I’d like to think that I understand symmetry groups. I know what the elements of a symmetry group are – they are transformations that preserve an object or its relevant features – and I know what the group operation is – composition of transformations. Given a polyhedron or wallpaper tiling or whatever, I could probably start spotting the symmetries, which would entail listing out elements of the symmetry group, and then I could start filling in the multiplication table.

Penrose attaches a group to impossible figures to capture their inherent ambiguity, and I’d like to grok these groups like I do symmetry groups. Take a prototypical example, the tribar:

He names the “ambiguity group” G=R+ (positive numbers under multiplication) to describe possible distances of points. We can split the figure up into three components, as above, and interpret them as being disconnected from each other in three-space but from our perspective they seem to make a single figure. For convenience, I think we should let Aij denote points on the figures as well as represent their distances from the origin, interchangeably.

One can define the relative distances by dij=Aij/Aji. Since dji=d−1ij, there are only three relevant proportions: d12, d23, and d31. According to Penrose, the dijs do not actually depend on our choice of overlapping points Aij, but this seems wrong to me: varying the points Aij through the overlap regions will change them linearly and so any ratio dij will only remain invariant if dij=1 to begin with. But probably this quibble is unimportant.

One can scale the distances the components Q1,Q2,Q3 are from the origin without affecting our perception of them. (Perhaps consider our “perception” of them to be their radial projection onto the unit sphere, or something.) The effect of scaling one of these Qi by a factor of λ on the d12,d23,d31 is to scale one of them by λ, a second by λ−1, and leave the third unchanged.

If the Q1,Q2,Q3 were compatible and could be combined into a single figure, then such a configuration would have (d12,d23,d31)=(1,1,1). If they were compatible but the components were separated by independent scalings q1,q2,q3 (respectively) then we’d have

(d12,d23,d31)=(q1q2,q2q3,q3q1).

Note that τ=d12d23d31 is an

invariant, in the sense that scaling the components independently does not change the value of τ. The compatibility situation (1) occurs precisely when τ=1.Penrose defines the group H to be the tuples (d12,d23,d31)∈(R+)3 modulo the rescalings by λ and modulo the elements of the form (1). As I understand it, the invariant (d12,d23,d31)↦τ is a bijection H→R+. But now here are my questions.

(A)

How do we know what the ambiguity group is?The tribar’s ambiguity group is G=R+. With the Necker cubes above, Penrose says the ambiguity group is G=Z2. Is the ambiguity group meant to parametrize the possible positions of the individual pieces of the figure? Where does the group operation of G actually come into play?(B)

How do we know what pieces to cut a figure up into?Since the tribar has obvious threefold symmetry, that kind of inspires the choice of three pieces. But it seems that with the congruence relation used to define H, we could choose to put any two of those pieces together into one component and fix its position, only letting the last component vary (which would be one degree of freedom, exactly as H≅R+ predicts). So we could have cut into two pieces. Or we ould cut into six pieces, or any number of pieces. Will it never matter how many pieces we choose? Why would we break apart the tribar’s corners but not break apart the faces of the Necker cubes? What figures would we cut into pieces, and what figures would we do something else to? And with the latter figures, what would we do to find their H group?(C)

What are the group elements and what is the group operation?As I mentioned with symmetry groups, it’s intuitive what their elements are and what the operation is. But what about with H? It seems the elements are physically realizable configurations consistent with our perception, modulo altering the configuration in a way that wouldn’t change our perception. The identity element would be the configurations in which our perception is actually correct and sensible. And the group operation seems to be … I don’t know. Presumably we could use componentwise multiplication of the representative tuples (d12,d23,d31), or equivalently multiplication of the invariants τ, which would make H→R+ a group isomorphism, but how would this operation be meaningful or relevant?(D)

Why is this called a cohomology group?Yes, our H1(Q,G)s are being called cohomology groups. I deliberately put off using that word as long as possible. (And so you’ve read this far. Suckers.) But in what sense are these groups cohomological? Are there higher cohomology groups Hn(Q,G) and coboundary operators? Is this cohomology dual to some kind of homology of impossible figures? Probably I will be unable to understand answers to this question, as I don’treallyknow what cohomology is in the first place. No time like the present?Ultimately, I’d like to be able to look at an impossible figure and systematically derive its cohomology group, just like I can derive a figure’s symmetry group. Or alternately, create impossible figures with given cohomology group. But perhaps the analogy isn’t tenable, as cohomology groups aren’t really symmetry groups at all.

**Answer**

To answer (A), one has to understand the meaning of the ambiguity group clearly first – indeed, your interpretation of the group is not entirely correct.

