I think that's a bit sloppy, your use of the "=" there, because the thing on the left is a set of ordered pairs, the thing on the right is not. What the whole thing really says is there is a morphism from one object to the other in the category Grp . . .? And a question related to the quote from ncatlab, what's the difference between the free group and the external direct product, what I've dubbed a fundamental group (really what I mean there is it's a fundamental part of the pattern, a kind of tile). What does the notation used in the quote mean to say about the two groups, \( \mathbb Z_2,\; \mathbb Z_3 \)?
One consideration and a way around the problem of trying to continuously rotate half a sphere, is the tiling in the image. Each tile is green or white, but there is no reason not to number them all. There are 8 triangles on each cubic face, so 6x8 = 48 triangles. There is an algebra of compositions of reflections that sends tiles to tiles, reflections are through the sides of each triangle. Please Register or Log in to view the hidden image! The algebra has vertical and horizontal reflections in it where a pair of blue lines intersect, and where red and blue lines intersect.
Re a,b,c,d being integers or whatever, the Wiki article is good. They define the Mobius group as I did, where a, b, c, d are complex numbers with ad - bc = 1. (That last part's not on the Wiki page, the video I watched noted that you can always normalize by a constant without changing the transformation). In general it's sufficient for ad - bc to be invertible, ie these are 2x2 nonsingular matrices. If a,b,c,d are real, we get PSL(2, R), and if a,b,c,d are integers we get PSL(2, Z), the modular group. So that explains all the terminology. I gather you're interested in this latter case, but that's NOT (in my opinion) directly on point to the question of rotating the top of a sphere. That's the question I understand, which is why I'm focussed on it. I understand it's not the question you're most interested in but I'm latching on to the parts I can understand. https://en.wikipedia.org/wiki/Möbius_transformation#Subgroups_of_the_Möbius_group Of course they're isomorphic. I wouldn't call that sloppy, it's perfectly clear and standard. They ARE the same group, just two different representations of it. Nobody would object but you're correct that they're different as sets. So are the integer 5 and the real number 5, but nobody claims that ignoring that fact is sloppy! Set theory is not to be used for idle pedantry! Unless I'm the pedant. I suppose I deserve to be on the receiving end of some pedantry so ok. The free group on a set is the set of all finite linear combinations of the elements of the set, interpreted as formal symbols ... oh wait, you have a little inaccuracy. Free groups are WAY too complicated for us at the moment. Let's talk about free Abelian groups, those are easier and not at all the same thing as free groups. The free Abelian group is the set of finite linear combinations over the integers (that is, the coefficients are integers) of the original set, whose elements are interpreted as formal symbols. That is, the free Abelian group on the set \(\{x, y\}\) is the set of all expressions \(n x + m y\) where n and m are integers, and '+' is a commutative operation. It's not hard to see that the free Abelian group on a finite set is just the direct sum of that many copies of the integers. That is, the free Abelian group on \(\{x, y\}\) is just \(\mathbb Z \oplus \mathbb Z\). Note that I said "is" and not "is isomorphic to," because there is only ONE UNIQUE free Abelian group on a two-element set, no matter what you call the elements of the set. That follows from the universal property of free Abelian groups. In other words yes, we are thinking categorically and isomorphism is as good as equality. Free groups in general are the same idea but we have noncommutative products instead of linear combinations, which gives the theory a very different flavor. I don't think we are concerned with free groups at the moment, only free Abelian groups. But for completeness, the free group on two letters is the set of all "reduced words" like \(x^{-3} y^2 x^6 y^{-8}\). The free group on two letters comes up in the proof of the Banach-Tarski paradox, because it can be decomposed into the union of one of its subsets with a translation of one of its other subsets, in two different ways. I'm going a little far afield but just know that free Abelian groups and free groups are two entirely different things. Can't comment. The fundamental group is a thing in algebraic topology that sounds related but I'm not sure. It's the set of homotopy classes of some object. https://en.wikipedia.org/wiki/Fundamental_group Well first, the notation \(\mathbb Z_n\) is really a shorthand in limited contexts for the group \(\mathbb Z / n \mathbb Z \). That's the integers "modded out" by the multiples of n, what we call the integers mod n or "clock arithmetic" mod n. I only mention this because in the most general context, the notation \(\mathbb Z_n\) stands for the n-adic integers, an entirely unrelated subject. Since writing \(\mathbb Z / n \mathbb Z \) is tedious, it's starting to become standard to notate this as \(\mathbb Z / (n)\) in cases where there might be any ambiguity. But for us, I assume \(\mathbb Z_n \) is the integers mod n. But you know that, I'm not sure why you're asking. Are you back to asking whether they're isomorphic or equal? It makes no difference at all. There is only one cyclic group of 6 elements no matter how we notate it. We're not in beginning group theory, where students have to be introduced to the idea of isomorphism gradually. I assume we can take the categorical point of view that there's only one such group and that it has many equivalent representations, just as 1/2 and 2/4 represent the same rational number. I hope this is perfectly sensible and not "sloppy." You have been adopting the categorical point of view which is a bit more sophisticated than undergrad algebra, so there should be no more pedantry about equality versus isomorphism I hope!
