Symmetry of the cube

Discussion in 'Physics & Math' started by arfa brane, Jan 22, 2019.

1. arfa branecall me arfValued Senior Member

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So before I get too lost again and stuff up my notation, what the notation lets me do is define an abstract topological slice, apart from the real slices meeting at the centre, or partly meeting--it doesn't matter because you see little of the inner surfaces, the relations between these are expressed (algebraically!) on a surface.

The slice is the one each octant's inner surfaces is on as it rotates three times, under a $\mathbb Z_3$ action on vertex orientations.
Its multiplicative version is the equivalent of addition modulo 3, but with complex roots of unity. These roots are tied to a well-known set of polynomials, or to the cyclotomic numbers. But I seem to have a lot of these with prime degree. The graph gets to x = 11 along that axis. So how many primes are there less than 11 (there's a well known formula that tells you), and what can I do with a set of primes less than or equal to 11?

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3. arfa branecall me arfValued Senior Member

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. . . a way to write set products, such as $\{\omega e_j,\, \omega \bar e_j \}^2$, or $\{\omega e_j,\, \omega \bar e_j \}^3$, or say, $\{\omega e_j,\, \omega \bar e_j \}^n$, with complex coefficients . . .
And I have this "*" or (Kleene) closure of the set of all finite strings over the characters in the sets under concatenation, such that:

$\{\omega e_j,\, \omega \bar e_j \}^* = \{\omega e_j,\, \omega \bar e_j \}^0\{\omega e_j,\, \omega \bar e_j \}^*$.

Just a review, I'm going to get rid of the set notation here and use ordered tuples:

$(\omega e_j,\, \omega \bar e_j )^* = (\omega e_j,\, \omega \bar e_j )^0(\omega e_j,\, \omega \bar e_j )^* = \omega ^0(e_j,\, \bar e_j )(\omega e_j,\, \omega \bar e_j )^*$.

Because a) it's less hassle, and (b) there's no loss of . . . generality.

I can emphasize that I mean a product of sets (as tuples) anytime, and in an FSA context (an automaton which accepts a string, but does nothing else), tuples have that meaning and products of tuples are accepted notation.

But this gives me a way to define a "zeroth" fibre in the set, partitioned along y = 0 "by" a function of x. This is the uncharacterised as yet, partition function or relation (relation as function); it must be transitive, reflexive, and antisymmetric to fit.

That is to say, I know it has to be like this, for a, b, c, each a set, a relation R is:
1. reflexive if aRa (the set is related to itself)
2. transitive if aRb and bRc both exist, aRc exists (is in the set of relations)
3. antisymmetric iff a = b and aRb and bRa both exist.

The relation "≤" is reflexive on sets of numbers, since a ≤ a; transitive since for all a, b, c, a ≤ b ≤ c is the relation a ≤ c; antisymmetric since a ≤ b and b ≤ a iff a = b. Meaning the permutation metric is the relation between the disjoint sets (points along lines of constant x, or n-pointed fibres). It factors the sets into equivalence classes, and I can make it look like a complex function, and map polynomials to the cyclotomic numbers.

Ok. Greg Egan shows a connection to the Fano plane, from his method of foliating the 3-sphere into a family of tori, each embedded inside the other, to solve Klein's quartic in homogenous coordinates (where I generalise the indices of the metric with $\alpha \in \{1,2,3\}, \beta \in (-1)\{1,2,3\}$ and fix an index set).

Last edited: Feb 20, 2019

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5. arfa branecall me arfValued Senior Member

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So It looks like I need to show how the indices in $g_{\alpha}, g_{\beta} = g_{\alpha}^{-1}$ are related.

I suppose I might be allowed to do $g_{\alpha}, g^{\beta} = g_{\alpha}^{-1}$, see if I can use a raised "generalised" coordinate and get away with it.

But it looks like a useful thing to use, because it characterises: $g_{\alpha}g_{\alpha}^{-1}$, which is the identity in each of three dimensions. So I now have $g_{\alpha}g_{\alpha}^{-1} = g_{\alpha}g^{\beta} \equiv \delta^{\alpha}_{\beta}g_{\alpha}g^{\beta} = \{I_x,I_y,I_z\}$

Last edited: Feb 20, 2019

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7. arfa branecall me arfValued Senior Member

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Can anyone spot the fairly obvious mistake in the last post? no?