The punchline here is that the ambiguity group measures ambiguity, not quite impossibility (which is the job of the cohomology group defined later), of the given figure. For example, consider the tribar, and piece it up into Q1,Q2,Q3, as done by Penrose. If you stretch Q1 by a factor of λ>0, shrink Q2 by a factor of λ (that is to say, stretch by λ−1), and leave Q3 as it is, the resulting object is really the same tribar, which is what Penrose was trying to convey through the following picture from the paper.

That is to say, modifying Qi‘s in this way gives you a symmetry of X. And you have such a symmetry for each λ>0. Of course, collection of such symmetries G is a group: consider one symmetry which stretches Q1 by λ, shrinks Q2 by λ and leaves Q3 as is and another one which does the same thing with λ′ this time instead. You can of course compose them, and the resulting symmetry stretches Q1 by λλ′, shrinks Q2 by λλ′, and leaves Q3 invariant. This is the ambiguity group. This partially answers (C). Note that the origin of the name comes from the fact that the tribar is *ambiguous* in the sense that you can have different Q1,Q2,Q3‘s matching up to produce the same tribar. You can never tell the length of a bar in the tribar – it’ll depend on the point of view you’re seeing it from. And the “group of symmetries” G captures all the ambiguity there is.

It is now clear that there is a bijection G→R+, sending a symmetry to the corresponding λ>0 in R+ (group under multiplication), which is in fact a homomorphism. Thus, G≅R+.

That said, given the Necker cubes, the ambiguity is in the vertices. Pick any vertex v from the figure. There are two point of views here – v can either point upwards, or point downwards. If v points upwards, then so does all the other vertices, and similar for the other choice. So two point of views, two choices – hence, the ambiguity group is Z/2Z. More formally, the symmetry is obtained from reflecting the whole figure (imagined as sitting in R3) along an appropriate hyperplane. Obviously, the symmetry group then is the cyclic group of order 2.

To answer (D). Yes, this is indeed a variant of Čech cohomology.

First, let me explain what cohomology really is. If X is a sufficiently nice topological space, you can always set up a homeomorphism X≅T of X with a simplicial complex T. Such a homeomorphism is called a triangulation of X. Note that there need not be a unique such homeomorphism, so there may be lots of ways to triangulate a given nice topological space.

Declare Δn(T) to be the free abelian group generated by n-simplices of T. Let Δn(T) be hom(Δn(T),Z), the dual of Δn(T). Elements of Δn(T) are called n-cochains. There is a natural map ∂:Δn−1(T)→Δn(T) taking an n-cochain ψ:Δn−1(T)→Z to the n-cochain ψ∘bd:Δn(T)→Δn−1(T)→Z where bd sends an n-simplex in T to sum of signed faces of that n-simplex (signs chosen appropriately to take care of orientation). ∂ is called the coboundary map. One can check that the map ∂∂:Δn−1(T)→Δn(T)→Δn+1(T) is the zero map. The sequence of maps

⋯→Δn−1(T)→Δn(T)→Δn+1(T)→⋯

is then a cochain complex, that is, going two steps in the sequences lands you into 0. As a consequence of ∂∂=0, we have that im∂⊂ker∂. Elements of im∂ are called coboundaries, and elements of ker∂ are called cocycles. As everything is abelian, im∂ is also normal in ker∂. We set Hn(X;Z):=ker∂/im∂. This is called the simplicial cohomology of X, and is indeed independent of the triangulation T we chose. It’s a topological invariant of X. We’ll see something similar happening in Penrose’s paper.

If X is the given ambiguous figure, let {Ui} be a good cover of X. That is, X=⋃Ui and each of Ui, Ui∩Uj, Ui∩Uj∩Uk and further intersections, are homeomorphic to balls Bn (a more general condition would be that they are all contractible). Let each intersection Ui1∩⋯∩Uin be denoted as Ui1⋯in. Treat these sets as symbols (i.e., formal intersection), and distinguish Uij from Uji even though they are the same set.

G be the ambiguity group of X. Define an n-dimensional cochain to be a function φn:{Ui1⋯in}→G such that φn(Ui1⋯ik⋯il⋯in)=φn(Ui1⋯il⋯ik⋯in)−1 holds for any k,l, where {Ui1⋯in} is the collection of all possible intersection of n+1 many sets from the cover. When n=0, it’s just an assignment of a an element of G to each Ui.

Let Cn(X) denote the group of all n-dimensional cochains, where we multiply two cochains by multiplying their values in G. Define the coboundary map ∂:Cn−1(X)→Cn(X) given by sending φn−1 to the n-cochain φn assigning to each Uii⋯in the number ∏jφn−1(Ui1⋯^ij⋯in)εj where εj=(−1)n if j is even and (−1)n+1 if j is odd, and ^ij means that term is missing from the indices. So for example, in the tribar example, assigning to each Qi the number qi is a 0-cochain and applying the coboundary map one gets an assignment of each (oriented) intersection Qij=Qi∩Qj to the number qi/qj, a 1-cochain in our language.