Oh I see your point now. Yes sure, but that's highly discontinuous. If that's ok with you then that looks right. I haven't any insight into discrete operations on a tiled sphere, but you're succeeding in educating me. I can see how free Abelian groups would come into the picture since rotations commute and we can combine integer linear combinations of elementary tile motions.
So some more considerations, or, things I've thunk. The two points (1,1) and (1,2), are both elements of a group and parts of a path a single 2x2x2 "permutation cube" follows. The complete graph is a quotient of a graph which is branched in a third dimension, this means my graph (I've shown a nice symmetric part of it) is a "flattened" version of the graph of the permutation group, it's missing quite a lot of information but nonetheless is like a coordinate map with a metric, the permutation metric. The graph is in the positive x, positive y quadrant because there are no negative cubes, reflections are excluded. However in say \( \mathbb Z_2 \), 1 is congruent to -1, so we can "glue" these values together, so we can glue the sides of \( \mathbb Z_2 \oplus \mathbb Z_2 \) together. Algorithmically, we can consider strings of letters corresponding to movement through a set of points in a graph; the set of all finite strings is the Kleene closure of the "free group".
About that tesselation, or tiling of the 2-sphere (i.e. the surface of the Riemann sphere): So we have (π/4, π/3, π/2) since each triangle bisects an angle, for the angles defining each green or white triangle. The link to the source of the image, and the quote is here, but it's mostly about hyperbolic surfaces and their tilings with (p,q,2) triangles (note the author isn't using a conventional notation, usually the numbers are ordered left to right).
A system of coordinates-the basis is a finite rectangular section of \( \mathbb Z^2 \); here's some more of it: Please Register or Log in to view the hidden image! Really it's a 2-d projection of a thing that looks like a tree in one more dimension. That is, any path through the lattice of points eventually gets to a boundary--a leaf node. Another maybe interesting detail is that the section of the upper boundary of the lattice from (0,0) to (4,8), is a partition of the symmetric group \( S_4 \). There are three copies of \( S_4 \), one in each dimension, so the partition function isn't something you can find with cycle notation, say. Notice there is a break after (4,8), there is no path to (5,10) . . . here \( S_4 \) is exhausted, but you can 'gauge' your way out and go to another latitude (?). But the tensor multiplication rule (for horizontal composition) is also broken here. What to do? I'll use a simpler notation, where the "+" operation is addition of points horizontally, the vertical 'lifting' is then automatic: So if I represent the direct product Δ, as <2,3>, and then write <2,3> + <2,3> = <3,5> <3,5> + <2,3> = <4,7> <4,7> + <2,3> = <5,9> . . . then <5,9> + <2,3> doesn't hold, it has no coordinates for its maximum.
A hint from Wikipoedians: Here, in the context of a diagram monoid, the objects are points in (the external direct product) \( \mathbb Z_p \oplus \mathbb Z_q \) and the arrows are edges in a graph--the graph of the equivalence classes, or cosets, of the permutation group in question. This is heuristically eight geometrically identical cubes (unit cubes) stacked together such that "layers" can be rotated (coloring the faces immediately changes how "free" the stack is). Rubik thought up an ingenious mechanical solution to holding such a freely stacked object together, such that its "liquid" properties, rotations, are preserved. Otherwise it tends to degenerate into a more "gaseous" state . . . One other thing about the posted graph. I can transform the whole thing by stretching (allowed in topology) the "x" and "y" axes apart by multiplying (i.e. scaling) the angle between them, and shrinking the complementary angle. The families of lines stay parallel, but I can make the angle between the upper and lower boundaries equal to π/2, or whatever I like, without altering the algebra in any way. One other observation: stacking solid cubes together in the absence of a gravitational field without a tube of glue is a difficult problem to solve. That's about the only connection, but it's kind of axiomatic and has nothing to to with topological slicing or gluing, gravity acts on massive cubes, therefore they tend to stay where you put them. So the algebra here isn't connected to physics except that you can derive it with a physical model. And it's helpful to have gravity in the frame.