Well, good thing I did. It's the mistake I made using the Kronecker delta on indices, and on negative exponents (the sign of the inverse permutation).

Ok, I need to fix this, a do-over. So start with the identity elements in $\mathbb R^3$, one for each of the three dimensions each turning circle in the projective plane has. So we have to distinguish between $I_n$ and using indices 1,2,3 for each of the three 1+1 dimensions.

Yah. Well maybe change the symbol $I_n$ to $\iota_n$

Then I have $\iota_j = g_jg^{-1}_j$, so I need $\delta_j^j g_jg_j^{-1}$

Or in general $(\iota_{\alpha})^{\beta} = (g_{\alpha}g_{\alpha}^{-1} \delta^{\alpha}_{\alpha})^{\beta}$.

And I have that $\beta = -{\alpha}$

Last edited: Feb 20, 2019
8. Write4UValued Senior Member

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It's really interesting to follow a scientist prepare a thesis. Good stuff.......

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9. arfa branecall me arfValued Senior Member

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Huh. A Scientist.

Well I guess what today's scientific lesson is, is that futzing with mixed indices and the Kronecker delta can be a problem.
But perhaps I'm trying to do things I don't need to, I should keep it all as simple as I can manage, not start overloading symbols and get confused about what they still mean etc.

But I'm reasonable certain that the basis vectors in my $2 + \bar 2$ 'space', with complex coefficients is a goer. It looks like I've simplified it enough (the notation) so I can at least write polynomials with complex roots of unity as coefficients, and that it's some more notation connected to lots of other notation describing cyclotomic number fields.

But this space with its basis is then the basis for the permutation space of 8 x 8 matrices (I know this, I already checked you need at least that size to keep track of positions over a vertex set of size 8, it kind of follows, probably some kind of law of numbers or what-have-you).

The whole mathematical decomposition-of-a-graph thing is starting to look like it has legs, but can I slap a kilt on it and still cull it a Scuotsmun?

10. arfa branecall me arfValued Senior Member

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Loop until $\alpha = \beta$ and use some bloudy paurentheses will ye!
1. It looks like I need to show how the indices in $g_{\alpha}, (g_{\beta} = g_{\alpha}^{-1})$ are related.

2. I suppose I might be allowed to do $g_{\alpha}, (g^{\beta} = g_{\alpha}^{-1})$, see if I can use a raised "generalised" coordinate and get away with it.

But $g_{\alpha}g_{\alpha}^{-1}$, is the identity in each of three dimensions.

3. So I now have $g_{\alpha}g_{\alpha}^{-1} = g_{\alpha}g^{\beta} \equiv \delta^{\alpha}_{\beta}g_{\alpha}g^{\beta} = \{I_x,I_y,I_z\}$

This flies if I only define $\alpha = \beta = \{1,2,3\}$. And if I recognise the negative exponentiations are tied to both these sets if I multiply one of them by -1 and put them up there. In other words I can put one of them there so it might as well be $\beta = \{1,2,3\}$, then the set doubles itself (spontaneously!) to $\beta = \{1,2,3,-1,-2,-3\}$.

So the $\alpha$ stay positive (in the upper half plane), but $\beta$ is given the freedom to go -ve. It copies itself , forms an ordered pair and multiplies one copy by -1. Then it forgets, it's just a set of six numbers.

Thanks, laws of matrix composition.

4. So that the sets of numbers, as indices, are also fibers over each place I can put them (a set of places). They are isotopic up to a factor of -1, --the top one tells me how many times I've iterated over an element $g_{\alpha}$ , $g^{\beta}$, in the polynomial 'expansion'

.. in which I make larger polynomials out of smaller ones, which then are factors of the larger--the key principle in cyclotomic polynomials, or ordinary algebra (quadratics, cubics, . . .); that number is tied to the vertical columns of points.

