 3A2 222 0020c

rhombic pyramid 1A2, 2m mm 002m0

rhombic dipyramid i, 3A2, 3m mmm 002mc

tetragonal pyramid 1A4 4 04000

tetragonal disphenoid 1R4 4 0400c

ditetragonal pyramid 1A4, 4m 4mm 040m0

tetragonal dipyramid i, 1A4, 1m 4/m 040mc

tetragonal trapzohedron 1A4, 4A2 422 04200

gyroid 3A4, 4A3, 6A2 432 0420c

tetragonal scalenohedron 1R4, 2A2, 2m 42m 042m0

ditetragonal dipyramid i, 1A4, 4A2, 5m 4/mmm 042mc

trigonal pyramid 1A3 3 30000

rhombohedron i,1A3 3 3000c

ditrigonal pyramid 1A3,3m 3m 300m0

trigonal dipyramid 1R6 6 300mc

hexagonal pyramid 1A6 6 30200

trigonal trapezohedron 1A3, 3A2 32 3020c

ditrigonal dipyramid 1R6, 3A2, 3m 6m2 302m0

hexagonal scalenohedron i, 1A3, 3A2, 3m 32/m 302mc

hexagonal trapezohedron 1A6, 6A2 622 34000

hexagonal dipyramid i, 1A6, 1m 6/m 3400c

dihexagonal pyramid 1A6, 6m 6mm 340m0

dihexagonal dipyramid i, 1A6, 6A2, 7m 6/mmm 340mc

tetartoid 3A2, 4A3 23 34200

diploid i, 3A2, 4A3, 3m m3 3420c

hextetrahedron 3A2, 4A3, 6m 43m 342m0

hexoctahedron i, 3A4, 4A3, 6A2, 9m m3m 342mc

**Giuliano Bettini****23rd May 2015 7:22pm**In order to leave comments to this article, you must be registered

Giuliano Bettini9th Jan 2014 8:17pm