The final thing to check is that ∂∂:Cn−1(X)→Cn(X)→Cn+1(X) is the zero map. This is a rather tedious and technical thing to check, but if the readers try verifying it, they will understand the motivation behind the rather bizarre definition of ∂ above. To sum up, we have that

⋯→Cn−1(X)∂→Cn(X)∂→Cn+1(X)→⋯

is a cochain complex. We can thus take it’s cohomology Hn(X;G):=ker∂/im∂ (the G is there to indicate that the cochains are G-valued. Indeed, we could have put any group to be the coefficient group there, but it seems Penrose finds it natural to use the ambiguity group of X). For n=1, this is precisely the 1st cohomology group of Penrose.

Let us compute the cohomology group of the tribar. H1(X;R+) is the group of 1-cocycles modulo 1-coboundaries. As there are no nontrivial 2-cochains, every 1-cochain is a cocycle. That is to say, we’re computing 1st cohomology of the two-term cochain 0→C0(X)∂→C1(X)→0 which is just C1(X)/im∂. Note that this has a slight similarity with your interpretation (the definition of H), but is not quite the same. In any case, I can find a 1-cocycle in C1(X) which is not a coboundary using the clever technique devised by Penrose: the 1-cochain ψ1 sending Qij=Qi∩Qj to dij=Ai/Aj is independent of the choice of Ak‘s, and I claim that it is a nontrivial cochain (i.e., is not a boundary). Well, if it was a boundary, then the equality dij=ψ1(Qij)=ψ0(Qi)/ψ0(Qj) would hold for some 0-cochain ψ0. But then we’d have τ(ψ1)=d12d23d31=1.

However, if we choose the placement of tribar appropriately so that A12 and A21 has the same distance from origin (i.e., d12=1) and A23 is farther away from origin than A32 (so that d23>1, then A13 would automatically be closer than A31 (that is, d31=d−113>1), which implies τ(ψ1) is strictly greater than 1, forcing ψ1 to be nontrivial. Once we have this nontrivial cochain, we can get |R+|-many nontrivial cochains by multiplying ψ1(Q12) by 1, ψ1(Q23) by λ, and ψ1(Q31) by 1/λ for each λ>0. This gives a natural identification with G. I think with some work you can show that these are all the nontrivial 1-cocycles there are, and conclude H1(X;R+)≅R+. In general, it need not be true that H1(Q;G) is the same as the ambiguity group G, but the 1-st cohomology group H1(Q;G) would be a module over the ambiguity group G.

I picked the open cover {Ui} of X to be good, i.e., all sets and all intersections are contractible. The reason is that by the nerve theorem, X would have the homotopy type of the nerve N of {Ui}. Čech cohomology just computes cohomology of the nerve N, but since this is homotopy equivalent to X, that is the same as taking cohomology of X, so it’s independent of whatever good cover we chose. I think something similar must be at work for these ambiguous/impossible figures. This probably answers (B), but only Penrose knows what he had in mind.

One more thing I haven’t addressed yet.

Ultimately, I’d like to be able to look at an impossible figure and systematically derive its cohomology group, just like I can derive a figure’s symmetry group. Or alternately, create impossible figures with given cohomology group. But perhaps the analogy isn’t tenable, as cohomology groups aren’t really symmetry groups at all.

The first part of this answers gives a rough overview of how you could recognize the ambiguity group from a given ambiguous picture, so I presmume I have answered that bit. I do not know what a rigorous definition of the ambiguity group might be, but I presume it’s some sort of inherent symmetry group of the chosen cover {Ui}. Also, certainly the analogy is tenable: if you read my answer to (A), you’d see that this ambiguity group is indeed some sort of symmetry groups which captures symmetries “induced from ambiguity”. However, I do not know if you can construct an ambiguous figure such that the ambiguity group G is isomorphic to a given group. Here’s some food for thought: Assume G is a finitely presented abelian group. Pick generators x1,⋯,xn of G. F be the free abelian group generated by these. If Y is some ambiguous figure with infinite cycle ambiguity group, then disjoint union of n copies of Y gives you a figure with ambiguity group isomorphic to F. Now given relators ri=1 for G, try identifying the copies of Y appropriately to kill the symmetries corresponding to ri. Can this be done?

PS: I wanted to add an explicit computation of the Penrose cohomology H1(X;Z/2) of the Necker cube as a summary, but I’d need some pictures for that and I’m too ashamed of my drawing skills to add it. Maybe if OP (and others reading this) are really interested.

**Attribution***Source : Link , Question Author : whacka , Answer Author : Balarka Sen*