Last edited: Feb 21, 2019
11. arfa branecall me arfValued Senior Member

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So maybe I can do what I said, as long as I apply the rule that a lower index is positive, but when I shift it upstairs, it can be negative over the same set of fixed values, an index set.

What I want to do is find where sums of the $e_j, e_i$ exist, because then I can rewrite $e_i + e_j = I$, in two dimensions.
I should start to see sums of $I$ after the first appearance, but in the polynomials in g, with what degree?

12. arfa branecall me arfValued Senior Member

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3a. So I now have $g_{\alpha}g_{\alpha}^{-1} = g_{\alpha}g^{\beta} \equiv \delta^{\alpha}_{\beta}g_{\alpha}g^{\beta} = \{I_x,I_y,I_z\}$

This flies if ... $\alpha = \beta = \{1,2,3\}$. And if I recognise . . . that the 'pseudosum' happens when I lower the raised index in question and leave a negative copy behind.

Thanks again, laws of matrix composition. The way I write the Kronecker delta indices isn't relevant, it works as long as there are two of them, so doesn't indicate anything really.

13. arfa branecall me arfValued Senior Member

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The way I've developed all this has been . . . iterative. I can keep going back to a starting point, where I know I'm on solid ground so to speak, with notation, which I want to be as hassle-free as I can get it.

Along the way, although the puzzles are physical, and the graph is just lines through a finite set of points (not much to see here), complex numbers, roots of unity show up; an additive cyclic group is a multiplicative cyclic group depending on a choice of representation.

But when I do use what are also known as primitive roots, their conjugates have to be accounted for:
$\{\bar\omega e_i, \bar\omega\bar e_j \}^k = (\bar\omega)^k\{e_i, \bar e_j \} = \{e_i, \bar e_j \}|\omega^k =1$. More negative exponents, in an index set.

I think what I'm trying to do here is treat all the indices like topological spaces (with points in them), a graph-like object, IOW. I need to graph the relations between the indices on elements of a permutation group with inverses (as exponents), two different boundaries.

I don't need to get too worried about Kronecker's delta or the permutation symbol, or even that excited. I should be able to turn any cyclic element into a complex number or polynomial in a space with (not sure yet) at least one, perhaps two, complex dimensions.

I can arbitrarily choose to make one of the x or y axes imaginary and go to $\mathbb Z[{i}]$ the Gaussian integers. I can most of all, turn the metric as g (the group element in one of three spacelike dimensions) into a 4th root of unity, which is not multiplication by a scalar, now is it?

And the inverse of $\omega$ is $\omega^{-1}$, such that $\omega \omega^{-1} = 1$

Last edited: Feb 21, 2019
14. arfa branecall me arfValued Senior Member

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Dealing with the complex numbers as scalars that multiply, first of all a vector basis, leads to certain representational problems, but I'm reasonably sure I can use set products and the *-closure operation to sort them out.

And I needn't use the imprimitive roots $\omega \in \{-1,1\}$, i.e. the roots of polynomials $\{x^2 + 1,\, x^2 -1\}$.

Since I know that the rotation group of the square is a subgroup of $Dih_4$, namely $\{r_0,\,r^1,\,r^2,\,r^3\}$,

then I have with the order 2 reflection group, $s^2 = 1$, the following multiplicative relation: $s\{r_0,\,r^1,\,r^2,\,r^3\} = \{sr_0,\,sr^1,\,sr^2,\,sr^3\}$.

Now factor the two polynomials (just do it).

$x^2 + 1 = (x + {i})(x - {i}),\, x^2 -1 = (x + 1) (x - 1)$.

Ok, iteration. The *-closure of both the rotation subgroup, and of the same subgroup multiplied by an order 2 'transposition', a reflection of pairs of edges of a square, vertically or horizontally, and of pairs of vertices, diagonally. That is, two topologically distinct actions, one on pairs of vertices of the square, and one on pairs of edges.

So we have $\{r_0,\,r^1,\,r^2,\,r^3\}^* = \{r_0,\,r^1,\,r^2,\,r^3\}$, and the * of the subgroup under composition is idempotent.

