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" $G=\Bbb R^+$ (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 $A_{ij}$ denote points on the figures as well as represent their distances from the origin, interchangeably.

One can define the relative distances by $d_{ij}=A_{ij}/A_{ji}$. Since $d_{ji}=d_{ij}^{-1}$, there are only three relevant proportions: $d_{12}$, $d_{23}$, and $d_{31}$. According to Penrose, the $d_{ij}$s do not actually depend on our choice of overlapping points $A_{ij}$, but this seems wrong to me: varying the points $A_{ij}$ through the overlap regions will change them linearly and so any ratio $d_{ij}$ will only remain invariant if $d_{ij}=1$ to begin with. But probably this quibble is unimportant.

One can scale the distances the components $Q_1,Q_2,Q_3$ 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 $Q_i$ by a factor of $\lambda$ on the $d_{12},d_{23},d_{31}$ is to scale one of them by $\lambda$, a second by $\lambda^{-1}$, and leave the third unchanged.

If the $Q_1,Q_2,Q_3$ were compatible and could be combined into a single figure, then such a configuration would have $(d_{12},d_{23},d_{31})=(1,1,1)$. If they were compatible but the components were separated by independent scalings $q_1,q_2,q_3$ (respectively) then we'd have

$$(d_{12},d_{23},d_{31})=\left(\frac{q_1}{q_2},\frac{q_2}{q_3},\frac{q_3}{q_1}\right). \tag{1}$$

Note that $\tau=d_{12}d_{23}d_{31}$ is an *invariant*, in the sense that scaling the components independently does not change the value of $\tau$. The compatibility situation $(1)$ occurs precisely when $\tau=1$.

Penrose defines the group $H$ to be the tuples $(d_{12},d_{23},d_{31})\in(\Bbb R^+)^3$ modulo the rescalings by $\lambda$ and modulo the elements of the form $(1)$. As I understand it, the invariant $(d_{12},d_{23},d_{31})\mapsto\tau$ is a bijection $H\to\Bbb R^+$. But now here are my questions.

$\sf \color{Fuchsia}{(A)}$ **How do we know what the ambiguity group is?** The tribar's ambiguity group is $G=\Bbb R^+$. With the Necker cubes above, Penrose says the ambiguity group is $G=\Bbb Z_2$. 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?

$\sf \color{Fuchsia}{(B)}$ **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 $H\cong\Bbb R^+$ 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?

$\sf \color{Fuchsia}{(C)}$ **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 $(d_{12},d_{23},d_{31})$, or equivalently multiplication of the invariants $\tau$, which would make $H\to\Bbb R^+$ a group isomorphism, but how would this operation be meaningful or relevant?

$\sf \color{Fuchsia}{(D)}$ **Why is this called a cohomology group?** Yes, our $H^1(Q,G)$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 $H^n(Q,G)$ 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't *really* know 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.