I don't follow any of the geometry but I wonder if you can explain the algebra. What objects are being direct producted? Do you mean direct sum? In the finite case it makes no difference but coproducts are direct sums and not direct products. But I don't follow the arithmetic. Are you saying, in the first one for example, that 2 + 2 = 3? I don't know any modular arithmetic that makes that work out. What does it mean? Keep it simple please, I don't follow anything you're discussing although I do agree with your tiled sphere example.
Define <2,3> := <{1,1}, {1,2}> <2,3> + <2,3> := <{1,1}, {1,2}> + <{1,1}, {1,2}> = <{2,2}, {2,3}, {2,4}> =: <3,5> Remember I'm taking a subset of the product group, this generates the whole group so I'm ok with it. Leaving out the elements with zeros makes the operation easier too.
AHA!!!! I have solved the original mystery that drew me to this thread. It happens to be the case that the tensor product of graphs is defined and notated differently than the tensor product of groups or modules. The Wiki article does not note this discrepancy but I definitely understand now where @arfa is coming from. This is a totally different kind of tensor product and has nothing to do with bilinearity. I do wonder why they chose such confusing terminology. There must be more to this, I'll keep looking for a connection. I apologize to @arfa for my tone earlier, I am wrong and learned something. https://en.wikipedia.org/wiki/Tensor_product_of_graphs In graph theory (paraphrasing Wiki) if \(G\) and \(H\) are graphs, we can define a "tensor product" \(G \times H\). Now already this is totally interesting. The symbol for tensor product is \(\otimes\), markup 'otimes'. So the graph theorists (assuming the Wiki article represents standard practice) have already taken pains to distinguish their tensor product from the algebraic one. [I'd usually consider graph theory a subtopic of algebra, but in this context I'll distinguish them]. The "vertex set" (I don't know any graph theory but their meaning's clear enough) is the usual Cartesian product of the vertex sets of \(G\) and \(H\); and we define two vertices to be "adjacent" (again I'm working from common sense here and only learned this terminology five minutes ago) if their respective coordinates are adjacent. Now this is really interesting: "The tensor product is also called the direct product, categorical product, ..." [their emphasis]. I happen to know that in the categories of groups or modules, the categorical product is the direct product and not the tensor product! The tensor product has a different definition that includes bilinearity. Sometimes category theory is like that. The particularization of a categorical concept to a particular category can sometimes be counterintuitive, and different in spirit from the particularization to another category. For example the categorical product of topological spaces is the weak topology and not the box topology as most people would assume. In this case category theory actually picks up the more natural product and not the most obvious one, which is interesting. [Details not important here, only that categorical generalizations hide a lot of variety]. So anyway we have this abstract concept of categorical product. If we apply it to groups we get the usual direct product; and the tensor product is something else. But when we apply it to graphs we get what they call the tensor product. Which from their definition seems much more like a conventional direct product and not a tensor product! In other words the elements are ordered pairs, and the operations are defined component-wise. So why do they call it a tensor product yet notate it differently? This is a real mystery at the moment. The article even acknowledges this issue: "The notation G × H is also sometimes used to represent another construction known as the Cartesian product of graphs, but more commonly refers to the tensor product." Until I understand this better, I would say that the graph theorists have misnamed their tensor product. They do have a thing called the Cartesian product of graphs, and it's different from their tensor product, but until I find more evidence, I do not believe it should be called a tensor product. Tensor products are the most general expression of bilinearity. Maybe there's a connection I'll find.
Yeah, I think what happened there is you de-categorified the tensor algebra as the group-theoretic one, initially, where I've been using the term in the context of planar graphs of a group. But that's interesting, the difference. Seriously, I just decided to use it to mean composition in the horizontal direction, of an object that has both horizontal and vertical components. Then I had a closer look at diagram monoids (semigroups with identity). I've also realised the additive identity has co-ordinates at the origin--the "solved" state, but exists at every vertex in the connected graph, perhaps I can push that into being anywhere along any line, but if you aren't at a vertex you aren't strictly "in" the integers. (0,0) doesn't do anything though, additively.
BTW, here is a characterisation of the branching 'function': When I do <2,3> + <2,3> and instead form an ordered tuple, I have ((2,2),(2,3),(2,3),(2,4)), and the element (or coordinate) (2,3) has a multiplicity of 2, because it's the union of two disjoint paths each going to those coordinates, so the distance is the same too. In post 30, I take the union of sets, inside the < > is the disjoint remainder. So again, I'm just defining operations on sets, or on tuples, as I choose to.