And $\{sr_0,\,sr^1,\,sr^2,\,sr^3\}^* = \{sr_0,\,sr^1,\,sr^2,\,sr^3\}$

So then for $\omega^k = 1$,

$\omega\{r^0,\,r^1,\,r^2,\,r^3\}^* = \omega^k\{r^0,\,r^1,\,r^2,\,r^3\} \,∪\, \omega^{k-1}\{r^0,\,r^1,\,r^2,\,r^3\} \,∪ \,. . . \,∪ \,\{r_0,\,r^1,\,r^2,\,r^3\}$
$\omega \{sr^0,\,sr^1,\,sr^2,\,sr^3\}^* = \omega^k\{sr^0,\,sr^1,\,sr^2,\,sr^3\}\, ∪\, \omega^{k-1}\{sr^0,\,sr^1,\,sr^2,\,sr^3\} \,∪\, . . . \,∪\, s\{r^0,\,r^1,\,r^2,\,r^3\}$

Last edited: Feb 21, 2019
15. arfa branecall me arfValued Senior Member

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So the key ideas I'm using (or trying to) are based on the fundamental structure of a graph G(V,E), nominally a set of vertices such that |V| is the order of, and |E| is the g of, the graph G.
G then represents a structure imposed on V by E.

Switch context here to a graph G(V,R), a set of vertices V and a set of relations R between vertices.

Relations can be coloured, or directed (injective), and lastly relations can be functions from vertices to vertices. Edge colouring is a redundant operation because with a relational graph you really only need colours on vertices to preserve all the structure in G, and you can still colour edges or label them (weight them). But the Big Idea is that all the structure can be wrapped up in vertex weights, with relations (also weights, on relations as edges).

So I can construct as many graphs as I want to for the rotation group of the square. I also know I can send the edges and vertices of a square (along with its interior) to infinity in a projective space:

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The projection has a square boundary congruent with the sides of a square. There are lines connected to points on the boundary of the square. Notice there's a bit of thickening, but the lines on the sphere are thickened too. Think about removing the colours from some of the projected, but bounded, regions. You reduce the number of permutations, under reflection of triangles on the sphere if you "blank" the thickened, coloured discs on the sphere tiling (the regions between the triangular {4,3} tiles; one of them is shown on the left with orange, yellow and red "squares").

Reconnecting the lines, contracting pairs of lines to the same point, etc, is a sequence of 'topological' moves, up to what's called ambient isotopy.

A graph when you redraw it so its edges intersect, or don't, has this ambient isotopy, of abstract graph-moves. Contracting edges so you reduce the graph but leave it connected (such that a path exists--a cover of the vertex set), is ambient isotopic to the original graph, up to an equivalence relation between the two graphs.
The isotopy of the plane is that it's the "smallest" surface with minimal area, being 2-dimensional.

The smallest line with minimal length is an ambient isotopic element of the former.

Last edited: Feb 21, 2019
16. arfa branecall me arfValued Senior Member

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I'm not paying as much attention here to what I post (largely because I have other things I should do), and so another editing snafu.

The set of edges in a graph, E, has a cardinality |E| which is the size in edges of the graph. What the order and size of a graph mean is that, since each vertex must be at least degree 1 to be connected to at most one other vertex. In that case you can call it either a vertex (in G) of degree 1 and see if there are others, if you know the order and size of G. You can also call it a terminal vertex or terminator.

All disconnected vertices, though in V, are not in E, so you can know something about G, if you have the tuple to hand: (V,E), without drawing it.

In switching theory it might be an input or output for a signal, etc. But in G it's a vertex in a set V. So I have a set of contexts: graph theory, switching theory, group theory, number theory, information theory, . . . but in general a theory of vector spaces in each context.

I could start at the Euclidean embedding as a space of position vectors in the plane, say. But I'm more interested in the modular arithmetic (algebra!) and the roots of unity which are primitive. A primitive root in number theory is a bit different, however. So I see that when I do the multiplication table "operation" on a G-set, I can do it in a way that preserves a certain order; I then see the order that I've done it isn't relevant because, no matter how I arrange the elements of the set along a top row and a left hand column, I get the same set of products.

So it's a matter of the indices they inherit, and the order is restored (a tiling of the table products!) by applying my $2 + \bar 2$ vector basis. I know the basis belongs in a general linear group of order 2 because of the matrix size (2 x 2 "vertices").

So I write larger and larger tables as I increase the iterant of the G-set, so I take the rule of table composition (it falls out of the algebra) to the vertex at (11,11). In principle . . .

Last edited: Feb 22, 2019
17. arfa branecall me arfValued Senior Member

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Ok? So after I have a graph on say, the disk (or disc), which has a 1-sphere boundary (topologically a circle), and is a 2-ball:-

I redraw this graph on another 'blank' 2-ball, maybe after removing some edges, loops on vertices are redundant (you can always move a point to where it already is), multiple edges between the same pair of edges can be contracted pairwise, if the pair are say, equal but opposite.

So the second graph is homeomorphic after some ambient isotopic moves, to the first. So I shrink both disks to a point and weight them with (V,E), and (V',E'), as vertices in a new graph with one edge in it, which is weighted by some graph moves.

I know the abstract tile-of-a-group I've called $T_\vartriangle$ is a graph I can construct a new graph from, a star graph with a central vertex and maximum degree which is the minimal tiling that preserves all the structure (relations between points on boundaries).

Now I also take this idea, a star graph, to the point at infinity on the Riemann sphere.

For any circle of latitude below this point, a cone of projective lines onto the plane exists. There are some in the (image of the) projection of the sphere above. So there are that many cones. The star of each cone is the iteration of any finite set of moves from vertex to vertex in a graph that goes around the latitude. If you take some root of unity as the distance between iterations, it can be as small (or large) as you choose. So choose a big number, like the 3,674,160th root.

Then for a given latitude around the sphere, the star graph with the point at infinity its centre is a line bundle, each line has a point on the plane, a point on a line of latitude, and a point at infinity.

You factor this graph by counting all the equivalence classes and constructing new graphs on the sphere (line bundles) at different latitudes. The equivalence class of size 841,500 is a star graph on the sphere with terminals on the plane, each line in the bundle (a 2-point fibre if infinity is a fixed point), has a point on it which is a root of unity to a finite power less than 841,500.

You factor the first root and reduce the number of points on the initial latitude with a graph move.

Last edited: Feb 22, 2019
18. arfa branecall me arfValued Senior Member

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So, for example, I can take this model of a cone of discrete points on a circle (topologically a graph with a large size which is a 1-torus) with a polar angle = a line of latitude on the sphere, to say a cone over the set of pointlike objects with a thermodynamic total energy, then a graph with two vertices, one weighted with this total energy in Joules, the other with a temperature, and a single edge weighted with "/" represents entropy--but has to output a logarithm to be a measure.

Information entropy is all in the sets of walks which are or aren't cycles, are Hamiltonian or Eulerian, in the graph $\mathfrak G$

Last edited: Feb 22, 2019
19. arfa branecall me arfValued Senior Member

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So my idea is actually one of these (and much larger):

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But I might need extra edges, walking through a star graph means visiting the centre multiple times. If you can walk through a star graph without doing that, it isn't a star graph. Adding edges will turn it into a different graph, and you want them as a cyclic subgraph of the star one.

So just do it anyway, and make all the terminals into vertices of degree 3. Perhaps now colour the cyclic subgraph, the edges in it are also different types (in a strictly data-type sense) than the radial ones. These of course are the ones that go to infinity when it gets embedded on a cone, where the central point goes on the apex at infinity.

Another kind of graph, but still a star of terminals:

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The centre gets fixed at the origin of a Euclidean vector space, the red arc represents a rotation by a positive angle of the terminal$\alpha$. The star graph has a set *-product. The * of the set containing $\alpha$ is the set containing the six terminals in the graph. So. heuristically, if I continue the rotation I walk through the star graph, or around a latitude on the sphere.