What I'm claiming, to be perfectly clear, is that pending new information, my sense is that the graph theorists' usage of the term tensor product is morally wrong; for the reason that the conceptual essence of tensor products is bilinearity; and I haven't yet seen any bilinearity in the graph business. I hope that's clear. I thought you said that you have a group, and I am wondering what your group structure is. So it's not actually a group structure, and in fact it may not have any algebraic structure at all. Without knowing anything about your application that feels wrong, since there is clearly a lot of abstract algebra being bandied about by the graph theorists. I doubt they'd consider a random operation that had no algebraic structure. Ah your cryptic sums generate a group. Is that right now? I hope you can provide clarity in this area. I am trying to understand the algebraic structure you are claiming exists.
No my sums generate the elements of a product of groups which all lie on the same line. They can be defined by a relation on pairs (a,b). I can do it like so, for p = 2, q = 3; 2 := {0,1}, 3:= {0,1,2} 2 × 3 = { {0,1} × {0,1,2} } = { {0,0}, {0,1}, {0,2}, {1,0}, {1,1}, {1,2} } <2,3> = { (a,b) | a ∈ 2, b ∈ 3, a ≤ b, a,b ≥ 1, {a,b} ∈ 2 × 3 } So that <2> + <3> = <{1}> + <{2}> = <{3}> = <4>, <1> + <1> = <1>, <1> + <2> = <2>, <2> + <2> = <3>, . . . So if it isn't obvious, I'm redefining the numbers p,q as sets of points (vertices), but with "canonical" order. I could use bold numerals maybe. So generally p := {0, 1, 2, ..., p - 1}, where the p inside the braces is |p|. The algebra acts on Cartesian products of sets, adding points to points. I think the operations are reasonably well-defined.
I see that's all a bit sloppy. So if p and q are sets of points, labeled with numbers 0, 1, ..., p-1, and 0, 1, ..., q-1 respectively, then the Cartesian product p × q is {(0,0), (0,1), ... , (0,q-1), (1,0), (1,1), ... , (1,q-1), ... , (p-1,q-1)} = {(a,b) | 0 ≤ a ≤ p-1, 0 ≤ b ≤ q-1 }. Then <p,q> = <p × q> = { (a,b) | a = p-1, p-1 ≤ b ≤ q-1 } The algebra does not include local operations on products like p × q, there is "+" as my definition of the horizontal addition of a vertical column of points, and there is "×" as the Cartesian product of sets of points. That is, operations like (1,1) + (1,2) in 2 × 3 aren't defined, although it's still true that (1,1) = (1,0) + (0,1) "internally", you generate (1,1) with a single action. Here, I don't distinguish the operation "+" when it's used in different contexts, hence the need for different notations.
Nyways, back to the question of what happens to the upper boundary of the graph, at (4,8). First clue, the point (4,8) is an element of <3,5>. Also the point (1,2) is order 6 as an element of <2,3>, but 5(1,2) = (5,10) ≡ (1,1) internally (i.e. "working in" <2,3>). Perhaps that's why the transition from (4,8) to (5,9) is a "1,1" transition, albeit the details (the mechanism) is still a bit obscure (you don't say?). Another detail, the partition of \( S_4 \) has 5 elements at (4,8), but these can be further subdivided by their order in \( S_4 \) (the five have mixed order). \( S_4 \) is 'localised' along a particular boundary--it gets glued there by the algebra--each vertical column of points in the graph is a section or slice (a foliation?) of the larger algebraic object, which since it obviously acts on eight distinct elements (the vertices of a cube, or the faces of the octahedral tiling of a 2-sphere under the action of a reflection group), is a subset of the larger symmetric group \( S_8 \).
Another comment, on the isomorphism between \( \mathbb Z_2 \oplus \mathbb Z_3 \) and \( \mathbb Z_6 \). This is true when the internal algebra of each object is compared, under 'local' addition and multiplication they have the same structure, a commutative ring. But the graphs are distinct. \( \mathbb Z_2 \oplus \mathbb Z_3 \) is a graph with no connections (edges), composed as two columns of two and three points, respectively. \( \mathbb Z_6 \), however, graphs as a single column (or row) of six points.
Then the notation's simply wrong. There's only one cyclic group of order 6. It matters not whether you notate it as \( \mathbb Z_2 \oplus \mathbb Z_3 \), or \( \mathbb Z_6 \), or the set of the sixth complex roots of 1 under multiplication. It's exactly the same group; or, if we're in undergrad algebra class, they're all isomorphic to one another. There is no counternarrative to this.
Again, try to lift your perspective out of the idea of a group with local symmetries, multiplication and addition, to the idea of graphing the symmetries. The graphs of ℤ2⊕ℤ3 and ℤ6 do not commute, they can be distinguished by applying a global perspective--the graph itself.