Last edited: Feb 22, 2019
20. arfa branecall me arfValued Senior Member

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So the whole projectors -> projective space thing then is what you see (embedded) in the projective plane (with plenty of room for it to happen), when each of the full set products, i.e. $\{ g_{\alpha},\,g_{\beta} \}^*$ which is 3,674,160 projections like this one:

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. . . are arranged around the respective latitude boundaries on the sphere. Since you know what this one looks like you might as well contract it to a point, but retain the graph moves (the edge) that lets you uncontract it if you ah, choose to so you can park it on a small (like really small) circle of latitude near the pole (either one will do).

Then you see that contracting the graph is a contraction of a 2-sphere, the whole construction is points along circles of latitude which are actually spheres contracted to points, and the distance between them is a root of unity.

Last edited: Feb 22, 2019
21. arfa branecall me arfValued Senior Member

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Another construction is via not contracting square projections, as in the above and for the reason we've "identified" it as the identity map of the -- identity permutation in the set of all N = 3,674,160 square maps, which have 'algebraic' representations as 8 x 8 square matrices,

Each tile after the above identity tile in this *new* graph, has to embed in another plane perpendicular to the 1st, or primary tile. It's iterated over the construction because the permutation metric partitions the tiling. That is, to step from the identity permutation and contract it, you have to shift to the next available latitude by moving to another perpendicular (or orthogonal basis in the vector space) plane "tiling" dimension, and there are two left in the $2 + \bar 2$ d.f. after you've occupied, so to say, the first one.

I'll let you guys figure out what the connection between permutation matrices and the points on the boundary of the above "identity tile" is. But might could be it's the Levi-Civita symbol with more dimensions than 2.

Last edited: Feb 22, 2019
22. arfa branecall me arfValued Senior Member

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Ok well I guess I can at least "count" the place I put the first tile, by mapping it back to an "actual" 2-sphere. I take the square projection (in the previous post) to this graph:

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I embed the square projection so it goes in the orange square at the top of this one as a square section, taking a 2-sphere's worth of the cube group to a spherical cap with a square boundary.

Then I rotate (via a composition of reflections) part of the sphere as a valid graph-move and repeat, iterating over all the permutations in, piecewise linear fashion. The iterant is the nth set product with "degree n" products.

So you tile an n-sphere with 2-spheres as tiles, which are projections from the sphere to itself.

So where do I put the next tile in this iterated construction? On which sphere?

Last edited: Feb 22, 2019
23. arfa branecall me arfValued Senior Member

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Sorry, I should use more than an "=" or maybe use "->", an arrow instead; the following as written aren't equalities. So I'll redo it:

So we have $\{r_0,\,r^1,\,r^2,\,r^3\}^* = \{r_0,\,r^1,\,r^2,\,r^3\}$, and the * of the rotation subgroup (in $Dih_4$) ... is idempotent.

Check.

And $\{sr_0,\,sr^1,\,sr^2,\,sr^3\}^* = \{sr_0,\,sr^1,\,sr^2,\,sr^3\}$. Nope it should be $\{sr_0,\,sr^1,\,sr^2,\,sr^3\}^* = \{sr_0,\,sr^1,\,sr^2,\,sr^3\} ∪ \{r_0,\,r^1,\,r^2,\,r^3\}$

Check, you get the full dihedral group.

So then for $\omega^k = 1$,

$\omega\{r^0,\,r^1,\,r^2,\,r^3\}^* = \omega^k\{r^0,\,r^1,\,r^2,\,r^3\} \,∪\, \omega^{k-1}\{r^0,\,r^1,\,r^2,\,r^3\} \,∪ \,. . . \,∪ \,\{r_0,\,r^1,\,r^2,\,r^3\}$
$\omega \{sr^0,\,sr^1,\,sr^2,\,sr^3\}^* = \omega^k\{sr^0,\,sr^1,\,sr^2,\,sr^3\}\, ∪\, \omega^{k-1}\{sr^0,\,sr^1,\,sr^2,\,sr^3\} \,∪\, . . . \,∪\, s\{r^0,\,r^1,\,r^2,\,r^3\} ∪\, \{r^0,\,r^1,\,r^2,\,r^3\}$

Whew. So if you see I haven't written out a set product correctly, just add "+ mixed products" mentally.

Last edited: Feb 22, 2019